At the end of the hoist there is a gross load on the loaded side of 11,000 pounds, made up as follows:
| Pounds | |
| Weight of material | 4,000 |
| Weight of mine car | 3,000 |
| Weight of cage | 3,000 |
| Friction, 10 per cent. | 1,000 |
| Total | 11,000 |
This is equal to a gross load moment of 11,000 pounds × 3½ feet = 38,500 foot-pounds, but there is a counterbalancing load of 8,100 pounds, made up as follows:
| Pounds | |
| Weight of mine car | 3,000 |
| Weight of cage | 3,000 |
| Weight of rope | 3,000 |
| Total | 9,000 |
| Less friction, 10 per cent. of 6,000 | 600 |
| 8,400 |
This is equal to a counterbalancing load moment of 8,400 pounds × 3½ feet = 29,400 foot-pounds, and leaves a net load moment against the engine of 38,500-29,400 = 9,100 foot-pounds. In other words, the load moment that the engine has to overcome varies from 30,100 foot-pounds at the beginning of the hoist to 9,100 foot-pounds at the end of the hoist.
15. Hoisting With Conical Drums.—Conical drums are designed to make the work of the engine as nearly uniform as possible throughout the hoist. To accomplish this, when the cage is at the bottom of the shaft, and the load is therefore heaviest, the rope winds on that part of the drum having the smallest diameter. As hoisting continues, the rope winds on a gradually increasing diameter of drum, and when the cage is at the top of the hoist, and the load therefore least, the rope is winding on that part of the drum having the greatest diameter; in this way, the moment of the load at every point of the hoist is approximately the same. The great difference in the loads at different parts of the hoist is due mainly to the variation in the weight of the rope hanging from the drum; hence, the less the weight of the rope in proportion to the total load on the engine, the more nearly uniform is the load on the engine.
Fig. 9
16. [Fig. 9 (a)] shows the condition at the beginning of the hoist when conical drums are used. Cage a is at the bottom and carries a loaded car; cage b is at the top and carries an empty car. The net moment that the engine must overcome is the sum of the weight of the material to be hoisted, weight of the cage and car at a, and the weight of the rope attached to a, multiplied by the small radius r of the drum, minus the weight of the car and cage at b, multiplied by the large radius R of the drum.
[Fig. 9 (b)] shows the condition of things at the end of the hoist, when the cage a is at the top and cage b at the bottom. The loaded car and cage a, whose rope in [Fig. 9 (a)] was winding on the smallest diameter of the drum, is now at the top and the rope is winding on the largest diameter of the drum. The cage b with the empty car is now at the bottom and the rope is unwinding from the smallest diameter of the drum. The net moment that the engine must overcome in this position is equal to the sum of the weight of the material hoisted, the weight of the cage a and the car, multiplied by the larger radius R of the drum, minus the sum of the weights of the cage b, the car, and the rope, multiplied by the small radius r of the drum.