rcc - phd, inc · for this application the 8 mm rcc will provide adequate torque at 87 psi. 4)...
TRANSCRIPT
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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
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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
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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
k dim isradiusof rod
a Fg1
Fg2
b
kFg
RCC ACTUATOR
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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
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SIZING EXAmPLE 1
1) Determine the load information
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
2) Determine minimum actuator based on radial or axial load
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 (rad/sec) = .035 x
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
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SIZING EXAmPLE 2
1) Determine the load information
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
2) Determine minimum actuator based on radial or axial load
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 (rad/sec) = .035 x
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.
Rotational Angle (deg)[Rotational Time(sec)]2
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)
Rotational Angle (deg)[Rotational Time(sec)]2
180°1.0 sec
Fg1
g
Fg
g
Fg2
g
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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.
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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
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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]
Center of Gravity Distance in [mm]
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]
Center of Gravity Distance in [mm]
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]
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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
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 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
k dim isradiusof rod
aa
b
c
Fg2
aFg1
b
k dim isradiusof rod
a Fg1
Fg2
bk
Fg
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
RF ROTARY ACTUATOR
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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
(Time of Rotation in Seconds)2
180
.035 x Rotation Angle in Degrees
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
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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 4 Calculate Peak Velocity
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
(Time of Rotation in Seconds)2
180
.035 x Rotation Angle in Degrees
Average Velocity =.5
= 360 degrees/sec
Peak Velocity = .035 x 360 = 12.6 rad/sec
Peak Velocity = .035 x Average Velocity (deg/sec)
180
RF ROTARY ACTUATOR
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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 4 Calculate Peak Velocity
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
(Time of Rotation in Seconds)2
180
.035 x Rotation Angle in Degrees
Average Velocity =1.0
= 180 degrees/sec
Peak Velocity = .035 x Average Velocity (deg/sec)
Peak Velocity = .035 x 180 = 6.3 rad/sec
180
RF ROTARY ACTUATOR
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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
CUSHIONANGLE ADJUSTMENT
CUSHIONANGLE ADJUSTMENT
CUSHIONANGLE ADJUSTMENT
CUSHIONANGLE ADJUSTMENT
CUSHIONANGLE ADJUSTMENT
CUSHIONANGLE ADJUSTMENT
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
THEORETICALTORQUE OUTPUT
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)
Factory lubricated for rated lifeField repairable
Rl ROTARY ACTUATOR
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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)
(Time of Rotation in Seconds)2
.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.
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RIES
Jm = x k2Fg
g
kFg
k dim isradiusof rod
Jm = x12
Fg
ga2 + 3k2
( ) ( )Fg1
gJm = x (4a2 + 3k2)
12+ Fg2
g(4b2 + 3k2)
12
( )k2Jm = +L2
3Fg
gxx 1
4
k dim isradiusof rod
b-a2
x
kk
Jm = x k2
2Fg
g
DiskEnd mounted on center
DiskMounted on center
L
k
a
aFg1
Fg2
b
a2 + b2
12Jm = xFg
g
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
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
Tg = Rotating Vertically(with gravity)
T = Rotating Horizontally(without gravity)
UNBALANCED LOADS
a
b
Tg = [(Jm x α) + (Fg x k)] x SF
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
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
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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
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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
Jm = .0146 in-lb-sec2 Jm = .00165 N-m-sec2
Total Jm Total Jm = .0146+.0104=.025 in-lb-sec2 = .00165+.00117=.00282N-m-sec2
b) Determine required acceleration of the load:
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
rotation angle (deg)rotational time (sec)
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.
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 = .035 x ω = .035 x = 12.6 rad/sec
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 .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.
Numbers in [ ] are for metric units and are in mm.
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
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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
DISPLACEmENTVOLUmE/DEG
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
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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)
(Time of Rotation in Seconds)2
.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
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SIZE08
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ROTA
RIES
Jm = x k2Fg
g
k dim isradiusof rod
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
k dim isradiusof rod
Tg = [(Jm x α) + [(Fg2 - Fg1) x (a + ( ))]] x SF
kFg
k
c Fg2
a
Fg1b
a
L
kk
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
aFg1
Fg2
b
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
a
b
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
RA ROTARY ACTUATOR
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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
b) Determine required acceleration of the load:
c) Calculate required torque:
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.
3) Determine the stopping capacity of the actuator for the application:
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
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
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
Numbers in [ ] are for metric units and are in mm.
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 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
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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
Jm = .0104 in-lb-sec2 Jm = .00117 N-m-sec2
RECTANGULAR PLATE ImPERIAL mETRIC
Jm = .0146 in-lb-sec2 Jm = .00165 N-m-sec2
Total Jm Total Jm = .0146+.0104=.025 in-lb-sec2 = .00165+.00117=.00282N-m-sec2
b) Determine required acceleration of the load:
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 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)
Numbers in [ ] are for metric units and are in mm.
T = [(Jm x α)+(Fg x k)] x SF
T = [(.00117 x 25.2) + (4.45 x .0508)] x 2
T = .51 N-m
rotation angle (deg)rotational time (sec)
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.
3) Determine the stopping capacity of the actuator for the application:
a) Determine the estimated peak rotational velocity using Equation A on page 199.
ω = .035 x ω = .035 x = 12.6 rad/sec
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 .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
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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
THEORETICALTORQUE OUTPUT
ROTATIONALVELOCITY mAX
mAX AXIALBEARING LOAD
lb
292
511
697
N
1300
2275
3100
mAX RADIALBEARING LOAD
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
RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50
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)
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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.”
(Time of Rotation in Seconds)2
.035 x Rotation Angle in Degrees
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
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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
RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50
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
)
moment of Inertia (in-lb-sec2) [Nms2]
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
)
moment of Inertia (in-lb-sec2) [Nms2]
RIDx25
RISx25
mAX ALLOWABLE CHART (Em)
ACCEPTABLE mOTION CHART (Ea)UNITSIZE
RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50
in-lb6696154213527754
VELOCITYrad/sec
57.724.258.527.628.919.7
Nm7.4610.817.424.159.585.2
ET*
UNITSIZE
RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50
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
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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
RISxx25RIDxx25RISxx32RIDxx32RISxx50RIDxx50
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
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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
k dim isradiusof rod
347
8
k
L
k k
a
b
c
Fg2
a Fg1
b
k dim isradiusof rod
a Fg1
Fg2
b
Tg = [(Jm x α) + [(Fg2 - Fg1) x (a + ( ))]] x SF
kFg
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 LOADSUNBALANCED LOADS
Jm = x k2
2Fg
g
a2 + b2
12Jm = xFg
g
a
Tg = [(Jm x α) + (Fg x k)] x SF
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
RI ROTARY ACTUATOR
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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)
STEP 3 Starting Torque
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)
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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
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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)
STEP 3 Starting Torque
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)
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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.
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SIZE08
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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
DISPLACEmENTVOLUmE/DEG
in-lb/psi.39.781.112.222.364.7210.6021.20
Nm/bar.641.281.213.643.877.7417.3734.75
THEORETICALTORQUE OUTPUT
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)
Factory lubricated for rated lifeField repairable
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
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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
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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
DISPLACEmENTVOLUmE/DEG
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)
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SIZE08
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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
DISPLACEmENTVOLUmE/DEG
in-lb/psi.391.112.3610.60
Nm/bar.641.823.8717.37
THEORETICALTORQUE OUTPUT
mAX AXIALBEARING LOAD
lb120240370800
N534
106816463558
mAX RADIALBEARING LOAD
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)
Factory lubricated for rated lifeField repairable
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°
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SIZE08
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ROTA
RIES
C2
E
C1
A
S2 S1 S3
�
T�
I� R�
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PORTS PRESSURIZEDC1 & C2
PORT PRESSURIZED - EFULL CCW POSITION
PORT PRESSURIZED - AFULL CW POSITION
�
T�
J� K�
��
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PORTS PRESSURIZEDD1 & D2
PORT PRESSURIZED - EFULL CCW POSITION
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.
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SIZE08
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ROTA
RIES
PLUMbING SCHEMATICS: ROTARY ACTUATORS
4 POSITION UNITSSERIES 2000-8000
5 POSITION UNITSSERIES 2000-8000
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PORT PRESSURIZED - EFULL CCW POSITION
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
�
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PORT PRESSURIZED - EFULL CCW POSITION
PORT PRESSURIZED - AFULL CW POSITION�
PORTS PRESSURIZEDD1 & D2
PORTS PRESSURIZEDB1 & B2
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