PciCustomProjection

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    So...in delving deeper into the dark world that is projections:

    1) I found the following in the pci help (go figure):

    Note
    - If the "flattening" (F) is known for the earth model, it may be used to derive the eccentricity squared (E**2) by the following formula: E**2 = (2 * F) - (F * F) [Which is the same as F = 1.0 - sqrt(1.0 - E**2) ]

    It's this last line that is so important.

    2) If we can determine F, it makes everything simple. Fortunately, PCI has rearranged the equation...I had it figured out as:

    0.081816153 = sqrt[(2*(1/f))-((1/f)2)]

    I couldn't rearrange the above formula to isolate f, which was hardly the point. It was one of the components of f that I couldn't figure out.

    I had one radius, but as the formula for f (at least the only formula I have) was: f = (semiMajor - semiMinor)/semiMajor

    I was unsure if the radius that I was give was the semi-major (equitorial) or semi-minor (polar) axis. Though, it turns out that the radius that I was given was most likely the equitorial axis. In the end, we don't have to use the equation first listed in #2.

    3) Using the PCI equation from 1.

    E2 = e^squared    Where e=0.081816153
    E2 = .00669388289
    
    f = 1-sqrt(1-.00669388289)
    f = .00335256
    
    The normal way (aka the forward way) to derive f is: f = (semiMajor - semiMinor)/semiMajor so, if: f = .00335256 and semiMajor = 6378273

    then to get semiMinor it's:

    semiMinor = -1 * (.00335256*6378273 - 6378273)
    semiMinor = 6356586.8718
    
    There...everything you needed to know about defining your own projection. This was the Hughes 1980 ellipsoid - which is used for SSM/I ice concentration data and probably anything else that is collected by the DMSPs. I reran the values using the semi major and semi minor axes and everything came out kosher. I've added this to the eccentricity page within the "distance_calculations_inExcel3.xls" sheet that was passed round (as mentioned at the geomatics meeting), so I won't have to suffer this painful process again - at least not for this ellipse!

    Thanks to all who provided hints (unfortunately the Net wasn't working here, so I didn't get to the web for it - Adrian D'hont had an example which confirmed the radius I had was the major axis - which would of helped this process along).

    -- Slacker

     

     

     

     

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