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    Re: Dip and the refractive invariant
    From: Bruce J. Pennino
    Date: 2013 Apr 19, 20:21 -0400
    
    Hello Paul, Frank and All:
     
    First, thank you Frank for pictures...very interesting.  In a week or two I'm going to call you so we can arrange to go to Tower Hill  and location in photos.  I want to get a little more data elsewhere before I visit you. 
     
    Next. Paul --- I've done 3 or 4 dip angle measurements with my Topcon Optical theodolite/EDM GTS 2B , averaging 8 to 10 angles in forward/normal view, then immediately doing 8-10  in the inverted/ reverse mode, always sighting a real reasonably crisp horizon.  To my utter amazement, the mean dip angles measured down from 90 or 270 , differ by  only 2 -3 seconds ON THE AVERAGE.  Realizing that the theodolite can be read directly to 6 seconds on scale, the index can be reasonably   interpreted(read)  to 2 or 3 seconds, my collimation error is the same as I can read the instrument.  Hard to believe, but I guess this is the "beauty" of "wrapping " and averaging  angles.  My reading and adjustment errors remain constant.  So I conclude, that all things considered, with a good crisp horizon, my error band is better than I first thought. Maybe on a good day, error band is +/- 5 seconds.   I believe the data, but still surprised.
     
    Paul,  I have one set of dip data that indicates a value of dip significantly higher than would be predicted in the NA or that I've measured elsewhere.  Interestingly, it is  a rushed data set at one of my lowest HoE.   The wind was blowing smog in my face from Long Island and NYC. I was sighting over Long Island Sound. Considering Frank's pictures and your analysis, how does smog influence refraction.  I would imagine I was in the same smog as the horizon.... your thoughts?

    Bruce
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
    ----- Original Message -----
    Sent: Friday, April 19, 2013 5:51 PM
    Subject: [NavList] Dip and the refractive invariant


    One way to calculate dip is to use the "refractive invariant", a
    principle of optics which says
    
    n * r * cos dip
    
    is constant at all points on a light ray. (Usually the expression
    uses sine of z, the zenith distance. I rewrote that to cosine of dip.)
    
    n = refractive index of the air = ca. 1.00028 at sea level
    r = distance from the center of Earth
    
    At the point where a ray is tangent to the sea, dip is zero. Since the
    cosine of 0 is 1, the invariant there is n * r. What is the refractive
    index n? At standard atmosphere conditions of 15° C and 1013.25
    millibars, it's 1.000277523. And at sea level r is simply Earth's
    radius, which I will call 6371000 meters. The refractive invariant n * r
    * cos dip at the tangent point (the sea horizon) is therefore
    1.000277523 * 6371000.
    
    That's also the value of n * r * cos a at the observer. Assume height of
    eye is 10 meters, and a standard atmosphere. Then refractive index at 10
    meters is 1.000277258. Equate the invariant at the horizon (left side)
    to the value at the observer:
    
    1.000277523 * 6371000 = 1.000277258 * (6371000 + 10) * cos dip
    
    rearranging the equation gives
    
    (1.000277523 / 1.000277258) * (6371000 / 6371010) = cos dip
    
    and solving on a calculator gives 5.55' = dip. This agrees with the
    almanac table for 10 m height of eye.
    
    Because both quotients are very close to 1, a more elegant and
    interesting solution is possible. Let's look at (6371000 / 6371010)
    first. After rewriting it in terms of Earth radius Re and height above
    sea level h, the expression is Re / (Re + h).
    
    Divide numerator and denominator by Re to get 1 / (1 + h/Re). It appears
    that does nothing but make the expression more complicated. But note
    that h/Re, the height of eye in Earth radii, is a tiny number. When you
    take the reciprocal of (1 + tiny number), the result is almost exactly
    (1 - tiny number). For example, try 1 / 1.001. Note that it's almost
    exactly .999. The closer to 1, the more accurate this approximation.
    
    What this means is that Re / (Re + h) simplifies to 1 - h, where h is
    height of eye in units of Earth radii.
    
    A similar technique can simplify the term (1.000277523 / 1.000277258).
    The numerator (index of refraction at the sea surface) we will call n0,
    and let dn be the increase in refractive index at the observer. In this
    example it's 1.000277258 - 1.000277523 = -2.65e-7 (that is, -2.65 times
    10 to the -7). Then the expression may be written as n0 / (n0 + dn). As
    before, when you divide both sides of the fraction by the numerator, the
    result is the reciprocal of (1 + tiny number), so the quotient of the
    refractive indices is practically 1 - dn. (Because the index of
    refraction at the observer is less than at the sea surface, dn is negative.)
    
    By simplifying both quotients in this manner, the equation becomes
    
    (1 - dn) * (1 - h) = cos dip
    
    Multiplying the terms on the left yields
    
    1 - dn - h + (dn * h) = cos dip
    
    where
    
    h = 10 m / 6371000 m = 1.57e-6
    dn = 1.000277258 - 1.000277523 = -2.65e-7
    
    The term (dn * h) is very small (about 4e-13) compared to the other two
    terms, so ignore it. Then the formula becomes
    
    1 - (dn + h) = cos dip
    
    Or in this example,
    
    1 - 1.31e-6 = cos dip
    
    Apply one more approximation. If the cosine of an angle is 1 - (tiny
    number), then the angle itself (in radians) is very close to √(2 * tiny
    number). So a good approximation of dip is √(2 * 1.31e-6) = .00162
    radians = 5.56', which agrees with the almanac table.
    
    Combining all the above, dip (radians) = √(2 * (dn + h))
    
    where dn is the difference in the refractive indices of the air at the
    sea surface and at the observer (negative if the latter is less), and h
    is the height of eye in earth radii.
    
    In minutes of arc, dip = 3438′ * √(2 * (dn + h)).
    
    Test that equation at 40 m height of eye. Assume standard atmosphere.
    Sea level refractive index = 1.000277523 as before. At 40 m it's
    1.000276462. The difference = dn = -1.06e-6. Height of eye = h = 40 /
    6371000 = 6.28e-6. Dip = 3438′ * √(2 * (-1.06e-6 + 6.28e-6)) = 11.1′.
    This agrees perfectly with the almanac table.
    
    In a vacuum, the refractive index is exactly 1 at any point, so dn
    equals zero. The formula simplifies to 3438′ * √(2 * h). That's the
    formula for geometric dip. In other words, the refractive invariant is
    true even when there is no refraction!
    
    
    REFERENCES
    
    I obtained refractive index values from this online calculator at NIST,
    which implements the Ciddor formulas:
    http://emtoolbox.nist.gov/Wavelength/Ciddor.asp
    For my computations I changed the wavelength of light to 550 nanometers
    (the approximate peak sensitivity of the eye) since the default is the
    red He-Ne laser wavelength.
    
    The calculator requires air pressure and temperature, which I obtained
    from this Javascript standard atmosphere calculator:
    http://www.luizmonteiro.com/StdAtm.aspx
    
    Andrew T. Young discusses the refractive invariant:
    http://mintaka.sdsu.edu/GF/explain/atmos_refr/invariant.html
    
    --
    I filter out messages with attachments or HTML.
    
    

    View and reply to this message: http://fer3.com/arc/m2.aspx?i=123613

       
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