The pitch distance of a propeller is given by:

$$p = 2\pi r \tan{\beta}$$

where $\beta$ is the pitch angle and $r$ is the radius.

I cannot find how this expression was derived.

Could you please explain how can we derive this expression?



1 Answer 1


I assume $\beta$ is the pitch angle at radius $r$ from the axis of the propeller, and $r$ is most likely at the tip.

Picture that propeller turning inside a horizontal cylindrical tube made of a material like wax or jello, so that as the propeller turns it has to move forward down the tube.

(Pardon my attempt at drawing this. I'm trying to show the tip of the propeller blade tracing a helical path down the tube.)

enter image description here

As the tip of the propeller blade travels a small vertical distance $d$ in the material of the tube, it has to also move horizontally a small distance, because of that angle $\beta$. The distance it has to move is $\tan\beta$ times $d$. Is that clear? Picture the little triangle.

OK, what if $d$ is not small? What if it goes completely around the circumference of the tube? Then won't the propeller have to move forward $\tan\beta$ times the circumference of the tube? OK, what's the circumference of the tube? Diameter times $\pi$. What's the diameter? Twice the radius. Multiply all that together, and you get $2\pi r \tan\beta$.

Of course, this pitch distance may not be the same distance the propeller actually travels forward on an actual airplane at a particular time, because it's not flying through jello. In fact, when the plane is stationary on the ground with the propellers turning at takeoff speed, the forward distance is zero, but there's lots of thrust, hundreds of pounds.

  • $\begingroup$ Sorry, I tried to imagine that but unfortunately, I couldn't understand, could you please provide some figures? $\endgroup$ Commented Mar 23, 2019 at 17:39
  • $\begingroup$ What do you mean by vertical distance $\endgroup$ Commented Mar 23, 2019 at 17:40
  • $\begingroup$ @IamNotaMathematician: I tried to draw it. $\endgroup$ Commented Mar 23, 2019 at 19:37
  • $\begingroup$ Using differentials that means:the elemental vertical distance $dh =r\mathrm d\theta$ and pitch elemental distance, $dx = \mathrm dh\cdot \tan{\beta}$ by integration: $p = \int_0^{2\pi}\mathrm dx = r\cdot \tan{\beta}\int_{0}^{2\pi}\mathrm d\theta = 2\pi r\cdot\tan{\beta}$ $\endgroup$ Commented Mar 23, 2019 at 22:36
  • 1
    $\begingroup$ @IamNotaMathematician: For a non-mathematician you got it pretty well! $\endgroup$ Commented Mar 24, 2019 at 3:16

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