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Hi,
I have to resample images taken from camera, whose target is a spherical object, onto a regular grid of 2 spherical coordinates: the polar and azimutal angles (θ, Φ). For best accuracy, I need to be aware of, and visualise, the "footprints" of the small angle differences onto the original images, so I tried to represent the ellipses of transformation at 3 different positions, which in the output space (θ, Φ) are at p1=(0, 0), p2=(0, 60) and p3=(30, 0). I tried using a linearization with the Jacobian matrix but I am not sure I did it right, or if that's enough.
First, consider my screenshot:
It shows (left) the spherical grid with a unitsphere, with the three points p1, p2, p3 that I consider. On the righthand side, we have my ellipses of transformation, which I interpret as the footprint onto the input space (the original image coordinates ) from small angle differences in the input space. While I expected the shape of the blue and red ellipse as they are here, I do not understand why the green one is not the 90degreesrotated red circle. Instead it is just a smaller circle. The original image is taken from a viewpoint in the Xaxis, using the present coordinate space and X, Y,Z axis of the lefthand side figure.
I used this formula for the Jacobian of the transformation from Spherical to Cartesian coordinates for the unitsphere:
[itex] \textbf{J} = \begin{pmatrix}
\sin(\theta)\cos(\phi) & \cos(\theta)\cos(\phi) & \sin(\theta)\sin(\phi) \\
\sin(\theta)\sin(\phi) & \cos(\theta)\sin(\phi) & \sin(\theta)\cos(\phi) \\
\cos(\theta) & \sin(\theta) & 0
\end{pmatrix} [/itex]
associated with the following transformation from spherical to cartesian (with a unit radius):
[itex]x = \sin(\theta) \cos(\phi) \\
y = \sin(\theta) \sin(\phi) \\
z = \cos(\theta) [/itex]
Then the ellipses of transformation are obtained by multiplying J by a vector of coordinate (dθ, dΦ) that satisfies the equation of a circle of a radius smaller than 1 (0.2 in this example) so I can see the footprints in all direction.
Am I missing something obvious ?
Thanks
I have to resample images taken from camera, whose target is a spherical object, onto a regular grid of 2 spherical coordinates: the polar and azimutal angles (θ, Φ). For best accuracy, I need to be aware of, and visualise, the "footprints" of the small angle differences onto the original images, so I tried to represent the ellipses of transformation at 3 different positions, which in the output space (θ, Φ) are at p1=(0, 0), p2=(0, 60) and p3=(30, 0). I tried using a linearization with the Jacobian matrix but I am not sure I did it right, or if that's enough.
First, consider my screenshot:
It shows (left) the spherical grid with a unitsphere, with the three points p1, p2, p3 that I consider. On the righthand side, we have my ellipses of transformation, which I interpret as the footprint onto the input space (the original image coordinates ) from small angle differences in the input space. While I expected the shape of the blue and red ellipse as they are here, I do not understand why the green one is not the 90degreesrotated red circle. Instead it is just a smaller circle. The original image is taken from a viewpoint in the Xaxis, using the present coordinate space and X, Y,Z axis of the lefthand side figure.
I used this formula for the Jacobian of the transformation from Spherical to Cartesian coordinates for the unitsphere:
[itex] \textbf{J} = \begin{pmatrix}
\sin(\theta)\cos(\phi) & \cos(\theta)\cos(\phi) & \sin(\theta)\sin(\phi) \\
\sin(\theta)\sin(\phi) & \cos(\theta)\sin(\phi) & \sin(\theta)\cos(\phi) \\
\cos(\theta) & \sin(\theta) & 0
\end{pmatrix} [/itex]
associated with the following transformation from spherical to cartesian (with a unit radius):
[itex]x = \sin(\theta) \cos(\phi) \\
y = \sin(\theta) \sin(\phi) \\
z = \cos(\theta) [/itex]
Then the ellipses of transformation are obtained by multiplying J by a vector of coordinate (dθ, dΦ) that satisfies the equation of a circle of a radius smaller than 1 (0.2 in this example) so I can see the footprints in all direction.
Am I missing something obvious ?
Thanks
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