0
$\begingroup$

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: $\phi=180^\circ-2\tan^{-1}\left(\frac{d/2}{h}\right)×\frac{180^\circ}{\pi}$

Figure

New contributor
Khalil.h is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$

Your Answer

Khalil.h is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.