SOLUTION:

1. Draw a circle with the specified pitch diameter.

From A, at any convenient angle to the pitch diameter

Boltholes

Constant

(AB), draw a line of indefinite length (AC). Adjust the

dividers at any convenient setting and step off nine

3.........

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..

0.866

spaces along line AC. Draw line B9.

4.........

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..

0.7071

2. From points 1, 2, 3, 4, 5, 6, 7, and 8, construct

5.........

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0.5 878

lines parallel to B9, which intersect line AB. This

6.........

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.

..

0.5

procedure divides line AB into nine equal parts. The

same principle can be used to divide a line of a specified

7.........

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..

0.4338

length into any number of equal parts.

8.........

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0.3827

3. With distance AB as a radius, scribe arcs from

9.........

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.

..

0.342

A and B, establishing point D. From D, draw a line that

10 . . . . . . . . .

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..

0.309

passes through 2´ (on line AB) to the circumference at

11 . . . . . . . . .

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.

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0.2817

E. Distance AE is the pitch chord and the length of arc

12 . . . . . . . . .

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0.2588

AE is equal to one-ninth of the pitch circle.

13 . . . . . . . . .

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0.2394

The principles illustrated in figure 16-7 and

14 . . . . . . . . .

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0.2225

discussed in problem III may be used to lay out a flange

15 . . . . . . . . .

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0.2079

with any number of holes. The division of line AC is

the variable of the layout. Just set the dividers to a

16 . . . . . . . . .

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0.195

convenient distance and step off spaces equal to the

17 . . . . . . . . .

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0.184

number of boltholes desired. If you want 13 boltholes,

18 . . . . . . . . .

.

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0.1736

divide line AC into 13 equal spaces. The second space

19 . . . . . . . . .

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0.1645

point (2´ in fig. 16-7) is al ways used for the construction

20 . . . . . . . . .

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0.1564

of line D2'E and the location of point E. If a pitch circle

is difficult to divide into the required number of spaces,

this method can be used to advantage. However, it is a

lengthy construction. The layouts for five- and

seven-hole flanges are easier to develop and have

diameter of 10 inches. From the table, select the

greater accuracy.

constant value for an 1 l-hole flange. The pitch diameter

Developments for four-, six-, eight-, and ten-hole

(10 inches) multiplied by the constant (0.28 17) equals

flanges are relatively simple. To determine the length of

the length of the pitch chord (2.817). Set your dividers

the pitch chord for a four-hole flange, construct two

to measure 2.817 inches, from point to point. Now step

bisectors that intersect at 90-degree angles. For six-hole

off the circumference of the pitch circle to locate the

flanges, the radius of the pitch circle is equal to the

centers of the flange boltholes.

length of the pitch chord. Step off that distance on the

You will find the mathematical method using table

pitch circle to lay out the flange boltholes. For eight- or

16-2 to be a rapid and convenient way to lay out flange

ten-hole flanges, bisect the pitch chords of the four- or

five-hole layouts.

boltholes. However, you should also be able to use the

geometric construction method. You will not always

The exact distance to set dividers for a certain

have a table of constant values available, and some

number of holes on a specified pitch circle can be

problems are not easily solved by the simple

determined by simple multiplication. However, a

multiplication process.

constant value for the desired number of boltholes must

be known. The diameter of the pitch circle multiplied

Boltholes are not the only holes you will have to lay

by the constant value equals the length of the pitch

out. At times, pipe will pass through decks and

chord. The constant value for a specified number of

bulkheads at an angle other than 90°. When this

holes is given in table 16-2.

happens, you will need to develop an ellipse. This can

Here is an example of the use of table 16-2. Assume

be done in several ways; however, problem IV

that you need a flange with 11 boltholes and a pitch

illustrates the geometric development.

16-8