Gear Train Ratio & Speed Simulator

Load Simple Idler Train: 12, 36, 24 and 36 teeth, 1,200 rpm in at 10 N·m. The panel says a ratio of 3.00:1. Two of those four tooth counts are doing nothing to it. Prove which, and then switch the staging and see what those same four gears can actually give.

Open this problem in the tool

1. One rule covers every mesh: where two gears touch, the teeth pass the contact point at the same rate, so the smaller gear turns faster in exact proportion to the tooth counts.

   $$\frac{\omega_1}{\omega_2} = \frac{N_2}{N_1} \qquad \text{at every mesh}$$

2. A simple train is three meshes in a row, so multiply the three.

   $$\frac{\omega_{\text{in}}}{\omega_{\text{out}}} = \frac{N_2}{N_1}\cdot\frac{N_3}{N_2}\cdot\frac{N_4}{N_3}$$

3. Put the numbers in and each stage looks like it matters: a 3.00, then a 0.667, then a 1.50.

   $$= \frac{36}{12}\cdot\frac{24}{36}\cdot\frac{36}{24} = 3 \times \tfrac{2}{3} \times \tfrac{3}{2} = 3.000$$

4. Now cancel. Every middle count appears once on top and once underneath, so the whole product collapses to the last over the first, and the gears between them leave no trace in it.

   $$\frac{N_2}{N_1}\cdot\frac{N_3}{N_2}\cdot\frac{N_4}{N_3} = \frac{N_4}{N_1} = \frac{36}{12} = 3$$

5. The rest follows from the ratio alone. Speed divides, torque multiplies.

   $$\omega_{\text{out}} = \frac{1200}{3} = 400\ \mathrm{rpm}, \qquad \tau_{\text{out}} = 10 \times 3 = 30\ \mathrm{N\,m}$$

6. And the product is untouched. Gears move power between speed and force; they never make any, and the tool holds this at every setting.

   $$\omega\tau\ (\mathrm{rpm}\cdot\mathrm{N\,m}) : \quad 1200 \times 10 = 400 \times 30 = 12{,}000$$

7. To make the middle gears count, put two of them on one shaft so they are forced to turn together. Then the stages multiply instead of cancelling, and the same four tooth counts give 4.50 instead of 3.00.

   $$\text{compounded instead: } \frac{36}{12}\cdot\frac{36}{24} = 3.00 \times 1.50 = 4.50$$

Load Compound Multiplier (12:1): 12, 36, 12 and 48 teeth, 1,200 rpm in at 10 N·m, 100 rpm and 120 N·m out. Between those two shafts sits a third one the panel never mentions. Work out what it is carrying, then decide which of the two stages should do the larger reduction.

Open this problem in the tool

1. The two stages multiply because gears 2 and 3 are keyed to one shaft and cannot turn independently of each other. Three times four is twelve, and the speed and torque rows follow from that number alone.

   $$i = \frac{N_2}{N_1}\cdot\frac{N_4}{N_3} = \frac{36}{12}\cdot\frac{48}{12} = 3.00 \times 4.00 = 12.00 \quad\Longrightarrow\quad 100.0\ \mathrm{rpm},\ 120.0\ \mathrm{N\,m}$$

2. That shaft is where the first stage ends and the second begins. It turns at the input speed divided by the first ratio and carries the input torque multiplied by it.

   $$n_{\mathrm{mid}} = \frac{n_{\mathrm{in}}}{r_1} = \frac{1200}{3.00} = 400\ \mathrm{rpm}, \qquad T_{\mathrm{mid}} = T_{\mathrm{in}}\,r_1 = 10 \times 3.00 = 30\ \mathrm{N\,m}$$

3. Which leaves the product untouched a second time. The panel already shows it holding between input and output; it holds in the middle too, because a gear train has nowhere to store anything.

   $$n\,T\ (\mathrm{rpm}\cdot\mathrm{N\,m}): \quad 1200 \times 10 \;=\; 400 \times 30 \;=\; 100 \times 120 \;=\; 12{,}000$$

4. Now swap the stages over — four first, three second. The overall ratio is still 12.00:1 and the output is still 100 rpm at 120 N·m, so nothing a reader can see on the page has changed. The middle shaft has: it is slower and carrying a third more torque.

   $$\text{stages swapped } (12/48/12/36): \quad i = 4.00 \times 3.00 = 12.00, \qquad n_{\mathrm{mid}} = 300\ \mathrm{rpm},\ T_{\mathrm{mid}} = 40\ \mathrm{N\,m}$$

5. The general statement is short. The first ratio fixes the middle shaft entirely, and the two ratios have to multiply to whatever overall ratio you wanted, so the middle torque lands somewhere between the input’s and the output’s and never outside them.

   $$T_{\mathrm{mid}} = T_{\mathrm{in}}\,r_1, \quad n_{\mathrm{mid}} = \frac{n_{\mathrm{in}}}{r_1}, \qquad r_1 r_2 = i \ \text{fixed} \;\Longrightarrow\; T_{\mathrm{in}} \le T_{\mathrm{mid}} \le T_{\mathrm{out}}$$

6. The choice is worth making because a shaft is sized by torque and not by speed. At one allowable shear stress the diameter goes as the cube root of the torque.

   $$d \propto T^{1/3} \quad (\text{torsion, one allowable shear stress}): \qquad \left(\frac{40}{30}\right)^{1/3} = 1.10$$