Finding smallest circle that fits over a trapezoid

.342 plus or minus a thou or so.

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

Xactly ...... Bobcad has it at .3414

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

Nick nailed it. Crosscheck: AutoCAD sez diameter = .34148, center is .07161 above the midpoint of the base. .

Crosscheck #3... Solidworks says .34148 also.

Ahhh, CAD.