3D coordinates on a sphere to Latitude and Longitude

Question

I've got the following information:

There exists a sphere with origin (0,0,0) and radius R. After doing a ray-sphere intersection I know a point (XYZ) in 3D space that is on the sphere (the exact position in 3D space where the line pierces the sphere hull).

For my program I'd like to calculate the Latitude and Longitude of the XYZ point on the sphere, but I can't think (or Google) up a way to do this easily.

So in short, the function that I'm trying to write is this:

public static LatLon FromVector3(Vector3 position, float sphereRadius)
{
    return Latitude and Longitude
}

Does anybody know how to do this? As a reference this Wiki SVG file might be helpful:

Update:

Thanks for all the helpful answers, so in the end I went with this code:

 public static LatLon FromVector3(Vector3 position, float sphereRadius)
    {
        float lat = (float)Math.Acos(position.Y / sphereRadius); //theta
        float lon = (float)Math.Atan(position.X / position.Z); //phi
        return new LatLon(lat, lon);
    }

Now I've got to think of which answer helped me the most to accept :P.

Solution 1

I guess it should not be difficult to find the spherical polar coordinates from x,y,z (3d-coordinate system).

1. r is always constant if it's on surface.

   

2. (90 - θ) your latitude (negative means it's on the bottom) as it's measured from top.

   

3. φ is your longitude. (but not quite sure about longitude system)

   

Also check this diagram from wikipedia.

Solution 2

lat=atan2(z,sqrt(x*x+y*y))
lng=atan2(y,x)

Using formulas with atan2() is more convenient. You don't have to add/subtract pi/2 or care about sign issues in different quadrants or division by zero.

lat will be >0 in the northern hemisphere
lat will be <0 in the southern hemisphere
lng will be >0 in the eastern hemisphere
lng will be <0 in the western hemisphere

Solution 3

This helped using Javascript/THREE.js:

var lat = 90 - (Math.acos(y / RADIUS_SPHERE)) * 180 / Math.PI;
var lon = ((270 + (Math.atan2(x , z)) * 180 / Math.PI) % 360) -180;

Comments

[PKCLsoft]

I found that to get the lat/long to have the correct sign using SceneKit, I needed: float latitude = -((float)acosf(result.localCoordinates.y / sphereRadius) - M_PI_2); //theta float longitude = M_PI - ((float)atan2f(result.localCoordinates.z, result.localCoordinates.x));/ //phi if (longitude > M_PI) { longitude = longitude - (2.0 * M_PI); }

[Roy T.]

Wow, thanks for the neatly layouted answer, I'll test this right now. And if everything works out I'll accept it in a bit :).

[Roy T.]

I'm getting very strange results with these formula, so I'm afraid this doesn't seem to be the solution.

[Roy T.]

I'm using this a spherical coordinates for a perfect sphere (in a game) so I don't have to use real-world complexities :)

[Jaydee]

I realised that after posting, I really should read the question properly sounding off. I've done a fair bit of this in the past so I see lat/long and start pontificating.

[Roy T.]

Your latitude seems to work, but your longitude is very erratic

[Roy T.]

nvm, you are correct. I didnt look at your picture, you use Z = up, I used Y = up. Also you swapped lat and lon.

[Curd]

@Roy T.: probably depends on the definition which one of the two axis is x- and -y. Try it with x- and y-axis exchanged (You didn't specify the coordinate system exactly).

[Motionharvest]

Would you be willing to help me figure out the reverse of this. I have the lat, lon, and radius, and I'm trying to figure out the {x,y,z} of a THREE.Vector3.

[ppareja]

In my case I had to change the calculation of lon to the following: var lon = ((270 + (Math.atan2(x , z)) * 180 / Math.PI) % 360) -360;

[Donny V.]

What is the RADIUS_SPHERE = ?

[Donny V.]

Is it the earths radius? 6378140