# Thread:

1. ## Radius from Arc

If the arc length s and the distance h from the centre point of the associated chord to the arc is known, how do you calculate the radius?

Thank You

If the arc length s and the distance h from the centre point of the associated chord to the arc is known, how do you calculate the radius?

Thank You
I don't think you will find a closed form solution for this problem - though
I am prepared to be proven wrong.

RonL

By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
But, $n=r\sin(s/2r)$
Thus, we have,
$r^2\sin(s/2r)=r(2r-h)$[/tex]
Thus,
$r\sin(s/2r)=2r-h$
----------------------
If you know calculus
Thus, you need to find the zero's of the function,
$f(x)=x\sin(s/2x)-2x+h$
Use Newton's method
Attached Thumbnails

4. Originally Posted by ThePerfectHacker
By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
Just as well that I checked my note on this problem

The intersection chord theorem would give in this case:

$<br /> n^2=h(2r-h)<br />$

surly?

RonL

5. Originally Posted by CaptainBlack
Just as well that I checked my note on this problem

The intersection chord theorem would give in this case:

$<br /> n^2=h(2r-h)<br />$

surly?

RonL
One thing that confuses me is how an immortal (me) makes such a mistake

6. Originally Posted by ThePerfectHacker
By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
But, $n=r\sin(s/2r)$
Thus, we have,
$r^2\sin(s/2r)=r(2r-h)$
...
Hello,

I'm a little bit confused: When you plug in the value of n, shouldn't be there a squared sine value too?:

$r^2\left(\sin(s/2r)\right)^2=r(2r-h)$

Greetings

EB

7. Originally Posted by earboth
Hello,

I'm a little bit confused: When you plug in the value of n, shouldn't be there a squared sine value too?:

$r^2\left(\sin(s/2r)\right)^2=r(2r-h)$

Greetings

EB
There should be, now how did I miss that (its in my notes)

RonL

8. Originally Posted by ThePerfectHacker
One thing that confuses me is how an immortal (me) makes such a mistake
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