On Thursday, February 2, 2017 at 11:33:07 AM UTC+3, wrote:
I worked some math out on page 15 of this PDF. The section is called drag..

http://spekje.snt.utwente.nl/~roeles/maccready.pdf

Ok, good. What exactly do you wish to point to there?

I see that in general there are a lot of approximations, assuming small angles etc. I didn't check if that always looks like a good simplification, but this did jump out at me:

"Using this formula we can calculate both CD0 and e if we know some data
about the best L/D point of the polar. Since at this point the induced drag
should equal the parasitic drag"

That is certainly wrong.

It's true that in practice where you are looking for a minimum of two different ways to lose, the minimum is likely to be at a point where they are roughly equal, but it's only very rough. The two things could easily be different by, say, a factor or two, or more.

What *is* true at the minimum is that the *derivatives* of the two things are equal in magnitude, and opposite in sign.

Take, for example, the (positive) minimum of 1/x + x^2 (which is something similar in form to induced plus form drag). They are equal when x = 1, and both halves are 1 (so the total is 2).

But the minimum is at x = 1/cubeRoot(2) ~= 0.793701. At this point 1/x is 1.26 and x^2 is 0.63. So in fact the "induced drag" is twice the "form drag" at the minimum.

As you'll know, the derivative of 1/x is -1/x^2, while the derivative of x^2 is 2x. So at x = 0.793701 the derivatives are -1.5874 and +1.5874 (cube root of 4) -- equal and opposite.

At the minimum of a sum, the *slopes* are equal (and opposite), not the *values*.