Since the diameter of a drum is generally given instead of the radius, it follows that if D = larger diameter, d = smaller diameter, and then, since D = 2R and d = 2r, formula 3 may be written
| (Wₘ + 2Wₖ + 2Wᵣ) | ||
| D = | d ———————— | (4) |
| (Wₘ + 2Wₖ) |
Formula 4 gives only approximate results, which are, however, sufficiently accurate for the mine superintendent’s use, and for this reason friction has been omitted, as it would make the formula much more complex. It may be expressed as a rule as follows:
Rule.—To find the large diameter of a conical drum, multiply the small diameter by the sum of the weight of the material to be hoisted, twice the weight of the cage and car, and twice the weight of the rope; divide this product by the sum of the weight of the material, and twice the weight of the cage and car.
Fig. 10
Applying this rule to the problem given in [Art. 14] and omitting friction,
| 7(4,000 + 12,000 + 6,000) | ||
| D = | ———————————— | = 9.6 feet |
| (4,000 + 12,000) |
The drum would then be 7 feet in diameter at the small end and 9 feet 7¼ inches at the larger end.