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. .