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This question: Sum and average length of chords (1) has already been sufficiently well answered, and What is the average length of 2 points on a circle, with generalizations (2) seems to be the same question, but malformed, although the responses seem to indicate the same results as (1).

However, if the end points of the chord are constrained, for example, if all the chords are constrained to be vertical, and no other chords are allowed, then how does one estimate the average length of chords constrained in such a manner using geometric probability? Are there any textbooks or references for this?

Lee David Chung Lin
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ASG
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2 Answers2

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So, I finally got around to doing this.

One has to use the formula for length of a chord given its distance to the center. The average length of a constrained chord (l) of a circle of radius (r), given its distance from center (h) varying from 0 to r is: $$\ l = {\int_0^r 2{\sqrt{r^2-h^2}}dh\over{r}}$$ Averaged over r, the integral is: $$ l = {\biggl[h\sqrt{h^2-r^2}+r^2tan^{-1}\Bigl(\frac{h}{\sqrt{h^2-r^2}}\Bigr)\biggr]_0^r\over{r}} $$

After applying the limits, and since, $tan^{-1}$ of large numbers is $\pi/2$, the average length of a vertical (or horizontal chord) would be: $$\ l = {\pi\over2}r$$ For a unit circle this would be: 1.57. I used a CAD software and very painstakingly verified this as well.

ASG
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Consider the unit circle $\rho=r$ in polar coordinates.

the generic vertical chord is $AB=2r \sin\theta$

The average chord is $$x=\frac{1}{\pi}\int_0^{\pi } 2 r \sin \theta \, d\theta=\frac{4r}{\pi}$$

In the unit circle the average vertical chord is $1.27324$

Looks weird :)

Raffaele
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  • The formula for a chord is the same whether it is vertical or not, result proposed is same as ref (1). Here, once the 1st point is selected, the other point is constrained to be at a particular location only (since the chord has to be vertical). So average chord should be larger than 1.27324 for a unit circle. I get 4, dividing by pi may not be required, but that cant possibly be correct? I'll try to do this problem "numerically", i.e. draw vertical chords with uniform spacing, measure the lengths and converge on a result. This is an application related to tribology (chords are lines of wear). – ASG Aug 03 '17 at 13:03
  • I've come up with an answer that seems to work, your inputs would be appreciated. – ASG Nov 20 '17 at 10:44