## How To Obtain The Index Of An Engine Lathe

If you will note what thread the lathe will cut when two given gears are in place, you can easily construct a table that will show you just what thread any two gears will cause the lathe to cut. Suppose that two sixty-threes cause 12 threads to the inch. Then place 12 in the space A in the diagram below.

 Stud. Screw 28 33 35 42 49 56 63 70 77 84 91 98 105 112 28 33 35 42 49 b 56 a C 63 B A D 70 E c 77 d 84 91 98 105 112
 Now, 63 : 56 :: A : C Direct proportion. 63 : 70 :: A : E Also, 56 : 63 :: A : B Inverse proportion. 70 : 63 :: A : D

The spaces may all be filled except a, b, c, d, etc., which it is useless to fill, as only your 63 gear is duplicated. A half-day's time will be sufficient for a good mathematician to fill out the table.

## Nails, Memoranda Concerning

This table will show at a glance the length of the various sizes, and the number of nails in a pound. They are rated from "3-penny" up to "20-penny." The first column gives the name, the second the length in inches, and the third the number per pound:

 3-penny, 1 in. long, 557 per lb. 4-penny, 1¼ in. long, 353 per lb. 5-penny, 1¾ in. long, 232 per lb. 6-penny, 2 in. long, 167 per lb. 7-penny, 2¼ in. long, 141 per lb. 8-penny, 2½ in. long, 101 per lb. 10-penny, 2¾ in. long, 98 per lb. 12-penny, 3 in. long, 54 per lb. 20-penny, 3½ in.long, 34 per lb. Spikes, 4 in. long, 16 per lb. Spikes, 4½ in. long, 12 per lb. Spikes, 5 in. long, 10 per lb. Spikes, 6 in. long, 7 per lb. Spikes, 7 in. long, 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 formulae or expansion table, multiply the boiler pressure for 5/8 cut-off by 0.91; for ½ cut-off by 0.85, 3/8 cut-off by 0.75, 3/10 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.

 Inch. Ins. of length. In thin brass castings....... 1/8 in 9 In thick " " ...................... 1/8 in 10 In zinc castings...................... 5/16 in 12 In lead, according to purity. 3/16 to 5/16 in 12 In copper " " " . 3/16to 7/32 in 12 In tin, " " " . 1/32 to 1/16 in 12 In silver, " " 1/8 in 12 In cast iron, according to purity, small castings. . . . 1/10 in 12 In cast steel, according to purity, pipes.................... 1/8 in 12

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.X 8 = 64, number of teeth.

## To Find The Pitch

Divide the number of teeth by the pitch diameter: 64 ÷ 8 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 = 8¼ in. O. D.

## To Find Circular Pitch

Divide the decimal 3.1416 by the diametrical pitch: 3.1416 ÷ 8 = 0.3927 in.

## 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 = 128 ÷ 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.

### 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

 Size of Room. Height of Ceiling. Number of Doors. Number of Windows. Rolls of Paper. Yards of Border. 7X9........... 8 1 1 6 11 7X9........... 9 1 1 7 11 7X9.......... 10 1 1 8 11 7X9........... 12 1 1 10 11 8 X 10........... 8 1 1 7 12 8X10........... 9 1 1 8 12 8X10........... 10 1 1 9 12 8X10........... 12 1 1 11 12 9X11........... 8 1 1 8 14 9X11........... 9 1 1 10 14 9X11.......... 10 1 1 11 14 9X11........... 12 1 1 13 14 10X12........... 8 1 1 9 15 10X12........... 9 1 1 10 15 10X12........... 10 1 1 11 15 10X12........... 12 1 1 13 15 11X12........... 8 2 2 8 16 11X12........... 9 2 2 9 16 11X12........... 10 2 2 10 16 11X12........... 12 2 2 13 16 12X13........... 8 2 2 8 17 12X13........... 9 2 2 10 17 12X13........... 10 2 2 11 17 12X13........... 12 2 2 14 17 12X15 or 13X14. . . 8 2 2 10 18 12X15 or 13X14.. . 9 2 2 11 18 12X15 or 13X14. . . 10 2 2 12 18 12X15 or 13X14. . . 12 2 2 15 18 13X15........... 8 2 2 10 19 13X15........... 9 2 2 11 19 13X15........... 10 2 2 13 19 13X15........... 12 2 2 16 19 14X16........... 9 2 2 12 20 14X16........... 10 2 2 14 20 14X16........... 12 2 2 17 20 14X18........... 9 2 2 13 22 14X18........... 10 2 2 15 22 14X18........... 12 2 2 19 22 15X16........... 10 2 2 15 21 15X17........... 12 2 2 19 22

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