Machinery's Handbook, 31st Edition
MEASURING SCREW THREADS 2133 profile of a milled thread will have some curvature in both axial and normal sections; hence angles A and A n represent the angular approximations of these slightly curved profiles. The equations that follow give the values needed to solve the screw thread problem as a helical gear problem. (4) (4a) (4b) cos M G R 2 b = W + tan tan tan = F B A = cos R E F 2 b =
sin tan
B A n
T
a =
T
(4c)
tan cos tan H F H b # =
(4d)
tan
B
T a E
2 R b cos H b W
π S
(4e) The tables of involute functions starting on page 112 provide values for angles from 14 to 51 degrees, used for gear calculations. The formula for involute functions on page 111 may be used to extend this table as required. Example: To illustrate the application of Formula (4) and the supplementary formulas, assume that the number of starts S = 6; pitch diameter E = 0.6250; normal thread angle A n = 20°; lead of thread L = 0.864 inch; T = 0.072; W = 0.07013 inch. . . . . . . tan B E L B H 19635 0 864 0 44003 = 23751 90 23 751 66 249 ° ° ° = − = ° = π = = + inv F + − inv G =
Helix angle n
tan sin tan =
F B A
. . 040276 0 36397 0 90369 . =
. 42104
F
=
=
°
. 2 0 6250 0 74193 0 23185 044003 0 072 0 16362 . . . . . × = = =
cos R E F T 2 b = =
=
T
tan
B
a
. 0 74193 2 27257 1 68609 . . × =
. 59328
tan cos tan =
H
H F H
=
=
°
b
b
The involute function of G is found next by Formula (4e).
6 3.1416 − = 0.20351 0.16884 Since 0.20351 is outside the values for involute functions given in the tables on pages 112 through 115 use the formula for involute functions on page 111 to extend these tables as required. It will be found that 44 deg. 21 min. or 44.350 degrees is the angular equivalent of 0.20351; hence, G = 44.350 degrees. . . . . cos M G R 2 071508 2 0 23185 0 07013 0 71859 b × = + = + = W inch 2 × 0.23185 × 0.51012 0.07013 inv G = + + 0.625 0.16362 Accuracy of Formulas (3) and (4) Compared.— With the involute helicoid Formula (4) any wire size that makes contact with the flanks of the thread may be used; however, in the preceding example, the wire diameter W was obtained by Formula (3a) in order to compare Formula (4) with (3) . If Example (3) is solved by Formula (3) , M = 0.71912; hence the difference between the values of M obtained with Formulas (3) and (4) equals 0.71912 - 0.71859 = 0.00053 inch. The included thread angle in this case is 40 degrees. If
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