For the trig/math guys here...

Replies to This Discussion

Permalink Reply by David Rutten on May 24, 2013 at 12:26pm

Proof is coming shortly.

--

David Rutten

david@mcneel.com

Poprad, Slovakia

Permalink Reply by David Rutten on May 24, 2013 at 12:43pm

This is how I solved it:

1. First off we draw a polygon with the same radius R as the circle. It's easy to construct this polygon, you don't need any equations for it.
2. The points on the corners of this polygon are L units apart where L = |PQ|. Again, you can measure this or -if you want- you can solve it mathematically.
3. As the polygon is scaled down to zero at the circle center, the points P and Q move linearly inwards to P' and Q'. We need to scale the polygon in such a way that |P'Q'| is twice as large as |PP'|.
4. This will result in two linear equations that both depend on the same variable, lets' call it f for scale factor. When f=1.0 the polygon has it's origin length, when f=0.0 the polygon has fully collapsed onto M.
5. In maths notation:
     |PP'| = R * (1-f)
     |P'Q'| = L * f
6. Now equate them to each other:
     2*|PP'| = |P'Q'|
   or
     2 * (R * (1 - f)) = L * f
7. Solve for f and you get:
     f = (2 * R) / (L + 2 * R)

--

David Rutten

david@mcneel.com

Poprad, Slovakia

Permalink Reply by Yoann Mescam (Systemiq) on May 24, 2013 at 5:11pm

Or with the intercept theorem, using David's notations:

|MP'| / |MP| = |P'Q'| / |PQ|

let say r=radius of inside circles,  and L = |PQ|

  so |MP'|= R-r, |MP|=R and |P'Q'|= 2*r

=> (R-r) / R = 2*r/L

=>...

=> r = R*L / (2*R+L)

Permalink Reply by Mitch Heynick on May 25, 2013 at 4:34am

Thanks for another approach!

--Mitch

Permalink Reply by Mitch Heynick on May 25, 2013 at 4:33am

Thanks David!  You da man!  (of course you know that...)

--Mitch