The size of drum gear evidently depends upon the method of fastening to the drum, and, other things being equal, should be kept as small as possible. One way would be to key the gear on the outside of the drum, another to bolt the gear to the end of the drum. The latter has the advantage that a standard gear pattern can be used with the slight change of addition of bolt flange on the arms. 'This makes a simple, direct, and strong drive, the bolts being in shear.
Sketching this arrangement as the preferred one (Fig. 2A), it is evident that the diameter of the gear should be at least as large as the drum in order to keep the tooth load down to a reasonable figure. On the other hand, if made too large, it spreads out the machine and destroys its compactness. As a diameter of 36 inches is not excessive, let us assume this, and see if a desirable proportion of gear tooth can be found to carry the load.
For a pitch diameter of 36 inches there will be a theoretical load of 5,000 x 27/ 36 = 3,750 pounds at the pitch line. But the load on the tooth must not only impart a pull of 6,000 pounds to the rope, but must overcome friction between the gear teeth in action, also between the drum shaft and its bearings. Assuming the efficiency between the rope and tooth load to be 95 per cent, the net load, therefore, which the tooth must take is 3,750/.95 = 3,947, say 4,000 pounds.
Assuming involute teeth, and applying the "Lewis" formula, (Part II, "Gears"):
W=s X p X f X y W= 4,000 s = 6,000 4,000=6,000 X p X f X .116 y=.116 (number of teeth assumed at 75) px f = 4,000/6,000 x .116 =5.7 inches p = circular Pitch f = face of gear Let f =3_p (a reasonable proportion for machine-cut teeth). Then 3 X p2 =5.7 p2 =1.9___ p = √ 1.9=1.378 inches.
The diametral pitch corresponding to this is:
3.1416/1.378 = 2.28 which is just between the regular standard pitches, 2 and 2½, for which stock cutters are made. To be safe, let us take the coarser pitch, which is 2. The circular pitch corresponding to this is:
3.1416/2 = 1.57, and making the face about three times the circular pitch gives:
3 X 1.57 = 4.71, say 4½ inches.
The number of teeth in the gear is then 36 X 2 = 72. Referring to the value assumed for the tooth factor in calculation above, it is seen that y was based on 75 as the number of teeth, which is near enough to 72 to avoid the necessity of further checking the result.
The pinion to mesh with this gear should be as small as possible in order to get a high-speed ratio between pinion shaft and drum, otherwise an excessive ratio will be required in the pulleys, making the large one of inconvenient size. Small pinions have the teeth badly undercut and therefore weak, 13 teeth being the lowest limit usually considered desirable, for that reason. Choosing that number, we have a pitch diameter of 13/2 = 6.5 in., which is probably ample to take the shaft and key, and still leave sufficient stock under the tooth for strength. If made of cast iron, however, the pinion teeth, on account of the low number, will be narrower at the root than those of the gear of 72 teeth. Yet it was upon the basis of the latter that the pitch was chosen, for it will be remembered that the value of y in the formula was taken at .116. Hence the pinion will be weaker than the gear unless we make it of stronger material than cast iron, of which the large gear is supposed to be made. Steel lends itself very readily to this requirement; and in practice, pinions of less than 20 teeth are usually made of this material, hence we shall specify the pinion to be of steel.
The question now is whether or not we can get a suitable ratio in the pulleys without making the large one of inconvenient size, or giving the motor too slow speed for an economical proportion.
Suppose we limit ourselves to a diameter of 42 inches for the large pulley, and try a ratio of 4 to 1; this will give a diameter for the small pulley of 42/4 = 10½ inches. We shall then have:
Total ratio between drum and motor..... 72/13 X 4 =288/ 13 =22.2
Rev. per min. of drum to give 150 f. p. m. of rope................................... 150/7.07 = 21.2
Rev. per min. of motor......................22.2 X 21.2=470
Horse-power of motor at 80 per cent efficiency (150 x 5,000)/(33,000x.80)=30.
A 30 II. P. motor running 470 r. p. m. would be classed as a slow speed motor and would be a heavier machine and cost more than one of higher speed. It will be noticed, however, that the diameter of the small pulley is already quite reduced, and it is hardly desirable to decrease it still further. Neither can we increase the large pulley, as we have already set the limit at 42 inches. Hence, for our present problem we cannot improve matters much without increasing the size of the large gear, which is undesirable, or putting in another pair of gears, which is contrary to the conditions of the problem. As such a motor is perfectly reasonable, we shall assume it to be chosen for the purpose.
In commercial practice it would be well to pick out some standard make of motor of the required horse-power, note the speed as specified by the makers, and then, if possible, suit the ratio in the machine to this speed. It is always best to use standard machinery, if possible, both from the standpoint of first cost, as well as ease of replacing worn parts. Machinery ordered special is expensive in first cost of designing, patterns, and tools, and extra spare parts for emergency orders are not often kept on hand.