- #1

ricky.helgess

- 6

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**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.

* [tex]\tau[/tex][tex]\Delta[/tex]t = [tex]\Delta[/tex][tex]\omega[/tex]

_{1}I+[tex]\Delta[/tex][tex]\omega[/tex]

_{2}I+[tex]\Delta[/tex][tex]\omega[/tex]

_{3}I

* 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}

[tex]\Delta[/tex][tex]\omega[/tex]

_{2}=3[tex]\Delta[/tex][tex]\omega[/tex]

_{3}

* Meaning: [tex]\tau[/tex][tex]\Delta[/tex]t=9[tex]\Delta[/tex][tex]\omega[/tex]

_{3}I+3[tex]\Delta[/tex][tex]\omega[/tex]

_{3}I+[tex]\Delta[/tex][tex]\omega[/tex]

_{3}I

* Meaning: [tex]\tau[/tex][tex]\Delta[/tex]t=13[tex]\Delta[/tex][tex]\omega[/tex]

_{3}I

* 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???

/Ricky