Question 11.3: Contact Ratio of Meshing Gear and Pinion A gear set has N1 t......
Contact Ratio of Meshing Gear and Pinion
A gear set has N_1 tooth pinion, N_2 tooth gear, pressure angle \phi , and diametral pitch P (Figure 11.7).
Find:
a. The contact ratio.
b. The pressure angle and contact ratio, if the center distance is increased by 0.2 in.
Given: N_1=15, N_2=45, \phi=20^{\circ}, P=2.5 \text { in. }^{-1} .
a=\frac{1}{P}=\frac{1}{2.5}=0.4 in. (by Table 11.1)
Assumption: Standard gear sizes are considered.
| TABLE 11.1 Commonly Used Standard Tooth Systems for Spur Gears |
|||
| Item | 20° Full Depth | 20° Stub | 25° Full Depth |
| \text { Addendum } a | 1 / P | 0.8 / P | 1 / P |
| Dedendum b_d | 1.25 / P | 1 / P | 1.25 / P |
| Clearance f | 0.25 / P | 0.2 / P | 0.25 / P |
| Working depth h_k | 2 / P | 1.6 / P | 2 / P |
| Whole depth h | 2.25 / P | 1.8 / P | 2.25 / P |
| Tooth thickness t | 1.571 / P | 1.571 / P | 1.571 / P |
Applying Equation (11.2), the pitch diameter for the pinion and gear are found to be
p=\frac{N}{d} (11.2)
d_1=\frac{15}{2.5}=6 \text { in. }=152.4 mm \quad \text { and } \quad d_2=\frac{45}{2.5}=18 \text { in. }=457.2 mm
Hence, the gear pitch radii are
r_1=3 \text { in. }=76.2 mm \quad \text { and } \quad r_2=9 \text { in. }=228.6 mm
a. The center distance c is the sum of the pitch radii. So
c=3+9=12 \text { in. }=304.8 mm
The radii of the base circle, using Equation (11.9), are
r_b=r \cos \phi (11.9)
\begin{aligned} & r_{b 1}=3 \cos 20^{\circ}=2.819 \text { in. }=71.6 mm \\ & r_{b 2}=9 \cos 20^{\circ}=8.457 \text { in. }=214.8 mm \end{aligned}
Substitution of the numerical values into Equation (11.14) gives the contact ratio as
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)
\begin{aligned} C_r & =\frac{2.5}{\pi \cos 20^{\circ}}\left[\sqrt{(3+0.4)^2-(2.819)^2}+\sqrt{(9+0.4)^2-(8.457)^2}\right]-\frac{12 \tan 20^{\circ}}{\pi / 2.5} \\ & =1.61 \end{aligned}
Comment: The result, about 1.6, represents a suitable value.
b. For a case in which the center distance is increased by 0.2 in., we have c =12.2 in. It follows that
c=\frac{1}{2}\left(d_1+d_2\right), \quad d_1+d_2=2(12.2)=24.4 \text { in. } (b)
By Equation (11.2),
\frac{N_1}{d_1}=\frac{N_2}{d_2}, \quad \frac{15}{d_1}=\frac{45}{d_2} (c)
Solving Equations (b) and (c), we have d_1 =6.1 in. and d_2 =18.3 in., or r_1 =3.05 in. and r_2 =9.15 in.
The diametral pitch becomes P=N_1 / d_1=15 / 6.1=2.459 \text { in. }{ }^{-1} . The addendum is therefore a=a_1=a_2=1 / 2.459=0.407 in . Base radii of the gears will remain the same. The new pressure angle can now be obtained from Equation (11.9):
\phi_{\text {new }}=\cos ^{-1}\left(\frac{r_{b 1}}{r_1}\right)=\cos ^{-1}\left(\frac{2.819}{3.05}\right)=22.44
Through the use of Equation (11.14), the new contact ratio is then
\begin{aligned} C_{r, \text { new }}= & \frac{2.459}{\pi \cos 22.44^{\circ}}\left[\sqrt{(3.05+0.407)^2-(2.819)^2}+\sqrt{(9.15+0.407)^2-(8.457)^2}\right] \\ & -\frac{12.2 \tan 22.44^{\circ}}{\pi / 2.459} \\ = & 1.52 \end{aligned}
Comment: Results show that increasing the center distance leads to an increase in pressure angle, but a decrease in the contact ratio.