that this part must be solved first. 10×10 =100, 6×6=36, 100-36-64. A=.7854 x64 50.26 square inches. By adding l-length; to the formula we can find the cubical contents and weight of any hollow cylinder or pipe, and calling this one 12 inches long we have 50.26×12=602.6 cubic inches, from which weight can be found for any material.

As an example of working backwards, find the thickness of a cast iron cylinder whose outer diameter is 10 inches, length 15 inches, and which must weigh 200 pounds. Cast iron is given as .26 pound per cubic inch. So dividing 200 by .26 we find that (200÷.26-769.23) 769.23 cubic inches are necessary to make the required weight. Dividing this by the length, 15 inches, we have 51.28 square inches as the area of the ring whose outer diameter is 10 inches. Then we can say 51.28 (A)=.7854× (100—ɗ2) and transposing we have

Chapter V.

Mensuration.

This can be called the measuring or computing of surfaces, areas and volumes of bodies, and is very useful to the mechanic in many ways. Taking a triangle first, figure 1, as this has the least number of sides of any figure; we discover that all the sides and angles are equal, hence it is called an equilateral triangle. Either half of it, as laid out by the dotted vertical dividing line, is called a right angled triangle because it contains one right angle.

A right angle is one formed by two lines perpendicular to each other or with an opening of 90 degrees or one-quarter the number of degrees of a complete circle.

This can be more readily seen in figure 2, if we take O as a center, and note that the four angles GO E, E O H, HOF, and F O G are all equal, and all are right angles.

The base of a triangle or other figure can be defined as the side on which it rests, in the case of figure 1, the base of the equilateral triangle A B C is the line B C, and half the base will of course be CD or D B.

The vertical height is shown by the line A D, and can be defined as a line perpendicular to the base and connecting it to the point or apex.

It is plain that if we took the right hand side and placed it on the upper left hand, as shown by the dotted lines, we have made a rectangle and that the area

will be the height multiplied by the base C D. But as we only have the dimensions of the sides of the triangles and not the vertical height, we must find this before we can find the area.

In a right angled triangle the square of the hypothenuse or slanting side, equals

the square root of the base plus the square of the vertical height, or calling the vertical height a, the base b and the hypothenuse с we say c2= a2 + b2 or c= √ a2 + b2. But as we know c to be 2 inches and the base 6 1 inch (one half of c) we must find

a. Transposing the formula we have a2=c2-62 or a=√2-62 and 62-22-a2 or b= √c2 — a2.

Then a√4-1=1.733 inches.

The area then equals 1.733x1=1.733 square inches.

The usual rule is to multiply the height by the base and divide by two, and as the base of the whole triangle is

2 inches we say

1.733×2
2

reasoning shows this.

=1.733; a little

Taking the square we have a simple form to find the area, simply multiplying one side by the other, or squaring.

The diagonal or hypothenuse is found in the same way as with the triangle although it is often more convenient to multiply the square of one side by 2 than to add both together, and of course it brings the same result.

For example, having a 3-inch square to find the diagonal.

c2=b2+a2 or c= √b2+ a2 or c=√262. c= √99 or c=√2 × 9= √184.24 inches as diagonal.

We can work this backward and