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Jim Bell Q'n'a, July 3, 2006, Your questions answered!
centsworth_II
post Jul 7 2006, 02:31 PM
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"The reason that the simple "multiply 0.273 by 1024" equation works for Pancam is simply because there is *no* geometric distortion in the optical system." -- Jim Bell, quoted by DE

I'm impressed. As an innocent bystander to the various technical threads on imaging it has all seemed so complicated to me. Can it be? Finally... a simple truth?
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mcaplinger
post Jul 7 2006, 04:48 PM
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QUOTE (centsworth_II @ Jul 7 2006, 07:31 AM) *
"The reason that the simple "multiply 0.273 by 1024" equation works for Pancam is simply because there is *no* geometric distortion in the optical system." -- Jim Bell, quoted by DE

Well, Jim certainly knows more about Pancam than I do, but it's a little surprising to me. Maybe it's because the lenses are so optically slow. The radial distortion parameters in the CAHVOR model for Pancam are certainly not zero, but that may be an artifact of the way the CAHVOR pinhole model works.


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Disclaimer: This post is based on public information only. Any opinions are my own.
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Guest_AlexBlackwell_*
post Jul 7 2006, 07:29 PM
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QUOTE (djellison @ Jul 7 2006, 01:31 PM) *
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Jim Bell - Determining the total size of the field of view for these other cameras is not as simple as it is for Pancam, however. That's because most wider-field camera systems have appreciable geometric distortion. So you can't just take Navcam's 0.82 mrad and multiply by 1024 and convert to degrees. That will get you close, but not spot on, and the calculation is even worse for the Hazcams, which have an enormous amount of distortion. The details of those cameras' fields of view can be found in Justin's paper--which I hope is posted online somewhere for folks to access, but I am not sure.


From Maki et al. [2003]:

"The Hazard Avoidance Cameras (Hazcams) are shown in Figures 8a and 8b. The Hazcams are an f/15 optical system with a focal length of 5.58 mm. The Hazcam optics are f-theta fish-eye lenses with a 124 124 horizontal/vertical field of view and a 180 diagonal FOV. The angular resolution at the center of the image is 2.1 mrad/pixel."

[...]

"The Navigation Cameras (Navcams, Figures 4a and 4b) are optically identical to the Descent camera: f/12 cameras with a 14.67 mm focal length. Each Navcam camera has a 45 45 field of view (60.7 diagonal), which is roughly equivalent to a 40 mm lens on a 35 mm camera. The angular resolution at the center of the field of view is 0.82 mrad/pixel. The depth of field of the Navcam camera ranges from 0.5 m to infinity, with best focus at 1.0 m."
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fredk
post Jul 19 2006, 01:16 AM
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QUOTE (djellison @ Jul 7 2006, 01:31 PM) *
From JB

The reason that the simple "multiply 0.273 by 1024" equation works for Pancam is simply because there is *no* geometric distortion in the optical system. We tried hard to measure it so we could characterize and correct for it, if needed, but as we wrote in our 2003 JGR paper we couldn't detect *any* distortion down to a residual of 0.01% or so across the field--even in the corners.


Doug, thanks for directing me to the more appropriate thread.

I have to disagree with that figure for distortion of 0.01%. There is a fundamental intrinsic distortion present in any optical system, completely independent of the quality of the optics (indeed it exists for a pinhole camera!). It arises because the goal of imaging is always to represent a section of an imaginary sphere surrounding the camera on a flat image plane. In exactly the same way that you cannot represent a section of the earth without distortion on a flat map, any flat image contains fundamentally irreducible distortion (ie even image processing cannot eliminate it).

I suspect the distortion figure that Bell was quoting referred to the amount of pinhole or barrel distortion. These can be eliminated. Thus if we image a grid of straight lines on a card normal to the line of sight, the lines will be straight on the image. But distortions remain, in particular radial and azimuthal angular image scales will have to differ.

Just as with maps of earth, the irreducible distortions increase with the size of the section of the sphere we're imaging, ie with the field of view. It's straightforward to calculate the size of the irreducible distortion.
Suppose we're imaging a circle, normal to the line of sight, whose radius subtends angle theta(radians) at the camera. Then it's angular circumference will be 2 pi sin(theta). Lets suppose the angular image scale is constant in the radial direction. Then, working from the image, if that image scale gives us theta for the radius of the circle, it must give us 2 pi theta for the circumference, in disagreement with the actual circumference of 2 pi sin(theta).

The relative error is specified by the ratio, R = 2 pi sin(theta) / 2 pi theta = sin(theta)/theta. As expected this ratio approaches unity, ie distortion approaches zero, as fov theta approaches zero. For pancam we have theta = 8 degrees, which gives R = 0.9968. Therefore the radial and azimuthal angular image scales must differ by (1 - R) = 0.32%. (Actually the value will be even larger in the corners.) This irreducible distortion is much larger than the quoted value 0.01%.

This ratio R is important when converting pixel separations to angular separations, since the scale factor used to do this (0.273mrad/pixel) must vary by at least 0.32% across the frame and for different orientations. Eg, when comparing a line through the centre of the fov (radial) vs. a line across one edge (more or less azimuthal).

Of course in the end the effect is still small. 0.32% corresponds to an error of only 3 pixels for a 1024 pixel line! Indeed according to the Bell et al JGR (2003) paper, the fov's of the 4 pancams vary by a comparable amount! (It's not clear if this is just measurement uncertainty).
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