To reduce any number of degrees of Temperature on Fahrenheit's Scale, to the number of degrees of an equal Temperature on Reaumer's Scale; and also to the number of degrees of an equal Temperature on the Centigrade Scale, or otherwise.

1.-Above the Freezing Point.

Any number of degrees of Fahrenheit, minus 32, multiplied by 4 and divided by 9

Reaumur.

45, and 45 + 32 = 77.

2.-Below the Freezing Point.

Any number of degrees of Fahrenheit, plus 32, multiplied by 4 and divided by 9

Reaumur.

54 × 4

Ꭱ. = 24

3.-Above the Freezing Point.

Any number of degrees of Fahrenheit, minus 32, multiplied by 5 and divided by 9 = Centigrade.

4. Below the Freezing Point.

Any number of degrees of Fahrenheit, plus 32, multiplied by 5 and divided by 9 = Centigrade

Shewing the Quantity and Weight of a Superficial Foot of Brick-work, from half a brick, to two and a half bricks in thickness.

A TABLE

Of the Specific Gravities of those Bodies chiefly used in Machinery, Building, &c. showing, in Avoirdupois Ounces and Pounds, the Weight of a Cubic Foot of each Body; also the Weight of a Cubic Inch, and the number of Cubic Inches in a Pound, with Multipliers to each, for finding the Weight when the Dimensions are given.

A TABLE

Of Specific Gravities, &c.--(continued)

The 1st, 2nd, 3rd, and 4th columns require no farther explanation than the titles they bear; the fifth column is to find the weight of any number of cubic inches, in avoirdupois pounds, of any of the different bodies required.

EXAMPLE 1.-Suppose a piece of cast iron to be 562 inches long, 16 inches broad, and of an inch in thickness, required its weight.

56.75 x 16.5 × .75 = 702.28125 cubic inches, × .263 = 184.7 lbs. nearly.

EXAMPLE 2.-Required the weight of an imperial gallon of proof spirits.

277.274 × .03352 = 9.294 lbs. nearly.

EXAMPLE 3.-Required the thickness of metal for a concave copper ball, 8 inches diameter without, so as to sink to its centre in common water.

83 x .5236 = 268.0832 cubic inches in the ball, 2 = 134.0416 cubic inches to be immersed, or cubic inches of water to be removed. -Then 134.0416 × .578 weight of a cubic inch of water = 77.4760448 ounces weight of water displaced, or, the weight of the copper ball; which divide by 5.159, the weight of a cubic inch of copper, = 15.0176 cubic inches of copper in the ball.

Again, 82 x 7854 × 4 = 202.0624 square inches, the superficies of the ball; and 15.0176 202.0624 = .0743 inches, the required thickness of the copper nearly.

EXAMPLE 4.-Required the weight necessary to counterpois a float of paving-stone proper for a steam-engine boiler, &c., the float being 14 inches diameter and 24 inches thick.

142 x .7854 × 2.5 = 384.846 cubic inches. Then 384.846 x .0873 = 33.597 lbs. the weight of the stone. And, 384.846 x .03617 = 13.919 lbs. weight of water displaced; then, 33.597– 13.919 19.678 lbs. difference between the weight of the stone and weight of the water; and, 19.6782 = 9.839 lbs. for a counterpoise, leaving the float in the water with a tendency to fall equal to 9.839 lbs. nearly.