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If I know the length of a chord of a circle and the length of the corresponding arc, but do not know the circle's radius, is there a formula by which I can calculate the length of the sagitta?

Somos
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RogerMartin
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  • Thank you, Ted. It seems this 63 year-old will have to brush up on "numerical methods". Won't hurt the old brain, I guess. Trying to solve a geometry problem at work - buckling caused by thermal expansion. – RogerMartin Jul 08 '18 at 21:59
  • Ah, it would help to give context, as we couldn't tell if you want a Greek-geometry solution or a calculus/numerical solution. But you can easily implement what I said with a modern calculator. – Ted Shifrin Jul 08 '18 at 23:30
  • Closely related to MSE question 1313686 "How do I plot sagitta versus arc length" – Somos Jul 08 '18 at 23:54
  • Using @Ted Shifrin's formula and creating a table of values for theta, sin theta and and the ratio of the two, I cranked out the answer in Excel. Thank you to all contributors. Helpful and inspirational site. Amazing what people accomplish motivated by their joy in the subject. – RogerMartin Jul 09 '18 at 13:10
  • Roger, glad we could help. Please accept an answer so that the question doesn't stay on the “unanswered” list, :) – Ted Shifrin Jul 09 '18 at 14:02

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I hadn't even known the term before you made me look it up.:) I don't see a way to compute it explicitly, but it is in fact uniquely determined. This is because knowing $r\theta=A$ and $r\sin\theta=B$, there is a unique $\theta$ ($0<\theta<\pi/2$) with $\dfrac{\sin\theta}{\theta} = \dfrac BA$, and then $r=\dfrac A\theta$ is unique as well. The sagitta is, of course, $r(1-\cos\theta) = r-\sqrt{r^2-B^2}$.

Clarifying Comment: Here $A$ is half the arclength and $B$ is half the chord length.

Ted Shifrin
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  • The one limitation, of course, is that $(\sin \theta)/\theta=c$ does not have an analytic solution (at least in terms of elementary functions). You can't have everything. – Oscar Lanzi Jul 08 '18 at 21:46
  • @OscarLanzi: I'm not disagreeing. I was merely arguing that there does exist a solution (by the intermediate value theorem, if you like) and that it's unique (by a bit more). – Ted Shifrin Jul 08 '18 at 21:48
  • @TedShifrin You are essentially right, but a little clarification is needed. I guess you let $A$ denote the length of the arc, $B$ the lenght of the chord and $\Theta$ half the angle between the two line segments connecting the center of the circle with the two endpoints of the chord lying on the circle, We have $0 < \Theta \le \pi/2$ since for $\Theta = 0$ we do not obtain a chord. Then $A = 2\Theta r$ and $B/2 = r \cos(\pi/2 -\Theta) = r \sin (\Theta)$. This means $\sin(\Theta)/\Theta = B/A$ and now it goes as in your answer. – Paul Frost Jul 08 '18 at 23:17
  • Yes, I was using $A$ and $B$ as half the lengths ... – Ted Shifrin Jul 08 '18 at 23:29
  • @TedShifrin Of course +1. – Paul Frost Jul 09 '18 at 07:29
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Starting from Ted Shifrin's answer, I think that we can have a reasonable estimate of $\theta$ using the Padé approximant built at $\theta=0$ $$\dfrac{\sin(\theta)}{\theta}=\frac{1-\frac{53 }{396}\theta^2+\frac{551 }{166320}\theta^4 } {1+\frac{13 }{396}\theta^2+\frac{5 }{11088}\theta^4 }$$ which is quite good for $0 \le \theta \le \pi/2$ (the maximum error is $2\times 10^{-6}$ at the upper bound).

Using it,solving $$\dfrac{\sin(\theta)}{\theta} =a$$ reduces to a quadratic equation in $\theta^2$ the solution of which being $$\theta=\sqrt{6\,\frac{455 a+1855- \sqrt{35} \sqrt{-3985 a^2+130862 a+25583}}{551-75 a} }$$

If you look here, you would find some interesting numerical approaches for the solution of $\sin(\theta)=a{\theta} $.

Edit

If you want a much better approximation, use $$\dfrac{\sin(\theta)}{\theta}=\frac{ 1-\frac{9168 }{68821}\theta^2+\frac{433 }{133573}\theta^4} {1+\frac{11374 }{340015}\theta^2+\frac{85 }{175694}\theta^4 }$$

Claude Leibovici
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These other answers are too complicated. Here is a simple way to approximate it, where s=sagitta, a=arc length, c=chord length:

$$s = 0.42 \sqrt{ a^2 - c^2}$$

If the arc is more like a full semi-circle, then the constant will be a little closer to 0.41. If the arc is more flat, then the constant will be closer to 0.43. But these are minor differences for approximations.

If you have only the sagitta and arc length and want to approximate the chord length, then you can use:

$$c = \sqrt{ a^2 - \left(\frac{s}{0.42}\right)^2 }$$

If you have only the sagitta and chord length and want to approximate the arc length:

$$a = \sqrt{ c^2 + \left(\frac{s}{0.42}\right)^2 }$$

Siong Thye Goh
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paulwal222
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