Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Horizontal + Celestial Coordinates to Geographic Coordinates

  1. Aug 23, 2012 #1
    Hey everyone,

    I cannot seem to figure this out and I'm having a hard time finding any guides online for this stuff. All I can find are calculators. I was wondering if it would be possible to calculate my Geographic Coordinates on Earth if I had the Horizontal and Celestial coordinates of a celestial body as well as the date, time, etc. Could anybody give me hints on this? I've been trying to figure it out for awhile.

    Thank you!
     
  2. jcsd
  3. Aug 23, 2012 #2
    Well you have Right Ascension and Declination which are Celestial Coordinates.

    They are fixed in the Sky.

    Right Ascension is a product of: Local Sidereal Time + 15*(hour + min/60 + sec/3600 - Timezone - dst) + Longitude.

    Local Sidereal Time is the distance in degrees from the Vernal Equinox.

    DST is Daylight Savings Time 1 for on and 0 for off.

    Declination is just the Latitude.

    Hour Angle = Local Sidereal Time - Right Ascension

    From Hour Angle, Latitude and Declination you find Azimuth and Altitude or Horizontal Coordinates.
     
    Last edited: Aug 23, 2012
  4. Aug 24, 2012 #3

    Filip Larsen

    User Avatar
    Gold Member

    I can recommend "Practical Astronomy with your Calculator" by Peter Duffet-Smith. You should be able to find a copy in your local library or, if you know how to use google, on the net.
     
  5. Aug 27, 2012 #4
    Thanks guys, I'll check out the book. Philosophaie I think you misunderstood. I am looking for calculate MY Geographical coordinates (longitude and latitude) from KNOWING the Celestial coordinates (right ascension, declination) of a celestial body AND the Horizontal coordinates of that body (azimuth, angle). I can't figure out how to reverse the calculations. :(

    But thanks!
     
  6. Sep 6, 2012 #5
    I couldn't figure it out :( I have very very little experience with Astronomy. Is there anyone who could maybe help me out a little more? Thanks...
     
  7. Sep 6, 2012 #6
    Latitude = Declination.

    LongitudeEast = Right Ascension - (Sidereal Time + 15*(hour + min/60 + sec/3600 - TimezoneEast - dst) ).

    where hour, min and sec are in local time.

    TimezoneEast (Eastern=-5) where East is positive.

    dst =1 for Daylight Savings Time and dst = 0 for not.

    Sidereal Time is LMST = (18.697374558 + 24.0657098244191 * d) + LongitudeEast
    where d = JulianDate-2451545

    Do the math.
     
    Last edited: Sep 6, 2012
  8. Sep 6, 2012 #7

    Filip Larsen

    User Avatar
    Gold Member

    I kind of missed that you wanted to calculate latitude and longitude from azimuth and elevation of a known star at a known time, and the other way around.

    What you ask for is in general and in practice a bit more complicated to perform than the reference I suggested can hope to explain. Perhaps you can search for "celestial navigation" and "sight reduction"? It seems there are some good site that tries to explain this, like for instance [1], but as the subject is involved it may require some effort on your part. If you want a more simple approach you can perhaps follow the guide at [2].


    [1] http://www.celnav.de/
    [2] http://www.eaae-astronomy.org/WG3-SS/WorkShops/LongLatOneStar.html
     
  9. Sep 12, 2012 #8
    Note: LMST = 15* (18.697374558 + 24.0657098244191 * d) - LongitudeEast where d is the time in days since 1-1-2000 @00:00:00 and LongitudeEast is in Degrees.
     
  10. Sep 14, 2012 #9
    Declination is only equal to Latitude when Azimuth = 0 and Altitude = 90deg or straight up. Other than up:

    Declination = asin(sin(Lat)*sin(Alt) - cos(Lat)*cos(Alt)*cos(Azi)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Horizontal + Celestial Coordinates to Geographic Coordinates
  1. Celestial pole (Replies: 1)

  2. Celestial Sphere (Replies: 0)

Loading...