Given point of (latitude,longitude), distance and bearing, How to get the new latitude and longitude

Question

I found a piece of code on web. It calculates the Minimum bounding rectangle by a given lat/lon point and a distance.

private static void GetlatLon(double LA         


        

Answer 1

This is a prime example of why commenting your code makes it more readable and maintainable. Mathematically you are looking at the following:

double ec = 6356725 + 21412 * (90.0 - LAT) / 90.0; //why?

This is a measure of eccentricity to account for the equatorial bulge in some fashion. 21412 is, as you know, the difference in earth radius between the equator and pole. 6356725 is the polar radius. (90.0 - LAT) / 90.0 is 1 at the equator, and 0 at the pole. The formula simply estimates how much bulge is present at any given latitude.

double ed = ec * Math.Cos(LAT * Math.PI / 180); // why?

(LAT * Math.PI / 180) is a conversion of latitude from degrees to radians. cos (0) = 1 and cos(1) = 0, so at the equator, you are applying the full amount of the eccentricity while at the pole you are applying none. Similar to the preceding line.

dx / ed //why?

dy / ec //why?

The above seems to be the fractional additions to distance in both the x and y directions attributable to the bulge at any given lat/lon used in the newLon newLat computation to arrive at the new location.

I haven't done any research into the code snippet you found, but mathematically, this is what is taking place. Hopefully that will steer you in the right direction.

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Haversine Example in C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double m2ft (double l) {            /* convert meters to feet       */
    return l/(1200.0/3937.0);
}

double ft2smi (double l) {          /* convert feet to statute miles*/
    return l/5280.0;
}

double km2smi (double l) {          /* convert km to statute mi.    */
    return ft2smi(m2ft( l * 1000.0 ));
}

static const double deg2rad = 0.017453292519943295769236907684886;
static const double earth_rad_m = 6372797.560856;

typedef struct pointd {
    double lat;
    double lon;
} pointd;

/* Computes the arc, in radian, between two WGS-84 positions.
   The result is equal to Distance(from,to)/earth_rad_m
      = 2*asin(sqrt(h(d/earth_rad_m )))
   where:
      d is the distance in meters between 'from' and 'to' positions.
      h is the haversine function: h(x)=sin²(x/2)
   The haversine formula gives:
      h(d/R) = h(from.lat-to.lat)+h(from.lon-to.lon)+cos(from.lat)*cos(to.lat)
   http://en.wikipedia.org/wiki/Law_of_haversines
 */
double arcradians (const pointd *from, const pointd *to)
{
    double latitudeArc  = (from-> lat - to-> lat) * deg2rad;
    double longitudeArc = (from-> lon - to-> lon) * deg2rad;

    double latitudeH = sin (latitudeArc * 0.5);
    latitudeH *= latitudeH;

    double lontitudeH = sin (longitudeArc * 0.5);
    lontitudeH *= lontitudeH;

    double tmp = cos (from-> lat * deg2rad) * cos (to-> lat * deg2rad);

    return 2.0 * asin (sqrt (latitudeH + tmp*lontitudeH));
}

/* Computes the distance, in meters, between two WGS-84 positions.
   The result is equal to earth_rad_m*ArcInRadians(from,to)
 */
double dist_m (const pointd *from, const pointd *to) {
    return earth_rad_m * arcradians (from, to);
}

int main (int argc, char **argv) {

    if (argc < 5 ) {
        fprintf (stderr, "Error: insufficient input, usage: %s (lat,lon) (lat,lon)\n", argv[0]);
        return 1;
    }

    pointd points[2];

    points[0].lat = strtod (argv[1], NULL);
    points[0].lon = strtod (argv[2], NULL);

    points[1].lat = strtod (argv[3], NULL);
    points[1].lon = strtod (argv[4], NULL);

    printf ("\nThe distance in meters from 1 to 2 (smi): %lf\n\n", km2smi (dist_m (&points[0], &points[1])/1000.0) );

    return 0;
}

/* Results/Example.
    ./bin/gce 31.77 -94.61 31.44 -94.698
    The distance in miles from Nacogdoches to Lufkin, Texas (smi): 23.387997 miles
 */

Answer 2

I assume 6356725 has something to do with the radius of the earth. Check out this answer, and also take a look at the Haversine Formula.