Geometric Quantities of a V-Belt Drive
A V belt is to operate on sheaves of 8 and 12 in. pitch diameters (Figure 13.5(a)). Calculate:
a. The center distance.
b. The contact angle.
Assumption: A B-section V belt is used, having the actual pitch length of 69.8 in. (Table 13.2).
TABLE 13.2 Pitch Lengths (in Inches) of Standard V Belts |
|||||||
Cross-Section | |||||||
A | B | C | A | B | C | D | E |
27.3 | 113.3 | 113.8 | 114.9 | ||||
36.3 | 36.8 | 121.3 | 121.8 | 122.9 | 113.3 | ||
43.3 | 43.8 | 145.8 | 146.9 | 147.3 | |||
52.3 | 52.8 | 53.9 | 159.8 | 160.9 | 161.3 | ||
61.3 | 61.8 | 62.9 | 174.8 | 175.9 | 173.3 | ||
69.3 | 69.8 | 70.9 | 181.8 | 182.9 | 183.3 | 184.4 | |
76.3 | 76.8 | 77.9 | 211.8 | 212.9 | 213.3 | 214.5 | |
86.3 | 86.8 | 87.9 | 240.3 | 240.9 | 240.8 | 241.0 | |
91.3 | 91.8 | 92.9 | 270.3 | 270.9 | 270.8 | 271.0 | |
106.3 | 106.8 | 107.9 | 300.3 | 300.9 | 300.8 | 301.0 |
a. Through the use of Equation (13.11),
b=L-\pi\left(r_2+r_1\right) (13.11)
b=L-\pi\left(r_1+r_2\right)=69.8-\pi(6+4)=38.38 in.
Equation (13.10) is therefore
c=\frac{1}{4}\left[b+\sqrt{b^2-8\left(r_2-r_1\right)^2}\right] (13.10)
c=\frac{1}{8}\left[38.38+\sqrt{38.38^2-8(6-4)^2}\right]=19.09 in.
b. Applying Equation (13.7),
\phi=\pi-2 \alpha (13.7)
\phi=\pi-2 \alpha=180^{\circ}-2 \sin ^{-1}\left(\frac{6-4}{19.09}\right)=168^{\circ}