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

SignalFerret wrote:

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?


..342 plus or minus a thou or so.

SignalFerret wrote:

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

Jim Stewart wrote:

SignalFerret wrote:

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?


.342 plus or minus a thou or so.

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

A

Now where did I leave that 0.3414 drill bit?

On Thu, 15 Sep 2005 19:11:19 -0400, Albert Finley
wrote:

I'm trying to figure out how to calculate the smallest circle that
will fit over a trapezoid.
Bobcad has it at .3414

SignalFerret wrote:
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


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=B.

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

Thanks Nick! Exactly what I was looking for. And thanks to everyone else
who contributed!

"nick" wrote in message
news
SignalFerret wrote:
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

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=B.

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

"SignalFerret" wrote in message
news:XGqWe.23510$Hs6.8480@trnddc07...
D=sqrt[16*H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2]/(4*H)

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.


In your case ('big H'), this gives D~.34148.

Cain't argue with it if the numbers come out good, though.

Mike Young wrote:
"SignalFerret" wrote in message
news:XGqWe.23510$Hs6.8480@trnddc07...

D=sqrt[16*H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2]/(4*H)


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.


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


In your case ('big H'), this gives D~.34148.


Cain't argue with it if the numbers come out good, though.


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

On Fri, 16 Sep 2005 05:54:46 -0600, nick wrote:

Mike Young wrote:
"SignalFerret" wrote in message
news:XGqWe.23510$Hs6.8480@trnddc07...

D=sqrt[16*H^4 + 8*H^2*(B^2+T^2) + (B^2-T^2)^2]/(4*H)


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.


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


In your case ('big H'), this gives D~.34148.


Cain't argue with it if the numbers come out good, though.


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.