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Assume that a motorcycle of mass $m$ has two wheels that are equidistant from its centre i.e the force on each wheel is $mg/2$.

If the motorcycle accelerates forward, will the two forces on each wheel (measured instantaneously) remain the same? If not, how can one mathematically describe the change in forces measured on each wheel and will these forces oscillate before converging when the motorcycle reaches final velocity?

Now, if the weight of the motorcycle isn't uniformly distributed along it's length e.g force measured at one wheel is say $mg/3$ and $2mg/3$ at the other, how different would the vehicle dynamics (oscillations etc) be? Can one somehow infer the fact that the weight isn't uniformly distributed without actually measuring the forces at the wheels?

I know this question is very vague but any ideas are very welcome.

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  • $\begingroup$ Why is this tagged fourier-transform? $\endgroup$
    – ACuriousMind ♦
    Commented Mar 9, 2016 at 11:58
  • $\begingroup$ @ACuriousMind edit it please $\endgroup$
    – hxri
    Commented Mar 9, 2016 at 11:58
  • $\begingroup$ It depends on the cg height relative to the wheel centers. $\endgroup$
    – John Alexiou
    Commented Mar 9, 2016 at 14:28
  • $\begingroup$ Apologies (regarding fourier-transform tag). I'm curious to know, if the bike does oscillate when it accelerates, whether one can somehow make a very rough estimation of the forces at the wheels while having access to any other time-varying information required (besides actually measuring the force at the wheels) e.g tractive force/torque from the engine, velocity, waveform of the oscillation of the bike which can be decomposed using a fourier transform etc. $\endgroup$
    – user1889776
    Commented Mar 9, 2016 at 15:05

2 Answers 2

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There is an old trick that turns a dynamics problem into a statics problem. Apply equal an opposite inertial forces on the center of mass.

Lets look at a free body diagram of a motorcycle accelerating with $\ddot{x}$.

fbd

The balance of forces in the horizontal direction equate the tractive force $B_x$ to the acceleration $$ \left. m \ddot{x} - B_x = 0 \right\} B_x = m \ddot{x} $$

The balance of forces in vertical direction, together with the balance of moment about any point (I choose the rear contact point) give us the distribution of loads on the tires.

$$\left. \begin{align} A_y + B_y - m g & = 0 \\ -\ell A_y + c m g - h m \ddot{x} & = 0 \end{align} \right\} \begin{aligned} A_y &= \frac{c}{\ell} m g - \frac{h}{\ell} m \ddot{x} \\ B_y & = \left(1-\frac{c}{\ell}\right) m g + \frac{h}{\ell} m \ddot{x} \end{aligned} $$

ALSO, if you want to find the minimum acceleration for a wheelie then set $A_y=0$ to find $$\ddot{x} \ge \frac{c}{h} g$$

FINALLY, if traction is limited to $B_x \le \mu B_y$ then the equations above give us the maximum acceleration before traction is lost $$\ddot{x} \le \mu g \frac{\ell-c}{\ell-\mu h} $$

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  • $\begingroup$ This answer has exceeded my expectations, and I've learned from it. Much appreciated. My precise problem is that I want to estimate $c$, and optionally, $h$. If we now add the bike suspensions into the mix, each with a spring constant $k$, the bike will tilt at an angle dependent on the difference $d$ between the compressive displacements of two suspensions. Assume I can measure the tractive force $B_x$, the acceleration and tilt angle of the bike, $\theta$. Can we now add another relation, $A_y = B_y * (1 - tan (\theta) )$? $\endgroup$
    – user1889776
    Commented Mar 10, 2016 at 0:06
  • $\begingroup$ I am glad this was helpful. At this point, I suggest you post a follow up question with your new situation. $\endgroup$
    – John Alexiou
    Commented Mar 10, 2016 at 2:01
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First of all geometric center and the center of mass are two different things.

Since a motorbike would not have center of mass in the very middle due to the facts that each part of the bike weighs different as well as the riders position. You would need to do a free body diagram to start your analysis. Then you can analyze the moments around the points of contact of each wheel then get the correct force value. If it was uniform object with wheels with equal distance to center of mass, your mg/2 assumption would be correct, however, it would be best to analyze the moments around the point of contact to find correct reaction forces.

Also, as the bike starts to accelerate, there will be forces applied through the front or rear wheel. This would change the balance of forces on the wheels as there are more forces present.

My advice to you is to, draw a FBD for each scenarios and then analyze the reaction forces. Through finding reaction forces, you can find shear and normal stresses which would help you to find oscilations. However this is not as simple forward as you put it in the question.

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  • $\begingroup$ Thanks. The first part of my question assumes center of mass is at center of gravity. This is a hypothetical motorbike with uniformly distributed weight. $\endgroup$
    – user1889776
    Commented Mar 9, 2016 at 14:50
  • $\begingroup$ @user1889776 center of mass is the same thing as center of gravity. I mentioned geometric center. You might want to clarify on that. Regardless, my answer should suffice. $\endgroup$
    – Mert Karakaya
    Commented Mar 9, 2016 at 14:52
  • $\begingroup$ Sorry, meant geometric center. $\endgroup$
    – user1889776
    Commented Mar 9, 2016 at 15:06

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