How do you calculate the DEC and RA of the sun?

I think what you’re wanting is a way to calculate the Sun’s altitude and azimuth for a given time and location. The calculations are bit involved—you’d need to determine the Julian (Ephemeris) Date, solve Kepler’s equation to determine eccentric anomaly, then true anomaly, the Sun’s radius vector to obtain the Sun’s longitude, the obliquity of the ecliptic to determine latitude, then transform the RA and dec coordinates to altitude and azimuth. I’m probably leaving something out. Anyway, I do have the formulae, but thought I’d see if that’s what you’re really wanting before I do a lot of typing.

Perhaps a more sane approach would be to modify a telescope (a Meade or other) and let its mounting do the tracking for you? Just an idea.

[edit]

My apologies for the delay in updating my answer. Here’s how to calculate the Sun’s RA and dec:

First, calculate the Julian Day (JD). (note: Julian Day begins at Greenwich mean noon, or 12h Universal Time).

Let

YYYY = year

MM = month

DD.dd = day (fraction of day derived from hours, minutes, seconds past 00:00:00 UT; convert local time to Universal Time before doing this)

If MM > 2, then

y = YYYY

m = MM

If MM = 2 or 1, then

y = YYYY – 1

m = MM + 12

If date >= October 15, 1582 (in your case it will be), calculate

A = INT(y / 100)

B = 2 – A + INT(A / 4)

else (not in your case)

B = 0

Julian Day is then:

JD = INT(365.25 * y) + INT(30.6001 * (m + 1)) + DD.dd + 1720994.5 + B

Next, calculate the time T (measured in Julian centuries of 36525 ephemeris days from the epoch 1900 January 0.5 ET):

T = (JD – 2415020.0) / 36525

Note: since T is expressed in centuries, it should be calculated with a sufficient number of decimals; an error of 0.0001 in T corresponds to an error of 0.37 day in the time.

Then, geometric mean longitude of the Sun (in degrees):

L = 279.69668 + 36000.76892 * T + 0.0003025 * T^2

Sun’s mean anomaly (in degrees):

M = 358.47583 + 35999.04975 * T – 0.000150 * T^2 – 0.0000033 * T^3

Eccentricity of the Earth’s orbit:

e = 0.01675104 – 0.0000418 * T – 0.000000126 * T^2

Here, I opted for an easier method that doesn’t require the solution of Kepler’s equation; the Sun’s radius vector isn’t needed:

Sun’s equation of center C (in degrees):

C = + (1.919460 – 0.004789 * T – 0.000014 * T^2) * sin(M) + (0.020094 – 0.000100 * T) * sin(2 * M) + 0.000293 * sin(3 * M)

Sun’s true longitude (stl) is:

stl = L + C

(The true anomaly (M + C) is not needed.)

Sun’s apparent longitude (sal) is then:

sal = stl - 0.00569 – 0.00479 * sin(259.18 – 1934.142 * T)

Then calculate the obliquity of the ecliptic (oe):

oe = 23.452294 – 0.0130125 * T – 0.00000164 * T^2 + 0.000000503 * T^3

To correct for apparent position:

oe = oe + 0.00256 * cos(259.18 – 1934.142 * T)

Then, for RA and dec:

tan(RA) = cos(oe) * sin(sal) / cos(sal)

sin(dec) = sin(oe) * sin(sal)

or

RA = arctan(cos(oe) * sin(sal) / cos(sal))

dec = arcsin(sin(oe) * sin(sal))

You can convert RA from degrees to hours by dividing by 15.

Assuming I didn’t overlook something or typo, this should get you the Sun’s RA and dec for a given time. Do you also need a method to transform RA and dec to altitude and azimuth for a given time and location? (I’m thinking yes)

A simple solar tracking device can be purchased from WattSun that you could probably modify for your purposes.

If you really do want to use the numbers, you can find them here

http://eosweb.larc.nasa.gov/sse

But the sun doesn't follow the same path through the sky every day and there are slight changes year to year.

Google...u can get everythin on google...(srry if thts not wot u wantd to here)