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.
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.
Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made a
misteak! :) )
Well, we were allowed to count any math theory course as a liberal arts
Of course, that was in the era when a programming language qualified as
the foreign language requirement in a PhD program, too...
It's been a while for me too, but I think a diameter of 3 9/16 is close
to what you want. It seems small to me, but I did it twice and checked,
so like I said it should be close (I got 3.57 inches for the diameter,
which is just a hair over 3 9/16.)
The setup is to draw the chord and radii from each endpoint and one from
the center of the chord - that gives four right triangles, two big ones
with a radius as hypotenuse and two little ones. Solve the small
triangle (you know the two sides, 1/8 and 21/32), then figure out how
the little triangle is related to the larger one with which it shares a
side. I figured out that the central angle of the larger right triangle
is just twice the smaller angle in the little triangle. I got
r sin 2(arctan ) = 21/32
and solved for r. You can check me on this, but I think it's right.
You got the answer (and I didn't until somebody pointed out the
algebraic error in sign I made, so take this for what that's worth :) ),
but in this particular case you can solve for the radius directly from
one of the large (right) triangles as two sides are known in terms of
one unknown (the radius) without even having to solve the quadratic as
the quadratic terms end up canceling...
> base length.
Life is short, the problem is very straight forward using a graphical
camber layout using the info given.
I'm reminded of the time I took the PE exam oh those many years ago.
I'm a young, pimple faced, smart assed engineering guy with the
fastest slide rule on the planet.
Seated at a desk a few rows in front of me was a gray haired man,
perhaps 50, with a small drafting board complete with triangles,
pencils and scales.
This old man was going to try to compete with the fastest slide rule
on the planet using graphical solutions.
Never heard if he passed the exam or not, but he completed the exam
before I did.
The older I get, the more respect I gain for that man and his
approach. He has has probably long since departed.
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