Womack Data Sheet 4: Force Required to Accelerate a Load
The load on a hydraulic cylinder (or motor) consists of these three components: 
This power due to extra pressure is carried as kinetic energy while the load is moving at a constant velocity, and may come back into the system as shock and heat when the load is stopped, unless it can be absorbed by the load in the form of work. The purpose of this data sheet is to show how to calculate the extra pressure or torque needed in a hydraulic system to accelerate an inertia load, Item (3), from standstill to its final velocity in a given time, assuming the pressure needed for Items (1) and (2), the work load and the friction resistance has already been calculated or assumed. 
Calculating for Inertia Load ... Use the following formula to calculate the extra PSI for acceleration to a final velocity in a specified time: (a). F = (V x W) ÷ (g x t) Lbs, in which: If the cylinder bore is known, the accelerating force for its piston can be found directly from the formula: (b). PSI = V x W ÷ (A x g x t), in which:A is piston area in square inches. Other symbols are same as above. 

Problem Data  Vertical Cylinder with Inertia Load 
Steady Load= 35,000 lbs. Cylinder Bore= 4" (Piston Area= 12.57 sq.ins.) 
Initial Velocity= 0; Final Velocity= 12ft/sec. Time Required to Accelerate= 2 Seconds 
Example of Inertia Calculation... Use the problem data in the box to solve for the total PSI needed on the vertically moving cylinder not only to lift the given load, but to accelerate it to its final velocity in the specified time. Or to accelerate it from a lower to a higher velocity.
PSI for Steady Movement... 35,000 lbs. (load weight) ÷ 12.57 (piston area) = 2784 PSI needed to raise the load.
PSI for Acceleration... PSI = (12 x 35,000) ÷ (12.57 x 32.16 x 2) = 520 PSI.
Total PSI... The cylinder must be provided with 2784 + 520 = 3304 PSI to meet all conditions of the problem.
NonInertia Loads... No significant extra PSI is needed to accelerate work loads which consist almost entirely of frictional resistance and negligible mass.
Moment of Inertia of Rotating Load... HOLLOW CYLINDER (Pipe)... SOLID CYLINDER... PRISM... 
Rotating Loads  Driven With Hydraulic Motors (c) T = J x 7 π x RPM ÷ 360 x t inchlbs., in which: Example of Pipe Spinning... First, solve for the moment of inertia using the formula: J = W x (R^{2} + r^{2}) ÷ 2g lnchLbsSecs^{2} J (moment of inertia) = 500 lbs x (52 + 3.52) ÷ 2 x 32.16 Next, use this calculated value of J in Formula (c) above to determine torque required just for acceleration. T = 289.6 X 3.14 X 700 ÷ 360 X 5 = 353.6 inch lbs. Remember, this calculated value of torque is in addition to steady torque required to keep the pipe spinning at a constant RPM. Also remember that the GPM oil flow to the motor must be sufficient to produce the desired 700 RPM. Acceleration From a Lower to a Higher Speed 
© 1988 by Womack Machine Supply Co. This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.