Calculate the force to hold the load for 30°.
30°
500 lbs. / 2 = 250 lbs. / cylinder 2224 N / 2 = 1112 N / side
250 / sine of 30° = 500 lbs. 112 / sine of 30°= 2,224 N
Sizing of cylinder:
500 # / 100 psi = 5 sq. in. 2224 / .69 = 3223 mm²
Since it is retracting, add the rod area to the required area.
5 + .7854 = 5.7854 sq. in. 3223 + 499 = 3722 mm²
5.7854 / .7854 = 7.366 3722 / .7854 = 4803
√ 7.366 = 2.71” actual size √4803 = 69 mm required
3” = standard size 80 mm standard
45° Force = 250 / sine of 45° = 353.5 lbs. 1,573 N
353.5 /100 psi = 3.54 sq. in.
3.54 + .7854 = 4.33 sq. in
/.7854 = 5.5
√5.5 = 2.3” 58.42 mm required
Standard size = 2.5” 60 mm standard
60° Force = 250 / sine of 60° = 288 lbs. 1,284 N
288 / 100 = 2.88 sq. in
2.88 + .7854 = 3.67 sq. in
3.67 / .7854 = 4.67
√4.67 = 2.16” required 54.8 mm required
Standard size = 2.5” 60 mm standard
Notes:
- When lifting anything in a sling, from a common hook out to the end of perhaps a beam, the load on the chains will be greater than the actual load if the angle is less than 30°. This is true even when hanging a picture on a wall.
- 250# / sine of 90° = 250 lbs. per cylinder. That’s the same as lifting straight up per cylinder.
