Question 9.3: Reverted Gear Train Design. Design a reverted compound train...
Reverted Gear Train Design.
Design a reverted compound train for an exact train ratio of 18:1.
1 Though it is not at all necessary to have integer gearset ratios in a compound train (only integer tooth numbers), if the train ratio is an integer, it is easier to design with integer ratio gearsets.
2 The square root of 18 is 4.2426, well within our 10:1 limitation. So two stages will suffice in this gearbox.
3 If we could form two identical stages, each with a ratio equal to the square root of the overall train ratio, the train would be reverted by default. Table 9‑8 shows that there are no reasonable combinations of tooth ratios that will give the exact square root needed. Moreover, this square root is not a rational number, so we cannot get an exact solution by this approach.
4 Instead, let’s factor the train ratio. All numbers in the factors 9 x 2 and 6 x 3 are less than 10, so they are acceptable on that basis. It is probably better to have the ratios of the two stages closer in value to one another for packaging reasons, so the 6 x 3 choice will be tried.
5 Figure 9‑31 shows a two-stage reverted train. Note that, unlike the nonreverted train in Figure 9‑29, the input and output shafts are now in-line and cantilevered; thus each must have double bearings on one end for moment support and a good bearing ratio as was defined in Section 2.18.
6 Equation 9.8 states the relationship for its compound train ratio. In addition, we have the constraint that the center distances of the stages must be equal. Use equation 9.9c and set it equal to an arbitrary constant K to be determined.
m_{V}=\left(-\frac{N_{2}}{N_{3}}\right)\left(-\frac{N_{4}}{N_{5}}\right) (9.8a)
N_{2}+N_{3}=N_{4}+N_{5} (9.9c)
N_{2}+N_{3}=N_{4}+N_{5}=K (a)
7 We wish to solve equations 9.8 and 9.9c simultaneously. We can separate the terms in equation 9.8 and set them each equal to one of the stage ratios chosen for this design.
\begin{aligned}&\frac{N_{2}}{N_{3}}=\frac{1}{6} \\&N_{3}=6 N_{2} \\&\frac{N_{4}}{N_{5}}=\frac{1}{3}\end{aligned} (b)
N_{5}=3 N_{4} (c)
8 Separating the terms in equation (a):
\begin{aligned}&N_{2}+N_{3}=K (d)\\&N_{4}+N_{5}=K (e)\end{aligned}
9 Substituting equation (b) in (d) and equation (c) in (e) yields:
\begin{aligned}&N_{2}+6 N_{2}=K=7 N_{2} (f)\\&N_{4}+3 N_{4}=K=4 N_{4} (g)\end{aligned}
10 To make equations (f) and (g) compatible, K must be set to at least the lowest common multiple of 7 and 4, which is 28. This yields values of N_{2}=4 \text { teeth and } N_{4}=7 \text { teeth } .
11 Since a four-tooth gear will have unacceptable undercutting, we need to increase our value of K sufficiently to make the smallest pinion large enough.
12 A new value of K = 28 x 4 = 112 will increase the four-tooth gear to a 16-tooth gear, which is acceptable for a 25° pressure angle (Table 9-4b). With this assumption of K =112, equations (b), (c), (f), and (g) can be solved simultaneously to give:
\begin{array}{ll}N_{2}=16 & N_{3}=96 \\N_{4}=28 & N_{5}=84\end{array} (h)
which is a viable solution for this reverted train.
| TABLE 9-8 Example 9-3 |
| Possible Gearsets for 18:1 Two-Stage Reverted Compound Train |
| \begin{array}{ccc}\begin{array}{c}\text { Gearset } \\\text { Ratio }\end{array} & \begin{array}{c}\text { Pinion } \\\text { Teeth }\end{array} & \begin{array}{c}\text { Gear } \\\text { Teeth }\end{array} \\\hline 4.2426 &\times 12 & =50.91 \\4.2426 &\times 13 & =55.15 \\4.2426 &\times 14 & =59.40 \\4.2426 &\times 15 & =63.64 \\4.2426 &\times 16 & =67.88 \\4.2426 &\times 17 & =72.12 \\4.2426 &\times 18 & =76.37 \\4.2426 &\times 19 & =80.61 \\4.2426 &\times 20 & =84.85\end{array} |
