A gearset consists of a 16-tooth pinion driving a 40-tooth gear. The diametral pitch is 2, and the addendum and dedendum are 1/P and 1.25/P, respectively. The gears are cut using a pressure angle of 20°.
(a) Compute the circular pitch, the center distance, and the radii of the base circles.
(b) In mounting these gears, the center distance was incorrectly made \frac{1}{4} in larger.
Compute the new values of the pressure angle and the pitch-circle diameters.
Question 13.1: A gearset consists of a 16-tooth pinion driving a 40-tooth g...
(a) P=\frac{\pi }{P}=\frac{\pi }{2}=1.571 in
The pitch diameters of the pinion and gear are, respectively,
d_P=\frac{N_P}{P}=\frac{16}{2}=8 in d_G=\frac{N_G}{P}=\frac{40}{2}=20 in
Therefore the center distance is
\frac{d_P+d_G}{2}=\frac{8+20}{2} =14 in
Since the teeth were cut on the 20° pressure angle, the base-circle radii are found to be, using r_b = r cos Φ,
r_b(pinion)=\frac{8}{2}\cos20°=3.759 in
r_b(gear)=\frac{20}{2}\cos20°=9.397 in
(b) Designating d'_P and d'_G as the new pitch-circle diameters, the \frac{1}{4} -in increase in the center distance requires that
\frac{d'_P+d'_G}{2}=14.250 (1)
Also, the velocity ratio does not change, and hence
\frac{d'_P}{d'_G}=\frac{16}{40} (2)
Solving Eqs. (1) and (2) simultaneously yields
d'_P = 8.143 in d'_G = 20.357 in
Since r_b = r cos Φ, using either the pinion or gear, the new pressure angle is
\phi'=\cos ^{-1}\frac{_rb(pinion)}{d'/2}= \cos ^{-1}\frac{3.759}{8.143/2}=22.59°
Related Answered Questions
(a) The axial pitch is the same as the transverse ...
The pitch diameters of gears 2 and 3 are
d_...
From Eq. (13–19)
\cos\psi =\frac{\tan\phi _...
The pitch angles are
\gamma =\tan ^{-1}\lef...
Since the ratio is greater than 10:1, but less tha...
The previous example demonstrated the difficulty w...
(a) d=\frac{N}{P...