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I have a bicycle with one of those O-locks on it and too often when I park the bike and I want to lock it, the lock hits one of the spokes of the rim. This can be frustrating and surprises me that it occurs so often. I mean, the spokes are so thin and not that many really so one would think that this should not happen that often (like every day or so). And so every time this happened I reminded myself to calculate the probability of this happening. I just did it (after a year) and now I want to know what you guys think about my calculation, is it OK?

This is not a textbook example so there is no answer to look up or anything, that's why I need your feedback.

I have modeled the situation as shown in the figure below, where I have included one spoke only.

Let the O-lock have diameter $d_{L}$ and the spoke have a diameter $d_S$. Let $R$ denote the "radius" from center of wheel to the point where the O-lock comes and goes (this is approximately equal to the radius of the rim). Then, we have that the corresponding angles are given by $w_S=d_S/R$ and $w_L = d_L/R$ so that the spoke will hit the lock (the shaded disk in figure below) when $\theta$ is in the interval $[0, w_L+w_S]$ modulo $2\pi$ (See figure: $R$ is fixed so the point $(R,\theta)$ is on a circle). Then the probability for hit (wheel with one spoke) is given by $$ P(1) = \frac{w_L+w_S}{2\pi}\times 1 = \frac{1}{2\pi R}(d_L+d_S).$$

Generalizing to $N$ equally spaced spokes we get $$ P(N) =\begin{cases} \frac{N}{2\pi R}(d_L+d_S)& \text{for } N=0,1\dots\\1 & \text{for } N\geq M= \Big\lceil\frac{2\pi R}{d_L+d_S}\Big\rceil\end{cases}. $$

Example: plugging in some typical numbers; $N=36, R\approx 0.3$m, $d_S\approx 2\times 10^{-3}$m , $d_L\approx \times 10^{-2}$m we find

$$P(36) \approx 0.23$$ which kind of agrees with my everyday experience of this problem.

I guess one could obtain a (Monte-Carlo)-value for $\pi$ this way?

Could someone tell me what I have done wrong above? Or perhaps derive the correct expression for the probability?

Cross section of the O-lock (the disk) and the cross section of one spoke in the plane. $(R,\theta)$ denotes the position on one "side" of the spoke.

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  • $\begingroup$ Hi Love Learning, this is a neat question but it doesn't have anything to do with physics. Also, if it was physics related, you'd run aground of our anti-"verify my work" homework policy. Perhaps you should ask this over at Math.SE? $\endgroup$
    – Brandon Enright
    Feb 7, 2014 at 21:15
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    $\begingroup$ So the physics guys send "verify my homework" questions over here so we can do it? :) $\endgroup$
    – TooTone
    Feb 7, 2014 at 21:23
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    $\begingroup$ Haha dis no homework $\endgroup$ Feb 7, 2014 at 21:26
  • $\begingroup$ What do you mean by "the lock hits one of the spokes of the rim" $\endgroup$
    – Rustyn
    Feb 7, 2014 at 21:44
  • $\begingroup$ It hits them perpendicularly, so you cannot lock the bike. $\endgroup$ Feb 7, 2014 at 21:51

1 Answer 1

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A simpler method is as follows:

Find out the diameter of the cross section of your lock. Now calculate the area of the tyre in which the lock locks the tyre. (this can be done by the formula: $π(D^2-d^2)$,with d being the distance from the center of the rim to the lower end of the lock's cross section and $D$ being the length between the centre and the upper end of the cross section)

now find out the area of all the spokes lying in that region by $(D-d)x$(thickness of spokes) and subtract with the area found earlier. Find the area of the cross section of the lock and you can get the probabilty by dividing it by the figure found earlier. You can also find the anti probability too. Suggestions welcome

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