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I The declination at the horizon in the east at a latitude

  1. Aug 15, 2016 #1
    I am trying to find the declination at the horizon in the east at a particular latitude.
    The equation is:

    delta = asin(sin(lat)*sin(Alt) - cos(lat)*cos(Alt)*cos(Azi))

    where Azi is the azimuth and is equal to 90 degrees in the eastern most direction.
    Alt is the Altitude and is equal to 0 degrees at the horizon.
    From this equation delta is always equal to 0 but intuitively the horizon plane changes with the latitude.
    Could this equation be misinterpreted?
     
    Last edited: Aug 15, 2016
  2. jcsd
  3. Aug 15, 2016 #2

    Filip Larsen

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  4. Aug 15, 2016 #3
    At different latitudes the horizon intersects the Celestial Meridian at different points. The declination only intersects at one as shown in the above drawing. Is there an alternate declination equation to calculate this divergence?
     
    Last edited: Aug 15, 2016
  5. Aug 15, 2016 #4

    Filip Larsen

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    I am not really sure what you are asking.

    In the diagram, the equator is the great circle marked with blue, the horizon is the great circle in black containing the points S, W, N and E, and the meridian is the great circle marked grey containing the point S, Z, P and N. Your original question then reads if it is correct that the horizon at point E has zero declination, and that is true as every point on the equator, and thus also point E, has zero declination.

    Can you elaborate which part of this you think should work differently?
     
  6. Aug 15, 2016 #5
    The picture above I believe is for latitude=0 or colatitude=90. At different latitudes the horizon shifts. And my question is: does that affect the declination being zero at the horizon or if so by what value?
     
  7. Aug 15, 2016 #6

    Filip Larsen

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    I suspect you are suffering from some misunderstanding of something. In the diagram latitude is indicated as the angle along the meridian from N to P, and seems to have been drawn around 60 deg or so in this example (and thus nowhere near zero). Also, in case you wonder, the diagram shows how the equatorial system "looks like" from horizontal system, so you should imagine the observer in the center of the sphere (where the imaginary line NS and EW crosses) with zenith (Z) being vertical up and looking out at the sky represented as the sphere placed "infinitely" far away.

    Perhaps you can explain what it is you are trying to achieve or understand, and what steps you have done so far?
     
  8. Aug 15, 2016 #7

    Filip Larsen

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    It suddenly occurred to me that perhaps you are asking about the declination of a point on the visible horizon for an observer at a particular height over the spheroid? Like if you are standing on a high tower and look straight east an wonder what declination a star there will have? If so you need to calculate the (negative) elevation the actual horizon as a function of the observers height over the spheroid and then use that negative elevation in your original equation for declination.
     
  9. Aug 15, 2016 #8
    I can calculate the Geoid distance from the center of the Earth to my observation point but putting that onto the Azimuth and Elevation is difficult. Could you expound on negative elevation.
     
  10. Aug 16, 2016 #9

    Filip Larsen

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    If you look at the Wikipedia entry for Horizon you can see the diagram repeated below that relates the actual horizon H for an observer O situated at height h over a sphere with radius R. For the observer, the zero altitude corresponds to the direction of the tangent at the observers foot point or, similar, a direction perpendicular to R+h line. Analyzing the triangle in the diagram you can see that this means the actual horizon is depressed the angle γ. In your case you then have Alt = -y.

    If you want you should be able to insert the trigonometric expression for γ in the Wikipedia page directly into the equation you used for declination, arriving at a very simple expression for the declination of the actual horizon in the east/west direction for an elevated observer over a sphere.

    All this is of course only valid if the sphere models your situation well, like if you are at sea or live in a flat region. If you have a mountain or similar blocking your view then the altitude of the actual eastern skyline may of course differ from this.

    GeometricDistanceToHorizon.png
     
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