Ahad's constant - the Milky Way part = -5.0?

Found this on Google via the Astronomy Physics forum:

Kerela Wilton wrote:

[In defining Ahad's constant]
There are two things that one has to get right :: (i) brightness of
Milky Way glow and (ii) adgregate of individual stars in solar neighbor
hood.

(i) came to -5.0 as he
http://astronomyphysics.com/read.php?f=1&i=349&t=349

(ii) came to -6.0 (to a limit of 15th magnitude)

Part (i) Ahad derived as -5.0 using this method:

The Milky Way galaxy's absolute magnitude, accepted in most official
journals as an astronomical constant, is -20.5. That figure is based on
the assumption that the *entire* galaxy is viewed face on, as one
integrated object, from a standard distance of 10 parsecs (32.6 light
years). Now, since we are located in one of the spiral arms of the
Milky Way not far from the galactic plane and only get an "edge-on
view" looking inwards towards the centre of the Milky Way, we see only
50% of the galaxy's total brightness stretching across the night sky
(since the remaining 50% is on the *other side* of the dense galactic
core, and not directly in view to us).

Now, the standard formula for evaluating the brightness ratio, R,
between any two objects of magnitudes M1 and M2 is given by:-

R = 10^[0.4*(M1-M2)]

Hence, this formula can be used to "reduce" the Milky Way galaxy's
total absolute magnitude of -20.5 by 50% to give a figure of -19.7,
representing the "portion" that we see stretching across our night sky.
Since we are located at a distance, d, of about 8,200 parsecs from the
galactic centre [Source: Handbook of the British Astronomical
Association], the apparent magnitude, m, of the bulk of this "portion"
can be calculated from:-

m = M - [5 - 5 * log10(d)] = -19.7 - [5 - 5 * log10 (8200)] = - 5.1

Hence, the net integrated magnitude of the "visible" Milky Way
stretching across our night skies ought to be about -5.1.

However, there are various dark, intervening clouds of interstellar gas
and dust, such as the "Cygnus Rift", the "Coal Sack" near Crux, many
dark clouds in Sagittarius looking towards the centre of the galaxy,
etc. which contribute to dimming the overall light reaching Earth from
the broader Milky Way. Hence, if one makes a 10% (0.1-magnitude)
allowance for light extinctions owing to such obscuring interstellar
media, one will arrive at a net magnitude of -5.0.

This would be one way that I would *analytically* estimate the Milky
Way's total integrated brightness as -5.0 magnitudes.

This figure of course relates to a full 360-degree view of the whole
Milky Way. In actual practice, from a particular location on the
Earth's surface, only a fraction of this total brightness will be
experienced by an observer depending on various factors such as how low
the horizons are, which particular quadrant of the galaxy is on view
(e.g. the Cygnus region is much brighter than the Auriga region),
airglow and light extinctions due to the Earth's own atmosphere which
depends on the observer's elevation above mean sea level, etc...



Is this technically valid? Anyone have any thoughts here?
Cheers me deers!
S-S