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Sunrise and sunset correction for astronomical and geographical factors

  1. Mar 23, 2012 #1
    Hi
    I've been meaning to find a correction method for the true location of the sun so when I add the correction to the true sun it would give me the apparent sun. To clarify, I want to be able to calculate the difference of time between sunsets or sunrises if:
    There were no geographical formations whatsoever, elevated or depressed, the entire planet being a perfect geoid in that case (observer's altitude would contribute to day length for increased height would move the horizon further away from the observer, allowing him to observe the sun longer),
    There was no refraction of light (that would probably be comparable to having no atmosphere),
    The sun was a point light source (a necessary comparison to compensate for the angular radius of the sun)
    The parallax of the sun, however minute, did not contribute to the difference of time (the angle that the Earth's equatorial radius would subtend if observed from the center of the Sun)
    And sunrises and sunsets if the four above-mentioned factors were as they are on our planet Earth. I know the question is phrased a bit noobish but it's been several years since I've graduated from Astronomy department and haven't studied since and all of a sudden I'm required to know the difference between these two situations. Please help me out here, how do I compensate for these factors?
     
  2. jcsd
  3. Mar 24, 2012 #2
    I've stumbled upon a set of formulae for my question but still I'd really appreciate an expert's opinion on the matter. The Formulae are as below:
    cos a = r / (r + x)
    r = 6373000 (Mean Earth radius in meters)
    x = 625 (I've yet to figure out what this is and why the author of this method gave this value but I'm guessing it's the observer's altitude in meters)

    R = Radius of the Sun (angular)
    a = Angle of horizon's descent (The angular difference between the sunset or sunrise from the viewing point of an observer at an elevated location and those from the viewing point of an observer from the lowest-altitude area of the same location)
    N = Amount of refraction at the horizon (This is to be calculated by keeping in account the observer's coordinates and altitude and atmospheric pressure)
    P = Horizontal Parallax of the Sun (the angle that the Earth's equatorial radius would subtend if observed from the center of the Sun)
    ∆H = (180/∏)*(1/15)*((sin ∆h )/(Cos φ . cos δ . Sin t))
    Cos H= (Cosz-sin φ.sin δ)/(Cos φ . cos δ)
    Δh = R+a+N-P The angular difference between the center of the Mean Sun at sunset or sunrise at 0 altitude and the setting or rising edge of the Apparent Sun at the observer's altitude
    ∆H= Correction for the Hour Angle H= Hour Angle of the Sun z= Zenith Angle of the Sun φ= Latitude of the observer δ= Declination of the Sun
    With the formulae above would H+∆H provide me with the correction for the factors I've mentioned in my previous post? If it would, how do I include the refraction and the horizontal parallax in this calculation? What method would be accurate to calculate the refraction at varying altitudes, zenith angles of the sun and latitudes?
     
  4. Mar 26, 2012 #3
    Please, anyone? Not a response? This is quite important for a large number of people I'm working with. Any professional insight would be truly appreciated.
     
  5. May 25, 2012 #4
    A formula for angular height (h) of a celestial body, e.g. Sun, is:

    Sin h = (sin lat)(sin dec) + (cos lat)(cos dec)(cos hour angle).

    Disregarding refraction, the Sun's height when its upper limb is at the horizon is about -[16' + 1.06'√(observer elevation in feet).

    For elevations at or above 10,000 feet use -[16' + 1.07'√(observer elevation in feet)].

    The semidiameter varies within about ±.3' of 16'. There is a table showing what it is throughout any year, and it is specified in the air and nautical almanacs.

    [The height considering atmospheric refraction would be about -[16' + 34' + 1.15'√(observer elevation in feet)]. 16' is an average semidiameter of the Sun disk, 34' is a nominal amount of refraction of a Sunlight ray reaching the horizon. 1.15'√(observer elevation in feet) is an estimate of the arcdistance between the observer and the horizon; it takes into account dip and refraction; it comes from a (Thomas Young) formula for nautical mile distance to the horizon].

    Sin h can be written as Sin -[16' + 1.06'√(observer elevation in feet)], and the equation can be rewritten as a formula solving for hour angle.

    Combine that local hour angle with longitude to get the Greenwich Hour Angle of the Sun.

    The Air Almanac, Nautical Almanac, and Online Nautical
    Almanac tabulate the Sun's GHA. By interpolation the moment of time can be found when the GHA of the Sun is predicted to have that GHA.

    Mark Prange
     
    Last edited: May 25, 2012
  6. Jun 8, 2012 #5
    Most common formulae for refraction break down at and below the horizon. Here's one that doesn't.

    For an object with a true (unrefracted) altitude H in degrees:

    compute v (in degrees) = H + (9.23 / (H + 4.59))
    then refraction r in arcsec is :
    r = 58.7 * cos v / sin v
    and apparent altitude h = H + r/3600

    For an object on the horizon, H=0, v=9.23/4.59 = 2.0109, r = 1671.8 arcsec, h=0.4644 degrees.

    Computing corrections to sunrise/set times depends on the Sun's declination and the observer's latitude.
     
  7. Jun 8, 2012 #6
    Perhaps this is simpler:

    For an airless world, distance to the horizon is

    D = 3571.6 √h

    ... where D is the distance to the horizon in meters, and h is eye height in meters.

    For Earth with a standard atmosphere:

    D ≈ 3924.8 √h

    This distance (and the differences between atmosphere and no-atmosphere conditions) can be converted to a time difference for a rising or setting object by:

    T = cos θ * (D/40075029) * 1440

    where θ is your latitude and T is in minutes.
     
  8. Jun 12, 2012 #7
    Thanks a lot for the replies, friends. I really appreciate it. I'll get to trying out these formulae as soon as I can and see if I can't implement them in our calculations. Much obliged.
     
  9. Jun 12, 2012 #8
    I do hope you know that's just an approximation, and you didn't even get your units right.
     
  10. Jun 12, 2012 #9
    Perhaps you would care to supply a correction.
     
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