382 Theory of Machines - brainzorp.files.wordpress.com Theory of Machines 12.4. Classification of...

382 � Theory of Machines

382

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12FFFFFeaeaeaeaeaturturturturtureseseseses1. Introduction.2. Friction Wheels.3. Advantages and

Disadvantages of Gear Drive.4. Classification of Toothed

Wheels.5. Terms Used in Gears.6. Gear Materials.7. Law of Gearing.8. Velocity of Sliding of Teeth.9. Forms of Teeth.

10. Cycloidal Teeth.11. Involute Teeth.12. Effect of Altering the Centre

Distance.13. Comparison Between Involute

and Cycloidal Gears.14. Systems of Gear Teeth.15. Standard Proportions of Gear

Systems.16. Length of Path of Contact.17. Length of Arc of Contact.18. Contact Ratio19. Interference in Involute

Gears.20. Minimum Number of Teeth on

the Pinion.21. Minimum Number of Teeth on

the Wheel.22. Minimum Number of Teeth on

a Pinion for Involute Rack inOrder to Avoid Interference.

23. Helical Gears.24. Spiral Gears.25. Centre Distance For a Pair of

Spiral Gears.26. Efficiency of Spiral Gears.

12.1.12.1.12.1.12.1.12.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoductionWe have discussed in the previous chapter, that the

slipping of a belt or rope is a common phenomenon, in thetransmission of motion or power between two shafts. Theeffect of slipping is to reduce the velocity ratio of the system.In precision machines, in which a definite velocity ratio is ofimportance (as in watch mechanism), the only positive driveis by means of gears or toothed wheels. A gear drive is alsoprovided, when the distance between the driver and the fol-lower is very small.

12.2.12.2.12.2.12.2.12.2. Friction WheelsFriction WheelsFriction WheelsFriction WheelsFriction WheelsThe motion and power transmitted by gears is kine-

matically equivalent to that transmitted by friction wheels ordiscs. In order tounderstand howthe motion can betransmitted bytwo toothedwheels, considertwo plain circularwheels A and Bmounted onshafts, having sufficient rough surfaces and pressing againsteach other as shown in Fig. 12.1 (a).

Chapter 12 : Toothed Gearing � 383Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, to be rotated. A

little consideration will show, that when the wheel A is rotated by a rotating shaft, it will rotate thewheel B in the opposite direction as shown in Fig. 12.1 (a).

The wheel B will be rotated (by the wheel A ) so long as the tangential force exerted by thewheel A does not exceed the maximum frictional resistance between the two wheels. But when thetangential force (P) exceeds the *frictional resistance (F), slipping will take place between the twowheels. Thus the friction drive is not a positive drive.

(a) Friction wheels. (b) Toothed wheels.

Fig. 12.1

In order to avoid the slipping, a number of projections (called teeth) as shown inFig. 12.1 (b), are provided on the periphery of the wheel A , which will fit into the correspondingrecesses on the periphery of the wheel B. A friction wheel with the teeth cut on it is known as toothedwheel or gear. The usual connection to show the toothed wheels is by their **pitch circles.

Note : Kinematically, the friction wheels running without slip and toothed gearing are identical. But due to thepossibility of slipping of wheels, the friction wheels can only be used for transmission of small powers.

12.3. Advantages and Disadvantages of Gear Drive

The following are the advantages and disadvantages of the gear drive as compared to belt,rope and chain drives :

Advantages1. It transmits exact velocity ratio.2. It may be used to transmit large power.3. It has high efficiency.4. It has reliable service.5. It has compact layout.

Disadvantages1. The manufacture of gears require special tools and equipment.2. The error in cutting teeth may cause vibrations and noise during operation.

* The frictional force F is equal to µ. RN, where µ = Coefficient of friction between the rubbing surface oftwo wheels, and RN = Normal reaction between the two rubbing surfaces.

** For details, please refer to Art. 12.4.


12.4. Classification of Toothed Wheels

The gears or toothed wheels may be classified as follows :

1. According to the position of axes of the shafts. The axes of the two shafts between whichthe motion is to be transmitted, may be

(a) Parallel, (b) Intersecting, and (c) Non-intersecting and non-parallel.

The two parallel and co-planar shafts connected by the gears is shown in Fig. 12.1. Thesegears are called spur gears and the arrangement is known as spur gearing. These gears have teethparallel to the axis of the wheel as shown in Fig. 12.1. Another name given to the spur gearing ishelical gearing, in which the teeth are inclined to the axis. The single and double helical gears con-necting parallel shafts are shown in Fig. 12.2 (a) and (b) respectively. The double helical gears areknown as herringbone gears. A pair of spur gears are kinematically equivalent to a pair of cylindricaldiscs, keyed to parallel shafts and having a line contact.

The two non-parallel or intersecting, but coplanar shafts connected by gears is shown in Fig.12.2 (c). These gears are called bevel gears and the arrangement is known as bevel gearing. Thebevel gears, like spur gears, may also have their teeth inclined to the face of the bevel, in which casethey are known as helical bevel gears.

The two non-intersecting and non-parallel i.e. non-coplanar shaft connected by gears is shownin Fig. 12.2 (d). These gears are called skew bevel gears or spiral gears and the arrangement isknown as skew bevel gearing or spiral gearing. This type of gearing also have a line contact, therotation of which about the axes generates the two pitch surfaces known as hyperboloids.

Notes : (a) When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpen-dicular, then the bevel gears are known as mitres.

(b) A hyperboloid is the solid formed by revolving a straight line about an axis (not in the sameplane), such that every point on the line remains at a constant distance from the axis.

(c) The worm gearing is essentially a form of spiral gearing in which the shafts are usually at rightangles.

(a) Single helical gear. (b) Double helical gear. (c) Bevel gear. (d) Spiral gear.

Fig. 12.2

2. According to the peripheral velocity of the gears. The gears, according to the peripheralvelocity of the gears may be classified as :

(a) Low velocity, (b) Medium velocity, and (c) High velocity.

The gears having velocity less than 3 m/s are termed as low velocity gears and gears havingvelocity between 3 and 15 m/s are known as medium velocity gears. If the velocity of gears is morethan 15 m/s, then these are called high speed gears.

Chapter 12 : Toothed Gearing � 385

3. According to the type of gearing. The gears, according to the type of gearing may beclassified as :

(a) External gearing, (b) Internal gearing, and (c) Rack and pinion.

In external gearing, the gears of the two shafts mesh externally with each other as shown in Fig.12.3 (a). The larger of these two wheels is called spur wheel and the smaller wheel is called pinion. Inan external gearing, the motion of the two wheels is always unlike, as shown in Fig. 12.3 (a).

(a) External gearing. (b) Internal gearing.

Fig. 12.3 Fig. 12.4. Rack and pinion.

In internal gearing, the gears of the two shafts mesh internally with each other as shown inFig. 12.3 (b). The larger of these two wheels is called annular wheel and the smaller wheel is calledpinion. In an internal gearing, the motion of the two wheels is always like, as shown in Fig. 12.3 (b).

Spiral Gears

Helical Gears

Double helical gears


Sometimes, the gear of a shaft meshes externally and internally with the gears in a *straightline, as shown in Fig. 12.4. Such type of gear is called rack and pinion. The straight line gear is calledrack and the circular wheel is called pinion. A little consideration will show that with the help of arack and pinion, we can convert linear motion into rotary motion and vice-versa as shown in Fig.12.4.

4. According to position of teeth on the gear surface. The teeth on the gear surface may be

(a) straight, (b) inclined, and (c) curved.

We have discussed earlier that the spur gears have straight teeth where as helical gears havetheir teeth inclined to the wheel rim. In case of spiral gears, the teeth are curved over the rim surface.

12.5. Terms Used in Gears

The following terms, which will be mostly used in this chapter, should be clearly understoodat this stage. These terms are illustrated in Fig. 12.5.

Fig. 12.5. Terms used in gears.

1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the samemotion as the actual gear.

Internal gears

* A straight line may also be defined as a wheel of infinite radius.

Rack and pinion

Chapter 12 : Toothed Gearing � 3872. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually

specified by the pitch circle diameter. It is also known as pitch diameter.

3. Pitch point. It is a common point of contact between two pitch circles.

4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replacedat the pitch circle.

5. Pressure angle or angle of obliquity. It is the angle between the common normal to twogear teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ.

The standard pressure angles are 1214 ° and 20°.

6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth.

7. Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom of the tooth.

8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric withthe pitch circle.

9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also calledroot circle.

Note : Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle.

10. Circular pitch. It is the distance measured on the circumference of the pitch circle froma point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc.Mathematically,

Circular pitch, pc = π D/T

where D = Diameter of the pitch circle, and

T = Number of teeth on the wheel.

A little consideration will show that the two gears will mesh together correctly, if the twowheels have the same circular pitch.

Note : If D1 and D2 are the diameters of the two meshing gears having the teeth T1 and T2 respectively, then forthem to mesh correctly,

1 2 1 1

1 2 2 2

orcD D D T

pT T D T

π π= = =

11. Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter in millimetres.It is denoted by pd . Mathematically,

Diametral pitch, dc

Tp

D p

π= = ... cD

pT

π = �

where T = Number of teeth, and

D = Pitch circle diameter.

12. Module. It is the ratio of the pitch circle diameter in millimeters to the number of teeth.It is usually denoted by m. Mathematically,

Module, m = D /T

Note : The recommended series of modules in Indian Standard are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16,and 20. The modules 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14 and 18 are of second choice.

13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, ina meshing gear. A circle passing through the top of the meshing gear is known as clearance circle.

14. Total depth. It is the radial distance between the addendum and the dedendum circles ofa gear. It is equal to the sum of the addendum and dedendum.


* For details, see Art. 12.16.** For details, see Art. 12.17.

15. Working depth. It is the radial distance from the addendum circle to the clearance circle.It is equal to the sum of the addendum of the two meshing gears.

16. Tooth thickness. It is the width of the tooth measured along the pitch circle.17. Tooth space . It is the width of space between the two adjacent teeth measured along the pitch

circle.18. Backlash. It is the difference between the tooth space and the tooth thickness, as mea-

sured along the pitch circle. Theoretically, the backlash should be zero, but in actual practice somebacklash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion.

19. Face of tooth. It is the surface of the gear tooth above the pitch surface.20. Flank of tooth. It is the surface of the gear tooth below the pitch surface.21. Top land. It is the surface of the top of the tooth.22. Face width. It is the width of the gear tooth measured parallel to its axis.23. Profile. It is the curve formed by the face and flank of the tooth.24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth.25. Path of contact. It is the path traced by the point of contact of two teeth from the

beginning to the end of engagement.26. *Length of the path of contact. It is the length of the common normal cut-off by the

addendum circles of the wheel and pinion.

27. ** Arc of contact. It is the path traced by a point on the pitch circle from the beginningto the end of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e.

(a) Arc of approach. It is the portion of the path of contact from the beginning of theengagement to the pitch point.

(b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of theengagement of a pair of teeth.Note : The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e. number of pairsof teeth in contact.

12.6. Gear Materials

The material used for the manufacture of gears depends upon the strength and service condi-tions like wear, noise etc. The gears may be manufactured from metallic or non-metallic materials.The metallic gears with cut teeth are commercially obtainable in cast iron, steel and bronze. The non-metallic materials like wood, raw hide, compressed paper and synthetic resins like nylon are used forgears, especially for reducing noise.

The cast iron is widely used for the manufacture of gears due to its good wearing properties,excellent machinability and ease of producing complicated shapes by casting method. The cast irongears with cut teeth may be employed, where smooth action is not important.

The steel is used for high strength gears and steel may be plain carbon steel or alloy steel. Thesteel gears are usually heat treated in order to combine properly the toughness and tooth hardness.

The phosphor bronze is widely used for worm gears in order to reduce wear of the wormswhich will be excessive with cast iron or steel.

12.7. Condition for Constant Velocity Ratio of Toothed Wheels–Law ofGearing

Consider the portions of the two teeth, one on the wheel 1 (or pinion) and the other on the

Chapter 12 : Toothed Gearing � 389wheel 2, as shown by thick line curves in Fig. 12.6. Let the two teethcome in contact at point Q, and the wheels rotate in the directions asshown in the figure.

Let T T be the common tangent and M N be thecommon normal to the curves at the point of contact Q. From thecentres O1 and O2 , draw O1M and O2N perpendicular to MN. Alittle consideration will show that the point Q moves in the directionQC, when considered as a point on wheel 1, and in the directionQD when considered as a point on wheel 2.

Let v1 and v2 be the velocities of the point Q on the wheels1 and 2 respectively. If the teeth are to remain in contact, then thecomponents of these velocities along the common normal MN mustbe equal.

∴ 1 2cos cosv vα = β

1 1 2 2

1 21 1 2 2 1 1 2 2

1 2

( ) cos ( ) cos

( ) ( ) or

O Q O Q

O M O NO Q O Q O M O N

O Q O Q

ω × α = ω × β

ω × = ω × ω × = ω ×

∴ 1 2

2 1

O N

O M

ω =ω

…(i)

Also from similar triangles O1MP and O2NP,

2 2

1 1

O N O P

O M O P= ...(ii)

Combining equations (i) and (ii), we have

1 2 2

2 1 1

O N O P

O M O P

ω = =ω

...(iii)

From above, we see that the angular velocity ratio is inversely proportional to the ratio of thedistances of the point P from the centres O1 and O2, or the common normal to the two surfaces at thepoint of contact Q intersects the line of centres at point P which divides the centre distance inverselyas the ratio of angular velocities.

Therefore in order to have a constant angular velocity ratio for all positions of the wheels, thepoint P must be the fixed point (called pitch point) for the two wheels. In other words, the commonnormal at the point of contact between a pair of teeth must always pass through the pitch point.This is the fundamental condition which must be satisfied while designing the profiles for the teeth ofgear wheels. It is also known as law of gearing.

Notes : 1. The above condition is fulfilled by teeth of involute form, provided that the root circles from whichthe profiles are generated are tangential to the common normal.

2. If the shape of one tooth profile is arbitrarily chosen and another tooth is designed to satisfy theabove condition, then the second tooth is said to be conjugate to the first. The conjugate teeth are not in commonuse because of difficulty in manufacture, and cost of production.

3. If D1 and D2 are pitch circle diameters of wheels 1 and 2 having teeth T1 and T2 respectively, thenvelocity ratio,

1 2 2 2

2 1 1 1

O P D T

O P D T

ω= = =

ω

Fig. 12.6. Law of gearing.

or


12.8. Velocity of Sliding of Teeth

The sliding between a pair of teeth in contact at Q occurs along the common tangent T T tothe tooth curves as shown in Fig. 12.6. The velocity of sliding is the velocity of one tooth relative toits mating tooth along the common tangent at the point of contact.

The velocity of point Q, considered as a point on wheel 1, along the common tangent T T isrepresented by EC. From similar triangles QEC and O1MQ,

1 11

or .EC v

EC MQMQ O Q

= = ω = ω

Similarly, the velocity of point Q, considered as a point on wheel 2, along the common tan-gent T T is represented by ED. From similar triangles QCD and O2 NQ,

22 2

2

or .vED

ED QNQN O Q

= = ω = ω

Let S Velocity of sliding at .v Q=

∴ S 2 1. .v ED EC QN MQ= − = ω − ω

2 1( ) ( )QP PN MP QP= ω + − ω −

1 2 2 1( ) . .QP PN MP= ω + ω + ω − ω ...(i)

Since 1 21 2

2 1

or . . ,O P PN

MP PNO P MP

ω = = ω = ωω

therefore equation (i) becomes

S 1 2( )v QP= ω + ω ...(ii)

Notes : 1. We see from equation (ii), that the velocity of sliding is proportional to the distance of the pointof contact from the pitch point.

2. Since the angular velocity of wheel 2 relative to wheel 1 is (ω1 + ω2 ) and P is the instantaneouscentre for this relative motion, therefore the value of vs may directly be written as vs (ω1 + ω2 ) QP, without theabove analysis.

12.9. Forms of TeethWe have discussed in Art. 12.7 (Note 2)

that conjugate teeth are not in common use.Therefore, in actual practice following are the twotypes of teeth commonly used :

1. Cycloidal teeth ; and 2. Involute teeth.

We shall discuss both the above mentionedtypes of teeth in the following articles. Both theseforms of teeth satisfy the conditions as discussedin Art. 12.7.

12.10. Cycloidal TeethA cycloid is the curve traced by a point on the circumference of a circle which rolls without

slipping on a fixed straight line. When a circle rolls without slipping on the outside of a fixed circle,the curve traced by a point on the circumference of a circle is known as epi-cycloid. On the otherhand, if a circle rolls without slipping on the inside of a fixed circle, then the curve traced by a pointon the circumference of a circle is called hypo-cycloid.

Chapter 12 : Toothed Gearing � 391In Fig. 12.7 (a), the fixed line or pitch line of a rack is shown. When the circle C rolls without

slipping above the pitch line in the direction as indicated in Fig. 12.7 (a), then the point P on the circletraces epi-cycloid PA . This represents the face of the cycloidal tooth profile. When the circle D rollswithout slipping below the pitch line, then the point P on the circle D traces hypo-cycloid PB, whichrepresents the flank of the cycloidal tooth. The profile BPA is one side of the cycloidal rack tooth.Similarly, the two curves P' A' and P'B' forming the opposite side of the tooth profile are traced bythe point P' when the circles C and D roll in the opposite directions.

In the similar way, the cycloidal teeth of a gear may be constructed as shown in Fig. 12.7 (b).The circle C is rolled without slipping on the outside of the pitch circle and the point P on the circleC traces epi-cycloid PA , which represents the face of the cycloidal tooth. The circle D is rolled on theinside of pitch circle and the point P on the circle D traces hypo-cycloid PB, which represents theflank of the tooth profile. The profile BPA is one side of the cycloidal tooth. The opposite side of thetooth is traced as explained above.

The construction of the two mating cycloidal teeth is shown in Fig. 12.8. A point on the circleD will trace the flank of the tooth T1 when circle D rolls without slipping on the inside of pitch circleof wheel 1 and face of tooth T2 when the circle D rolls without slipping on the outside of pitch circleof wheel 2. Similarly, a point on the circle C will trace the face of tooth T1 and flank of tooth T2. Therolling circles C and D may have unequal diameters, but if several wheels are to be interchangeable,they must have rolling circles of equal diameters.

Fig. 12.8. Construction of two mating cycloidal teeth.

A little consideration will show, that the common normal X X at the point of contact betweentwo cycloidal teeth always passes through the pitch point, which is the fundamental condition for aconstant velocity ratio.

(a) (b)

Fig. 12.7. Construction of cycloidal teeth of a gear.


12.11. Involute Teeth

An involute of a circle is a plane curve generated by apoint on a tangent, which rolls on the circle without slipping orby a point on a taut string which in unwrapped from a reel asshown in Fig. 12.9. In connection with toothed wheels, the circleis known as base circle. The involute is traced as follows :

Let A be the starting point of the involute. The basecircle is divided into equal number of parts e.g. AP1, P1P2,P2P3 etc. The tangents at P1, P2, P3 etc. are drawn and thelength P1A1, P2A 2, P3A 3 equal to the arcs AP1, AP2 and AP3 areset off. Joining the points A , A1, A2, A3 etc. we obtain the involutecurve A R. A little consideration will show that at any instantA 3, the tangent A 3T to the involute is perpendicular to P3A 3 and P3A 3 is the normal to the involute. Inother words, normal at any point of an involute is a tangent to the circle.

Now, let O1 and O2 be the fixed centres of the two base circles as shown in Fig. 12.10 (a). Letthe corresponding involutes A B and A 1B1 be in contact at point Q. MQ and NQ are normals to theinvolutes at Q and are tangents to base circles. Since the normal of an involute at a given point is thetangent drawn from that point to the base circle, therefore the common normal MN at Q is also thecommon tangent to the two base circles. We see that the common normal MN intersects the line ofcentres O1O2 at the fixed point P (called pitch point). Therefore the involute teeth satisfy thefundamental condition of constant velocity ratio.

(a) (b)

Fig. 12.10. Involute teeth.

From similar triangles O2NP and O1MP,

1 1 2

2 2 1

O M O P

O N O P

ω= =ω ... (i)

which determines the ratio of the radii of the two base circles. The radii of the base circles is given by

1 1 2 2cos , and cosO M O P O N O P= φ = φAlso the centre distance between the base circles,

1 2 1 21 2 1 2 cos cos cos

O M O N O M O NO O O P O P

+= + = + =φ φ φ

Fig. 12.9. Construction of involute.

Chapter 12 : Toothed Gearing � 393where φ is the pressure angle or the angle of obliquity. It is the angle which the common normal to thebase circles (i.e. MN) makes with the common tangent to the pitch circles.

When the power is being transmitted, the maximum tooth pressure (neglecting friction at theteeth) is exerted along the common normal through the pitch point. This force may be resolved intotangential and radial or normal components. These components act along and at right angles to thecommon tangent to the pitch circles.

If F is the maximum tooth pressure as shown in Fig. 12.10 (b), thenTangential force, FT = F cos φ

and radial or normal force, FR = F sin φ.

∴ Torque exerted on the gear shaft

= FT × r, where r is the pitch circle radius of the gear.

Note : The tangential force provides the driving torque and the radial or normal force produces radial deflectionof the rim and bending of the shafts.

12.12.Effect of Altering the Centre Distance on the Velocity Ratio forInvolute Teeth Gears

In the previous article, we have seen that the velocity ratio for the involute teeth gears is given by

1 1 2

2 2 1

O M O P

O N O P

ω= =ω

...(i)

Let, in Fig. 12.10 (a), the centre of rotation of one of the gears (say wheel 1) is shifted fromO1 to O1' . Consequently the contact point shifts from Q to Q '. The common normal to the teeth at thepoint of contact Q ' is the tangent to the base circle, because it has a contact between two involutecurves and they are generated from the base circle. Let the tangent M' N' to the base circles intersects

1O′ O2 at the pitch point P' . As a result of this, the wheel continues to work* correctly.Now from similar triangles O2NP and O1MP,

1 1

2 2

O M O P

O N O P= ...(ii)

and from similar triangles O2N'P' and O1'M'P',

1 1

2 2

O M O P

O N O P

′ ′′′ =′ ′ ...(iii)

But O2N = O2N', and O1M = O1' M'. Therefore from equations (ii) and (iii),

1 1

2 2

O P O P

O P O P

′′=

′ ...[Same as equation (i)]

Thus we see that if the centre distance is changed within limits, the velocity ratio remainsunchanged. However, the pressure angle increases (from φ to φ′) with the increase in the centredistance.

Example 12.1. A single reduction gear of 120 kW with a pinion 250 mm pitch circle diameterand speed 650 r.p.m. is supported in bearings on either side. Calculate the total load due to thepower transmitted, the pressure angle being 20°.

Solution. Given : P = 120 kW = 120 × 103 W ; d = 250 mm or r = 125 mm = 0.125 m ;N = 650 r.p.m. or ω = 2π × 650/60 = 68 rad/s ; φ = 20°

* It is not the case with cycloidal teeth.


Let T = Torque transmitted in N-m.

We know that power transmitted (P),

120 × 103 = T.ω = T × 68 or T = 120 × 103/68 = 1765 N-m

and tangential load on the pinion,

FT = T /r = 1765 / 0.125 = 14 120 N

∴ Total load due to power transmitted,

F = FT / cos φ = 14 120 / cos 20° = 15 026 N = 15.026 kN Ans.

12.13. Comparison Between Involute and Cycloidal GearsIn actual practice, the involute gears are more commonly used as compared to cycloidal

gears, due to the following advantages : Advantages of involute gears

Following are the advantages of involute gears :1. The most important advantage of the involute gears is that the centre distance for a pair of

involute gears can be varied within limits without changing the velocity ratio. This is not true forcycloidal gears which requires exact centre distance to be maintained.

2. In involute gears, the pressure angle, from the start of the engagement of teeth to the endof the engagement, remains constant. It is necessary for smooth running and less wear of gears. But incycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero atpitch point, starts decreasing and again becomes maximum at the end of engagement. This results inless smooth running of gears.

3. The face and flank of involute teeth are generated by a single curve where as in cycloidalgears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respec-tively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, thebasic rack has straight teeth and the same can be cut with simple tools.Note : The only disadvantage of the involute teeth is that the interference occurs (Refer Art. 12.19) with pinionshaving smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of themating teeth or the angle of obliquity of the teeth.

Advantages of cycloidal gearsFollowing are the advantages of cycloidal gears :1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than

the involute gears, for the same pitch. Due to this reason, the cycloidal teeth are preferred speciallyfor cast teeth.

2. In cycloidal gears, the contact takes place between a convex flank and concave surface,whereas in involute gears, the convex surfaces are in contact. This condition results in less wear incycloidal gears as compared to involute gears. However the difference in wear is negligible.

3. In cycloidal gears, the interference does not occur at all. Though there are advantages ofcycloidal gears but they are outweighed by the greater simplicity and flexibility of the involutegears.

12.14. Systems of Gear TeethThe following four systems of gear teeth are commonly used in practice :

1. 1214 ° Composite system, 2. 1

214 ° Full depth involute system, 3. 20° Full depth involute

system, and 4. 20° Stub involute system.

The 1214 ° composite system is used for general purpose gears. It is stronger but has no inter-

Chapter 12 : Toothed Gearing � 395changeability. The tooth profile of this system has cycloidal curves at the top and bottom and involutecurve at the middle portion. The teeth are produced by formed milling cutters or hobs. The tooth

profile of the 1214 ° full depth involute system was developed for use with gear hobs for spur and

helical gears.The tooth profile of the 20° full depth involute system may be cut by hobs. The increase of

the pressure angle from 1214 ° to 20° results in a stronger tooth, because the tooth acting as a beam is

wider at the base. The 20° stub involute system has a strong tooth to take heavy loads.

12.15. Standard Proportions of Gear Systems

The following table shows the standard proportions in module (m) for the four gear systemsas discussed in the previous article.

Table 12.1. Standard proportions of gear systems.

S. No. Particulars 12

°14 composite or full 20° full depth 20° stub involute

depth involute system involute system system1. Addenddm 1 m 1 m 0.8 m2. Dedendum 1.25 m 1.25 m 1 m3. Working depth 2 m 2 m 1.60 m4. Minimum total depth 2.25 m 2.25 m 1.80 m5. Tooth thickness 1.5708 m 1.5708 m 1.5708 m6. Minimum clearance 0.25 m 0.25 m 0.2 m

7. Fillet radius at root 0.4 m 0.4 m 0.4 m

12.16. Length of Path of ContactConsider a pinion driving the wheel as shown in Fig. 12.11. When the pinion rotates in

clockwise direction, the contact between a pair of involute teeth begins at K (on the flank near thebase circle of pinion or the outer end of the tooth face on the wheel) and* ends at L (outer end of thetooth face on the pinion or on the flank near the base circle of wheel). MN is the common normal atthe point of contacts and the common tangent to the base circles. The point K is the intersection of theaddendum circle of wheel and the common tangent. The point L is the intersection of the addendumcircle of pinion and common tangent.

Fig. 12.11. Length of path of contact.

* If the wheel is made to act as a driver and the directions of motion are reversed, then the contact betweena pair of teeth begins at L and ends at K .


We have discussed in Art. 12.4 that the lengthof path of contact is the length of common normal cut-off by the addendum circles of the wheel and the pinion.Thus the length of path of contact is KL which is the sumof the parts of the path of contacts KP and PL. The partof the path of contact KP is known as path of approachand the part of the path of contact PL is known as pathof recess.

Let rA = O1L = Radius of addendumcircle of pinion,

RA = O2K = Radius of addendumcircle of wheel,

r = O1P = Radius of pitch circle ofpinion, and

R = O2P = Radius of pitch circle of wheel.

From Fig. 12.11, we find that radius of the base circle of pinion, O1M = O1P cos φ = r cos φ

and radius of the base circle of wheel, O2N = O2P cos φ = R cos φ

Now from right angled triangle O2KN,

( )22 2 2 22 2 A( ) ( ) cosKN O K O N RR= − = − φ

and 2 sin sinPN O P R= φ = φ∴ Length of the part of the path of contact, or the path of approach,

( )2 2 2A cos sinKP KN PN R RR= − = − φ − φ

Similarly from right angled triangle O1ML,

and 2 2 2 2 21 1 A( ) ( ) ( ) cosML O L O M r r= − = − φ

1 sin sinMP O P r= φ = φ ∴ Length of the part of the path of contact, or path of recess,

2 2 2A( ) cos sinPL ML MP r r r= − = − φ − φ

∴ Length of the path of contact,

2 2 2 2 2 2A A( ) cos ( ) cos ( )sinKL KP PL R R r r R r= + = − φ + − φ − + φ

12.17. Length of Arc of ContactWe have already defined that the arc of contact is the path traced by a point on the pitch circle

from the beginning to the end of engagement of a given pair of teeth. In Fig. 12.11, the arc of contactis EPF or GPH. Considering the arc of contact GPH, it is divided into two parts i.e. arc GP and arcPH. The arc GP is known as arc of approach and the arc PH is called arc of recess. The anglessubtended by these arcs at O1 are called angle of approach and angle of recess respectively.

Bevel gear

Chapter 12 : Toothed Gearing � 397We know that the length of the arc of approach (arc GP)

Length of path of approach

cos cos

KP= =φ φ

and the length of the arc of recess (arc PH)

Length of path of recess

cos cos

PL= =φ φ

Since the length of the arc of contact GPH is equal to the sum of the length of arc of approachand arc of recess, therefore,

Length of the arc of contact

arc arccos cos cos

Length of path of contact

cos

KP PL KLGP PH= + = + =

φ φ φ

=φ

12.18. Contact Ratio (or Number of Pairs of Teeth in Contact)

The contact ratio or the number of pairs of teeth in contact is defined as the ratio of thelength of the arc of contact to the circular pitch. Mathematically,

Contact ratio or number of pairs of teeth in contact


cp=

where Circular pitch , andcp m= = π

Module.m =Notes : 1. The contact ratio, usually, is not a whole number. For example, if the contact ratio is 1.6, it does notmean that there are 1.6 pairs of teeth in contact. It means that there are alternately one pair and two pairs of teethin contact and on a time basis the average is 1.6.

2. The theoretical minimum value for the contact ratio is one, that is there must always be at least onepair of teeth in contact for continuous action.

3. Larger the contact ratio, more quietly the gears will operate.

Example 12.2. The number of teeth on each of the two equal spur gears in mesh are 40. Theteeth have 20° involute profile and the module is 6 mm. If the arc of contact is 1.75 times the circularpitch, find the addendum.

Solution. Given : T = t = 40 ; φ = 20° ; m = 6 mm

We know that the circular pitch,

pc = π m = π × 6 = 18.85 mm

∴ Length of arc of contact

= 1.75 pc = 1.75 × 18.85 = 33 mm

and length of path of contact

= Length of arc of contact × cos φ = 33 cos 20° = 31 mm

Let RA = rA = Radius of the addendum circle of each wheel.

We know that pitch circle radii of each wheel,

R = r = m.T / 2 = 6 × 40/2 = 120 mm


and length of path of contact

2 2 2 2 2 2A A31 ( ) cos ( ) cos ( ) sinR R r r R r= − φ + − φ − + φ

2 2 2A

2 ( ) cos sinR R R = − φ − φ ...(∵ R = r, and RA = rA)

2 2 2A

31( ) (120) cos 20 120 sin 20

2R= − ° − °

215.5 ( ) 12 715 41AR= − −

2 2A(15.5 41) ( ) 12 715R+ = −

2A A3192 12 715 ( ) or 126.12 mmR R+ = =

We know that the addendum of the wheel,

= A 126.12 120 6.12 mmR R− = − = Ans.

Example 12.3. A pinion having 30 teeth drives agear having 80 teeth. The profile of the gears is involutewith 20° pressure angle, 12 mm module and 10 mmaddendum. Find the length of path of contact, arc of contactand the contact ratio.

Solution. Given : t = 30 ; T = 80 ; φ = 20° ;m = 12 mm ; Addendum = 10 mm

Length of path of contactWe know that pitch circle radius of pinion,

r = m.t / 2 = 12 × 30 / 2 = 180 mmand pitch circle radius of gear,

R = m.T / 2 = 12 × 80 / 2 = 480 mm

∴ Radius of addendum circle of pinion,

rA = r + Addendum = 180 + 10 = 190 mm

and radius of addendum circle of gear,

RA = R + Addendum = 480 + 10 = 490 mmWe know that length of the path of approach,

2 2 2A( ) cos sinKP R R R= − φ − φ ...(Refer Fig.12.11)

2 2 2(490) (480) cos 20 480 sin 20= − ° − ° 191.5 164.2 27.3 mm= − =

and length of the path of recess,

2 2 2A( ) cos sinPL r r r= − φ − φ

2 2 2(190) (180) cos 20 180 sin 20= − ° − ° 86.6 61.6 25 mm= − =

We know that length of path of contact,

KL = KP + PL = 27.3 + 25 = 52.3 mm Ans.

Worm.

Chapter 12 : Toothed Gearing � 399Length of arc of contact

We know that length of arc of contact

Length of path of contact 52.355.66 mm

cos cos 20= = =

φ ° Ans.

Contact ratio

We know that circular pitch,

pc = π.m = π × 12 = 37.7 mm

∴ Length of arc of contact 55.66Contact ratio = 1.5 say 2

37.7cp= = Ans.

Example 12.4. Two involute gears of 20° pressure angle are in mesh. The number of teethon pinion is 20 and the gear ratio is 2. If the pitch expressed in module is 5 mm and the pitch linespeed is 1.2 m/s, assuming addendum as standard and equal to one module, find :

1. The angle turned through by pinion when one pair of teeth is in mesh ; and

2. The maximum velocity of sliding.

Solution. Given : φ = 20° ; t = 20; G = T/t = 2; m = 5 mm ; v = 1.2 m/s ; addendum = 1 module= 5 mm

1. Angle turned through by pinion when one pair of teeth is in mesh

We know that pitch circle radius of pinion,

r = m.t / 2 = 5 × 20 / 2 = 50 mm

and pitch circle radius of wheel,

R = m.T / 2 = m.G.t / 2 = 2 × 20 × 5 / 2 = 100 mm ... ( . )T G t=�


rA = r + Addendum = 50 + 5 = 55 mm

and radius of addendum circle of wheel,

RA = R + Addendum = 100 + 5 = 105 mm

We know that length of the path of approach (i.e. the path of contact when engagementoccurs),

2 2 2A( ) cos sinKP R R R= − φ − φ ...(Refer Fig.12.11)

2 2 2= (105) (100) cos 20 100 sin 20− ° − °

46.85 34.2 12.65 mm= − =

and the length of path of recess (i.e. the path of contact when disengagement occurs),


2 2 2(55) (50) cos 20 50 sin 20 28.6 17.1 11.5 mm= − ° − ° = − =


KL = KP + PL = 12.65 + 11.5 = 24.15 mm


and length of the arc of contact


cos cos 20= = =

φ °We know that angle turned through by pinion

Length of arc of contact × 360° 25.7 36029.45

Circumference of pinion 2 50

× °= = = °π ×

Ans.

2. Maximum velocity of sliding

Let ω1 = Angular speed of pinion, and

ω2 = Angular speed of wheel.

We know that pitch line speed,

v = ω1.r = ω2.R

∴ ω 1 = v/r = 120/5 = 24 rad/s

and ω2 = v/R = 120/10 = 12 rad/s

∴ Maximum velocity of sliding,

vS = (ω1 + ω2) KP ...( )KP PL>�

= (24 + 12) 12.65 = 455.4 mm/s Ans.

Example 12.5. A pair of gears, having 40 and 20 teeth respectively, are rotating in mesh,the speed of the smaller being 2000 r.p.m. Determine the velocity of sliding between the gear teethfaces at the point of engagement, at the pitch point, and at the point of disengagement if the smallergear is the driver. Assume that the gear teeth are 20° involute form, addendum length is 5 mm and themodule is 5 mm.

Also find the angle through which the pinion turns while any pairs of teeth are in contact.

Solution. Given : T = 40 ; t = 20 ; N1 = 2000 r.p.m. ; φ = 20° ; addendum = 5 mm ; m = 5 mm

We know that angular velocity of the smaller gear,

11

2 2 2000209.5 rad/s

60 60

Nπ π ×ω = = =

and angular velocity of the larger gear,

2 120

209.5 104.75 rad/s40

t

Tω = ω × = × = 2

1

...t

T

ω = ω �

Pitch circle radius of the smaller gear,

r = m.t / 2 = 5 × 20/2 = 50 mm

and pitch circle radius of the larger gear,

R = m.t / 2 = 5 × 40/2 = 100 mm

∴ Radius of addendum circle of smaller gear,

rA = r + Addendum = 50 + 5 = 55 mm

and radius of addendum circle of larger gear,

RA = R + Addendum = 100 + 5 = 105 mm

The engagement and disengagement of the gear teeth is shown in Fig. 12.11. The point K isthe point of engagement, P is the pitch point and L is the point of disengagement. MN is the commontangent at the points of contact.

Chapter 12 : Toothed Gearing � 401We know that the distance of point of engagement K from the pitch point P or the length of

the path of approach,

2 2 2A( ) cos sinKP R R R= − φ − φ

2 2 2(105) (100) cos 20 100 sin 20= − ° − °

46.85 34.2 12.65 mm= − =and the distance of the pitch point P from the point of disengagement L or the length of the path ofrecess,


2 2 2(55) (50) cos 20 50 sin 20 28.6 17.1 11.5 mm= − ° − ° = − =Velocity of sliding at the point of engagement

We know that velocity of sliding at the point of engagement K,

SK 1 2( ) (209.5 104.75) 12.65 3975 mm/sv KP= ω + ω = + = Ans.

Velocity of sliding at the pitch point

Since the velocity of sliding is proportional to the distance of the contact point from the pitchpoint, therefore the velocity of sliding at the pitch point is zero. Ans.

Velocity of sliding at the point of disengagement

We know that velocity of sliding at the point of disengagement L,

SL 1 2( ) (209.5 104.75) 11.5 3614 mm/sv PL= ω + ω = + = Ans.

Angle through which the pinion turns

We know that length of the path of contact,

KL = KP + PL = 12.65 + 11.5 = 24.15 mm

and length of arc of contact 24.15

25.7 mmcos cos 20

KL= = =φ °

Circumference of the smaller gear or pinion

= 2 π r = 2π × 50 = 314.2 mm

∴ Angle through which the pinion turns

360Length of arc of contact

Circumference of pinion

°= ×

36025.7 29.45

314.2

°= × = ° Ans.

Example 12.6. The following data relate to a pair of 20° involute gears in mesh :

Module = 6 mm, Number of teeth on pinion = 17, Number of teeth on gear = 49 ; Addendaon pinion and gear wheel = 1 module.

Find : 1. The number of pairs of teeth in contact ; 2. The angle turned through by the pinionand the gear wheel when one pair of teeth is in contact, and 3. The ratio of sliding to rolling motionwhen the tip of a tooth on the larger wheel (i) is just making contact, (ii) is just leaving contact withits mating tooth, and (iii) is at the pitch point.


Solution. Given : φ = 20° ; m = 6 mm ; t = 17 ; T = 49 ; Addenda on pinion and gear wheel= 1 module = 6 mm

1. Number of pairs of teeth in contact


r = m.t / 2 = 6 × 17 / 2 = 51 mm

and pitch circle radius of gear,

r = m.T / 2 = 6 × 49 / 2 = 147 mm


rA = r + Addendum = 51 + 6 = 57 mm

and radius of addendum circle of gear,

RA = R + Addendum = 147 + 6 = 153 mm

We know that the length of path of approach (i.e. the path of contact when engagementoccurs),

2 2 2A( ) cos sinKP R R R= − φ − φ ...(Refer Fig. 12.11)

2 2 2(153) (147) cos 20 147 sin 20= − ° − °

65.8 50.3 15.5 mm= − =and length of path of recess (i.e. the path of contact when disengagement occurs),


2 2 2(57) (51) cos 20 51 sin 20= − ° − °

30.85 17.44 13.41 mm= − =∴ Length of path of contact,

15.5 13.41 28.91 mmKL KP PL= + = + =

Racks


and length of arc of contact Length of path of contact 28.91

30.8 mmcos cos 20

= = =φ °


. 6 18.852 mmcp m= π = π × =∴ Number of pairs of teeth in contact (or contact ratio)

Length of arc of contact 30.81.6 say 2

Circular pitch 18.852= = = Ans.

2. Angle turned through by the pinion and gear wheel when one pair of teeth is in contact

We know that angle turned through by the pinion

Length of arc of contact 360° 30.8 36034.6

Circumference of pinion 2 51

× ×= = = °π ×

Ans.

and angle turned through by the gear wheel

Length of arc of contact 360° 30.8 36012

Circumference of gear 2 147

× ×= = = °π ×

Ans.

3. Ratio of sliding to rolling motion

Let ω1 = Angular velocity of pinion, and

ω2 = Angular velocity of gear wheel.

We know that 1 2 2 1 1 1/ / or / 17 / 49 0.347T t t Tω ω = ω = ω × = ω × = ω

and rolling velocity, R 1 2 1 1. . 51 51 mm/sv r R= ω = ω = ω × = ω(i) At the instant when the tip of a tooth on the larger wheel is just making contact with its

mating teeth (i.e. when the engagement commences), the sliding velocity

S 1 2 1 1 1( ) ( 0.347 ) 15.5 20.88 mm/sv KP= ω + ω = ω + ω = ω

∴ Ratio of sliding velocity to rolling velocity,

S 1

R 1

20.880.41

51

v

v

ω= =ω

Ans.

(ii) At the instant when the tip of a tooth on the larger wheel is just leaving contact with itsmating teeth (i.e. when engagement terminates), the sliding velocity,

S 1 2 1 1 1( ) ( 0.347 ) 13.41 18.1 mm/sv PL= ω + ω = ω + ω = ω

∴ Ratio of sliding velocity to rolling velocity

S 1

R 1

18.10.355

51

v

v

ω= =ω

Ans.

(iii) Since at the pitch point, the sliding velocity is zero, therefore the ratio of sliding velocityto rolling velocity is zero. Ans.

Example 12.7. A pinion having 18 teeth engages with an internal gear having 72 teeth. Ifthe gears have involute profiled teeth with 20° pressure angle, module of 4 mm and the addenda onpinion and gear are 8.5 mm and 3.5 mm respectively, find the length of path of contact.

Solution. Given : t = 18 ; T = 72 ; φ = 20° ; m = 4 mm ; Addendum on pinion = 8.5 mm ;Addendum on gear = 3.5 mm


Fig. 12.12 shows a pinion with centre O1, in mesh with internal gear of centre O2. It may benoted that the internal gears have the addendum circle and the tooth faces inside the pitch circle.

We know that the length of path of contact is the length of the common tangent to the twobase circles cut by the addendum circles. From Fig. 12.12, we see that the addendum circles cut thecommon tangents at points K and L. Therefore the length of path of contact is KL which is equal to thesum of KP (i.e. path of approach) and PL (i.e. path of recess).

Fig. 12.12

We know that pitch circle radius of the pinion,

1 . / 2 4 18/ 2 36 mmr O P mt= = = × =and pitch circle radius of the gear,

2 . / 2 4 72 / 2 144 mmR O P mT= = = × =∴ Radius of addendum circle of the pinion,

A 1 1r O L O P= = + Addendum on pinion = 36 + 8.5 = 44.5 mm

and radius of addendum circle of the gear,

A 2 2 Addendum on wheel = 144 – 3.5 = 140.5 mmR O K O P= = −From Fig. 12.12, radius of the base circle of the pinion,

1 1 cos cos 36 cos 20 33.83 mmO M O P r= φ = φ = ° =and radius of the base circle of the gear,

2 2 cos cos 144 cos 20 135.32 mmO N O P R= φ = φ = ° =We know that length of the path of approach,

2 22 2 2sin 20 ( ) ( )KP PN KN O P O K O N= − = ° − −

= 144 × 0.342 – 2 2(140.5) (135.32)− = 49.25 – 37.8 = 11.45 mm

Chapter 12 : Toothed Gearing � 405and length of the path of recess,

2 21 1 1( ) ( ) sin 20PL ML MP O L O M O P= − = − − °

2 2(44.5) (33.83) 36 0.342 28.9 12.3 16.6 mm= − − × = − =


11.45 16.6 28.05 mmKL KP PL= + = + = Ans.

12.19. Interference in Involute Gears

Fig. 12.13 shows a pinion with centre O1, in mesh with wheel or gear with centre O2. MN isthe common tangent to the base circles and KL is the path of contact between the two mating teeth.

Fig. 12.13. Interference in involute gears.

A little consideration will show, that if the radius of the addendum circle of pinion isincreased to O1N, the point of contact L will move from L to N. When this radius is further increased,the point of contact L will be on the inside of base circle of wheel and not on the involute profile oftooth on wheel. The tip of tooth on the pinion will then undercut the tooth on the wheel at the root andremove part of the involute profile of tooth on the wheel. This effect is known as interference, andoccurs when the teeth are being cut. In brief, the phenomenon when the tip of tooth undercuts theroot on its mating gear is known as interference.

Similarly, if the radius of the addendum circle of the wheel increases beyond O2M, then thetip of tooth on wheel will cause interference with the tooth on pinion. The points M and N are calledinterference points. Obviously, interference may be avoided if the path of contact does not extendbeyond interference points. The limiting value of the radius of the addendum circle of the pinion is*O1N and of the wheel is O2M.

From the above discussion, we conclude that the interference may only be avoided, if thepoint of contact between the two teeth is always on the involute profiles of both the teeth. In other

* From Fig. 12.13, we see that

2 2 2 21 1( ) ( ) ( ) [ ) sin ]bO N O M MN r r R= + = + + φ

where rb = Radius of base circle of pinion = O1P cos φ = r cos φ

and 2 2 2 22 2( ) ( ) ( ) [ ) sin ]bO M O N MN R r R= + = + + φ

where Rb = Radius of base circle of wheel = O2P cos φ = R cos φ


words, interference may only be prevented, if the addendum circles of the two mating gears cut thecommon tangent to the base circles between the points of tangency.

When interference is just avoided, the maximum length of path of contact is MN when themaximum addendum circles for pinion and wheel pass through the points of tangency N and M re-spectively as shown in Fig. 12.13. In such a case,

Maximum length of path of approach,

MP = r sin φand maximum length of path of recess,

PN = R sin φ∴ Maximum length of path of contact,

MN = MP + PN = r sin φ + R sin φ = (r + R) sin φand maximum length of arc of contact

( ) sin

( ) tancos

r Rr R

+ φ= = + φφ

Note : In case the addenda on pinion and wheel is such that the path of approach and path of recess are half oftheir maximum possible values, then

Path of approach, 12

KP MP=

or 2 2 2

Asin

( ) cos sin2

rR R R

φ− φ − φ =

and path of recess, 12

PL PN=

or 2 2 2

Asin

( ) cos sin2

Rr r r

φ− φ − φ =

∴ Length of the path of contact

1 12 2

( ) sin

2

r RKP PL MP PN

+ φ= + = + =

Example 12.8. Two mating gears have 20 and 40 involute teeth of module 10 mm and 20°pressure angle. The addendum on each wheel is to be made of such a length that the line of contacton each side of the pitch point has half the maximum possible length. Determine the addendumheight for each gear wheel, length of the path of contact, arc of contact and contact ratio.

Solution. Given : t = 20 ; T = 40 ; m = 10 mm ; φ = 20°

Addendum height for each gear wheel

We know that the pitch circle radius of the smaller gear wheel,

r = m.t / 2 = 10 × 20 / 2 = 100 mm

and pitch circle radius of the larger gear wheel,

R = m.T / 2 = 10 × 40 / 2 = 200 mm

Let RA = Radius of addendum circle for the larger gear wheel, and

rA = Radius of addendum circle for the smaller gear wheel.

Since the addendum on each wheel is to be made of such a length that the line of contact oneach side of the pitch point (i.e. the path of approach and the path of recess) has half the maximumpossible length, therefore


Path of approach, 12

KP MP= ...(Refer Fig. 12.13)

or 2 2 2

A.sin

( ) cos sin2

rR R R

φ− φ − φ =

or 2 2 2

A100 sin 20

( ) (200) cos 20 200 sin 20 50 sin 202

R× °

− ° − ° = = °

2A( ) 35 320 50 sin 20 200 sin 20 250 0.342 85.5R − = ° + ° = × =

2 2A( ) 35 320 (85.5) 7310R − = = ...(Squaring both sides)

(RA)2 = 7310 + 35 320 = 42 630 or RA = 206.5 mm

∴ Addendum height for larger gear wheel

A 206.5 200 6.5 mmR R= − = − = Ans.

Now path of recess, 12PL PN=

or 2 2 2A

.sin( ) cos sin

2

Rr r r

φ− φ − φ =

or 2 2 2A

200 sin 20( ) (100) cos 20 100 sin 20 100 sin 20

2r

°− ° − ° = = °

2 2 2A( ) (100) cos 20 100 sin 20 100 sin 20 200 0.342 68.4r − ° = ° + ° = × =

2 2A( ) 8830 (68.4) 4680r − = = ...(Squaring both sides)

2A A( ) 4680 8830 13 510 or 116.2 mmr r= + = =

∴ Addendum height for smaller gear wheel

A 116.2 100 6.2 mmr r= − = − = Ans.

Length of the path of contact

We know that length of the path of contact

1 1 ( )sin

2 2 2

r RKP PL MP PN

+ φ= + = + =

(100 200) sin 20

51.3 mm2

+ °= = Ans.


We know that length of the arc of contact

Length of the path of contact 51.3

54.6 mmcos cos 20

= = =φ °

Ans.

Contact ratio


Pc = π m = π × 10 = 31.42 mm

∴ Length of the path of contact 54.6

Contact ratio = 1.74 say 231.42cp

= = Ans.


12.20. Minimum Number of Teeth on the Pinion in Order to AvoidInterference

We have already discussed in the previous article that in order to avoid interference, theaddendum circles for the two mating gears must cut the common tangent to the base circles betweenthe points of tangency. The limiting condition reaches, when the addendum circles of pinion andwheel pass through points N and M (see Fig. 12.13) respectively.

Let t = Number of teeth on the pinion,,

T = Number of teeth on the wheel,

m = Module of the teeth,

r = Pitch circle radius of pinion = m.t / 2

G = Gear ratio = T / t = R / r

φ = Pressure angle or angle of obliquity.

From triangle O1NP,

2 2 21 1 1 1

2 2 2

( ) ( ) ( ) 2 cos

sin 2 . sin cos (90 )

O N O P PN O P PN O PN

r R r R

= + − × ×

= + φ − φ ° + φ

2...( sin sin )PN O P R= φ = φ�

2 2 2 2

2 2 22 2 2

2

sin 2 . sin

sin 2 sin1 1 2 sin

r R r R

R R R Rr r

r r rr

= + φ + φ

φ φ = + + = + + φ ∴ Limiting radius of the pinion addendum circle,

2 21

.1 2 sin 1 2 sin

2

R R m t T TO N r

r r t t = + + φ = + + φ

Let AP.m = Addendum of the pinion, where A P is a fraction by which the standardaddendum of one module for the pinion should be multiplied in orderto avoid interference.

We know that the addendum of the pinion

= O1N – O1P

∴ 2P

. .. 1 2 sin

2 2

m t T T m tA m

t t = + + φ − 1...( . / 2)O P r m t= =�

2.1 sin 12

2

m t T T

t t

= + φ −+

or 2P 1 sin 12

2

t T TAt t

= + φ −+

∴ P P

22

2 2

1 + ( + 2)sin 11 sin 12

A At

T T G Gt t

= = φ −+ φ −+

Chapter 12 : Toothed Gearing � 409This equation gives the minimum number of teeth required on the pinion in order to avoid

interference.

Notes : 1. If the pinion and wheel have equal teeth, then G = 1. Therefore the above equation reduces to

2

2

1 3 sin 1

pAt =

+ φ −2. The minimum number of teeth on the pinion which will mesh with any gear (also rack) without

interference are given in the following table :

Table 12.2. Minimum number of teeth on the pinion

S. No. System of gear teeth Minimum number of teeth on the pinion

1. 12

14 ° Composite 12

2. 12

14 ° Full depth involute 32

3. 20° Full depth involute 18

4. 20° Stub involute 14

12.21. Minimum Number of Teeth on the Wheel in Order to AvoidInterference

Let T = Minimum number of teeth required on the wheel in order to avoidinterference,

and AW.m = Addendum of the wheel, where AW is a fraction by which the standardaddendum for the wheel should be multiplied.

Using the same notations as in Art. 12.20, we have from triangle O2MP2 2 2

2 2 2 22 2 2

( ) ( ) ( ) 2 cos

sin 2 . sin cos (90 )

O M O P PM O P PM O PM

R r R r

= + − × ×= + φ − φ ° + φ

1...( sin )PM O P r= φ =�

= R2 + r2 sin2 φ + 2R.r sin2 φ2 2

2 2 22

sin 2 sin1 1 2 sin

r r r rR R

R R RR

2 φ φ = + + = + + φ ∴ Limiting radius of wheel addendum circle,

2 22

.1 sin 1 sin2 2

2

r mT tr tO M R

R TR T = + φ = + φ+ +

We know that the addendum of the wheel

= O2M – O2P

∴ 2W

. .1 sin2

2 2

mT t mTtA m

T T = + φ −+

2...( . / 2)O P R mT= =�

2.1 sin 12

2

mT t tT T

= + φ −+

or2

W 1 sin 122

T t tAT T

= + φ −+


∴ W W

2 2

2 2

1 11 sin 1 1 sin 12 2

A AT

t tT GT G

= = + φ − + φ −+ +

Notes : 1. From the above equation, we may also obtain the minimum number of teeth on pinion.

Multiplying both sides by ,t

T

W

2

2

1 11 sin 12

tAt TT

T

G G

×× =

+ φ −+

W

2

2

1 11 sin 12

At

GG G

= + φ −+

2. If wheel and pinion have equal teeth, then G = 1, and

W

2

2

1 3 sin 1

AT =

+ φ −

Example 12.9. Determine the minimum number of teeth required on a pinion, in order toavoid interference which is to gear with,1. a wheel to give a gear ratio of 3 to 1 ; and 2. an equal wheel.

The pressure angle is 20° and a standard addendum of 1 module for the wheel may beassumed.

Solution. Given : G = T / t = 3 ; φ = 20° ; A W = 1 module1. Minimum number of teeth for a gear ratio of 3 : 1

We know that minimum number of teeth required on a pinion,

W

2

2

1 11 sin 12

At

GG G

×=

+ φ −+

2

2 1 215.04 or 16

0.1331 13 1 sin 20 123 3

×= = = + ° −+

Ans.

2. Minimum number of teeth for equal wheelWe know that minimum number of teeth for equal wheel,

W

2 2

2 2 1 2

0.1621 3 sin 1 1 3 sin 20 1

At

× ×= = =+ φ − + ° −

12.34 or 13= Ans.

Example 12.10. A pair of spur gears with involute teeth is to give a gear ratio of 4 : 1. Thearc of approach is not to be less than the circular pitch and smaller wheel is the driver. The angle ofpressure is 14.5°. Find : 1. the least number of teeth that can be used on each wheel, and 2. theaddendum of the wheel in terms of the circular pitch ?

Chapter 12 : Toothed Gearing � 411Solution. Given : G = T/t = R/r = 4 ; φ = 14.5°

1. Least number of teeth on each wheelLet t = Least number of teeth on the smaller wheel i.e. pinion,

T = Least number of teeth on the larger wheel i.e. gear, andr = Pitch circle radius of the smaller wheel i.e. pinion.

We know that the maximum length of the arc of approach

Maximum length of the path of approach sintan

cos cos

rr

φ= = = φφ φ

and circular pitch,2

cr

p mt

π= π =2

...r

mt

= �

Since the arc of approach is not to be less than the circular pitch, therefore

2 2 2tan or 24.3 say 25

tan tan 14.5

rr t

t

π π πφ = = = =φ °

Ans.

and . 4 25 100T G t= = × = Ans. ...( / )G T t=�

2. Addendum of the wheelWe know that addendum of the wheel

2.1 sin 12

2

mT t tT T

= + φ −+

2100 25 251 sin 14.5 12

2 100 100

m × = + ° −+

50 0.017 0.85 0.85 / 0.27c cm m p p= × = = × π = Ans.

...( / )cm p= π�

Example 12.11. A pair of involute spur gears with 16° pressure angle and pitch of module6 mm is in mesh. The number of teeth on pinion is 16 and its rotational speed is 240 r.p.m. When thegear ratio is 1.75, find in order that the interference is just avoided ; 1. the addenda on pinion andgear wheel ; 2. the length of path of contact ; and 3. the maximum velocity of sliding of teeth on eitherside of the pitch point.

Solution. Given : φ = 16° ; m = 6 mm ; t = 16 ; N1 = 240 r.p.m. or ω1 = 2π × 240/60= 25.136 rad/s ; G = T / t = 1.75 or T = G.t = 1.75 × 16 = 28

1. Addenda on pinion and gear wheel

We know that addendum on pinion

2

2

.1 sin 12

2

6 16 28 281 sin 16 12

2 16 16

m t T T

t t

= + φ −+ × = + ° −+

48 (1.224 1) 10.76 mm= − = Ans.

2.and addendum on wheel 1 sin 12

2

mT t tT T

= + φ −+


26 28 16 16

1 sin 16 122 28 28

× = + ° −+

84 (1.054 1) 4.56 mm= − = Ans.

2. Length of path of contact

We know that the pitch circle radius of wheel,

. / 2 6 28 / 2 84 mmR m T= = × =and pitch circle radius of pinion,

. / 2 6 16 / 2 48 mmr m t= = × =∴ Addendum circle radius of wheel,

A Addendum of wheel 84 10.76 94.76 mmR R= + = + =and addendum circle radius of pinion,

rA = r + Addendum of pinion = 48 + 4.56 = 52.56 mm

We know that the length of path of approach,


2 2 2(94.76) (84) cos 16 84 sin16= − ° − °

49.6 23.15 26.45 mm= − =and the length of the path of recess,


2 2 2(52.56) (48) cos 16 48 sin16= − ° − °

25.17 13.23 11.94 mm= − =∴ Length of the path of contact,

26.45 11.94 38.39 mmKL KP PL= + = + = Ans.

3. Maximum velocity of sliding of teeth on either side of pitch point

Let ω2 = Angular speed of gear wheel.

We know that 1

2

1.75T

t

ω = =ω

or 12

25.13614.28 rad/s

1.75 1.75

ωω = = =

∴ Maximum velocity of sliding of teeth on the left side of pitch point i.e. at point K

1 2( ) (25.136 14.28) 26.45 1043 mm/sKP= ω + ω = + = Ans.

and maximum velocity of sliding of teeth on the right side of pitch point i.e. at point L

1 2( ) (25.136 14.28) 11.94 471 mm/sPL= ω + ω = + = Ans.

Example 12.12. A pair of 20° full depth involute spur gears having 30 and 50 teeth respec-tively of module 4 mm are in mesh. The smaller gear rotates at 1000 r.p.m. Determine : 1. slidingvelocities at engagement and at disengagement of pair of a teeth, and 2. contact ratio.

Solution. Given: φ = 20° ; t = 30 ; T = 50 ; m = 4 ; N1 = 1000 r.p.m. or ω1 = 2π × 1000/60= 104.7 rad/s

Chapter 12 : Toothed Gearing � 4131. Sliding velocities at engagement and at disengagement of pair of a teeth

First of all, let us find the radius of addendum circles of the smaller gear and the larger gear.We know that

Addendum of the smaller gear,

2.1 sin 12

2

m t T Tt t

= + φ −+

24 30 50 501 sin 20 12

2 30 30

× = + ° −+

60(1.31 1) 18.6 mm= − =

and addendum of the larger gear,

2

2

.1 sin 12

2

4 50 30 301 sin 20 12

2 50 50

100(1.09 1) 9 mm

mT t tT T

= + φ −+ × = + ° −+

= − =

Pitch circle radius of the smaller gear,

. / 2 4 30 / 2 60 mmr m t= = × =

∴ Radius of addendum circle of the smaller gear,

rA = r + Addendum of the smaller gear = 60 + 18.6 = 78.6 mm

Pitch circle radius of the larger gear,

R = m.T / 2 = 4 × 50 / 2 = 100 mm

∴ Radius of addendum circle of the larger gear,

RA = R + Addendum of the larger gear = 100 + 9 = 109 mm

We know that the path of approach (i.e. path of contact when engagement occurs),


2 2 2(109) (100) cos 20 100 sin 20 55.2 34.2 21 mm= − ° − ° = − =

and the path of recess (i.e. path of contact when disengagement occurs),


2 2 2(78.6) (60) cos 20= − ° – 60 sin 20° = 54.76 – 20.52 = 34.24 mm

Let ω2 = Angular speed of the larger gear in rad/s.

We know that 1 12

2

10.47 30or 62.82 rad/s

50

tT

t T

ω ω × ×= ω = = =ω

∴ Sliding velocity at engagement of a pair of teeth

1 2( ) (104.7 62.82)21 3518 mm/sKP= ω + ω = + =

= 3.518 m/s Ans.


and sliding velocity at disengagement of a pair of teeth

1 2( ) (104.7 62.82)34.24 5736 mm/sPL= ω + ω = + =

= 5.736m/s Ans.

2. Contact ratio

We know that the length of the arc of contact

Length of thepath of contact 21 34.24

cos cos cos 2058.78 mm

KP PL+ += = =φ φ °

=

and Circular pitch = π × m = 3.142 × 4 = 12.568 mm

∴ Length of arc of contact 58.78Contact ratio = 4.67 say5

Circular pitch 12.568= = Ans.

Example 12.13. Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1.The teeth are of involute form ; module = 6 mm, addendum = one module, pressure angle = 20°. Thepinion rotates at 90 r.p.m. Determine : 1. The number of teeth on the pinion to avoid interference onit and the corresponding number of teeth on the wheel, 2. The length of path and arc of contact,3.The number of pairs of teeth in contact, and 4. The maximum velocity of sliding.

Solution. Given : G = T / t = 3 ; m = 6 mm ; A P = A W = 1 module = 6 mm ; φ = 20° ;N1 = 90 r.p.m. or ω1 = 2π × 90 / 60 = 9.43 rad/s

1. Number of teeth on the pinion to avoid interference on it and the corresponding number of teeth on the wheel

We know that number of teeth on the pinion to avoid interference,

P

2 2

2 2 6

1 ( 2) sin 1 1 3 (3 2) sin 20 1

At

G G

×= =+ + φ − + + ° −

= 18.2 say 19 Ans.

and corresponding number of teeth on the wheel,

T = G.t = 3 × 19 = 57 Ans.

2. Length of path and arc of contact


r = m.t / 2 = 6 × 19/2 = 57 mm


rA = r + Addendum on pinion (A P) = 57 + 6 = 63 mm

and pitch circle radius of wheel,

R = m.T / 2 = 6 × 57 / 2 = 171 mm

∴ Radius of addendum circle of wheel,

A WAddendum on wheel ( ) 171 6 177 mmR R A= + = + =We know that the path of approach (i.e. path of contact when engagement occurs),


2 2 2(177) (171) cos 20= − ° – 171 sin 20° = 74.2 – 58.5 = 15.7 mm

Chapter 12 : Toothed Gearing � 415and the path of recess (i.e. path of contact when disengagement occurs),

2 2 2A

2 2 2

( ) cos sin

(63) (57) cos 20 57sin 20 33.17 19.5 13.67 mm

PL r r r= − φ − φ

= − ° − ° = − =

∴ Length of path of contact,

15.7 13.67 29.37 mmKL KP PL= + = + = Ans.

We know that length of arc of contact


cos cos 20= = =

φ °Ans.

3. Number of pairs of teeth in contactWe know that circular pitch,

6 18.852 mmcp m= π × = π × =∴ Number of pairs of teeth in contact

Length of arc of contact 31.251.66 say 2

18.852cp= = = Ans.

4. Maximum velocity of sliding

2

12 1

2

Let Angular speed of wheel in rad/s.

19We know that or 9.43 3.14 rad/s

57

T t

t T

ω =ω

= ω = ω × = × =ω

∴ Maximum velocity of sliding,

S 1 2( )v KP= ω + ω ...( )KP PL>�

(9.43 3.14) 15.7 197.35 mm/s= + = Ans.

12.22. Minimum Number of Teeth on a Pinion for Involute Rack in Order toAvoid Interference

A rack and pinion in mesh is shown in Fig. 12.14.

Fig. 12.14. Rack and pinion in mesh.

Let t = Minimum number of teeth on the pinion,


r = Pitch circle radius of the pinion = m.t / 2, and

φ = Pressure angle or angle of obliquity, and

A R.m = Addendum for rack, where A R is the fraction by which the standard

addendum of one module for the rack is to be multiplied.

We know that a rack is a part of toothed wheel of infinite diameter. Therefore its base circlediameter and the profiles of the involute teeth are straight lines. Since these straight profiles aretangential to the pinion profiles at the point of contact, therefore they are perpendicular to the tangentPM. The point M is the interference point.

Addendum for rack,

R . sinA m LH PL= = φ2( sin ) sin sinOP OP= φ φ = φ ...(∵ PL = OP sin φ)

2 2sin.

sin2

rm t

= φ = × φ

∴ R2

2

sint

A=

φExample 12.14. A pinion of 20 involute teeth and 125 mm pitch circle diameter drives a

rack. The addendum of both pinion and rack is 6.25 mm. What is the least pressure angle which canbe used to avoid interference ? With this pressure angle, find the length of the arc of contact and theminimum number of teeth in contact at a time.

Solution. Given : T = 20 ; d = 125 mm or r = OP = 62.5 mm ; LH = 6.25 mm

Least pressure angle to avoid interference

Let φ = Least pressure angle to avoid interference.

We know that for no interference, rack addendum,

2 2 6.25sin or sin 0.1

6.25

LHLH r

r= φ φ = = =

∴ sin 0.3162φ= or 18.435φ= ° Ans.Length of the arc of contact

We know that length of the path of contact,

2 2( ) ( )KL OK OL= − ...(Refer Fig. 12.14)

2 2

2 2

( 6.25) ( cos )

(62.5 6.25) (62.5 cos 18.435 )

4726.56 3515.62 34.8 mm

OP OP=

=

+ − φ

= + − °

− =∴ Length of the arc of contact

Length of the path of contact 34.836.68 mm

cos cos18.435= = =

φ °Ans.

Minimum number of teeth


/ 125 / 20 19.64 mmcp d T= π = π × =


Fig. 12.15. Helical gear.

Crossed helical gears.

and the number of pairs of teeth in contact

Length of the arc of contact 36.681.87

Circular pitch ( ) 19.64cp= = =

∴ Minimum number of teeth in contact

= 2 or one pair Ans.

12.23. Helical Gears

A helical gear has teeth in the formof helix around the gear. Two such gearsmay be used to connect two parallel shaftsin place of spur gear. The helixes may beright handed on one wheel and left handedon the other. The pitch surfaces are cylin-drical as in spur gearing, but the teeth in-stead of being parallel to the axis, windaround the cylinders helically like screwthreads. The teeth of helical gears withparallel axis have line contact, as in spurgearing. This provides gradual engage-ment and continuous contact of the engaging teeth. Hence helical gears give smooth drive with ahigh efficiency of transmission.

We have already discussed that the helical gears may beof single helical type or double helical type. In case of singlehelical gears, there is some axial thrust between the teeth, whichis a disadvantage. In order to eliminate this axial thrust, doublehelical gears are used. It is equivalent to two single helical gears,in which equal and opposite thrusts are produced on each gearand the resulting axial thrust is zero.

The following definitions may be clearly understood inconnection with a helical gear as shown in Fig. 12.15.

1. Normal pitch. It is the distance between similar faces of adjacent teeth, along a helix onthe pitch cylinder normal to the teeth. It is denoted by pN.

2. Axial pitch. It is the distance measured parallel to the axis, between similar faces of adja-cent teeth. It is the same as circular pitch and is therefore denoted by pc. If α is the helix angle, thencircular pitch,

N

coscp

p =α

Note : The helix angle is also known as spiral angle of the teeth.

12.24. Spiral Gears

We have already discussed that spiral gears (also known as skew gears or screw gears) areused to connect and transmit motion between two non-parallel and non-intersecting shafts. The pitchsurfaces of the spiral gears are cylindrical and the teeth have point contact. These gears are onlysuitable for transmitting small power. We have seen that helical gears, connected on parallel shafts,are of opposite hand. But spiral gears may be of the same hand or of opposite hand.


12.25. Centre Distance for a Pair of Spiral Gears

The centre distance, for a pair of spiral gears,is the shortest distance between the two shafts makingany angle between them. A pair of spiral gears 1 and 2,both having left hand helixes (i.e. the gears are of thesame hand) is shown in Fig. 12.16. The shaft angle θ isthe angle through which one of the shafts must be rotatedso that it is parallel to the other shaft, also the two shaftsbe rotating in opposite directions.

Let α1 and α2 = Spiral angles of gear teeth for gears 1 and 2 respectively,

pc1 and pc2 = Circular pitches of gears 1 and 2,

T1 and T2 = Number of teeth on gears1 and 2,

d1 and d2= Pitch circle diameters of gears 1 and 2,

N1 and N2 = Speed of gears 1 and 2,

G = Gear ratio = 2 1

1 2

,T N

T N=

pN = Normal pitch, and

L = Least centre distance between the axes of shafts.

Since the normal pitch is same for both the spiral gears, therefore

N N1 2

1 2

, andcos cosc c

p pp p= =

α α

Fig. 12.16. Centre distance fora pair of spiral gears.

Helical gears


We know that1 11

1 11

, or cc

p Tdp d

T

×π= =π

and2 22

2 22

, or cc

p Tdp d

T

×π= =π

∴ 1 2 1 1 2 21

2 2c cd d p T p T

L+ × × = = + π π

N N21 11 2

1 1 2

N 1

1 2

cos cos2 2

1

cos cos2

c cp pTT T

p p GT

GP T

+ × + ×= = α απ π × += α απ

Notes : 1. If the pair of spiral gears have teeth of the same hand, then

θ = α1 + α2

and for a pair of spiral gears of opposite hand,

θ = α1 – α2

2. When θ = 90°, then both the spiral gears must have teeth of the same hand.

12.26. Efficiency of Spiral Gears

A pair of spiral gears 1 and 2 in mesh is shown in Fig. 12.17. Let the gear 1 be the driver andthe gear 2 the driven. The forces acting on each of a pair of teeth in contact are shown in Fig. 12.17.The forces are assumed to act at the centre of the width of each teeth and in the plane tangential to thepitch cylinders.

Fig. 12.17. Efficiency of spiral gears.

Let F1 = Force applied tangentially on the driver,

F2 = Resisting force acting tangentially on the driven,

Fa1 = Axial or end thrust on the driver,


Fa2 = Axial or end thrust on the driven,

RN = Normal reaction at the point of contact,

φ = Angle of friction,

R = Resultant reaction at the point of contact, and

θ = Shaft angle = α1+ α2

...( Both gears are of the same hand)�

From triangle OPQ, F1 = R cos (α1 – φ)

∴ Work input to the driver

= F1× π d1.N1 = R cos (α1 – φ) π d1.N1

From triangle OST, F2 = R cos (α2 + φ)

∴ Work output of the driven

= F2 × π d2.N2 = R cos (α2 + φ) π d2.N2

∴ Efficiency of spiral gears,

2 2 2

1 1 1

cos( ) .Work output

Work input cos( ) .

R d N

R d N

α + φ πη = =α − φ π

2 2 2

1 1 1

cos ( ) .

cos ( ) .

d N

d N

α + φ=α − φ

...(i)

We have discussed in Art. 12.25, that pitch circle diameter of gear 1,

1 1 N 11

1coscp T P T

d×

= = ×π α π

and pitch circle diameter of gear 2,

2 2 N 22

2coscp T P T

d×

= = ×π α π

∴ 2 2 1

1 1 2

cos

cos

d T

d T

α=α

...(ii)

We know that 2 1

1 2

N T

N T= ...(iii)

Multiplying equations (ii) and (iii), we get,

2 2 1

1 1 2

. cos

. cos

d N

d N

α=α

Substituting this value in equation (i), we have

2 1

1 2

cos ( ) cos

cos ( ) cos

α + φ αη =α − φ α

...(iv)

1 2 1 2

2 1 2 1

cos ( ) cos ( )

cos ( ) cos ( )

α + α + φ + α − α − φ=α + α − φ + α − α + φ

1... cos cos [cos( ) cos ( )]

2A B A B A B

= + + − �


1 2

2 1

cos ( ) cos ( )

cos ( ) cos ( )

θ + φ + α − α − φ=θ − φ + α − α + φ

...(v)

1 2...( )θ = α + α�

Since the angles θ and φ are constants, therefore the efficiency will be maximum, whencos (α1 – α2 – φ) is maximum, i.e.

cos (α1 – α2 – φ) = 1 or α1 – α2 – φ = 0

∴ α 1 = α2 + φ and α2 = α1 – φSince α1 + α2 = θ, therefore

1 2 1 1or2

θ + φα = θ − α = θ − α + φ α =

Similarly, 2 2

θ − φα =

Substituting α1 = α2 + φ and α2 = α1 – φ, in equation (v), we get

cos ( ) 1

cos ( ) 1maxθ + φ +η =θ − φ +

...(vi)

Note: From Fig. 12.17, we find that 1 2N

1 2cos cos

F FR = =

α α

∴ Axial thrust on the driver, Fa1 = RN.sin α1 = F1.tan α1

and axial thrust on the driven, Fa2 = RN.sin α2 = F2.tan α2

Example 12.15. A pair of spiral gears is required to connect two shafts 175 mm apart, theshaft angle being 70°. The velocity ratio is to be 1.5 to 1, the faster wheel having 80 teeth and a pitchcircle diameter of 100 mm. Find the spiral angles for each wheel. If the torque on the faster wheel is75 N-m ; find the axial thrust on each shaft, neglecting friction.

Solution. Given : L = 175 mm = 0.175 m ; θ = 70° ; G = 1.5 ; T2 = 80 ; d2 = 100 mm = 0.1 mor r2 = 0.05 m ; Torque on faster wheel = 75 N-m

Spiral angles for each wheel

Let α1 = Spiral angle for slower wheel, and

α2 = Spiral angle for faster wheel.

We know that velocity ratio, 2 1

1 2

1.5N T

GN T

= = =

∴ No. of teeth on slower wheel,

T1 = T2 × 1.5 = 80 × 1.5 = 120

We also know that the centre distance between shafts (L),

1 2 1 0.10.175

2 2

d d d+ += =

∴ d1 = 2 × 0.175 – 0.1 = 0.25 m

and 2 2 1 1 1

1 1 2 2 2

cos 80 cos 2 cos0.1or

cos 0.25 120 cos 3 cos

d T

d T

α α α= = =α α α


∴ 11 2

2

cos 0.1 30.6 or cos 0.6 cos

cos 0.25 2

α ×= = α = αα ×

...(i)

We know that, α1 + α2 = θ = 70° or α2 = 70° – α1

Substituting the value of α2 in equation (i),

cos α1 = 0.6 cos (70° – α1) = 0.6 (cos 70° cos α1 + sin 70° sin α1)

[ ]... cos ( ) cos cos sin sinA B A B A B− = +�

1 10.2052 cos 0.5638 sin= α + αcos α1 – 0.2052 cos α1 = 0.5638 sin α1

1 10.7948 cos 0.5638 sinα = α

∴ 11 1

1

sin 0.7948tan 1.4097 or 54.65

cos 0.5638

αα = = = α = °α

and 2 70 54.65 15.35α = ° − ° = °Ans.

Axial thrust on each shaft

We know that Torque = Tangential force × Pitch circle radius

∴ Tangential force at faster wheel,

22

Torque on the faster wheel 751500 N

Pitch circle radius ( ) 0.05F

r= = =

and normal reaction at the point of contact,

RN = F2 / cos α2 = 1500/cos 15.35° = 1556 N

We know that axial thrust on the shaft of slower wheel,

Fa1 = RN. sin α1 = 1556 × sin 54.65° = 1269 N Ans.

and axial thrust on the shaft of faster wheel,

Fa2 = RN. sin α2 = 1556 × sin 15.35° = 412 N Ans.

Example 12.16. In a spiral gear drive connecting two shafts, the approximate centre distance is 400 mm and the speed ratio = 3. The angle between the two shafts is 50° and the normalpitch is 18 mm. The spiral angle for the driving and driven wheels are equal. Find : 1. Number ofteeth on each wheel, 2. Exact centre distance, and 3. Efficiency of the drive, if friction angle = 6°.

Solution. Given : L = 400 mm = 0.4 m ; G = T2 / T1 = 3 ; θ = 50° ; pN = 18 mm ; φ = 6°

1. Number of teeth on each wheel

Let T1 = Number of teeth on wheel 1 (i.e. driver), and

T2 = Number of teeth on wheel 2 (i.e. driven).

Since the spiral angle α1 for the driving wheel is equal to the spiral angle α2 for the drivenwheel, therefore

α1 = α2 = θ/2 = 25° 1 2...( 50 )α + α = θ = °�

We know that centre distance between two shafts (L),

N. 1 N 1

1 2 1

1 1.400

cos cos cos2 2

G Gp T p T + += = α α απ π 1 2...( )α = α�


11

1 31812.64

cos 252

TT

+× = = °π ∴ T1 = 400/12.64 = 31.64 or 32 Ans.

and T2 = G.T1 = 3 × 32 = 96 Ans.2. Exact centre distance

We know that exact centre distance,

N 1 N 11

1 2 1

1. . 1cos cos2 2 cos

Gp T p T GL

+ += = α απ π α 1 2...( )α = α�

1 318 32404.5 mm

cos 252

+× = = °π AAns.

3. Efficiency of the drive

We know that efficiency of the drive,

2 1 1

1 2 1

cos ( ) cos cos ( )

cos ( ) cos cos ( )

α + φ α α + φη = =α − φ α α − φ 1 2...( )α = α�

cos (25 6 ) cos 31 0.85720.907 90.7%

cos (25 6 ) cos 19 0.9455

° + ° °= = = = =° − ° ° Ans.

Example 12.17. A drive on a machine tool is to be made by two spiral gear wheels, thespirals of which are of the same hand and has normal pitch of 12.5 mm. The wheels are of equaldiameter and the centre distance between the axes of the shafts is approximately 134 mm. The anglebetween the shafts is 80° and the speed ratio 1.25. Determine : 1. the spiral angle of each wheel,2. the number of teeth on each wheel, 3. the efficiency of the drive, if the friction angle is 6°, and4. the maximum efficiency.

Solution. Given : pN = 12.5 mm ; L = 134 mm ; θ = 80° ; G = N2 / N1 = T1 / T2 = 1.25

1. Spiral angle of each wheel

Let α1 and α2 = Spiral angles of wheels 1 and 2 respectively, and

d1 and d2 = Pitch circle diameter of wheels 1 and 2 respectively.

We know that2 2 1

1 2 2 11 1 2

cosor cos cos

cos

d TT T

d T

α= α = αα 1 2...( )d d=�

∴ 1 11 2

2 2

cos1.25 or cos 1.25 cos

cos

T

T

α = = α = αα

...(i)

We also know that

1 2 2 180 or 80α + α = θ = ° α = ° − αSubstituting the value of α2 in equation (i),

cos α1 = 1.25 cos (80° – α1) = 1.25 (cos 80° cos α1 + sin 80° sin α1)

= 1.25 (0.1736 cos α1 + 0.9848 sin α1)

= 0.217 cos α1 + 1.231 sin α1

cos α1 – 0.217 cos α1 = 1.231 sin α1 or 0.783 cos α1 = 1.231 sin α1

∴ tan α1 = sin α1 / cos α1 = 0.783 / 1.231 = 0.636 or α1 = 32.46° Ans.

and α2 = 80° – 32.46° = 47.54° Ans.


2. Number of teeth on each wheelLet T1 = Number of teeth on wheel 1, and

T2 = Number of teeth on wheel 2. We know that centre distance between the two shafts (L),

1 21 2134 or 134 mm

2

d dd d

+= = = 1 2...( )d d=�

We know that1 1 1

11

. .

cosc Np T p T

d = =π π α

∴ 1 11

N

.cos 134 cos 32.4628.4 or 30

12.5

dT

p

π α π × × °= = = Ans.

and 12

3024

1.25 1.25

TT = = = Ans.

3. Efficiency of the driveWe know that efficiency of the drive,

2 1

1 2

cos ( ) cos cos (47.54 6 ) cos 32.46

cos ( ) cos cos (32.46 6 ) cos 47.54

α + φ α ° + ° °η = =α − φ α ° − ° °

0.5943 0.8437

0.83 or 83%0.8952 0.6751

×= =×

Ans.

4. Maximum efficiencyWe know that maximum efficiency,

cos ( ) 1 cos (80 6 ) 1 1.0698

cos ( ) 1 cos (80 6 ) 1 1.2756maxθ + φ + ° + ° +η = = =θ − φ + ° − ° +

0.838 or 83.8%= Ans.

EXERCISES1. The pitch circle diameter of the smaller of the two spur wheels which mesh externally and have

involute teeth is 100 mm. The number of teeth are 16 and 32. The pressure angle is 20° and theaddendum is 0.32 of the circular pitch. Find the length of the path of contact of the pair of teeth.

[Ans. 29.36 mm]2. A pair of gears, having 40 and 30 teeth respectively are of 25° involute form. The addendum length is

5 mm and the module pitch is 2.5 mm. If the smaller wheel is the driver and rotates at 1500 r.p.m., findthe velocity of sliding at the point of engagement and at the point of disengagement.

[Ans. 2.8 m/s ; 2.66 m/s]3. Two gears of module 4mm have 24 and 33 teeth. The pressure angle is 20° and each gear has a

standard addendum of one module. Find the length of arc of contact and the maximum velocity ofsliding if the pinion rotates at 120 r.p.m. [Ans. 20.58 mm ; 0.2147 m/s]

4. The number of teeth in gears 1 and 2 are 60 and 40 ; module = 3 mm ; pressure angle = 20° andaddendum = 0.318 of the circular pitch. Determine the velocity of sliding when the contact is at the tipof the teeth of gear 2 and the gear 2 rotates at 800 r.p.m. [Ans. 1.06 m/s]

5. Two spur gears of 24 teeth and 36 teeth of 8 mm module and 20° pressure angle are in mesh. Adden-dum of each gear is 7.5 mm. The teeth are of involute form. Determine : 1. the angle through which thepinion turns while any pair of teeth are in contact, and 2. the velocity of sliding between the teethwhen the contact on the pinion is at a radius of 102 mm. The speed of the pinion is 450 r.p.m.

[Ans. 20.36°, 1.16 m/s]

Chapter 12 : Toothed Gearing � 4256. A pinion having 20 involute teeth of module pitch 6 mm rotates at 200 r.p.m. and transmits 1.5 kW to

a gear wheel having 50 teeth. The addendum on both the wheels is 1/4 of the circular pitch. The angleof obliquity is 20°. Find (a) the length of the path of approach ; (b) the length of the arc of approach;(c) the normal force between the teeth at an instant where there is only pair of teeth in contact.

[Ans. 13.27 mm ; 14.12 mm ; 1193 N]7. Two mating involute spur gear of 20° pressure angle have a gear ratio of 2. The number of teeth on the

pinion is 20 and its speed is 250 r.p.m. The module pitch of the teeth is 12 mm.If the addendum on each wheel is such that the path of approach and the path of recess on each side arehalf the maximum possible length, find : 1. the addendum for pinion and gear wheel ; 2. the length ofthe arc of contact ; and 3. the maximum velocity of sliding during approach and recess.Assume pinion to be the driver. [Ans. 19.5 mm, 7.8 mm ; 65.5 mm ; 807.5 mm/s, 1615 mm/s]

8. Two mating gears have 20 and 40 involute teeth of module 10 mm and 20° pressure angle. If theaddendum on each wheel is such that the path of contact is maximum and interference is just avoided,find the addendum for each gear wheel, path of contact, arc of contact and contact ratio.

[Ans. 14 mm ; 39 mm ; 102.6 mm ; 109.3 mm ; 4]9. A 20° involute pinion with 20 teeth drives a gear having 60 teeth. Module is 8 mm and addendum of

each gear is 10 mm.1. State whether interference occurs or not. Give reasons.2. Find the length of path of approach and arc of approach if pinion is the driver.

[Ans. Interference does not occur ; 25.8 mm, 27.45 mm]10. A pair of spur wheels with involute teeth is to give a gear ratio of 3 to 1. The arc of approach is not to

be less than the circular pitch and the smaller wheel is the driver. The pressure angle is 20°. What isthe least number of teeth that can be used on each wheel ? What is the addendum of the wheel in termsof the circular pitch ? [Ans. 18, 54 ; 0.382 Pc]

11. Two gear wheels mesh externally and are to give a velocity ratio of 3. The teeth are of involute formof module 6. The standard addendum is 1 module. If the pressure angle is 18° and pinion rotates at 90r.p.m., find : 1. the number of teeth on each wheel, so that the interference is just avoided, 2. the lengthof the path of contact, and 3. the maximum velocity of sliding between the teeth.

[Ans. 19, 57 ; 31.5 mm ; 213.7 mm/s]12. A pinion with 24 involute teeth of 150 mm of pitch circle diameter drives a rack. The addendum of the

pinion and rack is 6 mm. Find the least pressure angle which can be used if under cutting of the teethis to be avoided. Using this pressure angle, find the length of the arc of contact and the minimumnumber of teeth in contact at one time. [Ans. 16.8° ; 40 mm ; 2 pairs of teeth]

13. Two shafts, inclined at an angle of 65° and with a least distance between them of 175 mm are to beconnected by spiral gears of normal pitch 15 mm to give a reduction ratio 3 : 1. Find suitable diam-eters and numbers of teeth. Determine, also, the efficiency if the spiral angles are determined by thecondition of maximum efficiency. The friction angle is 7°.

[Ans. 88.5 mm ; 245.7 mm ; 15, 45 ; 85.5 %]14. A spiral wheel reduction gear, of ratio 3 to 2, is to be used on a machine, with the angle between the

shafts 80°. The approximate centre distance between the shafts is 125 mm. The normal pitch of theteeth is 10 mm and the wheel diameters are equal. Find the number of teeth on each wheel, pitch circlediameters and spiral angles. Find the efficiency of the drive if the friction angle is 5°.

[Ans. 24, 36 ; 128 mm ; 53.4°, 26.6° ; 85.5 %]15. A right angled drive on a machine is to be made by two spiral wheels. The wheels are of equal

diameter with a normal pitch of 10 mm and the centre distance is approximately 150 mm. If the speedratio is 2.5 to 1, find : 1. the spiral angles of the teeth, 2. the number of teeth on each wheel, 3.the exactcentre distance, and 4. transmission efficiency, if the friction angle is 6°.

[Ans. 21.8°, 68.2° ; 18 , 45 ; 154 mm ; 75.8 %]

DO YOU KNOW ?1. Explain the terms : (i) Module, (ii) Pressure angle, and (iii) Addendum.2. State and prove the law of gearing. Show that involute profile satisfies the conditions for correct

gearing.3. Derive an expression for the velocity of sliding between a pair of involute teeth. State the advantages

of involute profile as a gear tooth profile.


4. Prove that the velocity of sliding is proportional to the distance of the point of contact from the pitchpoint.

5. Prove that for two involute gear wheels in mesh, the angular velocity ratio does not change if thecentre distance is increased within limits, but the pressure angle increases.

6. Derive an expression for the length of the arc of contact in a pair of meshed spur gears.7. What do you understand by the term ‘interference’ as applied to gears?8. Derive an expression for the minimum number of teeth required on the pinion in order to avoid

interference in involute gear teeth when it meshes with wheel.9. Derive an expression for minimum number of teeth required on a pinion to avoid interference when it

gears with a rack.10. Define (i) normal pitch, and (ii) axial pitch relating to helical gears.11. Derive an expression for the centre distance of a pair of spiral gears.12. Show that, in a pair of spiral gears connecting inclined shafts, the efficiency is maximum when the

spiral angle of the driving wheel is half the sum of the shaft and friction angles.

OBJECTIVE TYPE QUESTIONS1. The two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the

shaft. This arrangement is called(a) spur gearing (b) helical gearing (c) bevel gearing (d) spiral gearing

2. The type of gears used to connect two non-parallel non-intersecting shafts are(a) spur gears (b) helical gears (c) spiral gears (d) none of these

3. An imaginary circle which by pure rolling action, gives the same motion as the actual gear, is called(a) addendum circle (b) dedendum circle (c) pitch circle (d) clearance circle

4. The size of a gear is usually specified by(a) pressure angle (b) circular pitch (c) diametral pitch (d) pitch circle diameter

5. The radial distance of a tooth from the pitch circle to the bottom of the tooth, is called(a) dedendum (b) addendum (c) clearance (d) working depth

6. The product of the diametral pitch and circular pitch is equal to(a) 1 (b) 1/π (c) π (d) 2π

7. The module is the reciprocal of(a) diametral pitch (b) circular pitch (c) pitch diameter (d) none of these

8. Which is the incorrect relationship of gears?(a) Circular pitch × Diametral pitch = π (b) Module = P.C.D/No.of teeth(c) Dedendum = 1.157 module (d) Addendum = 2.157 module

9. If the module of a gear be m, the number of teeth T and pitch circle diameter D, then(a) m = D/T (b) D = T/m (c) m = D/2T (d) none of these

10. Mitre gears are used for(a) great speed reduction (b) equal speed(c) minimum axial thrust (d) minimum backlash

11. The condition of correct gearing is(a) pitch line velocities of teeth be same(b) radius of curvature of two profiles be same(c) common normal to the pitch surface cuts the line of centres at a fixed point(d) none of the above

12. Law of gearing is satisfied if(a) two surfaces slide smoothly(b) common normal at the point of contact passes through the pitch point on the line joining the

centres of rotation(c) number of teeth = P.C.D. / module(d) addendum is greater than dedendum

Chapter 12 : Toothed Gearing � 42713. Involute profile is preferred to cyloidal because

(a) the profile is easy to cut(b) only one curve is required to cut(c) the rack has straight line profile and hence can be cut accurately(d) none of the above

14. The contact ratio for gears is(a) zero (b) less than one (c) greater than one

15. The maximum length of arc of contact for two mating gears, in order to avoid interference, is(a) (r + R) sin φ (b) (r + R) cos φ (c) (r + R) tan φ (d) none of thesewhere r = Pitch circle radius of pinion,

R = Pitch circle radius of driver, andφ = Pressure angle.

16. When the addenda on pinion and wheel is such that the path of approach and path of recess are half oftheir maximum possible values, then the length of the path of contact is given by

(a)( ) sin

2

r R+ φ(b)

( ) cos

2

r R+ φ(c)

( ) tan

2

r R+ φ(d) none of these

17. Interference can be avoided in involute gears with 20° pressure angle by(a) cutting involute correctly(b) using as small number of teeth as possible(c) using more than 20 teeth(d) using more than 8 teeth

18. The ratio of face width to transverse pitch of a helical gear with α as the helix angle is normally(a) more than 1.15/tan α (b) more than 1.05/tan α(c) more than 1/tan α (d) none of these

19. The maximum efficiency for spiral gears is

(a)sin ( ) 1

cos ( ) 1

θ + φ +θ − φ + (b)

cos ( ) 1

sin ( ) 1

θ − φ +θ + φ +

(c)cos ( ) 1

cos ( ) 1

θ + φ +θ − φ + (d)

cos ( ) 1

cos ( ) 1

θ − φ +θ + φ +

where θ = Shaft angle, and φ = Friction angle.20. For a speed ratio of 100, smallest gear box is obtained by using

(a) a pair of spur gears(b) a pair of helical and a pair of spur gear compounded(c) a pair of bevel and a pair of spur gear compounded

(d) a pair of helical and a pair of worm gear compounded

ANSWERS1. (a) 2. (c) 3. (c) 4. (d) 5. (a)

6. (c) 7. (a) 8. (d) 9. (a) 10. (b)

11. (c) 12. (b) 13. (b) 14. (c) 15. (c)

16. (a) 17. (c) 18. (a) 19. (c) 20. (d)


Gear Gear Gear Gear Gear TTTTTrainsrainsrainsrainsrains

13FFFFFeaeaeaeaeaturturturturtureseseseses1. Introduction.

2. Types of Gear Trains.

3. Simple Gear Train.

4. Compound Gear Train.

5. Design of Spur Gears.

6. Reverted Gear Train.

7. Epicyclic Gear Train.

8. Velocity Ratio of EpicyclicGear Train.

9. Compound Epicyclic GearTrain (Sun and PlanetWheel).

10. Epicyclic Gear Train WithBevel Gears.

11. Torques in Epicyclic GearTrains.

13.1.13.1.13.1.13.1.13.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoductionSometimes, two or more gears are made to mesh with

each other to transmit power from one shaft to another. Sucha combination is called gear train or train of toothed wheels.The nature of the train used depends upon the velocity ratiorequired and the relative position of the axes of shafts. Agear train may consist of spur, bevel or spiral gears.

13.2.13.2.13.2.13.2.13.2. TTTTTypes of Gear ypes of Gear ypes of Gear ypes of Gear ypes of Gear TTTTTrainsrainsrainsrainsrainsFollowing are the different types of gear trains, de-

pending upon the arrangement of wheels :

1. Simple gear train, 2. Compound gear train, 3. Re-verted gear train, and 4. Epicyclic gear train.

In the first three types of gear trains, the axes of theshafts over which the gears are mounted are fixed relative toeach other. But in case of epicyclic gear trains, the axes ofthe shafts on which the gears are mounted may move relativeto a fixed axis.

13.3.13.3.13.3.13.3.13.3. Simple Gear Simple Gear Simple Gear Simple Gear Simple Gear TTTTTrainrainrainrainrainWhen there is only one gear on each shaft, as shown

in Fig. 13.1, it is known as simple gear train. The gears arerepresented by their pitch circles.

When the distance between the two shafts is small,the two gears 1 and 2 are made to mesh with each other to

428

Chapter 13 : Gear Trains � 429transmit motion from one shaft to the other, as shown in Fig. 13.1 (a). Since the gear 1 drives the gear2, therefore gear 1 is called the driver and the gear 2 is called the driven or follower. It may be notedthat the motion of the driven gear is opposite to the motion of driving gear.

(a) (b) (c)

Fig. 13.1. Simple gear train.

Let N1 = Speed of gear 1(or driver) in r.p.m.,

N2 = Speed of gear 2 (or driven or follower) in r.p.m.,

T1 = Number of teeth on gear 1, and

T2 = Number of teeth on gear 2.

Since the speed ratio (or velocity ratio) of gear train is the ratio of the speed of the driver tothe speed of the driven or follower and ratio of speeds of any pair of gears in mesh is the inverse oftheir number of teeth, therefore

Speed ratio 1 2

2 1

N T

N T= =

It may be noted that ratio of the speed of the driven or follower to the speed of the driver isknown as train value of the gear train. Mathematically,

Train value 2 1

1 2

N T

N T= =

From above, we see that the train value is the reciprocal of speed ratio.

Sometimes, the distance between the two gears is large. The motion from one gear to another,in such a case, may be transmitted by either of the following two methods :

1. By providing the large sized gear, or 2. By providing one or more intermediate gears.

A little consideration will show that the former method (i.e. providing large sized gears) isvery inconvenient and uneconomical method ; whereas the latter method (i.e. providing one or moreintermediate gear) is very convenient and economical.

It may be noted that when the number of intermediate gears are odd, the motion of both thegears (i.e. driver and driven or follower) is like as shown in Fig. 13.1 (b).

But if the number of intermediate gears are even, the motion of the driven or follower will bein the opposite direction of the driver as shown in Fig. 13.1 (c).

Now consider a simple train of gears with one intermediate gear as shown in Fig. 13.1 (b).

Let N1 = Speed of driver in r.p.m.,

N2 = Speed of intermediate gear in r.p.m.,


N3 = Speed of driven or follower in r.p.m.,

T1 = Number of teeth on driver,

T2 = Number of teeth on intermediate gear, and

T3 = Number of teeth on driven or follower.

Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio forthese two gears is

1 2

2 1

=N T

N T...(i)

Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore speed ratiofor these two gears is

32

3 2

=TN

N T...(ii)

The speed ratio of the gear train as shown in Fig. 13.1 (b) is obtained by multiplying theequations (i) and (ii).

∴ 31 2 2

2 3 1 2

× = ×TN N T

N N T Tor 31

3 1

=TN

N T

i.e. Speed of driver No. of teeth on drivenSpeed ratio = =

Speed of driven No. of teeth on driver

andSpeed of driven No. of teeth on driver

Train value = =Speed of driver No. of teeth on driven

Similarly, it can be proved that theabove equation holds good even if there areany number of intermediate gears. Fromabove, we see that the speed ratio and thetrain value, in a simple train of gears, is in-dependent of the size and number of inter-mediate gears. These intermediate gears arecalled idle gears, as they do not effect thespeed ratio or train value of the system. Theidle gears are used for the following two pur-poses :

1. To connect gears where a largecentre distance is required, and

2. To obtain the desired direction ofmotion of the driven gear (i.e. clockwise oranticlockwise).

13.4. Compound Gear TrainWhen there are more than one gear on a shaft, as shown in Fig. 13.2, it is called a compound

train of gear.

We have seen in Art. 13.3 that the idle gears, in a simple train of gears do not effect the speedratio of the system. But these gears are useful in bridging over the space between the driver and thedriven.

Gear trains inside a mechanical watch

Chapter 13 : Gear Trains � 431But whenever the distance between the driver and the driven or follower has to be bridged

over by intermediate gears and at the same time a great ( or much less ) speed ratio is required, thenthe advantage of intermediate gears is intensified by providing compound gears on intermediate shafts.In this case, each intermediate shaft has two gears rigidly fixed to it so that they may have the samespeed. One of these two gears meshes with the driver and the other with the driven or followerattached to the next shaft as shown in Fig.13.2.

Fig. 13.2. Compound gear train.In a compound train of gears, as shown in Fig. 13.2, the gear 1 is the driving gear mounted on

shaft A , gears 2 and 3 are compound gears which are mounted on shaft B. The gears 4 and 5 are alsocompound gears which are mounted on shaft C and the gear 6 is the driven gear mounted on shaft D.

Let N1 = Speed of driving gear 1,T1 = Number of teeth on driving gear 1,

N2 ,N3 ..., N6 = Speed of respective gears in r.p.m., andT2 ,T3..., T6 = Number of teeth on respective gears.

Since gear 1 is in mesh with gear 2, therefore its speed ratio is

1 2

2 1

N T

N T= ...(i)

Similarly, for gears 3 and 4, speed ratio is

3 4

4 3

N T

N T= ...(ii)

and for gears 5 and 6, speed ratio is

5 6

6 5

N T

N T= ...(iii)

The speed ratio of compound gear train is obtained by multiplying the equations (i), (ii) and (iii),

∴ 3 5 61 2 4

2 4 6 1 3 5

N N TN T T

N N N T T T× × = × × or

*2 4 61

6 1 3 5

T T TN

N T T T

× ×=

× ×

* Since gears 2 and 3 are mounted on one shaft B, therefore N2 = N3. Similarly gears 4 and 5 are mounted onshaft C, therefore N4 = N5.


i.e. Speed of the first driverSpeed ratio =

Speed of the last driven or follower

Product of the number of teeth on the drivens=

Product of the number of teeth on the drivers

andSpeed of the last driven or follower

Train value = Speed of the first driver

Product of the number of teeth on the drivers=

Product of the number of teeth on the drivens

The advantage of a compound train over a simple gear train is that a much larger speedreduction from the first shaft to the last shaft can be obtained with small gears. If a simple gear trainis used to give a large speed reduction, the last gear has to be very large. Usually for a speed reductionin excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed.

Note: The gears which mesh must have the same circular pitch or module. Thus gears 1 and 2 must have thesame module as they mesh together. Similarly gears 3 and 4, and gears5 and 6 must have the same module.

Example 13.1. The gearing of a machine tool is shownin Fig. 13.3. The motor shaft is connected to gear A and rotatesat 975 r.p.m. The gear wheels B, C, D and E are fixed to parallelshafts rotating together. The final gear F is fixed on the outputshaft. What is the speed of gear F ? The number of teeth oneach gear are as given below :

Gear A B C D E F

No. of teeth 20 50 25 75 26 65

Solution. Given : NA = 975 r.p.m. ;TA = 20 ; TB = 50 ; TC = 25 ; TD = 75 ; TE = 26 ;TF = 65

From Fig. 13.3, we see that gears A , Cand E are drivers while the gears B, D and F aredriven or followers. Let the gear A rotates inclockwise direction. Since the gears B and C aremounted on the same shaft, therefore it is acompound gear and the direction or rotation ofboth these gears is same (i.e. anticlockwise).Similarly, the gears D and E are mounted on thesame shaft, therefore it is also a compound gearand the direction of rotation of both these gearsis same (i.e. clockwise). The gear F will rotate inanticlockwise direction.

Let NF = Speed of gear F, i.e. last driven or follower.

We know that

Speed of the first driver Product of no. of teeth on drivens=

Speed of the last driven Product of no. of teeth on drivers

Fig. 13.3

Battery Car: Even though it is run by batteries,the power transmission, gears, clutches,brakes, etc. remain mechanical in nature.Note : This picture is given as additional informationand is not a direct example of the current chapter.

Chapter 13 : Gear Trains � 433

or A B D F

F A C E

50 75 6518.75

20 25 26

N T T T

N T T T

× × × ×= = =× × × ×

∴ AF

97552 r. p. m.

18.75 18.75

NN = = = Ans.

13.5. Design of Spur GearsSometimes, the spur gears (i.e. driver and driven) are to be designed for the given velocity

ratio and distance between the centres of their shafts.

Let x = Distance between the centres of two shafts,

N1 = Speed of the driver,

T1 = Number of teeth on the driver,

d1 = Pitch circle diameter of the driver,

N2 , T2 and d2 = Corresponding values for the driven or follower, andpc = Circular pitch.

We know that the distance between the centres of two shafts,

1 2

2

d dx

+= ...(i)

and speed ratio or velocity ratio,

1 2 2

2 1 1

N d T

N d T= = ...(ii)

From the above equations, we can conveniently find out the values of d1 and d2 (or T1 and T2)and the circular pitch ( pc ). The values of T1 and T2, as obtained above, may or may not be wholenumbers. But in a gear since the number of its teeth is always a whole number, therefore a slightalterations must be made in the values of x, d1 and d2, so that the number of teeth in the two gears maybe a complete number.

Example 13.2. Two parallel shafts, about 600 mm apart are to be connected by spur gears.One shaft is to run at 360 r.p.m. and the other at 120 r.p.m. Design the gears, if the circular pitch isto be 25 mm.

Solution. Given : x = 600 mm ; N1 = 360 r.p.m. ; N2 = 120 r.p.m. ; pc = 25 mm

Let d1 = Pitch circle diameter of the first gear, andd2 = Pitch circle diameter of the second gear.

We know that speed ratio,

1 2

2 1

3603

120

N d

N d= = = or d2 = 3d1 ...(i)

and centre distance between the shafts (x),

1 21

600 ( )2

d d= + or d1 + d2 = 1200 ...(ii)

From equations (i) and (ii), we find that

d1 = 300 mm, and d2 = 900 mm

∴ Number of teeth on the first gear,

21

30037.7

25c

dT

p

π π ×= = =


and number of teeth on the second gear,

22

c

900113.1

25

dT

p

π π ×= = =

Since the number of teeth on both the gears are to be in complete numbers, therefore let usmake the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teethon the second gear should be 38 × 3 = 114.

Now the exact pitch circle diameter of the first gear,

11

38 25302.36 mmcT p

d× ×′ = = =π π

and the exact pitch circle diameter of the second gear,

22

114 25907.1 mmcT p

d× ×′ = = =π π

∴ Exact distance between the two shafts,

1 2 302.36 907.1604.73 mm

2 2

d dx

′ ′+ +′ = = =

Hence the number of teeth on the first and second gear must be 38 and 114 and their pitchcircle diameters must be 302.36 mm and 907.1 mmrespectively. The exact distance between the two shaftsmust be 604.73 mm. Ans.

13.6. Reverted Gear TrainWhen the axes of the first gear (i.e. first driver)

and the last gear (i.e. last driven or follower) are co-axial,then the gear train is known as reverted gear train asshown in Fig. 13.4.

We see that gear 1 (i.e. first driver) drives thegear 2 (i.e. first driven or follower) in the opposite direc-tion. Since the gears 2 and 3 are mounted on the sameshaft, therefore they form a compound gear and the gear3 will rotate in the same direction as that of gear 2. Thegear 3 (which is now the second driver) drives the gear 4(i.e. the last driven or follower) in the same direction asthat of gear 1. Thus we see that in a reverted gear train,the motion of the first gear and the last gear is like.

Let T1 = Number of teeth on gear 1,

r1 = Pitch circle radius of gear 1, and

N1 = Speed of gear 1 in r.p.m.

Similarly,

T2, T3, T4 = Number of teeth on respective gears,

r2, r3, r4 = Pitch circle radii of respective gears, and

N2, N3, N4 = Speed of respective gears in r.p.m.

Fig. 13.4. Reverted gear train.

Chapter 13 : Gear Trains � 435Since the distance between the centres of the shafts of gears 1 and 2 as well as gears 3 and 4

is same, therefore

r1 + r2 = r3 + r4 ...(i)

Also, the circular pitch or module of all the gears is assumed to be same, therefore number ofteeth on each gear is directly proportional to its circumference or radius.

∴ *T1 + T2 = T3 + T4 ...(ii)

and Product of number of teeth on drivensSpeed ratio =

Product of number of teeth on drivers

or 1 2 4

4 1 3

×=×

N T T

N T T... (iii)

From equations (i), (ii) and (iii), we can determine the number of teeth on each gear for thegiven centre distance, speed ratio and module only whenthe number of teeth on one gear is chosen arbitrarily.

The reverted gear trains are used in automotive trans-missions, lathe back gears, industrial speed reducers, and inclocks (where the minute and hour hand shafts are co-axial).

Example 13.3. The speed ratio of the reverted geartrain, as shown in Fig. 13.5, is to be 12. The module pitch ofgears A and B is 3.125 mm and of gears C and D is 2.5 mm.Calculate the suitable numbers of teeth for the gears. Nogear is to have less than 24 teeth.

Solution. Given : Speed ratio, NA/ND = 12 ;mA = mB = 3.125 mm ; mC = mD = 2.5 mm

Let NA = Speed of gear A ,

TA = Number of teeth on gear A ,

rA = Pitch circle radius of gear A ,

NB, NC , ND = Speed of respective gears,

TB, TC , TD = Number of teeth on respective gears, and

rB, rC , rD = Pitch circle radii of respective gears.

Fig. 13.5

* We know that circular pitch,

2c

rp m

T

π= = π or .2

mTr = , where m is the module.

∴ 11

.2

m Tr = ; 2

2.2

m Tr = ; 3

3.2

mTr = ; 4

4.2

m Tr =

Now from equation (i),

31 2 4.. . .2 2 2 2

m Tm T m T m T+ = +

T1 + T2 = T3 + T4


Since the speed ratio between the gears A and B and between the gears C and D are to besame, therefore

* CA

B D

12 3.464NN

N N= = =

Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth,therefore

B D

A C

3.464T T

T T= = ...(i)

We know that the distance between the shafts

x = rA + rB = rC + rD = 200 mm

or C CA A B B D D.. . .200

2 2 2 2

m Tm T m T m T+ = + =

....

2

m Tr

= �

3.125 (TA + TB) = 2.5 (TC + TD) = 400 ...(∵ mA = mB, and mC = mD)

∴ TA + TB = 400 / 3.125 = 128 ...(ii)

and TC + TD = 400 / 2.5 = 160 ...(iii)

From equation (i), TB = 3.464 TA. Substituting this value of TB in equation (ii),

TA + 3.464 TA = 128 or TA = 128 / 4.464 = 28.67 say 28 Ans.

and TB = 128 – 28 = 100 Ans.

Again from equation (i), TD = 3.464 TC. Substituting this value of TD in equation (iii),

TC + 3.464 TC = 160 or TC = 160 / 4.464 = 35.84 say 36 Ans.

and TD = 160 – 36 = 124 Ans.

Note : The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is

A B D

D A C

100 12412.3

28 36

N T T

N T T

× ×= = =× ×

13.7. Epicyclic Gear TrainWe have already discussed that in an epicyclic gear train, the axes of the shafts, over which

the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown inFig. 13.6, where a gear A and the arm C have a common axis at O1 about which they can rotate. Thegear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the

* We know that speed ratio A

Dv

Speed of first driver12

Speed of last dri enN

N= = =

Also CA A

D B D

NN N

N N N= × ...(NB = NC, being on the same shaft)

For A

B

N

N and C

D

N

N to be same, each speed ratio should be 12 so that

CA A

D B D

12 12 12NN N

N N N= × = × =

Chapter 13 : Gear Trains � 437arm is fixed, the gear train is simple and gear A can drive gear Bor vice- versa, but if gear A is fixed and the arm is rotated aboutthe axis of gear A (i.e. O1), then the gear B is forced to rotateupon and around gear A . Such a motion is called epicyclic andthe gear trains arranged in such a manner that one or more oftheir members move upon and around another member areknown as epicyclic gear trains (epi. means upon and cyclicmeans around). The epicyclic gear trains may be simple or com-pound.

The epicyclic gear trains are useful for transmittinghigh velocity ratios with gears of moderate size in a compara-tively lesser space. The epicyclic gear trains are used in theback gear of lathe, differential gears of the automobiles, hoists,pulley blocks, wrist watches etc.

13.8. Velocity Ratioz of Epicyclic Gear TrainThe following two methods may be used for finding out the velocity ratio of an epicyclic

gear train.

1. Tabular method, and 2. Algebraic method.

These methods are discussed, in detail, as follows :

1. Tabular method. Consider an epicyclic gear train as shown in Fig. 13.6.

Let TA = Number of teeth on gear A , and

TB = Number of teeth on gear B.

First of all, let us suppose thatthe arm is fixed. Therefore the axes ofboth the gears are also fixed relative toeach other. When the gear A makes onerevolution anticlockwise, the gear B willmake *TA / TB revolutions, clockwise.Assuming the anticlockwise rotation aspositive and clockwise as negative, wemay say that when gear A makes + 1revolution, then the gear B will make(– TA / TB) revolutions. This statementof relative motion is entered in the firstrow of the table (see Table 13.1).

Secondly, if the gear A makes+ x revolutions, then the gear B willmake – x × TA / TB revolutions. Thisstatement is entered in the second rowof the table. In other words, multiplythe each motion (entered in the first row) by x.

Thirdly, each element of an epicyclic train is given + y revolutions and entered in the thirdrow. Finally, the motion of each element of the gear train is added up and entered in the fourth row.

* We know that NB / NA = TA / TB. Since NA = 1 revolution, therefore NB = TA / TB.

Fig. 13.6. Epicyclic gear train.

Inside view of a car engine.Note : This picture is given as additional information and is not

a direct example of the current chapter.


Arm fixed-gear A rotates through + 1revolution i.e. 1 rev. anticlockwise

Arm fixed-gear A rotates through + xrevolutions

Add + y revolutions to all elements

Total motion

Table 13.1. Table of motions

Revolutions of elements

Step No. Conditions of motion Arm C Gear A Gear B

1. 0 + 1A

B–

T

T

2. 0 + xA

B–

Tx

T×

3. + y + y + y

4. + y x + yA

B–

Ty x

T×

A little consideration will show that when two conditions about the motion of rotation of anytwo elements are known, then the unknown speed of the third element may be obtained by substitut-ing the given data in the third column of the fourth row.

2. Algebraic method. In this method, the motion of each element of the epicyclic train relativeto the arm is set down in the form of equations. The number of equations depends upon the number ofelements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. someelement is fixed and the other has specified motion. These two conditions are sufficient to solve all theequations ; and hence to determine the motion of any element in the epicyclic gear train.

Let the arm C be fixed in an epicyclic gear train as shown in Fig. 13.6. Therefore speed of thegear A relative to the arm C

= NA – NCand speed of the gear B relative to the arm C,

= NB – NCSince the gears A and B are meshing directly, therefore they will revolve in opposite directions.

∴ B C A

A C B

––

–

N N T

N N T=

Since the arm C is fixed, therefore its speed, NC = 0.

∴ B A

A B

–N T

N T=

If the gear A is fixed, then NA = 0.

B C A

C B

––

0 –

N N T

N T= or B A

C B

1N T

N T= +

Note : The tabular method is easier and hence mostly used in solving problems on epicyclic gear train.

Example 13.4. In an epicyclic gear train, an arm carriestwo gears A and B having 36 and 45 teeth respectively. If the armrotates at 150 r.p.m. in the anticlockwise direction about the centreof the gear A which is fixed, determine the speed of gear B. If thegear A instead of being fixed, makes 300 r.p.m. in the clockwisedirection, what will be the speed of gear B ?

Solution. Given : TA = 36 ; TB = 45 ; NC = 150 r.p.m.(anticlockwise)

The gear train is shown in Fig. 13.7. Fig. 13.7


Arm fixed-gear A rotates through + 1revolution (i.e. 1 rev. anticlockwise)

Arm fixed-gear A rotates through + xrevolutions


Total motion

We shall solve this example, first by tabular method and then by algebraic method.

1. Tabular method

First of all prepare the table of motions as given below :

Table 13.2. Table of motions.Revolutions of elements

Step No. Conditions of motion Arm C Gear A Gear B

1. 0 + 1A

B–

T

T

2. 0 + xA

B–

Tx

T×

3. + y + y + y

4. + y x + yA

B–

Ty x

T×

Speed of gear B when gear A is fixed

Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,

y = + 150 r.p.m.

Also the gear A is fixed, therefore

x + y = 0 or x = – y = – 150 r.p.m.

∴ Speed of gear B, AB

B

36– 150 150 270 r.p.m.

45

TN y x

T= × = + × = +

= 270 r.p.m. (anticlockwise) Ans.Speed of gear B when gear A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,

x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.

∴ Speed of gear B,

A

BB

36– 150 450 510 r.p.m.

45

TN y x

T= × = + × = +

= 510 r.p.m. (anticlockwise) Ans.2. Algebraic method

Let NA = Speed of gear A .

NB = Speed of gear B, and

NC = Speed of arm C.

Assuming the arm C to be fixed, speed of gear A relative to arm C

= NA – NC

and speed of gear B relative to arm C = NB – NC


Since the gears A and B revolve in opposite directions, therefore

B C A

A C B

––

–

N N T

N N T= ...(i)

Speed of gear B when gear A is fixed

When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.

NA = 0, and NC = + 150 r.p.m.

∴ B – 150 36– – 0.8

0 – 150 45

N = = ...[From equation (i)]

or NB = – 150 × – 0.8 + 150 = 120 + 150 = 270 r.p.m. Ans.

Speed of gear B when gear A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m. clockwise, therefore

NA = – 300 r.p.m.

∴ B – 150 36– – 0.8

–300 – 150 45

N = =

or NB = – 450 × – 0.8 + 150 = 360 + 150 = 510 r.p.m. Ans.

Example 13.5. In a reverted epicyclic geartrain, the arm A carries two gears B and C and acompound gear D - E. The gear B meshes with gear Eand the gear C meshes with gear D. The number of teethon gears B, C and D are 75, 30 and 90 respectively.Find the speed and direction of gear C when gear B isfixed and the arm A makes 100 r.p.m. clockwise.

Solution. Given : TB = 75 ; TC = 30 ; TD = 90 ;NA = 100 r.p.m. (clockwise)

The reverted epicyclic gear train isshown in Fig. 13.8. First of all, let us find thenumber of teeth on gear E (TE). Let dB , dC , dDand dE be the pitch circle diameters of gears B,C, D and E respectively. From the geometry ofthe figure,

dB + dE = dC + dD

Since the number of teeth on each gear,for the same module, are proportional to theirpitch circle diameters, therefore

TB + TE = TC + TD

∴ TE = TC + TD – TB = 30 + 90 – 75 = 45

The table of motions is drawn asfollows :

Fig. 13.8

A gear-cutting machine is used to cut gears.Note : This picture is given as additional informationand is not a direct example of the current chapter.


Arm fixed-compound gear D-Erotated through + 1 revolution ( i.e.1 rev. anticlockwise)

Arm fixed-compound gear D-Erotated through + x revolutions


Total motion

Table 13.3. Table of motions.


Step Conditions of motion Arm A Compound Gear B Gear CNo. gear D-E

1. 0 + 1E

B–

T

TD

C–

T

T

2. 0 + xE

B–

Tx

T× D

C–

Tx

T×

3. + y + y + y + y

4. + y x + yE

B–

Ty x

T× D

C–

Ty x

T×

Since the gear B is fixed, therefore from the fourth row of the table,

E

B

– 0T

y xT

× = or45

– 075

y x × =

∴ y – 0.6 = 0 ...(i)

Also the arm A makes 100 r.p.m. clockwise, therefore

y = – 100 ...(ii)

Substituting y = – 100 in equation (i), we get

– 100 – 0.6 x = 0 or x = – 100 / 0.6 = – 166.67

Model of sun and planet gears.

INPUTSpline to AcceptMotor Shaft

Housing OD Designed to meetRAM Bore Dia, and Share MotorCoolant Supply

OUTPUT- External Spline toSpindle

Ratio Detection SwitchesHydraulic or Pneumatic SpeedChange Actuator

Round Housing With O-ringSeated Cooling Jacket

Motor Flange

Hollow Through Bore forDrawbar Integration


From the fourth row of the table, speed of gear C,

DC

C

90– – 100 166.67 400 r.p.m.

30

TN y x

T= × = + × = +

= 400 r.p.m. (anticlockwise) Ans.

13.9. Compound Epicyclic Gear Train—Sun and Planet GearA compound epicyclic

gear train is shown in Fig. 13.9.It consists of two co-axial shaftsS1 and S2, an annulus gear A whichis fixed, the compound gear (orplanet gear) B-C, the sun gear Dand the arm H. The annulus gearhas internal teeth and the com-pound gear is carried by the armand revolves freely on a pin of thearm H. The sun gear is co-axialwith the annulus gear and the armbut independent of them.

The annulus gear Ameshes with the gear B and thesun gear D meshes with the gearC. It may be noted that when theannulus gear is fixed, the sun gearprovides the drive and when thesun gear is fixed, the annulus gearprovides the drive. In both cases, the arm acts as a follower.

Note : The gear at the centre is called the sun gear and the gears whose axes move are called planet gears.

Fig. 13.9. Compound epicyclic gear train.

Sun and Planet gears.

Speed ChangeShift Axis

Bearing HousingOutput Belt Pulley

Slide DogClutch

Output SunGear

MotorFlange

Input SunGear

PlanetGears

OilCollector


Arm fixed-gear C rotates through+ 1 revolution (i.e. 1 rev.anticlockwise)

Arm fixed-gear C rotates through+ x revolutions

Add + y revolutions to allelements

Total motion

Arm fixed-gear D rotatesthrough + 1 revolution

Arm fixed-gear D rotatesthrough + x revolutions


Total motion

Let TA , TB , TC , and TD be the teeth and NA, NB, NC and ND be the speeds for the gears A , B,C and D respectively. A little consideration will show that when the arm is fixed and the sun gear D isturned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwisedirection.

The motion of rotations of the various elements are shown in the table below.



Step Conditions of motion Arm Gear D Compound gear Gear ANo. B-C

1. 0 + 1D

C–

T

TD B

C A–

T T

T T×

2. 0 + xD

C–

Tx

T× D B

C A

–T T

xT T

× ×

3. + y + y + y + y

4. + y x + yD

C–

Ty x

T× D B

C A–

T Ty x

T T× ×

Note : If the annulus gear A is rotated through one revolution anticlockwise with the arm fixed, then thecompound gear rotates through TA / TB revolutions in the same sense and the sun gear D rotates through TA / TB × TC / TD revolutions in clockwise direction.

Example 13.6. An epicyclic gear consists of three gears A, B and C as shown in Fig. 13.10.The gear A has 72 internal teeth and gear C has 32 external teeth. The gear B meshes with both Aand C and is carried on an arm EF which rotates about the centre of A at 18 r.p.m.. If the gear A isfixed, determine the speed of gears B and C.

Solution. Given : TA = 72 ; TC = 32 ; Speed of arm EF = 18 r.p.m.

Considering the relative motion of rotation as shown in Table 13.5.



Step No. Conditions of motion Arm EF Gear C Gear B Gear A

1. 0 + 1C

B–

T

TC B C

B A A– –

T T T

T T T× =

2. 0 + xC

B–

Tx

T× C

A–

Tx

T×

3. + y + y + y + y

4. + y x + yC

B–

Ty x

T× C

A–

Ty x

T×


Speed of gear C

We know that the speed of the arm is 18 r.p.m. therefore,

y = 18 r.p.m.

and the gear A is fixed, therefore

C

A

– 0T

y xT

× = or 3218 – 0

72x × =

∴ x = 18 × 72 / 32 = 40.5

∴ Speed of gear C = x + y = 40.5 + 18

= + 58.5 r.p.m.

= 58.5 r.p.m. in the directionof arm. Ans.

Speed of gear B

Let dA, dB and dC be the pitch circle diameters of gearsA , B and C respectively. Therefore, from the geometry of Fig. 13.10,

C AB 2 2

d dd + = or 2 dB + dC = dA

Since the number of teeth are proportional to their pitch circle diameters, therefore

2 TB + TC = TA or 2 TB + 32 = 72 or TB = 20

∴ Speed of gear B C

B

32– 18 – 40.5 – 46.8 r.p.m.

20

Ty x

T= × = × =

= 46.8 r.p.m. in the opposite direction of arm. Ans.

Example 13.7. An epicyclic train of gears is arranged as shown inFig.13.11. How many revolutions does the arm, to which the pinions B andC are attached, make :

1. when A makes one revolution clockwise and D makes half arevolution anticlockwise, and

2. when A makes one revolution clockwise and D is stationary ?

The number of teeth on the gears A and D are 40 and 90respectively.

Solution. Given : TA = 40 ; TD = 90

First of all, let us find the number of teeth on gears B and C (i.e. TB and TC). Let dA, dB, dCand dD be the pitch circle diameters of gears A , B, C and D respectively. Therefore from the geometryof the figure,

dA + dB + dC = dD or dA + 2 dB = dD ...(� dB = dC)

Since the number of teeth are proportional to their pitch circle diameters, therefore,

TA + 2 TB = TD or 40 + 2 TB = 90

∴ TB = 25, and TC = 25 ...(� TB = TC)

Fig. 13.10

Fig. 13.11


The table of motions is given below :



Step No. Conditions of motion Arm Gear A Compound Gear Dgear B-C

1. 0 – 1A

B

T

T+ A B A

B D D

T T T

T T T+ × = +

2. 0 – xA

B

Tx

T+ × A

D

Tx

T+ ×

3. – y – y – y – y

4. – y – x – yA

B–

Tx y

T× A

D–

Tx y

T×

1. Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise

Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table,

– x – y = –1 or x + y = 1 ...(i)

Also, the gear D makes half revolution anticlockwise, therefore

A

D

1–

2

Tx y

T× = or 40 1

–90 2

x y× =

∴ 40 x – 90 y = 45 or x – 2.25 y = 1.125 ...(ii)

From equations (i) and (ii), x = 1.04 and y = – 0.04

∴ Speed of arm = – y = – (– 0.04) = + 0.04

= 0.04 revolution anticlockwise Ans.

2. Speed of arm when A makes 1 revolution clockwise and D is stationary

Since the gear A makes 1 revolution clockwise, therefore from the fourth row of thetable,

– x – y = – 1 or x + y = 1 ...(iii)

Also the gear D is stationary, therefore

A

D

– 0T

x yT

× = or 40– 0

90x y× =

∴ 40 x – 90 y = 0 or x – 2.25 y = 0 ...(iv)

From equations (iii) and (iv),

x = 0.692 and y = 0.308

∴ Speed of arm = – y = – 0.308 = 0.308 revolution clockwise Ans.

Arm fixed , gear A rotatesthrough – 1 revolution (i.e. 1rev. clockwise)

Arm fixed, gear A rotatesthrough – x revolutions

Add – y revolutions to allelements

Total motion


Example 13.8. In an epicyclic gear train, the internal wheels A and B and compound wheelsC and D rotate independently about axis O. The wheels E and F rotate on pins fixed to the arm G. Egears with A and C and F gears with B and D. All the wheels havethe same module and the number of teeth are : TC = 28; TD = 26;TE = TF = 18.

1. Sketch the arrangement ; 2. Find the number of teeth onA and B ; 3. If the arm G makes 100 r.p.m. clockwise and A is fixed,find the speed of B ; and 4. If the arm G makes 100 r.p.m. clockwiseand wheel A makes 10 r.p.m. counter clockwise ; find the speed ofwheel B.

Solution. Given : TC = 28 ; TD = 26 ; TE = TF = 18

1. Sketch the arrangement

The arrangement is shown in Fig. 13.12.

2. Number of teeth on wheels A and B

Let TA = Number of teeth on wheel A , and

TB = Number of teeth on wheel B.

If dA , dB , dC , dD , dE and dF are the pitch circle diameters of wheels A , B, C, D, E and Frespectively, then from the geometry of Fig. 13.12,

dA = dC + 2 dE

and dB = dD + 2 dF

Since the number of teeth are proportional to their pitch circle diameters, for the samemodule, therefore

TA = TC + 2 TE = 28 + 2 × 18 = 64 Ans.

and TB = TD + 2 TF = 26 + 2 × 18 = 62 Ans.

3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A is fixed

First of all, the table of motions is drawn as given below :



Step Conditions of Arm Wheel Wheel Compound Wheel F Wheel BNo. motion G A E wheel C-D

1. 0 + 1A

E

T

T+ A E

E C–

T T

T T× A D

C F

T T

T T+ × A D F

C F B

T T T

T T T+ × ×

2. 0 + xA

E

Tx

T+ × A

C–

Tx

T× A D

C F

T Tx

T T+ × × A D

C B

T Tx

T T+ × ×

3. + y + y + y + y + y + y

4. + y x + yA

E

Ty x

T+ × A

C–

Ty x

T× A D

C F

T Ty x

T T+ × × A D

C B

T Ty x

T T+ × ×

Fig. 13.12

Arm fixed- wheel Arotates through + 1revolution (i.e. 1 rev.anticlockwise)

Arm fixed-wheel Arotates through + xrevolutions

Add + y revolutionsto all elements

Total motion

A

C

–T

T= A D

C B

T T

T T= + ×

Chapter 13 : Gear Trains � 447Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,

y = – 100 ...(i)

Also, the wheel A is fixed, therefore from the fourth row of the table,

x + y = 0 or x = – y = 100 ...(ii)

∴ Speed of wheel A D

C B

64 26– 100 100 – 100 95.8 r.p.m.

28 62

T TB y x

T T= + × × = + × × = +

= – 4.2 r.p.m. = 4.2 r.p.m. clockwise Ans.4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter

clockwiseSince the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the

tabley = – 100 ...(iii)

Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of thetable,

x + y = 10 or x = 10 – y = 10 + 100 = 110 ...(iv)

∴ Speed of wheel A D

C B

64 26– 100 110 – 100 105.4 r.p.m.

28 62

T TB y x

T T= + × × = + × × = +

= + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise Ans.Example 13.9. In an epicyclic gear of the ‘sun and planet’ type shown

in Fig. 13.13, the pitch circle diameter of the internally toothed ring is to be224 mm and the module 4 mm. When the ring D is stationary, the spider A,which carries three planet wheels C of equal size, is to make one revolution inthe same sense as the sunwheel B for every five revolutions of the drivingspindle carrying the sunwheel B. Determine suitable numbers of teeth for allthe wheels.

Solution. Given : dD = 224 mm ; m = 4 mm ; NA = NB / 5Let TB , TC and TD be the number of teeth on the sun wheel B ,

planet wheels C and the internally toothed ring D. The table of motions is given below :



Step No. Conditions of motion Spider A Sun wheel B Planet wheel C Internal gear D

1. 0 + 1B

C–

T

TB C B

C D D– –

T T T

T T T× =

2. 0 + xB

C–

Tx

T× B

D–

Tx

T×

3. + y + y + y + y

4. + y x + yB

C–

Ty x

T× B

D–

Ty x

T×

Fig. 13.13

Spider A fixed, sun wheelB rotates through + 1revolution (i.e. 1 rev.anticlockwise)Spider A fixed, sun wheelB rotates through + xrevolutionsAdd + y revolutions to allelements

Total motion


We know that when the sunwheel B makes + 5 revolutions, the spi-der A makes + 1 revolution. Thereforefrom the fourth row of the table,

y = + 1 ; and x + y = + 5

∴ x = 5 – y = 5 – 1 = 4

Since the internally toothed ringD is stationary, therefore from the fourthrow of the table,

B

D

– 0T

y xT

× =

or B

D

1 – 4 0T

T× =

∴ B

D

1

4

T

T= or TD = 4 TB ...(i)

We know that TD = dD / m = 224 / 4 = 56 Ans.

∴ TB = TD / 4 = 56 / 4 = 14 Ans. ...[From equation (i)]

Let dB, dC and dD be the pitch circle diameters of sun wheel B, planet wheels C and internallytoothed ring D respectively. Assuming the pitch of all the gears to be same, therefore from the geom-etry of Fig. 13.13,

dB + 2 dC = dD

Since the number of teeth are proportional to their pitch circle diameters, therefore

TB + 2 TC = TD or 14 + 2 TC = 56

∴ TC = 21 Ans.

Example 13.10. Two shafts A and B are co-axial. A gear C (50 teeth) is rigidly mountedon shaft A. A compound gear D-E gears with C and an internal gear G. D has 20 teeth and gearswith C and E has 35 teeth and gears with an internal gear G. The gear G is fixed and is concen-tric with the shaft axis. The compound gear D-E is mounted on a pin which projects from an armkeyed to the shaft B. Sketch the arrangement and find the number of teeth on internal gear Gassuming that all gears have the same module. If the shaft A rotates at 110 r.p.m., find the speedof shaft B.

Solution. Given : TC = 50 ; TD = 20 ; TE = 35 ; NA = 110 r.p.m.


Number of teeth on internal gear G

Let dC , dD , dE and dG be the pitch circle diameters of gears C, D, E and G respectively. Fromthe geometry of the figure,

G C D E

2 2 2 2

d d d d= + +

or dG = dC + dD + dE

Power transmission in a helicopter is essentially throughgear trains.

Note : This picture is given as additional information and is not adirect example of the current chapter.

Main rotor Tail rotor

Tail boom

Landing skids Engine, transmis-sion fuel, etc.

Cockpit

Drive shaft

Chapter 13 : Gear Trains � 449Let TC , TD , TE and TG be the number of teeth on gears C, D, E and G respectively. Since all

the gears have the same module, therefore number of teeth are proportional to their pitch circlediameters.

∴ TG = TC + TD + TE = 50 + 20 + 35 = 105 Ans.

Fig. 13.14

Speed of shaft B




Step Conditions of motion Arm Gear C (or Compound Gear GNo. shaft A) gear D-E

1. 0 + 1C

D–

T

TC E

D G–

T T

T T×

2. 0 + xC

D–

Tx

T× C E

D G–

T Tx

T T× ×

3. + y + y + y + y

4. + y x + yC

D–

Ty x

T× C E

D G–

T Ty x

T T× ×

Since the gear G is fixed, therefore from the fourth row of the table,

C E

D G

– 0T T

y xT T

× × = or 50 35– 0

20 105y x × × =

∴ 5– 0

6y x = ...(i)

Arm fixed - gear C rotates through + 1revolution

Arm fixed - gear C rotates through + xrevolutions


Total motion


Since the gear C is rigidly mounted on shaft A , therefore speed of gear C and shaft A is same.We know that speed of shaft A is 110 r.p.m., therefore from the fourth row of the table,

x + y = 100 ...(ii)

From equations (i) and (ii), x = 60, and y = 50

∴ Speed of shaft B = Speed of arm = + y = 50 r.p.m. anticlockwise Ans.

Example 13.11. Fig. 13.15 shows diagrammatically a compoundepicyclic gear train. Wheels A , D and E are free to rotate independentlyon spindle O, while B and C are compound and rotate together on spindleP, on the end of arm OP. All the teeth on different wheels have the samemodule. A has 12 teeth, B has 30 teeth and C has 14 teeth cut externally.Find the number of teeth on wheels D and E which are cut internally.

If the wheel A is driven clockwise at 1 r.p.s. while D is drivencounter clockwise at 5 r.p.s., determine the magnitude and direction ofthe angular velocities of arm OP and wheel E.

Solution. Given : TA = 12 ; TB = 30 ;TC = 14 ; NA = 1 r.p.s. ; ND = 5 r.p.s.

Number of teeth on wheels D and E

Let TD and TE be the number of teeth on wheels D and E respectively. Let dA , dB , dC , dD and dEbe the pitch circle diameters of wheels A , B, C, D and E respectively. From the geometry of the figure,

dE = dA + 2dB and dD = dE – (dB – dC)

Since the number of teeth are proportional to their pitch circle diameters for the same module,therefore

TE = TA + 2TB = 12 + 2 × 30 = 72 Ans.

and TD = TE – (TB – TC) = 72 – (30 – 14) = 56 Ans.

Magnitude and direction of angular velocities of arm OP and wheel E

The table of motions is drawn as follows :



Step Conditions of motion Arm Wheel A Compound Wheel D Wheel ENo. wheel B-C

1. 0 – 1A

B

T

T+ A C

B D

T T

T T+ × A B

B E

T T

T T+ ×

A

E

T

T= +

2. 0 – xA

B

Tx

T+ × A C

B D

T Tx

T T+ × × A

E

Tx

T+ ×

3. – y – y – y – y – y

4. – y – x – yA

B–

Tx y

T× A C

B D–

T Tx y

T T× × A

E–

Tx y

T×

Fig. 13.15

Arm fixed A rotated through– 1 revolution (i.e. 1 revolu-tion clockwise)

Arm fixed-wheel A rotatedthrough – x revolutions

Add – y revolutions to all ele-ments

Total motion

Chapter 13 : Gear Trains � 451Since the wheel A makes 1 r.p.s. clockwise, therefore from the fourth row of the table,

– x – y = – 1 or x + y = 1 ...(i)Also, the wheel D makes 5 r.p.s. counter clockwise, therefore

CA

B D

– 5TT

x yT T

× × = or 12 14– 5

30 56x y× × =

∴ 0.1 x – y = 5 ...(ii)From equations (i) and (ii),

x = 5.45 and y = – 4.45

∴ Angular velocity of arm OP

= – y = –(– 4.45) = 4.45 r.p.s

= 4.45 × 2 π = 27.964 rad/s (counter clockwise) Ans.

and angular velocity of wheel A

E

12– 5.45 – (– 4.45) 5.36 r.p.s.

72

TE x y

T= × = × =

= 5.36 × 2 π = 33.68 rad/s (counter clockwise) Ans.Example 13.12. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal

wheel C with 82 teeth is concentricwith B. A compound wheel D-Egears with the two internal wheels;D has 28 teeth and gears with Cwhile E gears with B. The compoundwheels revolve freely on a pin whichprojects from a disc keyed to a shaftA co-axial with F. If the wheels havethe same pitch and the shaft A makes800 r.p.m., what is the speed of theshaft F ? Sketch the arrangement.

Solution. Given : TB = 80 ; TC= 82 ; TD = 28 ; NA = 500 r.p.m.


Fig. 13.16

First of all, let us find out the number of teeth on wheel E (TE). Let dB , dC , dD and dE be thepitch circle diameter of wheels B, C, D and E respectively. From the geometry of the figure,

dB = dC – (dD – dE )

HelicopterNote : This picture is given as additional information and is not a

direct example of the current chapter.


or dE = dB + dD – dC

Since the number of teeth are proportional to their pitch circle diameters for the same pitch,therefore

TE = TB + TD – TC = 80 + 28 – 82 = 26




Step Conditions of motion Arm (or Wheel B (or Compound Wheel CNo. shaft A) shaft F) gear D-E

1. 0 + 1B

E

T

T+ B D

E C

T T

T T+ ×

2. 0 + xB

E

Tx

T+ × B D

E C

T Tx

T T+ × ×

3. + y + y + y + y

4. + y x + yB

E

Ty x

T+ × B D

E C

T Ty x

T T+ × ×

Since the wheel C is fixed, therefore from the fourth row of the table,

B D

E C

0T T

y xT T

+ × × = or80 28

026 82

y x+ × × =

∴ y + 1.05 x = 0 ...(i)

Also, the shaft A (or the arm) makes 800 r.p.m., therefore from the fourth row of the table,

y = 800 ...(ii)

From equations (i) and (ii),

x = – 762

∴ Speed of shaft F = Speed of wheel B = x + y = – 762 + 800 = + 38 r.p.m.


Example 13.13. Fig. 13.17 shows an epicyclic geartrain known as Ferguson’s paradox. Gear A is fixed to theframe and is, therefore, stationary. The arm B and gears Cand D are free to rotate on the shaft S. Gears A, C and D have100, 101 and 99 teeth respectively. The planet gear has 20teeth. The pitch circle diameters of all are the same so that theplanet gear P meshes with all of them. Determine therevolutions of gears C and D for one revolution of the arm B.

Solution. Given : TA = 100 ; TC = 101 ; TD = 99 ;TP = 20 Fig. 13.17

Arm fixed - wheel B rotatedthrough + 1 revolution (i.e. 1revolution anticlockwise)

Arm fixed - wheel B rotatedthrough + x revolutions


Total motion

Chapter 13 : Gear Trains � 453The table of motions is given below :



Step No. Conditions of motion Arm B Gear A Gear C Gear D

1. 0 + 1A

C

T

T+ A C A

C D D

T T T

T T T+ × = +

2. 0 + xA

C

Tx

T+ × A

D

Tx

T+ ×

3. + y + y + y + y

4. + y x + yA

C

Ty x

T+ × A

D

Ty x

T+ ×

The arm B makes one revolution, thereforey = 1

Since the gear A is fixed, therefore from the fourth row of the table,x + y = 0 or x = – y = – 1

Let NC and ND = Revolutions of gears C and D respectively.From the fourth row of the table, the revolutions of gear C,

AC

C

100 11 – 1

101 101

TN y x

T= + × = × =+ Ans.

and the revolutions of gear D,

AD

D

100 11 –

99 99

TN y x

T= + × = = – Ans.

From above we see that for one revolution of the arm B, the gear C rotates through 1/101revolutions in the same direction and the gear D rotates through 1/99 revolutions in the oppositedirection.

Example 13.14. In the gear drive as shown in Fig.13.18, the driving shaft A rotates at 300 r.p.m. in the clock-wise direction, when seen from left hand. The shaft B is thedriven shaft. The casing C is held stationary. The wheels Eand H are keyed to the central vertical spindle and wheel Fcan rotate freely on this spindle. The wheels K and L arerigidly fixed to each other and rotate together freely on apin fitted on the underside of F. The wheel L meshes withinternal teeth on the casing C. The numbers of teeth on thedifferent wheels are indicated within brackets in Fig. 13.18.

Find the number of teeth on wheel C and the speedand direction of rotation of shaft B.

Solution. Given : NA = 300 r.p.m. (clockwise) ;TD = 40 ; TB = 30 ; TF = 50 ; TG = 80 ; TH = 40 ; TK = 20 ; TL = 30

In the arrangement shown in Fig. 13.18, the wheels D and G are auxillary gears and do notform a part of the epicyclic gear train.

Fig. 13.18

Arm B fixed, gear A rotatedthrough + 1 revolution (i.e. 1revolution anticlockwise)

Arm B fixed, gear A rotatedthrough + x revolutions


Total motion


Speed of wheel E, DE A

E

40300 400 r.p.m. (clockwise)

30

TN N

T= × = × =

Number of teeth on wheel CLet TC = Number of teeth on wheel C.Assuming the same module for all teeth and since the pitch circle diameter is proportional to

the number of teeth ; therefore from the geometry of Fig.13.18,TC = TH + TK + TL = 40 + 20 + 30 = 90 Ans.

Speed and direction of rotation of shaft BThe table of motions is given below. The wheel F acts as an arm.



Step Conditions of motion Arm or Wheel E Wheel H Compound Wheel CNo. wheel F wheel K-L

1. 0 – 1H

K

T

T+ H L

K C

T T

T T+ ×

2. 0 – x – xH

K

Tx

T+ × H L

K C

T Tx

T T+ × ×

3. – y – y – y – y – y

4. – y – x – y – x – yH

K–

Tx y

T× H L

K C–

T Tx y

T T× ×

Since the speed of wheel E is 400 r.p.m. (clockwise), therefore from the fourth row of the table,

– x – y = – 400 or x + y = 400 ...(i)Also the wheel C is fixed, therefore

H L

K C

– 0T T

x yT T

× × =

or 40 30

– 020 90

x y× × =

∴ 2

– 03

xy = ...(ii)


x = 240 and y = 160

∴ Speed of wheel F, NF = – y = – 160 r.p.m.

Since the wheel F is in mesh with wheel G, therefore speed of wheel G or speed of shaft B

FF

G

50– – – 160 100 r.p.m.

80

TN

T = × = × =

...(� Wheel G will rotate in opposite direction to that of wheel F.)

= 100 r.p.m. anticlockwise i.e. in opposite direction ofshaft A . Ans.

Arm fixed-wheel Erotated through – 1revolution (i.e. 1revolution clockwise)

Arm fixed-wheel Erotated through – xrevolutions

Add – y revolutions toall elements

Total motion

– 1(� E andH are on thesame shaft)

Chapter 13 : Gear Trains � 455Example 13.15. Fig. 13.19 shows a compound epicyclic gear in which the casing C contains

an epicyclic train and this casing is inside the larger casing D.Determine the velocity ratio of the output shaft B to the input shaft A when the casing D is

held stationary. The number of teeth on various wheels are as follows :Wheel on A = 80 ; Annular wheel on B = 160 ; Annular wheel on C = 100 ; Annular wheel

on D = 120 ; Small pinion on F = 20 ; Large pinion on F = 66.

Fig. 13.19

Solution. Given : T1 = 80 ; T8 = 160 ; T4 = 100; T3 = 120 ; T6 = 20 ; T7 = 66First of all, let us consider the train of wheel 1 (on A ), wheel 2 (on E), annular wheel 3 (on D)

and the arm i.e. casing C. Since the pitch circle diameters of wheels are proportional to the number ofteeth, therefore from the geometry of Fig. 13.19,

T1 + 2 T2 = T3 or 80 + 2 T2 = 120∴ T2 = 20The table of motions for the train considered is given below :



Step No. Conditons of motion Arm Wheel 1 Wheel 2 Wheel 3

1. 0 + 11

2–

T

T1 2 1

2 3 3– –

T T T

T T T× =

2. 0 + x1

2–

Tx

T× 1

3–

Tx

T×

3. + y + y + y + y

4. y x + y1

2–

Ty x

T× 1

3–

Ty x

T×

Arm fixed - wheel 1 rotatedthrough + 1 revolution(anticlockwise)

Arm fixed - wheel 1 rotatedthrough + x revolutions


Total motion


Let us assume that wheel 1 makes 1 r.p.s. anticlockwise.

∴ x + y = 1 ...(i)

Also the wheel 3 is stationary, therefore from the fourth row of the table,

1

3

– 0T

y xT

× = or 80– 0

120y x × =

∴ 2– 0

3y x = ...(ii)

From equations (i) and (ii), x = 0.6, and y = 0.4

∴ Speed of arm or casing C = y = 0.4 r.p.s.

and speed of wheel 2 or arm E 1

2

80– 0.4 – 0.6 – 2 r.p.s.

20

Ty x

T= × = × =

= 2 r.p.s. (clockwise)

Let us now consider the train of annular wheel 4 (on C), wheel 5 (on E), wheel 6 (on F) andarm E. We know that

T6 + 2 T5 = T4 or 20 + 2 T5 = 100

∴ T5 = 40




Step Conditions of motion Arm E or Wheel 6 Wheel 5 Wheel 4No. wheel 2

1. 0 + 16

5–

T

T6 5 6

5 4 4– –

T T T

T T T× =

2. 0 x1

61

5–

Tx

T× 6

14

–T

xT

×

3. + y1 + y1 + y1 + y1

4. + y1 x1 + y1

61 1

5–

Ty x

T× 6

1 14

–T

y xT

×

We know that speed of arm E = Speed of wheel 2 in the first train

∴ y1 = – 2 ...(iii)

Also speed of wheel 4 = Speed of arm or casing C in the first train

∴ 61 1

4

– 0.4T

y xT

× = or 120

–2 – 0.4100

x × = ...(iv)

or 1100

(–2 – 0.4) –1220

x = =

Arm fixed, wheel 6 rotatedthrough + 1 revolution

Arm fixed, wheel 6 rotatedthrough + x1 revolutions

Add + y1 revolutions to allelements

Total motion

Chapter 13 : Gear Trains � 457∴ Speed of wheel 6 (or F)

= x1 + y1 = – 12 – 2 = – 14 r.p.s. = 14 r.p.s. (clockwise)

Now consider the train of wheels 6 and 7 (both on F), annular wheel 8 (on B) and the arm i.e.casing C. The table of motions is given below :



Step No. Conditions of motion Arm Wheel 8 Wheel 7

1. 0 + 18

7

T

T+

2. 0 + x2

82

7

Tx

T+ ×

3. + y2 + y2 + y2

4. y2 x2 + y2

82 2

7

Ty x

T+ ×

We know that the speed of C in the first train is 0.4 r.p.s., therefore

y2 = 0.4 ...(v)

Also the speed of wheel 7 is equal to the speed of F or wheel 6 in the second train, therefore

82 2

7

–14T

y xT

+ × = or2

1600.4 –14

66x+ × = ...(vi)

∴ 266

( 14 0.4) 5.94160

x = − − = −

∴ Speed of wheel 8 or of the shaft B

x2 + y2 = – 5.94 + 0.4 = – 5.54 r.p.s. = 5.54 r.p.s. (clockwise)

We have already assumed that the speed of wheel 1 or the shaft A is 1 r.p.s. anticlockwise

∴ Velocity ratio of the output shaft B to the input shaft A

= – 5.54 Ans.Note : The – ve sign shows that the two shafts A and B rotate in opposite directions.

13.10. Epicyclic Gear Train with Bevel GearsThe bevel gears are used to make a more compact epicyclic system and they permit a very

high speed reduction with few gears. The useful application of the epicyclic gear train with bevelgears is found in Humpage’s speed reduction gear and differential gear of an automobile as discussedbelow :

1. Humpage’s speed reduction gear. The Humpage’s speed reduction gear was originallydesigned as a substitute for back gearing of a lathe, but its use is now considerably extended to allkinds of workshop machines and also in electrical machinery. In Humpage’s speed reduction gear, asshown in Fig. 13.20, the driving shaft X and the driven shaft Y are co-axial. The driving shaft carriesa bevel gear A and driven shaft carries a bevel gear E. The bevel gear B meshes with gear A (alsoknown as pinion) and a fixed gear C. The gear E meshes with gear D which is compound with gear B.

Arm fixed, wheel 8 rotated through+ 1 revolution

Arm fixed, wheel 8 rotated through+ x2 revolutions

Add + y 2 revolutions to allelements

Total motion


This compound gear B-D is mounted on the arm or spindle F which is rigidly connected with a hollowsleeve G. The sleeve revolves freely loose on the axes of the driving and driven shafts.

Fig. 13.20. Humpage’s speed reduction gear.2. Differential gear of an automobile. The differential gear used in the rear drive of an

automobile is shown in Fig. 13.21. Its function is

(a) to transmit motion from the engine shaft to the rear driving wheels, and(b) to rotate the rear wheels at different speeds while the automobile is taking a turn.As long as the automobile is running on a straight path, the rear wheels are driven directly by

the engine and speed of both the wheels is same. But when the automobile is taking a turn, the outerwheel will run faster than the * inner wheel because at that time the outer rear wheel has to cover moredistance than the inner rear wheel. This is achieved by epicyclic gear train with bevel gears as shownin Fig. 13.21.

The bevel gear A (known as pinion) is keyed tothe propeller shaft driven from the engine shaft throughuniversal coupling. This gear A drives the gear B (knownas crown gear) which rotates freely on the axle P. Twoequal gears C and D are mounted on two separate parts Pand Q of the rear axles respectively. These gears, in turn,mesh with equal pinions E and F which can rotate freelyon the spindle provided on the arm attached to gear B.

When the automobile runs on a straight path, thegears C and D must rotate together. These gears are rotatedthrough the spindle on the gear B. The gears E and F donot rotate on the spindle. But when the automobile is takinga turn, the inner rear wheel should have lesser speed thanthe outer rear wheel and due to relative speed of the inner and outer gears D and C, the gears E and Fstart rotating about the spindle axis and at the same time revolve about the axle axis.

Due to this epicyclic effect, the speed of the inner rear wheel decreases by a certain amountand the speed of the outer rear wheel increases, by the same amount. This may be well understood bydrawing the table of motions as follows :

Fig. 13.21. Differential gear of an automobile.

* This difficulty does not arise with the front wheels as they are greatly used for steering purposes and aremounted on separate axles and can run freely at different speeds.

Chapter 13 : Gear Trains � 459Table 13.17. Table of motions.


Step No. Conditions of motion Gear B Gear C Gear E Gear D

1. 0 + 1C

E

T

T+ C E

E D– – 1

T T

T T× =

( )C DT T=�

2. 0 + xC

E

Tx

T+ × –x

3. + y + y + y + y

4. + y x + yC

E

Ty x

T+ × –y x

From the table, we see that when the gear B, which derives motion from the engine shaft,rotates at y revolutions, then the speed of inner gear D (or the rear axle Q) is less than y by x revolu-tions and the speed of the outer gear C (or the rear axle P) is greater than y by x revolutions. In otherwords, the two parts of the rear axle and thus the two wheels rotate at two different speeds. We alsosee from the table that the speed of gear B is the mean of speeds of the gears C and D.

Example 13.16. Two bevel gears A and B (having 40 teeth and 30 teeth) are rigidly mountedon two co-axial shafts X and Y. A bevel gear C (having50 teeth) meshes with A and B and rotates freely on oneend of an arm. At the other end of the arm is welded asleeve and the sleeve is riding freely loose on the axes ofthe shafts X and Y. Sketch the arrangement.

If the shaft X rotates at 100 r.p.m. clockwise andarm rotates at 100 r.p.m.anitclockwise, find thespeed of shaft Y.

Solution. Given : TA = 40 ; TB = 30 ; TC = 50 ; NX= NA = 100 r.p.m. (clockwise) ; Speed of arm = 100 r.p.m.(anticlockwise)

The arangement is shown in Fig. 13.22.The table of motions is drawn as below :

Table 13.18. Table of motions.Revolutions of elements

Step No. Conditions of motion Arm Gear A Gear C Gear B

1. 0 + 1A

C

T

T± A C A

C B B– –

T T T

T T T× =

2. 0 + xA

C

Tx

T± × A

B–

Tx

T×

3. + y + y + y + y

4. + y x + yA

C

Ty x

T± × A

B–

Ty x

T×

Fig. 13.22

* The ± sign is given to the motion of the wheel C because it is in a different plane. So we cannot indicate thedirection of its motion specifically, i.e. either clockwise or anticlockwise.

Gear B fixed-Gear C rotatedthrough + 1 revolution (i.e.1 revolution anticlockwise )

Gear B fixed-Gear C rotatedthrough + x revolutions


Total motion

Arm B fixed, gear A rotatedthrough + 1 revolution (i.e. 1revolution anticlockwise)

Arm B fixed, gear A rotatedthrough + x revolutionsAdd + y revolutions to allelements

Total motion

*


Since the speed of the arm is 100 r.p.m. anticlockwise, therefore from the fourth row of thetable,

y = + 100

Also, the speed of the driving shaft X or gear A is 100 r.p.m. clockwise.

∴ x + y = – 100 or x = – y – 100 = – 100 – 100 = – 200

∴ Speed of the driven shaft i.e. shaft Y ,

NY = Speed of gear A

B

40– 100 – – 200

30

TB y x

T = × = ×

= + 366.7 r.p.m. = 366.7 r.p.m. (anticlockwise) Ans.

Example 13.17. In a gear train, asshown in Fig. 13.23, gear B is connected to theinput shaft and gear F is connected to the outputshaft. The arm A carrying the compound wheelsD and E, turns freely on the output shaft. If theinput speed is 1000 r.p.m. counter- clockwisewhen seen from the right, determine the speed ofthe output shaft under the following conditions :

1. When gear C is fixed, and 2. whengear C is rotated at 10 r.p.m. counter clockwise.

Solution. Given : TB = 20 ; TC = 80 ;TD = 60 ; TE = 30 ; TF = 32 ; NB = 1000 r.p.m.(counter-clockwise)




Step Conditions of motion Arm A Gear B Compound Gear C Gear F (orNo. (or input wheel D-E output shaft)

shaft)

1. 0 + 1B

D

T

T+ B D

D C–

T T

T T× B E

D F–

T T

T T×

B

C–

T

T=

2. 0 + xB

D

Tx

T+ × B

C–

Tx

T× B E

D F–

T Tx

T T× ×

3. + y + y + y + y + y

4. + y x + yB

D

Ty x

T+ × B

C–

Ty x

T× B E

D F–

T Ty x

T T× ×

Fig. 13.23

Arm fixed, gear B rotatedthrough + 1 revolution (i.e.1 revolution anticlockwise)

Arm fixed, gear B rotatedthrough + x revolutions


Total motion

Chapter 13 : Gear Trains � 4611. Speed of the output shaft when gear C is fixed

Since the gear C is fixed, therefore from the fourth row of the table,

B

C

– 0T

y xT

× = or 20– 0

80y x × =

∴ y – 0.25 x = 0 ...(i)We know that the input speed (or the speed of gear B) is 1000 r.p.m. counter clockwise,

therefore from the fourth row of the table,x + y = + 1000 ...(ii)

From equations (i) and (ii), x = + 800, and y = + 200

∴ Speed of output shaft = Speed of gear B E

D F

–T T

F y xT T

= × ×

20 30200 – 800 200 – 187.5 12.5 r.p.m.

80 32= × × = =

= 12.5 r.p.m. (counter clockwise) Ans.

2. Speed of the output shaft when gear C is rotated at 10 r.p.m. counter clockwise

Since the gear C is rotated at 10 r.p.m. counter clockwise, therefore from the fourth row of the table,

B

C

– 10T

y xT

× = + or20

– 1080

y x × =

∴ y – 0.25 x = 10 ...(iii)From equations (ii) and (iii),

x = 792, and y = 208

∴ Speed of output shaft

= Speed of gear B E

D F

20 30– 208 – 792

80 32

T TF y x

T T= × × = × ×

= 208 – 185.6 = 22.4 r.p.m. = 22.4 r.p.m. (counter clockwise) Ans.Example 13.18. Fig. 13.24 shows a differential

gear used in a motor car. The pinion A on the propellershaft has 12 teeth and gears with the crown gear B whichhas 60 teeth. The shafts P and Q form the rear axles towhich the road wheels are attached. If the propellershaft rotates at 1000 r.p.m. and the road wheel attachedto axle Q has a speed of 210 r.p.m. while taking a turn,find the speed of road wheel attached to axle P.

Solution. Given : TA = 12 ; TB = 60 ; NA = 1000r.p.m. ; NQ = ND = 210 r.p.m.

Since the propeller shaft or the pinion A rotates at1000 r.p.m., therefore speed of crown gear B,

A

B AB

121000

60

TN N

T= × = ×

= 200 r.p.m.


Fig. 13.24




Step No. Conditions of motion Gear B Gear C Gear E Gear D

1. 0 + 1C

E

T

T+ C E

E D– – 1

T T

T T× =

C D( )T T=�

2. 0 + xC

E

Tx

T+ × – x

3. + y + y + y + y

4. + y x + yC

E

Ty x

T+ × y – x

Since the speed of gear B is 200 r.p.m., therefore from the fourth row of the table,

y = 200 ...(i)

Also, the speed of road wheel attached to axle Q or the speed of gear D is 210 r.p.m., there-fore from the fourth row of the table,

y – x = 210 or x = y – 210 = 200 – 210 = – 10

∴ Speed of road wheel attached to axle P

= Speed of gear C = x + y

= – 10 + 200 = 190 r.p.m. Ans.

13.11. Torques in Epicyclic Gear Trains

Fig. 13.25. Torques in epicyclic gear trains.

When the rotating parts of an epicyclic gear train, as shown in Fig. 13.25, have no angularacceleration, the gear train is kept in equilibrium by the three externally applied torques, viz.

1. Input torque on the driving member (T1),

2. Output torque or resisting or load torque on the driven member (T2),

3. Holding or braking or fixing torque on the fixed member (T3).

Gear B fixed-Gear C rotatedthrough + 1 revolution (i.e. 1revolution anticlockwise)

Gear B fixed-Gear C rotatedthrough + x revolutions


Total motion

Chapter 13 : Gear Trains � 463The net torque applied to the gear train must be zero. In other words,

T1 + T2 + T3 = 0 ...(i)

∴ F1.r1 + F2.r2 + F3.r3 = 0 ...(ii)

where F1, F2 and F3 are the corresponding externally applied forces at radii r1, r2 and r3.

Further, if ω1, ω2 and ω3 are the angular speeds of the driving, driven and fixed membersrespectively, and the friction be neglected, then the net kinetic energy dissipated by the gear trainmust be zero, i.e.

T1.ω1 + T2.ω2 + T3.ω3 = 0 ...(iii)

But, for a fixed member, ω3 = 0

∴ T1.ω1 + T2.ω2 = 0 ...(iv)

Notes : 1. From equations (i) and (iv), the holding or braking torque T3 may be obtained as follows :

12 1

2

–T Tω= ×ω ...[From equation (iv)]

and T3 = – (T1+ T2 ) ...[From equation (i)]

1 11 1

2 2

– 1 – 1N

T TN

ω= = ω 2. When input shaft (or driving shaft) and output shaft (or driven shaft) rotate in the same direction,

then the input and output torques will be in opposite directions. Similarly, when the input and output shaftsrotate in opposite directions, then the input and output torques will be in the same direction.

Example 13.19. Fig. 13.26 shows an epicyclic gear train. PinionA has 15 teeth and is rigidly fixed to the motor shaft. The wheel B has 20teeth and gears with A and also with the annular fixed wheel E. PinionC has 15 teeth and is integral with B (B, C being a compound gearwheel). Gear C meshes with annular wheel D, which is keyed to themachine shaft. The arm rotates about the same shaft on which A is fixedand carries the compound wheel B, C. If the motor runs at 1000 r.p.m.,find the speed of the machine shaft. Find the torque exerted on themachine shaft, if the motor develops a torque of 100 N-m.

Solution. Given : TA = 15 ; TB = 20 ; TC = 15 ; NA = 1000 r.p.m.; Torque developed by motor (orpinion A) = 100 N-m

First of all, let us find the number of teeth on wheels D and E. Let TD and TE be the number ofteeth on wheels D and E respectively. Let dA, dB, dC, dD and dE be the pitch circle diameters of wheelsA , B, C, D and E respectively. From the geometry of the figure,

dE = dA + 2 dB and dD = dE – (dB – dC)

Since the number of teeth are proportional to their pitch circle diameters, therefore,

TE = TA + 2 TB = 15 + 2 × 20 = 55

and TD = TE – (TB – TC) = 55 – (20 – 15) = 50

Speed of the machine shaft


Fig. 13.26




Step Conditions of motion Arm Pinion Compound Wheel D Wheel ENo. A wheel B-C

1. 0 + 1A

B–

T

TA C

B D–

T T

T T× A B A

B E E

T T T

T T T− × = −

2. 0 + xA

B–

Tx

T× A C

B D–

T Tx

T T× × A

E

Tx

T− ×

3. + y + y + y + y + y

4. + y x + yA

B–

Ty x

T× A C

B D–

T Ty x

T T× × A

E

Ty x

T− ×

We know that the speed of the motor or the speed of the pinion A is 1000 r.p.m.Therefore

x + y = 1000 ...(i)

Also, the annular wheel E is fixed, therefore

A

E

– 0T

y xT

× = or A

E

150.273

55

Ty x x x

T= × = × = ...(ii)


x = 786 and y = 214

∴ Speed of machine shaft = Speed of wheel D,

CAD

B D

15 15– 214 – 786 37.15 r.p.m.

20 50

TTN y x

T T= × × = × × = +

= 37.15 r.p.m. (anticlockwise) Ans.

Torque exerted on the machine shaft

We know that

Torque developed by motor × Angular speed of motor

= Torque exerted on machine shaft × Angular speed of machine shaft

or 100 × ωA = Torque exerted on machine shaft × ωD

∴ Torque exerted on machine shaft

A A

D D

1000100 100 100 2692 N-m

37.15

N

N

ω= × = × = × =ω

Ans.

Arm fixed-pinion Arotated through + 1revolution(anticlockwise)

Arm fixed-pinion Arotated through + xrevolutions

Add + y revolutions toall elements

Total motion


Fig. 13.27

Example 13.20. An epicyclic gear train consists of a sun wheelS, a stationary internal gear E and three identical planet wheels Pcarried on a star- shaped planet carrier C. The size of different toothedwheels are such that the planet carrier C rotates at 1/5th of the speedof the sunwheel S. The minimum number of teeth on any wheel is 16.The driving torque on the sun wheel is 100 N-m. Determine : 1. num-ber of teeth on different wheels of the train, and 2. torque necessary tokeep the internal gear stationary.

Solution. Given : SC 5

NN =

1. Number of teeth on different wheels

The arrangement of the epicyclic gear train is shown in Fig. 13.27. Let TS and TE be thenumber of teeth on the sun wheel S and the internal gear E respectively. The table of motions isgiven below :



Step Conditions of motion Planet Sun Planet Internal gear ENo. carrier C wheel S wheel P

1. 0 + 1S

P–

T

TS P S

P E E– –

T T T

T T T× =

2. 0 + xS

P–

Tx

T× S

E–

Tx

T×

3. + y + y + y + y

4. + y x + yS

P–

Ty x

T× S

E–

Ty x

T×

We know that when the sunwheel S makes 5 revolutions, the planet carrier C makes 1revolution. Therefore from the fourth row of the table,

y = 1, and x + y = 5 or x = 5 – y = 5 – 1 = 4

Since the gear E is stationary, therefore from the fourth row of the table,

S

E

– 0T

y xT

× = or S

E

1 – 4 0T

T× = or S

E

1

4

T

T=

∴ TE = 4TS

Since the minimum number of teeth on any wheel is 16, therefore let us take the number ofteeth on sunwheel, TS = 16

∴ TE = 4 TS = 64 Ans.

Let dS, dP and dE be the pitch circle diameters of wheels S, P and E respectively. Now from thegeometry of Fig. 13.27,

dS + 2 dP = dE

Planet carrier C fixed, sunwheel Srotates through + 1 revolution (i.e.1 rev. anticlockwise)

Planet carrier C fixed, sunwheel Srotates through + x revolutions


Total motion


Assuming the module of all the gears to be same, the number of teeth are proportional to theirpitch circle diameters.

TS + 2 TP = TE or 16 + 2 TP = 64 or TP = 24 Ans.

2. Torque necessary to keep the internal gear stationary

We know that

Torque on S × Angular speed of S

= Torque on C × Angular speed of C

100 × ωS = Torque on C × ωC

∴ Torque on S S

C C

100 100 100 5 500 N-mN

CN

ω= × = × = × =

ω

∴ Torque necessary to keep the internal gear stationary

= 500 – 100 = 400 N-m Ans.

Example 13.21. In the epicyclic gear train, asshown in Fig. 13.28, the driving gear A rotating in clock-wise direction has 14 teeth and the fixed annular gear Chas 100 teeth. The ratio of teeth in gears E and D is 98 :41. If 1.85 kW is supplied to the gear A rotating at 1200r.p.m., find : 1. the speed and direction of rotation of gearE, and 2. the fixing torque required at C, assuming 100per cent efficiency throughout and that all teeth have thesame pitch.

Solution. Given : TA = 14 ; TC = 100 ; TE / TD= 98 / 41 ; PA = 1.85 kW = 1850 W ; NA = 1200 r.p.m.

Let dA, dB and dC be the pitch circle diameters of gears A , B and C respectively. From Fig.13.28,

dA + 2 dB = dC

Fig. 13.28

Gears are extensively used in trains for power transmission.


Arm fixed-Gear A rotatedthrough – 1 revolution (i.e.1 revolution clockwise)

Arm fixed-Gear A rotatedthrough – x revolutions

Add – y revolutions to allelements

Total motion

Since teeth of all gears have the same pitch and the number of teeth are proportional to theirpitch circle diameters, therefore

TA + 2TB = TC or C AB

– 100 – 1443

2 2

T TT = = =

The table of motions is now drawn as below :



Step Conditions of motion Arm Gear Compound Gear C Gear ENo. A gear B-D

1. 0 – 1A

B

T

T+ A B

B C

T T

T T+ × A D

B E

T T

T T+ ×

A

C

T

T= +

2. 0 – xA

B

Tx

T+ × A

C

Tx

T+ × A D

B E

T Tx

T T+ × ×

3. – y – y – y – y – y

4. – y – y – xA

B–

Ty x

T+ × A

C–

Ty x

T+ × A D

B E–

T Ty x

T T+ × ×

Since the annular gear C is fixed, therefore from the fourth row of the table,

A

C

– 0T

y xT

+ × = or 14– 0

100y x+ × =

∴ – y + 0.14 x = 0 ...(i)

Also, the gear A is rotating at 1200 r.p.m., therefore

– x – y = 1200 ...(ii)

From equations (i) and (ii), x = – 1052.6, and y = – 147.4

1. Speed and direction of rotation of gear E

From the fourth row of the table, speed of gear E,

A DE

B E

14 41– 147.4 – 1052.6

43 98

T TN y x

T T= + × × = × ×

= 147.4 – 143.4 = 4 r.p.m.


2. Fixing torque required at C

We know that torque on A A

A

60 1850 6014.7 N-m

2 2 1200

P

N

× ×= = =π π ×

Since the efficiency is 100 per cent throughout, therefore the power available at E (PE) willbe equal to power supplied at A (PA).


∴ Torque on E A

E

60 1850 604416 N-m

2 2 4

P

N

× ×= = =π × π ×

∴ Fixing torque required at C

= 4416 – 14.7 = 4401.3 N-m Ans.

Example 13.22. An over drive for a vehicle consists of anepicyclic gear train, as shown in Fig. 13.29, with compound planetsB-C. B has 15 teeth and meshes with an annulus A which has 60teeth. C has 20 teeth and meshes with the sunwheel D which is fixed.The annulus is keyed to the propeller shaft Y which rotates at 740rad /s. The spider which carries the pins upon which the planetsrevolve, is driven directly from main gear box by shaft X, this shaftbeing relatively free to rotate with respect to wheel D. Find thespeed of shaft X, when all the teeth have the same module.

When the engine develops 130 kW, what is the holdingtorque on the wheel D ? Assume 100 per cent efficiencythroughout.

Solution. Given : TB = 15 ; TA = 60 ; TC = 20 ; ωY = ωA = 740 rad /s ; P = 130 kW = 130 × 103 W

First of all, let us find the number of teeth on the sunwheel D (TD). Let dA , dB , dC and dD bethe pitch circle diameters of wheels A , B, C and D respectively. From Fig. 13.29,

CD B A

2 2 2 2

dd d d+ + = or dD + dC + dB = dA

Since the module is same for all teeth and the number of teeth are proportional to their pitchcircle diameters, therefore

TD + TC + TB = TA or TD = TA – (TC + TB) = 60 – (20 + 15) = 25




Step Conditions of motion Arm (or Wheel D Compound Wheel ANo. shaft X) wheel C-B (or shaft Y)

1. 0 + 1D

C–

T

TD B

C A–

T T

T T×

2. 0 + xD

C–

Tx

T× D B

C A–

T Tx

T T× ×

3. + y + y + y + y

4. + y x + yD

C–

Ty x

T× D B

C A–

T Ty x

T T× ×

Since the shaft Y or wheel A rotates at 740 rad/s, therefore

D B

C A

– 740T T

y xT T

× × = or 25 15– 740

20 60y x × × =

y – 0.3125 x = 740 ...(i)

Arm fixed-wheel D rotatedthrough + 1 revolution(anticlockwise)

Arm fixed-wheel D rotatedthrough + x revolutions

Add + y revolutions to all ele-ments

Total motion

Fig. 13.29

Chapter 13 : Gear Trains � 469Also the wheel D is fixed, therefore

x + y = 0 or y = – x ...(ii)


x = – 563.8 and y = 563.8

Speed of shaft X

Since the shaft X will make the same number of revolutions as the arm, therefore

Speed of shaft X , ωX = Speed of arm = y = 563.8 rad/s Ans.

Holding torque on wheel D

We know that torque on A = P/ωA = 130 × 103 / 740 = 175.7 N-m

and Torque on X = P/ωX = 130 × 103/563.8 = 230.6 N-m

∴ Holding torque on wheel D

= 230.6 – 175.7 = 54.9 N-m Ans.

Example 13.23. Fig. 13.30 shows some details of a compound epicyclic gear drive where Iis the driving or input shaft and O is the driven or output shaft which carries two arms A and Brigidly fixed to it. The arms carry planet wheels which mesh with annular wheels P and Q and thesunwheels X and Y. The sun wheel X is a part of Q. Wheels Y and Z are fixed to the shaft I. Z engageswith a planet wheel carried on Q and this planet wheel engages the fixed annular wheel R. Thenumbers of teeth on the wheels are :

P = 114, Q = 120, R = 120, X = 36, Y = 24 and Z = 30.

Fig. 13.30.

The driving shaft I makes 1500 r.p.m.clockwise looking from our right and the input at I is7.5 kW.

1. Find the speed and direction of rotation of the driven shaft O and the wheel P.

2. If the mechanical efficiency of the drive is 80%, find the torque tending to rotate the fixedwheel R.

Solution. Given : TP =144 ; TQ = 120 ; TR = 120 ; TX = 36 ; TY = 24 ; TZ = 30 ; NI = 1500r.p.m. (clockwise) ; P = 7.5 kW = 7500 W ; η = 80% = 0.8

First of all, consider the train of wheels Z,R and Q (arm). The revolutions of various wheelsare shown in the following table.


Arm fixed-wheel Z rotates through + 1revolution (anticlockwise)

Arm fixed-wheel Z rotates through + x revo-lutions


Total motion



Step No. Conditions of motion Q (Arm) Z (also I) R (Fixed)

1. 0 + 1Z

R–

T

T

2. 0 + xZ

R–

Tx

T×

3. + y + y + y

4. + y x + yZ

R–

Ty x

T×

Since the driving shaft I as well as wheel Z rotates at 1500 r.p.m. clockwise, therefore

x + y = – 1500 ...(i)

Also, the wheel R is fixed. Therefore

Z

R

– 0T

y xT

× = or Z

R

300.25

120

Ty x x x

T= × = × = ...(ii)


x = – 1200, and y = – 300

Now consider the train of wheels Y , Q, arm A , wheels P and X . The revolutions of variouselements are shown in the following table.



Step Conditions of motion Arm A, B Wheel Y Compound Wheel PNo. and Shaft O wheel Q-X

1. 0 + 1Y

Q–

T

TY X

Q P

T T

T T+ ×

2. 0 + x1

Y1

Q–

Tx

T× Y X

1Q P

T Tx

T T+ × ×

3. + y1 + y1 + y1 + y1

4. + y1 x1 + y1

Y1 1

Q–

Ty x

T× Y X

1 1Q P

T Ty x

T T+ × ×

Since the speed of compound wheel Q-X is same as that of Q, therefore

Y1 1

Q

– – 300T

y x yT

× = =

or 1 124

– – 300120

y x × =

Arm A f ixed-wheel Yrotates through + 1revolution (anticlockwise)

Arm A fixed-wheel Y rotatesthrough + x1 revolutions

Add + y1 revolutions to allelements

Total motion

Chapter 13 : Gear Trains � 471∴ y1 = 0.2 x1 – 300 ...(iii)

Also Speed of wheel Y = Speed of wheel Z or shaft I

∴ x1 + y1 = x + y = – 1500 ...(iv)

x1 + 0.2 x1 – 300 = – 1500 ...[From equation (iii)]

1.2 x1= – 1500 + 300 = – 1200

or x1 = – 1200/1.2 = – 1000

and y1 = – 1500 – x1 = – 1500 + 1000 = – 500

1. Speed and direction of the driven shaft O and the wheel PSpeed of the driven shaft O,

NO = y1 = – 500 = 500 r.p.m. clockwise Ans.

and Speed of the wheel P, Y XP 1 1

Q P

24 36– 500 – 1000

120 144

T TN y x

T T= + × × = × ×

= – 550 = 550 r.p.m. clockwise Ans.

2. Torque tending to rotate the fixed wheel RWe know that the torque on shaft I or input torque

11

60 7500 6047.74 N-m

2 2 1500

PT

N

× ×= = =π × π ×

and torque on shaft O or output torque,

2O

60 0.8 7500 60114.58 N-m

2 2 500

PT

N

η × × × ×= = =π × π ×

Since the input and output shafts rotate in the same direction (i.e. clockwise), therefore inputand output torques will be in opposite direction.

∴ Torque tending to rotate the fixed wheel R= T2 – T1 = 114.58 – 47.74 = 66.84 N-m Ans.

Example 13.24. An epicyclic bevel gear train (known as Humpage’s reduction gear) is shownin Fig. 13.31. It consists of a fixed wheel C, thedriving shaft X and the driven shaft Y. The compoundwheel B-D can revolve on a spindle F which canturn freely about the axis X and Y.

Show that (i) if the ratio of tooth numbersTB / TD is greater than TC / TE , the wheel E will ro-tate in the same direction as wheel A, and (ii) if theratio TB / TD is less than TC / TE, the direction of E isreversed.

If the numbers of teeth on wheels A, B, C, Dand E are 34, 120, 150, 38 and 50 respectively and7.5 kW is put into the shaft X at 500 r.p.m., what isthe output torque of the shaft Y, and what are theforces (tangential to the pitch cones) at the contactpoints between wheels D and E and between wheels B and C, if the module of all wheels is 3.5 mm ?

Solution. Given : TA = 34 ; TB = 120 ; TC = 150 ; TD = 38 ; TE = 50 ; PX = 7.5 kW = 7500 W ;NX = 500 r.p.m. ; m = 3.5 mm

Fig. 13.31


Spindle fixed, wheel Ais rotated through + 1revolution

Spindle fixed, wheel Ais rotated through + xrevolutions

Add + y revolutions toall elements

Total motion




Step Conditions of motion Spindle Wheel A Compound Wheel C Wheel E (orNo. F (or shaft X) wheel B-D shaft Y)

1. 0 + 1A

B

T

T+ A B

B C

T T

T T− × A D

B E

T T

T T− ×

A

C

–T

T=

2. 0 + xA

B

Tx

T+ × A

C–

Tx

T× A D

B E–

T Tx

T T× ×

3. + y + y + y + y + y

4. + y x + yA

B

Ty x

T+ × A

C–

Ty x

T× A D

B E–

T Ty x

T T× ×

Let us assume that the driving shaft X rotates through 1 revolution anticlockwise, thereforethe wheel A will also rotate through 1 revolution anticlockwise.

∴ x + y = + 1 or y = 1 – x ...(i)We also know that the wheel C is fixed, therefore

A

C

– 0T

y xT

× = or A

C

(1 – ) – 0T

x xT

× = ...[From equation (i)]

A

C

1 – 1 0T

xT

+ =

or C A

C

1T T

xT

+=

and C

C A

Tx

T T=

+...(ii)

From equation (i),

C A

C A C A

1 – 1 –T T

y xT T T T

= = =+ +

...(iii)

We know that speed of wheel E,

CA D A A DE

B E C A C A B E

– –TT T T T T

N y xT T T T T T T T

= × × = × ×+ +

CA D

C A B E

1 –TT T

T T T T

= × +

...(iv)

and the speed of wheel A ,NA = x + y = + 1 revolution

(i) If CB

D E

TT

T T> or TB × TE > TC × TD , then the equation (iv) will be positive. Therefore the

wheel E will rotate in the same direction as wheel A . Ans.


(ii) If CB

D E

TT

T T< or TB × TE < TC × TD , then the equation (iv) will be negative. Therefore the

wheel E will rotate in the opposite direction as wheel A. Ans.

Output torque of shaft Y

We know that the speed of the driving shaft X (or wheel A ) or input speed is 500 r.p.m.,therefore from the fourth row of the table,

x + y = 500 or y = 500 – x ...(v)

Since the wheel C is fixed, therefore

A

C

– 0T

y xT

× = or 34(500 – ) – 0

150x x × = ...[From equation (v)]

∴ 500 – x – 0.227 x = 0 or x = 500/1.227 = 407.5 r.p.m.

and y = 500 – x = 500 – 407.5 = 92.5 r.p.m.

Since the speed of the driven or output shaft Y (i.e. NY) is equal to the speed of wheel E(i.e. NE), therefore

A D

Y EB E

34 38– 92.5 – 407.5

120 50

T TN N y x

T T= = × × = × ×

= 92.5 – 87.75 = 4.75 r.p.m.

Assuming 100 per cent efficiency of the gear train, input power PX is equal to output power(PY), i.e.

PY = PX = 7.5 kW = 7500 W

∴ Output torque of shaft Y ,

Y

Y

60 7500 6015 076 N-m 15.076 kN-m

2 2 4.75

P

N

× ×= = = =π π ×

Ans.

Tangential force between wheels D and E

We know that the pitch circle radius of wheel E,

EE

3.5 5087.5 mm 0.0875 m

2 2

m Tr

× ×= = = =

∴ Tangential force between wheels D and E,

Torque on wheel 15.076172.3 kN

Pitch circle radius of wheel 0.0875

E

E= = = Ans.

...(∴ Torque on wheel E = Torque on shaft Y )

Tangential force between wheels B and C

We know that the input torque on shaft X or on wheel A

X

X

60 7500 60143 N-m

2 2 500

P

N

× ×= = =π π ×

∴ Fixing torque on the fixed wheel C

= Torque on wheel E – Torque on wheel A

= 15 076 – 143 = 14 933 N-m = 14.933 kN-m


Pitch circle radius of wheel C,

CC

3.5 150262.5 mm 0.2625 m

2 2

m Tr

× ×= = = =

Tangential force between wheels B and C

C

Fixing torque on wheel 14.93357 kN

0.2625

C

r= = = Ans.

EXERCISES1. A compound train consists of six gears. The number of teeth on the gears are as follows :

Gear : A B C D E F

No. of teeth : 60 40 50 25 30 24

The gears B and C are on one shaft while the gears D and E are on another shaft. The gear A drives gearB, gear C drives gear D and gear E drives gear F. If the gear A transmits 1.5 kW at 100 r.p.m. and the geartrain has an efficiency of 80 per cent, find the torque on gear F. [Ans. 30.55 N-m]

2. Two parallel shafts are to be connected by spur gearing. The approximate distance between the shaftsis 600 mm. If one shaft runs at 120 r.p.m. and the other at 360 r.p.m., find the number of teeth on eachwheel, if the module is 8 mm. Also determine the exact distance apart of the shafts.

[Ans. 114, 38 ; 608 mm]3. In a reverted gear train, as shown in Fig. 13.32, two shafts A and B are

in the same straight line and are geared together through an interme-diate parallel shaft C. The gears connecting the shafts A and C have amodule of 2 mm and those connecting the shafts C and B have amodule of 4.5 mm. The speed of shaft A is to be about but greater than12 times the speed of shaft B, and the ratio at each reduction is same.Find suitable number of teeth for gears. The number of teeth of eachgear is to be a minimum but not less than 16. Also find the exactvelocity ratio and the distance of shaft C from A and B. [Ans. 36, 126, 16, 56 ; 12.25 ; 162 mm]

4. In an epicyclic gear train, as shown in Fig.13.33, the number of teethon wheels A , B and C are 48, 24 and 50 respectively. If the arm rotates at 400 r.p.m., clockwise,find : 1. Speed of wheel C when A is fixed, and 2. Speed of wheel A when C is fixed.

[Ans. 16 r.p.m. (clockwise) ; 16.67 (anticlockwise)]

Fig. 13.33 Fig. 13.34

Fig. 13.32

Chapter 13 : Gear Trains � 4755. In an epicyclic gear train, as shown in Fig. 13.34, the wheel C is keyed to the shaft B and wheel F is

keyed to shaft A . The wheels D and E rotate together on a pin fixed to the arm G. The number of teethon wheels C, D, E and F are 35, 65, 32 and 68 respectively.If the shaft A rotates at 60 r.p.m. and the shaft B rotates at 28 r.p.m. in the opposite direction, findthe speed and direction of rotation of arm G. [Ans. 90 r.p.m., in the same direction as shaft A]

6. An epicyclic gear train, as shown in Fig. 13.35, is composed of a fixed annular wheel A having 150teeth. The wheel A is meshing with wheel B which drives wheel D through an idle wheel C, D beingconcentric with A . The wheels B and C are carried on an arm which revolves clockwise at 100 r.p.m.about the axis of A and D. If the wheels B and D have 25 teeth and 40 teeth respectively, find thenumber of teeth on C and the speed and sense of rotation of C. [Ans. 30 ; 600 r.p.m. clockwise]

Fig. 13.35 Fig. 13.36

7. Fig. 13.36, shows an epicyclic gear train with the following details :

A has 40 teeth external (fixed gear) ; B has 80 teeth internal ; C - D is a compound wheel having 20 and50 teeth (external) respectively, E-F is a compound wheel having 20 and 40 teeth (external) respec-tively, and G has 90 teeth (external).

The arm runs at 100 r.p.m. in clockwise direction. Determine the speeds for gears C, E, and B.[Ans. 300 r.p.m. clockwise ; 400 r.p.m. anticlockwise ; 150 r.p.m. clockwise]

8. An epicyclic gear train, as shown in Fig. 13.37, has a sun wheel S of 30 teeth and two planet wheelsP-P of 50 teeth. The planet wheels mesh with the internal teeth of a fixed annulus A . The driving shaftcarrying the sunwheel, transmits 4 kW at 300 r.p.m. The driven shaft is connected to an arm whichcarries the planet wheels. Determine the speed of the driven shaft and the torque transmitted, if theoverall efficiency is 95%. [Ans. 56.3 r.p.m. ; 644.5 N-m]

Fig. 13.37 Fig. 13.38

9. An epicyclic reduction gear, as shown in Fig. 13.38, has a shaft A fixed to arm B. The arm B has a pinfixed to its outer end and two gears C and E which are rigidly fixed, revolve on this pin. Gear Cmeshes with annular wheel D and gear E with pinion F. G is the driver pulley and D is kept stationary.

The number of teeth are : D = 80 ; C = 10 ; E = 24 and F = 18.

If the pulley G runs at 200 r.p.m. ; find the speed of shaft A .

[Ans. 17.14 r.p.m. in the same direction as that of G]


10. A reverted epicyclic gear train for a hoist block is shown inFig. 13.39. The arm E is keyed to the same shaft as the loaddrum and the wheel A is keyed to a second shaft which car-ries a chain wheel, the chain being operated by hand. Thetwo shafts have common axis but can rotate independently.The wheels B and C are compound and rotate together on apin carried at the end of arm E. The wheel D has internalteeth and is fixed to the outer casing of the block so that itdoes not rotate.

The wheels A and B have 16 and 36 teeth respectively with amodule of 3 mm. The wheels C and D have a module of 4mm. Find : 1. the number of teeth on wheels C and D whenthe speed of A is ten times the speed of arm E, both rotatingin the same sense, and 2. the speed of wheel D when thewheel A is fixed and the arm E rotates at 450 r.p.m.anticlockwise.[Ans. TC = 13 ; TD = 52 ; 500 r.p.m. anticlockwise]

11. A compound epicyclic gear is shown diagrammatically in Fig. 13.40. The gears A , D and E are free torotate on the axis P. The compound gear B and C rotate together on the axis Q at the end of arm F. Allthe gears have equal pitch. The number of external teeth on the gears A , B and C are 18, 45 and 21respectively. The gears D and E are annular gears. The gear A rotates at 100 r.p.m. in the anticlockwisedirection and the gear D rotates at 450 r.p.m. clockwise. Find the speed and direction of the arm andthe gear E. [Ans. 400 r.p.m. clockwise ; 483.3 r.p.m. clockwise]

12. In an epicyclic gear train of the ‘sun and planet type’ as shown in Fig. 13.41, the pitch circle diameterof the internally toothed ring D is to be 216 mm and the module 4 mm. When the ring D is stationary,the spider A , which carries three planet wheels C of equal size, is to make one revolution in the samesense as the sun wheel B for every five revolutions of the driving spindle carrying the sunwheel B.Determine suitable number of teeth for all the wheels and the exact diameter of pitch circle of the ring.

[Ans. TB = 14 , TC = 21 , TD = 56 ; 224 mm]

Fig. 13.40 Fig. 13.41

13. An epicyclic train is shown in Fig. 13.42. Internal gear A is keyed to the driving shaft and has 30 teeth.Compound wheel C and D of 20 and 22 teeth respectively are free to rotate on the pin fixed to the armP which is rigidly connected to the driven shaft. Internal gear B which has 32 teeth is fixed. If thedriving shaft runs at 60 r.p.m. clockwise, determine the speed of the driven shaft. What is the directionof rotation of driven shaft with reference to driving shaft? [Ans. 1980 r.p.m. clockwise]

Fig. 13.39


Fig. 13.42 Fig. 13.4314. A shaft Y is driven by a co-axial shaft X by means of an epicyclic gear train, as shown in Fig. 13.43.

The wheel A is keyed to X and E to Y . The wheels B and D are compound and carried on an arm Fwhich can turn freely on the common axes of X and Y . The wheel C is fixed. If the numbers of teethon A , B, C, D and E are respectively 20, 64, 80, 30 and 50 and the shaft X makes 600 r.p.m.,determine the speed in r.p.m. and sense of rotation of the shaft Y .

[Ans. 30 r.p.m. in the same sense as shaft X]

15. An epicyclic bevel gear train, as shown in Fig. 13.44, has fixed gear B meshing with pinion C. Thegear E on the driven shaft meshes with the pinion D. The pinions C and D are keyed to a shaft,which revolves in bearings on the arm A . The arm A is keyed to the driving shaft. The number ofteeth are : TB = 75, TC = 20, TD = 18, and TE = 70. Find the speed of the driven shaft, if 1. the drivingshaft makes 1000 r.p.m., and 2. the gear B turns in the same sense as the driving shaft at 400r.p.m., the driving shaft still making 1000 r.p.m.

[Ans. 421.4 r.p.m. in the same direction as driving shaft]

16. The epicyclic gear train is shown in Fig. 13.45. The wheel D is held stationary by the shaft A and thearm B is rotated at 200 r.p.m. The wheels E (20 teeth) and F (40 teeth) are fixed together and rotatefreely on the pin carried by the arm. The wheel G (30 teeth) is rigidly attached to the shaft C. Find thespeed of shaft C stating the direction of rotation to that of B.

If the gearing transmits 7.5 kW, what will be the torque required to hold the shaft A stationary, neglect-ing all friction losses?

[Ans. 466.7 r.p.m. in opposite direction of B; 511.5 N-m in opposite direction of B]

Fig. 13.44 Fig. 13.45

17. An epicyclic gear train, as shown in Fig. 13.46, consists of two sunwheels A and D with 28 and 24teeth respectively, engaged with a compound planet wheels B and C with 22 and 26 teeth. The sunwheel


D is keyed to the driven shaft and the sunwheel A is a fixed wheel co-axial with the driven shaft. Theplanet wheels are carried on an arm E from the driving shaft which is co-axial with the driven shaft.

Find the velocity ratio of gear train. If 0.75 kW is transmitted and input speed being 100 r.p.m.,determine the torque required to hold the sunwheel A . [Ans. 2.64 ; 260.6 N-m]

Fig. 13.46 Fig. 13.47

18. In the epicyclic reduction gear, as shown in Fig. 13.47, the sunwheel D has 20 teeth and is keyedto the input shaft. Two planet wheels B , each having 50 teeth, gear with wheel D and are carriedby an arm A fixed to the output shaft. The wheels B also mesh with an internal gear C which isfixed. The input shaft rotates at 2100 r.p.m. Determine the speed of the output shaft and the torquerequired to fix C when the gears are transmitting 30 kW.

[Ans. 300 r.p.m. in the same sense as the input shaft ; 818.8 N-m]

19. An epicyclic gear train for an electric motor is shown in Fig. 13.48. The wheel S has 15 teeth and isfixed to the motor shaft rotating at 1450 r.p.m. The planet P has 45 teeth, gears with fixed annulus Aand rotates on a spindle carried by an arm which is fixed to the output shaft. The planet P also gearswith the sun wheel S. Find the speed of the output shaft. If the motor is transmitting 1.5 kW, find thetorque required to fix the annulus A . [Ans. 181.3 r.p.m. ; 69.14 N-m]

Fig. 13.48 Fig. 13.49

20. An epicyclic gear consists of bevel wheels as shown in Fig. 13.49. The driving pinion A has 20 teethand meshes with the wheel B which has 25 teeth. The wheels B and C are fixed together and turn freelyon the shaft F. The shaft F can rotate freely about the main axis X X. The wheel C has 50 teeth andmeshes with wheels D and E, each of which has 60 teeth. Find the speed and direction of E when Arotates at 200 r.p.m., if

1. D is fixed, and 2. D rotates at 100 r.p.m., in the same direction as A .

In both the cases, find the ratio of the torques transmitted by the shafts of the wheels A and E, thefriction being neglected.

[Ans. 800 r.p.m. in the opposite direction of A ; 300 r.p.m. in the oppositedirection of A ; 4 ; 1.5]


DO YOU KNOW ?1. What do you understand by ‘gear train’? Discuss the various types of gear trains.

2. Explain briefly the differences between simple, compound, and epicyclic gear trains. What are thespecial advantages of epicyclic gear trains ?

3. Explain the procedure adopted for designing the spur wheels.

4. How the velocity ratio of epicyclic gear train is obtained by tabular method?

5. Explain with a neat sketch the ‘sun and planet wheel’.

6. What are the various types of the torques in an epicyclic gear train ?

OBJECTIVE TYPE QUESTIONS1. In a simple gear train, if the number of idle gears is odd, then the motion of driven gear will

(a) be same as that of driving gear

(b) be opposite as that of driving gear

(c) depend upon the number of teeth on the driving gear

(d) none of the above

2. The train value of a gear train is

(a) equal to velocity ratio of a gear train (b) reciprocal of velocity ratio of a gear train

(c) always greater than unity (d) always less than unity

3. When the axes of first and last gear are co-axial, then gear train is known as

(a) simple gear train (b) compound gear train

(c) reverted gear train (d) epicyclic gear train

4. In a clock mechanism, the gear train used to connect minute hand to hour hand, is

(a) epicyclic gear train (b) reverted gear train

(c) compound gear train (d) simple gear train

5. In a gear train, when the axes of the shafts, over which the gears are mounted, move relative to a fixedaxis, is called

(a) simple gear train (b) compound gear train

(c) reverted gear train (d) epicyclic gear train

6. A differential gear in an automobile is a

(a) simple gear train (b) epicyclic gear train

(c) compound gear train (d) none of these

7. A differential gear in automobilies is used to

(a) reduce speed (b) assist in changing speed

(c) provide jerk-free movement of vehicle (d) help in turning

ANSWERS1. (a) 2. (b) 3. (c) 4. (b) 5. (d)

6. (b) 7. (d)


CamsCamsCamsCamsCams

20FFFFFeaeaeaeaeaturturturturtureseseseses

1. Introduction.2. Classification of Followers.3. Classification of Cams.4. Terms used in Radial cams.5. Motion of the Follower.6. Displacement, Velocity and

Acceleration Diagramswhen the Follower Moveswith Uniform Velocity.

7. Displacement, Velocity andAcceleration Diagramswhen the Follower Moveswith Simple HarmonicMotion.

8. Displacement, Velocity andAcceleration Diagramswhen the Follower Moveswith Uniform Accelerationand Retardation.

9. Displacement, Velocity andAcceleration Diagramswhen the Follower Moveswith Cycloidal Motion.

10. Construction of CamProfiles.

11. Cams with SpecifiedContours.

12. Tangent Cam withReciprocating RollerFollower.

13. Circular Arc Cam with Flat-faced Follower.

20.1.20.1.20.1.20.1.20.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoductionA cam is a rotating machine element which gives

reciprocating or oscillating motion to another element knownas follower. The cam and the follower have a line contactand constitute a higher pair. The cams are usually rotated atuniform speed by a shaft, but the follower motion is pre-determined and will be according to the shape of the cam.The cam and follower is one of the simplest as well as oneof the most important mechanisms found in modernmachinery today. The cams are widely used for operatingthe inlet and exhaust valves of internal combustion engines,automatic attachment of machineries, paper cutting machines,spinning and weaving textile machineries, feed mechanismof automatic lathes etc.

20.2.20.2.20.2.20.2.20.2. Classification of FollowersClassification of FollowersClassification of FollowersClassification of FollowersClassification of FollowersThe followers may be classified as discussed below :

1. According to the surface in contact. The followers,according to the surface in contact, are as follows :

(a) Knife edge follower. When the contacting end ofthe follower has a sharp knife edge, it is called aknife edge follower, as shown in Fig. 20.1 (a). Thesliding motion takes place between the contactingsurfaces (i.e. the knife edge and the cam surface). Itis seldom used in practice because the small area ofcontacting surface results in excessive wear. In knifeedge followers, a considerable side thrust existsbetween the follower and the guide.774

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Chapter 20 : Cams � 775(b) Roller follower. When the contacting end of the follower is a roller, it is called a roller

follower, as shown in Fig. 20.1 (b). Since the rolling motion takes place between thecontacting surfaces (i.e. the roller and the cam), therefore the rate of wear is greatly reduced.In roller followers also the side thrust exists between the follower and the guide. Theroller followers are extensively used where more space is available such as in stationarygas and oil engines and aircraft engines.

(c) Flat faced or mushroom follower. When the contacting end of the follower is a perfectlyflat face, it is called a flat-faced follower, as shown in Fig. 20.1 (c). It may be noted thatthe side thrust between the follower and the guide is much reduced in case of flat facedfollowers. The only side thrust is due to friction between the contact surfaces of the followerand the cam. The relative motion between these surfaces is largely of sliding nature butwear may be reduced by off-setting the axis of the follower, as shown in Fig. 20.1 (f ) sothat when the cam rotates, the follower also rotates about its own axis. The flat facedfollowers are generally used where space is limited such as in cams which operate thevalves of automobile engines.

Note : When the flat faced follower is circular, it is then called a mushroom follower.

(d) Spherical faced follower. When the contacting end of the follower is of spherical shape,it is called a spherical faced follower, as shown in Fig. 20.1 (d). It may be noted that whena flat-faced follower is used in automobile engines, high surface stresses are produced. Inorder to minimise these stresses, the flat end of the follower is machined to a sphericalshape.

(a) Cam with knife (b) Cam with roller (c) Cam with flat edge follower. follower. faced follower.

(d) Cam with spherical (e) Cam with spherical (f) Cam with offset

faced follower. faced follower. follower.

Fig. 20.1. Classification of followers.

2. According to the motion of the follower. The followers, according to its motion, are of thefollowing two types:

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(a) Reciprocating or translating follower. When the follower reciprocates in guides as thecam rotates uniformly, it is known as reciprocating or translating follower. The followersas shown in Fig. 20.1 (a) to (d) are all reciprocating or translating followers.

(b) Oscillating or rotating follower. When the uniform rotary motion of the cam is convertedinto predetermined oscillatory motion of the follower, it is called oscillating or rotatingfollower. The follower, as shown in Fig 20.1 (e), is an oscillating or rotating follower.

3. According to the path of motion of the follower. The followers, according to its path ofmotion, are of the following two types:

(a) Radial follower. When the motion of the follower is along an axis passing through thecentre of the cam, it is known as radial follower. The followers, as shown in Fig. 20.1 (a)to (e), are all radial followers.

(b) Off-set follower. When the motion of the follower is along an axis away from the axis ofthe cam centre, it is called off-set follower. The follower, as shown in Fig. 20.1 ( f ), is anoff-set follower.

Note : In all cases, the follower must be constrained to follow the cam. This may be done by springs, gravityor hydraulic means. In some types of cams, the follower may ride in a groove.

20.3.20.3.20.3.20.3.20.3. Classification of CamsClassification of CamsClassification of CamsClassification of CamsClassification of CamsThough the cams may be classified in many ways, yet the following two types are important

from the subject point of view :

(a) Cylindrical cam with reciprocating (b) Cylindrical cam with oscillating follower.

follower.

Fig. 20.2. Cylindrical cam.

1. Radial or disc cam. In radialcams, the follower reciprocates oroscillates in a direction perpendicular tothe cam axis. The cams as shown in Fig.20.1 are all radial cams.

2. Cylindrical cam. In cylindricalcams, the follower reciprocates oroscillates in a direction parallel to the camaxis. The follower rides in a groove at itscylindrical surface. A cylindrical groovedcam with a reciprocating and an oscillatingfollower is shown in Fig. 20.2 (a) and (b)respectively.Note : In actual practice, radial cams are widelyused. Therefore our discussion will be onlyconfined to radial cams.

In IC engines, cams are widely used tooperate valves.

Chapter 20 : Cams � 777

20.4.20.4.20.4.20.4.20.4. TTTTTerererererms Used in Radial Camsms Used in Radial Camsms Used in Radial Camsms Used in Radial Camsms Used in Radial CamsFig. 20.3 shows a radial cam with reciprocating roller follower. The following terms are

important in order to draw the cam profile.

1. Base circle. It is the smallest circle that can be drawn to the cam profile.

2. Trace point. It is a reference point on the follower and is used to generate the pitch curve.In case of knife edge follower, the knife edge represents the trace point and the pitch curvecorresponds to the cam profile. In a roller follower, the centre of the roller represents the trace point.

3. Pressure angle. It is the angle between the direction of the follower motion and a normalto the pitch curve. This angle is very important in designing a cam profile. If the pressure angle istoo large, a reciprocating follower will jam in its bearings.

4. Pitch point. It is a point on the pitch curve having the maximum pressure angle.

5. Pitch circle. It is a circle drawn from the centre of the cam through the pitch points.

6. Pitch curve. It is the curve generated by the trace point as the follower moves relative tothe cam. For a knife edge follower, the pitch curve and the cam profile are same whereas for aroller follower, they are separated by the radius of the roller.

7. Prime circle. It is the smallest circle that can be drawn from the centre of the cam andtangent to the pitch curve. For a knife edge and a flat face follower, the prime circle and the basecircle are identical. For a roller follower, the prime circle is larger than the base circle by the radiusof the roller.

8. Lift or stroke. It is the maximum travel of the follower from its lowest position to thetopmost position.

Fig. 20.3. Terms used in radial cams.

20.5.20.5.20.5.20.5.20.5. Motion of the FollowerMotion of the FollowerMotion of the FollowerMotion of the FollowerMotion of the FollowerThe follower, during its travel, may have one of the following motions.

1. Uniform velocity, 2. Simple harmonic motion, 3. Uniform acceleration and retardation,and 4. Cycloidal motion.


We shall now discuss the displacement, velocity and acceleration diagrams for the cam whenthe follower moves with the above mentioned motions.

20.6.20.6.20.6.20.6.20.6. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theFollower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform Velocity

The displacement, velocity and acceleration diagrams when a knife-edged follower moveswith uniform velocity are shown in Fig. 20.4 (a), (b) and (c) respectively. The abscissa (base)represents the time (i.e. the number of seconds required for the cam to complete one revolution) orit may represent the angular displacement of the cam in degrees. The ordinate represents the dis-placement, or velocity or acceleration of the follower.

Since the follower moves with uniform velocity during its rise and return stroke, thereforethe slope of the displacement curves must be constant. In other words, AB1 and C1D must bestraight lines. A little consideration will show that the follower remains at rest during part of thecam rotation. The periods during which the follower remains at rest are known as dwell periods, asshown by lines B1C1 and DE in Fig. 20.4 (a). From Fig. 20.4 (c), we see that the acceleration orretardation of the follower at the beginning and at the end of each stroke is infinite. This is due tothe fact that the follower is required to start from rest and has to gain a velocity within no time.This is only possible if the acceleration or retardation at the beginning and at the end of each strokeis infinite. These conditions are however, impracticable.

Fig. 20.4. Displacement, velocity and Fig. 20.5. Modified displacement, velocity and acceleration diagrams when the acceleration diagrams when the follower follower moves with uniform velocity. moves with uniform velocity.

In order to have the acceleration and retardation withinthe finite limits, it is necessary to modify the conditions whichgovern the motion of the follower. This may be done byrounding off the sharp corners of the displacement diagramat the beginning and at the end of each stroke, as shown inFig. 20.5 (a). By doing so, the velocity of the followerincreases gradually to its maximum value at the beginningof each stroke and decreases gradually to zero at the end ofeach stroke as shown in Fig. 20.5 (b). The modified

Camshaft of an IC engine.

Chapter 20 : Cams � 779displacement, velocity and acceleration diagrams are shown in Fig. 20.5. The round corners of thedisplacement diagram are usually parabolic curves because the parabolic motion results in a verylow acceleration of the follower for a given stroke and cam speed.

20.7.20.7.20.7.20.7.20.7. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic Motion

The displacement, velocity and acceleration diagrams when the follower moves with simpleharmonic motion are shown in Fig. 20.6 (a), (b) and (c) respectively. The displacement diagram isdrawn as follows :

1. Draw a semi-circle on the follower stroke as diameter.2. Divide the semi-circle into any number of even equal parts (say eight).3. Divide the angular displacements of the cam during out stroke and return stroke into the

same number of equal parts.4. The displacement diagram is obtained by projecting the points as shown in Fig. 20.6 (a).The velocity and acceleration diagrams are shown in Fig. 20.6 (b) and (c) respectively. Since

the follower moves with a simple harmonic motion, therefore velocity diagram consists of a sinecurve and the acceleration diagram is a cosine curve. We see from Fig. 20.6 (b) that the velocity ofthe follower is zero at the beginning and at the end of its stroke and increases gradually to amaximum at mid-stroke. On the other hand, the acceleration of the follower is maximum at thebeginning and at the ends of the stroke and diminishes to zero at mid-stroke.

Fig. 20.6. Displacement, velocity and acceleration diagrams when the followermoves with simple harmonic motion.

Let S = Stroke of the follower,

Oθ and Rθ = Angular displacement of the cam during out stroke and return stroke of the

follower respectively, in radians, andω = Angular velocity of the cam in rad/s.


∴ Time required for the out stroke of the follower in seconds,

O O /t = θ ω

Consider a point P moving at a uniform speed Pω radians per sec round the circumference

of a circle with the stroke S as diameter, as shown in Fig. 20.7.The point ′P (which is the projection of a point P on the diam-eter) executes a simple harmonic motion as the point P rotates.The motion of the follower is similar to that of point P′.

∴ Peripheral speed of the point P ′ ,

PO O

1

2 2

S Sv

t

π π ω= × = ×θ

and maximum velocity of the follower on the outstroke,

POO O

.

2 2

S Sv v

π ω πω= = × =θ θ

We know that the centripetal acceleration of the point P,

22 2 2P

P 2O O

( ) . 2 .

2 2( )

v S Sa

OP S

πω π ω= = × = θ θ ∴ Maximum acceleration of the follower on the outstroke,

2 2

O P 2O

.

2( )

Sa a

π ω= =θ

Similarly, maximum velocity of the follower on the return stroke,

RR

.

2

Sv

πω=θ

and maximum acceleration of the follower on the return stroke,

2 2

R 2R

.

2 ( )

Sa

π ω=

θ

20.8.20.8.20.8.20.8.20.8. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and Retardation

The displacement, velocity and acceleration diagrams when the follower moves with uniformacceleration and retardation are shown in Fig. 20.8 (a), (b) and (c) respectively. We see that thedisplacement diagram consists of a parabolic curve and may be drawn as discussed below :

1. Divide the angular displacement of the cam during outstroke ( Oθ ) into any even number

of equal parts (say eight) and draw vertical lines through these points as shown in Fig.20.8 (a).

2. Divide the stroke of the follower (S) into the same number of equal even parts.

3. Join Aa to intersect the vertical line through point 1 at B. Similarly, obtain the other pointsC, D etc. as shown in Fig. 20.8 (a). Now join these points to obtain the parabolic curvefor the out stroke of the follower.

4. In the similar way as discussed above, the displacement diagram for the follower duringreturn stroke may be drawn.

Since the acceleration and retardation are uniform, therefore the velocity varies directly withthe time. The velocity diagram is shown in Fig. 20.8 (b).

Let S = Stroke of the follower,

Fig. 20.7. Motion of a point.


Oθ and Rθ = Angular displacement of the cam during out stroke and return stroke of the follower respectively, and

ω = Angular velocity of the cam.We know that time required for the follower during outstroke,

O O /t = θ ωand time required for the follower during return stroke,

R R /t = θ ωMean velocity of the follower during outstroke

= S/tOand mean velocity of the follower during return stroke

= S/tR

Fig. 20.8. Displacement, velocity and acceleration diagrams when the follower moveswith uniform acceleration and retardation.

Since the maximum velocity of follower is equal to twice the mean velocity, therefore maxi-mum velocity of the follower during outstroke,

OO O

2 2 .S Sv

t

ω= =θ

Similarly, maximum velocity of the follower during return stroke,

RR

2 .Sv

ω=θ


We see from the acceleration diagram, as shown in Fig. 20.8 (c), that during first half of theoutstroke there is uniform acceleration and during the second half of the out stroke there is uniformretardation. Thus, the maximum velocity of the follower is reached after the time tO / 2 (during outstroke) and tR /2 (during return stroke).

∴ Maximum acceleration of the follower during outstroke,

2O

O 2O O O O

2 2 . 4 .

/ 2 . ( )

v S Sa

t t

× ω ω= = =

θ θ . . . (∵ O O /t = θ ω)

Similarly, maximum acceleration of the follower during return stroke,

2

R 2R

4 .

( )

Sa

ω=

θ

20.9.20.9.20.9.20.9.20.9. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theFollower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal Motion

Fig. 20.9. Displacement, velocity and acceleration diagrams when thefollower moves with cycloidal motion.

The displacement, velocity and acceleration diagrams when the follower moves with cycloidalmotion are shown in Fig. 20.9 (a), (b) and (c) respectively. We know that cycloid is a curve tracedby a point on a circle when the circle rolls without slipping on a straight line.

In case of cams, this straight line is a stroke of the follower which is translating and thecircumference of the rolling circle is equal to the stroke (S) of the follower. Therefore the radius of


the rolling circle is / 2S π . The displacement diagram is drawn as discussed below :

1. Draw a circle of radius / 2S π with A as centre.

2. Divide the circle into anynumber of equal even parts(say six). Project these pointshorizontally on the verticalcentre line of the circle.

These points are shown by a′and b′ in Fig. 20.9 (a).

3. Divide the angular displace-ment of the cam during out-stroke into the same numberof equal even parts as thecircle is divided. Draw verti-cal lines through these points.

4. Join AB which intersects thevertical line through 3′ at c.

From a′ draw a line parallelto AB intersecting the verti-cal lines through 1′ and 2′at a and b respectively.

5. Similarly, from b′ draw aline parallel to AB intersect-ing the vertical lines through

4′ and 5′ at d and e respec-tively.

6. Join the points A a b c d e Bby a smooth curve. This is the required cycloidal curve for the follower during outstroke.

Let θ = Angle through which the cam rotates in time t seconds, and

ω = Angular velocity of the cam.We know that displacement of the follower after time t seconds,

O O

1 2sin

2x S

θ πθ= − θ π θ . . . (i)

∴ Velocity of the follower after time t seconds,

O O O

1 2 2cos

2

dx d dS

dt dt dt

θ π πθ θ= × − θ πθ θ . . . [Differentiating equation (i)]

O O O O

2 . 21 cos 1 cos

S d S

dt

θ πθ ω πθ= × − = − θ θ θ θ . . . (ii)

Cams are used in Jet and aircraft engines. Theabove picture shows an aircraft engine.


The velocity is maximum, when

O

2cos 1

πθ = − θ or

O

2πθ= πθ or O / 2θ = θ

Substituting O / 2θ = θ in equation (ii), we have maximum velocity of the follower duringoutstroke,

OO O

. 2 .(1 1)

S Sv

ω ω= + =θ θ

Similarly, maximum velocity of the follower during return stroke,

RR

2 .Sv

ω=θ

Now, acceleration of the follower after time t sec,

2

2O O O

. 2 2sin

d x S d

dtdt

ω π πθ θ= θ θ θ . . . [Differentiating equation (ii)]

2

2OO

2 . 2sin

( )

S πω πθ= θθ . . .

d

dt

θ = ω ∵ . . . (iii)

The acceleration is maximum, when

O

2sin 1

πθ = θ or

O

2

2

πθ π=θ

or O / 4θ = θ

Substituting O / 4θ = θ in equation (iii), we have maximum acceleration of the follower dur-

ing outstroke, 2

O 2O

2 .

( )

Sa

πω=

θ

Similarly, maximum acceleration of the follower during return stroke,

2

R 2R

2 .

( )

Sa

πω=

θ

The velocity and acceleration diagrams are shown in Fig. 20.9 (b) and (c) respectively.

20.10.20.10.20.10.20.10.20.10. Construction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamIn order to draw the cam profile for a radial cam, first of all the displacement diagram for the

given motion of the follower is drawn. Then by constructing the follower in its proper position ateach angular position, the profile of the working surface of the cam is drawn.

In constructing the cam profile, the principle of kinematic inversion is used, i.e. the cam isimagined to be stationary and the follower is allowed to rotate in the opposite direction to the camrotation.

The construction of cam profiles for different types of follower with different types ofmotions are discussed in the following examples.

Example 20.1. A cam is to give the following motion to a knife-edged follower :1. Outstroke during 60° of cam rotation ; 2. Dwell for the next 30° of cam rotation ;

3. Return stroke during next 60° of cam rotation, and 4. Dwell for the remaining 210° of camrotation.

The stroke of the follower is 40 mm and the minimum radius of the cam is 50 mm. Thefollower moves with uniform velocity during both the outstroke and return strokes. Draw the pro-file of the cam when (a) the axis of the follower passes through the axis of the cam shaft, and(b) the axis of the follower is offset by 20 mm from the axis of the cam shaft.

Chapter 20 : Cams � 785Construction

Fig. 20.10

First of all, the displacement diagram, as shown in Fig. 20.10, is drawn as discussed in thefollowing steps :

1. Draw a horizontal line AX = 360° to some suitable scale. On this line, mark AS = 60° torepresent outstroke of the follower, ST = 30° to represent dwell, TP = 60° to representreturn stroke and PX = 210° to represent dwell.

2. Draw vertical line AY equal to the stroke of the follower (i.e. 40 mm) and complete therectangle as shown in Fig. 20.10.

3. Divide the angular displacement during outstroke and return stroke into any equal numberof even parts (say six) and draw vertical lines through each point.

4. Since the follower moves with uniform velocity during outstroke and return stroke, there-fore the displacement diagram consists of straight lines. Join AG and HP.

5. The complete displacement diagram is shown by AGHPX in Fig. 20.10.(a) Profile of the cam when the axis of follower passes through the axis of cam shaft

The profile of the cam when the axis of the follower passes through the axis of the cam shaft,as shown in Fig. 20.11, is drawn as discussed in the following steps :

Fig. 20.11


1. Draw a base circle with radius equal to the minimum radius of the cam (i.e. 50 mm) withO as centre.

2. Since the axis of the follower passes through the axis of the cam shaft, therefore marktrace point A, as shown in Fig. 20.11.

3. From OA, mark angle AOS = 60° to represent outstroke, angle SOT = 30° to representdwell and angle TOP = 60° to represent return stroke.

4. Divide the angular displacements during outstroke and return stroke (i.e. angle AOS andangle TOP) into the same number of equal even parts as in displacement diagram.

5. Join the points 1, 2, 3 ...etc. and 0′ ,1′ , 2′ , 3′ , ... etc. with centre O and produce beyondthe base circle as shown in Fig. 20.11.

6. Now set off 1B, 2C, 3D ... etc. and 0′ H,1′ J ... etc. from the displacement diagram.7. Join the points A, B, C,... M, N, P with a smooth curve. The curve AGHPA is the complete

profile of the cam.Notes : The points B, C, D .... L, M, N may also be obtained as follows :

1. Mark AY = 40 mm on the axis of the follower, and set of Ab, Ac, Ad... etc. equal to the distances 1B,2C, 3D... etc. as in displacement diagram.

2. From the centre of the cam O, draw arcs with radii Ob, Oc, Od etc. The arcs intersect the producedlines O1, O2... etc. at B, C, D ... L, M, N.

(b) Profile of the cam when the axis of the follower is offset by 20 mm from the axis of the camshaftThe profile of the cam when the axis of the follower is offset from the axis of the cam shaft,

as shown in Fig. 20.12, is drawn as discussed in the following steps :

Fig. 20.12

1. Draw a base circle with radius equal to the minimum radius of the cam (i.e. 50 mm) withO as centre.

2. Draw the axis of the follower at a distance of 20 mm from the axis of the cam, whichintersects the base circle at A.

3. Join AO and draw an offset circle of radius 20 mm with centre O.4. From OA, mark angle AOS = 60° to represent outstroke, angle SOT = 30° to represent

dwell and angle TOP = 60° to represent return stroke.

Chapter 20 : Cams � 7875. Divide the angular displacement during outstroke and return stroke (i.e. angle AOS and

angle TOP) into the same number of equal even parts as in displacement diagram.

6. Now from the points 1, 2, 3 ... etc. and 0 ,1 , 2 ,3′ ′ ′ ′ ... etc. on the base circle, draw tangents

to the offset circle and produce these tangents beyond the base circle as shown in Fig.20.12.

7. Now set off 1B, 2C, 3D ... etc. and 0′ H,1′ J ... etc. from the displacement diagram.8. Join the points A, B, C ...M, N, P with a smooth curve. The curve AGHPA is the complete

profile of the cam.Example 20.2. A cam is to be designed for a knife edge follower with the following data :1. Cam lift = 40 mm during 90° of cam rotation with simple harmonic motion.2. Dwell for the next 30°.3. During the next 60° of cam rotation, the follower returns to its original position with

simple harmonic motion.4. Dwell during the remaining 180°.Draw the profile of the cam when(a) the line of stroke of the follower passes through the axis of the cam shaft, and(b) the line of stroke is offset 20 mm from the axis of the cam shaft.The radius of the base circle of the cam is 40 mm. Determine the maximum velocity and

acceleration of the follower during its ascent and descent, if the cam rotates at 240 r.p.m.

Solution. Given : S = 40 mm = 0.04 m; Oθ = 90° = π/2 rad = 1.571 rad ; Rθ = 60° =

π/3 rad = 1.047 rad ; N = 240 r.p.m.

Fig. 20.13First of all, the displacement diagram, as shown in Fig 20.13, is drawn as discussed in the

following steps :

1. Draw horizontal line AX = 360° to some suitable scale. On this line, mark AS = 90° torepresent out stroke ; SR = 30° to represent dwell ; RP = 60° to represent return strokeand PX = 180° to represent dwell.

2. Draw vertical line AY = 40 mm to represent the cam lift or stroke of the follower andcomplete the rectangle as shown in Fig. 20.13.

3. Divide the angular displacement during out stroke and return stroke into any equal num-ber of even parts (say six) and draw vertical lines through each point.

4. Since the follower moves with simple harmonic motion, therefore draw a semicircle withAY as diameter and divide into six equal parts.

5. From points a, b, c ... etc. draw horizontal lines intersecting the vertical lines drawn through1, 2, 3 ... etc. and 0′ ,1′ , 2′ ...etc. at B, C, D ... M, N, P.

6. Join the points A, B, C ... etc. with a smooth curve as shown in Fig. 20.13. This is therequired displacement diagram.


(a) Profile of the cam when the line of stroke of the follower passes through the axis of the camshaft

The profile of the cam when the line of stroke of the follower passes through the axis of thecam shaft, as shown in Fig. 20.14, is drawn in the similar way as is discussed in Example 20.1.

Fig. 20.14

(b) Profile of the cam when the line of stroke of the follower is offset 20 mm from the axisof the cam shaft

The profile of the cam when the line of stroke of the follower is offset 20 mm from the axisof the cam shaft, as shown in Fig. 20.15, is drawn in the similar way as discussed in Example 20.1.

Fig. 20.15


Maximum velocity of the follower during its ascent and descent

We know that angular velocity of the cam,

2 2 24025.14

60 60

π π×ω = = =N rad/s

We also know that the maximum velocity of thefollower during its ascent,

OO

. 25.14 0.04

2 2 1.571

Sv

πω π× ×= =θ ×

= 1 m/s Ans.

and maximum velocity of the follower during itsdescent,

RR

. 25.14 0.04

2 2 1.047

Sv

πω π× ×= =θ ×

= 1.51 m/s Ans.

Maximum acceleration of the follower during itsascent and descent

We know that the maximum acceleration of thefollower during its ascent,

2 2 2 2

O 2 2O

. (25.14) 0.04

2( ) 2 (1.571)

Sa

π ω π= =

θ = 50.6 m/s2 Ans.

and maximum acceleration of the follower during its descent,

2 2 2 2

R 2 2R

. (25.14) 0.04

2( ) 2(1.047)

Sa

π ω π= =θ

= 113.8 m/s2 Ans.

Example 20.3. A cam, with a minimum radius of 25 mm, rotating clockwise at a uniform speedis to be designed to give a roller follower, at the end of a valve rod, motion described below :

1. To raise the valve through 50 mm during 120° rotation of the cam ;

2. To keep the valve fully raised through next 30°;

3. To lower the valve during next 60°; and

4. To keep the valve closed during rest of the revolution i.e. 150° ;

The diameter of the roller is 20 mm and the diameter of the cam shaft is 25 mm.

Draw the profile of the cam when (a) the line of stroke of the valve rod passes through theaxis of the cam shaft, and (b) the line of the stroke is offset 15 mm from the axis of the cam shaft.

The displacement of the valve, while being raised and lowered, is to take place with simpleharmonic motion. Determine the maximum acceleration of the valve rod when the cam shaft rotatesat 100 r.p.m.

Draw the displacement, the velocity and the acceleration diagrams for one complete revolu-tion of the cam.

Solution. Given : S = 50 mm = 0.05 m ; Oθ = 120° = 2 π/3 rad = 2.1 rad ; Rθ = 60° =

π/3 rad = 1.047 rad ; N = 100 r.p.m.

Since the valve is being raised and lowered with simple harmonic motion, therefore the dis-placement diagram, as shown in Fig. 20.16 (a), is drawn in the similar manner as discussed in theprevious example.

Role of cams in piston movement.


(((((a) Profile of the cam when the line of stroke of the valve rod passes through the axis ofthe cam shaft

The profile of the cam, as shown in Fig. 20.17, is drawn as discussed in the following steps :

1. Draw a base circle with centre O and radius equal to the minimum radius of the cam( i.e. 25 mm ).

Fig. 20.16

2. Draw a prime circle with centre O and radius,

−OA = Min. radius of cam + 1

2 Dia. of roller =

125 20 35

2+ × = mm

3. Draw angle AOS = 120° to represent raising or out stroke of the valve, angle SOT = 30° torepresent dwell and angle TOP = 60° to represent lowering or return stroke of the valve.

4. Divide the angular displacements of the cam during raising and lowering of the valve (i.e.angle AOS and TOP ) into the same number of equal even parts as in displacement diagram.

5. Join the points 1, 2, 3, etc. with the centre O and produce the lines beyond prime circle asshown in Fig. 20.17.

6. Set off 1B, 2C, 3D etc. equal to the displacements from displacement diagram.7. Join the points A, B, C ... N, P, A. The curve drawn through these points is known as pitch

curve.


8. From the points A, B, C ... N, P, draw circles of radius equal to the radius of the roller.9. Join the bottoms of the circles with a smooth curve as shown in Fig. 20.17. This is the

required profile of the cam.

Fig. 20.17

Fig. 20.18


(b) Profile of the cam when the line of stroke is offset 15 mm from the axis of the cam shaft

The profile of the cam when the line of stroke is offset from the axis of the cam shaft, asshown in Fig. 20.18, may be drawn as discussed in the following steps :

1. Draw a base circle with centre O and radius equal to 25 mm.2. Draw a prime circle with centre O and radius OA = 35 mm.3. Draw an off-set circle with centre O and radius equal to 15 mm.4. Join OA. From OA draw the angular displacements of cam i.e. draw angle AOS = 120°,

angle SOT = 30° and angle TOP = 60°.5. Divide the angular displacements of the cam during raising and lowering of the valve into

the same number of equal even parts (i.e. six parts ) as in displacement diagram.

6. From points 1, 2, 3 .... etc. and 0′ ,1′ , 3′ , ...etc. on the prime circle, draw tangents to theoffset circle.

7. Set off 1B, 2C, 3D... etc. equal to displacements as measured from displacement diagram.8. By joining the points A, B, C ... M, N, P, with a smooth curve, we get a pitch curve.9. Now A, B, C...etc. as centre, draw circles with radius equal to the radius of roller.

10. Join the bottoms of the circles with a smooth curve as shown in Fig. 20.18. This is therequired profile of the cam.

Maximum acceleration of the valve rod

We know that angular velocity of the cam shaft,

2 2 10010.47

60 60


We also know that maximum velocity of the valve rod to raise valve,

OO

. 10.47 0.050.39

2 2 2.1

Sv

πω π× ×= = =θ ×

m/s

and maximum velocity of the valve rod to lower the valve,

RR

. 10.47 0.050.785

2 2 1.047

Sv

πω π× ×= = =θ ×

m/s

The velocity diagram for one complete revolution of the cam is shown in Fig. 20.16 (b).We know that the maximum acceleration of the valve rod to raise the valve,

2 2 2 2

O 2 20

. (10.47) 0.056.13

2( ) 2(2.1)

Sa

π ω π= = =θ

m/s2 Ans.

and maximum acceleration of the valve rod to lower the valve,

2 2 2 2

R 2 2R

. (10.47) 0.0524.67

2( ) 2(1.047)

Sa

π ω π= = =

θ m/s2 Ans.

The acceleration diagram for one complete revolution of the cam is shown in Fig. 20.16 (c).

Example 20.4. A cam drives a flat reciprocating follower in the following manner :

During first 120° rotation of the cam, follower moves outwards through a distance of 20 mmwith simple harmonic motion. The follower dwells during next 30° of cam rotation. During next120° of cam rotation, the follower moves inwards with simple harmonic motion. The followerdwells for the next 90° of cam rotation.

The minimum radius of the cam is 25 mm. Draw the profile of the cam.

Chapter 20 : Cams � 793ConstructionSince the follower moves outwards and inwards with simple harmonic motion, therefore the

displacement diagram, as shown in Fig. 20.19, is drawn in the similar manner as discussed earlier.

Fig. 20.19

Now the profile of the cam driving a flat reciprocating follower, as shown in Fig. 20.20, isdrawn as discussed in the following steps :

1. Draw a base circle with centre O and radius OA equal to the minimum radius of the cam(i.e. 25 mm).

2. Draw angle AOS = 120° to represent the outward stroke, angle SOT = 30° to representdwell and angle TOP = 120° to represent inward stroke.

3. Divide the angular displacement during outward stroke and inward stroke (i.e. angles AOSand TOP ) into the same number of equal even parts as in the displacement diagram.

Fig. 20.20


4. Join the points 1, 2, 3 . . . etc. with centre O and produce beyond the base circle.

5. From points 1, 2, 3 . . . etc., set off 1B, 2C, 3D . . . etc. equal to the distances measuredfrom the displacement diagram.

6. Now at points B, C, D . . . M, N, P, draw the position of the flat-faced follower. The axisof the follower at all these positions passes through the cam centre.

7. The curve drawn tangentially to the flat side of the follower is the required profile of thecam, as shown in Fig. 20.20.

Example 20.5. Draw a cam profile to drive an oscillating roller follower to the specifica-tions given below :

(a) Follower to move outwards through an angular displacement of 20° during the first 120°rotation of the cam ;

(b) Follower to return to its initial position during next 120° rotation of the cam ;

(c) Follower to dwell during the next 120° of cam rotation.

The distance between pivot centre and roller centre = 120 mm ; distance between pivotcentre and cam axis = 130 mm ; minimum radius of cam = 40 mm ; radius of roller = 10 mm ;inward and outward strokes take place with simple harmonic motion.

ConstructionWe know that the angular displacement

of the roller follower

20 20 /180 / 9= ° = × π = π rad

Since the distance between the pivotcentre and the roller centre (i.e. the radiusA1 A) is 120 mm, therefore length of the arcAA2, as shown in Fig. 20.21, along which thedisplacement of the roller actually takes place

120 / 9 41.88= × π = mm

. . . (∵ Length of arc = Radius of arc × Angle subtended by the arc at the centre in radians)

Since the angle is very small, therefore length of chord AA2 is taken equal to the length of arcAA2. Thus in order to draw the displacement diagram, we shall take lift of the follower equal tolength of chord AA2 i.e. 41.88 mm.

Fig. 20.22

The outward and inward strokes take place with simple harmonic motion, therefore the dis-placement diagram, as shown in Fig. 20.22, is drawn in the similar way as discussed in Example20.4.

Fig. 20.21

Chapter 20 : Cams � 795The profile of the cam to drive an oscillating roller follower, as shown in Fig. 20.23, is drawn

as discussed in the following steps :

1. First of all, draw a base circle with centre O and radius equal to the minimum radius ofthe cam (i.e. 40 mm)

2. Draw a prime circle with centre O and radius OA

= Min. radius of cam + radius of roller = 40 + 10 = 50 mm

3. Now locate the pivot centre A1 such that OA1 = 130 mm and AA1 = 120 mm. Draw apivot circle with centre O and radius OA1 = 130 mm.

Fig. 20.23

4. Join OA1. Draw angle A1OS = 120° to represent the outward stroke of the follower, angleSOT = 120° to represent the inward stroke of the follower and angle TOA1 = 120° torepresent the dwell.

5. Divide angles A1OS and SOT into the same number of equal even parts as in the displace-

ment diagram and mark points 1, 2, 3 . . . 4′ , 5′ , 6′ on the pivot circle.

6. Now with points 1, 2, 3 . . . 4′ , 5′ , 6′ (on the pivot circle) as centre and radius equal toA1A (i.e. 120 mm) draw circular arcs to intersect the prime circle at points 1, 2, 3 . . .

4′ , 5′ , 6′ .


7. Set off the distances 1B, 2C, 3D... 4 ,′L 5′M along the arcs drawn equal to the distances

as measured from the displacement diagram.

8. The curve passing through the points A, B, C....L, M, N is known as pitch curve.

9. Now draw circles with A, B, C, D....L, M, N as centre and radius equal to the radius ofroller.


Example 20.6. A cam, with a minimum radius of 50 mm, rotating clockwise at a uniformspeed, is required to give a knife edge follower the motion as described below :

1. To move outwards through 40 mm during 100° rotation of the cam ; 2. To dwell for next80° ; 3. To return to its starting position during next 90°, and 4. To dwell for the rest period of arevolution i.e. 90°.

Draw the profile of the cam

(i) when the line of stroke of the follower passes through the centre of the cam shaft, and

(ii) when the line of stroke of the follower is off-set by 15 mm.

The displacement of the follower is to take place with uniform acceleration and uniformretardation. Determine the maximum velocity and acceleration of the follower when the cam shaftrotates at 900 r.p.m.

Draw the displacement, velocity and acceleration diagrams for one complete revolution ofthe cam.

Solution. Given : S = 40 mm = 0.04 m; oθ =100° = 100 × π/180 = 1.745 rad ; Rθ = 90° =

π/2 = 1.571 rad ; N = 900 r.p.m.First of all, the displacement diagram, as shown in Fig. 20.24 (a), is drawn as discussed in

the following steps :

1. Draw a horizontal line ASTPQ such that AS represents the angular displacement of thecam during outward stroke (i.e. 100° ) to some suitable scale. The line ST represents thedwell period of 80° after outward stroke. The line TP represents the angular displacementof the cam during return stroke (i.e. 90°) and the line PQ represents the dwell period of90° after return stroke.

2. Divide AS and TP into any number of equal even parts (say six).

3. Draw vertical lines through points 0, 1, 2, 3 etc. and equal to the lift of the valve i.e. 40mm.

4. Divide the vertical lines 3-f and 3 - f′ ′ into six equal parts as shown by points a, b, c . . .

and ′a , ′b , ′c . . . in Fig. 20.24 (a).

5. Since the follower moves with equal uniform acceleration and uniform retardation, there-fore the displacement diagram of the outward and return stroke consists of a double pa-rabola.

6. Join Aa, Ab and Ac intersecting the vertical lines through 1, 2 and 3 at B, C and D respec-tively.

7. Join the points B, C and D with a smooth curve. This is the required parabola for the halfoutstroke of the valve. Similarly the other curves may be drawn as shown in Fig. 20.24.

8. The curve A B C . . . N P Q is the required displacement diagram.


Fig. 20.24

Fig. 20.25


(i) Profile of the cam when the line of stroke of the follower passes through the centre ofthe cam shaft

The profile of the cam when the line of stroke of the follower passes through the centre ofcam shaft, as shown in Fig. 20.25, may be drawn as discussed in the following steps :

1. Draw a base circle with centre O and radius 50 mm (equal to minimum radius of thecam).

2. Divide the base circle such that angle AOS = 100° ; angle SOT = 80° and angle TOP =90°.

3. Divide angles AOS and TOP into the same number of equal even parts as in displacementdiagram (i.e. six parts).

4. Join the points 1, 2, 3 . . . and 1′ , 2′ , 3′ , . . . with centre O and produce these lines beyondthe base circle.

5. From points 1, 2, 3 . . . and 1′ , 2′ , 3′ , . . . mark the displacements 1B, 2C, 3D . . . etc. asmeasured from the displacement diagram.

6. Join the points A, B, C . . . M, N, P with a smooth curve as shown in Fig. 20.25. This isthe required profile of the cam.

(ii) Profile of the cam when the line of stroke of the follower is offset by 15 mm

The profile of the cam when the line of stroke of the follower is offset may be drawn asdiscussed in Example 20.2. The profile of cam is shown in Fig. 20.26.

Fig. 20.26

Maximum velocity of the follower during out stroke and return stroke

We know that angular velocity of the cam shaft,

2 2 90094.26

60 60


Chapter 20 : Cams � 799We also know that the maximum velocity of the follower during out stroke,

OO

2 . 2 94.26 0.044.32

1.745

Sv

ω × ×= = =θ

m/s Ans.

and maximum velocity of the follower during returnstroke,

RR

2 . 2 94.26 0.044.8

1.571

Sv

ω × ×= = =θ

m/s Ans.

The velocity diagram is shown in Fig. 20.24 (b).

Maximum acceleration of the follower during outstroke and return stroke

We know that the maximum acceleration ofthe follower during out stroke,

2 2

O 2 2O

4 . 4(94.26) 0.04467

( ) (1.745)

Sa

ω= = =θ

m/s2 Ans.

and maximum acceleration of the follower during return stroke,

2 2

R 2 2R

4 . 4(94.26) 0.04

( ) (1.571)

Sa

ω= = =θ

576 m/s2 Ans.

The acceleration diagram is shown in Fig. 20.24 (c).

Example 20.7. Design a cam for operating the exhaust valve of an oil engine. It is requiredto give equal uniform acceleration and retardation during opening and closing of the valve eachof which corresponds to 60° of cam rotation. The valve must remain in the fully open position for20° of cam rotation.

The lift of the valve is 37.5 mm and the least radius of the cam is 40 mm. The follower isprovided with a roller of radius 20 mm and its line of stroke passes through the axis of the cam.

ConstructionFirst of all, the displacement diagram, as shown in Fig. 20.27, is drawn as discussed in the

following steps :

Fig. 20.27

1. Draw a horizontal line ASTP such that AS represents the angular displacement of the camduring opening (i.e. out stroke ) of the valve (equal to 60°), to some suitable scale. Theline ST represents the dwell period of 20° i.e. the period during which the valve remains

A type of roller follower.


fully open and TP represents the angular displacement during closing (i.e. return stroke)of the valve which is equal to 60°.

2. Divide AS and TP into any number of equal even parts (say six).3. Draw vertical lines through points 0, 1, 2, 3 etc. and equal to lift of the valve i.e. 37.5

mm.

4. Divide the vertical lines 3f and 3′ ′f into six equal parts as shown by the points a, b, c . ..

and ′a , ′b , ′c . . . in Fig. 20.27.5. Since the valve moves with equal uniform acceleration and retardation, therefore the dis-

placement diagram for opening and closing of a valve consists of double parabola.6. Complete the displacement diagram as shown in Fig. 20.27.Now the profile of the cam, with a roller follower when its line of stroke passes through the

axis of cam, as shown in Fig. 20.28, is drawn in the similar way as discussed in Example 20.3.

Fig. 20.28

Example 20.8. A cam rotating clockwise at a uniform speed of 1000 r.p.m. is required togive a roller follower the motion defined below :

1. Follower to move outwards through 50 mm during 120° of cam rotation,

2. Follower to dwell for next 60° of cam rotation,

3. Follower to return to its starting position during next 90° of cam rotation,

4. Follower to dwell for the rest of the cam rotation.

The minimum radius of the cam is 50 mm and the diameter of roller is 10 mm. The line ofstroke of the follower is off-set by 20 mm from the axis of the cam shaft. If the displacement of the

Chapter 20 : Cams � 801follower takes place with uniform and equal acceleration and retardation on both the outward andreturn strokes, draw profile of the cam and find the maximum velocity and acceleration during outstroke and return stroke.

Solution. Given : N = 1000 r.p.m. ; S = 50 mm = 0.05 m ; Oθ = 120° = 2 π/3 rad = 2.1 rad ;

Rθ = 90° = π/2 rad = 1.571 rad

Since the displacement of the follower takes place with uniform and equal acceleration andretardation on both outward and return strokes, therefore the displacement diagram, as shown inFig. 20.29, is drawn in the similar manner as discussed in the previous example. But in this case,the angular displacement and stroke of the follower is divided into eight equal parts.

Fig. 20.29

Now, the profile of the cam, as shown in Fig. 20.30, is drawn as discussed in the followingsteps :

1. Draw a base circle with centre O and radius equal to the minimum radius of the cam(i.e. 50 mm).

Fig. 20.30


2. Draw a prime circle with centre O and radius

OA = Minimum radius of the cam + radius of roller = 50 + 5 = 55 mm

3. Draw an off-set circle with centre O and radius equal to 20 mm.

4. Divide the angular displacements of the cam during out stroke and return stroke into eight

equal parts as shown by points 0, 1, 2 . . . and 0′ ,1′ , 2′ . . . etc. on the prime circle in Fig.20.30.

5. From these points draw tangents to the off-set circle.

6. Set off 1B, 2C, 3D . . . etc. equal to the displacements as measured from the displacementdiagram.

7. By joining the points A, B, C . . . T, U, A with a smooth curve, we get a pitch curve.

8. Now from points A, B, C . . . T, U, draw circles with radius equal to the radius of theroller.

9. Join the bottoms of these circles with a smooth curve to obtain the profile of the cam asshown in Fig. 20.30.

Maximum velocity of the follower during out stroke and return stroke

We know that angular velocity of the cam,

2 2 1000104.7

60 60

π π×ω = = =N rad/s.

We also know that the maximum velocity of thefollower during outstroke,

OO

2 . 2 104.7 0.055

2.1

Sv

ω × ×= = =θ

m/s Ans.

and maximum velocity of the follower during returnstroke,

RR

2 . 2 104.7 0.056.66

1.571

Sv

ω × ×= = =θ

m/s Ans.

Maximum acceleration of the follower during outstroke and return stroke

We know that the maximum acceleration of thefollower during out stroke,

2 2

O 2 2O

4 . 4 (104.7) 0.05497.2

( ) (2.1)

Sa

ω= = =θ

m/s2 Ans.

and maximum acceleration of the follower during return stroke,

2 2

R 2 2R

4 . 4(104.7) 0.05888

( ) (1.571)

Sa

ω= = =θ

m/s2 Ans.

Example 20.9. Construct the profile of a cam to suit the following specifications :

Cam shaft diameter = 40 mm ; Least radius of cam = 25 mm ; Diameter of roller = 25 mm;Angle of lift = 120° ; Angle of fall = 150° ; Lift of the follower = 40 mm ; Number of pauses aretwo of equal interval between motions.

A rocker using a cam.

Chapter 20 : Cams � 803During the lift, the motion is S.H.M. During the fall the motion is uniform acceleration and

deceleration. The speed of the cam shaft is uniform. The line of stroke of the follower is off-set12.5 mm from the centre of the cam.

ConstructionFirst of all the displacement diagram, as shown in Fig. 20.31, is drawn as discussed in the

following steps :1. Since the follower moves with simple harmonic motion during lift (i.e. for 120° of cam

rotation), therefore draw the displacement curve ADG in the similar manner as discussedin Example 20.2.

2. Since the follower moves with uniform acceleration and deceleration during fall (i.e. for150° of cam rotation), therefore draw the displacement curve HLP consisting of doubleparabola as discussed in Example 20.6.

Fig. 20.31Now the profile of the cam, when the line of stroke of the follower is off-set 12.5 mm from

the centre of the cam, as shown in Fig. 20.32, is drawn as discussed in the following steps :1. Draw a base circle with centre O and radius equal to the least radius of cam (i.e. 25 mm).

Fig. 20.32


2. Draw a prime circle with centre O and radius,

OA = Least radius of cam + radius of roller = 25 + 25/2 = 37.5 mm

3. Draw a circle with centre O and radius equal to 20 mm to represent the cam shaft.

4. Draw an offset circle with centre O and radius equal to 12.5 mm.

5. Join OA. From OA draw angular displacements of the cam, i.e. draw angle AOS = 120° torepresent lift of the follower, angle SOT = 45° to represent pause, angle TOP = 150° torepresent fall of the follower and angle POA = 45° to represent pause.

Note. Since the number of pauses are two of equal interval between motions (i.e. between lift andfall of the follower), therefore angular displacement of each pause

360 (120 150 )

452

° − ° + °= = °

6. Divide the angular displacements during lift and fall (i.e. angle AOS and TOP) into thesame number of equal even parts (i.e. six parts) as in the displacement diagram.

7. From points 1, 2, 3 . . . etc. and 0 , 1 , 2 , 3′ ′ ′ ′ . . . etc. on the prime circle, draw tangents to

the off-set circle.

8. Set off 1B, 2C, 3D . . . etc. equal to the displacements as measured from the displacementdiagram.

9. By joining the points A, B, C . . . M, N, P with a smooth curve, we get a pitch curve.

10. Now with A, B, C . . . etc. as centre, draw circles with radius equal to the radius of roller.


Example 20.10. It is required to set out the profile of a cam to give the following motion tothe reciprocating follower with a flat mushroom contact face :

(i) Follower to have a stroke of 20 mm during 120° of cam rotation ;

(ii) Follower to dwell for 30° of cam rotation ;

(iii) Follower to return to its initial position during 120° of cam rotation ; and

(iv) Follower to dwell for remaining 90° of cam rotation.

The minimum radius of the cam is 25 mm. The out stroke of the follower is performed withsimple harmonic motion and the return stroke with equal uniform acceleration and retardation.

ConstructionSince the out stroke of the follower is performed with simple harmonic motion and the return

stroke with uniform acceleration and retardation, therefore the displacement diagram, as shown inFig. 20.33, is drawn in the similar manner as discussed in the previous example.

Fig. 20.33

Chapter 20 : Cams � 805The profile of the cam with a flat mushroom contact face reciprocating follower, as shown in

Fig. 20.34, is drawn in the similar way as discussed in Example 20.4.

Fig. 20.34

Example 20.11. It is required to set out the profile of a cam with oscillating follower for thefollowing motion :

(a) Follower to move outward through an angular displacement of 20° during 90° of camrotation ; (b) Follower to dwell for 45° of cam rotation ; (c) Follower to return to its originalposition of zero displacement in 75° of cam rotation ; and (d) Follower to dwell for the remainingperiod of the revolution of the cam.

The distance between the pivot centre and the follower roller centre is 70 mm and the rollerdiameter is 20 mm. The minimum radius of the cam corresponds to the starting position of thefollower as given in (a). The location of the pivot point is 70 mm to the left and 60 mm above theaxis of rotation of the cam. The motion of the follower is to take place with S.H.M. during outstroke and with uniform acceleration and retardation during return stroke.

ConstructionWe know that the angular displacement of the

roller follower,

20 20 /180 / 9= ° = × π = π rad

Since the distance between the pivot centre andthe roller centre (i.e. radius A1A) is 70 mm, thereforelength of arc AA2, as shown in Fig. 20.35, along whichthe displacement of the roller actually takes place

70 / 9 24.5= × π = mm Fig. 20.35


Since the angle is very small, therefore length of chord AA2 is taken equal to the length of arcAA2. Thus in order to draw the displacement diagram, we shall take lift of the follower equal to thelength of chord AA2 i.e. 24.5 mm.

Fig. 20.36

Fig. 20.37


The follower moves with simple harmonic motion during out stroke and with uniform accel-eration and retardation during return stroke. Therefore, the displacement diagram, as shown in Fig.20.36, is drawn in the similar way as discussed in the previous example.

The profile of the cam, as shown in Fig. 20.37, is drawn as discussed in the following steps :

1. First of all, locate the pivot point A1 which is 70 mm to the left and 60 mm above the axisof the cam.

2. Since the distance between the pivot centre A1 and the follower roller centre A is 70 mmand the roller diameter is 20 mm, therefore draw a circle with centre A and radius equal tothe radius of roller i.e. 10 mm.

3. We find that the minimum radius of the cam

60 10 50= − = mm

∴ Radius of the prime circle,

OA = Min. radius of cam + Radius of roller = 50 + 10 = 60 mm

4. Now complete the profile of the cam in the similar way as discussed in Example 20.5.

Example 20.12. Draw the profile of the cam when the roller follower moves with cycloidalmotion during out stroke and return stroke, as given below :

1. Out stroke with maximum displacement of 31.4 mm during 180° of cam rotation,

2. Return stroke for the next 150° of cam rotation,

3. Dwell for the remaining 30° of cam rotation.

The minimum radius of the cam is 15 mm and the roller diameter of the follower is 10 mm.The axis of the roller follower is offset by 10 mm towards right from the axis of cam shaft.

Construction

First of all, the displacement diagram, as shown in Fig. 20.38, is drawn as discussed in thefollowing steps :

Fig. 20.38

1. Draw horizontal line ASP such that AS = 180° to represent the out stroke, SN = 150° torepresent the return stroke and NP = 30° to represent the dwell period.

2. Divide AS and SN into any number of even equal parts (say six).3. From the points 1, 2, 3 . . . etc. draw vertical lines and set-off equal to the stroke of the

follower.4. From a point G draw a generating circle of radius,

Stroke 31.4

2 2= =

π πr = 5 mm


5. Divide the generating circle into six equal parts and from these points draw horizontal

lines to meet the vertical diameter at ′a , G and ′b .

6. Join AG and GN. From point ′a , draw lines parallel to AG and GN to intersect the verti-

cal lines drawn through 1, 2, 4′ and 5′ at B, C, L and M respectively. Similarly draw

parallel lines from ′b intersecting the vertical lines through 4, 5, 1′ and 2′ at E, F, H andJ respectively.

Fig. 20.39

7. Join the points A, B, C . . . L, M, N with a smooth curve.

8. The curve A B C . . . L M N is the required displacement diagram.

Now the profile of the cam, as shown in Fig. 20.39, may be drawn in the similar way asdiscussed in Example 20.9.

20.11.20.11.20.11.20.11.20.11. Cams with Specified ContoursCams with Specified ContoursCams with Specified ContoursCams with Specified ContoursCams with Specified ContoursIn the previous articles, we have discussed about the design of the profile of a cam when the

follower moves with the specified motion. But, the shape of the cam profile thus obtained may bedifficult and costly to manufacture. In actual practice, the cams with specified contours (cam pro-files consisting of circular arcs and straight lines are preferred) are assumed and then motion of thefollower is determined.

20.12.20.12.20.12.20.12.20.12. Tangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller Follower

When the flanks of the cam are straight and tangential to the base circle and nose circle, thenthe cam is known as a tangent cam, as shown in Fig. 20.40. These cams are usually symmetricalabout the centre line of the cam shaft. Such type of cams are used for operating the inlet andexhaust valves of internal combustion engines. We shall now derive the expressions for displace-ment, velocity and acceleration of the follower for the following two cases :

Chapter 20 : Cams � 8091. When the roller has contact with the straight flanks ; and

2. When the roller has contact with the nose.

Let r1 = Radius of the base circle or minimum radius of the cam,

r2 = Radius of the roller,

r3 = Radius of nose,

α = Semi-angle of action of cam or angle of ascent,

θ = Angle turned by the cam from the beginning of the roller displacement,

φ = Angle turned by the cam for contact of roller with the straight flank, and

ω = Angular velocity of the cam.

1. When the roller has contact with straight flanks. A roller having contact with straightflanks is shown in Fig. 20.40. The point O is the centre of cam shaft and the point K is the centreof nose. EG and PQ are straight flanks of the cam. When the roller is in lowest position, (i.e. whenthe roller has contact with the straight flank at E ), the centre of roller lies at B on the pitch curve.Let the cam has turned through an angle* θ ( less than φ) for the roller to have contact at any point(say F) between the straight flanks EG. The centre of roller at this stage lies at C. Thereforedisplacement (or lift or stroke) of the roller from its lowest position is given by

1 cos

cos cos

− θ = − = − = θ θ OB

x OC OB OB OB

1 21 cos

( )cos

− θ = + θ r r . . . 1 2( )OB OE EB r r= + = +∵ . . . (i)

Fig. 20.40. Tangent cam with reciprocating roller follower having contact with straight flanks.

* Since the cam is assumed to be stationary, the angle θ is turned by the roller.


Differentiating equation (i) with respect to t, we have velocity of the follower,

1 2 2

sin( )

cos

θ θ θ = = × = + θ θ dx dx d d

v r rdt d dt dt

1 2 2

sin( )

cos

θ = ω + θ

r r . . . ( / )θ = ω∵ d dt . . . (ii)

From equation (ii), we seethat when θ increases, sin θincreases and cos θ decreases. In

other words, sin θ / cos2

θ increases. Thus the velocity is

maximum where θ is maximum.

This happens when θ = φ i.e.

when the roller just leaves contactwith the straight flank at G or whenthe straight flank merges into acircular nose.

∴ Maximum velocity of thefollower,

1 2 2

sin( )

cosmaxv r r

φ= ω + φ Now differentiating equation

(ii) with respect to t, we haveacceleration of the follower,

θ= = ×θ

dv dv da

dt d dt

2

1 2 4

cos .cos sin 2cos sin( )

cos

θ θ − θ× θ× − θ θ= ω + θ

dr r

dt

2 22

1 2 3

cos 2sin( )

cos

θ + θ= ω + θ r r . . .

d

dt

θ = ω ∵

2 22

1 2 3

cos 2(1 cos )( )

cos

θ + − θ= ω + θ

r r

22

1 2 3

2 cos( )

cos

− θ= ω + θ r r . . . (iii)

A little consideration will show that the acceleration is minimum when 2

3

2 cos

cos

− θθ

is mini-

mum. This is only possible when ( 22 cos− θ ) is minimum and cos3 θ is maximum. This happens

A car in the assembly line.Note : This picture is given as additional information and is not a

direct example of the current chapter.


when θ = 0°, i.e. when the roller is at the beginning of its lift along the straight flank (or when theroller has contact with the straight flank at E ).

∴ Minimum acceleration of the follower,

21 2( )mina r r= ω +

The acceleration is maximum when θ = φ, i.e. when the roller just leaves contact with the

straight flank at G or when the straight flank merges into a circular nose.

∴ Maximum acceleration of the follower,

1

22

2 3

2 cos( )

cosmaxa r r

− φ= ω + φ

2. When the roller has contact with the nose. A roller having contact with the circular noseat G is shown in Fig 20.41. The centre of roller lies at D on the pitch curve. The displacement isusually measured from the top position of the roller, i.e. when the roller has contact at the apex ofthe nose (point H) and the centre of roller lies at J on the pitch curve.

Fig. 20.41. Tangent cam with reciprocating roller follower having contact with the nose.

Let 1θ = Angle turned by the cam measured from the position when the roller

is at the top of the nose.


The displacement of the roller is given by

( ) ( ) ( )= − = − + = + − +x OJ OD OJ OA AD OK KJ OA AD

Substituting =OK r and 3 2= + = + =KJ KH HJ r r L , we have

1( ) ( cos cos )= + − × θ + βx r L OK DK

1( ) ( cos cos )= + − θ + βr L r L . . . 3 2( )DK KJ r r L= = + =∵

1cos cos= + − θ − βL r r L . . . (i)

Now from right angled triangles OAK and DAK,

1sin sin= β = θAK DK OK

or 1sin sinβ = θL r

Squaring both sides,

2 2 2 21sin sinβ = θL r or 2 2 2 2

1(1 cos ) sin− β = θL r

2 2 2 2 21cos sin− β = θL L r or 2 2 2 2 2

1cos sinβ = − θL L r

∴ 1

2 2 2 21cos ( sin )β = − θL L r

Substituting the value of cosβL in equation (i), we get

1

2 2 2 21 1cos ( sin )= + − θ − − θx L r r L r . . . (ii)

Differentiating equation (ii) with respect to t, we have velocity of the follower,

1

1

θ= = ×θ

ddx dxv

dt d dt

1

2 2 2 21 121 1 1 1

1sin ( sin ) ( 2sin cos )

2

−θ θ= − ×− θ × − − θ − × θ θd dr L r r

dt dt

1

2 2 2 21 121 1 1

1sin ( sin ) sin 2

2

−θ θ= θ × + − θ × θ ×d dr L r r

dt dt

1

1 12 2 2 2

1

sin 2. sin

2( sin )

θ

= ω θ +

− θ

rr

L r . . . 1Substituting

d

dt

θ = ω . . . (iii)

Now differentiating equation (iii) with respect to t, we have acceleration of the follower,

1

1

θ= = ×θ

ddv dva

dt d dt

1

12 2 2 2

1 11

2 2 2 221 1 1 1

1 2 2 21

( sin ) ( 2cos 2

1sin 2 ( sin ) ( 2sin cos )

2. cos2( sin )

L r r

dr L r r dt

rL r

−

− θ × θ + θ θ × − θ × θ θ = ω θ + − θ


Substituting 1θ= ω

d

dtand multiplying the numerator and denominator of second term by

12 2 2 2

1( sin )− θL r , we have

2 2 2 3 21 1 1

21 2 2 2 3/ 2

1

1( sin ) (2 cos 2 ) sin 2

2. cos2( sin )

− θ θ + × θ = ω θ +

− θ

L r r ra r

L r

2 3 2 3 21 1 1 1 1

21 2 2 2 3/ 2

1

12 cos 2 2 sin .cos 2 (2sin cos )

2. cos2( sin )

× θ − θ θ + × θ θ = ω θ +

− θ

L r r rr

L r

2 3 2 2 3 2 22 1 1 1 1 1

1 2 2 2 3/ 21

2 . cos 2 2 .sin (1 2sin ) 2 sin (1 sin ). cos

2( sin )

θ − θ − θ + θ − θ= ω θ + − θ

L r r rr

L r

2 3 42 1 1

1 2 2 2 3/ 21

. cos 2 sin. cos

( sin )

θ + θ= ω θ + − θ

L r rr

L r

Notes : 1. Since 1θ is measured from the top position of the roller, therefore for the roller to have contact at

the apex of the nose (i.e. at point H),then 1θ = 0, and for the roller to have contact where straight flank

merges into a nose (i.e. at point G), then 1θ = α − φ.

2. The velocity is zero at H and maximum at G.

3. The acceleration is minimum at H and maximum at G.

4. From Fig 20.41, we see that the distances OK and KD remains constant for all positions of the rollerwhen it moves along the circular nose. In other words, a tangent cam operating a roller follower and havingcontact with the nose is equivalent to a slider crank mechanism (i.e. ODK ) in which the roller is assumedequivalent to the slider D, crank OK and connecting rodDK. Therefore the velocity and acceleration of the rollerfollower may be obtained graphically as discussed inChapters 7 and 8.

Example 20.13. In a symmetrical tangent camoperating a roller follower, the least radius of thecam is 30 mm and roller radius is 17.5 mm. The angleof ascent is 75° and the total lift is 17.5 mm. Thespeed of the cam shaft is 600 r.p.m. Calculate : 1. theprincipal dimensions of the cam ; 2. the accelerationsof the follower at the beginning of the lift, wherestraight flank merges into the circular nose and atthe apex of the circular nose. Assume that there is nodwell between ascent and descent.

Solution. Given : r1 = 30 mm ; r2 = 17.5 mm ;α = 75° ; Total lift = 17.5 mm ; N = 600 r.p.m. orω= 2 π× 600/60 = 62.84 rad/s

Fig. 20.42


1. Principal dimensions of the camLet r = OK = Distance between cam centre and nose centre,

r3 = Nose radius, and

φ = Angle of contact of cam with straight flanks.

From the geometry of Fig. 20.42,

3 1+ =r r r + Total lift

= 30 + 17.5 = 47.5 mm

∴ 347.5= −r r . . . (i)

Also, = +OE OP PE or 1 3= +r OP r

∴ 1 3 330= − = −OP r r r . . . (ii)

Now from right angled triangle OKP,

cosOP OK= × α . . . ( cos / )OP OKα =∵

or 3 3 3 330 (47.5 )cos75 (47.5 )0.2588 12.3 0.2588− = − ° = − = −r r r r

. . . ( )OK r=∵

∴ r3 = 23.88 mm Ans.

and 347.5 47.5 23.88 23.62= = − = − =r OK r mm Ans.

Again, from right angled triangle ODB,

1 2

sin 23.62sin 75tan 0.4803

30 17.5

α °φ = = = = =+ +

DB KP OK

OB OB r r

∴ 25.6φ = °Ans.

2. Acceleration of the follower at the beginning of the lift

We know that acceleration of the follower at the beginning of the lift, i.e. when the roller hascontact at E on the straight flank,

2 2 21 2( ) (62.84) (30 17.5) 187 600mina r r= ω + = + = mm/s2

= 187.6 m/s2 Ans.

Acceleration of the follower where straight flank merges into a circular nose

We know that acceleration of the follower where straight flank merges into a circular nosei.e. when the roller just leaves contact at G,

2 22 2

1 2 3 3

2 cos 2 cos 25.6( ) (62.84) (30 17.5)

cos cos 25.6maxa r r

− φ − °= ω + = + φ °

2 0.813187600 303 800

0.733

− = = mm/s2 = 303.8 m/s2 Ans.


Acceleration of the follower at the apex of the circular noseWe know that acceleration of the follower for contact with the circular nose,

2 3 42 1 1

1 2 2 2 3/ 21

. cos 2 sin. cos

( sin )

θ + θ= ω θ + − θ

L r ra r

L r

Since 1θ is measured from the top position of the follower, therefore for the follower to havecontact at the apex of the circular nose (i.e. at point H ), 1θ = 0.

∴ Acceleration of the follower at the apex of the circular nose,

22 2 2

32 3

.. 1 . 1 . 1

= ω + = ω + = ω + +

L r r ra r r r

L r rL

2 23.62

(62.84) 23.62 117.5 23.88

= + + = 146 530 mm/s2 . . . 2 3( )L r r= +∵

= 146.53 m/s2 Ans.Example 20.14. A cam has straight working faces which are tangential to a base circle of

diameter 90 mm. The follower is a roller of diameter 40 mm and the centre of roller moves alonga straight line passing through the centre line of the cam shaft. The angle between the tangentialfaces of the cam is 90° and the faces are joined by a nose circle of 10 mm radius. The speed ofrotation of the cam is 120 revolutions per min.

Find the acceleration of the roller centre 1. when during the lift, the roller is just about toleave the straight flank ; and 2. when the roller is at the outer end of its lift.

Solution. Given : d1 = 90 mm or r1 = 45 mm ; d2 = 40 mm or r2 = 20 mm ; 2 α = 90° orα = 45° ; r3 = 10 mm ; N = 120 r.p.m. or ω = 2 π × 120/60 = 12.57 rad/s

The tangent cam operating a roller follower is shown in Fig. 20.43.

Fig. 20.43

First of all, let us find the *angle turned by the cam ( φ) when the roller is just about to leave

the straight flank at G. The centre of roller at this position lies at D.

* Since the cam is assumed to be stationary, φ is the angle turned by the roller when it is just about to leavethe straight flank at G.


From the geometry of the figure,

BD PK OP OE PE= = = −

1 3 45 10 35 mm

OE KG

r r

= −= − = − =

Now from triangle OBD,

1 2

tan

350.5385

45 20

BD BD

OB OE EBBD

r r

φ = =+

= = =+ +

∴ 28.3φ = °

1. Acceleration of the roller centre when roller is just about to leave the straight flank

We know that acceleration of the roller centre when the roller is just about to leave thestraight flank,

2 22 2

1 2 3 3

2 cos 2 cos 28.3( ) (12.57) (45 20)

cos cos 28.3

− φ − °= ω + = + φ ° a r r

= 18 500 mm/s2 = 18.5 m/s2 Ans.

2. Acceleration of the roller centre when the roller is at the outer end of the lift

First of all, let us find the values of OK and KD. From the geometry of the figure,

2 2( ) ( ) 2= = + = ×OK r OP PK OP . . . ( )OP PK=∵

2( ) 2(45 10) 49.5= − = − =OE EP mm

3 2 10 20 30= = + = + = + =KD L KG GD r r mm

We know that acceleration of the roller centre when the roller is at the outer end of the lift,i.e. when the roller has contact at the top of the nose,

2 3 42 21 1

1 2 2 2 3/ 21

. cos 2 sin. cos . 1

( sin )

θ + θ = ω θ + = ω + − θ

L r r ra r r

LL r

. . . (∵ At the outer end of the lift, 1 0θ = )

2 49.5(12.57) 49.5 1 20 730

30 = + =

mm/s2 = 20.73 m/s2 Ans.

20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced FollowerWhen the flanks of the cam connecting the base circle and nose are of convex circular arcs,

then the cam is known as circular arc cam. A symmetrical circular arc cam operating a flat-facedfollower is shown in Fig. 20.44, in which O and Q are the centres of cam and nose respectively. EF

and GH are two circular flanks whose centres lie at P and P′ respectively. The centres * P and P′

* The centres P and P′ may also be obtained by drawing arcs with centres O and Q and radii equal to OP

and PQ respectively. The circular flanks EF and GH are now drawn with centres P and P′ and radiusequal to PE.

In aircraft engines roller followers arewidely used.

Chapter 20 : Cams � 817lie on lines EO and GO produced.

Let r1 = Minimum radius of the cam or radius of the base circle = OE,r2 = Radius of nose,

R = Radius of circular flank = PE, 2α = Total angle of action of cam = angle EOG,

α = Semi-angle of action of cam or angle of ascent = angle EOK, and

φ = Angle of action of cam on the circular flank.

Fig. 20.44. Circular arc cam with flat face of the follower having contact with the circular flank. We shall consider the following two cases :

1. When the flat face of the follower has contact on the circular flank, and

2. When the flat face of the follower has contact on the nose.

In deriving the expressions for displacement, velocity and acceleration of the follower for theabove two cases, it is assumed that the cam is fixed and the follower rotates in the opposite senseto that of the cam. In Fig. 20.44, the cam is rotating in the clockwise direction and the followerrotates in the counter-clockwise direction.

1. When the flat face of the follower has contact on the circular flank. First of all, let usconsider that the flat face of the follower has contact at E (i.e. at the junction of the circular flankand base circle). When the cam turns through an angle θ (less than φ) relative to the follower, thecontact of the flat face of the follower will shift from E to C on the circular flank, such that flat faceof the follower is perpendicular to PC. Since OB is perpendicular to BC, therefore OB is parallel toPC. From O, draw OD perpendicular to PC.

From the geometry of the figure, the displacement or lift of the follower (x) at any instant forcontact on the circular flank, is given by

= = − = −x BA BO AO CD EO . . . (i)We know that

cos= − = − θCD PC PD PE OP

cos (1 cos )= + − θ = + − θOP OE OP OE OP


Substituting the value of CD in equation (i),

(1 cos ) (1 cos )x OE OP EO OP= + − θ − = − θ

1( )(1 cos ) ( ) (1 cos )PE OE R r= − − θ = − − θ . . . (ii)

Differentiating equation (ii) with respect to t, we have velocity of the follower,

dx dx d dxv

dt d dt d

θ= = × = ×ωθ θ . . . substituting

d

dt

θ = ω

1 1( )sin ( )sinR r R r= − θ×ω = ω − θ . . . (iii)

From the above expression, we see that at the beginning of the ascent (i.e. when θ = 0 ), thevelocity is zero (because sin 0 = 0 ) and it increases as θ increases. The velocity will be maximumwhen θ = φ, i.e. when the contact of the follower just shifts from circular flank to circular nose.Therefore maximum velocity of the follower,

1( )sinmaxv R r= ω − φNow differentiating equation (iii) with respect to t, we have acceleration of the follower,

dv dv d dv

adt d dt d

θ= = × = ×ωθ θ

21 1( )cos ( )cosR r R r= ω − θ×ω = ω − θ . . . (iv)

From the above expression, we see that at the beginning of the ascent (i.e. when θ = 0 ), theacceleration is maximum (because cos 0 = 1 ) and it decreases as θ increases. The accelerationwill be minimum when θ = φ .

∴ Maximum acceleration of the follower,

21( )maxa R r= ω −

and minimum acceleration of the follower,

21( ) cosmina R r= ω − φ

2. When the flat face of the follower has contact on the nose. The flat face of the followerhaving contact on the nose at C is shown in Fig. 20.45. The centre of curvature of the nose lies atQ. In this case, the displacement or lift of the follower at any instant when the cam has turnedthrough an angle θ (greater than φ ) is given by

x AB OB OA CD OA= = − = − . . .( ∵ OB = CD ) . . . (i)

But cos ( )CD CQ QD CQ OQ= + = + α − θSubstituting the value of CD in equation (i), we have

cos( )x CQ OQ OA= + α − θ − . . . (ii)

The displacement or lift of the follower when the contact is at the apex K of the nose i.e.when α − θ = 0 is

* 2 1x CQ OQ OA r OQ r= + − = + −

* From the geometry of Fig. 20.45, we also find that lift of the follower when the contact is at the apex Kof the nose is

x = JK = OQ + QK – OJ = OQ + r2 – r1

Chapter 20 : Cams � 819Differentiating equation (ii) with respect to t, we have velocity of the follower,

dx dx d dx

vdt d dt d

θ= = × = ×ωθ θ

sin ( ) sin ( )OQ OQ= α − θ ω = ω× α − θ . . . (iii). . .(∵ CQ, OQ, OA and α are constant)

From the above expression, we see that the velocity is zero when α − θ = 0 or α = θ i.e.when the follower is at the apex K of the nose. The velocity will be maximum when ( α − θ ) ismaximum. This happens when the follower changes contact from circular flank to circular nose atpoint F, i.e. when ( α − θ ) = φ .

Now differentiating equation (iii) with respect to t, we have acceleration of the follower,

dv dv d dv

adt d dt d

θ= = × = ×ωθ θ

2cos ( ) cos ( )OQ OQ= −ω× α − θ ω = −ω × α − θ . . . (iv)The negative sign in the above expression shows that there is a retardation when the

follower is in contact with the nose of the cam.From the above expression, we see that retardation is maximum when α − θ = 0 or θ = α ,

i.e. when the follower is at the apex K of the nose.

∴ Maximum retardation = 2 OQω ×The retardation is minimum when α − θ is maximum. This happens when the follower changes

contact from circular flank to circular nose at point F i.e. when θ = φ.

∴ Minimum retardation = 2 cos( )OQω × α − φ

Fig. 20.45. Circular arc cam with flat face of the follower having contact on the nose.


Example 20.15. A symmetrical circular cam operating a flat-faced follower has thefollowing particulars :

Minimum radius of the cam = 30 mm ; Total lift = 20 mm ; Angle of lift = 75° ; Nose radius= 5 mm ; Speed = 600 r.p.m. Find : 1. the principal dimensions of the cam, and 2. the accelera-tion of the follower at the beginning of the lift, at the end of contact with the circular flank , at thebeginning of contact with nose and at the apex of the nose.

Solution. Given : r1 = OE = 30 mm ; x = JK = 20 mm ; α = 75° ; r2 = QF = QK = 5 mm ;

N = 600 r.p.m. or ω = 2 600 / 60 62.84π× = rad/s

1. Principal dimensions of the cam

A symmetrical circular cam operating a flat faced follower is shown in Fig. 20.46.

Let OQ = Distance between cam centre and nose centre,

R = PE = Radius of circular flank, and

φ = Angle of contact on the circular flank.

We know that lift of the follower (x),

2 120 5 30 25= + − = + − = −OQ r r OQ OQ

∴ OQ = 20 + 25 = 45 mm Ans.

We know that PQ PF FQ PE FQ OP OE FQ= − = − = + −

30 5 ( 25)OP OP= + − = + mm

Now from a triangle OPQ,

2 2 2( ) ( ) ( ) 2 cosPQ OP OQ OP OQ= + − × × β

2 2 2( 25) ( ) 45 2 45cos (180 75 )OP OP OP+ = + − × × ° − °

2 2( ) 50 625 ( ) 2025 23.3OP OP OP OP+ + = + +

50 23.3 2025 625OP OP− = −or 26.7 OP = 1400and OP = 1400/26.7 = 52.4 mm

∴ Radius of circular flanks, R PE OP OE= = + = 52.4 + 30

= 82.4 mm Ans.and PQ = OP + 25 = 52.4 + 25

= 77.4 mm Ans.

In order to find angle φ , consider a triangle OPQ. We

know that

sin sin

OQ PQ=φ β

or sin 45 sin (180 75 )

sin 0.561677.4

OQ

PQ

× β × ° − °φ = = =

∴ 34.2φ = °Ans.

Fig. 20.46


2. Acceleration of the follower

We know that acceleration of the follower at the beginning of the lift,

2 21 1( ) cos ( )a R r R r= ω − θ = ω − . . . (∵ At the beginning of lift, 0θ = ° )

= (62.84)2 (82.4 – 30) = 206 930 mm/s = 206.93 m/s2 Ans.

Acceleration of the follower at the end of contact with the circular flank,

2 21 1( )cos ( )cosa R r R r= ω − θ = ω − φ

. . . (∵ At the end of contact with the circular flank, θ = φ )

= – (62.84)2 (82.4 – 30) cos 34.2° = 171 130 mm/s2 = 171.13 m/s2 Ans.

Acceleration of the follower at the beginning of contact with nose,

2 2cos ( ) cos ( )a OQ OQ= −ω × α − θ = −ω × α − φ

. . . (∵ At the beginning of contact with nose, θ = φ )

= – (62.84)2 45 cos (75°– 34.2°) = – 134 520 mm/s2 = – 134.52 m/s2

= 134.52 m/s2 (Retardation) Ans.

and acceleration of the follower at the apex of nose,

2 2cos ( )a OQ OQ= −ω × α − θ = −ω × ... (∵At the apex of nose, α − θ = 0 )

= – (62.84)2 45 = – 177 700 mm/s2 = – 177.7 m/s2

= 177.7 m/s2 (Retardation) Ans.Example 20.16. A symmetrical cam with convex flanks operates a flat-footed follower. The

lift is 8 mm, base circle radius 25 mm and the nose radius 12 mm. The total angle of the camaction is 120°.

1. Find the radius of convex flanks, 2. Draw the profile of the cam, and 3. Determine themaximum velocity and the maximum acceleration when the cam shaft rotates at 500 r.p.m.

Solution. Given : x = JK = 8 mm ; r1 = OE = OJ = 25 mm ; r2 = QF = QK = 12 mm ;

2 120EOGα = ∠ = ° or 60EOKα = ∠ = ° ; N = 500 r.p.m. or 2ω = π × 500/60 = 52.37 rad/s

1. Radius of convex flanks

Let R = Radius of convex flanks = PE = P G′

A symmetrical cam with convex flanks operating a flat footed follower is shown in Fig.20.47. From the geometry of the figure,

OQ OJ JK QK= + − 1 2r x r= + −

= 25 + 8 – 12 = 21 mm

PQ PF QF PE QF= − = − = (R – 12) mm

and ( 25)OP PE OE R= − = − mm


Now consider the triangle OPQ. We know that

2 2 2( ) ( ) ( ) 2 cosPQ OP OQ OP OQ= + − × × β

2 2 2( 12) ( 25) (21) 2 ( 25)21cos(180 60 )R R R− = − + − − ° − °

2 224 144 50 625 441 21 525R R R R R− + = − + + + − – 24 R + 144 = – 29 R + 541 or 5 R = 397

∴ R = 397/5 = 79.4 mm Ans.

Fig. 20.47

2. Profile of the cam

The profile of the cam, as shown in Fig. 20.47, is drawn as discussed in the following steps :(a) First of all, draw a base circle with centre O and radius OE = r1 = 25 mm.(b) Draw angle EOK = 60° and angle KOG = 60° such that the total angle of cam action is

120°.(c) On line OK mark OQ = 21 mm (as calculated above). Now Q as centre, draw a circle of

radius equal to the nose radius r2 = QK = QF = 12 mm. This circle cuts the line OK at J.Now JK represents the lift of the follower (i.e. 8 mm).

(d) Produce EO and GO as shown in Fig. 20.47. Now with Q as centre and radius equal toPQ = R – r2 = 79.4 – 12 = 67.4 mm, draw arcs intersecting the lines EO and GO producedat P and P′ respectively. The centre P′ may also be obtained by drawing arcs withcentres O and Q and radii OP and PQ respectively.

(e) Now with P and P′as centres and radius equal to R = 79.4 mm, draw arcs EF and GHwhich represent the convex flanks. EFKHGAE is the profile of the cam.

3. Maximum velocity and maximum acceleration

First of all, let us find the angle φ . From triangle OPQ,

sin sin

OQ PQ=φ β


or 21

sin sin sin (180 60 ) 0.269879.4 12

OQ

PQφ = × β = × ° − ° =

−

. . . (∵ PQ = R – 12 )

∴ 15.65φ = °

We know that maximum velocity,

1( )sin 52.37(79.4 25)sin15.65 770maxv R r= ω − φ = − ° = mm/s

= 0.77 m/s Ans.

and maximum acceleration,

2 21( ) (52.37) (79.4 25) 149200maxa R r= ω − = − = mm/s2= 149.2 m/s2 Ans.

Example 20.17. The following particulars relate to a symmetrical circular cam operating aflat faced follower :

Least radius = 16 mm, nose radius = 3.2 mm, distance between cam shaft centre and nosecentre = 25 mm, angle of action of cam = 150°, and cam shaft speed = 600 r.p.m.

Assuming that there is no dwell between ascent or descent, determine the lift of the valve,the flank radius and the acceleration and retardation of the follower at a point where circularnose merges into circular flank.

Solution. Given : r1 = OE = OJ = 16 mm ; r2 = QK = QF = 3.2 mm ; OQ = 25 mm ;2 α = 150° or α = 75° ; N = 600 r.p.m. or ω = 2 π× 600/60 = 62.84 rad/s

Lift of the valve

A symmetrical circular cam operating a flat facedfollower is shown in Fig. 20.48.

We know that lift of the valve,

x JK OK OJ= = −

2 1OQ QK OJ OQ r r= + − = + −

= 25 + 3.2 – 16 = 12.2 mm Ans.

Flank radius

Let R = PE = Flank radius.

First of all, let us find out the values of OP andPQ. From the geometry of Fig. 20.48,

16OP PE OE R= − = −

and 3.2PQ PF FQ R= − = −

Now consider the triangle OPQ. We know that

2 2 2( ) ( ) ( ) 2 cosPQ OP OQ OP OQ= + − × × β

Substituting the values of OP and PQ in the above expression,

2 2 2( 3.2) ( 16) (25) 2( 16) 25cos(180 75 )R R R− = − + − − × ° − °

Fig. 20.48


2 26.4 10.24 32 256 625 (50 800) ( 0.2588)R R R R R− + = − + + − − −

6.4 10.24 19.06 673.96R R− + = − + or 12.66 R = 663.72

∴ R = 52.43 mm Ans.

Acceleration and retardation of the follower at a point where circular nose merges into circularflank

From Fig. 20.48 we see that at a point F, the circular nose merges into a circular flank. Let φbe the angle of action of cam at point F. From triangle OPQ,

sin sin

OQ PQ=φ β

or sin sin (180 75 ) sin105OQ OQ

PQ PF FQφ = × ° − ° = × °

−

25

0.966 0.490752.43 3.2

= × =−

∴ 29.4φ = °

We know that acceleration of the follower,

2 21cos ( )cosa OP R r= ω × × θ = ω − φ . . . (∵ θ = φ)

= (62.84)2 (52.43 – 16) cos 29.4° = 125 330 mm/s2

= 125.33 m/s2 Ans.

We also know that retardation of the follower,

2 2cos( ) cos ( )a OQ OQ= ω × α − θ = ω × α − φ . . . (∵ θ = φ)

= (62.84)2 25 cos (75° – 29.4°) = 69 110 mm/s2

= 69.11 m/s2 Ans.

Example 20.18. A flat ended valve tappet is operated by a symmetrical cam with circulararc for flank and nose. The straight line path of the tappet passes through the cam axis. Totalangle of action = 150°. Lift = 6 mm. Base circle diameter = 30 mm. Period of acceleration is halfthe period of retardation during the lift. The cam rotates at 1250 r.p.m. Find : 1. flank and noseradii ; 2. maximum acceleration and retardation during the lift.

Solution. Given : 2 α = 150° or α = 75° ; x = JK = 6 mm ;d1 = 30 mm or r1 = OE = OJ = 15 mm ; N = 1250 r.p.m. orω = 2 π × 1250/60 = 131 rad/s

1. 1. 1. 1. 1. Flank and nose radii

The circular arc cam operating a flat ended valve tappetis shown in Fig. 20.49.

Let R = PE = Flank radius, and

2r QF QK= = = Nose radius. Fig. 20.49

Chapter 20 : Cams � 825First of all, let us find the values of OP, OQ and PQ. The acceleration takes place while the

follower is on the flank and retardation while the follower is on nose. Since the period of accelera-tion is half the period of retardation during the lift, therefore

1

2φ = γ . . . (i)

We know that 180 180 75 105β = °− α = °− ° = °

∴ 75 180 180 105 75φ + γ = ° = ° − β = ° − ° = ° . . . (ii)


25φ = ° , and 50γ = °Now from the geometry of Fig. 20.49,

1 2 2 215 6 21OQ OJ JK QK r x r r r= + − = + − = + − = − . . . (iii)

and 2( ) 15PQ PF FQ PE FQ OP OE FQ OP r= − = − = + − = + − . . . (iv)

Now from triangle OPQ,

sin sin sin

OP OQ PQ= =γ φ β

or 2 221 15

sin 50 sin 25 sin105

r OP rOP − + −= =

° ° °

∴ 2 22

21 21sin 50 0.766 38 1.8

sin 25 0.4226

r rOP r

− −= × ° = × = −

° . . . (v)

Also 2 215 15sin 50 0.766

sin105 0.966

OP r OP rOP

+ − + −= × ° = ×

°

= 0.793 × OP + 11.9 – 0.793 r2

∴ 0.207 OP = 11.9 – 0.793 r2 or OP = 57.5 – 3.83 r2 . . . (vi)

From equations (v) and (vi),

38 – 1.8 r2 = 57.5 – 3.83 r2 or 2.03 r2 = 19.5

∴ r2 = 9.6 mm Ans.

We know that 238 1.8 38 1.8 9.6 20.7OP r= − = − × = mm . . . [From equation (v)]

∴ R = PE = OP + OE = 20.7 + 15 = 35.7 mm Ans.

2. Maximum acceleration and retardation during the lift

We know that maximum acceleration

2 2 21( ) (131) 20.7 355230R r OP= ω − = ω × = = mm/s2

= 355.23 m/s2 Ans.

and maximum retardation, 2 22(21 )OQ r= ω × = ω − . . . [From equation (iii) ]

= (131)2 (21 – 9.6) = 195 640 mm/s2 = 195.64 m/s2 Ans.


Example 20.19. A cam consists of a circular disc of diameter75 mm with its centre displaced 25 mm from the camshaft axis. Thefollower has a flat surface (horizontal) in contact with the cam andthe line of action of the follower is vertical and passes through theshaft axis as shown in Fig. 20.50. The mass of the follower is 2.3 kgand is pressed downwards by a spring which has a stiffness of 3.5N/mm. In the lowest position the spring force is 45 N.

1. Derive an expression for the acceleration of the follower interms of the angle of rotation from the beginning of the lift.

2. As the cam shaft speed is gradually increased, a value isreached at which the follower begins to lift from the cam surface.Determine the camshaft speed for this condition.

Solution. Given : d = 75 mm or r = OA = 37.5 mm ;OQ = 25 mm ; m = 2.3 kg ; s = 3.5 N/mm ; S = 45 N

1. Expression for the acceleration of the follower

The cam in its lowest position is shown by full lines in Fig. 20.51 and by dotted lines whenit has rotated through an angle θ .

From the geometry of the figure, the displacement of the follower,

x AB OS OQ QS= = = −

cosOQ PQ= − θ

cosOQ OQ= − θ . . . ( )PQ OQ=∵

(1 cos ) 25(1 cos )OQ= − θ = − θ . . . (i)

Differentiating equation (i) with respect to t, weget velocity of the follower,

dx dx d dxv

dt d dt d

θ= = × = ×ωθ θ

. . .(Substituting /d dtθ = ω)

25sin 25 sin= θ×ω = ω θ . . . (ii)

Now differentiating equation (ii) with respect tot, we get acceleration of the follower,

25 cosdv dv d

adt d dt

θ= = × = ω θ× ωθ

225 cos= ω θ mm/s2 = 20.025 cosω θ m/s2 Ans.

2. Cam shaft speed

Let N = Cam shaft speed in r.p.m.We know that accelerating force

2 2. 2.3 0.025 cos 0.0575 cosm a= = × ω θ = ω θ N

Now for any value of θ , the algebraic sum of the spring force, weight of the follower and theaccelerating force is equal to the vertical reaction between the cam and follower. When this reac-tion is zero, then the follower will just begin to leave the cam.

Fig. 20.50

Fig. 20.51


∴ . . . 0S s x m g m a+ + + =

245 3.5 25(1 cos ) 2.3 9.81 0.0575 cos 0+ × − θ + × + ω θ =

245 87.5 87.5cos 22.56 0.0575 cos 0+ − θ + + ω θ =

2155.06 87.5cos 0.0575 cos 0− θ + ω θ =

22697 1522cos cos 0− θ + ω θ = . . . ( Dividing by 0.0575 )

2 cos 1522cos 2697ω θ = θ − or 2 1522 2697secω = − θ

Since sec 1θ ≥ + or , 1≤ − , therefore the minimum value of 2ω occurs when θ = 180° there-fore

2 1522 ( 2697) 4219ω = − − = . . . [Substituting sec θ = – 1 ]

∴ 65ω = rad/s

and maximum allowable cam shaft speed,

60 65 60

2 2N

ω× ×= =π π = 621 r.p.m. Ans.

EXERCISESEXERCISESEXERCISESEXERCISESEXERCISES1. A disc cam is to give uniform motion to a knife edge follower during out stroke of 50 mm during

the first half of the cam revolution. The follower again returns to its original position with uniformmotion during the next half of the revolution. The minimum radius of the cam is 50 mm and thediameter of the cam shaft is 35 mm. Draw the profile of the cam when 1. the axis of follower passesthrough the axis of cam shaft, and 2. the axis of follower is offset by 20 mm from the axis of thecam shaft.

2. A cam operating a knife-edged follower has the following data :

(a) Follower moves outwards through 40 mm during 60° of cam rotation.

(b) Follower dwells for the next 45°.

(c) Follower returns to its original position during next 90°.

(d) Follower dwells for the rest of the rotation.

The displacement of the follower is to take place with simple harmonic motion during both theoutward and return strokes. The least radius of the cam is 50 mm. Draw the profile of the cam when1. the axis of the follower passes through the cam axis, and 2. the axis of the follower is offset 20mm towards right from the cam axis. If the cam rotates at 300 r.p.m., determine maximum velocityand acceleration of the follower during the outward stroke and the return stroke.

[Ans. 1.88 m/s, 1.26 m/s ; 177.7 m/s2, 79 m/s2]3. A disc cam rotating in a clockwise direction is used to move a reciprocating roller with simple

harmonic motion in a radial path, as given below :

(i) Outstroke with maximum displacement of 25 mm during 120° of cam rotation,

(ii) Dwell for 60° of cam rotation,

(iii) Return stroke with maximum displacement of 25 mm during 90° of cam rotation, and

(iv) Dwell during remaining 90° of cam rotation.

The line of reciprocation of follower passes through the camshaft axis. The maximum radius of camis 20 mm. If the cam rotates at a uniform speed of 300 r.p.m. find the maximum velocity andacceleration during outstroke and return stroke. The roller diameter is 8 mm.


Draw the profile of the cam when the line of reciprocation of the follower is offset by 20 mmtowards right from the cam shaft axis. [Ans. 0.59 m/s, 0.786 m/s ; 27.8 m/s2, 49.4 m/s2]

4. Design a cam to raise a valve with simple harmonic motion through 50 mm in 1/3 of a revolution,keep if fully raised through 1/12 revolution and to lower it with harmonic motion in 1/6 revolution.The valve remains closed during the rest of the revolution. The diameter of the roller is 20 mm andthe minimum radius of the cam is 25 mm. The diameter of the camshaft is 25 mm. The axis of thevalve rod passes through the axis of the camshaft. If the camshaft rotates at uniform speed of 100r.p.m. ; find the maximum velocity and acceleration of a valve during raising and lowering.

[ Ans. 0.39 m/s, 0.78 m/s ; 6.17 m/s2, 24.67 m/s2]5. A cam rotating clockwise with a uniform speed is to give the roller follower of 20 mm diameter with

the following motion :

(a) Follower to move outwards through a distance of 30 mm during 120° of cam rotation ;

(b) Follower to dwell for 60° of cam rotation ;

(c) Follower to return to its initial position during 90° of cam rotation ; and

(d) Follower to dwell for the remaining 90° of cam rotation.

The minimum radius of the cam is 45 mm and the line of stroke of the follower is offset 15 mmfrom the axis of the cam and the displacement of the follower is to take place with simple harmonicmotion on both the outward and return strokes. Draw the cam profile.

6. A cam rotating clockwise at a uniform speed of 100 r.p.m. is required to give motion to knife-edgefollower as below :

(a) Follower to move outwards through 25 mm during 120° of cam rotation,

(b) Follower to dwell for the next 60° of cam rotation,

(c) Follower to return to its starting position during next 90° of cam rotation, and

(d) Follower to dwell for the rest of the cam rotation.

The minimum radius of the cam is 50 mm and the line of stroke of the follower passes through theaxis of the cam shaft. If the displacement of the follower takes place with uniform and equal accel-eration and retardation on both the outward and return strokes, find the maximum velocity andacceleration during outstroke and return stroke. [Ans. 0.25 m/s, 0.33 m/s ; 2.5 m/s2 , 4.44 m/s2]

7. A cam with 30 mm as minimum diameter is rotating clockwise at a uniform speed of 1200 r.p.m.and has to give the following motion to a roller follower 10 mm in diameter:

(a) Follower to complete outward stroke of 25 mm during 120° of cam rotation with equal uniformacceleration and retardation ;

(b) Follower to dwell for 60° of cam rotation ;

(c) Follower to return to its initial position during 90° of cam rotation with equal uniform accelera-tion and retardation ;

(d) Follower to dwell for the remaining 90° of cam rotation.

Draw the cam profile if the axis of the roller follower passes through the axis of the cam.

Determine the maximum velocity of the follower during the outstroke and return stroke andalso the uniform acceleration of the follower on the out stroke and the return stoke.

[Ans. 3 m/s , 4 m/s ; 360.2 m/s2, 640.34 m/s2]8. A cam rotating clockwise at a uniform speed of 200 r.p.m. is required to move an offset roller

follower with a uniform and equal acceleration and retardation on both the outward and returnstrokes. The angle of ascent, the angle of dwell (between ascent and descent) and the angle ofdescent is 120°, 60° and 90° respectively. The follower dwells for the rest of cam rotation. The leastradius of the cam is 50 mm, the lift of the follower is 25 mm and the diameter of the roller is 10mm. The line of stroke of the follower is offset by 20 mm from the axis of the cam. Draw the camprofile and find the maximum velocity and acceleration of the follower during the outstroke.

9. A flat faced reciprocating follower has the following motion :

Chapter 20 : Cams � 829(i) The follower moves out for 80° of cam rotation with uniform acceleration and retardation, the

acceleration being twice the retardation.

(ii) The follower dwells for the next 80° of cam rotation.

(iii) It moves in for the next 120° of cam rotation with uniform acceleration and retardation, theretardation being twice the acceleration.

(iv) The follower dwells for the remaining period.

The base circle diameter of the cam is 60 mm and the stroke of the follower is 20 mm. The lineof movement of the follower passes through the cam centre.

Draw the displacement diagram and the profile of the cam very neatly showing all construc-tional details.

10. From the following data, draw the profile of a cam in which the follower moves with simple har-monic motion during ascent while it moves with uniformly accelerated motion during descent :

Least radius of cam = 50 mm ; Angle of ascent = 48° ; Angle of dwell between ascent anddescent = 42° ; Angle of descent = 60° ; Lift of follower = 40 mm ; Diameter of roller = 30 mm ;Distance between the line of action of follower and the axis of cam = 20 mm.

If the cam rotates at 360 r.p.m. anticlockwise, find the maximum velocity and acceleration ofthe follower during descent. [Ans. 2.88 m/s ; 207.4 m/s2]

11. Draw the profile of a cam with oscillating roller follower for the following motion :

(a) Follower to move outwards through an angular displacement of 20° during 120° of cam rota-tion.

(b) Follower to dwell for 50° of cam rotation.

(c) Follower to return to its initial position in 90° of cam rotation with uniform acceleration andretardation.

(d) Follower to dwell for the remaining period of cam rotation.

The distance between the pivot centre and the roller centre is 130 mm and the distance betweenthe pivot centre and cam axis is 150 mm. The minimum radius of the cam is 80 mm and thediameter of the roller is 50 mm.

12. Draw the profile of the cam when the roller follower moves with cycloidal motion as given below :

(a) Outstroke with maximum displacement of 44 mm during 180° of cam rotation.

(b) Return stroke for the next 150° of cam rotation.

(c) Dwell for the remaining 30° of cam rotation.

The minimum radius of the cam is 20 mm and the diameter of the roller is 10 mm. The axis ofthe roller follower passes through the cam shaft axis.

13. A symmetrical tangent cam operating a roller follower has the following particulars :

Radius of base circle of cam = 40 mm, roller radius = 20 mm, angle of ascent = 75°, total lift = 20mm, speed of cam shaft = 300 r.p.m.

Determine : 1. the principal dimensions of the cam, 2. the equation for the displacement curve,when the follower is in contact with the straight flank, and 3. the acceleration of the follower whenit is in contact with the straight flank where it merges into the circular nose.

[Ans. r3 = 33 mm ; θ θ θ θ θ = 23.5° ; 89.4 m/s2]14. A cam profile consists of two circular arcs of radii 24 mm and 12 mm, joined by straight lines,

giving the follower a lift of 12 mm. The follower is a roller of 24 mm radius and its line of actionis a straight line passing through the cam shaft axis. When the cam shaft has a uniform speed of 500rev/min, find the maximum velocity and acceleration of the follower while in contact with the straightflank of the cam. [Ans. 1.2 m/s ; 198 m/s2]

15. The following particulars relate to a symmetrical tangent cam operating a roller follower :-

Least radius = 30 mm, nose radius = 24 mm, roller radius = 17.5 mm, distance between cam shaftand nose centre = 23.5 mm, angle of action of cam = 150°, cam shaft speed = 600 r.p.m.

Assuming that there is no dwell between ascent and descent, determine the lift of the valve and the


acceleration of the follower at a point where straight flank merges into the circular nose.

[Ans. 17.5 mm ; 304.5 m/s2]16. Following is the data for a circular arc cam working with a flat faced reciprocating follower :

Minimum radius of the cam = 30 mm ; Total angle of cam action = 120° ; Radius of the circular arc= 80 mm ; Nose radius = 10 mm.

1. Find the distance of the centre of nose circle from the cam axis ; 2. Draw the profile of the camto full scale; 3. Find the angle through which the cam turns when the point of contact moves fromthe junction of minimum radius arc and circular arc to the junction of nose radius arc and circulararc ; and 4. Find the velocity and acceleration of the follower when the cam has turned through an

angle of θ = 20°. The angle θ is measured from the point where the follower just starts movingaway from the cam. The angular velocity of the cam is 10 rad/s.

[Ans. 30 mm ; 22°; 68.4 mm/s ; 1880 mm/s2]17. The suction valve of a four stroke petrol engine is operated by a circular arc cam with a flat faced

follower. The lift of the follower is 10 mm ; base circle diameter of the cam is 40 mm and the noseradius is 2.5 mm. The crank angle when suction valve opens is 4° after top dead centre and whenthe suction valve closes, the crank angle is 50° after bottom dead centre. If the cam shaft rotates at600 r.p.m., determine: 1. maximum velocity of the valve, and 2. maximum acceleration and retarda-tion of the valve.

[Ans. 1.22 m/s ; 383 m/s2, 108.6 m/s2][Hint. Total angle turned by the crankshaft when valve is open

= 180° – 4° + 50° = 226°

Since the engine is a four stroke cycle, therefore speed of cam shaft is half of the speed of the crankshaft.

∴ Total angle turned by the cam shaft during opening of valve, 2 α = 226/2 = 113° orα = 56.5°].

18. The following particulars relate to a symmetrical circular cam operating a flat-faced follower :

Least radius = 25 mm ; nose radius = 8 mm, lift of the valve = 10 mm, angle of action of cam =120°, cam shaft speed = 1000 r.p.m.

Determine the flank radius and the maximum velocity, acceleration and retardation of the follower. Ifthe mass of the follower and valve with which it is in contact is 4 kg, find the minimum force to beexerted by the spring to overcome inertia of the valve parts.

[Ans. 88 mm ; 1.93 m/s, 690.6 m/s2, 296 m/s2 ; 1184 N]

DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?1. Write short notes on cams and followers.

2. Explain with sketches the different types of cams and followers.3. Why a roller follower is preferred to that of a knife-edged follower ?4. Define the following terms as applied to cam with a neat sketch :-

(a) Base circle, (b) Pitch circle, (c) Pressure angle, and (d) Stroke of the follower.

5. What are the different types of motion with which a follower can move ?6. Draw the displacement, velocity and acceleration diagrams for a follower when it moves with simple

harmonic motion. Derive the expression for velocity and acceleration during outstroke and returnstroke of the follower.

7. Draw the displacement, velocity and acceleration diagrams for a follower when it moves with uni-form acceleration and retardation. Derive the expression for velocity and acceleration during out-stroke and return stroke of the follower.

8. Derive expressions for displacement, velocity and acceleration for a tangent cam operating on aradial-translating roller follower :

Chapter 20 : Cams � 831(i) when the contact is on straight flank, and(ii) when the contact is on circular nose.

9. Derive the expressions for displacement, velocity and acceleration for a circular arc cam operating aflat-faced follower(i) when the contact is on the circular flank, and(ii) when the contact is on circular nose.

OBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONS1. The size of a cam depends upon

(a) base circle (b) pitch circle (c) prime circle (d) pitch curve

2. The angle between the direction of the follower motion and a normal to the pitch curve is called

(a) pitch angle (b) prime angle

(c) base angle (d) pressure angle

3. A circle drawn with centre as the cam centre and radius equal to the distance between the camcentre and the point on the pitch curve at which the pressure angle is maximum, is called

(a) base circle (b) pitch circle

(c) prime circle (d) none of these

4. The cam follower generally used in automobile engines is

(a) knife edge follower (b) flat faced follower

(c) spherical faced follower (d) roller follower

5. The cam follower extensively used in air-craft engines is

(a) knife edge follower (b) flat faced follower

(c) spherical faced follower (d) roller follower

6. In a radial cam, the follower moves

(a) in a direction perpendicular to the cam axis

(b) in a direction parallel to the cam axis

(c) in any direction irrespective of the cam axis

(d) along the cam axis

7. A radial follower is one

(a) that reciprocates in the guides (b) that oscillates

(c) in which the follower translates along an axis passing through the cam centre of rotation.

(d) none of the above

8. Offet is provided to a cam follower mechanism to

(a) minimise the side thrust (b) accelerate

(c) avoid jerk (d) none of these

9. For low and moderate speed engines, the cam follower should move with

(a) uniform velocity (b) simple harmonic motion

(c) uniform acceleration and retardation (d) cycloidal motion

10. For high speed engines, the cam follower should move with

(a) uniform velocity (b) simple harmonic motion

(c) uniform acceleration and retardation (d) cycloidal motion

11. Which of the following displacement diagrams should be chosen for better dynamic performance ofa cam-follower mechanism ?

(a) simple hormonic motion (b) parabolic motion

(c) cycloidal motion (d) none of these


12. For a given lift of the follower of a cam follower mechanism, a smaller base circle diameter isdesired.

(a) because it will give a steeper cam and higher pressure angle.(b) because it will give a profile with lower pressure angle(c) because it will avoid jumping(d) none of the above.

13. The linear velocity of the reciprocating roller follower when it has contact with the straight flanks ofthe tangent cam, is given by

(a) 1 2( )sinr rω − θ (b) 1 2( )cosr rω − θ

(c) 1 22( )sin secr rω + θ θ (d) 1 2

2( )cos cosecr rω + θ θwhere ω = Angular velocity of the cam shaft,

r1 = Minimum radius of the cam, r2 = Radius of the roller, and

θ = Angle turned by the cam from the beginning of the displacement for contact

of roller with the straight flanks.14. The displacement of a flat faced follower when it has contact with the flank of a circular arc cam, is

given by

(a) (1 cos )− θR (b) (1 sin )− θR

(c) 1( )(1 cos )R r− − θ (d) 1( )(1 sin )R r− − θwhere R = Radius of the flank,

r1 = Minimum radius of the cam, and θ = Angle turned by the cam for contact with the circular flank.

15. The retardation of a flat faced follower when it has contact at the apex of the nose of a circular arccam, is given by

(a) 2ω ×OQ (b) 2 sinω × θOQ

(c) 2 cosω × θOQ (d) 2 tanω × θOQ

where OQ = Distance between the centre of circular flank and centre of nose.

ANSWERSANSWERSANSWERSANSWERSANSWERS1. (a) 2. (d) 3. (b) 4. (c) 5. (d)6. (a) 7. (a) 8. (a) 9. (b) 10. (d)

11. (c) 12. (d) 13. (c) 14. (c) 15. (a)

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