Question 8.11: Devise a gear train for a conveyor drive. The drive motor ro...

We use the getting started recommendations given earlier to start the solution.
Permissible Train Values
First let’s compute the nominal train value that will produce an output speed of 26.0 rpm at the middle of the allowable range:

TV_{nom} = (input speed)/(nominal output speed) = 1150/26 = 44.23
Now we can compute the minimum and maximum allowable speed ratio:
TV_{min} = (input speed)/(maximum output speed) = 1150/28 = 41.07
TV_{max} = (input speed)/(minimum output speed) = 1150/24 = 47.92
Possible Ratio for Single Pair
The maximum ratio that any one pair of gears can produce occurs when the gear has 150 teeth and the pinion has 17 teeth (see Table 8–7). Then

VR_{max} = N_G/N_P = 150/17 = 8.82 (too low)
Possible Train Value for Double-Reduction Train
TV = (VR_1)(VR_2)
But the maximum value for either VR is 8.82. Then the maximum train value is
TV_{max} = (8.82)(8.82) = (8.82)2 = 77.9
A double-reduction train is practical.

Optional Designs
The general layout of the proposed train is shown in Figure 8–47. Its train value is
TV = (VR_1)(VR_2) = (N_B/N_A)(N_D/N_C)
We need to specify the number of teeth in each of the four gears to achieve a train value within the range just computed. Our approach is to specify two ratios, VR_1 and VR_2, such that their product is within the desired range. If the two ratios were equal, each would be the square root of the target ratio, 44.23. That is,

But, as described in Item 16 of the General Principles for Devising Gear Trains, we want the first ratio to be somewhat larger than the second. Let’s specify
VR_1 = 8.0 = (N_B/N_A)
If we let pinion A have 17 teeth, the number of teeth in gear B must be
N_B = (N_A)(8) = (17)(8) = 136

Then the second ratio should be approximately
VR_2 = TV/(VR_1) = 44.23/8.0 = 5.53
This is the residual ratio left after the first ratio has been specified. Now if we specify 17 teeth for pinion C, gear D must be

VR_2 = 5.53 = N_D/N_C = N_D/17
N_D = (5.53)(17) = 94.01
Rounding this off to 94 is likely to produce an acceptable result. Finally,
N_A = 17  N_B = 136  N_C = 17  N_D = 94
We should check the final design:
TV = (136/17)(94/17) = 44.235 = n_A/n_D
The actual output speed is
n_D = n_A/TV = (1150 rpm)/44.235 = 26.0 rpm
This is right in the middle of the desired range.