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I have a segment defined by 2 LatLon points(A and B) and a third LatLon point (C). I need an alghorithm, formulae, or some code to find the projection point of C on the segment AB (call this point D) so that CD segment is perpendicular to AB...

i few word I've to find the minimum distance between C and AB segment or in other word the height of the triangle ABC.

I'found this question: Find the nearest point on a line and this reply: http://gis.stackexchange.com/a/39176/16839

I've translated the algorithm in Java (GeoPoint is a simple implementation LatLon class and VectorN is a class that mantain an array of doubles and implement basic vector operation like dot product, sum etc...) but it don't seems to work well...

My source for Geopoints coordinates is googleMap... so they are in WGS84 system... and this may be the source of problems... but in my debug I found that results are very inconsistent... a point that stay at 30meters is given as staying at 0.5meters and a point at 18meters is given as 3m away...

Can you help me?

public static GeoPoint GetOrthographicProjection(GeoPoint start, GeoPoint end, GeoPoint offsetpoint) {
        double EarthRadius=6371;


    // Start node conversion
    double sLat = degToRad(start.getLat());
    double sLon = degToRad(start.getLng());

    double sX = EarthRadius * Math.cos(sLat) * Math.cos(sLon);
    double sY = EarthRadius * Math.cos(sLat) * Math.sin(sLon);
    double sZ = EarthRadius * Math.sin(sLat);

    //End node conversion
    double eLat = degToRad(end.getLat());
    double eLon = degToRad(end.getLng());

    double eX = EarthRadius * Math.cos(eLat) * Math.cos(eLon);
    double eY = EarthRadius * Math.cos(eLat) * Math.sin(eLon);
    double eZ = EarthRadius * Math.sin(eLat);

    //Offsetpoint conversion
    double  oLat = degToRad(offsetpoint.getLat());
    double  oLon = degToRad(offsetpoint.getLng());

    double  oX = EarthRadius * Math.cos(oLat) * Math.cos(oLon);
    double  oY = EarthRadius * Math.cos(oLat) * Math.sin(oLon);
    double  oZ = EarthRadius * Math.sin(oLat);



    // Create vectors
    VectorN p1 = new VectorN(new double[] {sX,sY,sZ});

    VectorN p2 = new VectorN(new double[] {eX,eY,eZ});

    VectorN o = new VectorN(new double[] {oX,oY,oZ});



    // Calculate
    /*Vector<Double> u = p2 - p1;
    Vector<Double> po = o - p1;
    Vector<Double> w2 = po - (u * Vector3.DotProduct(po, u) / Math.Pow(u.Magnitude, 2));
    Vector<Double> point = o - w2;*/

    VectorN u = p2.minus(p1);
    VectorN po = o.minus(p1);
    VectorN w2 = po.minus( (u.times(po.dot(u)).times(1 / Math.pow(u.magnitude(), 2)))  );
    VectorN point = o.minus(w2);

    // Convert to latlon
    double rlat = radToDegrees(Math.asin(point.cartesian(2) / EarthRadius));
    double rlon = radToDegrees(Math.atan2(point.cartesian(1), point.cartesian(0)));

    return new GeoPoint(rlat, rlon);

  }

Example of result: gp1= 40.8880224061686 , 17.391275949776173 gp2= 40.888020378444516 , 17.39116732031107

output point (that should stay on the segment gp1-gp2) 40.88801433354826 , 17.390843482462408

On gMaps you can see the point (gp1 and gp2 are blue, the output is the red marker... it should be between blue markers... instead is far away...) see this pic: http://i.stack.imgur.com/bBMd6.png

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One comment is that Google Maps uses Web Mercator (EPSG:3857) rather than WGS 84. You should try using that projection and see if that helps. –  Jay Guarneri Apr 4 '13 at 18:02
    
Perhaps i can't actually find a proof that gMaps user WGS84, what you say seems strange to me as the formulae i've used in my software to calculate distance between 2 point use WGS84 system and for various other features implemented by my software that mix and remix data i can assert that results don't suffer of any problem... eg. if i take 2 point from gMaps A and B, calculate distance, than calculate another point C at the same distance in the same direction of B... Well, B match perfectly with C graphically and numerically (difference or error is 0.0) –  Alex Apr 4 '13 at 22:11
    
The semi-major axis for WGS84 is 6378 (6,378,137m) –  Mintx Apr 4 '13 at 23:00
    
Do you think this may solve the problem? Anyway guys, please, if you know someone can help me with a definitive solution, share the link to this question! ;-) –  Alex Apr 5 '13 at 9:43
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1 Answer 1

See my answer to this problem at

WGS point to WGS line segment (great circle) distance

The paper referred to there has now been published and is available at

http://dx.doi.org/10.1007/s00190-012-0578-z

(no subscription required).

Code to do this problem (using GeographicLib) is available at

http://sourceforge.net/p/geographiclib/discussion/1026621/thread/21aaff9f

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