WEIGHT OF A SUPERFICIAL FOOT OF PLATE OR SHEET IRON,
COPPER, AND BRASS, IN POUNDS.

Note.-No. 1 wire gauge equal ths of an inch.

The great variety of thicknesses into which

copper

whereby to determine the thickness required, the unit is manufactured, cause in trade the weight to be named

being that of a common sheet, so designated, viz., 4 feet by 2 feet, in lbs., thus:

A 70 b. plate is ths of an inch in thickness.

The thickness of lead is also in common determined or understood by the weight, the unit being that of a square or superficial foot; thus,

4 lb. lead is th of an inch in thickness.

1. Suppose I have an article of plate iron, the weight of which is 728 lbs., but want the same of copper, and of similar dimensions, what will be its weight?

728 × 1.16 844.48 lbs.

2. A model of dry pine weighing 32 tbs., and in which the iron for its construction forms no material portion of the weight, what may I anticipate its weight to be in cast iron?

32.5 x 11 357.5 lbs.

Note.-It frequently occurs in the formation or construction

of models, that neither the quality nor condition of the timber can be properly estimated, and in such cases it may be a near enough approximation to reckon 10 tbs. of cast iron to each fb. of model.

WEIGHTS OF 9-FEET LENGTHS OF CAST IRON PIPES, OF VARIOUS DIAMETERS.

TO ASCERTAIN THE WEIGHTS OF PIPES OF VARIOUS METALS, AND ANY DIAMETER REQUIRED.

Rule. To the interior diameter of the pipe, in inches, add the thickness of the metal; multiply the sum by the decimal numbers opposite the required thickness and under the metal's name; also by the length of the pipe in feet, and the product is the weight of the pipe in tbs.

1. Required the weight of a copper pipe whose interior diameter is 7 inches, its length 61 feet, and the metal of an inch in thickness.

7.5+125=7·625 × 1·52 × 6.25=72.4 lbs.

2. What is the weight of a leaden pipe 181 feet in length, 3 inches interior diameter, and the metal of an inch in thickness?

3+25=3.25 × 3·867 x 18.5=232.5 tbs.

1. What will be the weight of a hollow ball or shell of cast iron, the external diameter being 9 and internal diameter 83 inches?

Opposite 9 are 118.06, and
Opposite 8 are 92.24, subtract

-25.82 lbs., weight required.

2. Requiring to remove a cast iron ball 37.8 lbs. in weight, and in diameter 6 inches, and replace it by one of lead of an equal weight, what must be the diameter of the leaden ball?

Weight of lead to that of cast iron = 1.56 (see Table, page 61).

6.53

Then =3/176 5.6 inches, the diameter.

1.56