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Conservation of momentum through a gear-train

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I'm having some problems with my gear-train code. I have two rotating masses connected together via a clutch and gears. Lets say the clutch is initally disengaged so sides A and B are turning independently. I need it to work so that when they clutch is engaged the two rotating masses will achieve the proper speeds (given the gears in the middle!) From conservation of momentum I have that: (wa*Ia + wb*Ib) = (wfinal*Itotal) w final will be composed of two speeds wa' and wb'. And Itotal is Ia+Ib wa' and wb' are linked together via gears (of ratio R). So wa' = R*wb', agreed? wa*Ia + wb*Ib = wa'Ia + wb'Ib replacing wa' with R*wb' yields: wa*Ia + wb*Ib = R*wb'Ia + wb'Ib wa*Ia + wb*Ib = wb'(R*Ia+Ib) solving for wb' = (wa*Ia + wb*Ib) / (R*Ia + Ib) and wa' = R*wb' This seems to work as intended for positive values of R. (the two masses are rotating in the same direction). However it fails for negative R values (-1 in particular will cause a divide by zero error!) The resultant speeds of the masses after being 'collided together' with the clutch tend to become divergant (bigger and bigger in opposing directions) when they should converge on an inbetween value. For instance 100RPM and 0RPM should converge on +50 and -50RPM's given a ratio of -1. This doesnt happen..at all. It is more likely to end up around 200 and -200! What am I doing wrong? (besides everything?)

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This is a problem in trying to solve for the angular momentums
La and Lb,
where Lt = La + Lb and is constant.

Your two gears are rotating independantly, with rotational velocities of
wa and wb.
These are what will change when you engage the clutch.

Two constants are the inertias Ia and Ib, which are a property of the shape of the gears, and their masses. These won't change no matter what.

To solve the system, find the reduced inertia:
U=(Ia*Ib)/(Ia+Ib)

The rotational speed of either gear is:
Lt*U/Ib2 for gear A and Lt*U/Ia2 for gear B

[Edited by - erissian on December 19, 2005 12:27:57 AM]

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Ok, I have setup the above in Excel to test things and I'm getting odd results.

I use the values of Ia and Ib = 10. U comes out to be (10*10)/(20) = 5.

If i put in starting velocities of 100 and 0, the wa' and wb' end up being 25 and 25 respectively. This seems odd. I would expect 50 and 50 given the same inertias. Also, what happens if the gear ratio is NOT 1:1? Then clearly they cannot both be 25.

If seems no matter what value I put in for wa and wb, I get an identical pair of wa' and wb'. They are never different. Is this set of equations solved for where the gear ratio is always 1:1?

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Quote:
Original post by erissian
Sorry, I did the last step backwards. I've fixed it now.
The U you got was right.
In this case, wa and wb should both be 50.


Ok, it behaves properly for a ratio of +1:1. What happens if it is not 1:1, or a 'negative' ratio (reversing the direction of the two masses?)

For instance, if I have wa and wb of 100 and -100, with a ratio of -1, they should stay at 100 and -100. This code sets them both to zero.

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Maybe I misunderstood the problem. If two gears of equal size are spinning at the same speed, but in opposite directions while separate, then when coupled, their motions would be in opposition to each other. Under these conditions, the wheels should stop moving.

Can you post a sketch of your problem?

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Ok, heres a quick sketch


The two rotational masses at the ends can have different inertias. They can also have any angular velocity (both directions are valid). I am representing a clockwise direction as positive for the sake of this discussion.

The gearbox in the middle can have any ratio. Meaning it can be 2 gears (producing rotation in opposite directions) or it could have an extra idler gear to produce rotation in the same direction. It can have any gear ratio (so the speeds of the two ends do not have to be the same, this also provides a torque multiplication which is sort of the bit that is confusing me in regards to finding the final velocities of these things).

The clutch initially starts disengages and then engages instantaneously. (meaning I cant really use an integration method to solve this over time). I need to solve for the beginning and end states without calculating the torques and such acting in the clutch in the middle.

Does that make it more clear?

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It does make it more clear. This uses the same type of equations, except with the constraint that wa/wb = R.

Solving for wb

La,i + Lb,i = La,f + Lb,f
=wa,fIa + wb,fIb
=Rwb,fIa + wb,fIb
=(RIa + Ib)wb,f

wb,f=(wa,iIa + wb,iIb)/(RIa + Ib)

If you were solving for wa, then the denominator becomes:
(Ia + R-1Ib)

Using the example where I = Ia = Ib = 10
and wa = 100
and wb = 0
and R = 1:1 = 1/1 = 1

wa = (10*100 + 10*0)/(1*10 + 10) = 1000/20 = 50
wb = (10*100 + 10*0)/(10 + 10/1) = 1000/20 = 50

That seems to work.

You will alway get errors with the second example, because it's an incorrect representation of the system. R should always be positive. If the mystery gearbox forces the gears to turn in the opposite direction, then the velocities should be positive as long as they are turning "cooperatively". If they are turning in opposition to each other, then one should be negative. Clockwise really isn't the way to go to define positive or negative velocity in this case.

If they are opposing each other, where wa = -wb, where R=1, then they will cancel out their motion. If you redefine positive for the second wheel, so that wa = wb, then they remain constant like they should.

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Ok, interesting that you end up deriving what I was already using. I guess that's ok then.

The problem I am having with this, is that my wheels can be going either direction. (the car can go forewards and back, obviously).
The engine should always be going its usual direction...
the problem comes when the transmission is in reverse!

You are saying it would be sufficient to switch the sign of wb-intial when the transmission is in reverse?

edit: just tried it, and it appears to be working.

thanks. rating++.

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