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I want to find 3D equation of a falling droplet that I have considered it as an ellipsoid. I put two cameras, one in xy plane and another in zy plane to capture two projected views of the droplet and then analyze the frames by image processing to find the related equations for each ellipse and finally the equation for ellipsoid using Finding equation of an ellipsoid. I have only two cameras (i.e. two projected views), so I need one more projected view or another equation(s) to be added to my set of equations.

To test the method, I chose a hypothetical ellipsoid (so I know all 6 coefficients of it) $$3x^2+3y^2+2z^2+2zy+2xz+2xy=1 \ \tag{1}$$ Then extract the corresponding equations of two projected ellipses. So, I have 6 eqs. and 6 unknowns. When I solved this set of eqs, it gave me 3 different answers (two of them are ellipsoids, while I need just one unique ellipsoid equation) Therefore, in order not to need the 3rd camera, I supposed that my ellipsoid is a spheroid: $$\alpha_1=\alpha_2 \ \tag{2}$$ and add this as my 7th equation: $$ \begin{cases} 2/3= \beta_1 - \frac{\beta_2\beta_3}{\alpha_1}\\ 8/3= \alpha_2 - \frac{\beta_3^2}{\alpha_1}\\ 5/3= \alpha_3 - \frac{\beta_2^2}{\alpha_1}\\ 5/2= \alpha_1 - \frac{\beta_2^2}{\alpha_3}\\ 5/2= \alpha_2 - \frac{\beta_1^2}{\alpha_3}\\ 1/2= \beta_3 - \frac{\beta_1\beta_2}{\alpha_3}\\ \alpha_1=\alpha_2 \end{cases} \ \tag{3} $$ I anticipated that adding another equation compensates the dependency we have for two of those 6 equations. But, surprisingly I understood that even when I add the 7th equ, the equ. set still returns two ellipsoids: $$3x^2+3y^2+2z^2+2zy+2xz+2xy=1$$ $$2.7x^2+2.7y^2+1.8z^2+1.2zy-1.2xz+0.6xy=1 \ \tag{4}$$ These two ellipsoids have the same projections both on XY and ZY planes.

I am enthusiast to know your opinion that for finding a unique ellipsoid by only two projected views, whether it is possible to get any extra required piece of information from somewhere else rather than from adding the 3rd camera. Do you have any idea for doing that?

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vorujak
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  • Do you know the three semiaxis lengths of those last two ellipsoids? – David K Sep 18 '14 at 21:35
  • @DavidK Thanks for comment. No, I do not have this information. But would you please let me know how the axes length could help me find a unique ellipsoid? – vorujak Sep 18 '14 at 23:16
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    Based on the graph, I'm wondering if the pair consist of one oblate spheroid and one prolate spheroid. I'd expect a falling droplet to be oblate. Actually I'd expect that among all possible ellipsoids, the one most closely resembling the droplet would be an oblate spheroid whose short axis is vertical, but you seem to be getting a different shape. – David K Sep 19 '14 at 01:17
  • @DavidK thanks for your comment. If we suppose that the original ellipsoid should be always oblate (or to say accurately, almost oblate), how I can understand whether the other one is 'always' prolate or at least far enough from oblate state that will be recognizable from the original one? (e.g. is it possible through the angle between major axis of the red ellipsoid and major axis of the blue one? or ...) – vorujak Sep 19 '14 at 02:37

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