The same rule and calculation would apply if the first shaft had carried a 40-teeth cog-wheel, and the second a 16-teeth wheel. In taking the dimensions of a pulley for calculations the diameter is often taken; it does not matter, though, if the circumference be taken, but care must be exercised in taking the same dimension for the driven as is taken for the driver. If the diameter is taken of one, the diameter must be taken of the other. _To get Speed of Loom from Engine._--Multiply the engine speed by all the driving pulleys, and divide by the driven ones. If the engine make 46 strokes per minute, spur-wheel 105 teeth, second motion pinion 52 teeth; also on same a 52 driving a 49 on line shaft in shed. Pulley on line shaft on which is a 15-inch drum driving a loom pulley on the crank-shaft of 8 inches. The driving and driven pulleys are always alternate; then as the first must be a driver-- (46 × 105 × 52 × 15)/(52 × 49 × 8) = 185 nearly. The answer gives the _calculated_ picks per minute. About 4 per cent. must, however, be allowed for slippage, reducing the 185 to an actual speed of about 177. _To find the Size of Pulley for any required Speed._--Find the ratio of the given speed and arrange size of pulley accordingly. Suppose a shaft running at 100 revolutions per minute has to drive a loom-shaft at a speed of 180 picks per minute the ratio of speed is as 100 to 180 or as 5 to 9; arrange the pulleys in this proportion--say 10 inches and 18 inches, the larger pulley being on the driving shaft. _To alter Speeds._--Calculate in proportion to the alteration. If a twill-shaft is driven by a 30-cog wheel, and revolves at a speed of 45 revolutions per minute for a 4-leaf twill, and it is desirable to change this to a 3-leaf twill with the twill-shaft at 60 revolutions per minute, then, as 45 is to 60, so 30: _x_-- _x_ = (60 × 30)/45 = 40 A 40 wheel must now _drive_ the wheel on the twill-shaft, and the speed will be increased one-third more. ENGINES. The strength of a steam engine is indicated in horse-powers. A horse-power is taken as the capacity of performing 33,000 foot-pounds of work in one minute; lifting 3300lb. 10 feet high, or 10lb. 3300 feet high would be 33,000 foot-pounds of work. _To obtain the Indicated Horse-power--the most usual Standard._--A diagram is taken from the engine by a small apparatus, and this diagram, when measured and averaged at different points of its length, gives the mean pressure of steam in the cylinder. Multiply this by the speed of the piston, by the area of the piston, and divide by 33,000, and the I.H.P. is to hand. 39·81 average pressure per square inch, area of piston 400 square inches, length of stroke 5-1/2 feet, strokes per minute 40 (or 11 feet both ways)-- (39·81 × 400 × 11 × 40)/33,000 = 212·32 I.H.P. Nominal horse-power (condensing) = area of piston divided by 22; ditto high pressure = area of piston divided by 11.