Referring to Table relating to flat wire ropes in Hoisting, Part 2, it is found that a flat steel rope 6 inches by ½ inch in size and with a breaking strength of 150,000 pounds weighs 5.1 pounds per foot; hence, 2,000 feet of it weighs 2,000 × 5.1 = 10,200 pounds. The total load on the rope will then be 19,000 pounds, made up as follows:
| Pounds | |
| Weight of material | 5,000 |
| Weight of skip | 3,000 |
| Friction, 10 per cent. | 800 |
| Weight of rope | 10,200 |
| Total | 19,000 |
This rope gives a factor of safety of 150,000/19,000 = 7.8, which is not quite enough when figured from the dead load without that due to acceleration.
An 8" × ½" rope with a breaking strength of 200,000 pounds weighs 6.9 pounds per foot; hence, 2,000 feet of it weighs 2,000 × 6.9 = 13,800 pounds. The load on the rope will then be 22,600 pounds, made up as follows:
| Pounds | |
| Weight of material | 5,000 |
| Weight of skip | 3,000 |
| Friction, 10 per cent. | 800 |
| Weight of rope | 13,800 |
| Total | 22,600 |
This rope gives a factor of safety of 200,000/22,600 = 8.8.
Substituting the foregoing weights of material, skip, and rope in formula 4, in [Art. 17], gives
| d(5,000 + 6,000 + 27,600) | |
| D = | ———————————— |
| (5,000 + 6,000) |
Hence, the equation of moments is D = 3.5d. In other words, the large diameter, or that of the last coil of rope, should be 3.5 times the small diameter, or that of the reel hub.