If you compute the circumference, multiply the

shape, which we will say is 12 inches. The second is the

diameter by pi (š), or 3.1416. This formula is read as

circumference, which is computed as follows. Use the

formula C = šD + 2W. D is the diameter of the circle

C = šD. If the measurement is not critical, you can

that you would have if both curved ends of the shape

round off pi to 3.14 or 3 l/7. Either way, the formula

were put together. W is shown in figure 12-14. We will

will give you one dimension of your stretchout. The

assume that D = 5 inches, and W = 6 inches.

height or length is your other dimension.

C = šD+2W

Another method is by the use of the circumference

rule. The upper edge of the circumference rule is

C = 3.14 × 5+2 ×6

graduated in inches in the same manner as a regular

layout scale, but the lower edge is graduated, as shown

C = 15.7 + 12

in figure 12-13.

C = 27.7

The lower edge gives you the approximate

We find that the circumference is 27.7 inches.

circumference of any circle within the range of the rule.

Therefore, our stretchout measures 27.7 inches by 12

You will notice in figure 12-13 that the reading on the

inches.

lower edge directly below the 3-inch mark is a little over

9 3/8 inches. This reading would be the circumference

of a circle with a diameter of 3 inches and would be the

length of a stretchout for a cylinder of that diameter. The

dimensions for the stretchout of a cylindrical object,

Following are the procedures in using geometry for

then, are the height or length of the cylinder and the

making various types of layouts.

circumference. Do not forget that you will have to allow

for the scams.

CONSTRUCT A 90-DEGREE OR RIGHT

ANGLE. This is no problem at all if you have a true

A VARIATION OF THE CYLINDRICAL JOB is

steel square. We will describe three methods that you

a flat-sided structure with rounded ends (fig. 12-14).

may use to erect a perpendicular to produce a right angle

To figure the stretchout for this shape, you will need

when you do not have a usable true steel square.

two dimensions. The first is simply the length of the

1. For the first method, break out your dividers, a

scriber, and a straightedge. Draw a base line like the one

labeled AB in figure 12-15. Set the dividers for a

distance greater than one-half AB, then, with A as a

center, scribe arcs like those at C and D. Then, without

changing the setting of the dividers, use B as a center,

and scribe another set of arcs at C and D. Draw a line

through the points where the arcs intersect and you will

have perpendiculars to line AB, forming four 90-degree

or right angles. Not only have you constructed a

12-5