Re: more trigonometry help


"dpb" <none@xxxxxxx> wrote in message news:fedrlq$91d$1@xxxxxxxxxxx
Joe wrote:
"dpb" <none@xxxxxxx> wrote in message news:fedq8u$391$2@xxxxxxxxxxx
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm a
long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection
above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

That's easier than a couple of alternatives... :)

No real way for ASCII art for the circle, so I'll try just the algebra
from a verbal description. Consider the triangle of the radius
perpendicular to the chord and another radius to one end of the chord.
That's a right triangle whose hypotenuse is R, one leg is your desired L/2
and the other side is the radius R-1/8.

So, letting x = L/2, r = R and h the projection desired, Pythagorus says

x^2 + (r-h)^2 = r^2

x^2 + r^2 - 2rh + h^2 = r^2

x^2 - 2rh + h^2 = 0

r = (x^2 - h^2)/2h

Substituting in your values I get r = 6.828 --> 6-53/64", approx.

--

Don't have AutoCad; so did about the same calculation as you did.
In your last formula, it should be "+h^2". Then the result matches
AutoCad's.
Kerry


.

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