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How do I get momentum to assist rotational force from overcoming traction force?

  1. Jan 28, 2010 #1
    (video game) I have a car and it is accelerating wonderfully.
    But I need to know when my acceleration is too great from my traction force and my tires should slip reducing acceleration.

    Currently what I'm doing is calculating rotational force = EngineTorque * transmision gear ratio * rearend gear ratio / TireRadius.
    I compute traction force (friction) = TireCoefficient of Friction * mass * gravity.

    when rotation force is greater than traction force I slip the tires.

    Problem: I believe In real life as you move forward (momenturm = v*m) plays into the above forces so if you speed up slowly the rotational force won't over power the traction force. Currently if I accelerate slow or fast I slip. I don't know if its momentum or what the forces are but I think newton's second law of motion kinda points to it. I'm a software guy and physics is interesting as all get out... but it's also really difficult. I hope this makes some kind of sense.

    And thank you for your time and consideration.

  2. jcsd
  3. Jan 28, 2010 #2
    I actually didn't understand what your problem was. First you said that the car was "accelerating wonderfully", and then that "Currently if I accelerate slow or fast I slip"?

    Can you adjust the engine torque continuously? E.g. put it at some small value, and the let the car accelerate slowly? If I understand you correctly, the tires slip in this situation?

    Secondly (unrelated to your problem), as the tires are slipping you should take into account that they still provide some frictional force. It points in the direction opposite to the rubber's movement relative to the ground. The value will be determined by the same formula as you already use, but with a reduced friction coefficient, e.g. a third of the one you use for static friction.

  4. Jan 29, 2010 #3
    " I have a car and it is accelerating wonderfully" yes that is a bit confusing. But what I meant is I take the actually torque curve specification of an actual engine convert it to a fifth degree polynomial and can calculate Torque N-m from rpm. Use a fourth-order rungetta to get a very accurate acelleratioin using gear ratios, mass, gravity, tire radius and the the car accellerates positively and negatively very accurately.

    And yes I can use my gamepad to accelerate up and down very slowly and it works nice.

    But then I want to simulate tire slip when I accelerate too much. So I then calculate rotational force = EngineTorque (from polynomial) * transmision gear ratio * rearend gear ratio / TireRadius.
    I compute traction force (friction) = TireCoefficient of friction * mass * gravity and I do this step independently of my car accelleration(above) to see if I should be slipping my tires. So when the rotational force is greater than the traction force slipping begins. If I back off the throttle slipping stops. I move the throttle up and the slip begins again. The problem is in real life when this is done with a car the momentum allows the driver to move past this point maintining traction due to momentum (i think its momentum). I think the problem is I don't know how to integrate the two steps so that the rotation force has the benefit of the momentum from the first calculation for acceleration.

    I hope this clarifies things. I've been stuck on this for about two weeks and I'm hoping you can help me. Again thanks a bunch for you reply.

  5. Jan 31, 2010 #4
    The way I'm interpreting your claim is (and please correct me if I'm wrong): You claim that as the speed increases you should be able to get more traction, because of the momentum of the car.

    I don't agree with this statement. The friction coefficient determines the maximum possible friction forces between the tire and the ground, and this should be independent of the car's velocity at least as a reasonable approximation.

    What I thing could be related to this is the fact that you should allow for a fixed (as a first approx) friction force while spinning, that is determined by the dynamic friction coeffient (much less than the static friction coefficient which is used while rolling without slipping). This fixed friction will also keep the engine's RPM from diverging towards infinity, and keep your car from actually loosing speed while spinning (because of wind resistance). Maybe you already are doing this?

    I'm assuming that your throttle acts as a torque control? The graph of engine torque vs rpm gives the maximum possible torque at each value of RPM. You throttle control will adjust the amount of torque linearly between 0 and the maximum allowable value, for a given RPM?

    Of course, there is one way the momentum can give you more traction, and this is if your car has a bigger downforce at large speeds. But I don't know if you are doing this?

  6. Jan 31, 2010 #5
    "The way I'm interpreting your claim is (and please correct me if I'm wrong): You claim that as the speed increases you should be able to get more traction, because of the momentum of the car." I wasn't clear here at all. Traction is constant, but when I climb my torque curve with RPM my rotational force is overcome my traction force and slip begins.

    It seems to me that when I accelered very quickly (open the throttle) the momentum (p = m*v) would have not much effect on assisting rotational force ( EngineTorque (from polynomial) * transmision gear ratio * rearend gear ratio / TireRadius) because velocity (v) is small because I'm moving or displacing slowly, just starting out.

    If I accelerate slowly and reach the same rotational force ( EngineTorque (from polynomial) * transmision gear ratio * rearend gear ratio / TireRadius) at a higher speed my momentum (p=m*v) is larger because my velocity has had time to increase. That momentum would come into play somehow (I don't know how) possibly and not directly becuase my velocity is higher it would reduce the rotational force not the traction force. But the rotational force should not be as strong when momentum is involved. That's my guess.

    Again thanks for answering and i hope I'm getting closer to explaining what my problem is.

    Last edited: Jan 31, 2010
  7. Feb 2, 2010 #6
    It looks like your are correctly calculating the forces between the wheels and the ground, so I guess I just don't understand why there is a problem. In other words, I'm don't quite agree with the claim that momentum should change anything regarding the forces/slipping.

    Here is what I think your are saying:

    Assume that the effective gear ratio in first gear is twice that of the second gear. Assume also that the maximum torque at 4000RPM is twice that of at 2000RPM.

    In first gear, the maximum possible force from the wheels at 2000RPM is called F. Assume that this is just above the limit for slipping. On the other hand, in second gear at 4000RPM, the maximum possible force from the wheels is also F, since the gear ratio is halved, but the maximum possible torque at the engine is twice because of the higher RPM. In this case the momemtum is much larger. You are saying that the consequence of the increased momentum might be that the wheels are not slipping in this case?

    In that case, I disagree with this. I would say that the wheels will slip, independently of the momentum. The reason a real car isn't able to slip at higher speeds is only because of the gear ratios that subdue decrease the maximum possible torque at the wheels.

    The reason is: before slipping, independently of the momentum, at the point of contact between the tire and the road, the tire is always at rest with respect to the road since it is rolling. So the usual considerations of force and friction coefficients apply also when rolling fast. Of course some small corrections come in, like tire deformation because of the spin and so on, but I don't think that is what you are looking for here, and anyway that contributes in the wrong direction, since it makes the effective contact area between the ground and tire smaller.

  8. Feb 2, 2010 #7

    Thanks for your response. And based on more reading I don't think momentum is the affecting factor but rather jerk (the 3rd derivative of position with respect to time). Let me take your scenario and see if I can't explain myself more clearly.

    If we take the situation as you describe above, ignoring second gear I see the problem as follows: If we use first gear only and attempt to accelerate as quickly as possible (stump on the throttle, going from no throttle to full throttle very quickly) we should slip at 4000 RPM due to the fact that our rotational force with its jerk is greater than our friction (traction). I believe this is because there is a larger jerk.

    If we take the same scenario and attempt to acellerate very slowly in 1st gear moving the throttle from no throttle to full throttle in over a 1 minute when we reach 4000 RPM we should not slip when we reach the same rotational force because we have a much smaller jerk. While both the rotational forces and the friction (traction) are the same the jerk will be less, allowing the car to move forward without slip.

    And if this is correct I'm not sure how to factor in jerk.

    And I would like to thank you for hanging in there with me over all these questions your patience is appreciated.

    Thank you

  9. Feb 2, 2010 #8
    Let's start with the equation for kinetic energy:

    E = ½·m·v2

    So power is

    P = dE/dt = m·v·dv/dt = m·v·a

    where v = velocity and a = acceleration. If the static (non-slipping) friction coefficient is 0.8, and the car is 2-wheel drive, with the effective mass m/2 and effective weight mg/2 on the drive wheels, then the maximum acceleration is 0.4g.

    So the maximum power that can be used without slipping is then

    P ≤ 0.4·m·v·g

    which may be less than the power available, especially at low velocity.

    Power is related to torque and RPM by

    P = torque (Newton-meters) x 2·π·RPM /60

    This is the power anywhere in the drive train, including rear axle.

    Bob S
  10. May 6, 2010 #9
    I think that you also should take into account the difference in speed between the wheel and the ground. What I do in my simulation is the following:

    * Find Maximum Engine Torque based on the current engines torque curve and current rpm.
    * Find Actual Engine Torque = Multiply Maxumum Engine Torque with a Throttle Factor (between 1 and 0).
    * Multiply Actual Engine Torque with Gear Box Ratio and Final Drive Ratio.
    * Find the Drive Line Total Equivalent Inertia which is each drive line part's inertia multiplied by the gear ratio * gear ratio... that is Ieq = Ilocal*gr*gr.
    * Find the difference in speed between the wheel and the ground (maybe your vehicle was airborne before or whatever).
    * Use the difference in speed and equivalent inertia to find how much force is needed to change the drive line speed to match the ground.
    * Add the Gear Ratio-multiplied Actual Engine Torque.
    * Add any drive line friction (engine break due to vacuum, wheel break, hand break, efficiency loss in gearbox etc.)
    * Find how much force you WANT to add to the wheels.
    * Find Normal Force (mass, weight transfer, downforce...)
    * Find out how much the wheels can handle by using normal force and friction coefficient from last frame (did the tire have grip or not in last frame).
    * Use any force from the tire to update the velocities of each part in the drive line tree (the tricky part here is to simulate differentials and the clutch correctly).

    Also, a tip that is important. When turning the wheels when the player steers with the steering wheel, do NOT turn the wheels equally. Use Ackermann Condition for the wheels, or you will put too much power on the outer wheel and the vehicle will be hard to control.

    Another tip: In some physics engines, there is no way to find the normal force (like in PhysX). Instead, you could find out how much the suspension is compressed and calculate normal force based on that. This way, you can add downforce on the vehicle body... and you also get the weight transfer when turning automatically.

    Hope this helps! :)

    Last edited: May 6, 2010
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