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Frank wrote, about the midnight Suns photograph-

"The spacing is 5.6 and 5.7 degrees between the outer pairs while it's
5.1/5.2 right near the center of the image. Usually, this would imply that
the altitudes would be distorted by a similar amount in the same part of
the field of view. That is, if a measured horizontal angle is 5.7 degrees
when we have reason to believe it should be 5.1, then a measured vertical
angle of 5.7 degrees in the same part of the image should be corrected to
5.1 degrees."

What does "usually" mean, in this context? Why must the scale distortion be
the same in one direction (largely circumferential, about the centre of the
image) as it is radial? Far from it.

In a posting dated 11th June I wrote, about a perfect optical system.which
images a flat-plane object with no distortion at all--

If we put a spot on the film at the centre of symmetry of the optical
system, then the radius of a point from that spot is not proportional to
the angle A, subtended at the lens, but to tan A. And that implies an
expansion, of radial lines, by a factor of 1/cos-squared A. So at the outer
Sun images of that photo, radial lines will be magnified by 22%, compared
with the centre of the picture. Circles around that centre spot must always
correspond to 360-degrees, and to do that, all circumferential lines must
be magnified by a different factor, Tan A /A (where A is in radians), which
at that same radius works out as 7%.

It is, of course, quite possible that there are errors in that analysis,
though nobody has pointed to them yet. The figures given were for an angle,
taken at the edge of the film, subtending an angle of 25 degrees from the
centre of the image.

For a subtended angle of 20 degrees, the enhancement of a Sun disc (or
anything else) in a direction radial from the film centre would be 13%. In
the circumferential direction, it is 4.3%. That distortion is the
consequence of a system which images a flat-plane object without
distortion. That means that it has a transfer function between linear
displacement of a plane object from the central axis of the optics, and
linear displacement of the resulting image from the centre of the film,
that is precisely linear. It does NOT mean that the relation between
ANGULAR displacement from the axis, and position on the film, is linear. It
is not.

So the result is that near the film's outer edge a Sun image will become
elliptical, with its major axis pointing away from the centre of the film.
Spacings between those Sun images, equally-spaced in angle, and largely
radial in direction towards the edge, will be enhanced. However, Sun
altitudes, being largely circumferential about the film's centre, will be
also enhanced, but only by about a third as much.

The effect is nothing to do with large-angle lens distortion. It's a
consequence of an optical system that images plane objects, undistorted..

So Frank's assumption, that angular enhancement is equal in different
directions, is wrong. Unless my whole analysis is wrong. It may be. Can
anyone knock it down?

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: "Frank Reed" 
To: 
Sent: Saturday, June 12, 2010 8:19 PM
Subject: [NavList] Re: NG's "Midnight Fun"


Herbert, you wrote:
"You meant to say 78.5."

Yes. 90-11.5 is 78.5 ...not 87.5. :)

Unfortunately, that wasn't the end of my arithmetic errors while looking at
this image so I'll just start over (for the record, from my "back of the
envelope" notes, it appears that I subtracted 158 from 166 and got
6 --ouch).

I measured the diameter of the central Sun image in pixels. That's about 8
pixels at the scale displayed on the web page. Then I counted pixels from
the Sun's LL to the horizon. I get 66. Given the Sun's diameter of 32', the
altitude of the Sun's LL is about 264' or 4.5 degrees +/- 0.25 degrees (my
estimate of the uncertainty in the scaling based on the Sun's diameter). We
can also count pixels from the right-hand limbs of the two central Sun
images and for that I get an angle of 5.1 degrees which fits well with the
caption's statement that the exposures were about twenty minutes apart. I'm
figuring a net of near zero for dip, refraction and SD so the corrected
altitude of the Sun's center is again just about 4.5 degrees. That minimum
altitude of the Sun below the pole, along with the latitude of 78.5,
implies that the Sun's declination was 16.0 degrees +/- 0.25d. The date
then would be near 15h GMT August 9, 1947 +/- one day. Local midnight at
Refuge Harbor, Greenland would be around 0500 GMT. That rules out Aug. 8 so
the time was either 0500 on August 9 or 0500 on August 10 (Greenwich
dates). Again, I am assuming that this photo was taken in summer, after the
solstice. The same declination would have occurred around May 5. I would
expect more ice around that date, so I am going with the August date.

You asked:
"How do you explain the steadily increasing sun diameter towards the sides
of the picture?"

Clear sky. The images near the center are covered by some clouds or fog
which would make a nice sun filter (given the right kind of clouds). I
interpreted the images away from the center as over-exposed images of the
Sun in clear sky. Don't they look over-exposed?? The images of the Sun
fourth and fifth from the left appear to have been nicely dimmed by the
clouds which are clearly visible in the image (of course, being a multiple
exposure, the clouds could have been photographed at some other time but
that's not likely since they would need to be significantly backlit to show
up in a short exposure).

What about the earlier and later altitudes and azimuths? Here's where you
have to deal with some "plate constants" as they say in astrometry. If we
apply the scaling as above of 4 minutes of arc per pixel, the apparent
azimuths in degrees away from the fifth Sun image
are -21.3, -15.7, -10.3, -5.1, 0, 5.2, 10.5, 15.9, 21.6. The apparent
altitudes of the Sun's center corrected for refraction and dip are 5.3,
4.9, 4.6, 4.4, 4.4, 4.4, 4.6, 5.0, 5.4. Clearly the differences between the
azimuths are greater near the outer limits of the image. The spacing is 5.6
and 5.7 degrees between the outer pairs while it's 5.1/5.2 right near the
center of the image. Usually, this would imply that the altitudes would be
distorted by a similar amount in the same part of the field of view. That
is, if a measured horizontal angle is 5.7 degrees when we have reason to
believe it should be 5.1, then a measured vertical angle of 5.7 degrees in
the same part of the image should be corrected to 5.1 degrees. So if I take
the corrected altitudes and scale them by a factor such that the
differences in azimuths are all the same (e.g. the first altitude is scaled
by 5.1/5.7) then I get this sequence of corrected altitudes: 4.8, 4.65,
4.45, 4.4, 4.4, 4.4, 4.5, 4.65, 4.8. If I calculate the altitudes for these
azimuths, I get 5.1, 4.8, 4.6, 4.4, 4.4, 4.4, 4.6, 4.8, 5.1. That's not too
bad. The calculated altitudes are just about half-way between the original
altitudes and the modified altitudes and within the margin of error in any
case. So the image seems consistent with a simple projection. Note that if
you take the Sun from the center of the image and increase its size by 12%
(the scale factor near the edge) you don't get anything like the images of
the Sun on the left and right. They are over-exposed images rather than
just enlarged by projection.

George, I think, asked about sunspots. At this scale only the very largest
of sunspots would show up. There was a huge one famously earlier in 1947.
Can anyone find a sunspot photo/drawing for August 9, 1947?

I'm attaching a screen capture of the NG image, cropped a bit, in case
anyone wants to play with it and could not figure out how to grab it from
the web page.

-FER


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