I just had my mind blown in answer to my previous question. I was going to rework that question, but decided it makes more sense to ask a new one.

I had been converting between ECEF and lat/lon, and ran into an issue with Geoid vs Ellipsoid height. I was looking for an easy way to have an intuitive sense of position on the earth as well as elevation without having to worry about MSL, ellipses and imperfect models.

I would like to define a newLatitude and a newLongitude that are based on a tiny, perfect sphere located at the center of the earth. This sphere rotates with the earth just like ECEF does.

To define a location in this system you must have a triplet {newLat, newLon, Radius}. Using only {newLat, newLon} defines a line radiating from the center of the sphere out into space. You have to say how far along that line you are with the Radius. A radius of 200 meters is 200 meters from the center of the earth. The radius I'm sitting at right now is 6,499,100 meters, etc.

Then we don't have to worry about the average level of the tide for the last 19 years to figure out MSL. We simply use the same reference point for Radius.

Essentially, this is using spherical coordinates for something that only approximates a sphere, but that's fine, because we always have Radius.

To me, this seems to simplify a lot of the problems I have with different coordinate systems. Yes, sure, it leaves unanswered how to define the surface of the earth in this system, but that's where DTED information comes in.

Is this a feasible coordinate system that could be used at all locations on or near the globe without ambiguity? Are there any problems with this scheme (ignoring inertia of existing schemes and reluctance to change... I'm only asking about the validity of the system.)

  • One thing that could be a bit complicated (albeit not impossible) is to calculate the distance (on ground or on the sphere) between two points with different radius, and even more so if you also want areas.
    – Martin
    Commented May 23, 2014 at 13:27
  • 1
    Check out the Minute Physics: What is Sea Level? video. The issue is that in almost every local application of mapping up and down is defined by gravity, but stepping back to global models gravity complicates everything because it's not consistant. It's also worth noting that "down" is rarely pointed towards the center of the earth.
    – Weston
    Commented Jun 26, 2014 at 14:36
  • @Weston Thank you. That was a super informative video. I appreciate it.
    – kmort
    Commented Jul 1, 2014 at 20:12

1 Answer 1


I admire your enthusiasm but have to say you're not defining anything new.

The simplest geographic/geodetic models of the earth are perfect spheres/globes. The only "difference" between that simple old system and your "new" system being that heights, instead of measured relative to the surface of the fixed-radius (R) sphere/globe, are now relative to the center. You replace R + h with new R.

The problem with a simple sphere is that it is a relatively weak approximation of the geoid/earth. Better approximations are made with rotational ellipses, or ellipsoids. And they do not have constant radii. So to position points relative to the surface of an ellipsoid requires h, height above ellipsoid.

The problem with height above ellipsoid, as mentioned in the answer to your previous question, is the important matter of gravity. An equipotential surface (surface where gravitational potential is equal) is not exactly ellipsoidal. All people and all of nature that exist on earth relate to vertical directions and horizontal surfaces defined by gravity. Thus we like to position points relative to an equipotential surface. The one we use is called the geoid, or mean sea level, and the height, H, is called orthometric height, or height above sea level.

Use google to see many diagrams illustrating ellipsoid and geoid height.

  • Great answer by martin. I'd just like to add that if you do any distance/routing calculations on the sphere, there's a good chance you'll come up with the wrong number.
    – Mintx
    Commented May 23, 2014 at 18:49
  • @Mintx - Yes, that's zactly what i was going to say, then settled for "weak approximation"...
    – Martin F
    Commented May 24, 2014 at 15:30

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