# Radius from Arc

• Apr 26th 2006, 06:19 AM

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
• Apr 26th 2006, 09:16 AM
CaptainBlack
Quote:

Originally Posted by askmemath

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
• Apr 28th 2006, 01:12 PM
ThePerfectHacker
Here is a picture to help you out.
By the theorem of chord in a circle we have that,
$\displaystyle n^2=r(2r-h)$
But, $\displaystyle n=r\sin(s/2r)$
Thus, we have,
$\displaystyle r^2\sin(s/2r)=r(2r-h)$[/tex]
Thus,
$\displaystyle r\sin(s/2r)=2r-h$
----------------------
If you know calculus
Thus, you need to find the zero's of the function,
$\displaystyle f(x)=x\sin(s/2x)-2x+h$
Use Newton's method
• Apr 28th 2006, 01:18 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Here is a picture to help you out.
By the theorem of chord in a circle we have that,
$\displaystyle n^2=r(2r-h)$

Just as well that I checked my note on this problem :D

The intersection chord theorem would give in this case:

$\displaystyle n^2=h(2r-h)$

surly?

RonL
• Apr 29th 2006, 06:49 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Just as well that I checked my note on this problem :D

The intersection chord theorem would give in this case:

$\displaystyle n^2=h(2r-h)$

surly?

RonL

One thing that confuses me is how an immortal (me) makes such a mistake :mad:
• Apr 29th 2006, 09:21 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
Here is a picture to help you out.
By the theorem of chord in a circle we have that,
$\displaystyle n^2=r(2r-h)$
But, $\displaystyle n=r\sin(s/2r)$
Thus, we have,
$\displaystyle r^2\sin(s/2r)=r(2r-h)$
...

Hello,

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

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

Greetings

EB
• Apr 29th 2006, 09:42 PM
CaptainBlack
Quote:

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?:

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

Greetings

EB

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

RonL
• Apr 30th 2006, 04:36 AM
earboth
Quote:

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