Catalog
58Pages

HYDRAULIC AND PNEUMATIC www.rotachydroac.com www.micromaticllc.com >

Welcome to the world of rotary motion. Weappreciateyour interest and are pleased to offer youour catalog, featuring Micromatics extensive line of rotary actuators. Micromatic actuators have earned worldwide renown under the trade names of Rotac Ү and Hyd-ro-ac TM . Micromatic, long recognized as anindustry leader, has been designing and manufacturing rotary actuators for over forty years.Micromatic provides actuators with either hydraulic orpneumatic capability. Your imagination alone limits the number of possible applications.We offer true A to ZӔ capability. Our standard units produce up...

Product: Back Hoes Actuator Used: SS Series Application: Back Hoe Swing > Product: Boat Launch Actuator Used: HS Series > Application: Davit Swing > Product: Wheelchair Lift Actuator Used: Special Application: Raise and LowerWheelchair Platform >

Product: Log Loaders Actuator Used: SS Series and HS Series Application: Boom Swing and Grapple Swing > Product: Paint Spraying Robot Actuator Used: MP Series Application: Rotate Spray Head > Product: Brake Rotor Gage Actuator Used: MP Series Application: Parts Transfer >

Product: Animated Character Actuator Used: MP Series Application: Animated Character Joint Movement (Muscle) > Product: Truck Mounted Booms Actuator Used: SS and HS Series Application: Boom Swingand Grapple Swing >

Product: Printing Presses Actuator Used: SS Series Application: Tensioning Device > Product: Press Brakes Actuator Used: HS Series Application: Press Brake Drive >

Air bleeding in hydraulic systems is usually not required if actuator ismounted with supply ports upward.In other positions,air will gradually dissolve in the oil and be carried away as the actuator is cycled.Special bleed connections are available as an optional feature on some actuators if specified when ordering.Internal by-pass flow is always present to a small degree,andincreases with increase of pressure.On air applications it must be recognized that on stall-out applications,under air pressure,there will be a small continuous by-pass flow.Pure torque out-put from the actuator without...

FIGURE 1. TYPICAL HARMONIC MOTION DRIVE ARRANGEMENT Applications requiring the linear transfer of a load under controlledacceleration and deceleration are quite common.Within limits,this type of motion can be achieved thru a harmonic motion drive.An actuator driven,scotch yoke arrangement as shown in Figure 1 imparts this type motion.The scotch yoke converts the constant speed rotating motion to a sinusoidal motion producing maximum linear force for acceleration,maximum linear speed thru the middle of the actuator stroke,and maximum decelerating forces to slow and stop the load.The...

Example #2 Problem 2: (assume r > 1 in example #1=12" all other parameters remain the same)Jm Find the torque required to rotate a thin hollow pipe about its transverse axis through a given angle in a specified time. > 2 2 = 20386.4 > 2 2 1 = m > 1 7.8 > 2 2 2 = m > 2 2 3 3 = .157 in-lb sec 玱 = same as previous (44.67)T = ∑ Jm α = (Jm > 1 + Jm > 2 ) α = (8.06 + .157) 44.67= 367 in-lb of torque required Solution: T = Jm α = (Jm Problem 1A: Find the torque required to lift a weight and rotate it vertically thru a specified arc in a specified time. > c + Jm > p ) α Solution: T = ∑ (Jm α + For...

Problem 3: Problem 4: Find the torque required to open or close a door through a given angle in a specified time. Find the torque required to rotate several plates of various thicknesses through a given angle in a specified time. Solution: T= Jm > CL Rotation α = ∑ [(Jm > 1 + Jm > 2 + Jm > 3 ) α + (w > 1 r > 1+ w > 2 r > 2+ w > 3 r > 3 )] Jm > 1 = (a M > 1 12 > 12 + b > 12 ) + m > 1 r > 12 Solution: T = Jm Jm > 2 2 = (a M 12 > 22 + b > 22 ) + m > 2 r > 22 CL hinge α M > 3 Jm > A-A = m (a > 2 + b > 2 )12Jm Jm > 3 = (a 12 > 32 + b > 32 ) + m > 3 r > 32 Assume: > CL hinge = Jm > A-A + mr > 2...

Problem 5: Problem 6: Find the torque required to produce a given force as shown in the figure below. Find the torque required to produce a given force in a typical die closer application. Solution: Solution: T = T = > Design Notes: Design Notes: 1.The design should be such that angles > θ 1 and > θ 2 are not permittedto go to zero degrees.2.Force,F,must be less than the bearing capacity of the actuator. 1.The design should be such that angles > θ 1 and > θ 2 are not permittedto go to zero degrees.2.Force,F,may be greater than the bearing capacity of the actuatorsince it is transmitted...

All axes pass through the center of gravity unless otherwise noted.W = total weight of the body. m =W386.4 size="-1">

All axes pass through the center of gravity unless otherwise noted.W = total weight of the body. m =W386.4 15 >

All axes pass through the center of gravity unless otherwise noted.W = total weight of the body. m =W386.4 size="-1">

ABBREVIATIONS:SYMBOLS: BTU.British Thermal Unit 1 BTU = Heat required to raiseAAreatemperature of one pound of water 1װF.aLinear acceleration (FPS > 2 ),rate of change of velocityCDegrees Centigrade ڎ f Angular acceleration (Radians per SEC. Coefficient of frictionFPS.Feet per secondgAcceleration of gravity (IPS > 2 )CAL.Calorie ї 1 CAL.= Heat required to raise temperatureCCompressibility of oil (CU.IN.)of one gram of water 1C.DDensity,mass per unit volumeC.C.Cubic CentimeterEEnergy CU.FT.Cubic FootFForce,(LB.) an influence which produces or tends to CU.IN.Cubic Inchproduce,motion or change...

ACCELERATIONa = =From F = Ma and M = α = Radians/SEC. F FgM W Wg > 2 2 = Degrees/SEC. 57.3 FORCEF = AP FRICTIONMf= W xf Note:Static (or breakaway) friction coefficient is greater than kinetic (or moving) friction coefficient GRAVITYg = 386.4 in./ SEC. > 2 (at sea level) HORSEPOWERHP = = = MASSM = or,at sea level,= ,or = or FULP TN550 1714 63,025 Wg W32.2 W (grams)980 W386.4 NOTE: Mass is constant regardless of altitude. ORIFICE AREASee pressure drop PRESSUREP = (consistent units) FA PRESSURE DROP > For oil hydraulic systems,the following will approximate pressure drop thru short...