Devise a gear train to reduce the speed of rotation of a drive from an electric motor shaft operating at 3450 rpm to approximately 650 rpm. Use Nmax = 150 teeth.

Question

Devise a gear train to reduce the speed of rotation of a drive from an electric motor shaft operating at 3450 rpm to approximately 650 rpm. Use [latex]N_{max}[/latex] = 150 teeth.

Accepted Answer

First we will compute the nominal train value:
TV = (input speed)/(output speed) = 3450/650 = 5.308
If a single pair of gears is used then the train value is equal to the velocity ratio for that pair. That is, [latex]TV = VR = N_G/N_P[/latex]

Let’s decide that spur gears having 20°, full-depth, involute teeth are to be used. Then we can refer to Table 8–7 and determine that no fewer than 16 teeth should be used for the pinion in order to avoid interference. We can specify the number of teeth in the pinion and use the velocity ratio to compute the number of teeth in the gear:

[latex]N_G = (VR)(N_P) = (5.308)(N_P)[/latex]
All possible examples are given in Table 8–11.

Conclusion and Comments
The combination of [latex]N_P[/latex] = 26 and [latex]N_G[/latex] = 138 gives the most ideal result for the output speed. But all of the trial values give output speeds reasonably close to the desired value. Only two are more than 2.0 rpm off the desired value. It remains a design decision as to how close the output speed must be to the stated value of 650 rpm. Note that the input speed is given as 3450 rpm, the full load speed of an electric motor. But how accurate is that? The actual speed of the input will vary depending on the load on the motor. Therefore, it is not likely that the ratio must be precise.

TABLE 8–7 Number of Pinion Teeth to Ensure No Interference
For a pinion meshing with a rack		For a 20°, full-depth pinion meshing with a gear
Tooth form	Minimum number of teeth	Number of pinion teeth	Maximum number of gear teeth	Maximum ratio
[latex]14\frac{1}{2}°[/latex], involute, full-depth	32	17	1309	77.00
20°, involute, full-depth	18	16	101	6.31
25°, involute, full-depth	12	15	45	3.00
		14	26	1.85
		14	16	1.23

TABLE 8–11 All Possible Values for [latex]N_P[/latex] and [latex]N_G[/latex] to Produce the Desired Velocity Ratio
[latex]N_P[/latex]	Computed [latex]N_G = (5.308)(N_P)[/latex]	Nearest integer [latex]N_G[/latex]	Actual [latex]VR: VR = N_G /N_P[/latex]	Actual output speed (rpm): [latex]n_G = n_P /VR = n_P (N_P /N_G)[/latex]
16	84.92	85	85/16 = 5.31	649.4
17	90.23	90	90/17 = 5.29	651.7
18	95.54	96	96/18 = 5.33	646.9
19	100.85	101	101/19 = 5.32	649
20	106.15	106	106/20 = 5.30	650.9
21	111.46	111	111/21 = 5.29	652.7
22	116.77	117	117/22 = 5.32	648.7
23	122.08	122	122/23 = 5.30	650.4
24	127.38	127	127/24 = 5.29	652
25	132.69	133	133/25 = 5.32	648.5
26	138	138	138/26 = 5.308	650.0 Exact
27	143.31	143	143/27 = 5.30	651.4
28	148.61	149	149/28 = 5.32	648.3
29	153.92	154 Too large

Question 8.10: Devise a gear train to reduce the speed of rotation of a dri...

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