Finding smallest circle that fits over a trapezoid

I'm trying to figure out how to calculate the smallest circle that will fit
over a trapezoid. No, it not home work -- I need to know what size to make
the hole for the thingy I'm building. I tried asking some coworkers, a PhD
ME, a MSME, and a physicist. I'm just a electronic tech, so assumed they'd
be smarter than me.
The particulars in this case are, the cross section is an trapezoid, 0.310"
bottom width, 0.200" top width, 0.210" height, symmetrical about the Y
axis -- What size hole will that fit in to?
Robert
Reply to
SignalFerret
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.342 plus or minus a thou or so.
Reply to
Jim Stewart
Just drawing it on the cad because it's fast and calculates all by itself, it looks like that trapezoid would fit into a .342" dia hole.
Of course I didn't verify this was true by calculation sooo....dont' bet the farm on it.
Koz
Reply to
Koz
Xactly ...... Bobcad has it at .3414
A
Reply to
Albert Finley
Now where did I leave that 0.3414 drill bit?
I'm trying to figure out how to calculate the smallest circle that will fit over a trapezoid. Bobcad has it at .3414
Reply to
Thomas Kendrick
Let B the bottom width, T the top width, and H the height. In general, there are two solutions (assume B >= T). If H and T are small compared to B, the smallest circle is one with diameter D=B. To be more specific:
If H sqrt(B^2-T^2)/2, then
D=sqrt[16*H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2]/(4*H)
In your case ('big H'), this gives D~.34148.
If you put the trapezoid with its base centered on the x-axis (middle of the base at (0,0)), the center of your circle will be
y=(4*
H^2 + T^2-B^2)/(8*H)
Nick
Reply to
nick
Thanks Nick! Exactly what I was looking for. And thanks to everyone else who contributed!
Reply to
SignalFerret
It don't smell right. I haven't run numbers to check, but a quick check on dimensions says it won't. D comes out in length^2 units, assuming it makes sense to add length^4 + length^3.
Cain't argue with it if the numbers come out good, though.
Reply to
Mike Young
For dimensions I see (sqrt[L^4])/L = L. Would have been clearer with a couple of extra parentheses:
D=(sqrt[16*H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2])/(4*H) or maybe
D= sqrt[16*
H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2] -------------------------------------------- 4*H
Checking the dimensions is a good idea. Always happy to have someone check my work. I've seen too many cases where the wrong formula gives close to the right answer.
Nick
Reply to
nick
Nick nailed it. Crosscheck: AutoCAD sez diameter = .34148, center is .07161 above the midpoint of the base. .
Reply to
Don Foreman
Crosscheck #3... Solidworks says .34148 also.
Ahhh, CAD.
Reply to
Adam

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