Hydraulic Cylinder Calculations

Cylinder Blind End Area (in square inches):

PI x (Cylinder Radius)2

Example: What is the area of a 6" diameter cylinder?

Diameter = 6"
Radius is 1/2 of diameter = 3"
Radius2 = 3" x 3" = 9"

π x (Cylinder Radius)2 = 3.14 x (3)2 = 3.14 x 9 = 28.26 square inches

 

Cylinder Rod End Area (in square inches):

Blind End Area - Rod Area

Example: What is the rod end area of a 6" diameter cylinder which has a 3" diameter rod?

Cylinder Blind End Area = 28.26 square inches
Rod Diameter = 3"
Radius is 1/2 of rod diameter = 1.5"
Radius2 = 1.5" x 1.5" = 2.25"
π x Radius2 = 3.14 x 2.25 = 7.07 square inches

Blind End Area - Rod Area = 28.26 - 7.07 = 21.19 square inches

 

Cylinder Output Force (in pounds):

Pressure (in PSI) x Cylinder Area

Example: What is the push force of a 6" diameter cylinder operating at 2,500 PSI?

Cylinder Blind End Area = 28.26 square inches
Pressure = 2,500 psi
Pressure x Cylinder Area = 2,500 X 28.26 = 70,650 pounds

What is the pull force of a 6" diameter cylinder with a 3" diameter rod operating at 2,500 PSI?

Cylinder Rod End Area = 21.19 square inches
Pressure = 2,500 psi

Pressure x Cylinder Area = 2,500 x 21.19 = 52,975 pounds

 

Fluid Pressure in PSI Required to Lift Load (in PSI):

Pounds of Force Needed ÷ Cylinder Area

Example: What pressure is needed to develop 50,000 pounds of push force from a 6" diameter cylinder?

Pounds of Force = 50,000 pounds
Cylinder Blind End Area = 28.26 square inches
Pounds of Force Needed ÷ Cylinder Area = 50,000 ÷ 28.26 = 1,769.29 PSI

What pressure is needed to develop 50,000 pounds of pull force from a 6" diameter cylinder which has a 3" diameter rod?

Pounds of Force = 50,000 pounds
Cylinder Rod End Area = 21.19 square inches

Pounds of Force Needed ÷ Cylinder Area = 50,000 ÷ 21.19 = 2,359.60 PSI

 

Cylinder Speed (in inches per second):

(231 x GPM) ÷ (60 x Net Cylinder Area)

Example: How fast will a 6" diameter cylinder with a 3" diameter rod extend with 15 gpm input?

GPM = 6
Net Cylinder Area = 28.26 square inches
(231 x GPM) ÷ (60 x Net Cylinder Area) = (231 x 15) ÷ (60 x 28.26) = 2.04 inches per second

How fast will it retract?

Net Cylinder Area = 21.19 square inches

(231 x GPM) ÷ (60 x Net Cylinder Area) = (231 x 15) ÷ (60 x 21.19) = 2.73 inches per second

 

GPM of Flow Needed for Cylinder Speed:

Cylinder Area x Stroke Length in Inches ÷ 231 x 60 ÷ Time in seconds for one stroke

Example: How many GPM are needed to extend a 6" diameter cylinder 8 inches in 10 seconds?

Cylinder Area = 28.26 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds
Area x Length ÷ 231 x 60 ÷ Time = 28.26 x 8 ÷ 231 x 60 ÷ 10 = 5.88 gpm

If the cylinder has a 3" diameter rod, how many gpm is needed to retract 8 inches in 10 seconds?

Cylinder Area = 21.19 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds

Area x Length ÷ 231 x 60 ÷ Time = 21.19 x 8 ÷ 231 x 60 ÷ 10 = 4.40 gpm

 

Cylinder Blind End Output (GPM):

Blind End Area ÷ Rod End Area x GPM In

Example: How many GPM come out the blind end of a 6" diameter cylinder with a 3" diameter rod when there is 15 gallons per minute put in the rod end?

Cylinder Blind End Area =28.26 square inches
Cylinder Rod End Area = 21.19 square inches
GPM Input = 15 gpm
Blind End Area ÷ Rod End Area x GPM In = 28.26 ÷ 21.19 x 15 = 20 gpm

 

This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.

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