because a ring slipped over the rod and caused to move around it to the right or to the left would always be the same distance from the center of the profile to which the rod was formed. A piece of bar iron formed to a circular profile would also be a solid of revolution. This solid could be regarded as being generated by a rectangle revolving about a center point. Figure 132 is drawn around two arcs whose center is the center of the elbow. It can also be seen that the more pieces "put in" the elbow, the nearer the straight lines come to the arcs about which they are drawn. If the elbow were made of a very great number of pieces, these would become so much like arcs that they could hardly be distinguished from them. This or any other elbow can be treated as a solid of revolution.
Solids of Revolution Have Three Diameters.—Every solid of revolution may be considered as having three diameters. The radius of the throat of Fig. 132 is 4½″. If four of these elbows were joined so as to make a complete ring, it would have a diameter of 4½″×2, or 9″. This would be the inside diameter of the solid of revolution. The radius of the back, Fig. 132, is 4½″+2½″, or 7″. The corresponding diameter for the whole ring would be 7″×2 or 14″. This would be the outside diameter of the solid of revolution. The third diameter is twice the center line radius of the elbow. In Fig. 132 the center line radius is 5¾″, and for the whole ring this would be 5¾″×2 = 11½″. This is called the neutral diameter of the solid of revolution, because when any rod or bar is formed into a circular profile the metal near this line stands still, that outside of the line stretches, and that inside of the line shrinks a like amount. This can be proved by drawing straight lines on a pencil eraser and bending the eraser, at the same time noting the distances between the lines.
Rule for Surface Area.—The surface area of a solid of revolution is equal to the circumference of its right section (profile) multiplied by the length of its neutral zone (diameter of the neutral×3.1416).
Sample Problem.— What is the surface area of the elbow shown in Fig. 132?
| Outside | diameter | of | ring | =14″ | |
| Inside | " | " | " | =9″. | |
| Neutral | " | " | =11½″. | ||
| Length of neutral zone (11½″×π)=36.128″. | |||||
| Perimeter of right section (length of line of stretchout)=20″. | |||||
