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    Re: Digital Image CN Exercise
    From: Marcel Tschudin
    Date: 2010 Apr 6, 15:09 +0300

    OK, Greg. I therefore proceeded with your Pixel scale of 0.097 Minutes
    of arc per Pixel corresponding to 0.001616667 degrees of arc per
    Pixel. The location obtained with the scale derived from the green
    disc will only be mentioned for reference.
    In the mean time the dip-error has been located and corrected. The
    formula was actually correct but only used to verify whether the
    object is above or below the visible horizon. I didn't remember that
    the results shown in apparent coordinates are provided relative to the
    astronomical horizon and not to the apparent one. The too high value
    which appeared in the output referred to the selected "standard
    location" which was different to those used for solving this problem.
    The rest seems now to be straight forward. The locations are shown in
    the attached screen shot of Google Earth and the location where the
    photo was taken in the NOAA chart.
    The horizon is claimed to be at 2468 Pixels below the UL (the image is
    actually cut at 2464 Px); the UL is therefore 3.989933333 degrees
    above the apparent horizon.
    At the given date and time:
    3 April 2010, 01:57:41 UTC
    the local meteorological conditions are found in the History data for Oxnard, CA
    Temp: 13.9°C and Press: 1017.2 hPa
    using these data, the apparent heights above the horizon are as follows:
    At Loc 0, the location initially provided (aLat=34°10'N,
    aLon=119°14'W, Height of eye 13 ft.= 4 m), the sun's position would
    have been:
    UL = 3.997120833 degrees above the horizon (instead of 3.989933333).
    The photo must have been taken slightly south of this location where
    the sun sets slightly earlier.
    At Loc 1 (34.274415 N, 119.288867 W) which has been found by trial and
    error the UL of the sun corresponds with sufficient accuracy to the
    one where the photo could have been taken:
    UL = 3.9896 degrees above the horizon (instead of 3.9899).
    Regarding the different locations mentioned here see attached file
    "_All LOCs.gif" and "_Chart.gif".
    Finding the off-shore oil rig:
    Measuring the distances in x-direction between centre of the sun to
    the oil rig in the original photo:
    Sun: Right limb = 744 Px, Left limb = 418 Px, the Centre is at 581 Px
    Oil rig (about):
    Right edge = 146 Px, Left edge = 52 Px, the Centre is at 99 Px
    Angular distance between sun centre and oil rig centre: 482 Px =
    0.779233333 degrees.
    Note: The azimuths are counted from South = 0 degrees to West = 90 degrees etc.
    The azimuth of the sun's centre is 93.90984 degrees (thus slightly
    north of west direction)
    The oil rig is about 0.779233333 degrees South of the sun's centre.
    The azimuth to the oil rig is therefore 93.13060667 degrees
    A line with this azimuth drawn from Loc 1 passes in about 14.2 km
    distance about 1.4 km south of the oil rig Gilda.
    By drawing a parallel line which crosses this oil rig one obtains at
    the coast the location "Loc Photo" which lies in the opposite
    direction of the initial Loc 0.
    So, something is still going wrong and I realise also what. My program
    calculates the apparent position of the centre of the sun. The
    refraction is calculated for this position to which I added half of
    the observed height of the flattened sun. This is likely to be
    slightly different from calculating the apparent position for the top
    of the sun. In an other part of my program I actually have provided
    the possibility to select a location relative to the body's shape from
    -1 (lower limb) to +1 (upper limb), but unfortunately not in this part
    where it would first have to be added.
    Regarding your "extra credit". The first could be done manually
    knowing the distance which is behind the apparent horizon. The second
    I have to pass since I'm not a bird watcher, I hardly can
    differentiate between a gull and a sparrow ;-)
    P.S: By using the different scale of 0.100 Minutes of arc per Pixel
    Loc 1 would be about 15 km further north, about 2.5 km after Ventura.



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