Posted in xn--kfs74mzzid01b.com edit by webmaster on January 7th, 2009
The goal is to predict the apparent position of the Sun in the Earth
sky at a given date, a given hour and a given latitude.
Let alpha be the angle between North and the direction of the Sun, and
beta then angle of the Sun above the horizon, L the latitude, l the
longitude, and t the date-time.
I'm looking for the function F :
( albha, beta ) = F( L , t )
For exemple, right now, I am at 48 49'N 02 17'23 E, and we are the
15th of november 2006, time is 1035 GMT.
**Roughly** observed, the Sun is at 165 (North being zero) and 22
(horizon being zero)
I want F that could have told me :
F( 48 49' , 2 17', 2006/11/15, 1035) = (165, 22)Hi!!
First of all, a direct function will give a very long and difficult to
compute equation. There are some intermediate steps that must be done,
so an algorithm (using several different formulas), instead of a
single formula, is the way to predict theposition of the Sun in the
Earth sky at a given date, a given hour and a given latitude.
Here is the result of my research:
Nice graph of what we will calculate:
http://www.srrb.noaa.gov/highlights/sunrise/azelzen.gif
First thing you need to compute is the fractional year g in degrees:
g = (360/365.25)*(N + hour/24)
where:
N = day number --> January 1 = day 1, January 2 = day 2, etc.
Note about hour: during the period when Daylight Saving Time (DST) is
in force one hour should be added to obtain clock time.
Note that hour is the local hour and it is expressed in fractions of
hours, that is 10h 35m = 10 + 35/60 = 10.58333
November 15, 2006 is the day number 319, then:
g = (360/365.25)*(319 + 10.58333/24) = 314.849
Follows the calculation of the declination of the sun, again you an
use a table or a formula:
"Table of the Declination of the Sun":
November 15 declination = -18 20' (= 18.33333 )
http://www.wsanford.com/~wsanford/exo/sundials/DEC_Sun.html
Or use the following formula:
D = 0.396372-22.91327*cos(g)+4.02543*sin(g)-0.387205*cos(2*g)+
+0.051967*sin(2*g)-0.154527*cos(3*g) + 0.084798*sin(3*g)
In this case:
D = -18.6162
Now calculate the time correction for solar angle:
TC = 0.004297+0.107029*cos(g)-1.837877*sin(g)-0.837378*cos(2*g)-
-2.340475*sin(2*g)
In your case:
TC = 3.72763
Now we can calculate the Solar Hour Angle (SHA)
SHA = (hour-12)*15 + Longitude + TC
Longitude in degrees (this figure should be negative Weast of
Greenwich and positive East of Greenwich).
Your longitude in degrees is:
02 17'23 E = 2+17/60+23/3600 = 2.28972
Then:
SHA = (10.58333-12)*15 + 2.28972 + 3.72763 = -15.2327
Note that if SHA is greater than 180, then you must add (-360) to the
result and if SHA is lower than -180, then you must add 360 to the
result.
Now we can calculate the Sun Zenith Angle (SZA):
cos(SZA) = sin(Latitude)*sin(D)+cos(Latitude)*cos(D)*cos(SHA)
Your Latitude is 48 49'N = 48 + 49/60 = 48.81667
cos(SZA) = sin(48.81667)*sin(-18.6162)+
+cos(48.81667)*cos(-18.6162)*cos(-15.2327) =
= 0.36184
SZA = arccos(0.36184) = 68.78676
Well SZA is the complementary angle of the Sun Elevation Angle or
Altitude(SEA), therefore SEA = 90-68.78676 = 21.213 (21 12'48'')
NOTE: if cos(SZA) formula gives a figure greater than 1 use 1 and if
it gives you a figure lower than -1 use -1.
To finish we will calculate the Azimuth Angle (AZ):
cos(AZ) = (sin(D)-sin(Latitude)*cos(SZA))/(cos(Latitude)*sin(SZA)) =
= -0.96367
AZ = arccos(-0.96367) = 164.50848 (164 30'30.5'')
As you can see this method gives you from the inputs ( 48 49';
2 17'23''; 2006/11/15; 10:35) the results (164 30'30.5''; 21 12'48'');
your rough figures gave a good aproximation.
You can use the above formulae to make an Excel spreadsheet.
Sources:
NOAA (National Oceanic and Atmospheric Administration) Surface
Radiation Research Branch:
"NOAA GVI GUIDE APPENDIX L: Software to Calculate Relative Azimuth
from Third Generation Weekly Composite GVI Date":
http://www2.ncdc.noaa.gov/docs/gviug/html/l/app-l.htm
"Horizontal coordinate system - Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Horizontal_coordinate_system
"Positional Astronomy":
http://star-www.st-and.ac.uk/~fv/webnotes/
"Basics of Positional Astronomy":
http://www.geoastro.de/elevaz/basics/index.htm
You can check other results using a Java applet here:
"Sun Calculator"
http://users.zoominternet.net/~matto/Java/Solar%20Calculator.htm
Search strategy:
solar time calculation
solar azimuth calculation
solar azimuth formula
I hope this helps you. Feel free to request for a clarification if you need it.
Best regards,
livioflores-gaDear Livioflores,
***********************************************************
Heartful thanks for your much accurate and detailed answer.
:)
***********************************************************
If I may, I'd like to add a couple of whats and whys :
1) regarding fractional year g :
I guess that the figure 365.25 you use to compute g is the actual
duration of a solar year, in days ?
2) Regarding solar declination D:
2.1) Is the table accurate regardless of latitude ? I'd guess yes, but
I'd like a confirmation.
2.2) The formula you give for D uses quite a few numeric values. Do
you know where they come from ? what they represent ? do they have a
name ?
<>
0.396372
22.91327
4.02543
0.387205
0.051967
0.154527
0.084798
3) Regarding time correction TC
3.1) same question as above
3.2) I understand this notion is sometimes known as "Equation of time"
: am I correct ?
---------------------------------------
Finaly : thank you very much again !Hi!!
Thank you for the good comments on my answer.
Now I will try to answer your questions:
1) regarding fractional year g :
I guess that the figure 365.25 you use to compute g is the actual
duration of a solar year, in days ?
YES.
2) Regarding solar declination D:
2.1) Is the table accurate regardless of latitude ? I'd guess yes, but
I'd like a confirmation.
YES. Recall that, despite of some complicated definitions, Declination
(D) is the latitude at which the sun is directly overhead at a given
time. It is always between 23.5 N and 23.5 S latitude.
2.2) The formula you give for D uses quite a few numeric values. Do
you know where they come from ? what they represent ? do they have a
name ?
As far as I know the coeficients have no specific name and they come
from some complicated geometrical calculations derived from the
conversion between different systems of coordinates (astronomical and
terrestrial coordinates) and the orbital equation of the planet. I am
unable to find their origin but at the following page you will find
another equation to estimate D (with its own set of coeficients), you
will see also how some coeficients are estimated or calculated:
"Astronomy Answers: Position of the Sun":
http://www.astro.uu.nl/~strous/AA/en/reken/zonpositie.html
See also "EQUATION OF TIME -- PROBLEM IN ASTRONOMY":
You can see here how complex are the calculations and coeficients
derivation from the coordinates conversion.
http://info.ifpan.edu.pl/firststep/aw-works/fsII/mul/mueller.html
3) Regarding time correction TC
3.1) same question as above
3.2) I understand this notion is sometimes known as "Equation of time"
: am I correct ?
TC is independent of the latitude but it is not the EoT, but I guess
it is related because both helps to determine the Solar Hour Angle.
Regarding the coeficients, the answer is the same.
See for further references the following pages:
"Chapter 3 - The Sun?s Position":
http://www.powerfromthesun.net/chapter3/Chapter3Word.htm
and
"Positional Astronomy - index":
Just follow the index of links.
http://star-www.st-and.ac.uk/~fv/webnotes/index.html
I hope this helps you.
Regards,
livioflores-ga#If you have any other info about this subject , Please add it free.# |
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