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Walt Pater
07-23-2004, 8:13 PM
Let's say you are given a crescent-shaped portion of a circle, formed by a line bisecting the circle in two points. Given the length of the line, 'a', formed within the circle by those two points, and the length of another line, 'b,' which runs perpendicular to 'a' to the apex of the circle, how do you determine the radius of the circle? I thought about this when I was scribing the arched tops to some French doors I was trimming. I am sorry if my geometrical descriptions are a little coarse- it's been awhile since my last nap in HS geometry. TIA, Walt.

Ted Shrader
07-23-2004, 8:39 PM
Walt -

Working on the Pythagorean theorem; The length of the hypotenuse (radius) is square root of the sum of the squares of the other two sides.

Where "A" is the length of the a chord on the circle, ½A is the length of one side of the triangle and (R-B) (as you described) is the length of the other side, square both of those and solve for R.

R<sup>2</sup> = (½A)<sup>2</sup> + (R-B)<sup>2</sup>

R = ½(A<sup>2</sup>/4B + B)

The proof is left to the reader. (I always wanted to write that) :rolleyes:

Regards,

Ted

Walt Pater
07-23-2004, 8:56 PM
Wow, that was quick, Ted. But I don't know if I was clear enough. I know that the circle and line figure I described would produce two mirrored right triangles, and that finding the hypotenuse would be a matter of the Pyth. theorem. but I don't think that the hypotenuse of those triangles would be the radius of the original circle. Let's say that you took the same circle, and simply moved the line vertically an inch or two. There would be smaller right triangles, but the radius of the circle would remain unchanged. Quite a puzzler, huh?

Ted Shrader
07-23-2004, 9:37 PM
. . . .. but I don't think that the hypotenuse of those triangles would be the radius of the original circle. Let's say that you took the same circle, and simply moved the line vertically an inch or two. There would be smaller right triangles, but the radius of the circle would remain unchanged. Quite a puzzler, huh?

Walt -

The vertical sides of the triangles would get and inch or two longer (depending on how high you moved the horizontal line). The other sides would get correspondingly shorter so the the sum of the squares would still be the same and the hypotenuse (radius) would rotate a few degrees (or fractions of a radian - but I never did like radians). The end of the hypotenuse would still remain on the circle.

Sometimes it is hard to visualize. It helps me when I use a compass and straightedge.

I just did the same thing on a head board and foot board for my daughter and son-in-law. It worked out that the radius/trammel for the router was 11' 9<sup>5</sup>/<sub>16</sub>" long to give a 3" rise in the middle (your B). I used two metal studs to make the trammel.

Ted

Walt Pater
07-23-2004, 11:05 PM
Bang on, Ted. You got it. I was visualizing the triangles above the bisecting line, not below (using sides A & B, not A & R-B). Your picture is worth one thousand words and makes up for my apparent lack of IQ points. Good to know that you also have put this formula to work for your own W'Wing- nothing like hands-on experience. Thanks a lot. Walt.
PS: I don't like radians either.
PPS: What exactly is a radian? Are they from Mars?

Michael Ballent
07-24-2004, 1:54 AM
Man this is creepy... I was just wondering the same thing, but that was for trying to calculate the radius for a guitar neck... Thanks Ted!!! That helped a bunch.

If you need a quick primer on these types of things you could go to http://www.purplemath.com One of my co-workers runs it and it is to help people out with simple geometry and algebra.