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I was reading a research paper and have had difficulty understanding something. How did the author get tan(phi) in these terms?

This is something I've been trying to figure out for a long time now.

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    $\begingroup$ This is a question about Trigonometry, not Aviation, $\endgroup$ – Ralph J Dec 3 '20 at 15:16
  • $\begingroup$ Yeah exactly. But by basic trigonometry this relationship can't be obtained. I figured it has to do something with velocity triangle of propeller wing so posted here! $\endgroup$ – Shivam Sundriyal Dec 3 '20 at 21:05
  • $\begingroup$ V_propeller, V2, and tan(phi) are related as a basic trig problem. What else are you looking for? $\endgroup$ – Ralph J Dec 3 '20 at 21:37

There is an obvious error in this paper. Just look at the final formula, and remember that for small angles $\tan(\phi) \sim \phi$. The formula says that if you increase $\phi$, $v_2$ decreases. Of course, the opposite happens.

By (trigonometric) definition,

$$\tan(\phi) = \frac{v_2}{v_{propeller}}$$


$$v_2 = v_{propeller} \cdot \tan(\phi)$$

The rest follows; the result is similar but you have multiplication instead of the fraction.

P.S. You can get a hint that this is a low quality paper by the fact that they obliviously confuse upper- and lowercase $v$ (on the picture and in the formulae). In math/science, case matters! Sometimes even the script matters.

  • $\begingroup$ Thanks for the answer. I got this from the paper DESIGN AND DEVELOPMENT OF A HYBRID DRONE by British Columbia Institute of technology. They had various professors with PhD's working on it so I thought they wouldn't have made such a basic mistake! Can you look up their report once? $\endgroup$ – Shivam Sundriyal Dec 4 '20 at 10:08

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