Expert: Philip Stahl - 8/17/2006

Question
Dear Mr. Stahl,

Maybe I used the wrong terminology. By solar model I do mean solar energy model. But the part I fail to realise is to calculate the postition of the sun standing on the earth. So I want to calculate the position of the sun in the sky, meaning altitude and azimuth using the locan time and GPS data of a certain place. I want to use this data to calculate where to point my solar cell to, to get the best performance out of the cells.

Thank you in advance,
Greetings Bart Aerts
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Followup To

Question -
Dear mister Stahl,

I'm currently busy with making a model of the sun to use in a solar program. This program would calculate where to point a solar cell to to get the best reception of solar energy. I have found many formula's on the Internet, but none of them are really formulating the solution.

I would use the local time or GMT, date and GPS data to calculate the position.

I hope you can help me.

Thank you in advance.

Greetings,

Bart Aerts

Answer -
Hello,

I would like to assist you but am afraid that I don't understand where you are coming from with your question. So I need more information.

If you are making a "solar model" why would you need solar cells? What would they tell you about the Sun *itself*? What sort of aspects of the model are you looking at? Can you identify all the parameters for me?

What exact formulas are you looking at? For what properties of the Sun?

I am assuming here that this is really a solar model - and not a solar energy model! (For which I am afraid I cannot be of much assistance, since I am not a solar energy expert or engineer)

Hopefully you can provide more info as indicated above - or - if this is really about solar energy, you can find an expert able to assist.

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This is an addendum for further reference:

When we generally speak of  “solar model” in astrophysics, we mean the mathematical construction of the Sun from the inside out, layer by layer – showing different physical parameters, conditions at each layer. To get this done we make use of specific initial conditions for composition, e.g.

X = 0.92     Y = 0.07  Z = 0.01

where X, Y and Z denote proportions for hydrogen, helium and heavier elements, respectively

in addition to a specific equation specifying energy generation, e.g.

E(pp) = 2.5 x 10^6 (rho) X^2 (10^6/ T)^2/3    exp [- 33.8 (10^6/T)^1/3]

where pp denotes proton-proton cycle

and ALSO a set of differential equations, e.g. such as the one for hydrostatic equilibrium ( outward gas pressure must balance inner gravitational force at every point):

dP/ dr  =   - GM rho(r)/ r^2


where rho(r) is the density for a given layer

The table headers for the model parameters is often given as:



r/ R  -----M_r/M-----L_r/L ----- log P---log T----log (rho)----k


where r/R is the fractional radius, e.g. 0.05 from the center or r = 0.05R. M_r/M is the fraction of mass to that point; L_r/L the fraction of luminosity to that point; log P is the logarithm of the pressure at that layer (r); log T is the logarithm of temperature, log (rho) is the logarithm of density and k is the opacity.  

Answer
Hello,

I kind of figured that this is what you wanted - but preferred to make sure on hearing back from you.

I think your best bet to obtain these coordinates (and the most painless) would be to obtain a Nautical Almanac or Astronomical Ephemeris. The Sun's azimuth and altitude for given times and locations ought to be provided therein.

It is possible, of course, to grind out the numbers yourself. Starting with say, the Sun's declination and right ascension.

To get the Sun's RA you will need to know when a specific star for which you already know *its* RA makes local transit.

Then one can use:

Local Time of Transit = Star's RA - Sun's RA

Or, re-arranging:

Sun's RA = Star's RA - LTT

So, assume LTT = 19h 30m and the Star's RA at that time is 23 h 00m, then the Sun's RA would be:

Sun's RA =  23 h 00 m -  19h 30 m =  3 h 30m

You will also need to obtain the hour angle (h) from this - by using

h =  LST  - Sun RA

The first can easily be worked out using a sidereal time calculator (which ought to be on the net).

For obtaining the Sun's declination, this method may be used

Sun's Declination (D) = Latitude + Meridian Zenith Distance (MZD)

E.g. say my latitude is 40 degrees (N) and MZD = -30 degrees

(i.e. 30 degrees SOUTH of my zenith or overhead point)

THEN: D =  40 degrees + (- 30 degrees) = 10 degrees.

call this 'z' for short

The azimuth of the Sun would then be (from the representative astronomical triangle):


tan (A) =  [- cos (D) sin (h)/ sin (D) cos (lat) - cos (D)cos(h) sin (lat)]


where 'lat' denotes the latitude of interest.

Of course, the altitude is found from the zenith distance (z) using:

a =  sin ^-1 (z)


Hopefully you will find this info useful!