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    Re: sight reduction with GPS receiver
    From: Frank Reed CT
    Date: 2005 Mar 21, 20:06 EST
    Herbert you wrote:
    "Reflecting the economic reality of modern s/w production for the end
    user market, the small importance of the issue is reflected in the want
    of documentation rather than quality of implementation. The manuals that
    come with my various Garmins just speak of "distance", never ever
    specifying what kind. Simple tests with models 45, 12, 12XL reveal that
    they all display distance along the geodesic line on the chosen
    reference ellipsoid For consistency, I would expect no less from a unit
    promising 15m positioning accuracy."
     
    It truly amazes me that the people making these devices would bother with this issue. But you're right, it's the "economic reality" that drives the outcome.
     
    And:
    "To achieve the 4 digit accuracy on a Garmin display, an approximative
    formula is sufficient. Even a full numerical integration would not be a
    big deal for an infrequent computation such as this. I see no reason not
    to do it right."
     
    And yet, there is no "right" distance. Great circle distances are only relevant for purposes of comparison. Corrected ellipsoidal differences wouldn't seem to offer any practical advantage, but, since the devices have spare memory and spare computing power, why not "do it right"? Thinking on it more, the only unambiguous "right" distance is the straight-line distance (through the Earth).
     
    And:
    "Paul Hirose obviously has a unit that displays GCD. I would like to hear
    which model it is and whether it does this automatically or has to be
    coerced into doing it. I have a faint memory of my first Magellan (as
    big as a shoe box) being able to display GCD."
     
    I would bet that it depends on the age of the receiver (just speculating).
     
    Maybe you can help me with something here. Years ago, I worked this out as puzzle. I posted some results on the maps-l list back in 1997 (ouch... time flies). A copy follows below. I've never had any confirmation that my approximate expressions are accurate (or how accurate they are). Here's a copy:
    >>>>>
    Step 0) Use e = 1/298 for the flattening of the Earth.
    Step 1) Convert the latitudes to co-latitudes and all angles to
    radians as usual:
       phi1' = pi*(90-lat1)/180
       phi2' = pi*(90-lat2)/180
       theta = pi*(lon2-lon1)/180
    Step 2) Calculate adjusted colatitudes:
       phi1 = phi1'+(3/4)*e*sin(2*phi1')
       phi2 = phi2'+(3/4)*e*sin(2*phi2')
    Step 3) Get the standard great circle angular distance:
       a' = arccos[cos(phi1)*cos(phi2)+sin(phi1)*sin(phi2)*cos(theta)]
    Step 4) Calculate the geodesic distance on the ellipsoid:
       a = a'*[1+(1/2)*e*(sin(phi1)*sin(phi2)*sin(theta)/sin(a'))^2]
    Step 5) Convert to miles
       D = 3955.5*a.
    <<<<<
    What do you think?
     
    -FER
    42.0N 87.7W, or 41.4N 72.1W.
    www.HistoricalAtlas.com/lunars
       
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