How to bound the great-circle distance of two points on a sphere, only given their euclidean distance?

Question

Suppose I have a great-circle of a sphere in $\mathbb{R}^n$, the chord length (the euclidean distance of any two points) is $L$, how can we upper bound the arc length $C$ (for any radius)?

I read somewhere $C\le \frac{\pi L}{2}$, is it correct? If so, how to prove it?

Thanks.

Is this the chord of a circle? – Mark Bennet Jan 02 '15 at 10:50 — Mark Bennet, Jan 02 '15 at 10:50
If it is a circle, your formula is correct. – Bumblebee Jan 02 '15 at 10:53 — Bumblebee, Jan 02 '15 at 10:53
Yes, it is the great-circle of a sphere in $\mathbb{R}^n$. – user3138073 Jan 02 '15 at 10:56 — user3138073, Jan 02 '15 at 10:56

score 1 · Answer 1 · answered Jan 02 '15 at 10:55

1

If the longer of the two arcs is allowed, then there is no bound: An arbitrarily short chord can correspond to an arbitrarily long arc.

Otherwise, let $\alpha\le \frac\pi2$ be half the subtending angle. Then the arc length is $C=2r\alpha$ and the chord length is $L=2r\sin\alpha$. Hence $\frac CL=\frac{\alpha}{\sin\alpha}$. This expression grows with $\alpha$ (tends to $1$ as $\alpha\to 0$), hence has its maximum at the right end, i.e., for $\alpha=\frac\pi2$, where the quotient is $\frac\pi2$.

answered Jan 02 '15 at 10:55

Hagen von Eitzen

374,180

Oh... I did n't see that. Thank you. – Bumblebee Jan 02 '15 at 11:00
Thanks for the answer. But the chord length should be $L=2r\sin\frac{\alpha}{2}$, am I right? – user3138073 Jan 02 '15 at 11:08
Also, $C=r\alpha$. But the idea is correct. Thank you! – user3138073 Jan 02 '15 at 11:16

How to bound the great-circle distance of two points on a sphere, only given their euclidean distance?

1 Answers1