I'm trying to calculate the angle Phi in the picture in the case where the droplet is the perfect sphere I have the correct formula but I'm not sure how they found it. and I want to know the formula is if the droplet isn't a perfect sphere (ellipsoid). the formula in the case if droplet is sphere is: $\phi=180^\circ-2\tan^{-1}\left(\frac{d/2}{h}\right)×\frac{180^\circ}{\pi}$


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  • $\begingroup$ Then you don't have enough information. You need to know something like the ratio of the semiaxes. $\endgroup$
    – Andrei
    Sep 14 at 22:49
  • $\begingroup$ @Khalil.h Normally as per new guidelines we do not reply directly. However your question has an error of sign that needs to be pointed out. $\endgroup$
    – Narasimham
    Sep 15 at 0:06

$$\varphi= \frac{\pi}{2}+2\tan^{-1}\left(\frac{h}{d/2}\right)$$

To convert into degrees multiply by $\dfrac{180}{\pi}.$

The factor $2$ in second part arises due to the angle at circle center being double that at top of droplet.

enter image description here


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