23. [Fig. 13] represents a coil of flat rope whose greater diameter D and smaller diameter d are to be determined. The area of the hub about which the rope is to coil is (¼)πd², while the area included by the outer coil of rope is (¼)πD² hence, the area of annular space occupied by the rope is
(¼)πD² - (¼)πd² = (¼)π(D² - d²).
Such values for D and d must be chosen that the equation of moments in [Art. 22] is satisfied, while the area (¼)π(D²-d²) must correspond to the space occupied by the given rope when rolled.
Fig. 13
Illustration.—2,000 feet of rope ½ inch thick requires
| 2,000 × 12 | |
| ————— | = 12,000 |
| 2 |
square inches in which to be coiled. To satisfy the equation of moments, D must equal 3.5 d; hence, to satisfy both these conditions
(¼)π[(3.5d)² - d²] = 12,000;
- d = 37 inches, or 3 feet 1 inch;
- D = 37 × 3.5 = 129.5 inches, or 10 feet 9½ inches.