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Drive Line Out Of Gear Correction

  1. Apr 4, 2010 #1
    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. :)


    * 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.


    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.
    * [tex]\tau[/tex][tex]\Delta[/tex]t = [tex]\Delta[/tex][tex]\omega[/tex]1I+[tex]\Delta[/tex][tex]\omega[/tex]2I+[tex]\Delta[/tex][tex]\omega[/tex]3I
    * Since the gears are interconnected, their velocities are linked to each other:
    [tex]\Delta[/tex][tex]\omega[/tex]1 = 3[tex]\Delta[/tex][tex]\omega[/tex]2
    * Meaning: [tex]\tau[/tex][tex]\Delta[/tex]t=9[tex]\Delta[/tex][tex]\omega[/tex]3I+3[tex]\Delta[/tex][tex]\omega[/tex]3I+[tex]\Delta[/tex][tex]\omega[/tex]3I
    * Meaning: [tex]\tau[/tex][tex]\Delta[/tex]t=13[tex]\Delta[/tex][tex]\omega[/tex]3I
    * Meaning: [tex]\Delta[/tex][tex]\omega[/tex]3=([tex]\tau[/tex][tex]\Delta[/tex]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???

  2. jcsd
  3. Apr 5, 2010 #2
    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).
    Radius (r).
    An external torque acting on the gear ([tex]\tau[/tex]).

    We have a time step ([tex]\Delta[/tex]t).

    The goal is to find the change of angular momentum of the gears ([tex]\Delta\omega[/tex]).

    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:
    [tex]\omega_{2} = \omega_{1}\frac{r_{1}}{r_{2}}[/tex]

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

    [tex]\tau_{total}\Delta t = \sum\Delta\omega_{i}I_{i}[/tex]

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

    [tex](\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}[/tex]

    ...which means that [tex]\Delta\omega_{1}[/tex] should be:

    [tex]\Delta\omega_{1} = \frac{(\tau_{1}+\tau_{2}\frac{r_{1}}{r_{2}})\Delta t}{I_{1}+\frac{r_{1}}{r_{2}}I_{2}}[/tex]

    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! :-)

  4. Apr 6, 2010 #3
    I found the error! :)

    The velocity of the second gear is [tex]\Delta\omega_{1}\frac{r_{1}}{r_{2}}[/tex] and the torque needed to rotate it is [tex]\tau \frac{r_{1}}{r_{2}}[/tex] 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.

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