Cylinder force and speed calc
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This document provides examples and calculations for determining the force generated by hydraulic cylinders based on bore diameter, piston area, and gauge pressure. It also provides examples and calculations for determining the speed of hydraulic cylinder extension and retraction based on bore diameter, piston area, flow rate, and whether the cylinder is extending or retracting. The document includes tables to look up cylinder force and speed values but also shows how to calculate them using formulas relating piston area, pressure, and flow rate.
Cylinder force and speed calc
- 1. Hydraulic Cylinder Force and Speed Calculations Calculation of Hydraulic Cylinder Force… EXAMPLE: A certain application requires a cylinder force of 25 tons. What should be the cylinder bore diameter used and at what gauge pressure? SOLUTION: The required force is 25 tons × 2000 = 50,000 pounds. Refer to the “Hydraulic Cylinder Force” table on pages 478 and 479 which shows several combinations of piston diameter and PSI pressure which will produce 50,000 pounds of force or more. For example, a 6 inch piston will produce 56,550 pounds at 2000 PSI; a 7 inch piston will produce 57,725 lbs at 1500 PSI; an 8 inch piston will produce 50,265 lbs at 1000 PSI, a 10 inch piston will produce 58,900 lbs. at 750 PSI, etc. So there are many combinations which could be used, and the final choice is a matter of preference or of matching the pressure and flow capability of other components, particularly the pump. In practice, choose a combination which will produce from 10% to 25% more than actually required by the load alone. This will provide a safety allowance which will take care of pressure losses in valves and piping, and mechanical losses in the cylinder. EXAMPLE: How many pounds of force will be developed on the extension stroke of a 3Zv˝ bore cylinder operating at 1500 PSI? If this cylinder has a 1Cv˝ diameter piston rod, how much force will be developed on the retraction stroke? SOLUTION: Refer to the “Hydraulic Cylinder Force” table on pages 478 and 479. The chart shows 12,444 lbs. A solution can also be obtained by using the piston area (8.296 square inches) and multiplying by the pressure (1500 PSI); 8.296 square inches × 1500 PSI = 12,444 lbs. On the retraction stroke the amount of force developed on the 2.41 square inch rod area must be subtracted: 12,444 – 3608 = 8836 lbs. EXAMPLE: What PSI gauge pressure is required for retraction of a 50,000 lb. load with an 8 inch bore cylinder having a 4 inch diameter rod? SOLUTION: The net piston area must be found which is the full piston area minus the rod area. 50.27 (piston area) – 12.57 (rod area) = 37.7 square inches. PSI = 50,000 ÷ 37.7 = 1326 PSI. The actual pressure will be slightly greater due to friction of the piston in the barrel. Calculation of Hydraulic Cylinder Speed… EXAMPLE: At what speed would the piston of a 4 inch bore cylinder extend on an oil flow of 12 GPM? SOLUTION: The table of “Hydraulic Cylinder Speeds” on pages 480 and 481 may be used or the speed figured with the formula which says that “speed is equal to the incoming flow of oil in cubic inches per minute, divided by the square inch area of the piston”. The speed will be in inches per minute. A flow of 12 GPM is 231 × 12 = 2772 cubic inches per minute. The speed is 2772 (flow rate) ÷ 12.57 (piston area) = 220.5 inches per minute. This checks very closely with the value shown in the table on page 480. EXAMPLE: Find the GPM flow necessary to cause a 5 inch bore cylinder to travel at a rate of 175 inches per minute while extending. How fast would this cylinder retract on the same oil flow if it had a 2 inch diameter piston rod? SOLUTION: Flow is determined by multiplying the piston area in square inches times the travel rate in inches per minute. This gives flow in cubic inches per minute. Divide by 231 to convert to GPM: 19.64 (piston area) × 175 = 3437 cubic inches per minute. 3437 ÷ 231 = 14.88 GPM. This checks very closely with 15 GPM at 174 inches per minute shown on the chart on page 480. To find the retraction speed on 14.88 GPM, the net piston area must be found. This is the full piston area minus the rod area: 19.64 (piston area) – 6.5 (rod area) = 16.5 square inches. The flow rate is 3437 cubic inches per minute (equivalent to 14.88 GPM) ÷ 16.5 (net area) = 208 inches per minute. Note that this is faster than the extension speed on the same oil flow. 477