Allen Bradley Drive Sizing

Guide to Drives

Allen-Bradley 353

Measuring Torque

If the amount of torque required to drive a machine cannot be determined from the machine builder, it can be easily measured. Fasten a pulley securely to the shaft which the motor is to drive. Secure one end of a cord to the outer surface of the pulley and wrap a few turns of the cord around the pulley. Tie the other end of the cord to a spring scale (see Figure 28).

Figure 28 Measuring Machine Torque Required

Pull on the scale until the shaft turns. The force in pounds indicated on the scale, multiplied by the radius of the pulley (from the centerline to the outside edge) in feet gives the torque value in lb-ft. On some machines, this torque may vary as the shaft rotates. The highest value of torque must be used in selecting a drive.

The running torque required by a machine will be approximately equal to the starting torque, if the load is composed almost entirely of friction. When the load is primarily inertia or windage, the characteristics of the inertia or windage producing elements must be determined.

The running torque of a machine can be accurately determined. An armature controlled DC drive of known horsepower rating can be used in making a test run (along with a shunt wound or permanent magnet DC motor). The DC drive should have an ammeter in the armature circuit so significant current readings can be observed and recorded throughout the speed range of the machine. Since armature current and torque are directly proportional, the current readings will provide accurate information for selecting the drive rating required by the machine.

Most machines require a higher torque value at breakaway. Once the machine is running, the torque requirement will decrease. Many drives have 150% load capability for 1 minute, which may allow the required additional breakaway torque to be obtained without increasing the drive horsepower rating.

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354 Allen-Bradley

General Rules

If the running torque is equal to or less than the breakaway torque divided by 1.5, use the breakaway torque divided by 1.5 as the full-load rated torque required to determine the motor horsepower.

If the running torque is greater than breakaway torque divided by 1.5, but less than the breakaway torque, use the running torque as the full-load rated torque required to determine the motor horsepower.

Calculating Torque

(Acceleration Torque Required for Rotating Motion)Some machines must be accelerated to a given speed in a certain period of time. The torque rating of the drive may have to be increased to accomplish this objective. The following equation may be used to calculated the average torque required to accelerate a known inertia (WK2). This torque must be added to all the other torque requirements of the machine when determining the drive and motors required peak torque output.

Equation 5

Where:T = Acceleration torque (lb-ft.)WK2 = total system inertia (lb-ft.2) that the motor must accelerate.

This value includes motor armature, reducer and load. N = Change in speed required (RPM)t = time to accelerate total system load (seconds)

Note: The number substituted for (WK2) in Equation 5 must be in units of lb-ft.2.

The same formula can also be used to determine the minimum acceleration time of a given drive, or if a drive can accomplish the desired change in speed within the required time period.

Equation 6Rearranged Equation:

Most drives have 150% load capability for 1 minute. Therefore additional acceleration torque will be available without increasing the drive horsepower rating.

T WK2 N( )308t

-----------------------------=

t WK2 N( )308T

--------------------------------=

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Allen-Bradley 355

General Rules

If the running torque is equal to or less than the accelerating torque divided by 1.5, use the accelerating torque divided by 1.5 as the full-load torque required to determine the motor horsepower.

If the running torque is greater than accelerating torque, use the running torque as the full-load rated torque required to determine the motor horsepower.

Calculating Horsepower

Note: The following equations for calculating horsepower are meant to be used for estimating purposes only. These equations do not include any allowance for machine friction, windage or other factors. These factors must be considered when selecting a drive for a machine application.

Once the machine torque is determined, the required horsepower is calculated using the formula:

Equation 7

Where:HP = HorsepowerT = Torque (lb-ft.)N = Speed of motor at rated load (RPM)

If the calculated horsepower falls between standard available motor ratings, select the higher available horsepower rating. It is good practice to allow some margin when selecting the motor horsepower.

For many applications, it is possible to calculate the horsepower required without actually measuring the torque required. The following equations will be helpful:

Conveyors

Equation 8

Equation 9

HP T N5252------------=

HP (Vertical) Force/Weight (lbs.) Velocity (FPM)33 000,

--------------------------------------------------------------------------------=

HP (Horizontal) Force/Weight (lbs.) Velocity (FPM) Coef. of Friction33 000,

-------------------------------------------------------------------------------------------------------------------------=

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356 Allen-Bradley

Web Transport Systems and Surface Winders

Equation 10

The tension value used in this calculation is the actual web tension for Surface Winder applications.

The above tension value is the tension differential (Downstream Tension Upstream Tension) for sectional drives.

Constant Tension Center Winders (Armature Control Only)

Equation 11

Note: The required HP (armature control tension) can be reduced by taper ratio.

Constant Tension Center Winders (Field Control Only)

Equation 12

Fans and Blowers

Equation 13

Equation 14

Equation 15

HP Tension (lbs.) Line Speed (FPM)33 000,

----------------------------------------------------------------------------=

HPMax. Tension (lbs.) Max. Line Speed (FPM)

33 000,---------------------------------------------------------------------------------------------------- Buildup=

HP Tension (lbs.) Line Speed (FPM)33 000,

----------------------------------------------------------------------------=

HP CFM Pressure lbs/ft.2( )

33 000, (Efficiency of Fan)--------------------------------------------------------------=

HP CFM Pressure lbs/in.2( )

229 (Efficiency of Fan)---------------------------------------------------------=

HP CFM (Inches of Water Gauge)6356 (Efficiency of Fan)

----------------------------------------------------------------------=

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Allen-Bradley 357

Pumps

Equation 16

PSI = pounds per square inchCFM = cubic feet per minuteGPM = gallons per minuteSpecific gravity of water = 1.01 Cubic foot per second = 448 GPM1 PSI = A head of 2,309 ft. for water weighing 62.36 lbs per cu. ft. at 62FEfficiency of fan or pump = %/100Displacement pump efficiency:Displacement pumps may vary between 50 and 80% efficiency depending on size of pumps.Centrifugal pump efficiency:

500 to 100 gal. per min = 70 to 75%1000 to 1500 gal. per min = 75 to 80%Larger than 1500 gal. per min. = 80 to 85%

Inertia

Inertia is a measure of a bodys resistance to changes in velocity, whether the body is a rest or moving at a constant velocity. The velocity can be either linear or rotational.

The movement of Inertia (WK2) is the product of the weight (W) of an object and the square of the radius of gyration (K2). The radius of gyration is a measure of how the mass of the object is distributed about the axis of rotation. Because of this distribution of mass, a small diameter cylindrical part has a much lower inertia than a large diameter part.

WK2 or WR2

WR2 refers to the inertia of a rotating member that was calculated by assuming the weight of the object was concentrated around its rim at a distance R (radius) from the center (e.g. flywheel).

WK2 refers to the inertia of a rotating member that was calculated by assuming the weight of the object was concentrated at some smaller radius, K (termed the radius of gyration). To determine the WK2 of a part, the weight is normally required (e.g. cylinder, pulley, gear).

HP 3960 (Efficiency of Pump)GPM Head (ft.) (Specific Gravity)---------------------------------------------------------------------------------=

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Calculations

When performing inertia calculations, the two terms commonly used are lb-ft.2 and in-lb-sec.2. Many standard inertia calculations are defined in lb-ft.2. However, in some motion control applications, such as a servo system, inertias are defined in in-lb-sec.2. When performing inertia calculations, be consistent with the formulas and units used. An answer in lb-ft.2 cannot be substituted for an answer in units of in-lb-sec.2. The following examples will show calculations using both sets of terms.

Cylinders

Inertia calculations of a cylinder.

Solid CylindersEquation 17

WK2 0.000681 L (D)4=

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Allen-Bradley 359

Hollow CylindersEquation 18

WK2 = inertia of a cylinder (lb-ft.2) = density of cylinder material in lb/in.3 (see density chart below)D1 = inside diameter of cylinder (inches)D2 = outside diameter of cylinder (inches)L = length of cylinder (inches)

Common Material Densities ()Aluminum = 0.0977Brass = 0.311Cast Iron = 0.2816Steel = 0.2816In motion control applications, the moment of inertia calculations (J) for cylinders can be performed using the following equations. Note that the answer will be in units of in-lb-sec.2.

Solid Cylinders:Inertia calculations of a cylinder (Motion Control Applications)

Equation 19

WK2 0.000681 L D2( )4 D1( )4[ ]=

J0.5 D( )

4

16------------- L

386.4----------------------------------------------------------=

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Hollow Cylinders:Equation 20

J = inertia of a cylinder (in-lb-sec.2) = 3.141 = density of cylinder material in lb/in.3 (see density chart below)D1 = inside diameter of cylinder (inches)D2 = outside diameter of cylinder (inches)L = length of cylinder (inches)

Common Material Weight Densities ()Aluminum = 0.0977Brass = 0.311Cast Iron = 0.2816Steel = 0.2816

Pulley/Gear

To calculate the inertia of a pulley or gear, divide up the piece as shown in Figure 29. Using the same equation for calculating hollow cylinders (Equation 21), perform the calculations of each separate part and add them together for a total inertia. Note: WK12 and WK22 are the separate inertia calculations.

J

0.5 D2( )4 D1( )4

16----------------------------------

L

386.4---------------------------------------------------------------------------------=

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Figure 29 Inertia of a Pulley or Gear

Equation 21

Note: To convert lb-ft.2 to in-lb-sec.2: divide above answer by 2.68.

Figure 30 shows the same examples as Figure 29, but shows calculations related to motion control applications. Because the resulting answer needs to be in in-lb-sec2, the hollow cylinder equation (Equation 22) is used.

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362 Allen-Bradley

Figure 30 Inertia of a Pulley of Gear (Motion Control Applications)

Equation 22

Total Inertia = J1 + J2 = 0.009 + 0.0051 = 0.0141 in-lb-sec2

Note: To convert in-lb-sec.2 to lb-ft 2: multiply above answer by 2.68

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Calculations Using Weight Or Dimensions

Inertia equations require weight and size of the particular object. The weight of an object will generally give the best value of inertia. However, in some motion control applications, the weight is not known. Therefore in those applications, the equation will depend on size, which provides a good estimate of inertia.

To calculate the inertia of a shaft (J), use the equations in Figure 31 when either the weight or dimensions of the shaft are known. Note: The resulting answer will be in units of in-lb-sec.2.

Figure 31 Inertia Calculations using Dimensions or Weight

Equation 23

Equation 24

Equation 25

Equation 26

J.5 D( )

4

16------------- L

386.4-------------------------------------------------------=

J.5 3.141 0.2816

1( )416

------------ 1386.4

--------------------------------------------------------------------------- .000072 in-lb-sec2==

Volume of the rod 3.141 (R)2

Length=

3.141 (.5)2

1=

.785 in3

=

Weight of the rod VOLUME DENSITY=

.785 in3

.2816=

.221 lbs=

lbs

in3

--------

JW

386.4---------------

R

2 -------------------- 2=

J.221

386.4---------------

.5

2 -------------------- 2 .000072 in-lb-sec2==

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364 Allen-Bradley

Speed Reducer Selection

Note: Adjustable speed drive motors can be damaged by continuous operation at low speed and rated torque. The motors cooling ability diminishes as speed is reduced. Servomotors, however, can supply full rated torque down to zero speed.

The motor should always be coupled to the driven machine by a power transmission that will permit maximum motor RPM at maximum machine speed. The power transmission may be a simple belt-sheave or sprocket-chain arrangement or a compact gear reducer. In applications requiring speed reductions greater than 5:1, the gear reducer may be the most economical choice.

Gear Reducer Selection

A gear reducer transmits power by an arrangement of various forms of gears. It provides an efficient, positive drive to change speed, direction and torque. This may mean a change of speed with a corresponding change in torque or a change in output direction or position. A common result is a combination of the above.

The gear reducer serves as a torque amplifier, increasing the torque by a factor proportional to the reducer ratio, less an efficiency factor. See Figure 32.

Figure 32 Gear Reducer Characteristics

A 1 HP, 1750 RPM motor has an output torque of 3 lb-ft. If a 30:1 ratio reducer with 85% efficiency is used, the reducer output torque will be 3 x 30 x 0.85 = 76.5 lb-ft.

A typical application involves selecting a gear reducer that permits the drive motor to operate at base speed when the driven machine is at maximum speed. The gear reducer should also provide adequate torque to drive the machine.

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Allen-Bradley 365

Application ExampleA 1750 RPM motor is selected for a machine which is to operate at a 57.5 RPM maximum speed and requires 70 lb-ft. of torque.

To find the answer, the following two steps must be accomplished.

1. Determine the required ratio:

Equation 27

Note: When the calculated reducer ratio is not a standard speed reducer ratio, a chain, belt or additional gears with further reduction are necessary (located on the input or output side).

2. Determine the motor torque and horsepower:

A 30:1 gear reducer is selected which is capable of supplying 70 lb-ft. of output torque. Since the machine torque requirement is known, this value is divided by the reduction ratio and efficiency factor to arrive at required motor torque (TM).

Equation 28

Since a 1 HP, 1750 RPM base motor delivers 3 lb-ft. of torque, it is chosen for this application along with a 30:1 gear reducer with a minimum of 70 lb-ft. output torque.

Where the reduction ratio permits the use of a chain or belt, the same formulas are used as for reducers.

Gear Reducer, Overhung Load

An overhung load (OHL) is defined as dead weight the gear reducer bearings can support on an output shaft at a distance equal to the shaft diameter. This distance is measured from the outside end of the bearing housing along the shaft (see the following figure). If the acting load is at a point different from the OHL point, it must be converted to the reference point and compared to the manufacturers catalogued value.

Reducer Ratio Maximum Motor RPMMaximum Driven Machine RPM------------------------------------------------------------------------------------=

Reducer Ratio 175057.5------------- 30.4 or a 30:1 ratio= =

TMRequired Torque (lb-ft.)

Reducer Ratio Efficiency Factor--------------------------------------------------------------------------=

TM70

30 0.85--------------------- 2.75 lb-ft.= =

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366 Allen-Bradley

When a gear reducer is driven by a belt, chain or gear drive, or when the gear reducer drives a driven unit through a belt chain or gear drive, an overhung load (radial thrust) is produced. The overhung load must not exceed the rating of the gear reducer as listed by the manufacturer. The magnitude of the overhung load should always be kept to a minimum. Excessive loads could lead to fatigue failure of either bearing or shaft. The sprocket or pulley should always be located as close to the gear housing as possible.

Increasing the sprocket or pulley diameter results in a reduced overhung load. Use the following equation to determine the overhung load:

Equation 29

Note: K is constant which is:

1.0 for chain drives1.25 for gears or gear-belt drives1.50 for V-belt drive2.50 for flat-belt drivesNo overhung loads are encountered when the gear reducer is directly coupled to the motor and/or driven machine shaft. However, care must be taken in aligning the shafts to avoid pre-loading bearings by misalignment.

OHL (lb) 2 Shaft Torque (lb-in.) KDiameter sprocket, sheave, pulley, gear (in.)------------------------------------------------------------------------------------------------=

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Tables and Formulas Introduction

The data contained in this section is provided for reference only. Many formulas are for estimating only because they cannot consider all factors in every machine application. Many formulas can assist the reader by demonstrating basic physical or electrical relationships needed to understand a more abstract concept in control or motor operation.

Other data such as conversion factors is included for your convenience to provide a more comprehensive resource when working in an international design environment.

Note: The following equations for calculating horsepower are meant to be used for estimating purposes only. These equations do not include any allowance for machine friction, windage or other factors. These factors must be considered when selecting a drive for a machine application.

Horsepower Formulas

Rotating Objects

Where:T = Torque (lb-in)N = Speed (RPM)

Where:T = Torque (lb-ft)N = Speed (RPM)

Objects in Linear Motion

Where:F = Force (lb.)V = Velocity (ft./min.)

Where:F = Force (lb.)V = Velocity (in./min.)

HP T N63 000,----------------=

HP T N5252------------=

HP F V33 000,----------------=

HP F V396 000,-------------------=

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For Pumps

Where:GPM = Gallons per MinuteHead = Height of Water (ft.)Efficiency of Pump = %/100PSI = Pounds per Square Inch

Specific Gravity of Water = 1.01 Cu. Ft. Per Sec. = 448 GPM1 PSI = A head of 2.309 ft. - for water weighing 62.36 lbs. per cu. ft. at 62F

For Fans and Blowers

Where:CFM = Cubic Feet per MinutePSF = Pounds per Square FootPIW = Inches of Water GaugePSI = Pounds per Square InchEfficiency of Fan = %/100

For Conveyors

Where:F = Force (lbs.)V = Velocity (ft./min.) Ball or Roller Slide = .02

Coef. of Friction:Dovetail Slide = .20Hydrostatic Ways = .01Rectangle Ways with Gib = .1 - .25

HP GPM Head Specific Gravity3960 Efficiency of Pump

---------------------------------------------------------------------=

HP GPM PSI Specific Gravity1713 Efficiency of Pump

-----------------------------------------------------------------=

HP CFM PSF33 000, Efficiency of Fan-----------------------------------------------------------=

HP CFM PIW6356 Efficiency of Fan-----------------------------------------------------=

HP CFM PSI229 Efficiency of Fan--------------------------------------------------=

HP (Vertical) F V33 000,----------------=

HP (Horizontal) F V Coef. of Friction33 000,

-----------------------------------------------------=

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Torque Formulas

Where:T = Torque (lb-ft.)HP = HorsepowerSpeed (RPM)

Where:T = Torque (lb-ft.)F = Force (lbs.)R = Radius (ft.)

Where:T = Torque (lb-ft.)WK2 = Inertia reflected to the Motor Shaft (lb-ft.2)RPM = Change in Speedt = Time to Accelerate (sec.)

Note: To change lb-ft.2 to in-lb-sec.2: Divide by 2.68To change in-lb-sec.2 to lb-ft2: Multiply by 2.68

AC Motor Formulas

Where:Sync Speed = Synchronous Speed (RPM)Freq. = Frequency (Hz)

Where:FL Speed = Full Load Speed (RPM)Sync Speed = Synchronous Speed (RPM)

Where:WK2 = Inertia (lb-ft.2)

T HP 5252N

------------------------=

T F R=

T (accelerating) WK2 RPM308 t

--------------------------------=

Sync Speed Freq. 120Number of Poles----------------------------------------=

% Slip Sync Speed FL Speed( ) 100Sync Speed

-------------------------------------------------------------------------=

Reflected WK2 WK2 of Load

(Reduction Ratio)2--------------------------------------------------=

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Electrical Formulas

Ohms Law

Where:I = Current (Amperes) E = EMF or Voltage (Volts)R = Resistance (Ohms)

Power In DC Circuits

Where:P = Power (Watts)I = Current (Amperes)E = EMF or Voltage (Volts)kW = KilowattskWH = Kilowatt-Hours

Power In AC Circuits

Where:kVA = Kilovolt-AmperesI = Current (Amperes)E = EMF or Voltage (Volts)

Where:KW = KilowattsI = Current (Amperes)E = EMF or Voltage (Volts)PF = Power Factor W = WattsV = VoltskVA = Kilovolt-Amperes

IER----= R

EI----= E I R=

P I E=

kW I E1000----------=

HP I E746----------=

kWH I E Hours1000

------------------------------=

kVA (1) I E1000----------= kVA (3) I E 1.73

1000--------------------------=

kW (1) I E PF1000

-----------------------=

kW (2) I E PF 1.421000

---------------------------------------=

kW (3) I E PF 1.731000

---------------------------------------=

PF WV I----------

kWkVA-----------= =

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Calculating Motor Amperes

Where:HP = HorsepowerE = EMF or Voltage (Volts)Eff = Efficiency of Motor (%/100)kVA = Kilovolt AmpereskW = KilowattsPF = Power Factor

Other Formulas

Calculating Accelerating Force For Linear Motion

Where:F = Force (lbs.)W = Weight (lb.)V = Change in Velocity (FPM)t = Time to accelerate weight (seconds)

Calculating Minimum Accelerating Time of a Drive

Where:t = Time required to accelerate load (seconds)WK2 = Total inertia that the motor must accelerate (lb-ft.2).

This includes motor rotor gear reducer and load.N = Change in speed required (RPM)T = Accelerating torque (lb-ft.)

Note: To change lb-ft.2 to in-lb-sec.2: Divide by 2.68To change in-lb-sec.2 to lb-ft.2: Multiply by 2.68

Where:RPM = Revolutions per MinuteFPM = Feet per MinuteD = Diameter (ft.)

Motor Amperes HP 746E 1.732 Eff PF----------------------------------------------=

Motor Amperes kVA 10001.73 E

--------------------------=

Motor Amperes kW 10001.73 E PF-------------------------------=

F (Acceleration) W V1933 t------------------=

t WK2 N

308 T------------------------=

RPM FPM.262 D-------------------=

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Where:WK2 = Inertia (lb-ft.2)RPM = Revolutions per Minute

Engineering Constants

Temperature

Length

Weight

WK2 Reflected to Motor Load WK2Load RPMMotor RPM------------------------------

2=

0C = Freezing point of water

32F = Freezing point of water = 0C

100C = Boiling point of water at atmospheric pressure

212F = Boiling point of water at atmospheric pressure

1.85DF = 15DC

0.252 Calories = 1 Btu

-270C = Absolute Zero

-459.6F = Absolute Zero

1760 Yards = 1 Mile

25.4 mm = 2.54 cm = 1 in.

3 ft. = 1 Yard

3.2808 ft = 1 m

39.37 in. = 1 m = 100 cm = 1000 mm

5280 ft. = 1 Mile

0.62137 Miles = 1 km

16 oz. = 1 lb.

2.2046 lb. = 1 kg.

2.309 ft. water at 62F = 1 PSI

28.35 gm = 1 oz.

59.76 lbs. = Weight of 1 cu. ft. of water at 212F

0.062428 lb. per cu = 1 kg/cu. m

62.355 lbs. = Weight of 1 cu. ft. water at 62F

8-1/3 (8.32675) lbs. = Weight 1 gallon water at 62F

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Power

Area

Mathematic

Pressure

1.3410 HP = 1 kW

2,545 Btu per hour = 1 HP

33,000 ft-lb. per minute = 1 HP

550 ft-lb. per second = 1 HP

745.7 W = 1 HP

10.764 sq. ft. = 1 sq. m

1,273,239 circular mils = 1 sq. in.

144 sq. in = 1 sq. ft.

645 mm2 = 1 sq. in.

9 sq. ft. = 1 sq. yard

0.0929 sq. m = 1 sq. ft.

1.4142 = square root of 2

1.7321 = square root of 3

3.1416 = = ratio of circumference of circle to diameter = ratio of area of a circle to square of radius

57.296 = 1 rad. (angle)

0.7854 x diameter squared = area of a circle

14.223 PSI = 1 kg per sq. cm = 1 metric atmosphere

2.0355 in. Hg at 32F = 1 PSI

2.0415 in. Hg at 62F = 1 PSI

2,116.3 PSF = atmospheric pressure

27.71 in. water at 62F = 1 PSI

29.921 in. Hg at 32F = atmospheric pressure

30 in. Hg at 62F = atmospheric pressure (approximate)

33.974 ft. water at 62F = atmospheric pressure

0.433 PSI = 1 ft. of water at 62F

5196 PSF = 1 in. water at 62F

760 mm Hg = atmospheric pressure at 0C

0.07608 lb. = weight 1 cu. ft. air at 625F and 14.7 PSI

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Volume

Standard Abbreviation Descriptions

Temperature

Length

Weight

1,728 cu. in. = 1 cu. ft.

231 cu. in = 1 gallon (U.S.)

277.274 cu. in = 1 gallon (British)

27 cu. ft. = 1 cu. yard

31 gallon (31.5 U.S. Gallons)

= 1 Barrel

35.314 cu. ft. = 1 cu. m

3.785 liters = 1 gallon

61.023 cu. in. = 1 liter

7.4805 gallon = 1 cu. ft.

deg = degrees

C = Celsius (Centigrade)

F = Fahrenheit

Btu = British Thermal Unit

yd. = yard

m = meter

mm = millimeter (1/1000 of a meter)

cm = centimeter 1/100 of a meter)

in. = inch

ft. = feet

km = kilometer

oz. = ounce

lb. = pound

kg. = kilogram

g = gram

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Electrical

Power/Energy

Work/Inertia

Area

Rotation Rate

= Ohms

= Phase

V = Volts

A = Amperes

a = milliamperes

ma = microamperes

mV = millivolts

KV = kilovolts

kVA = kilovolt-Ampere

HP = Horsepower

W = Watt

kW = Kilowatt

kWH = Kilowatt-Hours

ft-lb. = foot pounds

WK2 = moment of inertia

N-m2 = Newton meters2

sq. ft. = square feet

sq. m = square meters

mil = unit of length or angular measurement

mm2 = square millimeters

sq. in. = square inch

FPM = feet per minute

FPS = feet per second

m/s = meters per second

mph = miles per hour

cfm = cubic feet per minute

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Mathematic

Pressure

Volume

Conversion Factors

Length

Example: 10 Meters x 3.3281 = 32.81 Feet

= pi

rad. = radians

= Density

= Summation

= Change

kg. per sq. cm = kilograms per square centimeter

Hg = Mercury symbol

PSI = pounds per square inch

PSF = pounds per square foot

cu. = cubic

cu. in. = cubic inch

gal. = gallon

cu. ft. = cubic feet

ml = milliliter

fl. oz. = fluid ounce (U.S.)

To Convert: To: Multiply By:

Meters Feet 3.281

Meters Inches 39.37

Inches Meters 0.0254

Feet Meters 0.3048

Millimeters Inches 0.0394

Inches Millimeters 25.4

Thread/Inch Millimeter Pitch Divide Into 25.4

Yards Meters 0.914

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Area

(0.5 x 10-7 Circular Mils = .00000005 m2)

Example: 100 Circular Mils y 0.5 x 10-9 = 0.5 x 10-7 m2

Power

Example: 1500W x .00134 = 2.01 HP

Rotation/Rate

Example: 1800 RPM x 6.00 = 10800 deg./sec.


Circular mil Meter2 0.50 y 10-9

Yard2 Meter2 0.8361


Watts HP 0.00134

ft-lb./min. HP 0.0000303

HP kW 0.746


RPM deg./sec. 6.00

RPM Rad./sec. 0.1047

deg./sec. RPM 0.1667

Rad./sec. RPM 0.540

FPM m/s 0.00508

FPS m/s 0.3048

gal./min. cm3/sec. 63.09

in./sec. m/sec. 0.0254

km/hr. m/sec. 0.2778

mph m/sec. 0.447

mph km/hr. 1.609

RPM rad./sec. 0.1047

yd.3/min. m3/sec. 0.01274

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Moment of Inertia

Example: 25 Newton-Meters 2 x 2.42 = 60.5 lb-ft.2

Mass/Weight

Example: 50 oz. x 31.1 = 1555 deg./sec.

Torque

Example: 30 Newton-Meters x 0.7376 = 22.13 lb-ft.

Volume

Example: 250 cm3 x .000001 = .00025 m3


Newton-Meters2 lb-ft.2 2.42

oz-in.2 lb-ft.2 0.000434

lb-in2 lb-ft.2 0.00694

Slug-ft.2 lb-ft.2 32.17

oz-in-sec.2 lb-ft.2 0.1675

in-lb-sec.2 lb-ft.2 2.68


oz. g 31.1

kg lbs. 2.205

lb. kg 0.4536

Newtons lbs. 0.2248


Newton-Meters lb-ft. 0.7376

lb-ft. Newton-Meters 1.3558

lb-in. lb-ft. 0.0833

lb-ft. lb-in. 12.00


cm3(ml) m3 0.00001

fl. oz. cm3 29.57

ft.3 of water (39.2F) kg (or liter) 28.32

cfm m3/sec. 0.000472

liter m3 0.001

yd.3 m3 0.7646

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Temperature

Fractional Inch to Equivalent Millimeters and Decimals


deg. F deg. C deg. C = (deg. F - 32)1.8

deg. C deg. F deg. F = (deg. C x 1.8) + 32

Example: 62F 68 32( )1.8

-------------------------- 20C= =

20C 20 1.8( ) 32+ 68F= =

Inch

Equivalent

Inch

Equivalent

Inch

Equivalent

Inch

Equivalent

mm Decimal mm Decimal mm Decimal mm Decimal

1/64 .3969 .015617/64 6.7469 .2656

33/64 13.0969 .515649/64 19.4469 .7656

1/32 .7938 .03139/32 7.1438 .2813

17/32 13.4938 .531325/32 19.8438 .7813

3/64 1.1906 .046919/64 7.5406 .2969

35/64 13.8906 .546951/64 20.2406 .7969

1/16 1.5875 .06255/16 7.9375 .3125

9/16 14.2875 .562513/16 20.6375 .8125

5/64 1.9844 .078121/64 8.3344 .3181

37/64 14.6844 .578153/64 21.0344 .8281

3/32 2.3813 .093811/32 8.7313 .3438

19/32 15.0813 .593827/32 21.4313 .8438

7/64 2.7781 .109423/64 9.1281 .3594

39/64 15.4781 .609455/64 21.8281 .8594

1/8 3.1750 .12503/8 9.5250 .3750

5/8 15.8750 .62507/8 22.2250 .8750

9/64 3.5719 .140625/64 9.9219 .3906

41/64 16.2719 .640657/64 22.6219 .8906

5/32 3.9688 .156316/32 10.3188 .4063

21/32 16.6688 .656329/32 23.0188 .9063

11/64 4.3656 .171931/64 10.7156 .4219

43/64 17.0656 .671959/64 23.4156 .9219

3/16 4.7625 .18757/16 11.1125 .4375

11/16 17.4625 .687515/16 23.8125 .9375

13/64 5.1594 .203129/64 11.5094 .4531

46/64 17.8594 .703161/64 24.2094 .9531

7/32 5.5563 .218816/32 11.9063 .4688

23/32 18.2563 .718831/32 24.6063 .9688

15/64 5.9531 .234431/64 12.3031 .4844

47/64 18.6531 .734463/64 25.0031 .9844

1/4 6.3500 .25001/2 12.700 .5000

3/4 19.0500 .7500

Guide to Drives

380 Allen-Bradley

Inertia of a Solid Steel Shaft (lb-ft.2 per inch of length)

Diameter (Inches) Wk2






3/4 0.00006 8 0.791 141/4 7.97 37 360.70 62 2844.3 87 11028

1 0.00007 81/4 0.895 141/2 8.54 38 401.30 63 3032.3 88 11544

11/4 0.0005 81/2 1.000 14

3/4 9.15 39 445.30 64 3229.5 89 12077

11/2 0.001 83/4 1.130 15 9.75 40 492.78 65 3436.1 90 12629

13/4 0.002 9 1.270 16 12.59 41 543.90 66 3652.5 91 13200

2 0.003 91/4 1.410 17 16.04 42 598.80 67 3879.0 92 13790

21/4 0.005 91/2 1.550 18 20.16 43 658.10 68 4115.7 93 14399

21/2 0.008 93/4 1.750 19 25.03 44 721.40 69 4363.2 94 15029

23/4 0.011 10 1.930 20 30.72 45 789.30 70 4621.7 95 15679

3 0.016 101/4 2.130 21 37.35 46 861.80 71 4891.5 96 16349

31/2 0.029 101/2 2.350 22 44.99 47 939.30 72 5172 97 17041

33/4 0.038 103/4 2.580 23 53.74 48 1021.80 73 5466 98 17755

4 0.049 11 2.830 24 63.71 49 1109.60 74 5774 99 18490

41/4 0.063 111/4 3.090 25 75.02 50 1203.07 75 6090 100 19249

41/2 0.079 111/2 3.380 26 87.76 51 1302.20 76 6422

5 0.120 113/4 3.680 27 102.06 51 1407.40 77 6767 110 28183

51/2 0.177 12 4.000 28 118.04 53 1518.80 78 7125 120 39914

6 0.250 121/4 4.350 29 135.83 54 1636.70 79 7498 130 54978

61/4 0.296 121/2 4.720 30 155.55 55 1761.40 80 7885 140 73948

61/2 0.345 123/4 5.110 31 177.77 56 1893.10 81 8286 150 97449

63/4 0.402 13 5.58 32 201.80 57 2031.90 82 8703 160 126152

7 0.464 131/4 5.96 33 228.20 58 2178.30 83 9135 170 160772

71/4 0.535 131/2 6.42 34 257.20 59 2332.50 84 9584 180 20271

71/2 0.611 133/4 6.91 35 386.80 60 2494.70 85 10048 190 250858

73/4 0.699 14 7.42 36 323.20 61 2665.20 86 10529 200 307988

Guide to Drives

Allen-Bradley 381

Maintenance of Industrial Control Equipment

Periodic Inspection

Industrial control equipment should be inspected periodically. Inspection intervals should be based on environmental and operating conditions and adjusted as indicated by experience. An initial inspection within 3 to 4 months after installation is suggested. Applicable parts of the following guidelines should be used.

Contamination

If inspection reveals that dust, dirt, moisture or other contamination has reached the control equipment, the cause must be eliminated. This could indicate an incorrectly selected or ineffective enclosure, unsealed enclosure openings (conduit or other) or incorrect operating procedures. Replace any improperly selected enclosure with one that is suitable for the environmental conditions - refer to National Electrical Manufacturers Association (NEMA) Standards Publication No. 250 for enclosure type descriptions and test criteria. Replace any damaged or embrittled elastomer seals and repair or replace any other damaged or malfunctioning parts (e.g., hinges, fasteners, etc.). Dirty, wet, or contaminated control devices must be replaced unless they can be cleaned effectively by vacuuming or wiping.

!ATTENTION: Servicing energized industrial control equipment can be hazardous. Severe injury or death can result from electrical shock, burn, or unintended actuation of controlled equipment. Recommended practice is to disconnect and lockout control equipment from power sources, and release stored energy, if present. Refer to National Fire Protection Association Standard No. NFPA70E, Part II for safety related work practices including procedural requirements for lockout-tagout, and appropriate work practices, personnel qualifications and training requirements where it is not feasible to de-energize and lockout or tagout electric circuits and equipment before working on or near exposed circuit parts.

Guide to Drives

382 Allen-Bradley

Operating Mechanisms

Check for proper functioning and freedom from sticking or binding. Replace any broken, deformed or badly worn parts or assemblies according to individual product renewal parts lists. Check for and retighten securely and loose fasteners. Lubricate if specified in individual product instructions.

Note: Allen-Bradley magnetic starters, contactors, and relays are designed to operate without lubrication - do not lubricate these devices since oil or grease on the pole faces (mating surfaces) of the operating magnet may cause the device to stick in the ON mode. Some parts of other devices are factory lubricated - if lubrication during use or maintenance of these devices is needed, it will be specified in their individual instructions. If in doubt, consult the nearest Allen-Bradley sales office for information.

Contacts

Check contacts for excessive wear and dirt accumulations. Vacuum or wipe contacts with a soft cloth if necessary to remove dirt. Contact are not harmed by discoloration and slight pitting. Contacts should never be filed, as dressing only shortens contact life. Contact spray cleaners should not be used as their residues on magnet pole faces or in operating mechanisms may cause sticking, and on contacts can interfere with electrical continuity. Contacts should only be replaced after silver has become badly worn. Always replace contacts in complete sets to avoid misalignment and uneven contact pressure.

Terminals

Loose connections in power circuits can cause overheating that can lead to equipment malfunction or failure. Loose connections in control circuits can cause control malfunctions. Loose bonding or grounding connections can increase hazards of electrical shock. Check the tightness of all terminals and bus bar connections and tighten securely any loose connections. Replace any parts or wiring damage by overheating, and any broken wiring or bonding straps.

AC Hoods

Check for cracks, breaks or deep erosion. Arc hoods and arc chutes should be repaired or replaced if damaged or deeply eroded.

Guide to Drives

Allen-Bradley 383

Coils

If a coil exhibits evidence of overheating (cracked, melted or burned insulation), it must be replaced. In that event, check for and correct overvoltage or undervoltage conditions, which can cause coil failure. Be sure to clean any residues of melted coil insulation from other parts of the device or replace such parts.

Pilot Lights

Replace any burned out lamps or damaged lenses.

Photoelectric Switches

The lenses of photoelectric switches require periodic cleaning with a soft dry cloth. Reflective devices used in conjunction with photoelectric switches also require periodic cleaning. Do not use solvents or cleaning agents on the lenses or reflectors. Replace any damaged lenses and reflectors.

Solid State Devices

Solid state devices require little more than a periodic visual inspection. Printed circuit boards should be inspected to determine whether they are properly seated in the edge board connectors. Board locking tabs should also be in place. Necessary replacements should be made only at the PC board or plug-in component level. Solvents should not be used on printing circuit boards. Where blowers are used, air filters should be cleaned or changed periodically depending on the specific environmental conditions encountered.

!ATTENTION: Use of other than factory recommended test equipment for solid state controls may result in damage to the control or test equipment or unintended actuation of the controlled equipment.

Guide to Drives

384 Allen-Bradley

Locking and Interlocking Devices

Check these devices for proper working condition and capability of performing their intended functions. If necessary, readjust, repair or replace in accordance with individual product instructions.

Maintenance After a Fault Condition

Opening of the short circuit protective device (such as fuses or circuit breakers) in a properly coordinated motor branch circuit is an indication of a fault condition in excess of operating overload. Such conditions can cause damage to control equipment. Before restoring power, the fault condition must be corrected and any repairs or replacements must be made to restore the control equipment to good working order. Refer to NEMA standards Publication No. ICS-2, Part ICS2-302 for procedures.

Final Check Out

After maintenance or repair of industrial controls, always test the control system for proper functioning under controlled conditions that avoid hazards in the event of a control malfunction.

For additional information, refer to NFPA70B, RECOMMENDED PRACTICE FOR ELECTRICAL EQUIPMENT MAINTENANCE, published by the National Fire Protection Association.

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Allen Bradley Drive Sizing

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