Homework 6 – Instant CentersProblem 6.12 Find all the instant centers of the linkages shown in Figure 6-5.

3I23

I34

I12

I14I14I14

I13

I24

a) This problem is relatively straightforward – it is the same as the slider-crank linkage. Since it has four links, it has six instant centers. The instant center I14 lies at infinity.

I12, I24, I14

I34

I23

I13

b) This problem is a little trickier. The instant center between two gears is the point of contact between the two gears, since the point where the teeth mesh must have the same velocity. The instant center I13 is located where gear 3 makes rolling contact with the internal gear (body 1). The instant center between links 2 and 4 (the center gear and the bar) is at

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the ground pivot, since the center gear and the bar have the same velocity at that point. Finally, the center gear and the bar are pinned to ground at the ground pivot, so instant centers I12 and I14 are located there as well. Interestingly, all six instant centers lie on the same straight line.

I14I12

I23

I13 at infinity

I24 at infinityI34

I13 at infinity

I24 at infinity

c) The lines connecting instant centers I23-I34 and I12-I14 are parallel, since the heights of the pins on the wheels are equal. Since these lines are parallel, they intersect at infinity. Thus, instant center I24 exists on a horizontal line at infinity. Similarly, the lines between the contact points and the pins of each wheel are parallel, so the instant center I13 is also at infinity.

I12

I23

I34

I14 at infinity

I24

I14 at infinity

I13

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d) The instant center I14 is at infinity along a line perpendicular to the slider. Since the cam and follower make rolling contact, their point of contact is the instant center I23. The other two can be found using Kennedy’s rule.

I13

I12 I23

e) The two cams slide past each other at their point of contact. The instant center between them must lie along a line perpendicular to the direction of sliding, as shown in the figure above. Using Kennedy’s rule, we can find the exact location of I23 by finding the intersection between the perpendicular and the line between I12 and I13.

I24 at infinity

I14 at infinity

I12 at infinty

I23

I34

I13

f) The location of five of the six instant centers is relatively straightforward, but the sixth (I24) is a challenge. The textbook solution claims that it lies at infinity along the line between I34 and I23, but it is unclear how this would intersect the line between I14 and I12 (both of which also lie at infinity. Extra credit, anyone?

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6-16: The linkage in Figure P6-5 has O2A = 0.8, AB = 1.93, AC = 1.33, and offset = 0.38in. The crank angle in the position shown is 34.3° and angle BAC = 38.6°. Find ω3, VA, VB and VC for the position shown for ω2 = 15 rad/s in the direction shown.

V A=15rads·0.8∈¿12 ¿

s

ω3=12 ¿s

2.118∈¿=5.666rads

¿

V C=5.666rads·1.014∈¿5.745 ¿

s

V B=5.666rads·2.045∈¿11.587 ¿

s

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VA

VB

VC

6.18b: The linkage in Figure 6-5f has AB = 1.8 and AC = 1.44in. The angle of AB in the position shown is 128° and angle BAC = 49°. The slider at B is at an angle of 59°. Find ω3, VA, VB, and VC for the position shown for VA = 10in/sec in the direction shown.

ω3=10 ¿s

0.753∈¿=13.28rads

¿

V C=13.28rads·1.293∈¿17.17 ¿

s

V B=13.28rads·0.716∈¿9.51 ¿

s

5

VA

VC

I13

VB

6.21b: The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1mm. θ2 is 68.3° in the xy coordinate system, which is at -23.3° in the XY coordinate system. The X component of O2C is 59.2mm. For the position shown, find the velocity ratio VI56/VI23 and the mechanical advantage from link 2 to link 6.

First, find the needed instant centers, as shown in the diagram above. We’ll choose an arbitrary velocity for point A, and use this to find the angular velocity of link 3.

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Let us arbitrarily assume that

V A=1mms

The distance between instant center I13 and point A is 44.679mm. Thus

ω3=1mms

44.679mm=0.0224

rads

And the distance between instant center I13 and point B is 50.175mm. The speed at point B is

V B=0.0224rads·50.175mm=1.123mm

s

The figure above shows a magnified view of link 5, with the dimensions between instant center I15 and points B and C. We can use the velocity at point B to find the angular velocity of link 5.

ω5=1.123

mms

22.774mm=0.0493

rads

Finally, the velocity at point C is

V C=0.0493rads·11.12mm=0.5483 mm

s

Thus, the velocity ratio between point A and point C is

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V I 56V I 23

=0.5483

mms

1mms

=0.548

For the second part of the problem, we will use equations 6.12 and 6.13 to find the mechanical advantage between links 2 and 6. The definition of mechanical advantage is

mA=FoutF¿

The input power is

P¿=T¿ω¿=F¿ r¿ω2=F¿ L2ω2

Thus,

F ¿=P¿

L2ω2

The output power is

Pout=FoutV out=FoutV C

Thus

Fout=PoutV C

Substituting these into the expression for mechanical advantage gives

mA=P¿

L2ω2·V CPout

If we assume that input and output powers are equal (100% efficiency) then

mA=V CL2ω2

=V CV A

=1.82

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