Machinery's Handbook, 31st Edition
Lengths of Chords 711 Example 3(a), Determining Hole Coordinates for Fig. 3 : Write equations for the coordi nates of holes 1, 2, and 3 of Fig. 3. x 1 = 7.5 + 5cos(30 ° ) = 11.8301 y 1 = 6.0 + 5sin(30 ° ) = 8.5000 x 2 = 7.5 + 5cos(30 ° + 60 ° ) = 7.5 y 2 = 6.0 + 5sin(30 ° + 60 ° ) = 11.0000 x 3 = 7.5 + 5cos(30 ° + 120 ° ) = 3.1699 y 3 = 6.0 + 5sin(30 ° + 120 ° ) = 8.5000 Example 3(b), Modify Equation (2b) for Fig. 3 : In Fig. 3, hole numbering is rotated 90 ° ( p /2 radian) in the clockwise (negative) direction relative to Fig. 2b, and the direction of the + y coordinate axis is the reverse, or negative, of that given in Fig. 2b. Equations for Fig. 3 can be obtained from Equation (2b) by 1) subtracting 90 ° from the angle of each hole in the x and y equations of Equation (2b) 2) multiplying the y H equation by –1 to reverse the orientation of the y axis sin cos x D H X y D H Y 2 1 2 90 2 1 2 90 H H 0 0 i i i i =− − + − + =− − − + − + a ^ a a ^ h k h k k In Fig. 3, q = 360/ n = 60 ° for 6 holes, X O = 7.5, and Y O = –6.0. x 1 = –5sin(30 – 90) + 7.5 = 11.8301 y 1 = 5cos(30 – 90) + 6 = 8.5000 x 2 = –5sin(60 ° + 30 ° – 90 ° ) + 7.5 = 7.5 y 2 = 5cos(60 ° + 30 ° – 90 ° ) + 6 = 11.0000 x 3 = –5sin(120 ° + 30 ° – 90 ° ) + 7.5 = 3.1699 y 3 = 5cos(120 ° + 30 ° – 90 ° ) + 6 = 8.5000 Lengths of Chords on Hole Circle Circumference.— Table 3 on page 720 gives the lengths of chords for spacing off the circumferences of circles. The object of this table is to make possible the division of the periphery into a number of equal parts without trials with the dividers. Table 3 is calculated for circles having a diameter equal to 1. For circles of other diameters, the length of chord given in the table should be multiplied by the diameter of the circle. Table 3 may be used by toolmakers when setting “buttons” in circular forma tion and with inch or metric dimensions. See Coordinates for Hole Circles on page 708 for more information on this topic. Example: Assume that it is required to divide the periphery of a circle of 20 inches diam eter into thirty-two equal parts. Solution: From the table the length of the chord is found to be 0.098017 inch, if the diam eter of the circle were 1 inch. With a diameter of 20 inches the length of the chord for one division would be 20 3 0.098017 = 1.9603 inches. Another example, in metric units: For a 100 millimeter diameter requiring 5 equal divisions, the length of the chord for one division would be 100 × 0.587785 = 58.7785 millimeters. Example: Assume that it is required to divide a circle of 6 1 ⁄ 2 millimeter diameter into seven equal parts. Find the length of the chord required for spacing off the circumference. Solution: In Table 3, the length of the chord for dividing a circle of 1 millimeter diameter into 7 equal parts is 0.433884 mm. The length of chord for a circle of 6 1 ⁄ 2 mm diameter is 6 1 ⁄ 2 × 0.433884 = 2.820246 mm. Example: Assume that it is required to divide a circle having a diameter of 9 23 ⁄ 32 inches into 15 equal divisions. Solution: In Table 3, the length of the chord for dividing a circle of 1 inch diameter into 15 equal parts is 0.207912 inch. The length of chord for a circle of 9 inches diameter is 9 23 ⁄ 32 3 0.207912 = 2.020645 inches.
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