How to calculate the sunset/rise

Question

When I get to the azimuth point how to convert that into time?

same with the altitude...

Please help me. I am stuck at that point.

Don't worry I know how to get to that point so u don't have to explain that to me.

Answer 1

Time is related to Hour Angle (HA) and is not directly related to Azimuth and Altitude. When you know the HA of a celestial object, you can obtain the Local Sidereal Time (LST) from the equation LST = HA + RA, where RA is the Right Ascension of the object. You would then convert LST into Local Mean Time.

DEC is the declination of the object.

Additional notes:

I don't know what information you're suppied with initially, so I may be making some assumptions here. It looks as though you're calculating sunrise and sunset knowing the Sun's true longitude (lambda) measured along the ecliptic. I assume this figure is given to you, along with epsilon (the obliquity of the ecliptic). If lambda is between 0 and 90 degrees, then so is RA. Similarly if lambda is between 90 and 180 degrees, then so is RA and the same applies for 180 - 270 and 270 - 360 degrees.

You have the formulae to convert the Sun's longitude to RA and DEC (DEC is the same as delta). Sunrise/set is the time when the Sun's altitude is -0.833 degrees (not 0 degrees because it's necessary to allow for atmospheric refraction and the Sun's semidiameter). You then need to find the hour angle corresponding to this altitude and convert to LST and Mean Time as I mentioned before. I assume you're also supplied with a formula for this.

The value of lambda slowly changes, so for more accuracy, you may need to repeat the calculation by updating lambda to the time of sunset you found on your first try.

More notes:

You are correct to reduce lambda to the range 0 - 360 degrees, but I make it that 39374.12552 degrees reduces to 134.12552. You have simply divided by 360. If you took your remainder of .3725626 and multiplied it by 360, you would get the correct result. I did it on the Windows XP calculator by 39374.12552 mod 360.

I don't know all the workings of a TI-84, but many calculators have an 'inv' function, if so, 'inv' 'sin' would give the same result as 'arcsin'.

Even more notes:

You've quoted a formula for azimuth, but altitude is the important value in calculating sunrise and sunset times. The formula for altitude (which I assume you have anyway) is

ALT [degrees] = ARCSIN [ SIN(LAT) x SIN(DEC) + COS(LAT) x COS(DEC) x COS(HA) ].

As I mentioned before, taking ALT = 0 degrees at sunrise/set is approximate, but rearranging the above equation for ALT = 0 would give

HA = ARCCOS [-SIN(LAT) x SIN(DEC) / COS(LAT) x COS(DEC)] which reduces to

HA = ARCCOS [-TAN(LAT) x TAN(DEC)].

The more accurate ALT = -0.833 degrees gives

HA = ARCCOS [(SIN(-0.833) - (SIN(LAT) x SIN(DEC)) / COS(LAT) x COS(DEC)] which becomes

HA = ARCCOS [(-0.0145 - (SIN(LAT) x SIN(DEC)) / COS(LAT) x COS(DEC)].

You appear to have a value of lambda, so you can presumably obtain RA and DEC. Plugging DEC and LAT into either of the above equations will give you HA, the Sun's hour angle at sunrise/set.

You then calculate the local sidereal time (LST) from the equation I gave you at the start, LST = HA + RA.

You now need to convert LST to local mean time. I don't know if you have a similar formula to this, but during 2008, for any longitude (LON),

LST = 99.04 + 360.9856 x (D + TZ/24) - LON

where D is the day of the year and TZ is your time zone (hours west of Greenwich). For today (August 6th), D = 219, so for New York (Eastern Daylight Time, TZ = 4) and LON = 73.93,

LST at 0 hours = 99.04 + 360.9856 x 219.167 - 73.93 = 79141.241.

Reducing this to an angle less than 360 degrees gives LST = 301.24 degrees. So, if you've calculated the LST at midnight at the start of the day and at the location in question, you subtract this LST from the LST of sunrise/set that you've calculated, remembering to add 360 degrees if the result is less than zero. To convert from sidereal time to mean time, divide by 1.002738, then change from degrees to hours by dividing by 15.

To go back to the New York example, if you've calculated the LST of sunset as (say) 243 degrees, then 243 - 301.24 = -58.24. Add 360 to this and we get 301.76. Divide this by 1.002738 to give 300.94, then 300.94/15 = 20.062 hours, which is 20 hours 04 minutes, which is the time of sunset.