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    Re: Azimuth by Sun Ray
    From: Greg Rudzinski
    Date: 2011 Nov 5, 12:48 -0700

    Dave,

    I tried working in reverse to get time with the photographers position,Sun declination, and date known.

    Getting Hs required setting up a triangular ratio with the sub triangle pictured in the attached image. The ray of light goes corner to corner so that the 418px/189px = Hs px/1136px. Hs= 2512.5 px . I have already calibrated the 50mm lens used for this photo so the minutes of arc per pixel will be .3757'.

    (2512.5px) .3757' = 15* 43.9' (Sun center). Ho= 15* 38.0' (dip -2.5' ref -3.4')

    The standard sin cos formula can now be used to solve for Meridian angle (t) = 58* 33.5'
    Dec= 15* 11'
    Lat= 34* 9.8' N
    Hc= 15* 38.0'

    Convert t= 58* 33.5' to time = 3 hrs 54 min 14 sec. add to time of meridional passage (nearest minute) at photographers longitude (119* 13.8' W) 12 hrs 41 min to get time of photo 16 hrs 35 min 14 sec PDT(ZD+7). Actual time of photo is 16 hrs 35 min 11 sec PDT.

    So it looks like the Sun's rays will provide a reasonably accurate Hs altitude by triangulation.

    Using the vertical angle of the island peak to get minutes of arc per pixel is going to give a slightly higher number but good enough for an approximate time of photo.

    Greg Rudzinski

    [NavList] Re: Azimuth by Sun Ray
    From: waldendand---com
    Date: 4 Nov 2011 17:37

    can you find the altitude of the sun using the known height of the island and the intersect of the rays an get approximate time?
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