## Gear Torque Basics

$$\tau$$ is the torque applied to the gear shaft; either from a motor or some resistance; even friction. $$I \alpha$$ is the «inertia torque» that is not available to the other gears.

First, I realized that I've put an extra minus sign, the proper equations would be:

$$\frac{\tau_{1} - I_{1} \alpha_{1}}{r_{1}} = \frac{\tau_{2} - I_{2} \alpha_{2}}{r_{2}}$$

$$\alpha_{1} r_{1} = - \alpha_{2} r_{2}$$

If you solved those equations you get:

$$\tau_{1} - \left( \frac{r_{1}}{r_{2}} \right) \tau_{2} = \left[ I_{1} + \left( \frac{r_{1}}{r_{2}} \right) ^{2} I_{2} \right] \alpha_{1}$$

Or:

$$\tau_{eq} = I_{eq} \alpha_{1}$$

Where $$\tau_{eq}$$ and $$I_{eq}$$ are the equivalent torque and inertia of the 2 gears that need to be accelerated, based on the acceleration of gear 1.

If you assume that a motor is on gear 1 and nothing on gear 2, then $$\tau_{1}$$ is the torque of the motor and $$\tau_{2} = 0$$ since there is no exterior input or output torque. So:

$$\tau_{1} = \left[ I_{1} + \left( \frac{r_{1}}{r_{2}} \right) ^{2} I_{2} \right] \alpha_{1}$$

If there was a torque applied at gear 2 then it would be added to (or subtracted from) the torque of gear 1. For example if we assume that clockwise direction is positive and that $$\tau_{1}$$ is positive, then a motor putting a counterclockwise torque (negative) on gear 2 would increase the total torque. If it was a clockwise torque (positive), it would work against the motor at gear 1, hence the subtraction.

If $$\tau_{eq}$$ is zero, then $$\alpha_{1}$$ must be zero, meaning the gear set is turning at a constant rpm.

If you have more gears, you will still end up with an equation of the form $$\tau_{eq} = I_{eq} \alpha_{1}$$. For the 3-gear example you did not solve the equation with $$\tau_{2}$$ and $$\tau_{3}$$ which will give $$\tau_{2}$$ as a function of $$\alpha_{1}$$ (since $$\tau_{3} = 0$$).

As for the 5-gear setup, you won't have to assume anything if you know the torque of M1 and M2, because there are no redundant path. For example, if M1 has sufficient torque for the acceleration of gear A, C, H1 and half of E and M2 has sufficient torque for the acceleration of gear B, D, H2 and half of E, there will be no torque transfer from on side to the other. But if it is not the case (assume torque at M2 is zero), then some torque of M1 will have to through gear E to accelerate gear B, D and H2.

 Quote by jack action $$\tau$$ is the torque applied to the gear shaft; either from a motor or some resistance; even friction. $$I \alpha$$ is the «inertia torque» that is not available to the other gears.
So for any given gear, part of the torque applied to it is transferred to the connected gears, and part of it is used to accelerate the gear ($$I\alpha$$)?
 Quote by jack action $$\tau_{1} - \left( \frac{r_{1}}{r_{2}} \right) \tau_{2} = \left[ I_{1} + \left( \frac{r_{1}}{r_{2}} \right) ^{2} I_{2} \right] \alpha_{1}$$
I think I understand how this works for 2 gears now, but the three-gear example has me confused:
 Quote by jack action If you have more gears, you will still end up with an equation of the form $$\tau_{eq} = I_{eq} \alpha_{1}$$. For the 3-gear example you did not solve the equation with $$\tau_{2}$$ and $$\tau_{3}$$ which will give $$\tau_{2}$$ as a function of $$\alpha_{1}$$ (since $$\tau_{3} = 0$$).
If the only exterior source of torque was on gear 1, wouldn't the torque at gears 3 and 2 be zero? I don't understand the reason for finding $$\tau_{2}$$ as a function of $$\alpha_{1}$$. In fact I still don't understand the reason for solving $$\frac{\tau_{2}-I_{2}\alpha_{2}}{r_{2}} = \frac{\tau_{3}-I_{3}\alpha_{3}}{r_{3}}$$, when I have found $$\alpha_{2}$$ using $$\frac{\tau_{1}-I\alpha_{1}}{r_{1}} = \frac{\tau_{2}-I\alpha_{2}}{r_{2}}$$ and $$\alpha_{1}r_{1} = -\alpha_{2}r_{2}$$. If I know $$\alpha_{2}$$, surely I know $$\alpha_{3}$$ using $$\alpha_{2}r_{2} = -\alpha_{3}r_{3}$$? But like I say, I know I can't have done this right.

 Quote by Frimkron So for any given gear, part of the torque applied to it is transferred to the connected gears, and part of it is used to accelerate the gear ($$I\alpha$$)?
Yes.

 Quote by Frimkron I think I understand how this works for 2 gears now, but the three-gear example has me confused: If the only exterior source of torque was on gear 1, wouldn't the torque at gears 3 and 2 be zero?
You're right, there is something wrong, I'm trying to go too fast. Let's start again.

Assumption: 3-gear set; All torque applications and all gear rotations are clockwise.

For gear 1:

$$\frac{\tau_{1} - I_{1} \alpha_{1}}{r_{1}} = F_{12}$$

Where $$F_{12}$$ is the force common to gear 1 and 2.

For gear 2:

$$\frac{\tau_{2} - I_{2} \alpha_{2}}{r_{2}} = F_{12} + F_{23}$$

For gear 3:

$$\frac{\tau_{3} - I_{3} \alpha_{3}}{r_{3}} = F_{23}$$

Opposite and equal acceleration for gear 1 and 2:

$$\alpha_{1} r_{1} = - \alpha_{2} r_{2}$$

Opposite and equal acceleration for gear 2 and 3:

$$\alpha_{3} r_{3} = - \alpha_{2} r_{2}$$

So 5 equations, 5 unknowns ($$\alpha_{1}$$, $$\alpha_{2}$$, $$\alpha_{3}$$, $$F_{12}$$, $$F_{23}$$). Solving:

$$\frac{\tau_{2} - I_{2} \alpha_{2}}{r_{2}} = \frac{\tau_{1} - I_{1} \alpha_{1}}{r_{1}} + \frac{\tau_{3} - I_{3} \alpha_{3}}{r_{3}}$$

Replacing $$\alpha_{2}$$ and $$\alpha_{3}$$ with $$\alpha_{1}$$:

$$\frac{\tau_{2} + I_{2} \alpha_{1} \frac{r_{1}}{r_{2}}}{r_{2}} = \frac{\tau_{1} - I_{1} \alpha_{1}}{r_{1}} + \frac{\tau_{3} - I_{3} \alpha_{1} \frac{r_{1}}{r_{3}}}{r_{3}}$$

Rewriting:

$$\tau_{1} - \tau_{2} \frac{r_{1}}{r_{2}} + \tau_{3} \frac{r_{1}}{r_{3}} = \left[ I_{1} + \left( \frac{r_{1}}{r_{2}} \right) ^{2} I_{2} + \left( \frac{r_{1}}{r_{3}} \right) ^{2} I_{3} \right] \alpha_{1}$$

Or:

$$\tau_{eq} = I_{eq} \alpha_{1}$$

Note that $$\tau_{2}$$ is subtracted because we assumed it was applied clockwise, just like for gear 1 and 3; Though the rotation is opposite. But all the inertia terms are additive. If $$\tau_{2} = \tau_{3} = 0$$ then $$\tau_{eq} = \tau_{1}$$.

 Tags basic questions, gear ratio, gears, motors, torque