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Converting torque to force

  1. Jan 19, 2009 #1
    I've got a robot with four wheels, two of which are driven by 12V DC motors. I can vary the voltage sent the motors using a PWM speed controller. The coefficient of friction between the wheels and the surface is very low. My problem is: I need to limit the amount of voltage sent to the motors to avoid causing the wheels to slip against the low-friction surface.

    First off, I need to calculate the force of static friction between the wheel and the surface.

    Ff = [tex]\mu[/tex]FN
    FN = mg
    Ff = [tex]\mu[/tex]mg

    Where [tex]\mu[/tex] is the coefficient of static friction between the wheel and the surface, m is the mass supported by that particular wheel (~mass of robot / 4 in my case), and g is gravity. I know all three of these values, so I'm good so far.

    Starting from the other direction, I can redefine motor voltage in terms of rotational velocity and torque.

    V = ([tex]T[/tex] / Kt) + (KeN)
    Kt = [tex]T[/tex]s / Vref
    Ke = Vref / Nf
    V = (Vref(Nf[tex]T[/tex] + N[tex]T[/tex]s)) / (Nf[tex]T[/tex]s)

    Where Nf is the free rpm, [tex]T[/tex]s is the stall torque, N is the rpm of the motor, [tex]T[/tex] is the torque exerted by the motor on the load, and V is the voltage I am sending to the motor. I know all of the constants I just mentioned. I can measure rpm using optical encoders on the wheels. That just leaves torque in the equation.

    To avoid wheel slippage, I need to make sure the force exerted by the wheel on the surface as a result of the motor's torque is not greater than Ff calculated earlier. So, I somehow need to convert torque to force. I attempted to do this.

    [tex]T[/tex] = I[tex]\alpha[/tex]
    F = ma
    a = F / m
    af = Ff / m
    af = [tex]\mu[/tex]mg / m
    [tex]\alpha[/tex] = a / r
    [tex]\alpha[/tex] = [tex]\mu[/tex]mg / mr
    [tex]T[/tex] = I[tex]\mu[/tex]g / r

    Where I is the moment of inertia, [tex]\mu[/tex] is the coefficient of friction between the wheel and the surface, g is gravity, and r is the radius of the wheel.

    I know the latter three values. What I cannot figure out is how to calculate I. I know I = kmr2, but what should I use for m? Is that just the mass of the wheel itself? Or is it the mass of the wheel plus the mass of the robot that the wheel is supporting? If I use the mass of the wheel and assume k = 1 (the wheel has most of its mass concentrated on the outside), I get the final equation.

    V = (Vr(gmwNf[tex]\mu[/tex]r + N[tex]T[/tex]s))/(Nf[tex]T[/tex]s)

    When I plug in the numbers for my situation, I get a maximum safe voltage when accelerating from standstill of 0.09V - this is way too low and obviously wrong. If I use 1/4 of the robot mass for m (1/4 because the load is evenly distributed among four wheels), I get the more believable answer of 6V. Based on some rough testing with the actual robot a few days ago, 6V (half power) is approximately when the wheels start slipping. I would be happy with that answer, if it made sense. I just can't understand why I would use the mass of anything other than the wheel when calculating the moment of inertia of the wheel.

    Comments? Ideas?

  2. jcsd
  3. Jan 19, 2009 #2
    I think you have been a bit too careless with "m". There is "m" and then there is "M" to be considered in this problem.

    In the angular acceleration relation, it is true that [tex]T[/tex] = I[tex]\alpha[/tex] and I = kmr2 where that "m" is the mass of the wheel.

    But when you write F = ma, you really might want to reconsider and write that as F = M a, because the mass that is being accelerated is the wheel and some portion of the entire platform (one quarter if this is a four wheeled platform being uniformly accelerated by each wheel). Thus you need a very different M here as compared to the mass of just the wheel.
  4. Jan 20, 2009 #3


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    PVNRT4ME: First let me say, to get an accurate answer, we would need the wheel radii dimensions, wheel base dimensions, and the vehicle CG horizontal and vertical coordinates (excluding the mass of all wheels). And we would need to know whether the drive wheels are on the front or rear, and whether the non-drive wheel mass and mass moment of inertia are identical to a drive wheel. However, I will pretend the vehicle weight is evenly distributed to the four wheels in the following, and I will assume all four wheels are identical, until (unless) you post the dimensions, CG location, and non-drive wheel information.

    Your intuition is correct; the wheel mass moment of inertia uses only the wheel mass. However, there are some mistakes in your last group of equations. I currently get the following answers, assuming four identical wheels (two of which are drive wheels), neglecting axle bearing friction, drive system friction, and rolling resistance, and pretending the vehicle weight is evenly distributed to the four wheels, where m1 = vehicle mass excluding all wheels, m2 = mass of each wheel, I = mass moment of inertia of each wheel, r = radius of each wheel, mu = static coefficient of friction between wheel and road, Ff = frictional force that the ground exerts on one drive wheel, a = vehicle acceleration, and T = maximum possible drive torque applied to one wheel without slippage.

    Ff = mu*g*(0.25*m1 + m2).
    a = 0.5*mu*g.
    T = r*Ff + I*mu*g/r.

    Try that and see if it gets closer to your expected value. Don't forget to multiply T by 2, because T in the above equation is for each of the two drive wheels. Also, if you want to include rolling resistance, you could try to look up the coefficient of rolling resistance for your wheel/tire/road material combination.
  5. Feb 7, 2009 #4
    Just to let you know, nvn: as a test, we measured the amount of voltage required to make the wheels slip when starting from standstill and resolved your equation for mu. After plugging in the numbers, the result we got was almost exactly equal to the published coefficient of friction between our wheels and our surface. Thanks!
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