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I'm working on reconstucting movement of 3D objects based on 2D orthogonal projections of that object.

Till now I'm working with two orthogonal projections of the object, one on the z-x plane and one on the z-y plane.

(I want to neglect the projection to x-y plane. That information is not important to me. Should that be a problem then I can easily compute the 2x3 matrix for x-y, too and add that information).

By using two consecutive frames of those projections I can compute a 2x3 transformation matrices using OpenCV's estimateRigidTransform() function. The 2x3 transformation matrices include Rotation, Scaling and Translation:

$$T_{z,x} = \left[ \begin{array}{ccc} \cos(\theta)s & -\sin(\theta)s & tz \\ \sin(\theta)s & \cos(\theta)s & tx\end{array} \right]$$

$$T_{z,y} = \left[ \begin{array}{ccc} \cos(\theta)s & -\sin(\theta)s & tz \\ \sin(\theta)s & \cos(\theta)s & ty\end{array} \right]$$

(Here you can find more detailed information about what exactly estimateRigidTransform() does.)

Here is a sample visualization for clarification:

So after computing this for based on the two 2D projections I want to reconstruct a 3D transformation matrix out of this information the manipulate the 3D object. My aim is to undo the transformation that happend to the 3D object between Frame 1 and Frame 2. Therefore I want to apply the 3D transfrmation matrix to the object so that it is in the same spatial position it was at when Frame 1 was taken.

Sample 3D visualization:

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I understand that a 3D transformation matrix is build up by the following logic:

$$ S = \begin{bmatrix} S_{x}& 0& 0& 0\\ 0& S_{y}& 0& 0\\ 0& 0& S_{z}& 0\\ 0& 0& 0& 1 \end{bmatrix} $$

$$ T = \begin{bmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ t_{x}& t_{y}& t_{z}& 1\\ \end{bmatrix} $$

$$R_{x}(\theta) = \begin{bmatrix} 1& 0& 0& 0\\ 0& \cos\theta & −\sin\theta& 0\\ 0& \sin\theta & \cos\theta& 0\\ 0& 0& 0& 1\\ \end{bmatrix}$$

$$R_{y}(\theta) = \begin{bmatrix} \cos\theta& 0& \sin\theta& 0\\ 0& 1& 0& 0\\ −\sin\theta& 0& \cos\theta& 0\\ 0& 0& 0& 1\\ \end{bmatrix}$$

$$R_{z}(\theta) =\begin{bmatrix} \cos\theta & −\sin\theta & 0& 0\\ \sin\theta & \cos\theta & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1 \end{bmatrix}$$

(I do not want to include shearing.)

My question is how to use the information held by the two 2D 2x3 transformation matrices to construct a 3D transformation matrix that in total transforms the 3D object in the same way as the 2D matrices would if I apply them to their corresponding planes?

EDIT:

As suggested by amd I tried to find more constrains to reduce the number of free parameters.

I analysed the movement of the object and I found out that the 3 dominant parameters are:

  • Translation in z-direction
  • Rotation around the x-axis (Pitch)
  • Rotation around the y-axis (Yaw)

What can be neglected is:

  • Translation in negative z-direction
  • Rotation around the z-axis (Roll)

I hope that helps.

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  • $\begingroup$ On the face of it, a complete reconstruction of the 3-D transformation doesn’t seem possible. The transformations you’re considering have 9 degrees of freedom, but the two estimated transformations of the projections only supply 8 parameters. Do you have any constraints on the possible transformations. $\endgroup$ – amd Oct 11 '16 at 19:27
  • $\begingroup$ Also, what’s the nature of the two projections? Orthogonal vs. perspective, for instance, makes a big difference. $\endgroup$ – amd Oct 11 '16 at 19:47
  • $\begingroup$ The projections are orthogonal. I will think about the constrains to see if we can reduce the degrees of freedom. $\endgroup$ – Kevin Katzke Oct 11 '16 at 20:33
  • $\begingroup$ If you can see the same object feature in both views, you could potentially make use of that to correlate them. $\endgroup$ – amd Oct 12 '16 at 17:52
  • $\begingroup$ @amd I edited my question and provided more information about possible constrains to reduce the number of free parameters. $\endgroup$ – Kevin Katzke Oct 13 '16 at 12:10
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[Not an answer, but too long for comments.]

I wonder if you might be making the problem harder by “simplifying” it. Here’s what a few simple rigid motions look like when projected onto the $y$-$z$ plane:

  • Rotation about the $z$-axis: $[x\sin\gamma+y\cos\gamma,z]$, so equivalent to the planar transformation $$\begin{bmatrix}y&z&1\end{bmatrix}\begin{bmatrix}\cos\gamma&0\\0&1\\x\sin\gamma&0\end{bmatrix}$$ which looks like scaling+translation.
  • Rotation about the $y$-axis: $[y,z\cos\beta-x\sin\beta,1]$, equivalent to $$\begin{bmatrix}y&z&1\end{bmatrix}\begin{bmatrix}1&0\\0&\cos\beta\\0&-x\sin\beta\end{bmatrix}$$ which also looks like scaling+translation
  • Rotation about $y$, then $z$: $[y\cos\gamma+z\sin\beta\sin\gamma+x\cos\beta\sin\gamma,z\cos\beta-x\sin\beta]$, or $$\begin{bmatrix}y&z&1\end{bmatrix}\begin{bmatrix} \cos\gamma & 0 \\ \sin\beta\sin\gamma & \cos\beta \\ x\cos\beta\sin\gamma & -x\sin\beta \end{bmatrix}$$ which is more like a shear+translation. Trying to approximate this by a shearless affine transformation might not work so well.

Of concern, too, is the “free” factor of $x$ in the translation part of all of these transformations. This doesn’t really correspond to anything in the image plane and makes the resulting transformation non-affine. It’s reminiscent of the way that $z$ appears in the derivation of the perspective projection, so I wonder if you might get better results by computing a planar perspective transform for each view and trying to reconstruct the 3-D transformation from those. On the other hand, if the frame-to-frame difference is small, affine approximations shouldn’t be too bad.

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