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412 STRENGTH OF GEAR TEETH
The speed of the teeth is iooo feet per minute, for which
by interpolation in table No. 55, we find the value of s to be
2800 and by substituting these values we have:
w = 2800 x 2 x 5 x 0.071 x 3X3
3
I;8
8
a
3
x73
2
I
4
8
8
l34. 8) .
w = 1988 x 0.795.
W = 1580 pounds.
Above formula involves considerable labor in calculations
and Mr. Lewis gives a simpler approximate formula:
d
W= sX pX/X yX p-
This formula gives almost identically the same results as
the more complicated formula given above, when d is not less
than I D, as is the case in good practice
Example.
Find the safe working horse-power of a pair of cast iron
bevel gears of 2 inches circular pitch and 5 inches face, if one
gear has 60 teeth and the other 30 teeth. The speed is 900 feet
per minute, the form of the teeth is the common 14-i degree
involute. In this case d is more than two-thirds and, therefore,
we may use the simple approximate formula:
Solution:
As the gear having the least number of teeth is (when both
are of the same material) the weakest of the two, we make the
calculation for the gear having 30 teeth.
Examining the drawing and calculating we find
3 X 2 .
D - 3^6 = ^ l lnches*
d = 14! inches.
Measuring the back cone radius on the scale drawing and
calculating, we find the formative number of teeth to be between
33 and 34. The value of y, corresponding to 34 teeth, is in table
No. 54 given as 0.104. Inserting those values in the formula
we have:
x 4-5
W = 3000 X 2 X 5 X 0.104 X y^~f
= 2369 pounds.
At a speed of 900 feet per minute, this will give:
2369 X 900 , ,
= 64 horse power.
33000 * F
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