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Hydraulic Cylinder Force and Speed CalculationsCalculation of Hydraulic Cylinder Force… EXAMPLE: A certain application requires a cylinder force of 25 tons. What should be thecylinder 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 combinationsof piston diameter and PSI pressure which will produce 50,000 pounds of force or more. Forexample, a 6 inch piston will produce 56,550 pounds at 2000 PSI; a 7 inch piston will produce57,725 lbs at 1500 PSI; an 8 inch piston will produce 50,265 lbs at 1000 PSI, a 10 inch pistonwill produce 58,900 lbs. at 750 PSI, etc. So there are many combinations which could beused, and the final choice is a matter of preference or of matching the pressure and flowcapability of other components, particularly the pump. In practice, choose a combination which will produce from 10% to 25% more thanactually required by the load alone. This will provide a safety allowance which will take careof 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 muchforce will be developed on the retraction stroke? SOLUTION: Refer to the “Hydraulic Cylinder Force” table on pages 478 and 479. The chartshows 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 areamust be subtracted: 12,444 – 3608 = 8836 lbs. EXAMPLE: What PSI gauge pressure is required for retraction of a 50,000 lb. load withan 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 rodarea. 50.27 (piston area) – 12.57 (rod area) = 37.7 square inches. PSI = 50,000 ÷ 37.7 = 1326PSI. 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 oilflow of 12 GPM? SOLUTION: The table of “Hydraulic Cylinder Speeds” on pages 480 and 481 may beused or the speed figured with the formula which says that “speed is equal to the incomingflow of oil in cubic inches per minute, divided by the square inch area of the piston”. The speedwill be in inches per minute. A flow of 12 GPM is 231 × 12 = 2772 cubic inches per minute. The speed is 2772 (flowrate) ÷ 12.57 (piston area) = 220.5 inches per minute. This checks very closely with the valueshown in the table on page 480. EXAMPLE: Find the GPM flow necessary to cause a 5 inch bore cylinder to travel at arate 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 pistonrod? SOLUTION: Flow is determined by multiplying the piston area in square inches timesthe travel rate in inches per minute. This gives flow in cubic inches per minute. Divide by 231to 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 thechart on page 480. To find the retraction speed on 14.88 GPM, the net piston area must be found. This isthe full piston area minus the rod area: 19.64 (piston area) – 6.5 (rod area) = 16.5 squareinches. The flow rate is 3437 cubic inches per minute (equivalent to 14.88 GPM) ÷ 16.5 (netarea) = 208 inches per minute. Note that this is faster than the extension speed on the sameoil flow. 477
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