The formula I used, and can't now remember where I got it from is:

1/26 is .0384

.0384 x .866 is .033

However, .025 worked just right.

Alan

There are two methods one can use: the trig method and the Pythagorean theorem. Ultimately, they are exactly the same thing which can be proven mathematically but the Pythagorean theorem is easier to use.

Knowing that an isoceles triangle has all three sides being equal makes this very easy to use for the American style and metric styles which both use 60deg angles on all three sides. for the British 55 deg angles, this is a bit harder but ultimately the same formulation.

Construct an isoceles triangel with the base at the bottom, labelled A, the left side label B and the right side label C. Then cut the triangle from the tip to the center of A. Call this H for height. From the point where H meets A, which is A/2 and meets at a right angle, to either point to the left or right meeting either B or C could get a special name like "a" or maybe just continue to use "A/2" which is what I will do. Then useing the Pythagorean theorem: H^2 + (A/2)^2 = B^2 = C^2. We know the length of A (and thus A/2) and B which are 1/26 in inches which is .0385 and A/2 = .0192.

then using the formulation, H^2 + (A/2)^2 = B^2 we solve for H^2:

H^2 = B^2 - (A/2)^2 and pluggin in the numbers above we get:

H^2 = .0385^2 - .0192^2 = .00148225 - .0003705 = .00111169

simply take the square roots of each side: H = .03334

This is the mathematically correct solution, that is, it is as close enough "approximation" that we need in machining this. However, there are other factors as we machinists know, for instance, machining is never as clean as a mathematical formula. There are always factors of machinability that math does not address. If both the inside and outside diameters are "perfect" the items are actually not likely to fit, as we all know a bit of clearance is needed, Usually a couple thous minimum and the tops of the sharp threads reduced a small percentage. for instance, say one has an OD of 1" and 1/8th" threads (1-8tpi). It might be easiest to cut the 1" down to begin with by a few thou before you even begin your threads. so with IDs.

Am I wrong? Let me know. Also, using trig the very same number will come up for the above calculations. I hope this was not too long winded and I hope even more so that it is clear and easily understandable.