507·5 × 24 standard B15 picks
= 812
This 812, then, is the product of the two drivers, and any two convenient wheels which, multiplied together, give this number can be used—thus 81228 = 29. Therefore the two drivers may have 28 and 29 teeth respectively. The two wheels are found by experiment. If the dividend of the five wheels is 609 a 20 standard wheel is used, and the same drivers as in the preceding case will do. If it is required to change only one wheel, and to have the arrangement such as to give an exact number of picks, or half-picks, or quarter-picks, in the quarter-inch of cloth, by taking the two drivers A and C of such numbers that their product amounts to 507, the number of teeth in the driven wheel B will always equal the number of picks per quarter-inch exactly. Thus 507/13 = 39. Therefore if the drivers A and C have respectively 13 and 39 teeth, every tooth in the driven wheel B will represent one pick per quarter-inch.
Suppose half-picks are required exactly, the method of obtaining the wheels is as follows:—Multiply the 507·5 by 2, which equals 1015, then find two convenient wheels which, multiplied together, produce this number; 35 × 29 = 1015, and the two drivers A and C may be 35 and 29. This will cause every tooth in the driven wheel B to represent half a pick exactly.
Thus with a 35 wheel A, and a 29 wheel C, a 31 wheel B will give 15½ picks per quarter-inch, the other wheels being the same as in an ordinary 507 dividend motion.
The following examples will prove this:—
50 rack × 31 B × 120 stud × 75 beam wheel35 A × 29 C × 15 pinion × 60 quarter-inches
= 15·27
and