10 shows the simplest form of gear train, having but two gears. Fig. 11 shows the same gears A and B, as in Fig. 10, but with a third gear, usually called an intermediate gear, between them. The intermediate gear C can be used for either of two reasons:

1. To connect A and B and thus permit of a greater distance between the centers of A and B without increasing the size of the gears; or

2. To reverse the direction of rotation of either A or B. If A turns in a clockwise direction, as shown in both Figs. 10 and 11, B in Fig. 10 will turn in the opposite, or counter-clockwise direction, but in Fig. 11, B will turn in the same direction as A.

The introduction of the intermediate gear C has no effect on the speed ratio of A to B. If A has 48 teeth and B 96 teeth, the speed ratio of A to B will be 2 to 1 in either Fig. 10 or Fig. 11. In Fig. 10 suppose A to be the driver.

Hence, the speed ratio of A to B is 2 to 1.

In the case shown in Fig. 11 when A moves a distance of one tooth, the same amount of motion will be given to C, and C must at the same time move В one tooth. To move B 96 teeth, or one revolution, will require a motion of 96 teeth on A, or two revolutions of A. Hence, A will turn twice to each one turn of

B, or the speed ratio of A to B is 2 to 1, just as in the case of Fig. 10.

62. Compound Gear and Pulley Trains.-Quite often it is desired to make such a great change in speed that it is practically necessary to use two or more pairs of gears or pulleys to accomplish it. If a great increase or reduction of speed is made by a single pair of gears or pulleys, it means that the difference in the diameters will have to be very great. The belt drive of a lathe is an example of a compound train of pulleys, though here the train is used chiefly for other reasons. In the first step, the pulley on the lineshaft drives a pulley on the countershaft; then another pulley on the countershaft drives the lathe. The back gearing on a lathe is an example of compound gearing, two pairs of gears being used to make the speed reduction from the cone pulley to the spindle and face plate.

Fig. 12 shows a common arrangement of compound gearing. Here A drives B and causes a certain reduction of speed. B and C are fastened together and therefore travel at the same speed. A further reduction in speed is made by the two gears C and D. A and C are the driving gears of the two pairs and B and D are the driven gears.

In making calculations dealing with compound gear or pulley trains, we might make the calculations for each pair as explained in Chapter VIII and then proceed to the next pair, etc., but this can be shortened to form a much simpler process.

The speed ratio for a pulley or gear train is equal to the product of the ratios of all the separate pairs of pulleys or gears making up the train.

In using this principle for calculations, the ratios are written as fractions and we have the following formula:

R. P. M. of last driven gear
R. P. M. of first driver

Product of Nos. of teeth of all drivers
Product of Nos. of teeth of all driven gears

Or, if we want the ratio stated the other way around—

R. P. M. of first driver
R. P. M. of last driven gear

Example:

Product of Nos. of teeth of all driven gears
Product of Nos. of teeth of all drivers

Let us calculate the speed ratio for the train of gears in Fig. 12. This would be the ratio of the speed of A to the speed of D.

A is the first driver and D the last driven gear, and the ratio of their speeds is the ratio for the whole train.

Speed of A: Speed of D=10:1

In other words, A revolves 10 times as fast as D.

Problems in getting the speed ratios of pulley trains are solved in the same way except that diameters are used instead of numbers of teeth.

Speed of last driven pulley_Product of diameters of all driving pulleys Speed of first driving pulley Product of diameters of all driven pulleys

Example:

Let us take the pulley train of Fig. 13 and calculate the ratio of the speeds of pulleys A and D. A and C are the drivers and B and D are the driven pulleys.

Trains are frequently used having combinations of pulleys and gears. In nearly all machine tools, we will find both pulleys and gears between the lineshaft and the work. In wood-working

machinery, on the other hand, we usually find only pulleys and belts, on account of the high speeds at which the machines are In calculating the speed ratios of these combined trains, we can use the diameters of the pulleys and the numbers of teeth of the gears in the same formula.

run.

If a problem calls for the calculation of the size of one pulley or gear in a train, all the others and the speed ratio being known, start at one or both ends of the train and work toward the gear or pulley in question until you get a proportion which will give the desired quantity.

Example:

The punch shown in Fig. 14 is to be set up so that it will make 20 strokes per minute. (The 80 tooth gear must, therefore, run 20 R. P. M.) The punch is to be driven from a countershaft and we want to calculate the size of the pulley to put on the countershaft, to drive the punch at the desired speed. We find that the main lineshaft runs 240 R. P. M. and carries a 16-in. pulley which drives a 24-in. pulley on the countershaft.

Working from the lineshaft:

24 in. X R. P. M. of countershaft=16X240.

FLY WHEEL

This is also the R. P. M. of the 24-in. pulley and this pulley is driven by the unknown pulley on the countershaft, which we have found runs 160 R. P. M.

Another way to solve this would be to write out an equation for the entire train. using X to represent the pulley whose size we want to find.

Hence, X=12 in., Answer.