# Calculating acceleration, etc. of a gear train with multiple driven gears

I'm writing a basic gear train simulation, where it is possible for every gear to be attached to a source of torque/angular friction. All the online resources I've found only deal with systems where a single gear is powered and all others simply accept torque from that gear, so I've kind of had to build the equations from scratch. This is what I've come up with so far:

I started by modeling gears as levers, and looking at the force they exerted on one another.

$$F_n=\frac{\tau_n}{r_n}\\ F_{net12}=F_1+F_2=\frac{\tau_1}{r_1}+\frac{\tau_2}{r_2}$$

Then I converted to torque and found the angular acceleration: $$\tau_{net_1}=F_{net12}*r_1,\ \tau_{net_2}=F_{net12} * r_2\\ \alpha_n=\frac{\tau_{net_n}}{m_n*r_n^2}\\ a_n=\alpha_n*r_n$$ (I'm considering gears as perfect disks for simplification here)

But if you substitute in, the $r_n$'s in $\tau_n$ and $a_n$ cancel out those in $\alpha_n$, leaving you just with $$a_n=\frac{F_{net12}}{m_n}$$

And therefore the equation for a system of many gears is $$F_{net}=\sum_n \frac{\tau_n}{r_n}\\ a_0=a_1=a_2=\ ...\ =\frac{F_{net}}{\sum_n m_n}$$

I have two questions:

First of all, is my reckoning correct? It seems strange that the evolution of a rotational system is expressed only in linear units. But since the radius of each gear could be different, there can't be some global sum torque acting on all of them equally, which means there has to be a global sum force.

Secondly, if it is correct, how could I elegantly extend this model to a system that allows for multiple gears on an axle? And how could I (preferably numerically, rather than logically or analytically) check for impossible systems, like this one?

The equation $\alpha = \frac {\tau}{m\,r^2}$ is only valid for a point object, or something like a hoop where all the mass is at the same radius. For something like a gear where some of the mass is closer to the rotation point, you have to consider the mass elements that are at a different distance.

You'd normally replace it with $\alpha = \frac{\tau}{I}$, where $I$ is the moment of inertia for the disk. Because different gears can have different moments of inertia, you're not going to be able to drop out the $r^2$ term so easily.

You can consider each gear touching another and each gear sharing a shaft to be a different equation, and they all need to be solved simultaneously. If there is no solution (other than for $\omega = 0$), then you can't rotate the gears.

When the gears mesh, then the linear velocities at the rim are equal and opposite. $v_1 = -v_2$. When the gears share a shaft, then the angular rotation of each is equal. $\omega_1 = \omega_2$. If you set that up for your impossible set, then you'll find no non-zero rotation will be a solution.

• I should have put this in the original post, but I'm considering each gear as a disk for simplification. – a52 Oct 7 '17 at 2:14
• Then for a disk you can use $I = \frac 12 mr^2$ – BowlOfRed Oct 7 '17 at 2:19
• ... no non-zero rotation will be a solution. Possibly ambiguous use of the double negative. – sammy gerbil Oct 8 '17 at 8:32

I think I'll go ahead and answer this one myself, because I think I figured it out, but I had a very hard time finding resources for the questions I had. Hopefully this helps anyone else doing similar work.

(For this answer, I'll be using this system as an example. I'll also be ignoring the fact that gears alternate sign, just to make the equations a bit clearer.)

Lemma 1: All gears in a train have the same linear values (linear velocity/acceleration, force, etc.), while all gears that share an axle have the same angular values (angular velocity/acceleration, torque, etc.).

This means that to calculate the net force of a gear train, you just sum the torques divided by the radii, and to calculate the net torque on an axle, you just sum up contributing torques and forces from each gear, times their radius.

$$F_{NetAB}=\frac{\tau_a}{r_a}+\frac{\tau_{bc}}{r_b}\\ \tau_{NetBC}=\tau_{bc}+F_{AB}\cdot r_b+F_{DC}\cdot r_c$$

To get the sum for an entire system, you combine the two -- multiplying by the radius whenever you switch from an axle to a train, and dividing by it whenever you switch back. Note that because we're looking at both trains and axles, both torque and force depend on which gear you choose as your reference (but the results will be the same). I'll be using A from here on out.

$$F_{Net\ (A)}=\left(\left(\frac{\tau_g}{r_g}\right)r_e\frac{1}{r_d}+\frac{\tau_d}{r_d}\right)r_c\frac{1}{r_b}+\frac{\tau_{bc}}{r_b}+\frac{\tau_a}{r_a}$$

To find the acceleration of the system, you cannot just divide by mass, as @BowlOfRed pointed out, you have to convert to torque and then divide by the total moment of inertia.

$$a_A=\frac{\tau_{Net}}{I_{Net}}\cdot r_a=\frac{F_{Net}\cdot r_a^2}{I_{Net}}$$

This may initially seem like a problem, because we're multiplying by $r_a^2$ and then dividing by a constant, meaning that two gears with different radii in the same train will accelerate (linear) at different rates. (This stumped me for a long time, and although I didn't realize the impact it had, was part of the reason I asked the question in the first place.) But as it turns out, $I$ is not constant for every member of a gear train. If you take a closer look at the formula:

$$I_{Net}=I_a + \left(\frac{r_a}{r_b}\right)^2\left(I_b+I_c+\left(\frac{r_c}{r_d}\right)^2\left(I_d+I_e+\left(\frac{r_e}{r_g}\right)^2\left(I_g\right)\right)\right)$$ (Source)

Not only does $I$ change based on reference gear, but it's actually proportional to the missing $r_a^2$! The two factors above cancel out, linear acceleration stays constant, and the acceleration formula above is correct. Once you've figured out the acceleration of the reference gear, you can determine it for all other gears using simple ratios and lemma 1.