HD 145622
V = 5.730 [0.009] 
Lab Manual V: lambda=.55 um, Delta lambda=.09 um, F0=3.75E-8 W/m^2 um
diameter pupil .005 m

* @constants.cal
* set lambda=.55e-6 deltaL=.09 F0=3.75E-8 V=5.73
* ? F0*10^(-V/2.5)
    .1914393749907772E-09 !W/(m^2 um)
* ? res/(h*c/lambda)
    530050479.9260179     !#/(s m^2 um)
* ? res*deltaL
    47704543.19334161     !#/(s m^2)
* ? res*pi/4*(.005)^2
    936.6765152441187     !#/s

solid angles are measured in steradians=area/R^2
The small image square, while flat, can be considered (by radian measure)
a spherical area (theta*R)^2, and so a solid angle of
theta^2
4 Pi steradians in a sphere, so
N=4*pi/(28/60*pi/180)^2
* ? 4*pi/(28/60*pi/180)^2
    189426.8628799864    
FYI: google says there are 41253 square degrees in a sphere and we got apprx 4x that

At mag=5.5 I reported 2,822 stars and #6 says for every mag deeper you get 4x more
2,822*4^x=189426
4^x=189426/2,822
x*log(4)=log(189426/2,822)
? log(res/2822)/log(4)
    3.034389568845798    
so to get about N stars you need to go to mag 8.5
http://www.stargazing.net/david/constel/howmanystars.html

The density of stars on a sphere of radius R would be:
2822/4 pi R^2 = 1/(D^2 sin(pi/3))
(D/R)^2=4 pi/(2822 sin(pi/3))
radian theta=D/R=sqrt(4 pi/(2822 sin(pi/3)))
? sqrt(4*pi/(2822*sin(pi/3)))*180/pi
   4.108504281235549    !(degrees, seems big too me)
   
Swihart chapter 6: #5 & #6

consider N stars/volume each of absolute magnitude M
All the stars inside radius r would be brighter than apparent m, where
m-M=5 log(r/10)
10^((m-M)/5)*10=r
10^(m+1-M)/5)=r
by Eq 6.4-2: M=4.77-2.5 log(L)
10^(m/5+1-4.77/5+.5*log(L))=r
10^(m/5+.046+.5*log(L))=r
as: 1-4.77/5=.046

#6 for any fixed value of L, a huge cancelation occurs for r(m+1)/r(m):
r(m+1)/r(m)=10^.2
so n(m+1)/n(m)=(r(m+1)/r(m))^3=10^.6=3.98
* ? 10^.6
    3.981071705534972    
    
Now what if we have two different classes of stars, for one star
n(m+1)/n(m)=4/1
for the other
n(m+1)/n(m)=12/3
note if we look at totals: n(m+1)/n(m)=(4+12)/(1+3)=16/4=4

more generally if we have N classes of star labelled by i and we have
for each class: n(m+1)_i = f*n(m)_i
where f is a fixed value, then
total (m+1)=sum[ n(m+1)_i ] = f*sum[ n(m)_i ]
total (m) = sum[ n(m)_i ]
clearly the ratio is still f