# Drive Line Out Of Gear Correction

1. Apr 4, 2010

### ricky.helgess

I am working on a vehicle based computer game. The focus right now is to get the internal physics of each drive line component as physically correct as possible. Since a few weeks, I am stuck with a problem that I cannot get past by myself. I have narrowed it down to a specific issue. It should not be too hard to solve but I am stuck so please help. :)

Prerequisites

* I use a tree of physical components to represent the drive line. To simplify my question, let's see each component in the drive line tree as several two-disc-gears, welded together (I am not sure if I use the English language correctly here - I hope you can follow my reasoning anyway).
* Each two-disc-gear has an inertia of 0.1.
* The time step in calculating is 1/50 = 0.02.
* The first (left) gear's small disc is connected to the second gear's large disc and so on and so on.
* The radius of the large discs is 1 meter.
* The radius of the small discs is 1/3 meter.
* There are three gears in my example.
* The first (left (index 1 below)) and last (right (index 3 below)) gears have a weight attached to them at the same height as the fulcrum, both at 1 meter from the fulcrum.
* The weights generates 10 N each.
* The gears are at rest at t0.

Problem

What should my formula look like to find the angular velocity of each gear at t1 (0.02 seconds later)?

Let's take a look at the last (right (index 3 below)) gear and find how much it's angular velocity should change:

* The resulting torque multiplied with the time step should equal the system's change in angular momentum.
* $$\tau$$$$\Delta$$t = $$\Delta$$$$\omega$$1I+$$\Delta$$$$\omega$$2I+$$\Delta$$$$\omega$$3I
* Since the gears are interconnected, their velocities are linked to each other:
$$\Delta$$$$\omega$$1 = 3$$\Delta$$$$\omega$$2
$$\Delta$$$$\omega$$2=3$$\Delta$$$$\omega$$3
* Meaning: $$\tau$$$$\Delta$$t=9$$\Delta$$$$\omega$$3I+3$$\Delta$$$$\omega$$3I+$$\Delta$$$$\omega$$3I
* Meaning: $$\tau$$$$\Delta$$t=13$$\Delta$$$$\omega$$3I
* Meaning: $$\Delta$$$$\omega$$3=($$\tau$$$$\Delta$$t)/(13I)

Now, WHAT is the resulting torque? If I calculate from left to right I get 80 Nm. If I calculate from right to left, I get 8.888...

It seems I should use 8.888... but why? If the rightmost gear was connected to the ground, I would move the vehicle with 80 Nm/tire radius (thinking about the rightmost gear as a tire on the ground). But when not all the torque is added to the ground and I must use the excess torque to speed up the drive line tree, all of a sudden, I cannot use 80 Nm. Why? Where am I going wrong?

If I have torques added on different places in the drive line tree, how do I use these when speeding up the drive line tree?

The calculations I use work very well when the rightmost gear (tire) has full grip but when I have no grip, I just get too large values if I use the gear-ratio-multiplied torques... so, how should I think???

/Ricky

2. Apr 5, 2010

### ricky.helgess

Since I did not get any answers, I will change this into a simpler example:

You have two gears (not welded double gears as in the first example, just two simple gears). The gears are connected to eachother using a weightless chain belt.

Each of the two gears have the following quantities:

Inertia (I).
An external torque acting on the gear ($$\tau$$).

We have a time step ($$\Delta$$t).

The goal is to find the change of angular momentum of the gears ($$\Delta\omega$$).

What is the formula to find the angular momentum change of each of the gears?

I tried the following but somewhere I go wrong:

The angular velocities of the two gears have a relationship:
$$\omega_{2} = \omega_{1}\frac{r_{1}}{r_{2}}$$

The total torque multiplied with delta time should equal the total system's change of angular momentum:

$$\tau_{total}\Delta t = \sum\Delta\omega_{i}I_{i}$$

Now, looking from the first gears perspective, I try the following:

$$(\tau_{1}+\tau_{2}\frac{r_{1}}{r_{2}})\Delta t = \Delta\omega_{1}I_{1}+\Delta\omega_{1}\frac{r_{1}}{r_{2}}I_{2}$$

...which means that $$\Delta\omega_{1}$$ should be:

$$\Delta\omega_{1} = \frac{(\tau_{1}+\tau_{2}\frac{r_{1}}{r_{2}})\Delta t}{I_{1}+\frac{r_{1}}{r_{2}}I_{2}}$$

This does not seem to work though, because when I enter numbers in the equation, I always get the same change in angular velocities on both gears and that is obviously not right.

Please give me some direction! :-)

/Ricky

3. Apr 6, 2010

### ricky.helgess

I found the error! :)

The velocity of the second gear is $$\Delta\omega_{1}\frac{r_{1}}{r_{2}}$$ and the torque needed to rotate it is $$\tau \frac{r_{1}}{r_{2}}$$ which means that the formula should be:

$$(\tau_{1}+\tau_{2}\frac{r_{1}}{r_{2}})\Delta t = \Delta\omega_{1}I_{1}+\Delta\omega_{1}\left(\frac{r_{1}} {r_{2}}\right)^{2}I_{2}$$

Finally I can continue with other stuff in the game.

/Ricky