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I have a sphere or radius r that has a small point like object moving at a constant velocity inside it. Each time it hits the side it bounces in a random direction but always away from the tangent plane at the contact point. The longest distance it can travel is through the center to the other side $2r$ if it bounces at $45^\circ$) ($\alpha =\pi/4 $) it will hit the sphere a point on the sphere that is on the intersection of the tangent plane if moved to the center. The distance would be $2 \pi \cos(\alpha) $. I assume this works for all bounce directions for $0 < a \leq \pi$

I need to find the average distance traveled between bounces, but I am somewhat at a loss as to how find the solution. I assume its an integral of some type?? This should be a fixed value in terms as a simple multiple of the radius and the same not matter the value of $r$, and would be happy with that number.

But I would be very happy if I could get a detailed method, as I would like to be able to express the angle of reflection in terms of a curve. As the question is asked above that curve is a flat line (uniform distribution). $\alpha = \operatorname{random}(0\;\text{to}\;\pi)p(x)$ where $p(x) = 1$ but $p(x)$ could be any function. Is this possible?

Update

To clarify the angle of reflection $(\alpha_x)$ is expressed as the angle from the axis that is the line from the point of contact through the center and conforms to $0 < \alpha_x \leq \pi$ where pi is traveling through the center to the opposite side. The second angle around that axis ($\alpha_y$) can be $0 < \alpha_y < 2\pi $

Sorry I have very little knowledge of the vernacular used to express math concepts.

The random function should return an evenly distributed random value.

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  • $\begingroup$ I don't understand the last paragraph -- partly because it's ungrammatical -- could you rephrase that? Regarding the first paragraph, John Hughes has already pointed out the lack of clarity. $\endgroup$ – joriki Jun 29 '16 at 23:52
  • $\begingroup$ I'm going to suggest a possible improvement of the question: "A ray leaves the south pole in a direction $d$ such that the angle between $d$ and the south-to-north diameter of the sphere is chosen uniformly randomly from $0$ to $\pi$. What is the expected length of the ray?" Does that express it? $\endgroup$ – John Hughes Jun 29 '16 at 23:54
  • $\begingroup$ oops...I meant $\pi/2$. $\endgroup$ – John Hughes Jun 30 '16 at 0:02
  • $\begingroup$ @JohnHughes Yes that expresses the question correctly. Thankyou. $\endgroup$ – Blindman67 Jun 30 '16 at 0:16
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Here's my restatement of your problem, and a general answer:

Q1: "A ray leaves the south pole in a direction $d$ such that the angle between $d$ and the south-to-north diameter of the sphere is chosen uniformly randomly from $0$ to $\pi/2$. What is the expected length of the part of the ray within the sphere?"

First assume the sphere has radius $1$.

For angle $a$, the length $L(a)$ of the ray is $2 \cos(a)$. (This is an elementary fact from geometry).

The expected value of $L$ is computed as $$ E[L] = \int_0^{\pi/2} L(a) p(a) da, $$ where $p(a) = \frac{2}{\pi}$ is the probability density at angle $a$. This evaluates to \begin{align} E[L] &= \int_0^{\pi/2} L(a) p(a) da\\ &= \frac{4}{\pi} \int_0^{\pi/2} \cos(a)~ da\\ &= \frac{4}{\pi} \left. \sin(a)\right|_0^{\pi/2}\\ &= \frac{4}{\pi} (1 - 0)\\ &= \frac{4}{\pi}. \end{align}

Q2: "A ray leaves the south pole in a direction $d$ such that the angle $a$ between $d$ and the south-to-north diameter of the sphere is chosen according to some distribution $a \mapsto u(a)$ on the interval $[0, \pi/2]$. What is the expected length of the part of the ray within the sphere?"

The formula for the expected value is $$ E[L] = \int_0^{\pi/2} L(a) p(a)~ da = = \int_0^{\pi/2} 2 \cos(a) p(a)~ da. $$ Without knowing $p$, however, this as far as we can go.

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  • $\begingroup$ I think this should be a comment to help clarify the question, not an answer. $\endgroup$ – joriki Jun 29 '16 at 23:51
  • $\begingroup$ I sort of agree...but it's too long for a comment, and it IS an answer to the question that was asked. $\endgroup$ – John Hughes Jun 29 '16 at 23:52
  • $\begingroup$ Yes, but I think we should differentiate between salvageable questions and unsalvageable questions. I sometimes give answers like this when I feel that the question is hopelessly confused and the main thrust of an answer should be to point out the confusion. In the present case, by contrast, it's just that no distribution has been specified, and once one is specified, the question makes sense (at least the first part that's clearly formulated). So it seems more productive to ask for the distribution and then answer the question than to treat the question as unanswerable. $\endgroup$ – joriki Jun 29 '16 at 23:54
  • $\begingroup$ I've edited my answer to solve the problem I think that OP meant to ask. Thanks for the push in the right direction. $\endgroup$ – John Hughes Jun 30 '16 at 0:07
  • $\begingroup$ Great. My tendency would have been to assume a uniform distribution over the solid angle -- then we'd have $p(a)\propto\sin a$ and thus \begin{align} E[L] &= \frac{\int_0^\frac\pi22\cos a\sin a\,\mathrm da}{\int_0^\frac\pi2\sin a\,\mathrm da} \\ &= \frac{\int_0^\frac\pi2\sin2a\,\mathrm da}{\int_0^\frac\pi2\sin a\,\mathrm da} \\ &= 1\;. \end{align} $\endgroup$ – joriki Jun 30 '16 at 0:18

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