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I capture each of the projected views of a droplet through a high speed camera (one in xy plane and one in zy) and then analyze the frames by image processing to find the related equations for each ellipse and finally the equation for ellipsoid. I have only two cameras (i.e. two projected views), while based on the answer, having two projected views is not enough to find the ellipsoid equation and I will need one more equation. Is it possible to get this extra piece of information from somewhere else? (rather than from adding the 3rd camera or supposing it as a spheroid). Do you have any other idea for doing that?

vorujak
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    One thing to point out is that you may be able to get away with only two projections if you assume that two of the ellipsoid axes have the same length. This means modeling each droplet as a spheroid rather than a true ellipsoid. Whether this is sufficient for your purposes depends crucially on the setup (and raises a different question, namely how one validates this assumption...) – Semiclassical Aug 22 '14 at 21:11
  • Note that there's still an interesting question here, which is both practical and mathematical: If one models the droplet as a spheroid, how does one confirm the validity of that assumption? I think it would be effective as an average description of the droplet since its shape would be determined by gravity, but I would imagine there would be also be anisotropic fluctuations in its dimensions. Hmm... – Semiclassical Aug 22 '14 at 21:26
  • @Semiclassical Thank you for your reply. You are right! The droplet is not even an exact ellipsoid, not to mention a sphere. But, we want to model it and ellipsoid hypothesis will give us an appropriate level of accuracy. Regarding my question, apparently, there is no other way rather than using my last shot! (simplification) In that case, I think I should add just one of the following equations: $$\alpha_1 = \alpha_2$$ or $$\alpha_1 = \alpha_3$$ or $$\alpha_2 = \alpha_3$$ I think nothing should be applied for $\beta$s. Am I right? – vorujak Aug 22 '14 at 21:41
  • I would think so, yes. Though I think you'll only need to consider one of the three since $123$ are just labels for the principal axes. I'd have to work through the details to be completely sure of that, though. – Semiclassical Aug 22 '14 at 21:42
  • @Semiclassical I chose a hypothetical ellipsoid (so I know all 6 coefficients of my ellipsoid) and then extract the corresponding equations of two projected ellipses. Then, based on the answer provided here "http://math.stackexchange.com/questions/897720/finding-equation-of-an-ellipsoid" I have 6 eqs. and 6 unknowns. When I solved this set of eqs, I found 3 different answers for it and two of them are ellipsoids. Then, I supposed that alpha1=alpha2 and add this eq. as the 7th, but this time the equ. set returned no answers...(please continue in next comment) – vorujak Aug 27 '14 at 19:19
  • I tried it for alpha2=alpha3 and alpha1=alpha3 as well and I received the similar results (no answer) each time. Based on our previous discussion above, I anticipated that adding another equation will compensate the dependency we have for two of those 6 equations. But, apparently it is not true. I am enthusiast to know your opinion regarding the matter. (I have written whole the procedure in detail and plotted the graphs. If you prefer, I can email them for you as well). I appreciate your response in advance. – vorujak Aug 27 '14 at 19:32

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