Supposing the radius of the sail to be 30 feet, then the sail will commence at th, or 5 feet from the axis, where the angle of inclination will be 72°, at ths or 10 feet from the axis will be 71°, and so on.

In order to utilize the maximum effect of wind, therefore, it is necessary to load the wind-engine so that the number of revolutions of the wheel is proportional to the velocity of the wind.

To find proper number of revolutions of a wind-mill.

7 per lb. 5 per

lb.

From this table an estimate of quantity and suitable sizes for any job can be easily made.

The relative adhesion of nails in the same wood, driven transversely and longitudinally, is as 100 to 78, or about 4 to 3 in dry elm, and 2 to 3 in deal.

HORSE-POWER, VERY ROUGH WAY OF ESTIMATING. The power of a steam engine is calculated by multiplying together the area of the piston in inches, the mean steam pressure in pounds per square inch, the length of stroke in feet, and the number of strokes per minute, and dividing the product by 33,000. Or, multiply the square of the diameter of the cylinder in inches by 0.7854, and this product by the mean engine pressure, and the last product by the piston travel in feet per minute. Divide the last product by 33,000 for the indicated horse-power.

In

the absence of logarithmic formulæ or expansion table, multiply the boiler pressure for cut-off by 0.91; for cut-off by 0.85, cut-off by 0.75, cut-off by 0.68. This will give the mean engine pressure per square inch near enough for ordinary practice, for steam pressures between 60 and 100 lbs., always remembering that the piston travel is twice the stroke multiplied by the number of revolutions per minute.

CASTINGS, CONTRACTION OF.-By Messrs. Bowen & Co., brass founders, London.

The above values fluctuate with the form of pattern, amount of ramming, and temperature of metal when poured. Green sand castings contract less than loam or dry sand castings.

GEARING, SIMPLE RULES ON.-The following rules will apply to both bevel and spur gears. When the term pitch is used, it always signifies diametrical, not circular pitch. For illustrations we will use gears having 64 teeth and 8 pitch.

To Find Pitch Diameter.-Divide the number of teeth by the pitch: 64÷8-8 in. pitch diameter.

To Find Number of Teeth.-Multiply the pitch diameter by the pitch: 8 in. X8-64, number of teeth.

To Find the Pitch.-Divide the number of teeth by the pitch diameter: 648 in. =8, pitch.

To Find Outside Diameter of Spur Wheels.Add 2 to the number of teeth and divide by the pitch: 64+2=66÷8=84 in. O. D.

To Find Circular Pitch.-Divide the decimal 3.1416 by the diametrical pitch: 3.1416 +8=0.3927 in.

Ex.-Diameter of driver (as before) 24 in. X revolutions 150=3,600. Number of revolutions of driven required 300. Then 3,600

=

300 12 in. The rules following are but changes of the same, and will be readily understood from the foregoing examples.

To ascertain the size of the driver: Rule.-Multiply the diameter of the driven by the number of revolutions you wish to make, and divide the product by the required revolutions of the driver; the quotient will be the size of the driver.

To ascertain the size of pulleys for given speed:

Rule.-Multiply all the diameters of the drivers together and all the diameters of the driven together; divide the drivers by the driven; the answer multiply by the known revolutions of main shaft.

PAPER, WALL.-The following table from the New York Newsdealer shows how many rolls of wall-paper are required to cover a room of the dimensions indicated by the figures in the left-hand column, also the number of yards of border necessary

128

To Find the Distance between the Centers of Two Spur Gears.-Divide half the sum of the teeth of both gears by the pitch: 64+ 64 +2=64÷8-8 in. centers.

PULLEYS, RULES FOR CALCULATING THE SPEED OF. The diameter of the driven being given, to find its number of revolutions

Rule.-Multiply the diameter of the driver by its number of revolutions, and divide the product by the diameter of the driven; the quotient will be the number of revolutions of the driven.

Ex. Twenty-four in. diameter of driver X 150, number of revolutions, = 3,600 ÷ 12 in. diameter of driven = 300.

The diameter and revolutions of the driver being given, to find the diameter of the driven, that shall make any given number of revolutions in the same time.

Rule.-Multiply the diameter of the driver by its number of revolutions, and divide the product by the number of required revolutions of the driven; the quotient will be its diameter.

12 X 15 or 13 X 14. 12 X 15 or 13 X 14. 12 X 15 or 13 X 14. 12 X 15 or 13 X 14. 13 X 15. 13 X 15.

15 X 16.

15 X 17.

22222

Deduct one-half roll of paper for each ordinary door or window extra-size 4X7 feet.