NavList:

   

Reply

George Huxtable writes:

> I, for one, would very much like to see his photos.

I've put them on my web site at http://www.pmonta.com/lunars
including the original, full-resolution camera images in Canon RAW
and JPEG formats and a few derived images showing detected stars.

> First, we have to be aware that, if stars need a 1/5 sec exposure to show
> them up, there's no hope of taking such a picture from at sea, with the
> inevitable motion underfoot.

Yes, this is a problem.

> It would call for an improvement in imaging light-sensitivity by
> several orders of magnitude to make short-enough exposures practicable at
> sea. However, in time, that may come.

It strikes me that the L3CCD (low-light-level CCD) might help
here---they have on-chip image intensifiers---but they're likely to
be pretty expensive for the near future.  There are also the
classical night-vision electron-multiplier devices, but they're also
expensive and probably have resolution and distortion problems.  For
now, on a practical basis, we're stuck with ordinary CCD or CMOS
sensors (which are getting better all the time, of course).

> However, it's always possible to hit a bull'e eye with your first dart.

Quite right; I'll take some more images once I have clear skies
again.

> ... taken with different Moon phases

This is an important point---this image has a nearly-full Moon, and
I imagine the crescent moon will be a little tougher to get a good
estimate from.  Instead of 180 degrees of limb, a crescent might
have only 120 degrees of usable, high-quality limb unaffected by the
rest of the lunar image.

> And, to be absolutely fair, the
> Moon's position on the array should be deduced before its astronomical
> position is known, in a proper blind-test (not that I'm suggesting that
> Peter's position estimates are influenced by his expected result).

Certainly this would be an advantage of a totally automated
procedure.  In the case of this image, yes, I did look up the known
position last, after everything else was done.  Still, I'd like to
remove the subjective step of estimating the pixel coordinates of
the lunar limb.

> Perhaps
> he will consider passing full-resolution images to this list for others to
> have a go at, with details of location and precise time.

The JPEG images linked to above have the time of exposure in UTC in
their EXIF headers, but I'll put that information also in plain text
on the web page, as well as the location (my camera doesn't have GPS,
so it can't include that in the EXIF information).

> Peter wrote- " astrometry.net can take any image and find out where it's
> pointed in the sky with no prior information whatsoever.  Quite amazing."
> Indeed, from that description, it seems powerful, and clever. But how many
> stars need to appear in shot for it to do so? It has to determine many
> parameters: altitude, azimuth, orientation, angular scale, distortion, and
> arrive at an unambiguous solution which is unique to that bit of the sky.
> How many bright stars does it need to identify, to do all that?

My rough guess is that it needs about 10 stars spread over the
image.  More is always better, of course.  It might be possible to
constrain the search using the known image scale (and maybe known
distortion too), but I didn't do that: instead I just ran it in
"totally blind" mode.  Apparently the program is biased toward very
few false positives: it might return "can't match this image"
sometimes, but very seldom returns a position that's a false match.
It uses a "quad" of four stars for the match, then any additional
detected stars for confirmation.  Their papers are at
http://astrometry.net/biblio.html and a picture gallery at
http://astrometry.net/gallery.html.  The software is free and
open-source.

> And what's the angular size of the patch of sky, in the photo, from which it
> has to do so?. My guess would be about 5 degrees across, but, not being a
> photographic expert, I'm unsure what a 85mm lens implies when used with a
> CCD array.

My camera has a CMOS chip that's somewhat smaller than the 35mm film
frame; the field size with this lens turns out to be about 15.1
degrees by 10.1 degrees.

> If I understand it right, that patch has to contain sufficient
> bright-enough stars for the program to do its thing. There may be parts of
> the sky that are easily rich enough in stars for that to happen, but again,
> there may be many others around the ecliptic, in which bright stars are
> sparse. It depends on the brightness-threshold that is detectable; there's
> no shortage of faint stars! Peter's photos may answer the question.

Right.  It should be possible to simulate the worst case for star
coverage with a given field of view.

This is certainly one clear break from lunar-distance tradition:
instead of very bright sources far from the moon, use dimmer sources
close to the moon.  For that matter, maybe a wider-angle lens would
be a better tradeoff overall, despite the loss in angular resolution.
I used the 85mm lens only because it's the one I have.

> A question about overexposure. How sure can Peter be that his Moon image
> isn't overexposed, which, if it was, could significantly affect its edge? If
> he can make out the Moon's features, its "seas", that answers the question.

It still is considerably overexposed, even on the short exposure.
Lunar features are completely washed out and saturated except very
close to the edge which is not the limb.  Still, the limb is sharp,
and any effect on the limb position should be symmetrical, so the
estimate of the center would be relatively unaffected.  For example,
if overexposure increases the effective size of a full-moon image
from 90 pixels diameter to 94 pixels diameter, that should have
little effect on the estimated center location (certainly much less
than 4 pixels).  Crescent moons should still be okay as well
(okay in the sense of "unbiased estimate of the center under
varying degrees of overexposure") but the lunar diameter in pixels
will have to be carefully estimated from the image.

Canon, if you're listening:  would it have killed you to provide
more than two stops of exposure bracketing?  :-)

> Can he estimate the Moon's diameter from its pixels, and does that result in
> the correct answer? If it is less than full Moon, he has only one limb to
> work from in estimating its centre. How does he go about that? Does he use
> its tabulated semidiameter?

Well, this is the part that could use some improvement.  I did a
cheesy eyeball estimate of the coordinates of the right edge and
bottom edge to the nearest pixel, then estimated the lunar diameter
in pixels using the 180-degree chord of the limb.  (Multiplied by the
image scale it came out to about 32 arcminutes diameter, but I
didn't check that against the tabulated value.)  Finally, subtracting
half the diameter from the edge locations gives the center.

I'm sure this could be improved.  What is really wanted is a
least-squares estimate of the center using all the information
available in the image around the usable part of the limb.

Cheers,
Peter Monta

-------------------------------------------
[Sent from archive by: pmonta-AT-gmail.com]

--
NavList message boards: www.fer3.com/arc
Or post by email to: NavList@fer3.com
To , email NavList+@fer3.com

   

Reply