Question 9.1: Determining Gear Tooth and Gear Mesh Parameters. Find the ge...

1 The gear ratio is found from the tooth numbers on pinion and gear using equations 9.5a and 9.5c.

m_{V}=\pm \frac{d_{i n}}{d_{o u t}}=\pm \frac{N_{i n}}{N_{o u t}}                  (9.5a)

m_{G}=\left|m_{V}\right| \text { or } m_{G}=\left|m_{T}\right|, \text { for } m_{G} \geq 1                        (9.5c)

m_{G}=\frac{N_{g}}{N_{p}}=\frac{37}{19}=1.947                    (a)

2 The circular pitch can be found either from equation 9.4a or 9.4d.

p_{c}=\frac{\pi d}{N}                     (9.4a)

p_{d}=\frac{\pi}{p_{c}}                   (9.4d)

p_{c}=\frac{\pi}{p_{d}}=\frac{\pi}{6}=0.524 \text { in }                        (b)

3 The base pitch measured on the base circle is (from equation 9.4b):

p_{b}=p_{c} \cos \phi                      (9.4b)

p_{b}=p_{c} \cos \phi=0.524 \cos \left(20^{\circ}\right)=0.492 \text { in }                   (c)

4 The pitch diameters and pitch radii of pinion and gear are found from equation 9.4c.

d_{p}=\frac{N_{p}}{p_{d}}=\frac{19}{6}=3.167 \text { in, } \quad r_{p}=\frac{d_{p}}{2}=1.583 \text { in }               (d)

d_{g}=\frac{N_{g}}{p_{d}}=\frac{37}{6}=6.167 \text { in, } \quad r_{g}=\frac{d_{g}}{2}=3.083 \text { in }                    (e)

5 The nominal center distance C is the sum of the pitch radii:

C=r_{p}+r_{g}=4.667 \text { in }                    (f)

6 The addendum and dedendum are found from the equations in Table 9-1:

a=\frac{1.0}{p_{d}}=0.167 \text { in, } \quad b=\frac{1.25}{p_{d}}=0.208 \text { in }                  (g)

7 The whole depth h_{t} is the sum of the addendum and dedendum.

h_{t}=a+b=0.167+0.208=0.375 \text { in }                    (h)

8 The clearance is the difference between dedendum and addendum.

c=b-a=0.208-0.167=0.042 \text { in }                 (i)

9 The outside diameter of each gear is the pitch diameter plus two addenda:

D_{o_{p}}=d_{p}+2 a=3.500 \text { in, } \quad D_{o_{g}}=d_{g}+2 a=6.500 \text { in }                    (j)

10 The contact ratio is found from equations 9.2 and 9.6a.

\begin{aligned}Z=& \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}}-C \sin \phi \\=& \sqrt{(1.583+0.167)^{2}-\left(1.583 \cos 20^{\circ}\right)^{2}} \\&+\sqrt{(3.083+0.167)^{2}-\left(3.083 \cos 20^{\circ}\right)^{2}}-4.667 \sin 20^{\circ}=0.798 \text { in } \\m_{p}=& \frac{Z}{p_{b}}=\frac{0.798}{0.492}=1.62\end{aligned}                   (k)

11 If the center distance is increased from the nominal value due to assembly errors or other fac‑ tors, the effective pitch radii will change by the same percentage. The gears’ base radii will remain the same. The new pressure angle can be found from the changed geometry. For a 2% increase in center distance (1.02x):

\phi_{\text {new }}=\cos ^{-1}\left(\frac{r_{\text {base circle }_{p}}}{1.02 r_{p}}\right)=\cos ^{-1}\left(\frac{r_{p} \cos \phi}{1.02 r_{p}}\right)=\cos ^{-1}\left(\frac{\cos 20^{\circ}}{1.02}\right)=22.89^{\circ}                      (l)

12 The change in backlash as measured at the pinion is found from equation 9.3.

\theta_{B}=43200(\Delta C) \frac{\tan \phi}{\pi d}=43200(0.02)(4.667) \frac{\tan \left(22.89^{\circ}\right)}{\pi(3.167)}=171 \text { minutes of arc }                    (m)