Question 12.1: Geometric Quantities for Helical Gears Two helical gears hav......
Geometric Quantities for Helical Gears
Two helical gears have a center distance of c =10 in., width b =1.9 in., a pressure angle of \phi=25^{\circ} , a helix angle of \psi=30^{\circ}, and a diametral pitch of P=6 in.^{−1}. If the speed ratio is to be r_s =1/3, calculate:
a. The (transverse) circular, normal circular, and axial pitches.
b. The number of teeth of each gear.
c. The normal diametral pitch and normal pressure angle.
d. The total contact ratio.
a. Applying Equations (12.1) and (12.2),
p_n=p \cos \psi, \quad p_a=p \cot \psi=\frac{p_n}{\sin \psi} (12.1)
P p=\pi, \quad P_n p_n=\pi, \quad P_n=\frac{P}{\cos \psi}, \quad P=\frac{N}{d} (12.2)
p=\frac{\pi}{6}=0.5236 in.
p_n=0.5236 \cos 30^{\circ}=0.453 \text { in. }
p_a=0.5236 \cot 30^{\circ}=0.907 \text { in. }
b. Through the use of Equation (11.8),
r_s=\frac{\omega_2}{\omega_1}=\frac{n_2}{n_1}=\frac{N_1}{N_2}=\frac{d_1}{d_2} (11.8)
r_s=\frac{1}{3}=\frac{N_1}{N_2} \quad \text { or } \quad N_2=3 N_1
Equation (12.5) gives then
c=\frac{d_1+d_2}{2}=\frac{p}{2 \pi}\left(N_1+N_2\right)=\frac{N_1+N_2}{2 P_n \cos \psi} (12.5)
10=\frac{0.5236}{2 \pi}\left(N_1+3 N_1\right) \quad \text { or } \quad N_1=30
and hence, N_2 =90. Thus, d_1 =30/6=5 in.=127 mm and d_2 =15 in.=381 mm.
c. From Equation (12.2), we have
P_n=\frac{6}{\cos 30^{\circ}}=6.928 \text { in. }^{-1}
By Equation (12.3),
\tan \phi_n=\tan \phi \cos \psi (12.3)
\tan \phi_n=\tan 25^{\circ} \cos 30^{\circ}
from which
\phi_n=22^{\circ}
d. The addendum equals a=a_1=a_2=1 / P=1 / 6 in . Applying Equation (11.14), the contact ratio of spur gear is
C_r=\frac{1}{p \cos \phi}\left[\sqrt{\left(r_p+a_p\right)^2-\left(r_p \cos \phi\right)^2}+\sqrt{\left(r_g+a_g\right)^2-\left(r_g \cos \phi\right)^2}\right]-\frac{c \tan \phi}{p} (11.14)
C_r=\frac{1}{p \cos \phi}\left[\sqrt{\left(r_1+a\right)^2-\left(r_1 \cos \phi\right)^2}+\sqrt{\left(r_2+a\right)^2+\left(r_2 \cos \phi\right)^2}\right]-\frac{c \tan \phi}{p}
Introducing the data
\begin{aligned} & C_r-\frac{1}{0.5236 \cos 25^{\circ}} {\left[\sqrt{(2.5+1 / 6)^2-\left(2.5 \cos 25^{\circ}\right)^2}\right.} \\ \quad \quad &\left.+\sqrt{(7.5+1 / 6)^2-\left(7.5 \cos 25^{\circ}\right)^2}\right]-\frac{10 \tan 25^{\circ}}{0.5236} \\ &=1.53 \end{aligned}
The total contact ratio for the helical gear, by Equation (12.8), is
C_{r t}=C_r+C_{r a} (12.8)
C_{r t}=C_r+C_{r a}=1.53+\frac{1.9 \tan 30^{\circ}}{0.5236}=3.63
Comment: The result, about 3.63, is a reasonable value.