1

I have the longitude and latitude of a known point O(x,y) where x and y are obtained from transforming the longitude and latitude from EPSG 4326 to EPSG 32635. This is the observer point. I have another point A(x1, y1) whose coordinates I don't know, but I know the distance from O to A in meters and the azimuth from O to A in radians.

How can I compute the coordinates of A knowing only these?

A function in Qgis in C++ would be the most useful, or at least a mathematical formula would be great.

1

If QGIS Field calculator is an option;

enter image description here

INPUTS

  • x_32635: your input x-coordinate on EPSG:32645
  • y_32635: your input y-coordinate on EPSG:32645
  • d_m: your input distance in meters
  • az_rad: your input azimuth in radians (nb. clockwise from north)

Calculation

x_new (x-coordinate of new point on UTM35N)

x(translate(geom_from_wkt('POINT('||"x_32635"||' '||"y_32635"||')'),
            "d_m"*sin("az_rad"), "d_m"*cos("az_rad")))

y_new (y-coordinate of new point on UTM35N)

y(translate(geom_from_wkt('POINT('||"x_32635"||' '||"y_32635"||')'),
            "d_m"*sin("az_rad"), "d_m"*cos("az_rad")))

x_4326 (longitude of new point)

x(transform(translate(geom_from_wkt('POINT('||"x_32635"||' '||"y_32635"||')'), 
                       "d_m"*sin("az_rad"), "d_m"*cos("az_rad")), 
            'EPSG:32635', 'EPSG:4326'))

y_4326 (latitude of new point)

y(transform(translate(geom_from_wkt('POINT('||"x_32635"||' '||"y_32635"||')'), 
                       "d_m"*sin("az_rad"), "d_m"*cos("az_rad")), 
            'EPSG:32635', 'EPSG:4326'))
0

Thank you everyone for your suggestions. I found the solution to be

y1 = asin(sin(y1) * qCos(d/R) + cos(y) * sin(d/R) * cos(a));

x1 = x + atan2(sin(a) * sin(d/R) * cos(y), cos(d/R) - sin(y) * sin(y1));

where R is the radius of Earth , d is the distance OA and a is the azimuth.

  • 2
    This is only an approximation because the Earth is not a perfect sphere. – AndreJ Nov 14 '18 at 7:23
  • Besides that, the distance in that formula is supposed to be ellipsoidal, not planimetric. – Gabriel De Luca Jan 14 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.