NavList:

   

Reply

Marcel is correct: the function I was discussing was indeed, as he terms
it, the "conversion function", of pixels, measured from the array centre,
as a function of offset angle, in arc-minutes, measured from the optical
centre-line. That function is the starting-point of the dicsussion, in my
recent post.

It seems that we have been at cross-purposes, because Marcel's discussion
is mainly in terms of the SLOPE of that line, which he has measured at
different radii from the array centre, using a Sun-image as a yardstick.
One is simply the differential of the other. Or it would be, if that slope
ratio had been plotted the other way up, and if it had been plotted against
angle rather than pixel-count. If, instead of plotting moa / Px,  pixels
per arc-minute had been plotted on his "cal fig" page, that would have
become a parabola which was U-shaped rather than dome-shaped, with its
minimum at the centre-line. Such a function is exactly what's expected as
the differential of the conversion-function model I suggested, in which
Px = K1 A + K3 A^3. The slope of that plots as the parabola K1 + 3 K3 A^2,
a simple quadratic function of A. The minimum slope of the conversion
function occurs at the optical centre-line, and increases on either side,
just as expected. So, there's nothing wrong with fitting such a quadratic
to the slope function, and I wasn't objecting to that.

I do have some reservations, however, about Marcel's procedure for
calibration, using Sun diameter as the yardstick, and reckon that Greg's
procedure could be much more precise. But that's for another time.

What I was objecting to as inherently wrong and unphysical was to take the
conversion function itself as a quadratic, which would have resulted in a
linear variation of slope. This is the way that Greg has been trying to
force-fit his data, to the slope changing linearly.

If, instead, the conversion function had been modelled as Px = K tan A,
which is what one would get with a perfectly rectilinear lens, then the
slope of that would plot as K / (cos A)^2 , which would be a very similar
shape to the parabola, especially at small angles.

I think I can follow all Marcel's steps, in obtaining that parabolic fit,
and then integrating it up, to get a relation between angle-span and
pixel-span when the image is centred. But when arriving at the three
possible formulae on his "conversion" page, and his words under "conversion
function" in his "integral" page, I get confused. If he is trying to obtain
such a fit for the conversion function itself, then as explained in my
recent posting, the quadratic version is unphysical. And if this is
supposed to be a fit to the curve of slope, then only the quadratic version
is reasonable. These three models can't all be alternatives for doing the
same job, whatever that job may be. Would Marcel explain a bit further,
please?

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.
----- Original Message -----
From: "Marcel Tschudin" 
To: 
Sent: Monday, July 05, 2010 5:43 PM
Subject: [NavList] Re: Camera sextant? was: Re: On The Water Trial of
Digital Camera CN


| George,
|
| After some additional thoughts I do now understand that you referred
| here to the conversion function:
|
| > 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.
|
| You are right that theoretically this function has to go through the
| origin (0,0). However, a polynomial like
| a Px^3 + b Px
| shows only a slightly better correlation than just a linear regression
| with the "wrong" constant off-set. In order to get a real improvement
| one would have to use even higher terms. But how practical would then
| such function be for being tagged on the lens cover? The best fits
| have been obtained with the arc-tan-function. In order to obtain those
| parameters one has to use the Solver tool. For those who don't have
| this tool or are not familiar how to use it the next best option
| compared to the linear regression was thought to be the quadratic
| equation with the "wrong" constant which can be shown "automatically",
| i.e. without using Solver.
|
| Marcel
|
|
|

   

Reply