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Greg Rudzinski wrote-

"It is my understanding that the modern digital 50mm SLR camera produces an 
image equivalent to a 75 or 80mm film version."

He added- "Also the diagonal is probably not a good way to measure with 
sensors  arrayed in a rectangular grid.".

Well, assuming radial symmetry of the distortion about the frame centre, 
the measurement could be made just as well across a central axis, 
lengthways or heightways, as across a diagonal. My intention in checking on 
the diagonal was to explore distortion over the whole area, right into the 
corners, where the distortions are likely to be greatest.

and- "No clamping of the camera is needed with shutter speeds over 1/500 of 
a second. At these shutter speeds camera shake is not a factor and will 
allow use on a small craft. Sun limbs are surprisingly sharp even with a 
hand held telephoto. See attached 200mm example."

That Sun disc in Greg's attachment was very sharp, showing what a steady 
hand he has. But that wasn't the purpose of the clamping that I proposed. 
The intention was to fix the camera over the whole sequence of 
observations, just as in the Greenland example we've been looking at.

Presumably Greg's picture was taken from on land. I suggest that observing 
from the unsteady footing of a small craft would be a very different 
matter, even at 1/500 sec.

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

If we take a mid-value of 78 mm, between Greg's suggested equivalent of "75 
or 80mm", that would subtend angles about the centre point of +/13.0º about 
the width of the frame, +/- 8.7º about the height, and +/- 15.5º to the 
corners. This is significantly narrower than the angular width of the 
Greenland photo we have been looking at, so the distortions will be 
considerably less.

The "50mm" graph supplied by Marcel, and attached again here, showed 
altitudes amounting to nearly 22º, equally disposed about the centre, so 
that the horizon must be 11º below the centre and the Sun 11º above it, 
that must have been made with the camera in "portrait" mode. That's what 
Greg was referring to in "rotating the camera 90 degrees".

Marcel provided calibration plots for both 50mm and 100mm lenses, but the 
one of real interest is that of 50mm, which is attached once again.

Those graphs show remarkably careful observation, with very little scatter 
about the fitted line, and don't depart much from simple linearity. But 
depart they do, as shown by the computed line of fit not passing through 
the origin, but showing a small offset.

Two alternative fitted equations are shown, one linear, one second-order, 
with little to choose between them, but it looks to me as though only one 
line-of-fit has been drawn, the second-order fit.

I have used that graph to check out my suggested model, in which radius in 
the array varies as the tan of the subtended angle from the centre point. 
In this case, because each measurement has been equally disposed about that 
centre point (a sensible practice which minimises distortion) the relevant 
angle is half the altitude.

So I have tried a function as follows-

H(pixels) = 18140 Tan (altitude / 2) (in degrees)

or, for those that prefer it the other way,

Altitude in degrees = 2 arc-tan ( pixels / 18140)

And I can report that that function fits exactly on top of the data and the 
fit-line that's already drawn in, so that it's just not possible to see any 
difference.

And if, as is likely but unconfirmed, the array has been made with exactly 
the same spacing along the other axis, that will be the calibration curve 
on that direction also. And, indeed, along a radius in every other 
direction too, if such a radius is measured in pixel-equivalents using 
Pythagoras..

The constant of 18140, which depends on the pixel spacing (which we haven't 
been told) and the focal length, was arbitrarily chosen to get a fit.

Of course, it's a poor test of the correctness of the tangent model, simply 
because it's restricted to such a small  range of angles from the optic 
axis, so the resulting distortion is so low. All we can really say is that 
it doesn't discredit that model by showing any disagreement with 
observation.

However, from that model, I can now confidently predict that at the maximum 
altitude shown. of 22º, the distortion will enhance the observed diameter 
of the Sun, in the vertical direction, by 3%, over what it would have been 
if seen on the optic axis. And enhance its width by 1%. So if Sun diameter 
is being used as a yardstick for calibration, that would have to be taken 
into account.

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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