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This relates to Marcel Tschudin's post of 3 July, in which he shows, in an 
Excel spreadsheet, his fitting of three different functions (linear, 
quadratic, tangent) for Pixel-count versus angular offset from the optical 
centre-line.

Also to Greg Rudzinski's posting of 26 June with his linked plot at
http://www.fer3.com/arc/imgx/IMG_8464.JPG

in which he tries to fit his data with a straight-line, in plotting, the 
ratio of pixel-span to angular-span, as that ratio varies with angular 
span. This corresponds closely with Marcel's trial of a quadratic fit. I 
doubt if either of those procedures will really succeed, because they 
ignore two important aspects of the physical reality of such a camera 
system, caused by its symmetry about the central optical axis, which should 
pass through the mid-point of the array. However, if the optics of the 
system happen to be off-centre, my arguments will fail.

With centred optics, any radial plot, of pixel count from the centre point 
of the array, Px, against offset angle A from the optical centre-line, MUST 
pass through the origin, at (0,0), and MUST be antisymmetric about that 
origin. When the pixel count from the centre is expressed as a function of 
angle Px = f (A), then f (-A ) must equal - f (A), when we consider light 
arriving with opposite offsets from the centre-line.

That implies that any polynomial expansion must have no constant term: 
otherwise it would fail to pass through the origin. And to preserve the 
anti-symmetry, it can have only odd powers of offset angle, and the even 
terms must all be zero.

To sum up so far, then, a general polynomial fit of Px in terms of A, which 
we might originally think of as being-
Px = K0 + K1 A + K2 A^2 + K3 A^3 + K4 A^4 + K5 A^5 + ...

reduces to-

Px = K1 A + K3 A^3 + K5 A^5 ..., all other terms being zero.

We have got this far by considering only axial symmetry, and haven't even 
considered what form any radial distortion will take.

Note that a simple function Px = tan A meets that requirement, because the 
expansion of tan A is
tan A = A + (1/3)A^ 3 +(2/15)A^5  +...  (with A in radians).

However, I'm not claiming that a fit using a tan function will necessarily 
do all that's needed. It would certainly fit the geometrical distortion, 
caused by having to translate from imaging a flat-plane object (for which 
lenses are generally intended) to imaging a spherical object in terms of 
angle. But there can be some additional distortion, due to departure of the 
lens design from the "rectilinear" ideal, which by symmetry must follow a 
similar odd-powers-only law, so the end-product may differ from a simple 
tan A result. However, the terms K0, K2, K4, etc must still be zero.

In the expansion above, K1 could in theory be derived from the exact 
array-pitch and focal length, if both were known, but in general this will 
not be the case, or not to sufficient exactness. So K1 has to be derived 
empirically, from observation, as do the other odd terms.

===================

How is it, then, that both Greg and Marcel appear to be satisfied with a 
quadratic fit to their observations? One reason is that over the range of 
angles being considered, up to +/- 12� from the centre-line, the departure 
from linearity is small. Another is that the observations do not extend to 
the region within +/- 6� of the centre-line, and in particular do not 
include that vital point at the origin, through which any such calibration 
curve must pass. If they did, it would become clear that Greg's 
straight-line fit to the varying slope had become unacceptable. Even as it 
is, the observed points in Greg's plot, in 
http://www.fer3.com/arc/imgx/IMG_8464.JPG , can be seen to be curving away 
from his "best-fit" line.

What I'm suggesting here is that Marcel's attempts to fit observation with 
a quadratic are doomed to fail. Clearly, fits using a tan function are much 
better than a straight-line, as is to be expected, and he should proceed 
along those lines, either adjusting the parameters of the  tan dependence, 
or else try fitting a polynomial using

Px = K1 A + K3 A^3 and (only if necessary) adding in higher, odd powers of 
A.

===================

I have little doubt that there are texts on optical imaging which deal with 
this business more formally, and in greater detail. My arguments, above, 
are based on simple reasoning from first principles, and may or may not 
correspond with the textbooks.

George.

contact George Huxtable, at  george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. 

   

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