I need to know the length of one complete revolution on a helix if the helical angle = 21° and the diameter is 20 mm. I'm guessing around 220 mm?

There are arc-length integrals to compute the length of any arbitrary curve in 3 dimensions, but your problem is easier due to the coil's symmetry: Just unroll it.

This transforms each complete turn of the coil into a 21° right triangle with an adjacent leg L equal to the circumference of the enclosed cylinder. The problem is to find the length H of the hypotenuse.

From basic trig we have L/H = cosine(21°) = 0.93358+
With a diameter of 20 mm we have L = 20(pi) = 62.83+ mm.
So H = 62.83/0.99358 = 67.30 mm.

BTW, the length of the triangle's opposite side is H*sin(21°) = 24.1+ mm. This is the coil's pitch -- the distance between turns.

Well if you're saying that the length of one turn is 67.3 then something is wrong. My gear is only 35 mm long and one tooth doesn't even make it ¹/₄ of the way around. Are your functions correct? (The helical angle is 21° to the helix axis). Just looking at the gear with a steel rule I estimate that one tooth makes a ¹/₄ turn in approx 55 mm.

Oh, helical gears. Sorry, Kendor, I took 21° to be the angle w/ respect to a perpendicular to the axis. So in the above analysis just change 21° to 69° (or change cosine to sine).

Sine(21°) = cosine(69°) = 0.3584.
So the answer is 20(pi)/.3584 = 175.33 mm. per turn. If that still looks wrong then I guess I'll need a diagram or other clarification.

That's probably it Professor, thanks. I'll let you know if it works for me.