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    Re: dip, dip short, distance off with buildings, etc.
    From: Bill B
    Date: 2006 Jan 9, 23:20 -0500

    Frank wrote:
    
    > You can calculate dip or the altitude of a tall building peeking up from
    > beyond the horizon using straight Euclidean geometry and trigonometry
    > ignoring
    > refraction completely. Then to include refraction, you simply change the
    > radius of the Earth from R to R/(1-x) where x depends on the  temperature
    > gradient
    > of the atmosphere. On average it's about 0.15 but it can  easily be anywhere
    > in the range 0.13 to 0.17 and sometimes it's as low as 0 or  as high as 1.0
    > (temperature inversions yield higher values of x).
    
    Ed.  Frank suggest this approximates the data from Bowditch "coastal
    piloting" tables.
    
    Later Frank wrote:
    
    "... That is, if I fire a  beam of light
    horizontally (or even at some significant angle away from  horizontal) from
    my apartment in Chicago, when it reaches an observer in Gary, Indiana 25
    miles way, its direction will have rotated downward, away from a  straight
    line trajectory, by an angle that is directly proportional to the distance
    traveled. On average, the constant of proportionality is about 0.15 minutes
    of arc per nautical mile. Note that this is really a dimensionless result:
    it's 0.15 arcminutes bending per 1.0 arcminute traveled as measured from
    the center of the Earth...."
    =====================================
    This get interesting, especially the path Frank is leading me down.  Using
    Frank's 0.15 figure for x in R/(1-x) to approximate Bowditch, predicted lift
    at 20.55 nm would be 66 feet.  By my calculations with Bowditch, approx. 64
    ft lift.
    
    NOTE: I am using 21600 ft as earth's circumference, and 6076.11 ft per nm in
    calculations. Also right triangles.  As Frank noted, the difference is
    insignificant when using a right vs oblique triangle.
    
    Using Frank's later "real world" mean refraction of 0.15' rise per minute of
    distance, predicted lift due to refraction for 20.55 nm would be 3.08
    minutes or 0d 03' 05".
    
    Using the formula for 700 nanometer light in a vacuum: Distance in meters ^2
    * 2.55 * 10^-8, lift would be 121 ft.
    
    LIFT AT 20.55nm
    64 ft = 0d 01' 46"
    66 ft = 0d 01' 49"
    112 ft = 0d 03' 05"
    121 ft = 0d 03' 20"
    
    It would appear Bowditch significantly underestimates lift due to refraction
    by today's standards.
    
    Now the above is all well and good if we know the distance etc., but finding
    distance from an object whose base is below the horizon was the initial
    challenge.  That leaves us with the Bowditch formula for table 15:
    
    Distance in nm = square root ((tan A /0.0002419)^2 +((H-h) / 0.7349))) -
    tan A / 0.0002419
    
    I do not pretend to understand the geometry behind the formula at this point
    (anymore than I could duplicate the Mona Lisa with cans of dayglow orange
    and dayglow yellow spray paint), but looking at patterns is interesting:
    
    Tan A / 0.0002419 = tan A * 4134 = tan A / (R/(1 - 0.1684))
    
    Converting the second term to nautical miles where H and h were in feet:
    
    H and h in feet:   (H-h) / 0.7349
    H and h in nm:    (H-h) * 8268 = (H-h) * 2 * 4134
    
    Remember good old 4134 from the first term?  3438 / (1 - 0.1864).
    Coincidence?  (Radius = 3438 nm)
    
    I am most likely oversimplify the problem (when all you have is a hammer the
    whole world looks like a nail), but can we bring the Bowditch formula up to
    speed by solving for a new value of x so the calculation uses actual mean
    refraction as we currently understand it, therefore giving a more accurate
    distance?
    
    Bill
    
    
    

       
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