## Complicated Math question

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### Complicated Math question

I have an extremely complicated math question. I have a function that can transform the difference in latitude and longitude of two points into a distance in feet. It is a WGS84 to XY conversion. (WGS84 is the coordinate system used to measure lat and lon). This all works well. The thing is I want to be able to reverse it. I will know my starting lat and lon and how far away I want to go in feet, and then find out the lat and lon of the new point. Here is my function:

Code: Select all
`WGS84_to_XY(double curLat, double curLon, double origLat, double origLon,      double &x, double &y){  double coslat, sinlat, deltaLat, deltaLon, smallLat;  deltaLon = origLon - curLon;  deltaLat = curLat - origLat;  if (fabs(origLat) > fabs(curLat))    smallLat = origLat;  else    smallLat = curLat;  coslat = cos(smallLat*M_PI/180); //M_PI is 3.14159....  sinlat = sin(smallLat*M_PI/180);  x = -20925646.99508*(deltaLon*M_PI/180)*coslat*        (1/sqrt(1-0.00669437999014*sinlat*sinlat));  y = 20890581.85603409*(deltaLat*M_PI/180)       -(52619.77967973982*sin(2*deltaLat*M_PI/180))       +(55.22375743665538*sin(4*deltaLat*M_PI/180))       -(0.0715248258209759*sin(6*deltaLat*M_PI/180));}`

Even if I can get it to relatively good accuracy it would be helpful. This info came from inforamtion on the web. I essentially will have x, y, origLat, origLon. I want to solve for curLat, curLon. Lat is a lot easier to solve for. Just taking the circumference of the earth, about 24,859.82 miles, and divide by 360 degrees and you know how far one degree of latitude equals in miles. So essenentially all I need now is curLon, as I can caculate curLat. Thanks.
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OgreCoder

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Joined: Fri Oct 01, 2004 1:52 pm
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I will know my starting lat and lon and how far away I want to go in feet, and then find out the lat and lon of the new point.

You will also need a direction.

Darryl

Darryl

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