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Variable diameter bicycle wheel

  1. Jan 26, 2008 #1
    I'm attempting to write a simple model to describe the angular velocity profile (aka angular acceleration) of a bicycle wheel for different torque vs time profiles.

    Where this gets complex is that it's not a normal bicycle wheel. It is a normal wheel with weights on the spokes connected to the hub by springs. So as the wheel increases it's angular velocity, the weights move outward, changing the overall moment of inertia for the wheel.

    I'm going to assume that the weights are a second rim with fixed mass, but variable radius, and both 'rims' are thin shells (so I= mr^2). I'm also just (for now) looking at a torque being applied to the 'real' rim and not considering the inertia or accelerations on the whole wheel as if it were rolling down a hill.

    Any thoughts on how best to approach this? This is what I have so far:

    [tex]\dot{\omega}[/tex] = f(T,I)

    [tex]\dot{T}[/tex] = f([tex]\dot{F}[/tex],r1)

    [tex]\dot{I}[/tex] = f(m2,[tex]\dot{r}2[/tex])

    [tex]\dot{r}2[/tex] = f(m2,[tex]\omega[/tex],k)

    [tex]\dot{F}[/tex] = f(t)

    thoughts? and easier way to approach it?
  2. jcsd
  3. Jan 26, 2008 #2


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    Welcome to PF, Vortmax.
    It sounds like a pretty cool idea. I know absolutely nothing involving math, so I can't comment upon your posted formulae. I would expect, however, that the same calculations that were used for the old 'flyball governors' on steam engines would apply. They also increased their diameter with rising rpm's. Maybe there's something in the old literature than can help you.
  4. Jan 26, 2008 #3


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    Well, I'll make a shot at this, with a very simplified model:

    Let the radius of the wheel be R, the mass M.
    On one of the massless spokes of the wheel, there is a tiny "cylinder" of mass m that can move back and forth along the spoke without friction, having radial position r(t).

    Joining the hub and the tiny cylinder is a massless spring with spring constant k, capable of giving a radial force to the cylinder, and the springs rest length is L.

    For simplicity, I'll let the hub be at rest, but the wheel may rotate with respect to some externally applied torque T.

    We also assume that the angular velocity has a constant direction, albeit its strength may vary.

    We let [itex]\vec{i}_{r}[/itex] be the unit radial vector.

    The position of the cylinder is then:
    and its velocity:
    where [itex]\omega(t)[/itex] is the angular velocity of the wheel.

    The cylinder's contribution to the system's angular momentum is therefore:
    where [itex]\vec{k}[/itex] is the constant direction of the angular velocity.

    Hence, the system's total angular momentum is:
    With external torque [itex]\tau\vec{k}[/itex] applied, we get:
    [tex]\tau=(MR^{2}+mr^{2})\dot{\omega}+2mr\dot{r}\omega (1)[/tex]

    Now, consider the radial component of Newton's second law of motion for the cylinder, i.e, precisely that motion of the system not captured by the moment-of-momentum equation:
    We get:
    [tex]-k(r(t)-L)=m(\ddot{r}-r\omega^{2}) (2)[/tex]

    Equations (1) and (2) describe the motion of the system.
  5. Jan 28, 2008 #4
    Thanks, that did help a lot. Now I just need to get it into discrete form so I can code it.

    To make the calculation easier, I broke down the process. Tell me if this makes sense.

    For anyone not familiar with iterative methods, I just start with initial conditions, calculate any changes, update the original conditions and repeat.

    I start with my initial conditions: mass of the rim, mass of the weights, radius of the rim, radius of the weights, and current angular velocity. I accelerate the wheel in it's current configuration by applying a fixed torque for x seconds (the time step). This gives me a new angular velocity. From this angular velocity, I calculate the new position of the weights (based on centripetal acceleration). From this, I can get a new moment of inertia for the system. Since angular momentum is conserved, I apply this new moment of inertia to adjust the angular velocity.

    My problem now is calculating the radius of the weight based on angular velocity. Centripetal force as a function of angular velocity is m[tex]\omega[/tex]^2*r, and the displacement of a spring mass system is F=-kr. When you equate the forces, the radius term is eliminated completely. Am I missing something, or how do you equate the displacement of a mass in a 1D rotating mass-spring system.
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