I'm working with maritime sensors that give information of moving objects it detects, in azimuth, range and elevation, and we convert it to x,y coordinates for our application to be able to process it.

We found out that most naval systems consider the earth's center as the point of reference when doing position calculations, but since we needed x,y, we are considering that when our software starts, we note the position of our boat with reference to the earth's center and convert it to x,y coordinates, where the current position of our boat would be 0,0 (the origin).

The boat can move at upto 25 knots, so when we have traveled a good distance, we would convert all x,y coordinates in the software to radius, azimuth and elevation with respect to the earth's center and then convert it to current x,y position of the boat (with the current position being the origin of the x,y grid). Only problem would be, the ambiguity in position of all the moving objects that the sensor was tracking.

Two questions:

  1. Is this the best way to do it or is there a better way? We have to work with x,y coordinates.
  2. How many kilometers would a boat have to move so that the x,y coordinate system would become erroneous due to the earth's curvature?
  • 3
    To what precision? All projection of geographic coordinates will result in some distortion in scale, area, distance, and direction. How much depends on the projection. "Noticeable" and "erroneous" are relative terms. You might want to take a look at en.wikipedia.org/wiki/Horizon and en.wikipedia.org/wiki/Scale_(map). Particularly the latter's section titled Large-scale maps with curvature neglected. For your purposes, it sounds like an equidistant projection would be appropriate. Why x,y instead of polar?
    – Chris W
    Jul 25, 2015 at 18:46
  • Thank you very much. I was looking for accuracies of 10m. The large scale maps with curvature neglected section seems to have what I need. Couldnt use polar because the sensor calculations become more complex and innacurate for radial velocities. Please post as an answer and I'll mark it as the accepted answer.
    – Nav
    Jul 26, 2015 at 16:11
  • I think you need to clarify what you mean by terms like "radius, azimuth and elevation" and "the earth's center". You'd be surprised that they can have several different meanings. As it is, i don't understand your question.
    – Martin F
    Jul 27, 2015 at 18:49

1 Answer 1


"Erroneous" is relative to your accuracy needs. Let us therefore estimate the accuracy in terms of the distance the boat has traveled: by means of such a result, you can decide when positional calculations become "erroneous."

I understand the question as saying that mapping is carried out in an azimuthal equidistant coordinate system centered at the boat's origin. It is, in effect, using polar coordinates to chart the positions of the boat in terms of their distances from the origin and their angles made to some standard direction (such as true North) at that origin. The positions of all objects observed from the boat would be referenced to their bearings (relative to the boat's bearing) and (true) distances from the boat.

Over fairly large distances--perhaps up to a few thousand kilometers or more--the earth's shape is sufficiently close to a sphere that we shouldn't worry about the minor flattening of about one part in 300. This enables us to use simple calculations based on spherical geometry.

The calculations all involve three points: a sighted object's position A, the boat's position B, and the origin at C. These points form a triangle. Let the side opposite vertex A have length a and let the included angle at vertex A be alpha, with a similar convention using b, c, beta, and gamma for the other sides and angles. It is proposed to replace the true distance c between A and B, computed using the Spherical Law of Cosines via

S = cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma)

by distances computed in the mapped coordinates using the Euclidean Law of Cosines via

E = c^2 = a^2 + b^2 - 2ab cos(gamma).

(S gives distances in radians, which are converted to Euclidean distances upon multiplying by the effective radius of the approximating sphere. For a boat traveling 25 knots for a few hours, or even a few days, these distances will be small: 25 knots less than half a degree per hour, or about 0.0073 radians per hour. It would take almost six days to travel one radian.)

Let's consider the relative error of the distances,

r = Sqrt(E) / ACos(S) - 1,

in terms of some measure of the distances a and b between origin and the object or boat. Since the formulas are symmetric in a and b, we might as well take a greater than b and write b = ta where t is between 0 and 1. Expanding this ratio as a MacLaurin series in t through third order, and then replacing t by b/a, gives

r = b^4 sin(gamma)^2 / (6 * (a^2 + b^2 - 2ab cos(gamma)).

This expression depends on gamma (the angle between the boat and the object as seen from the origin). It is smallest when gamma is a multiple of 180 degrees: that is, the object and boat are along the same course. Then there is no error at all (because distances along lines radiating from the origin are correct). The relative error (which is always positive, by the way, proving the earth has positive curvature) is greatest at an intermediate angle where it can attain a value as large as b^2/6. This is a universal upper bound for the relative error, independent of the angle gamma. It will an excellent approximation provided b^4 is much smaller than b^2. This will happen whenever b is substantially smaller than 1 radian--around 57 degrees or over 3,000 nautical miles (nm).

The power of two in this simple formula is the crucial part: it shows that the maximum relative error in computing boat-to-object distances grows only quadratically with the closer of the two distances to the origin. We can make a simple table using only mental arithmetic, stopping our calculations at distances substantially smaller than one radian.

 Distance (nm)  Distance (radians)  Maximum relative error (%)
 -------------  ------------------  --------------------------
             1              0.0003                      1.4E-6
            10              0.003                       1.4E-4
           100              0.03                        0.014
          1000              0.29                        1.4

The relative error of 1.4% isn't bad even at 1000 nm (about 1850 km) out. Errors in computing object-to-object distances will be at most twice these amounts. Consequently, using this projection will be just fine up to distances where the relative errors are tolerable. In answer to question (1), for greater distances other forms of computation would be advisable.

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