While studying about terrestrial magnetism, references were made to north direction, and the geographic meridian and later magnetic meridian defined using that.

But what is actually the north direction? Without it, it was difficult to perceive the magnetic field, declination angle, etc.

When a person on the earth's surface points north, what direction is he referring to? Please answer using less of geographic terms and more of geometrical terms.( that is why the wikipedia article didn't give me my answer).

P.S.:I could not find any other appropriate tags. Add some suitable tags, if any.


1 Answer 1


The Earth rotates around an axis. Denote the unit vector directed along this axis (with the direction determined by the right hand rule) as $\hat z$. This rotational axis is key in finding "north". I'll assume a nice simple geometrically shaped Earth, either a sphere or an ellipsoid, and I'll assume a person who is not standing on the north or south pole.

Any point on this surface has a plane tangent to the surface at that point. Define "down" as the unit vector normal to the tangent plane, pointing into the Earth. Call this vector $\hat d$. The cross product $\hat d \times \hat z$ points eastward (but is not a unit vector). Normalizing gives the "east" unit vector $\hat e$.

Finally, "north" is simply the cross product of the downward and eastward unit vectors, $\hat n = \hat d \times \hat e$. These three unit vectors form an orthogonal basis for local coordinates. One standard representation is north, east, down, which is a right handed system. Another is east, north, up, which is also right handed. A third system that is rapidly falling out of favor is north, east, up, which is a left-handed coordinate system.

Note that these vectors will point in slightly different directions assuming a spherical versus ellipsoidal Earth. The standard is to use an ellipsoid Earth, typically the WGS84 ellipsoid. What if you aren't on the ellipsoid? There are two points on the ellipsoid where the upward or downward projection of the $\hat d$ vector will intersect your location. One of those points is on the other side of the Earth. It's the closer one that is of interest. The height above the ellipsoid is your geodetic altitude. Now raise that local $(\hat n, \hat e, \hat d)$ triple to your position. The $\hat n$ vector is geodetic north.

A couple of images to help understanding

The following image uses $\hat \omega$ instead of my $\hat z$, $\hat x$ instead of my $\hat n$, $\hat y$ instead of my $\hat e$, and $\hat z$ instead of my $\hat d$. The concept is the same.

(source: what-when-how.com)

The following image depicts (greatly exaggerated) the difference between geocentric latitude (spherical Earth) and geodetic latitude (ellipsoidal Earth). Latitude is directly related to where down (or up) points.

  • $\begingroup$ Note that instead of doing the double cross-product you can simply project the rotational velocity $\vec{\omega}$ onto the tangent plane. $\endgroup$ Commented Jan 19, 2015 at 23:23
  • $\begingroup$ I would still have needed a cross product to get $\hat e$. Besides, I would have used the vector triple product rule for the projection because $\hat d \times (\hat z \times \hat d) = (\hat d \cdot \hat d)\hat z - (\hat d \cdot \hat z)\hat d$ by the vector triple product rule, and that reduces to $\hat z - (\hat d \cdot \hat z)\hat d$, which is the projection. $\endgroup$ Commented Jan 19, 2015 at 23:44
  • $\begingroup$ I find the projection operator a little bit more intuitive than the triple product. After thinking about it a little, doesn't your solution work in general on any surface, not just ellipsoids? $\endgroup$ Commented Jan 20, 2015 at 0:05
  • $\begingroup$ Sure, but the question is where North is. Geodesists are the one who decide what north, east, south, west, latitude, longitude etc. mean, and those definitions are all based on the ellipsoid Earth. $\endgroup$ Commented Jan 20, 2015 at 1:52
  • $\begingroup$ @DavidHammen, Since the rotate-axis of the earth changes all the time (like two wobbling tops), do you mean that the direction of North changes accordingly? $\endgroup$
    – Pacerier
    Commented Jan 21, 2018 at 18:09

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