[Closed] Today's maths question - circle geometry

The diameter is the longest chord possible. If the (chord x pi)/2 > arc then you have less than a semi-circle so the chord is the longest distances between the two points. right?

Not sure about what to do if you have more than a semi-circle though.

https://en.wikipedia.org/wiki/Sagitta_(geometry)#

I did try to Google, but like that Wiki link, everything I find seems to assume I know the radius, which I don't. There must be a way?

So you know length of the arc, but you want the distance in a straight line yeah?

I don’t know the answer, but it seems like a fun thing to work out.

Can you measure or calculate the angle? That might make it easier?

Right, to be clear, from the diagram below I know L and CL, I want to know CH, I don't know r

I'm a bit rusty at this so bear with me, you know two facts the length of the chord (CL) and the length of the arc (L). There are formulae to calculate these when you know the radius and the angle (a) made by the two radius lines that join the centre of the circle and ends of the chord. We need the radius r

L = a*r*pi/180 and CL = 2*r*sin(a/2)

So rearrange the first equation to get a and replacing a with this in 2nd equation will give you an equation in 1 unknown that you can re-arrange to get r. Once you have r you can work out CH.

I’ve had a go at solving this. Thought I was getting somewhere, but was wrong ☹️. So, having a wee Google I’m inclined to think there’s no straightforward solution and that only approximation is possible.

On a brighter note, I found this website with a calculator. Plug in your numbers and abracadabra.

https://handymath.com/cgi-bin/arc18.cgi

This website describes the solution, similarly to rjmccann101. Unfortunately you end up with a non-tractable analytical equation, so you have to iterate (put in a first guess for theta and modify until you get something that works): radius from arc and chord length

In your notation, let theta be the angle from the centre to the chord in radians = rjmccann101's a*pi*180. One identity is L = r*theta; the other is from the right angled triangle CL/2 = r* sin(theta/2). Then sin(θ/2) = CL/2 * θ/L. To solve, sin(θ/2) - CL/2 * θ/L ~ 0: put in CL and L, start guessing theta as radians until you get close to zero. Then stick the theta giving you closest to zero back in r = theta/L; BC on the linked diagram = r* cos(theta/2) (right angled triangle) and work out the sagitta = r - BC.

I'm not actually convinced that there is a unique solution to this as you would need to solve simultaneously for r, CH & the triangle described by r, CL/2 and r-CH. I think there are too many unknowns and not enough independent equations.

Just use Pythagoras theorem: (CL/2)^2 = r^2 + (r - CH)^2

CL^2/4 = r^2 + r^2 - 2 r CH + CH^2

CH^2 - 2 r CH + (2 r^2 - CL/4) = 0

solve quadratic for CH: CH = (1/2)*(2r +/- sqrt(4 r^2 - 4*(2 r^2 - CL/4)))

r is defined by L.

Using your notation:

r = (CH^2 +(CL/2)^2) / (2 . CH)

Calculate angle subtended by the chord (A)

A/2 = acos((r-CH)/CL/2))

L = 2.pi.r.A/360

Just use Pythagoras theorem: (CL/2)^2 = r^2 + (r – CH)^2

There are three variables in that equation CL, CH, & r but only one is a known value CL. There isn't therefore a unique solution.

Sneaky edit there.

don't think there is one solution for r or θ given just CL and L and therefore no one solution for CH

As far as I can get you'd need to solve the following equations simultaneously

Theta = 2*arcsine(CL/2r)
L= 0.1745*Theta*r
CH=r-sqrt(r^2-(cl/2)^2)

but it wouldn't surprise me if those equations collapsed when you tried.

don’t think there is one solution

If you consider a chord of given length, there are infinite circles that can pass through the end points of the chord. However, there is only one circle that has the given arc length. Hence a unique solution (ignoring the congruent circle in the opposite direction).

Seeing as that the calculator posted by gauss1777 works from 2 inputs, there must be a formula solution

Seeing as that the calculator posted by gauss1777 works from 2 inputs, there must be a formula solution

There is. See my previous post.

This page gives you the arc to chord length ratio as a function of the angle. So plug in your ratio to find the angle and you should be able to work out the rest.

http://www.1728.org/cntlangl.htm#:~:text=ARC%20TO%20CHORD%20RATIO&text=Dividing%20the%20arc%20length%20by,ratio%20will%20always%20be%201.1107207345.

r = (CH^2 +(CL/2)^2) / (2 . CH)

CH is the unknown though.

And I'm not entirely convinced that that's correct in any case.

And I’m not entirely convinced that that’s correct in any case.

No, actually, I am.

CH is the unknown though.

Doh - I thought he was looking for CL

The formula is correct for calculating radius from chord length and sag

r = (s^2 + l^2)/2s where s = sag and l is half the chord width - proved by considering the triangle from half the chord and the centre of the circle.

Solve as a ratio:

rCos (theta/2) = r - CH -> cos(theta/2) = 1 - CH/r
rSin(theta/2) = CL -> sin(theta/2) = CL/r

so sin^2 + cos^2 = 1, hence

CH/r = 1 - sqrt(1 - (CL/r)^2)

is the solution as a ratio to the radius of the circle.

If ever there was evidence of the value of choosing good variable names, this question is it.