Problem: Standard Size Cylinder 2

Here we have 100 lbs. of pull on this roll of paper. Looking at the design, we have a 2:1 mechanical disadvantage on the end of the arm that equals a force of 200 lbs. The arm has a 3:1 mechanical disadvantage. That is equal to a 6:1 disadvantage.

Step ONE: Force x force distance = effort x effort distance. That is 200 lbs. x 9″ = 3 x load on the cylinder.  Two hundred x 9 / 3 = 600 lbs. on the cylinder.

Step TWO: Now use F = PA where the force is 600 lbs. and the pressure is 115 psi. Six hundred / 115 = 5.2 square inches needed for the cylinder.

Step THREE: Because we are pulling on the rod end of the cylinder, we will need to find the effective rod end area. The rod is 1 ½”.  Area equals 1.5² x 0.7854 = 1.767. Now we need to add the area of the cylinder and the area of the rod together to find the area of the piston. 5.2 + 1.767 = 6.98 in².

Step FOUR: Working backwards with A = D² x 0.7854 is 6.98 / 0.7854 = 8.89 and now we take the √  (square root) to find D and that equals 2.98 diameter. We will round that up to a standard 3-inch cylinder.

Note:  The “Lightning Reference” 8th edition on page 136 will show standard size cylinders. Go down the area column and look for the closes area equal to 5.2 or slightly greater. Be sure you are using the pull and not the push of any cylinder. Three inch cylinder with 1.5″ rod, you will see the area is 5.3 sq. in. – that is just slightly greater that the 5.2 sq. in. that we need.

With this design, one inch of movement on the cylinder will take up six inches of slack in the roll of paper.