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.
tia
jc
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"?
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There's those terms I can't remember!
Yes, that's exactly what I was trying to say.
Joe
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.
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The diameter would be 3.570". Thank God for AutoCAD.
todd
Hi Joe,
Radius: 1.7852" Diameter 3.5704"
Ed Bennett snipped-for-privacy@ts-aligner.com
Dang! I'll have to check the calculator, then...
Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"
Wonder where the difference is coming from?
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Well, the last bit should be r=(x^2 + h^2)/2h. Plug those numbers in and it works out.
todd
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
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! :) )
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Thank you everyone for the answer. I have a bending form to make.
be well, work wood
jc
Following this thread has reminded my why I was a liberal arts major.
"Joe" wrote in news:RVtOi.38972$ snipped-for-privacy@newsreading01.news.tds.net:
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 {(1/8)/(21/32)}) = 21/32 and solved for r. You can check me on this, but I think it's right.
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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...
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Drawing that on AutoCAD that comes to a diameter of 3-9/16 rounded to the nearest 1/16th.
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.
Lew
"Joe" wrote
I think that Mr Cole and his fellow geometricians would have called Joe's 'apex' the 'sagitta'.
Jeff
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.
Sorry to have arrived so late. This gets asked enough that I put up a web page with diagram and step-by-step algebra at the link below.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA
Decartes proved, oh so many years ago, that geometry and algebra were mathematically equivalent - anything doable in one was doable in the other.
Well, we were allowed to count any math theory course as a liberal arts elective... :)
Of course, that was in the era when a programming language qualified as the foreign language requirement in a PhD program, too...
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