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Re: Camera sextant? was: Re: On The Water Trial of Digital Camera CN
From: George Huxtable
Date: 2010 Jul 5, 10:39 +0100

```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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