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    Re: Timing Upper Limb Sunset by Camera
    From: Bill B
    Date: 2015 Apr 16, 13:35 -0400

    On 4/8/2015 2:01 PM, Greg Rudzinski wrote:
    ...The NC-77 uses the
    > Radau model. Not sure what the Palm Pilot uses. USNO doesn't provide 0°
    > refraction data :(  Hard to say what qualifies as good on these
    > sunset horizon sights..
    
    To finish up on "Find Greg:"
    
    Attached was my final go at the exercise. It is "wishful thinking." I
    placed an AP in the center of brackets that would be the upper and lower
    limits of rounding fractional GPS lat and lon up or down to whole tenths
    yielding N34 03!2 W118 57!7 displayed on your smartphone. (In essence
    anything within a box plus or minus 0!05 lat and lon from the smartphone
    yet on the highway, closely approximating the center point of a 0!05 COP
    intersecting the highway.) There is evidence to support a possible HE of
    55' or greater assuming the eye is 8' above the road when seated on a
    Greyhound on that stretch of road.
    
    Radau model? Had to Google that. Above my pay grade. :-)
    
    For refraction I used -49!7, the refraction average from being on the
    cusp of NA -49!8 and -49!6.
    
    What qualifies as a good sunset horizon sight? Approximately 1 nm from
    the AP on a bus with a camera sounds damned good to me. Nice shooting!
    
    When I started cel nav with HO229 I was trying to figure out how to use
    the tables for a UL sunset with binoculars. I turned to the group for
    help. The consensus was it was mostly fruitless as changes in refraction
    could alter the event by minutes, and the upper limb would be
    extinguished by the atmosphere so not visible at the moment it hit the
    horizon.
    
    Hypothetically if refraction was in fact -49!7 and your observation time
    was approx 5 seconds early,  then (all other things being equal) it
    would amount to an intercept increase of approx. 1!0 towards from AP.
    That would move the LOP up the coast from the AP (west and/or north when
    declination is N), which is exactly what I see.
    
    
    
    
    
    
    
    
    
    

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