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

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

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

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

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.

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

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.