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Greg's interpretation of his scale-factor as varying linearly is an 
illusion, caused by the restricted range of his observations, which do not 
include the important central portion of his display.

I have taken the liberty of doing a shrink, cut, and paste job, in the two 
attachments, of his linked files

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in such a way that I can extrapolate them down to small angles, around the 
central region of the array. And then I have stuck my neck out, in daring 
to predict the sort of shape that Greg is likely to find if he observed 
down to those smaller angles. The general shape, similar to a 
parabola-on-its side, is an inevitable consequence of the axial symmetry. 
It corresponds closely to one side of the plot of "cal fig" in Marcel's 
spreadsheet, but placed on its side because of the different way up it's 
been plotted.

It's clear that the observations made with the 50mm Pentax are very 
compatible with such a shape. Those made with the 100mm Nikon are not; they 
would be, except for one rogue point, at around 2700 pixels span, which 
doesn't show up clearly on my attachment, but can be seen better in his 
linked file. I would be a bit suspicious about that point, and suggest that 
Greg might give it a closer look. There are no error-bars shown on that 
plot, and it might be helpful if they were, on account of its 
highly-expanded scale. It would call for no more that a one-in-2000 change 
in the scale-factor at that point to bring it exactly on to my predicted 
curve. Is Greg confident of all his observations, down to that precision? 
Is the sinuous nature of the curve he has drawn real, or is it no more than 
the result of experimental scatter?

Greg seems confident that his linear variation of scale factor will serve 
his purposes, and perhaps it will, as long as he measures only within his 
calibrated spans. But if he starts to measure much smaller angles that the 
range he has calibrated for, which cover 7.5º to 12º for the 100 mm lens, 
and 12.5º to 22º for the 50mm lens, he will find large errors in the scale 
factor, if he tries to extrapolate his derived straight-line fit.

We still haven't been told the overall size of the array that Greg is 
dealing with, which would tell us where its central point would be. 
Clearly, it must be similar to the array that Marcel refers to, which I 
take to be 3880 pixels, so that its centre is at 1940. Are they identical?

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: "Greg Rudzinski" 
To: 
Sent: Tuesday, July 06, 2010 12:03 AM
Subject: [NavList] Re: Camera sextant? was: Re: On The Water Trial of 
Digital Camera CN


George,

See the linked graph images of 50mm and 100mm lens moa/px vs. px. I have 
hand drawn highlighted curves over the linear red lines for direct visual 
comparison. The maximum discrepancy for the 50mm lens calculates out to 
0.35' moa at 3500 pixels. The maximim discrepancy for the 100mm lens 
calculates out to 0.24' moa also at 3500 pixels. Chasing tenths of a minute 
of arc by graph or formula won't add much utility for conventional CN 
applications. If a photographer were trying to shoot a day time lunar then 
every tenth of a moa would be important. Photographing a sharp crescent 
Moon 24� from the Sun may not be possible. I'll have to give it a try.

Greg Rudzinski
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