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Math involved in reverse geocoding

This is a discussion on Math involved in reverse geocoding within the MapPoint Desktop Discussion forums, part of the Map Forums category; hi all i have read the article about reverse geocoding, very intersing of course .. but i don't have fully ...

  1. #1
    bule is offline Junior Member Yellow Belt
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    Math involved in reverse geocoding

    hi all i have read the article about reverse geocoding, very intersing of course ..

    but i don't have fully understood some math involved in those algorithms:

    - what about the expression 1\60\1852 ? (distance in meter)*(1/1852) = distance in sea miles...and what about the multiplication for 1/60, what does it mean ?

    - offset for startingLongitude= radius*cos(angle) / cos( startingLongitude in radius)... what about the mening of the division for:
    cos( startingLongitude in radius) ?

    i hope you can help me ...
    thanks

  2. #2
    Winwaed's Avatar
    Winwaed is offline Mapping-Tools.com Black Belt
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    Re: Math involved in reverse geocoding

    I don't know the algorithm you're looking at, but the
    "distance in meter" * 1/1852 looks to be asimple conversion from metres to nautical miles.
    A statute mile is about 1600metres (to 2sf), and a nautical mile is a bit bigger.

    The first one be metres converted to degrees latitude? 60 probably isn't far from the number of nautical miles in a degree latitude; and I know the definition is a nice round number...

    Richard
    Winwaed Software Technology LLC
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    bule is offline Junior Member Yellow Belt
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    Re: Math involved in reverse geocoding

    I'am using the algorithm found in the article of Wilfried Mestdagh i have found in this site. The algorithm starts making your lat and longit as the center of concentric circumferences with incremental radius, in this circumferences it selects new values of lat and longit..

    so do you think that after the conversion from meter to nauticakl mile the multiplication for 1/60 is used to find the number of degree of nautical miles in a degree of latitude? mmm... i don't think this is it's purpose because the 1\60\1852*distanceInMeter is the length of the radius ...

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    Wilfried is offline Senior Member Black Belt
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    Re: Math involved in reverse geocoding

    Hi,

    1 arc minute is exacly 1 nautical mile. 60 arc minute is 1 degree. 1852 meter is also 1 nautical mile. So the calculation converts degree to meters.

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    bule is offline Junior Member Yellow Belt
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    Re: Math involved in reverse geocoding

    Thanks...

    but probably i am a big donkey...

    so: Distance(in what unit of measure?) = (1/60)*1852*(Distance(in what unit of measure?))....

  6. #6
    bule is offline Junior Member Yellow Belt
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    Re: Math involved in reverse geocoding

    ...finally i have the solution

    distanceInDegree= (1/60)*(1/1852)*distanceINMeter..is it right?

    now my only doubt is about

    - offset for startingLongitude= radius*cos(angle) / cos( startingLatin radius)... what about the mening of the division for:
    cos( startingLat in radius) ?

  7. #7
    Wilfried is offline Senior Member Black Belt
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    Re: Math involved in reverse geocoding

    Hi,

    Yes you right, it convert to meter, because 1852 meter is exacly 1 nautical mile.

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    Wilfried is offline Senior Member Black Belt
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    Re: Math involved in reverse geocoding

    Hi,

    The Cos(latitude) calculation is done to correct the longitude for the right distance. On the equator 1 minute longitude is exacly 1 minute latitude, so both ar 1 exact nautical mile. But on the poles every longitude calculation is null. And between poles and equator it is a value depending on the Latitude.

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    bule is offline Junior Member Yellow Belt
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    Re: Math involved in reverse geocoding

    thanks a lot, very useful...

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