Peter,

When Astrometrica platesolves an image in a particular photometric band
it compares the catalogue magnitude of each star with the count value
for the same star extracted from the image. It then fits a straight line
to this and reports the zero point. In Astrometrica's definition the ZP
is the magnitude that corresponds to a count of 1 unit for that exposure
so the derived magnitude of any given star is then:

m = ZP - 2.5 log10(C)

where C is the count value. If the atmosphere was perfectly transparent
the ZP would be constant for any given optical system, filter, camera,
catalogue and exposure duration. Variations in the ZP are effectively a
measure of the amount of light lost in the path through the atmosphere.

With my setup I know that, on very transparent nights with the object
near the zenith, I get a ZP of around 24.75 for a 60s exposure with my
C11 and ASI6200MM camera when the camera gain is set to 100. I can
therefore judge how good the night is by looking at the ZP value. If the
atmosphere is not very transparent the ZP drops. In the example attached
it is 24.56. The comet was at an altitude of 42 deg when this image was
taken (a zenith distance of 48 deg) so some of the loss (around 0.07
mag) will be due to extinction but the rest will be due to the hazy
atmosphere.

Note that the ZP is not the same as the sky brightness. The ZP depends
only on the transparency, i.e. how much light is lost passing through
the atmosphere. Light pollution and sky glow raises the background level
in an image but this does not affect the ZP.

Nick.

On 02/01/2024 19:16, Peter Carson wrote:
> Hi Nick,
>
> Using Astrometrica’s zero point is an excellent way of comparing transparency
> from one set of observations to another.
>
> Could you elaborate a little for others in this group who might not be familiar
> with what a zero point is?
>
> Peter