Physics of gears
Author
I'm having trouble with torques, angular velocities and gears.
Let A and B be gears of respective radiuses rA and rB. Let wA be the the angular velocity of gear A, and wB be B's.
Then wA and wB are related by
wA*rA = -wB*rB.
If we derive with respect to time, we get that
aA*rA = -aB*rB, where aA and aB are the angular accelerations.
We also know that: Sum of external torques on a solid = angular acceleration * inertia of the solid.
If we apply that to our gears A and B we get:
Sum (Torques on A) = aA*Ia and Sum (Torques on B) = aB*Ib, where Ia and Ib are the inertias of A and B.
If we sum these two sums, we get:
Sum (Torques on A and B) = aB*(Ib - (rB/rA)*Ia)
If Ib and Ia are equal, and if rA and rB are also equal, then the right hand side is zero, which can't be right.
What I am doing wrong?
If you wonder where this comes from, I am trying to model the transmission of a go-kart, including the engine, the gearbox, the clutch, the brakes, the axle with the wheels and the ground.
Your equation is right, total angular momentum L is always null(Everything is symmetric and we have wA=-Wb).
dL/dt=Sum (Torques on A and B) => Sum (Torques on A and B)=0
The torques shouldn't be 0 unless the system stopped.
Also, 0 is not null, it's 0.
You're problem lies in your conceptualization of the two gears. If gear A and gear B are linked together and still, the resultant torque is 0. (nothing is moving)
If you being to rotate gear A only, it will also rotate gear B. gear B's rotation is purely based on gear A. so the magnitude of the torque of gear A = the magnitude of the torque of gear B.
If you want to turn both gears at one time, then tA + tB = tT, where tA is the torque on A, tB is the torque on B, and tT is the total torque on the system. Bare in mind, this mandates that there is a positive direction of motion and a negative direction. In other words, if the two gears are rotating in opposite directions, you need them to subtract (or well, they will have oppositely signed angular velocities)
Also, 0 is not null, it's 0.
You're problem lies in your conceptualization of the two gears. If gear A and gear B are linked together and still, the resultant torque is 0. (nothing is moving)
If you being to rotate gear A only, it will also rotate gear B. gear B's rotation is purely based on gear A. so the magnitude of the torque of gear A = the magnitude of the torque of gear B.
If you want to turn both gears at one time, then tA + tB = tT, where tA is the torque on A, tB is the torque on B, and tT is the total torque on the system. Bare in mind, this mandates that there is a positive direction of motion and a negative direction. In other words, if the two gears are rotating in opposite directions, you need them to subtract (or well, they will have oppositely signed angular velocities)
I'm confused at what the problem is exactly.
So here's a thread where I struggled with the same stuff. Maybe it will help you.
http://www.gamedev.net/community/forums/topic.asp?topic_id=364810&whichpage=1�
Or maybe not.
Good luck.
So here's a thread where I struggled with the same stuff. Maybe it will help you.
http://www.gamedev.net/community/forums/topic.asp?topic_id=364810&whichpage=1�
Or maybe not.
Good luck.
Author
Thanks all for your answers!
Kambiz is right, my equation was correct, but I misinterpreted it.
Sum(Torques on A) + Sum(Torques on B) is not the same as Sum(Torques on A and B).
The fact that Sum(Torques on A and B) is zero does not imply that the respective angular velocities of A and B are constant.
I did my calculations again, the way my teachers 10 years ago always insisted I should do:
System A is composed of gear A with radius rA and inertia Ia
System B is composed of gear B with radius rB and inertia Ib
External forces on A: force F due to B (I leave out the ground and brakes for clarity)
External forces on B: force -F due to A (action/reaction principle) and torque due to the engine.
For A we get: rA*F = Ia * d(wA)/dt
For B we get: rB*F + T(engine) = Ib * d(wB)/dt
Since we know that rA*d(wA)/dt = rB*d(wB)/dt, we can solve the problem.
The result I get is:
d(wB)/dt = T(engine)/(Ib + R^2*Ia) where R=rB/rA
I'm a bit suprised that R is still in the picture, but that does seem too shocking.
CombatWombat: Thanks for the link. Seems interesting, but I haven't digested it yet. Makes me wonder, would be nice if the forum allowed to write formulas using e.g. latex syntax. A "drawing board" using e.g. a Java applet would be nice too.
Kambiz is right, my equation was correct, but I misinterpreted it.
Sum(Torques on A) + Sum(Torques on B) is not the same as Sum(Torques on A and B).
The fact that Sum(Torques on A and B) is zero does not imply that the respective angular velocities of A and B are constant.
I did my calculations again, the way my teachers 10 years ago always insisted I should do:
System A is composed of gear A with radius rA and inertia Ia
System B is composed of gear B with radius rB and inertia Ib
External forces on A: force F due to B (I leave out the ground and brakes for clarity)
External forces on B: force -F due to A (action/reaction principle) and torque due to the engine.
For A we get: rA*F = Ia * d(wA)/dt
For B we get: rB*F + T(engine) = Ib * d(wB)/dt
Since we know that rA*d(wA)/dt = rB*d(wB)/dt, we can solve the problem.
The result I get is:
d(wB)/dt = T(engine)/(Ib + R^2*Ia) where R=rB/rA
I'm a bit suprised that R is still in the picture, but that does seem too shocking.
CombatWombat: Thanks for the link. Seems interesting, but I haven't digested it yet. Makes me wonder, would be nice if the forum allowed to write formulas using e.g. latex syntax. A "drawing board" using e.g. a Java applet would be nice too.
This topic is closed to new replies.
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