time	HA		tau	HA(calc)	tau(calc)
21:00	23:49:33	1.008	23.739		1.008		1688
23:00	1:49:52		1.073	25.739		1.067		1585
1	3:50:12		1.350	3.805		1.344		1525
2	4:50:21		1.634	4.805		1.625		1464
3	5:50:32		2.118	5.805		2.104		1397
3:30	6:20:37		2.493	6.305		2.476		1304

vega RA=18:36:56 dec=38:47:01 wiki
observator: 45:34:30 & 94:23:47
sept 1-> D=244

* @hmmss.cal
* let lat=f(45.3430)*pi/180
* let dec=f(38.4701)*pi/180
* let long=f(94.2347)/15
* let day=244
* let RA=f(18.3656)
* ? 21+.0657*day+(5-long)+f(6.37)-RA
    23.73881851851852    
* let HA=res*pi/12
* ? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
 1.008361097397202
? 23+.0657*day+(5-long)+f(6.37)-RA
let HA=res*pi/12
? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
    25.73881851851852    
    1.066761081564735    
? 1+.0657*(day+1)+(5-long)+f(6.37)-RA
let HA=res*pi/12
? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
    3.804518518518517    
   1.344126787154939    
? 2+.0657*(day+1)+(5-long)+f(6.37)-RA
let HA=res*pi/12
? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
    4.804518518518517    
    1.625118945785555    
? 3+.0657*(day+1)+(5-long)+f(6.37)-RA
let HA=res*pi/12
? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
    5.804518518518517    
    2.104142999214284    
? 3.5+.0657*(day+1)+(5-long)+f(6.37)-RA
let HA=res*pi/12
? 1/(cos(lat)*cos(dec)*cos(HA)+sin(lat)*sin(dec))
    6.304518518518517    
    2.475942012115166    
An analysis of data submitted by computer: on 14-SEP-2018 at 14:39 indicates that a function of the form:
--Exponential-- y=A exp(Bx)
can fit the 6 data points with a reduced chi-squared of 0.86

                FIT
 PARAMETER     VALUE        ERROR
   A =          1906.        63.    
   B =        -0.1538       0.20E-01
 
 NO x-errors
 y-errors based on formula: SQRT(Y)


                   ACTUAL     ERROR IN         CALCULATED   DEVIATION
 POINT    X          Y            Y                Y        FROM FIT
   1     1.01      0.169E+04 +-  41.             0.163E+04    55.9    
   2     1.07      0.158E+04 +-  40.             0.162E+04   -32.4    
   3     1.34      0.152E+04 +-  39.             0.155E+04   -24.9    
   4     1.62      0.146E+04 +-  38.             0.148E+04   -20.3    
   5     2.10      0.140E+04 +-  37.             0.138E+04    18.1    
   6     2.48      0.130E+04 +-  36.             0.130E+04    1.80    
 

see 364t104.pdf #1 (but with nu)
lets consider a bin between .49e-6 & .51e-6 m
there are 17 photons in this bin which has frequency width
* ? c*(1/.49e-6-1/.51e-6)
    23992993837534.98    
seek photons per second per area per bin-size
? 17/600/(pi/4*1^1)/res
    .1503568945102904E-14 !1/(sec*m^2*Hz)

it was not requested but might seek energy flux (17 photons each with E=h*nu)
? h*c/.5e-6*res
    .5973516504707997E-33
in terms of Jansky: Jy=(1E-26 W/m^2/Hz)
 ? res/1e-26
    .5973516504707997E-07 !or 60 nJy

Remark: while the flux of photons is of course much reduced light years from the
star, the distribution of the received photons should match what was emitted,
so we concentrate on measures of that distribution (mean, median, mode) rather
than the absolute flux. (Using the measured flux would be possible if we knew the 
distance --and hence how much the flux was reduced-- but we don't.)

from class: (where x=h nu/kT)
mode: x=1.59362
median: x=2.35676
average: x=2.70118

in spreadsheet calculate: h*c/k*(1/cells) =x*T for all 1000 cells, 
find median and/or average of these, equate to above
* ? h*c/(kb*1e-6)
    .1438777353827720E5

0.03284879803259635E6 =median  = median(x)*T; T= 0.03284879803259635E6/2.35676 = 13938
0.03833068594121622E6 =average = average(x)*T; T=0.03833068594121622E6/2.70118 = 14190

The statistical language R finds modes:
D=read.csv("1000_lambda.txt",sep="",header=F,skip=1)
d=as.vector(unlist(D))
di=1/d
> mean(di)
[1] 2.664115
> median(di)
[1] 2.283105
dd=density(di)
dd$x[dd$y==max(dd$y)]
1.302463
h*c*1e6*1.302463/k=mode(x)*T; T=h*c*1e6*1.302463/1.59362/kb; T=11759  (low)