RCC - PHD, Inc · For this application the 8 mm RCC will provide adequate torque at 87 psi. 4)...

www.phdinc.com/apps/sizing • (800) 624-8511

1�0

SIZE08

See Productivity Solutions (CAT-08) for ordering, dimensional, and options data.

ROTA

RIES

RCC ROTARY ACTUATOR

ROTATION90°180°90°180°90°180°

SIZE

08

12

16

BASEWEIGHTlb kg.32 .15.31 .14.65 .30.61 .281.20 .551.16 .53

DISPLACEmENTVOLUmE

in2 mm2

.146 240

.292 479

.263 432

.527 863

.514 8421.027 1683

ROTATIONALVELOCITY mAX.

deg/sec

180°/.16

180°/.24

180°/.24

THEORETICALTORQUE OUTPUTin-lb/psi Nm/bar

.018 .0021

.050 .0056

.137 .0155

mAX. AXIALBEARING LOAD

lb N

7.0 31.1

15.0 66.7

30.0 133.4

BOREDIAmETERin mm

.315 8

.472 12

.630 16

mAX. RADIALBEARING LOAD*

lb N

1.3 5.8

3.5 15.6

9.0 40.0

SPECIFICATIONSMIN. OPERATING PRESSUREMAX. OPERATING PRESSUREOPERATING TEMPERATURE RANGEROTATIONAL TOLERANCEBACKLASHDEGREES OF ROTATIONLUBRICATION

RCCx08 RCCx12 RCCx1630 psi [2.0 bar] 25 psi [1.7 bar] 20 psi [1.4 bar]

100 psi [7 bar]32 to 150°F [0 to 65°C]

Nominal +10° to -10° with angle adjustmentNo backlash at end of rotation

90° and 180°Permanent for non-lube air

NOTE: *At .5 in [12.7 mm] from hub face


1�1

SIZE08


ROTA

RIES

SERIES RCC SELECTIONTo select the appropriate Series RCC Rotary Actuator, it is

important to consider all factors that influence actuator life. The main factors for selecting the proper RCC Rotary Actuator are: radial bearing capacity, thrust bearing capacity, kinetic energy stopping capabilities, torque requirements, and rotation time. Follow the steps below to select the appropriate RCC actuator.

1) Determine the load information

Depending on the application, this may include the following information: a) Rotation angle and time to achieve full rotation b) Weight of load c) Radius of gyration d) Axis orientation e) Center of gravity (Cg) measured from the hub f) Operating pressure

2) Determine minimum actuator based on radial or axial load

a) Calculate moment created by the radial load. For radial load applications, the allowable load is based on the moment induced by the load. The Cg distance is shown below for a radial load application.

Moment = (Weight of Load) x (Cg Distance)

b) Select the minimum actuator based on the axial load capacity or calculate the moment induced by an unbalanced axial load.

For axial load applications where the load is on the hub centerline, the maximum load is based on the maximum allowable axial load, see the Maximum Bearing Capacity Table for the maximum allowable loads.

For unbalanced axial loads, see the Bearing Capacity Graph for the allowable loads.

For unbalanced axial loads with Cg distance greater than the hub radius, it is best to calculate the moment created by the off-center loads and size the actuator based on the maximum moment capacity.

c) Select the minimum actuator based on the maximum allowable loads by comparing the calculated moment to the values given in the Maximum Bearing Capacity Table and by taking axial load values from the Bearing Capacity Graph.

3) Determine torque requirements

a) Calculate Mass Moment of Inertia (Jm). Select the illustration from the application types on the following page.

b) Determine the required acceleration.

α = .035 x Rotational Angle (deg)/[Rotational Time (sec)]2

c) Calculate required torque. PHD recommends a minimum safety factor (SF) of 2 to account for friction loss, air line and valve size.

For balanced and unbalanced loads rotating without gravity, the following torque formula applies.

T = Jm x α x SF

For unbalanced loads rotating without gravity, see the unbalanced load application types for the appropriate torque formulas.

d) Calculate the Minimum Operating Pressure (see the Minimum Operating Pressure Table). This step will determine which actuator is capable of providing adequate torque. NOTE: When calculating minimum operating pressure, any unbalanced axial load with a Cg distance smaller than the hub radius will be treated as an axial load.

Using the theoretical torque values given in the Engineering Data section, select the minimum operating pressure. The moment calculated in 2a/2b and torque from 3c are used in the formulas.

If the calculated pressure is greater than or equal to the actual operating pressure, the next larger actuator should be used to provide adequate torque throughout the life of the actuator.

4) Determine the stopping capacity required for this application

a) Determine the impact velocity

ω (rad/sec) = .035 x

b) Using Jm calculated in step 3a and impact velocity from step 4a, determine the kinetic energy of the system by using the basic KE equation. Or you can select the appropriate actuator from the KE Capacity Chart using the Jm value and the impact velocity.

KE = x Jm x ω2

c) Use the Maximum Allowable Kinetic Energy Table to select appropriate RCC actuator.

12

Radial Load Cg Distance

Cg

Unbalanced Axial Load Cg Distance

Cg

Rotation Angle (deg)Rotation Time (sec)

RCC ACTUATOR


1��

SIZE08


ROTA

RIES

ImPERIAL UNITS:Jm = Rotational Mass Moment of Inertia (in-lb-sec2) (Dependent on physical size of object and weight)g = Gravitational Constant = 386.4 in/sec2 Fg = Weight of Load (lb) k = Radius of Gyration (in)T = Torque required to rotate load (in-lbs) α = Acceleration (rad/sec2) t = time (sec)SF = Safety Factor

mETRIC UNITS:Jm = Rotational Mass Moment of Inertia (N-m-sec2) (Dependent on physical size of object and weight)g = Gravitational Constant = 9.81 m/sec2 Fg = Weight of Load (N) k = Radius of Gyration (m)T = Torque required to rotate load (N-m) α = Acceleration (rad/sec2) t = time (sec)M = Mass = Fg / g (kg) SF = Safety Factor

BALANCED LOADST = Jm x α x SF

Jm = x k2Fg

g

Jm = x12

Fg

ga2 + 3k2

( ) ( )Fg1

gJm = x (4a2 + 3k2)

12+ xFg2

g(4b2 + 3k2)

12

b-a2

Tg = [(Jm x α) + (Fg x k)] x SFT = Jm x α x SF

( )k2Jm = +L2

3Fg

gxx 1

4

Tg = [(Jm x α) + [(Fg2 - Fg1) x (a + ( ))]] x SF

DiskEnd mounted on center

DiskMounted on center

Rectangular PlateMounted on center

Rectangular PlateMounted off center

Solid SphereMounted on center

RodMounted on center

RodMounted off center

Jm = x 4a2 + c2

12+ x 4b2 + c2

12Fg1

gFg2

g

Jm = x x k225

Fg

g

Point Load

LOAD ORIENTATION

Tg = Rotating Vertically(with gravity)

T = Rotating Horizontally(without gravity)

UNBALANCED LOADS

T = Jm x α x SF

UNBALANCED LOADS

Jm = x k2

2Fg

g

a2 + b2

12Jm = xFg

g

kL

k k

k dim isradiusof rod

aa

b

c

Fg2

aFg1

b


a Fg1

Fg2

b

kFg

RCC ACTUATOR


1��

SIZE08


ROTA

RIES

32

28

24

20

16

12

8

4

0

[142.3]

[124.6]

[106.8]

[89.0]

[71.2]

[53.4]

[35.6]

[17.8]

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2[5.08] [10.16] [15.25] [20.32] [25.4] [30.48] [35.56] [40.64] [45.72] [50.8]

LOAD

lb [N

]

CENTER OF GRAVITY DISTANCE in [mm]

BEARING CAPACITY

RCCx16

RCCx12

RCCx8

1800

1600

1400

1200

1000

800

600

400

200

00 0.001 0.002 0.003 0.004 0.005 0.006

[0.00011] [0.00023] [0.00034] [0.00045] [0.00057] [0.00068]

ALLO

WAB

LE Im

PACT

VEL

OCI

TY (d

eg/s

ec)

ATTACHED LOAD, mOmENT OF INERTIA (in-lb-sec2) [N-m-s2]

SHOCK PAD ENERGY CAPACITY

RCCx8 RCCx12RCCx16

SIZE81216

FORmULA2.86 x (Axial Load) + 15.4 x (Moment) + 54.3 x (Torque) + 30

1.0 x (Axial Load) + 5.7 x (Moment) + 20 x (Torque) + 250.5 x (Axial Load) + 1.56 x (Moment) + 7.3 x (Torque) + 20

mINImUm OPERATING PRESSURESIZE

81216

in-lb0.030.040.08

mAXImUm ALLOWABLEKINETIC ENERGY

Nm0.00340.00450.0090

SIZE81216

in-lb0.651.754.50

Nm0.0730.1980.508

lb7.015.030.0

N31.166.7133.4

mOmENT AXIAL LOADmAXImUm BEARING CAPACITY

RCC ACTUATOR


1��

SIZE08


ROTA

RIES

SIZING EXAmPLE 1


Load = Aluminum disk mounted on center Rotation Angle = 180° Pressure = 87 psi Rotation Time = 0.6 seconds Weight = 0.236 lb Load Radius = 0.875 in Axis Orientation = Horizontal Center of Gravity Distance = 0.50 in Safety Factor = 2


a) Calculate the moment created by the radial load:

Moment = (Weight of Load) x (Cg Distance) Moment = (0.236 lb) x (0.50 in) Moment = 0.118 in-lb

b) Select minimum actuator based on axial load:

Axial Load = 0 lb for this application 8 mm RCC satisfies the requirement

c) Select minimum actuator based on moment load:

Based on the moment load created by the horizontal load, the 8 mm RCC satisfies the requirement.

3) Determine torque requirements:

a) Calculate the mass moment of inertia: Disk mounted on center

Jm = x

Jm = x

Jm = .000234 in-lb-sec2

b) Determine the required angular acceleration:

α = .035 x

α = .035 x

α = 17.5 rad/sec2

c) Calculate the required torque:

T = Jm x α x SF T = .000234 x 17.5 x 2 T = .0082 in-lbs

Fg

gk2

2.875 in2

2

d) Calculate the minimum operating pressure: Based on theoretical values, the 8 mm RCC will provide adequate torque at 87 psi. Check if minimum operating pressure exceeds the operating pressure for this application.

P = 2.86 x (Axial Load) + 15.4 x (Moment) +54.3 x (Torque) + 30 P = 2.86 x (0) + 15.4 x (.118) + 54.3 x (.0082) + 30 P = 32.3 psi < 87 psi

For this application the 8 mm RCC will provide adequate torque at 87 psi.

4) Determine the stopping capacity required:

a) Calculate the impact velocity


ω = .035 x

ω = 10.5 rad/sec

b) Using Jm from step 3a and velocity from step 4a, determine the kinetic energy of the system:

KE = x Jm x ω2

KE = x .000234 x 10.52

KE = .0129 in-lb

c) Use the Maximum Allowable Kinetic Energy Table to select the appropriate RCC actuator.

The 8 mm RCC has sufficient KE capability and satisfies the requirments for torque and bearing capacity.

1.00

.875

Rotational Angle (deg)Rotational Time(sec)

180°.6 sec

1212

.236 lb386.4

Rotational Angle (deg)[Rotational Time(sec)]2

180°[.6 sec]2

RCC ACTUATOR


1��

SIZE08


ROTA

RIES

SIZING EXAmPLE 2


Load = Unbalanced bar with point load Rotation Angle = 180° Pressure = 60 psi Rotation Time = 1.0 seconds Point Load Weight = 0.300 lb Point Load Cg Radius = 2.0 in Bar Weight = .112 lb Axis Orientation = Vertical Safety Factor = 2


a) Calculate the moment created by the radial load:

Radial Load = 0 lb for this application

b) Select actuator based on axial load:

Centered Axial Load = 0 lb for this application because the load is unbalanced

c) Calculate the moment created by the unbalanced load:

Moment = (Weight of Point Load) x (Cg Distance)

For this application a portion of the bar adds to the moment while the remainder subtracts from the moment. The bar is treated as two parts in the moment equation. The portion that adds is 2.5 inches and weighs (.112 x (2.5/3.0)) = .093 lb. The remainder that subtracts will be .5 inches and weighs (.112 x (.5/3.0)) = .019 lb.

Moment = (.300) x (2.0) + (.093) x (1.25) - (.019) x (.25)

Moment = .712 in-lb

d) Select the minimum actuator based on moment load:

Based on the moment load created by the unbalanced load, the 12 mm RCC satisfies the requirement.

3) Determine torque requirements:

a) Calculate the mass moment of inertia:

Point load plus a rectangular plate mounted off center.

Point load:

Jm1 = x k2

Jm1 = x 2.02

Jm1 = .0031 in-lb-sec2

Rectangular plate mounted off center:

Jm2 = x + x

Jm2 = x + x

Jm2 = .0005 in-lb-sec2

Fg

g.300 lb386.4

Sum Jm1 and Jm2:

Jmtotal = Jm1 + Jm2

Jmtotal = .0031 + .0005 Jmtotal = .0036 in-lb-sec2

b) Determine the required angular acceleration:

α = .035 x

α = .035 x

α = 6.3 rad/sec2

c) Calculate the required torque:

T = Jm x α x SF T = .0036 x 6.3 x 2 T = .0454 in-lb

d) Calculate the minimum operating pressure:

Based on theoretical values, the 12 mm RCC will provide adequate torque at 60 psi. Check if minimum operating pressure exceeds the operating pressure for this application.

P = 1.0 x (Axial Load) + 5.7 x (Moment) + 20.0 x (Torque) + 25 P = 1.0 x (0) + 5.7 x (.712) + 20.0 x (.0454) + 25 P = 30.0 psi < 60 psi

4) Determine stopping capacity required:

a) Calculate the impact rotational velocity:


ω = .035 x

ω = 6.3 rad/sec

b) Using Jm from step 3a and velocity from step 4a, determine the kinetic energy of the system:

KE = x Jm x ω2

KE = x .0036 x (6.3)2

KE = .07 in-lb

c) Use the Maximum Allowable Kinetic Energy Table to select the appropriate RCC actuator.

The 16 mm RCC has sufficient KE capability and satisfies the requirements for torque and bearing capacity.


180°[1.0 sec]2

.750

3.000

2.000 .500

4a2 + c2

124b2 + c2

12.019 lb386.4

4(.5)2 + .752

12.093 lb386.4

4(2.5)2 + .752

12

1212

RCC ACTUATOR

Rotating load horizontally(without gravity)


180°1.0 sec

Fg1

g

Fg

g

Fg2

g


1��

SIZE08


ROTA

RIES

ANGLE OF ROTATIONStandard angle of rotation is 180°. Consult PHD for rotation

requirements above 180°. All units are supplied with angle adjustment which provides 90° adjustment from each end.

ROTATION RATESThe speeds given in the chart above reflect one cycle of 180°

with no load applied at 80 psi [5.5 bar]. Times given are average and include the deceleration time through to stopping.

Rf ROTARY ACTUATOR

SPECIFICATIONSOPERATING PRESSUREOPERATING TEMPERATURERATED LIFEROTATIONAL TOLERANCEBACKLASH*LUBRICATIONMAINTENANCE

SERIES RF20 to 100 psi max [1.4 to 6.8 bar]

-20° to 160°F [-29° to 71°C]5 million cycles

Nominal rotation +6° to -180° with angle adjustments0° at end of rotation

Factory lubricated for rated lifeField repairable

SIZE142025

ROTATION180°180°180°

lb.621.883.43

kg.28.851.56

BASEWEIGHT

in.551.787.984

mm142025

BOREDIAmETER

in3

.441.534.18

mm3

7.1725.0768.55

DISPLACEmENTVOLUmE

in-lb/psi.07.24.67

Nm/bar.11.401.09

THEORETICALTORQUE OUTPUT

deg/sec180°/.35180°/.43180°/.37

deg/sec90°/0.2490°/0.2690°/0.23

ROTATION RATES @ 80 psimAXImUm VELOCITY

BACKLASH AT mID-ROTATION± DegreesUNIT SIZE

142025

2.801.380.82

142025

BEARING LOADS TABLE

mAX RADIALBEARING LOAD

mAX AXIALBEARING LOADUNIT

SIZE N112236

lb2.54.98.1

mAXImUm COmBINEDRADIAL AND AXIAL

PAYLOADN

7.614.724.5

lb1.73.35.5

lb3.05.89.7

N132643

NOTE: *Angle adjustment screw must be engaged or adjusted to achieve 0° backlash.

ROTATION SPEED CONTROLSControl of output hub speed is extremely important as kinetic

energy generated by a rotating load is a function of rotational speed and distance from the load to output hub center. Flow controls should be considered to set speed so that the energy is within the limit of the unit.


1��

SIZE08


ROTA

RIES

RF ROTARY ACTUATORSIZING A SERIES RF UNIT BASED ON TORQUE OUTPUT AND STOPPING CAPACITY

SIZING A number of factors must be considered when selecting a Series RF Rotary Actuator. These include actuator orientation, total load attached and rotational speed. The process of selecting the proper Series RF rotary actuator consists of three main steps:

1) Size the actuator based on the torque requirements 2) Size the actuator based on stopping capacity 3) Size the actuator based on bearing capacityChoose the actuator which meets the requirements of your

application.

STEP 1 Determine Rotational mass moment of Inertia (Jm) Select the illustration from the application types on page

189 that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for the type of condition illustrated. The total mass moment of inertia is the sum of the individual calculations.

STEP 2 Determine Necessary Acceleration (αA) This equation calculates the acceleration necessary to move

through the required angle of rotation in the specified time. The results are given in radians/sec2.

STEP 3 Calculate the Required Starting Torque (TA) Select the illustration from the application types on page

189 that most resembles your specific application. Several separate calculations may be necessary to fully describe your

application. Using the appropriate application equation, calculate the torque for each for each type of condition illustrated that matches your application. The total torque will be the sum of the individual calculations. Note: Torque calculations are theoretical, an appropriate safety factor should be considered. PHD recommends a minimum safety factor of 2 to account for friction loss, air line and valve size, and attached accessories.

SIZING A SERIES RF UNIT BASED ON STOPPING CAPACITY

STEP 4 Calculate the Peak Velocity (ω) This formula estimates the peak velocity of the Series RF in operation, and is used to determine the stopping capacity of the rotary actuator. The result is given in radians/sec.

STEP 5 Compare Peak Velocity (ω) to Allowable Impact Compare the peak velocity to the maximum allowable velocity for the given Mass Moment of Inertia (Jm) of your application. The chart is labeled Shock Pad Energy Capacity. The charts represent the total amount of energy that is able to be absorbed and provide acceptable motion of the actuator. Acceptable motion is defined as a maximum of one degree of motion reversal when the load comes to the end of stroke. Note: The unit may be run at slightly higher velocities and loads than these charts indicate without damage; however, the motion profile may be unacceptable. Please contact PHD if the Series RF Rotary Actuator is to be used outside of the recommended energy range. If the shock pad does not provide enough stopping capacity for the application, the next larger size of actuator should be considered.

(Time of Rotation in Seconds)2

.035 x Rotation Angle in Degrees

Rotational Angle in DegreesTime of Rotation in Seconds

Average Velocity (deg/sec) =

Estimated Peak Velocity = .035 x Average Velocity (deg/sec)

SHOCK PAD ENERGY CAPACITY18.0

16.0

14.0

12.0

10.0

8.0

6.0

4.0

2.0

0.00 0.01

[.00113]0.02

[.00226]

Attached Load, moment Of Inertia (in-lb-sec2) [Nm-s2]

Allo

wab

le Im

pact

Vel

ocity

(rad

/sec

)

RFSx25RFSx20

RFSx14

0.03[.00339]

0.04[.00452]

Starting Torque (in/lb) = TA, TAg


1��

SIZE08


ROTA

RIES

SIZING A SERIES RF UNIT BASED ON mOmENT CAPACITY

STEP 6Use these charts to determine if your application is within

the allowable attached load for a specific size. The charts are used when the load is defined and the distance of the center of gravity of the load from the center of rotation or face of the rotary actuator is known. See the illustration below.

Horizontal Orientation (in)(CG)= Distance from Face of Hub to

Center of Gravity of Load

Vertical Orientation (in)(CG)= Distance from Centerline of Hub

to Center of Gravity of Load

CG

CG

RF ROTARY ACTUATOR

mAXImUm mOmENT CAPACITY

RFSx14

Center of Gravity Distance in [mm]

1.6 [7.12]

1.4 [6.23]

1.2 [5.34]

1 [4.45]

0.8 [3.56]

0.6 [2.67]

0.4 [1.78]

0.2 [0.89]

00 0.2 0.4 0.6 0.8 1 1.2 1.4

[5.08] [10.16] [15.24] [20.32] [25.4] [30.48] [35.56]

Load

lb

[N]

RFSx203.5 [15.58]

3 [13.35]

2.5 [11.12]

2 [8.90]

1.5 [6.67]

1 [4.45]

0.5 [2.22]

0

Load

lb

[N]


0 0.5 1 1.5 2 2.5[12.7] [25.4] [38.1] [50.8] [63.5]

RFSx255.5 [24.48]5 [22.25]

4.5 [20.02]4 [17.79]

3.5 [15.57]3 [13.35]

2.5 [11.12]2 [8.90]

1.5 [6.67]1 [4.45]

0.5 [2.22]0

Load

lb

[N]


0 0.5 1 1.5 2 2.5 3 3.5 4[12.7] [25.4] [38.1] [50.8] [63.5] [76.2] [88.9] [101.6]


1��

SIZE08


ROTA

RIES

Jm = x k2Fg

g

Jm = x12

Fg

ga2 + 3k2

( ) ( )Fg1

gJm = x (4a2 + 3k2)

12+ xFg2

g(4b2 + 3k2)

12

b-a2

( )k2Jm = +L2

3Fg

gxx 1

4









Jm = x 4a2 + c2

12+ x 4b2 + c2

12Fg1

gFg2

g

Jm = x x k225

Fg

g

Point Load

LOAD ORIENTATION



UNBALANCED LOADSUNBALANCED LOADS

Jm = x k2

2Fg

g

a2 + b2

12Jm = xFg

g

Tg = [(Jm x α) + (Fg x k)] x SF

k

L

k k


aa

b

c

Fg2

aFg1

b


a Fg1

Fg2

bk

Fg




RF ROTARY ACTUATOR


1�0

SIZE08


ROTA

RIES

APPLICATION INFORmATION - EXAmPLE 1Weight = .75 lb + .25 lb PlateRotation Angle = 180°Pressure = 65 psiOrientation = VerticalCenter of Gravity Distance = 1.125" for .75 lb .5" for .25 lbDesired Cycle Rate = .40 secSafety Factor = 2Axial Load = .75 lb + .25 lb

EXAmPLE 1STEP 1 Determine Jm of Plate mounted off center

gJm = 12

x

Jm = .0008633 in-lb-sec2

g+

12x

386.4 12x +

386.4 12x

(.0002148 x .6667) + (.000432195 x 1.6667)

.000143 + .000720326

Fg1 4a2 + c2 Fg2 4b2 + c2

.083 4(1)2 + (2)2 .167 4(2)2 + (2)2

Determine Point Load

gJm = x

386.4xJm = (1.125)2.75

xJm = .0019409 1.2656

Jm = .002456 in-lb-sec2

Jm Total = .00086 + .00245 = .00331 in-lb-sec2

FgK2

STEP 2 Determine Acceleration

(.40)2

39.38 rad/sec2

T =

T = .00331 x 39.38 x 2

.035 x

T = .261 in-lb


180


STEP 3 Starting Torque

STEP 4 Calculate Peak Velocity

Average Velocity =.40

= 450 deg/sec

Peak Velocity = .035 x 450 = 15.75 rad/sec

Peak Velocity = .035 x Average Velocity(deg/sec)

180

STEP 5 Compare the Peak Velocity

Compare this value to the Shock Pad Energy Capacity Graph on page 187 and the Maximum Velocity Table on page 186. We see the following:

• The size 14 will not handle the Jm value.

• The size 20 will not attain the cycle time required.

• The size 25 will perform the task in the desired time.

STEP 6 Determine the bearing capabilities of a Size 25 Since we know the axial loading but not the radial loading for this application, we compare it to the Maximum Moment Capacity Graph on page 188. At this loading condition the size 25 has the capability of 1 lb at around 2.75 inches off center. Our application is at 1.125 inches.

Therefore; the RFSx25 is suitable for this application.

1.125"

.25 lb

CG

.75 lb.5"

2"

RF ROTARY ACTUATOR


1�1

SIZE08


ROTA

RIES

APPLICATION INFORmATION - EXAmPLE 2Weight = 1.5 lbRotation Angle = 180°Pressure = 60 psiOrientation = HorizontalCenter of Gravity Distance = .5" Desired Cycle Rate = .5 secSafety Factor = 2Axial Load = ØCycles per minute = 20

EXAmPLE 2STEP 1 Determine Jm (Equation is from page 189, Disk Mounted on Center)

STEP 2 Determine Acceleration

STEP 3 Starting Torque (Equation is from page 189)


STEP 5 Compare the Peak Velocity Check this value against the Shock Pad Energy Capacity Graph on page 187.

• The size 14 cannot handle the Jm value of .0038. The size 14 cannot stop the load.

• The size 20 can stop the load and perform the task in the required time.

STEP 6 Determining the bearing capabilities of a Size 20.

We now check the loading condition against the Maximum Moment Capacity Graph for the size 20 on page 188.

We see that a size 20 can handle approximately 2.5 lbs at .5 inches from center of gravity distance.

Therefore; the RFSx20 is suitable for this application.

Jm = .00328 in-lb-sec2

386.4xJm =

(1.3)2

21.5

gJm = 2

xFg K2

(.5)2

25.2 rad/sec2

T =

T = .00328 x 25.2 x 2

.035 x

T = .165 in-lb


180


Average Velocity =.5

= 360 degrees/sec


Peak Velocity = .035 x Average Velocity (deg/sec)

180

RF ROTARY ACTUATOR


1��

SIZE08


ROTA

RIES

APPLICATION INFORmATION - EXAmPLE 3Weight = 1.75 lbRotation Angle = 180°Pressure = 80 psiOrientation = VerticalCenter of Gravity Distance = Ø Desired Cycle Rate = 1.0 secSafety Factor = 2Axial Load = 1.75 lb (weight)Cycles per minute = 30

EXAmPLE 3STEP 1Determine Jm (Equation is from page 189, Disk Mounted on Center)

STEP 2Determine Acceleration

STEP 3 Starting Torque (Equation is from page 189)


STEP 5 Compare the Peak Velocity

Compare this value to the Shock Pad Energy Capacity Graph on page 187.

• The size 14 will handle the Jm value at the rated Peak Velocity.

STEP 6 Determine the bearing capabilities of size 14.

Use the Bearing Load Table on page 186 for the RFSx14.

Since we have only an axial load we can support up to 2.5 lb.

Therefore; the Series RFSx14 is suitable for this application.

Jm = .0044 in-lb-sec2

386.4xJm =

(1.4)2

2

gJm = 2

xFg K2

1.75

(1.0)2

6.3 rad/sec2

T =

T = .0044 x 6.3 x 2

.035 x

T = .055 in-lb


180


Average Velocity =1.0

= 180 degrees/sec

Peak Velocity = .035 x Average Velocity (deg/sec)


180

RF ROTARY ACTUATOR


1��

SIZE08


ROTA

RIES

45° OR 90°

OPTION WEIGHT TABLENOmINAL ROTATION

lb0.40.40.50.60.90.91.41.42.02.43.23.66.06.810.410.6

kg0.180.180.230.270.410.410.640.640.911.071.451.632.723.084.714.81

lb0.40.50.60.70.91.01.51.52.32.74.04.36.77.611.812.0

kg0.180.220.270.320.410.450.680.681.041.221.811.953.043.455.355.44

lb0.50.50.70.71.01.11.61.72.73.04.95.37.78.513.513.7

kg0.220.220.320.320.450.500.700.801.221.362.222.403.493.856.126.21

TYPE OFUNIT

CUSHIONANGLE ADJUSTMENT








BORESIZE

12 mm

16 mm

20 mm

25 mm

32 mm

40 mm

50 mm

63 mm

135° OR 180° 225° OR 270°

NOTE: Units with shock pad options are the same approximate weight as plain units. Unitswith shock absorber options are the same approximate weight as units with angle adjustment.

SIZE

12

16

20

25

32

40

50

63

ROTATION45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°

lb.3.4.4.4.5.6.7.8.91.11.21.41.72.02.32.63.34.35.26.06.99.210.512.3

kg.13.18.18.18.22.27.32.36.41.50.54.64.77.911.041.171.491.952.362.723.134.174.765.57

BASEWEIGHT

in

.472

.630

.787

.984

1.260

1.575

1.969

2.480

mm

12

16

20

25

32

40

50

63

BOREDIAmETER

in3/°

.0005

.001

.002

.004

.008

.017

.032

.063

mm3/°

8.19

16.39

32.77

65.55

131.10

278.58

524.39

1032.38

DISPLACEmENTVOLUmE/DEG

in-lb/psi

.029

.062

.122

.228

.468

.974

1.826

3.624

Nm/bar

.05

.10

.20

.37

.77

1.60

2.99

5.94


deg/sec

180°/.03

180°/.03

180°/.05

180°/.05

180°/.05

180°/.06

180°/.075

180°/.075

ROTATIONALVELOCITY mAX

mAX AXIALBEARING

LOADlb

26

39

39

110

160

184

285

450

N

115

173

173

489

711

818

1267

2001

mAX RADIALBEARING

LOADlb

165

230

230

320

390

420

660

925

N

734

1023

1023

1423

1734

1868

2935

4114

DISTANCEBETWEENBEARINGSin

.65

.73

.89

1.11

1.28

1.60

1.93

2.52

mm

16.6

18.6

22.6

28.1

32.6

40.6

49.1

64.1

SPECIFICATIONSOPERATING PRESSUREOPERATING TEMPERATURERATED LIFEROTATIONAL TOLERANCEBACKLASH AT END OF ROTATION

LUBRICATIONMAINTENANCE

SERIES RL20 to 150 psi [1.4 to 10 bar]-20° to 180°F [-29° to 82°C]

5 million cyclesNominal rotation +10° to -0°

1° 30' (12/16mm), 1° 0' (20/25mm)0° 45' (32/40mm), 0° 30' (50/63mm)


Rl ROTARY ACTUATOR


1��

SIZE08


ROTA

RIES

3) Determine the stopping capacity of the actuator by using the equation given below.

a) Determine the rotational velocity by using equation A.

To select the appropriate RL rotary actuator, it is crucial to consider several factors including bearing capacity, torque requirements and stopping capacity of the actuator. The bearing capacities are listed on page 193. To determine the required torque to rotate the load in a given time, the rotational mass moments of inertia, gravity, time and acceleration must be taken into account. To stop an actuator, all of the same required information for torque is needed plus kinetic energy. Follow the steps below to select the appropriate RL actuator.

1) Review page 193 to make sure RL rotary actuator bearings can withstand axial and radial bearing loads.

2) Determine the torque requirements of the actuator.

a) Determine Mass Moment of Inertia. Select the illustration from the application types on the following page that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for each type of illustration. The total mass moment of inertia will be the sum of the individual calculations.

b) Determine the necessary acceleration.

c) Calculate the required torque. Select the illustration from the application types on

the following page that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for each type of illustration. The total torque will be the sum of the individual calculations.

Note: Torque calculations are theoretical, an appropriate safety factor should be considered. PHD recommends a minimum safety factor of 2 to account for friction loss, airline and valve size, and attached accessories.

Acceleration (α) =

RL ROTARY ACTUATOR

.035 x Degrees of RotationTime of Rotation in seconds

ROTATIONAL VELOCITY EQUATIONS

KINETIC ENERGY BASIC EQUATION

Estimated Peak Velocity (rad/sec)Uniformly accelerated from rest

radsec

EQUATION A

2 x (Rotation angle in radians)


.035 x (Rotation angle in degrees)

(Time of Rotation in Seconds)2Acceleration (α) =

KINETIC ENERGY TABLEKE mAX. WITH

SHOCK ABSORBERKE mAX. WITH

CUSHIONKE mAX. WITH

SHOCK PADKE mAX.

PLAIN UNITBORESIZE

12 mm16 mm20 mm25 mm32 mm40 mm50 mm63 mm

———

6.0012.0030.0048.0084.00

in-lb———

.6781.3563.3905.4239.491

Nm.35.53.60.791.663.606.259.21

in-lb.040.060.068.089.188.406.7061.040

Nm—.26.30.39.831.803.124.60

in-lb—.03.03.04.09.20.35.52

Nm.07.09.16.22.481.031.782.63

in-lb.008.011.018.025.054.116.202.297

Nm

b) Using Jm from step 2a and velocity from step 3a, calculate the kinetic energy of the application.

c) Use the KE Energy Table below to select appropriate RL actuator.


1��

SIZE08


ROTA

RIES

Jm = x k2Fg

g

kFg


Jm = x12

Fg

ga2 + 3k2

( ) ( )Fg1

gJm = x (4a2 + 3k2)

12+ Fg2

g(4b2 + 3k2)

12

( )k2Jm = +L2

3Fg

gxx 1

4


b-a2

x

kk

Jm = x k2

2Fg

g



L

k

a

aFg1

Fg2

b

a2 + b2

12Jm = xFg

g






Jm = x 4a2 + c2

12+ x 4b2 + c2

12Fg1

gFg2

g

c Fg2

a

Fg1b

Jm = x x k225

Fg

g

Point Load

UNBALANCED LOADSTg = [(Jm x α) + [(Fg2 - Fg1) x (a + ( ))]] x SF

LOAD ORIENTATION



UNBALANCED LOADS

a

b


RL ROTARY ACTUATORImPERIAL UNITS:Jm = Rotational Mass Moment of Inertia (in-lb-sec2) (Dependent on physical size of object and weight)g = Gravitational Constant = 386.4 in/sec2 Fg = Weight of Load (lb) k = Radius of Gyration (in)T = Torque required to rotate load (in-lbs) α = Acceleration (rad/sec2) t = time (sec)SF = Safety Factor




1��

SIZE08


ROTA

RIES

k1.00

[25.4]

1.75[44.45]

k

180°.1 sec

rotation angle (deg)rotational time (sec)

APPLICATION EXAmPLE A Disk rotating about centerline of unit.

xJm = Fg k2

g 2 Jm =

x

x Fg k2

g 2

x Jm = .236 lb (.875 in) 2

386.4 2

α = .035 x

Jm =

b) Using Jm from step 2a and velocity from step 3a, determine KE of the system from the basic KE equation:

ImPERIAL mETRIC KE = 1/2 x Jm x ω2 KE = 1/2 x Jm x ω2

KE = .5 x .000234 x 632 KE = .5 x 2.64 x 10-5 x 632

KE = .464 in-lbs KE = .052 N-m

c) Use the KE Energy Table on page 194 to select the appropriate RL actuator. The following units satisfy the requirements. 32 mm plain, 32 mm with shock pads, and a

16, 20, or 25 mm with cushions.

T = Jm x α x 2

T = 2.64 x 10-5 x 630 x 2 = .03 N-m

Numbers in [ ] are for metric units and are in mm.

ω = .035 x = 63 rad/sec

RL ROTARY ACTUATOR

1) Determine load information: ImPERIAL mETRICROTATION ANGLE / TImE 180°/.10 sec 180°/.10 secLOAD Aluminum Disk Aluminum DiskWEIGHT .236 lb 1.05 NmASS .107 KgPRESSURE 87 psi 6 barSAFETY FACTOR 2 2

2) Determine torque requirement for the application: a) Calculate Rotational mass moment of Inertia (Jm) using

equations given on page 195. ImPERIAL mETRIC

Jm = .000234 in-lb-sec2 Jm = 2.64 x 10-5 N-m-sec2

b) Determine required acceleration of the load:

c) Calculate required torque: ImPERIAL mETRIC

T = Jm x α x SF

T = .000234 x 630 x 2 = .29 in-lbs To select minimum actuator based on torque, calculate theoretical

torque for 87 psi [6 bar] by using table on page 193.

3) Determine the stopping capacity of the actuator for the application:

a) Determine the estimated peak rotational velocity using

Equation A on page 194. ω = rad/sec = .035 x

1.05 N (.0222m) 2

9.81 2

rotational angle (deg)[rotational time (sec)]2

180°(.1 sec)2α = .035 x = 630 rad/sec2


1��

SIZE08


ROTA

RIES

2"[50.8]

(k)

1lb[4.45 N]

2"[50.8]

(b)

6" [152.4] (a)1) Determine load information:

ImPERIAL mETRICROTATION ANGLE / TImE 180°/.5 sec 180°/.5 secRECTANGULAR PLATE Steel Plate Steel PlateWEIGHT 1.698 lb 7.55 NmASS .77 KgPOINT LOAD 1 lb 4.45 N (2" off center) (50.8 mm off center)PRESSURE 87 psi 6 barSAFETY FACTOR 2 2 2) Determine torque requirement for the application:

a) Calculate Rotational mass moment of Inertia (Jm) using equations given on page 195.

POINT LOAD ImPERIAL mETRIC

Jm = .0104 in-lb-sec2 Jm = .00117 N-m-sec2

RECTANGULAR PLATE ImPERIAL mETRIC


Total Jm Total Jm = .0146+.0104=.025 in-lb-sec2 = .00165+.00117=.00282N-m-sec2


c) Calculate required torque:

POINT LOAD ImPERIAL mETRICT = [(Jm x α)+(Fg x k)] x 2

T = [(.0140 x 25.2) + (1 x 2)] x 2

T = 4.5 in-lbs

Jm = x 7.55 (.1524)2+(.0508)2 9.81 12

Jm = x

x (.0508 m)2

APPLICATION EXAmPLE B Combination of rectangular plate mounted on center and a point load mounted off center.

4.45 N9.81

Jm =

Fg g x k2Jm =

Jm = x

1 lb386.4

Fg g x k2Jm =

x (2 in)2 Jm =

1.698 62+22 386.4 12

Fg a2+b2

g 12

Jm = x

α = .035 x rotational angle (deg) [time (sec)]2

α = .035 x = 25.2 rad/sec2180°(.5 sec)2


180°.5 sec

Fg a2+b2

g 12

T = Jm x α x SF

T = .00166 x 25.2 x 2 = .084 N-m

Total T = .51 + .084 = .594 N-m

RECTANGULAR PLATE ImPERIAL mETRICT = Jm x α x SF

T = .0146 x 25.2 x 2 = .74 in-lbs

Total T = 4.5 + .74 = 5.24 in-lbs

To select minimum actuator based on torque, calculate theoretical torque for 87 psi [6 bar] by using table on page 193.


a) Determine the estimated peak rotational velocity using Equation A on page 194.

ω = .035 x ω = .035 x = 12.6 rad/sec



KE = .5 x .025 x 12.62 KE = .5 x .00282 x 12.62 KE = 1.98 in-lbs KE = .224 N-m

c) Use the KE Energy Table on page 194 to select the appropriate RL actuator. The following units satisfy the requirements: 63 mm plain, 50 mm with shock pads, 40 mm with cushions, and a 25 mm with shock absorbers.


T = [(Jm x α)+(Fg x k)] x SF

T = [(.00117 x 25.2) + (4.45 x .0508)] x 2

T = .51 N-m

RL ROTARY ACTUATOR


1��

SIZE08


ROTA

RIES

CUSHION AND OUTPUT HUB WEIGHTS

BORESIZE

20 mm25 mm32 mm40 mm50 mm

ADDER WITHCUSHION OPTION -DB

kg.13.16.24.34.47

lb.3.4.6.81.1

ADDER WITHHUB OPTION -Q10 OR -Q19

lb.03.03.04.12.23

kg.01.01.02.05.11

SIZE

20

25

32

40

50

ROTATION45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°45°/90°

135°/180°225°/270°

lb1.801.802.302.402.803.604.304.906.507.708.8011.8011.6012.8017.70

kg.77.771.021.081.241.601.922.192.943.473.965.315.225.788.01

SERIES RA20 to 150 psi [1.4 to 10 bar]-20° to 180°F [-29° to 82°C]

10 million cyclesNominal rotation +10° to -45° with angle adjustments

0°Factory lubricated for rated life

Field repairable

BASEWEIGHT

in

.787

.984

1.260

1.575

1.969

mm

20

25

32

40

50

BOREDIAmETER

in3/°

.002

.004

.007

.014

.027

mm3/°

32.77

65.55

114.71

229.42

442.45


in-lb/psi

.097

.190

.415

.779

1.522

Nm/bar

.16

.31

.68

1.28

2.49

THEORETICALTORQUEOUTPUT

deg/sec

180°/.05

180°/.05

180°/.05

180°/.075

180°/.075

ROTATIONALVELOCITY

mAX

mAX AXIALBEARING

LOADlb

97

118

182

237

325

N

431

524

809

1054

1445

mAX RADIALBEARING

LOADlb

376

453

640

746

966

N

1672

2015

2846

3318

4296

in

1.34

1.61

1.94

2.56

2.90

mm

34.0

40.9

49.3

65.0

73.6

DISTANCEBETWEENBEARINGS

SPECIFICATIONSOPERATING PRESSUREOPERATING TEMPERATURERATED LIFEROTATIONAL TOLERANCEBACKLASH AT END OF ROTATION*LUBRICATIONMAINTENANCE

NOTE: *Angle adjustment screw must be engaged or adjusted to achieve 0° backlash

RA ROTARY ACTUATOR


1��

SIZE08


ROTA

RIES

RA ROTARY ACTUATOR

KINETIC ENERGY TABLE

BORESIZE

20 mm25 mm32 mm40 mm50 mm

KE mAX. WITHSHOCK ABSORBER

KE mAX.WITH CUSHION

KE mAX.PLAIN UNIT

.21

.46

.961.742.13

in-lb.0237.0519.1085.1966.2407

Nm.751.703.606.758.81

in-lb.0848.1921.4068.7628.9955

Nm3.309.3021.3045.0089.00

in-lb0.3731.0512.4075.08410.056

Nm

To select the appropriate RA rotary actuator, it is crucial to consider several factors including bearing capacity, torque requirements and stopping capacity of the actuator. The bearing capacities are listed on page 198. To determine the required torque to rotate the load in a given time, the rotational mass moments of inertia, gravity, time and acceleration must be taken into account. To stop an actuator, all of the same required information for torque is needed plus kinetic energy. Follow the steps below to select the appropriate RA actuator.

1) Review page 198 to make sure RA rotary actuator bearings can withstand axial and radial bearing loads.

2) Determine the torque requirements of the actuator.

a) Determine Mass Moment of Inertia. Select the illustration from the application types on the following page that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for each type of illustration. The total mass moment of inertia will be the sum of the individual calculations.

b) Determine the necessary acceleration.

c) Calculate the required torque. Select the illustration from the application types on

the following page that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for each type of illustration. The total torque will be the sum of the individual calculations.

Note: Torque calculations are theoretical, an appropriate safety factor should be considered. PHD recommends a minimum safety factor of 2 to account for friction loss, airline and valve size, and attached accessories.

Acceleration (α) =

.035 x Degrees of RotationTime of Rotation in seconds

ROTATIONAL VELOCITY EQUATIONS

KINETIC ENERGY BASIC EQUATION

Estimated Peak Velocity (rad/sec)Uniformly accelerated from rest

radsec

EQUATION A

3) Determine the stopping capacity of the actuator by using the equation given below.

a) Determine the rotational velocity by using equation A.

2 x (Rotation angle in radians)


.035 x (Rotation angle in degrees)

(Time of Rotation in Seconds)2Acceleration (α) =

b) Using Jm from step 2a and velocity from step 3a, calculate the kinetic energy of the application.

c) Use the KE Energy Table below to select appropriate RA actuator.

ROTARY ACTUATOR SELECTION


�00

SIZE08


ROTA

RIES

Jm = x k2Fg

g


Jm = x12

Fg

ga2 + 3k2

( ) ( )Fg1

gJm = x (4a2 + 3k2)

12+ xFg2

g(4b2 + 3k2)

12

b-a2

Tg = [(Jm x α) + (Fg x k)] x SFT = Jm x α x SF

( )k2Jm = +L2

3Fg

gxx 1

4



kFg

k

c Fg2

a

Fg1b

a

L

kk








Jm = x 4a2 + c2

12+ x 4b2 + c2

12Fg1

gFg2

g

Jm = x x k225

Fg

g

Point Load

aFg1

Fg2

b

LOAD ORIENTATION



UNBALANCED LOADS

T = Jm x α x SF

UNBALANCED LOADS

Jm = x k2

2Fg

g

a2 + b2

12Jm = xFg

g

a

b




RA ROTARY ACTUATOR


�01

SIZE08


ROTA

RIES

1) Determine load information: ImPERIAL mETRICROTATION ANGLE / TImE 180°/.10 sec 180°/.10 secLOAD Aluminum Disk Aluminum DiskWEIGHT .236 lb 1.05 NmASS .107 KgPRESSURE 87 psi 6 barSAFETY FACTOR 2 2

2) Determine torque requirement for the application: a) Calculate Rotational mass moment of Inertia (Jm) using

equations given on page 200. ImPERIAL mETRIC

Jm = .000234 in-lb-sec2 Jm = 2.64 x 10-5 N-m-sec2



ImPERIAL mETRICT = Jm x α x SF

T = .000234 x 630 x 2 = .29 in-lbs To select minimum actuator based on torque, calculate theoretical

torque for 87 psi [6 bar] by using table on page 198.


a) Determine the estimated peak rotational velocity using

Equation A on page 199. ω = rad/sec = .035 x

ω = .035 x = 63 rad/sec180°.1 sec


APPLICATION EXAmPLE A Disk rotating about centerline of unit.

xJm = Fg k2

g 2 Jm =

x

x Fg k2

g 2

1.05 N (.0222m) 2

9.81 2x Jm =

.236 lb (.875 in) 2

386.4 2

α = .035 x rotational angle (deg)[rotational time (sec)]2

α = .035 x = 630 rad/sec2180°(.1 sec)2

Jm =

T = Jm x α x 2

T = 2.64 x 10-5 x 630 x 2 = .03 N-m

k1.00

[25.4]

1.75[44.45]

k



ImPERIAL mETRIC KE = 1/2 x Jm x ω 2 KE = 1/2 x Jm x ω2

KE = .5 x .000234 x 632 KE = .5 x 2.64 x 10-5 x 632

KE = .464 in-lbs KE = .052 N-m

c) Use the KE Energy table on page 199 to select the appropriate RA actuator. The following units satisfy the requirements. 32 mm plain and a 25 or 20 mm with cushions.

RA ROTARY ACTUATOR


�0�

SIZE08


ROTA

RIES

1) Determine load information:

ImPERIAL mETRICROTATION ANGLE / TImE 180°/.5 sec 180°/.5 secRECTANGULAR PLATE Steel Plate Steel PlateWEIGHT 1.698 lb 7.55 NmASS .77 KgPOINT LOAD 1 lb 4.45 N (2" off center) (50.8 mm off center)PRESSURE 87 psi 6 barSAFETY FACTOR 2 2 2) Determine torque requirement for the application:

a) Calculate Rotational mass moment of Inertia (Jm) using equations given on page 200.

POINT LOAD ImPERIAL mETRIC


RECTANGULAR PLATE ImPERIAL mETRIC


Total Jm Total Jm = .0146+.0104=.025 in-lb-sec2 = .00165+.00117=.00282N-m-sec2



POINT LOAD ImPERIAL mETRICT = [(Jm x α)+(Fg x k)] x 2

T = [(.0140 x 25.2) + (1 x 2)] x 2

T = 4.5 in-lbs

Jm = x 7.55 (.1524)2+(.0508)2 9.81 12

Jm = x

x (.0508 m)2

APPLICATION EXAmPLE B Combination of rectangular plate mounted on center and a point load mounted off center.

4.45 N 9.81

Jm =

Fg g x k2Jm =

Jm = x

1 lb 386.4

Fg g x k2Jm =

x (2 in)2 Jm =

1.698 62+22 386.4 12

Fg a2+b2

g 12

Jm = x

α = .035 x rotational angle (deg) [time (sec)]2

α = .035 x = 25.2 rad/sec2180°(.5 sec)2

Fg a2+b2

g 12

2"[50.8]

(k)

1lb[4.45 N]

2"[50.8]

(b)

6" [152.4] (a)


T = [(Jm x α)+(Fg x k)] x SF

T = [(.00117 x 25.2) + (4.45 x .0508)] x 2

T = .51 N-m


180°.5 sec

T = Jm x α x SF

T = .00166 x 25.2 x 2 = .084 N-m

Total T = .51 + .084 = .594 N-m

RECTANGULAR PLATE ImPERIAL mETRICT = Jm x α x SF

T = .0146 x 25.2 x 2 = .74 in-lbs

Total T = 4.5 + .74 = 5.24 in-lbs

To select minimum actuator based on torque, calculate theoretical torque for 87 psi [6 bar] by using table on page 198.


a) Determine the estimated peak rotational velocity using Equation A on page 199.

ω = .035 x ω = .035 x = 12.6 rad/sec



KE = .5 x .025 x 12.62 KE = .5 x .00282 x 12.62 x 4 KE = 1.98 in-lbs KE = .224 N-m

c) Use the KE Energy Table on page 199 to select the appropriate RA actuator. The following units satisfy the requirements: 50 mm plain, 40 or 32 mm with cushions, and a 25 or 20 mm with shock absorbers.

RA ROTARY ACTUATOR


�0�

SIZE08


ROTA

RIES

Ri ROTARY ACTUATOR

mANIFOLD PINION SPECIFICATIONS

UNITSIZE

RISx25RIDx25RISx32RIDx32RISx50RIDx50

NUmBER OFPASSAGES

446688

FLOW THROUGHPASSAGES @ 87 psi [6 bar]

CFm11

1.31.31.51.5

Liter/min

CENTER THROUGH HOLEDIAmETER

in mm0.1970.1970.2760.2760.4330.433

55771111

28.328.336.836.842.542.5

SIZERISxx25RIDxx253RIDxx25RISxx32RIDxx323RIDxx32RISxx50RIDxx503RIDxx50

ROTATION/mID ROT

180°180°

180°/90°180°180°

180°/90°180°180°

180°/90°

lb3.03.54.17.68.09.614.315.017.6

kg1.361.591.863.443.634.366.486.807.98

BASEWEIGHT

in

.984

1.260

1.969

mm

25

32

50

BOREDIAmETER

in3

.0063

.0126

.0140

.0118

.0236

.0262

.0415

.0830

.0923

cm3

.103

.206

.233

.193

.387

.429

.6801.361.51

DISPLACEmENTVOLUmE/deg

in-lb/psi.37.74.37.731.45.732.384.762.38

Nm/bar.611.21.611.202.381.203.907.803.90

deg/sec180°/.13180°/.23180°/.23180°/.11180°/.28180°/.28180°/.13180°/.28180°/.28


ROTATIONALVELOCITY mAX

mAX AXIALBEARING LOAD

lb

292

511

697

N

1300

2275

3100


lb

572

1206

1850

N

2546

5365

8229

SPECIFICATIONSOPERATING PRESSUREOPERATING TEMPERATURERATED LIFEROTATIONAL TOLERANCEBACKLASH AT END OF ROTATION*LUBRICATIONMAINTENANCE

SERIES RI20 to 100 psi [1.4 to 6.8 bar]-20° to 160°F [-29° to 71°C]

5 million cyclesNominal rotation +13° to -180° with angle adjustment

0°Factory lubricated for rated life

Field repairable

.26

.26

.23

.23

.21

.21

RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50

BACKLASH SPECIFICATIONS

REPEATABILITY

BACKLASH

mID

ROTATIONUNITSIZE +/- (degrees)

0.140.530.420.940.120.35

BACKLASH

THREE POSITION

UNIT

—1.25—

0.65—

0.40

+/- (degrees)

REPEATABILITY

THREE POSITION

UNIT

—0.16—

0.10—

0.06

+/- (degrees)

ROTATION RATE TABLE

UNITSIZE


ROTATION RATES at 87 psi(seconds maximum)

SHOCKPAD SHOCK

SPEEDCONTROL

0.130.230.110.280.130.28

0.180.410.110.300.220.40

0.180.310.230.320.290.78

(No load conditions)

NOTE: *Angle adjustment screw must be engaged or adjusted to achieve 0° backlash

+/- (degrees)


�0�

SIZE08


ROTA

RIES

RI ROTARY ACTUATOR

SIZING AN RI UNIT WITH ANGLE ADJUSTmENTS

STEP 1 Determine Rotational mass moment of Inertia (Jm) Select the illustration from the application types on page

207 that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the mass moment of inertia for the type of condition illustrated. The total mass moment of inertia is the sum of the individual calculations.

STEP 2 Determine Necessary Acceleration (αs) This equation calculates the acceleration required to move the

desired rotation in the desired time. The solution is given in radians/sec2.

STEP 3 Calculate the Required Starting Torque (TA) Select the illustration from the application types on page

207 that most resembles your specific application. Several separate calculations may be necessary to fully describe your application. Using the appropriate application equation, calculate the torque for each for each type of condition illustrated that matches your application. The total torque will be the sum of the individual calculations. Note: Torque calculations are theoretical, an appropriate safety factor should be considered. PHD recommends a minimum safety factor of 2 to account for friction loss, airline and valve size, and attached accessories.

STEP 4 Calculate the Peak Velocity (ω) This formula estimates the peak velocity of the Series RIx in

operation, and is used to determine the stopping capacity of the rotary actuator. The solution is given in radians/sec.

STEP 5 Compare Peak Velocity (ω) to Allowable Impact Compare your peak velocity to the maximum allowable

velocity for the given Mass Moment of Inertia (Jm) of your application. The chart is labeled Shock Pad Energy Capacity. The charts represent the total amount of energy that is able to be absorbed and provide acceptable motion of the actuator. Acceptable motion is defined as a maximum of one degree of motion reversal when the load comes to end of stroke. Note: You may run slightly higher velocities and loads than these charts provide and not damage the unit; however, you may find the motion profile unacceptable. Please contact PHD if you are considering using the Series RIx actuator outside of the recommended energy range and shock absorbers are not a desired option. If the shock pad does not provide enough stopping capacity for your application, go to the next sizing section titled “Sizing a RIx Unit with Shocks.”



Starting Torque (in/lb) = TA, TAg

.035 x Rotational Angle in DegreesTime of Rotation in Seconds

Average Velocity (deg/sec) =

ALLO

WAB

LE Im

PACT

VEL

OCI

TY (r

ad/s

ec)

mOmENT OF INERTIA (in-lb sec2) [Nms2]

0 0.5[.0565]

1.0[.113]

1.5[.170]

2.0[.226]

2.5[.283]

3.0[.339]

3.5[.396]

4.0[.452]

4.5[.508]

5.0[.565]

12

10

8

6

4

2

0

50 mm max KE

SHOCK PAD ENERGY CAPACITY

50 mm Acceptable motion

32 mm max25 mm max & 32 mm Acceptable motion

25 mm Acceptable motion


�0�

SIZE08


ROTA

RIES

SIZING AN RI UNIT WITH SHOCKSSTEP 6 Compare Peak Velocity (ω) to Allowable Impact Compare your peak velocity to the maximum allowable

velocity for the given Mass Moment of Inertia (Jm) of your application. The chart is labeled Shock Energy Capacity. The charts represent the total amount of energy that is able to be absorbed and provide acceptable motion of the actuator. Acceptable motion is defined as a maximum of one degree of motion reversal when the load comes to end of stroke. Note: You may run slightly higher velocities and loads than these charts provide and not damage the unit; however, you may find the motion profile unacceptable. Please contact PHD if you are considering using the Series RIxxx actuator outside of the recommended energy and load range.

STEP 7 Calculate the Kinetic Energy (Ke) This formula calculates the kinetic energy of your application.

This value will be used to calculate the actual total energy to be compared to the maximum allowable total energy.

STEP 8 Calculate the Propelling Energy (Pe) These formulas calculate the additional amount of energy that

the shock will experience due to the piston force of the actuator.

STEP 9 Calculate the Total Energy (Et) This formula sums all of the energies that the shock

will experience.

STEP 10 Compare the Total Energy (Et) to the maximum Total Energy

(Em) and also Acceptable motion (Ea) If Acceptable Motion is desired as defined in STEP 6, the total

energy should be less than both of the charted values given below. If some additional bounce is acceptable, (Et) can be up to the same value as (Em). If not, go to a larger actuator or contact PHD for application assistance.

STEP 11 Calculate Energy per Hour (Eh) Compare your applications energy per hour requirement

against the charted maximum.

UNITSIZE


in-lb.3572 x psi.7144 x psi.935 x psi1.471 x psi2.769 x psi5.539 x psi

Pe = Propelling EnergyNm

.5852 x bar1.170 x bar1.5321 x bar2.409 x bar4.538 x bar9.0768 x bar

Total Energy Et (in/lb [Nm]) = Ke + Pe

Energy/Hour (in/lb [Nm]) = Cycles/Hour x Et

RIxx32

0 0.10 0.20 0.30 0.40 0.50 0.60 0.70[.011] [.023] [.034] [.045] [.056] [.068] [.079]

0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0[.113] [.226] [.339] [.452] [.565] [.678] [.791] [.904] [1.02]

0 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00[.023] [.045] [.068] [.090] [.113] [.136] [.158] [.181] [.203] [.226]

SHOCK ENERGY CAPACITY

60

55

50

45

40

35

30

25

20

15

10

5

0

Allo

wab

le Im

pact

Vel

ocity

(rad

/sec

)

moment of Inertia (in-lb-sec2) [Nms2]

RIxx25

RIDx25

RISx25

Allo

wab

le Im

pact

Vel

ocity

(rad

/sec

)


302826242220181614121086420

RIxx50

RIDx50

RISx50

60

55

50

45

40

35

30

25

20

15

10

5

0

Allo

wab

le Im

pact

Vel

ocity

(rad

/sec

)


RIDx25

RISx25

mAX ALLOWABLE CHART (Em)

ACCEPTABLE mOTION CHART (Ea)UNITSIZE


in-lb6696154213527754

VELOCITYrad/sec

57.724.258.527.628.919.7

Nm7.4610.817.424.159.585.2

ET*

UNITSIZE


in-lb80116175233577804

ENERGY/HOUR

300,000300,000400,000400,000600,000600,000

Nm9.0413.119.826.365.290.8

ET

in-lb/Hr33,89033,89045,19045,19067,79167,791

Nm/Hr

*Acceptable motion is defined as a maximum of one degree of motion reversal when the load comes to end of stroke.

RI ROTARY ACTUATOR


�0�

SIZE08


ROTA

RIES

DETERmINING ALLOWABLE ATTACHED LOAD WEIGHT Following are the steps required to determine the allowable

attached load weight on the Series RIx rotary actuator. You will need to know the weight of the attached load, the orientation of the rotary, and the center of gravity distance of the load from the hub face. Please refer to the supplied formulas to determine each of the allowable conditions.

STEP 12 Determine Allowable Attached Load Weight (Lf) The next step in determining the proper Series RIx actuator

size is to determine the bearing capacity of the unit according to your application requirements.

STEP 13 Calculate maximum Actuator Radial Loading (Lm) This formula calculates the maximum radial loading allowed

for the Series RI actuator based on 5,000,000 cycles and the axial load (La) that you are placing on the bearings. Note: Center of Gravity distance is different depending on if the unit is horizontal or vertical. In horizontal applications, (Cg) is the distance from the mounting face of the hub to the (Cg) of the load. In vertical applications, (Cg) is the distance from the centerline of the hub to (Cg) of the load.

STEP 15 Calculate the Deceleration (αd) This formula calculates the deceleration of the unit based on

the peak velocity of the individual actuator. The solution is given in radians/sec2

STEP 16 Calculate Stopping Torque (Td) This is the stopping torque energy used to stop a rotary load

to your application conditions. This formula is one of the components required when comparing reaction forces on the bearing. Using the identical illustrations and formulas on pages 204 and 207 used when calculating the required starting torque, replace the acceleration value with the deceleration value. This is the reaction torque required to stop the load. PHD recommends a safety factor of 1 to 1.25.

STEP 17 Calculate Radial Bearing Load At Stopping (LS) This formula converts the sum torque’s of the propelling torque

and stopping torque into the reaction force on the two bearings.

STEP 18 Calculate max. Fixed Radial Load (Lf) This formula will produce the maximum radial load weight that

can be safely attached to the rotary actuator, given the axial load weight and (Cg) distance of your application.

STEP 19 Compare (Lf) to Actual Load Affixed to Actuator (Lr) Compare the (Lf) value to the weight of the attached load. If

the attached load is less than the (Lf) value, the actuator is correct for your application. If the attached load is greater than the (Lf) value, go to the next size actuator and rerun the above calculations until the (Lf) value is greater than the attached load weight.

La = Axial Load Weight (lb)

Horizontal Orientation (in)(Cg) = Distance from Face of Hub to

Center of Gravity of Load

Vertical Orientation (in)(Cg) = Distance from Centerline of Hub

to Center of Gravity of Load

mAX ACTUATOR RADIAL LOADING (Lm)

Cg

Cg

ImPERIAL mETRIC

-1.4175 (La) + 1106.861.933 + Cg

Lm =

-1.8138 (La) + 3015.572.5 + Cg

Lm =

-2.699 (La) + 6573.923.553 + Cg

Lm =

-36.0024 (La) + 125042.449.1 + Cg

Lm =

-46.0702 (La) + 340706.263.5 + Cg

Lm =

-68.5696 (La) + 74265690.25 + Cg

Lm =

UNITSIZE

RIxxx25

RIxxx32

RIxxx50

UNITSIZE


in-lb.369 x psi.737 x psi.727 x psi1.454 x psi2.378 x psi4.755 x psi

Propelling Torque (Tp)Nm

.6047 x bar1.2077 x bar1.1913 x bar2.3827 x bar3.8969 x bar7.7921 x bar

UNITSIZE

RIxxx25RIxxx32RIxxx50

lb(Tp + Td)/.96875(Tp + Td)/1.1667(Tp + Td)/1.5625

Radial Bearing Load at Stopping (LS)N

(Tp + Td)/.0246(Tp + Td)/.0296(Tp + Td)/.0399

Max Fixed Radial Load (Lf) = Lm - Ls

Lr = Weight of Attached Load

UNIT SIZERIxxx25RISxx32RIDxx32RISxx50RIDxx50

STEP 14 Calculate Propelling Torque (Tp) This formula is one of the components required when

comparing reaction forces on the bearing. You may use the formula or simply look up the torque produced by the rotary actuator at a specified pressure.

Stopping Torque (in-lb) = TA, TAg

RI ROTARY ACTUATOR


�0�

SIZE08


ROTA

RIES

Jm = x k2Fg

g

Jm = x12

Fg

ga2 + 3k2

( ) ( )Fg1

gJm = x (4a2 + 3k2)

12+ xFg2

g(4b2 + 3k2)

12

b-a2

( )k2Jm = +L2

3Fg

gxx 1

4


347

8

k

L

k k

a

b

c

Fg2

a Fg1

b


a Fg1

Fg2

b


kFg








Jm = x 4a2 + c2

12+ x 4b2 + c2

12Fg1

gFg2

g

Jm = x x k225

Fg

g

Point Load

LOAD ORIENTATION



UNBALANCED LOADSUNBALANCED LOADS

Jm = x k2

2Fg

g

a2 + b2

12Jm = xFg

g

a





RI ROTARY ACTUATOR


�0�

SIZE08


ROTA

RIES

APPLICATION INFORmATION - EXAmPLE 1Weight = 32.2 lbRotation Angle = 180°Pressure = 87 psiOrientation = HorizontalCenter of Gravity Distance = 2"Desired Cycle Rate = .75 secSafety Factor: Acceleration = 2 Deceleration = 1 Axial Load (La) = 0Radial Load (Lr) = 32.2 lbCycles per Minute = 40

EXAmPLE 1Determine Required Starting Torque for Application STEP 1 Determine (Jm)

STEP 2 Determine (αA)


RISxx25 WILL PRODUCE SUFFICIENT TORQUE

Check for Stopping Capacity STEP 4 Calculate Peak Velocity (ω) RISxx25

STEP 5 Compare to Graph (refer to page 204)

SHOCK PAD WILL NOT PERFORM AS DESIREDThis velocity is greater than the shock pad allows, go to the section labeled

“Sizing an RIx Unit with Shocks”

STEP 6 Compare Peak Velocity to Allowable Impact Velocity for a given (Jm) Load using Shock Absorbers

Compare to graph on page 205.RISx is acceptable for this application.

STEP 7 Calculate Kinetic Energy (Ke)

gJm = 2

x =386.4 2

x

Jm = .0833 x 4.5 = .375 in-lb-sec2

.035(.75)2 = 11.2 rad/sec2

T =

T = .375 x 11.2 x 2 = 8.4 in-lb

.035Time of Rotation

in Seconds2

Fg k2 32.2 32

180

Angle Rotationin Degrees

.035 x.75

= 11.2 rad/sec

Ke = 1/2 x (.375) x (11.2)2 = 23.52 in-lb

180

STEP 8 Calculate Propelling Energy (Pe)

STEP 9 Calculate Total Energy (Et)

STEP 10 Compare Maximum Total Energy (Em) to Total Energy (Et) and Acceptable Motion Energy to Total Energy

SHOCKS WILL PERFORM AS DESIRED

STEP 11 Calculate Energy per Hour (Eh)

STEP 12 Calculate Allowable Attached Load Weight

Axial Load from Application = La

RISx25 = .3572 x psiPe = .3572 x 87 = 31.08 in-lb

Et = Ke + Pe

Et = 23.52 + 31.08 = 54.60 in-lb

Em ≥ Et 80 ≥ 54.60

Ea ≥ Et 66 ≥ 54.60

Cycles/Hr = Cycles/min x 60Cycles/Hr = 40 x 60 = 2400

La = 0

300,000 ≥ 131,040

Eh = 2400 x 54.60 in-lb = 131,040 in-lb/hr

k

STEP 13 Calculate Max Actuator Radial Loading (Lm)Determine Cg Distance = 2"

STEP 14 Calculate Propelling Torque (Tp)

STEP 15 Calculate Deceleration (αd)

Tp = .369 x 87 psi = 32.103 in-lbTp = .369 x psi

1.933 + CgLm =

Lm = 281.43 lb

-1.4175 (La) + 1106.86

RI ROTARY ACTUATOR

(refer to page 204)


�0�

SIZE08


ROTA

RIES

STEP 18 Calculate Max Fix Radial Load (Lf)

STEP 19 Compare Max Fix Radial Load (Lf) to Actual Load

Affixed to Actuator (Lr)

EXAmPLE 1 CONT. STEP 16 Calculate Stopping Torque (Td) (from STEP 16 on page 206)

Ls = (Tp + Td)/.96875Ls = (32.103 + 26.88)/.96875

Ls = 60.9 lb

STEP 17 Calculate Radial Bearing Load at Stopping (Ls) (from chart on page 206)

Lf = Lm - Ls

Lf = 281.43 - 60.9Lf = 220.53

Lf ≥ Lr

220.53 ≥ 32.2 lbRISxx25 FITS THIS APPLICATION

Td = .375 x 71.68 x 1 = 26.88

Td = 26.88

RI ROTARY ACTUATOR


�10

SIZE08


ROTA

RIES

Check for Stopping Capacity STEP 4 Calculate Peak Velocity (ω) RIDxx32

STEP 5 Compare Peak Velocity to Allowable Impact Graph (page 204)

This velocity is in the range of shock pads but not with the attached load Jm of 6.055.

Go to “Sizing an RIxx Unit with Shocks”

APPLICATION INFORmATION - EXAmPLE 2Weight = 15 lb mounting plate & two - 8 lb grippers Rotation Angle = 180°Pressure = 65 psiOrientation = Vertical (grippers facing down)Center of Gravity Distance = 10"Desired Cycle Rate = 1.25 secSafety Factor: Acceleration = 2 Deceleration = 1Cycles per Minute = 20 cyc/min = 1200 cyc/hrAxial Load (La) = 31 lbRadial Load (Lr) = 0

EXAmPLE 2Determine Required Starting Torque for Application STEP 1 Determine (Jm) for mounting Plate

Jm for 2 Point Loads (Gripper)

STEP 2 Determine (αA)


RIDxx32 WILL PRODUCE SUFFICIENT TORQUE

gJm = x

12

Jm =386.4

x12

Jm = .0388198 x 49.333

Jm = 1.9151 in-lb-sec2

gJm = x k2

386.4x 102 = 2.0704 in-lb-sec2

Total Jm = 1.9151 + 2(2.0704)

Jm = 6.056 in-lb-sec2

.035(1.25)2 = 4.032 rad/sec2

Fg a2 + b2

15 (24)2 + (4)2

Fg

8

180

TA =

TA = 6.056 x 4.032 x 2

TA = 48.836 in-lb

.035 x1.25

= 5.04 rad/sec180

Ke = 1/2 x Jm x ω2

Ke = 1/2 (6.056) x (5.04)2 = 76.9 in-lb

10"

4"

24"

gripper

gripper

mountingplate

NOTE: Picture rotated up for clarity.

RI ROTARY ACTUATOR

STEP 6 Compare Peak Velocity to Allowable Impact Velocity for a given (Jm) Load using Shock Absorbers

Compare to graph on page 205.RIDxx32 is not acceptable for this application.Use larger size RISxx50 for this application.

STEP 7 Calculate Kinetic Energy (Ke)

STEP 12 Calculate Allowable Attached Load WeightAxial Load Weight = 31 lb = (La)

STEP 13 Calculate Max Actuator Radial Loading (Lm)Determine Cg Distance = 10"

RISx50 = 2.769 x psiPe = 2.769 x (65 psi) = 179.99 in-lb

Et = Ke + Pe

Et = 76.9 + 179.99 = 256.88 in-lb

Em ≥ Et 577 ≥ 256.9Ea ≥ Et 527 ≥ 256.9

Cycles/Hr = Cycles/min x 60Cycles/Hr = 20 x 60 = 1200

Eh = 1200 x 172.5 in-lb = 207,018 in-lb/Hr

207,018 ≤ 600,000

Lm = = 478.90 lb3.553 + 10

3.553 + CgLm =

-2.699 (31) + 6573.92

-2.699 (La) + 6573.92

STEP 8 Calculate Propelling Energy (Pe)

STEP 9 Calculate Total Energy (Et)

STEP 10 Compare Max. Total Energy (Em) to Total Energy (Et)

SHOCK WILL PERFORM AS DESIRED

STEP 11 Calculate Energy per Hour (Eh)


�11

SIZE08


ROTA

RIES

STEP 14 Calculate Propelling Torque (Tp) RISx50 =

Tp = 2.378 x psiTp = 2.378 x 65 psi = 154.57 in-lb

STEP 15 Calculate Deceleration (αd)

STEP 16 Calculate Stopping Torque (Td) (from STEP 16 on page 206)

Td = 6.056 x 10.368 x 1Td = 62.79 in-lb

STEP 17 Calculate Radial Bearing Load at Stopping (Ls) (refer to chart on page 206)

Ls = (Tp + Td)/1.5625

Ls = (154.57 + 62.79) / 1.5625

Ls = 217.36/1.5625

Ls = 139.11 lb

STEP 18 Calculate Max Fix Radial Load (Lf)

Lf = Lm - Ls

Lf = 478.90 - 139.11

Lf = 339.76 lb

STEP 19 Compare Max Fix Radial Load (Lf) to Actual Load Affixed to Actuator (Lr)

Lf ≥ Lr

339.76 lb ≥ 31 lb

RISxx50 FITS THIS APPLICATION

2.452.45

10.368 rad/sec2

RI ROTARY ACTUATOREXAmPLE 2 CONT.


�1�

SIZE08


ROTA

RIES

1000-8000 ROTARY ACTUATOR

SIZE1(000)2(000)3(000)4(000)5(000)6(000)7(000)8(000)

BASElb/°

.0022

.0043

.0064

.0127

.0093

.0185

.0289

.0578

kg/°.0010.0020.0029.0058.0042.0084.0131.0262

WEIGHT

in1.0001.0001.3751.3752.0002.0003.0003.000

mm25.425.434.934.950.850.876.276.2

BOREDIAmETER

in3/°.007.014.019.038.041.082.185.370

cm3/°.115.229.312.623.6721.3443.0326.064


in-lb/psi.39.781.112.222.364.7210.6021.20

Nm/bar.641.281.213.643.877.7417.3734.75


mAX AXIALBEARING

LOADlb

120

240

370

800

N534

1068

1646

3558

mAX RADIALBEARING

LOADlb

300

600

925

2000

N1334

2669

4114

8896

lb2.33.36.99.710.715.734.442.2

kg1.01.53.14.44.87.115.619.1

ADDER

DISTANCEBETWEEN SHAFT

BEARINGSin

1.375

2.188

2.235

3.750

mm34.9

55.6

56.8

95.3

SPECIFICATIONSPNEUMATIC OPERATING PRESSUREHYDRAULIC OPERATING PRESSURE**OPERATING TEMPERATUREROTATIONAL TOLERANCEBACKLASH AT ANY MID-ROTATION POINT AND

AT END OF ROTATION WITHOUT -A (DOUBLE RACK)BACKLASH AT END OF ROTATION WITH -A* (DOUBLE RACK)BACKLASH ON ALL SINGLE RACK UNITS

(END AND ANY MID-ROTATION)LUBRICATIONMAINTENANCE

SERIES 1000-800020 to 150 psi [1.4 to 10 bar]

40 to 1500 psi [2.8 to 103 bar]-20° to 180°F [-29° to 82°C]Nominal rotation +10° to -0°

1° (2000), 0°30' (4000, 6000), 0°15' (8000)

0° (2000, 4000, 6000, 8000)

1° (1000), 0°30' (3000, 5000), 0° 15' (7000)


PRESSURE RATINGS FOR OPTIONSAll pneumatic rotary actuators have a maximum pressure rating

of 150 psi [10 bar] air. Most hydraulic rotary actuators have amaximum pressure rating of 1500 psi [100 bar], except as noted inthe chart.

Minimum factor of safety at maximum rated hydraulic pressurefor output shaft is 2:1, and for hydraulic chambers is 3:1. ConsultPHD for proof pressure data. Hydraulic ratings based on non-shock,hydraulic service.

HYDSERIES10002000300040005000600070008000

—1000——————

-P -DPLAIN -E OR -mOPTION psi [bar]

—[69]——————

—750—750—750—750

—[52]—

[52]—

[52]—

[52]

—750—750—750—750

—[52]—

[52]—

[52]—

[52]

————750750500500

————

[52][52][35][35]

NOTE: **All hydraulic ratings are based on non-shock hydraulic service.

NOTE: *-A angle adjustment screw must be engaged or adjusted to achieve 0° backlash


�1�

SIZE08


ROTA

RIES

1000-8000 ROTARY ACTUATOR

45° 90° 180° 270° 360° 450°

1000

2000

3000

4000

5000

6000

7000

8000

POWERpneumatichydraulic

pneumatichydraulic

pneumatichydraulic

pneumatichydraulic

pneumatichydraulic

pneumatichydraulic

pneumatichydraulic

pneumatichydraulic

per 90°Adder forDouble

lb2.42.63.53.77.27.410.210.811.112.516.619.235.740

44.853.4

kg1.091.181.591.683.273.364.634.905.035.677.538.7116.1918.1420.3224.22

lb2.52.73.63.97.57.810.811.411.513.217.420.737

42.247.457.8

kg1.131.221.631.773.403.544.905.175.225.997.899.3916.7819.1421.5026.22

lb2.72.94

4.48.18.512

12.812.414.619.123.639.646.552.666.4

kg1.221.321.812.003.673.865.445.815.626.628.6610.7017.9621.0923.8630.12

lb2.93.14.44.88.69.213.114.213.216.120.726.542.250.957.875.1

kg1.321.412.002.183.904.175.946.445.997.309.3912.0219.1423.0926.2234.06

lb3.13.44.85.39.29.914.215.614

17.522.429.444.855.263

83.7

kg1.411.542.182.404.174.496.447.086.357.9410.1613.3420.3225.0428.5837.97

lb3.33.65.25.89.810.515.417

14.919

24.132.347.459.568.292.4

kg1.501.632.362.634.454.766.997.716.768.6210.9314.6521.5026.9930.9341.91

lb0.200.230.390.470.580.691.141.390.841.441.672.912.604.345.208.66

kg0.090.100.180.210.260.310.520.630.380.650.761.321.181.972.363.93

lb0.060.060.060.060.380.380.380.380.50.50.50.52.42.42.42.4

kg0.030.030.030.030.170.170.170.170.230.230.230.231.091.091.091.09

SERIES

WEIGHT TABLE - SINGLE SHAFT EXTENSION ACTUATORS

KINETIC ENERGY TABLEKE mAX. WITH

CUSHIONKE mAX. WITH

SHOCK PADKE mAX.

PLAIN UNITUNIT

RxxA1RxxA2RxxA3RxxA4RxxA5RxxA6RxxA7RxxA8

8.078.0723.1323.1326.3226.3270.5370.53

in-lb0.910.912.612.612.972.977.977.97

Nm3.53.51010

11.511.53131

in-lb0.400.401.141.141.301.303.493.49

Nm2.022.025.785.786.586.5817.6317.63

in-lb0.230.230.650.650.740.741.991.99

Nm


�1�

SIZE08


ROTA

RIES

A

B

2000-8000AIR/OIL TANDEM ROTARY ACTUATOR

45° 90° 180° 270° 360° 450°

SERIES2000400060008000

per 90°Adder forDouble

lb4.712.219.244.7

kg2.135.538.7120.28

lb5

12.920.347.3

kg2.275.859.2121.45

lb5.514.422.552.5

kg2.496.5310.2123.81

lb6.115.824.757.8

kg2.777.1711.2026.22

lb6.617.326.963

kg2.997.8512.2028.58

lb7.1

18.729.168.2

kg3.228.48

13.2030.93

lb0.531.452.205.23

kg0.240.661.002.37

lb0.060.380.52.4

kg0.030.170.231.09

WEIGHT TABLE - SINGLE SHAFT EXTENSION ACTUATORS

OPERATING PRINCIPLEThis feature is available on Series 2000, 4000, 6000, and 8000.

One end functions as a control member only, reducing the effective output torque to match 1000, 3000, 5000, and 7000 respectively.

The illustration shows a tandem actuator with built-in Port Controls®, crossover manifold and oil reservoir. The latter serves as an accumulator to compensate for oil volume changes due to temperature variation.

NOTE: The reservoir should have 20 psi [1.4 bar] pressure at all times to ensure the system remains purged.

3-POSITION mID-POSITIONTOLERANCES & BACKLASH

SERIES2000

4000 & 60008000

TOLERANCE±1°

±0°30'±0°15'

BACKLASH1°30'1°15'

1°

SPECIFICATIONSPNEUMATIC OPERATING PRESSUREOPERATING TEMPERATUREFULL (TOTAL) ROTATIONAL TOLERANCEMID-ROTATIONAL TOLERANCES (3-POSITION UNIT)BACKLASH

AT ANY MID-ROTATION POINT AND ATEND OF ROTATION WITHOUT -A OPTION

AT END OF ROTATION WITH -A OPTION*(DOUBLE RACK)

AT MID-POSITION LOCATION (3 POSITION UNIT)LUBRICATIONMAINTENANCE

SIZE2(000)4(000)6(000)8(000)

lb4.511.518.141.0

kg2.05.28.218.6

BASEin

1.0001.3752.0003.000

mm25.434.950.876.2

BOREDIAmETER

in3/°.007.019.041.185

cm3/°.115.312.6723.032


in-lb/psi.391.112.3610.60

Nm/bar.641.823.8717.37

THEORETICALTORQUEOUTPUT

deg/sec366°348°216°156°

mAX SPEEDAT 80 psi

mAX AXIALBEARING

LOADlb

120240370800

N534106816463558

mAX RADIALBEARING

LOADlb

3006009252000

N1334266941148896

lb/°.0059.0161.0244.0581

kg/°.0027.0073.0111.0264

ADDERWEIGHT

DISTANCEBETWEEN SHAFT

BEARINGSin

1.3752.1882.2353.750

mm34.955.656.895.3

TANDEm SERIES 2000-800020 to 150 psi [1.4 to 10 bar]-20° to 180°F [-29° to 82°C]Nominal rotation +10°/-0°

(see chart below for mid-position tolerance)

1° (2000), 0° 30' (4000, 6000) 0° 15' (8000)

0° (2000, 4000, 6000, 8000)

(see chart below for mid-position backlash)Factory lubricated for rated life

Field repairable

NOTE: *Angle adjustment screw must be engaged or adjusted to achieve 0° backlash. (-A standard on 3-position units)


�1�

SIZE08


ROTA

RIES

2000-8000MULTI-POSITION ROTARY ACTUATOR

HYDSERIES2000400060008000

OPTION psi [bar]-P -D

––––

––

750500

-E OR -m––

[52][35]

––––

––––

––––

NOTE: **All hydraulic ratings are based onnon-shock hydraulic service.

All pneumatic rotary actuators have a maximum pressure rating of 150 psi [10 bar] air. Most hydraulic rotary actuators have a maximum pressure rating of 1500 psi [100 bar], except as noted in chart below.

Minimum factor of safety at maximum rated hydraulic pressure for output shaft is 2:1, and for hydraulic chambers is 3:1. Consult PHD for proof of pressure data. All ratings based on non-shock hydraulic service and with full rotation tubes not being double powered.

PRESSURE RATINGS FOR OPTIONS

SIZE2000400060008000

in1.0001.3752.0003.000

mm25.434.950.876.2

BOREDIAmETER

in3/°.014.038.082.370

cm3/°.229.62313.446.06


in-lb/psi.391.112.3610.60

Nm/bar.641.823.8717.37


mAX AXIALBEARING LOAD

lb120240370800

N534

106816463558


lb3006009252000

N1334266941148896

DISTANCE BETWEENSHAFT BEARINGS

in1.3752.1882.2353.750

mm34.955.656.895.3

SPECIFICATIONSPNEUMATIC OPERATING PRESSUREHYDRAULIC OPERATING PRESSURE**OPERATING TEMPERATUREFULL (TOTAL) ROTATIONAL TOLERANCEMID-POSITION ROTATIONAL TOLERANCES (ALL MID-POSITIONS 2, 3, 4)BACKLASH

AT ANY MID-ROTATION POINT, ALL UNITS AND 4 POSITION,END OF ROTATIONS

AT END OF ROTATIONS ON 3 AND 5 POSITIONS*AT MID-POSITION LOCATIONS (ALL MID-POSITIONS 2, 3, 4)

LUBRICATIONMAINTENANCE

mULTI-POSITION SERIES 2000-800020 to 150 psi [1.4 to 10 bar]

40 to 1500 psi [2.8 to 103 bar] (see option table below)-20° to 180° F [-29° to 82° C]

Nominal rotation +10°/-0°(see chart below for mid-position tolerance)

1° (2000), 0° 30' (4000, 6000), 0° 15' (8000)

0° (2000, 4000, 6000, 8000)(see chart below for mid-position backlash)


NOTE: *Angle adjustment screw must be engaged or adjusted to achieve 0° backlash.

***Rotational position from one intermediate position to another (measured at centers of backlash).

BACKLASH & INTERmEDIATE POSITION TOLERANCES

ROTATIONAL SERIES TOLERANCE*** BACKLASH 2000 ±1° 1° 30' 4000 & 6000 ±0° 30' 1° 15' 8000 ±0° 15' 1°


�1�

SIZE08


ROTA

RIES

C2

E

C1

A

S2 S1 S3

�

T�

I� R�

��

��

PORTS PRESSURIZEDC1 & C2

PORT PRESSURIZED - EFULL CCW POSITION

PORT PRESSURIZED - AFULL CW POSITION

�

T�

J� K�

��

��

PORTS PRESSURIZEDD1 & D2


PORT PRESSURIZED - C2FULL CW POSITION

D1

D2

S2 S1 S3

E

C2

PLUMbING SCHEMATICS: ROTARY ACTUATORS3 POSITION UNITSSERIES 2000-8000

3 POSITION TANDEm UNITSSERIES 2000-8000

CAUTION:Rotary actuators require back pressure in opposite ports before rotating from one position to another. Lack of back pressure or governing media causes uncontrolled angular velocity and improper function of cushions and port controls.

Review the following for typical valve sequencing for controlling standard multi-position actuators. (Disregard for tandem units.)

Starting at full CW position (port A pressurized):

3 POSITION UNITS:

• Rotate from CW to CCW (S3 valve is activated). Energize S1 and S2 valves, then de-energize S2 and S3 valves. Unit will rotate to full CCW position.

• Rotate from CCW to CW (S1 valve is activated). Energize S2 and S3 valves, then de-energize S2 and S1 valves. Unit will rotate to full CW position.

• Rotate from CW to mid-position (S3 valve is activated). Energize S1 and S2 valves, then de-energize S1 and S3 valves. Unit will rotate to mid-position.

• Rotate from mid-position to full CCW (S2 valve is activated). Energize S1 and S3 valves, then de-energize S2 and S3. Unit will rotate to full CCW.

4 POSITION UNITS (same as above plus):

• Rotate from CCW to intermediate position II. (S1 valve is activated). Energize S2, S3, and S4 valves, then de-energize S1, S2, and S3. Unit will rotate to intermediate position II.

5 POSITION UNITS (same as above plus):

• Rotate from CCW to intermediate position IV. (S1 valve is activated). Energize S2, S3, and S5 valves, then de-energize S1, S2, and S3. Unit will rotate to intermediate position IV.


�1�

SIZE08


ROTA

RIES

PLUMbING SCHEMATICS: ROTARY ACTUATORS

4 POSITION UNITSSERIES 2000-8000

5 POSITION UNITSSERIES 2000-8000

�

��


PORT PRESSURIZED - AFULL CW POSITION

J� I�R�

K�

��

T�PORTS PRESSURIZED

D1 & D2PORTS PRESSURIZEDC1 & C2

A

S2 S4 S1

E

C2

S3

C1D1

D2

T��

PORTS PRESSURIZEDC1 & C2

A

S2

S4 S1

E

C2

S3

B1D1

D2

S5

C1

B2

�

��


PORT PRESSURIZED - AFULL CW POSITION�

PORTS PRESSURIZEDD1 & D2

PORTS PRESSURIZEDB1 & B2

I�

J�

R�

K�

P�

N�

Date post:	20-Jan-2019
Category:	Documents
Upload:	lydang
View:	212 times
Download:	0 times

RCC - PHD, Inc · For this application the 8 mm RCC will provide adequate torque at 87 psi. 4)...

Documents