GPS: reconsidering logic behing this answer

Question

this is a follow up question that I think might help those who are wondering in details of Distance Computing: a follow up to this question : here The best answer gives us distance between 2 given point in Meter.

public static float distFrom(float lat1, float lng1, float lat2, float lng2) {
    double earthRadius = 3958.75;
    double dLat = Math.toRadians(lat2-lat1);
    double dLng = Math.toRadians(lng2-lng1);
    double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
               Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) *
               Math.sin(dLng/2) * Math.sin(dLng/2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    double dist = earthRadius * c;

    int meterConversion = 1609;

    return (float) (dist * meterConversion);
}

but my question starts here:

1. as commented under the chosen answer " The earth is not a perfect sphere, so this solution results in a slight error ". is this true? if so then how precise is it? ( in Meter )

2. In solution like this, I see different inputs as the earth radius from 6360 to 6379 KM. is this gap really excusable in output ?

Edit: Let's Imagine our two points are both located within a 5 KM distance.

Answer 1

(1) We can't really tell you the accuracy in meters, because it depends where on Earth the two points you're computing the distance between lie - e.g., for some formulas, accuracy is decent unless the points are approximately opposite eachother on the Earth, in which case accuracy diminishes quickly.

(Note: according to this, "So as long as we're assuming a spherical Earth, any single formula for distance on the Earth is only guaranteed correct within 0.5%" - this gives you an idea of how much error is involved.)

For the haversine formula, according to both Wikipedia and this page (which has a ton of useful information), the margin of error you're looking at is up to 0.55%, but depending on latitude it is typically as low as 0.3%. This means that, for two points 5 KM apart, the error should be approximately 15 M (again, depending on latitude and how good your approximation of the Earth's radius is).

(2) Whether or not the level of error you have is acceptable is entirely dependant on what you're doing with this code. It might be acceptable, or it might not; it depends.

For more information, read this page. There are a lot of ways to do this, with varying degrees of accuracy, and Wikipedia has a pretty good description of quite a few of them.

Answer 2

One thing in the code you posted is not very "stylish":

It defines the earth radius in miles, and later converts to meter.
This is not a good idea. Use meters, this is the corect SI unit. And it avoids one multiplication.

The exakt radius to use: Has not much influence (try it out). But correct would be to use the radius of the WGS84 reference ellipsoid (6378137 meter), because WGS84 most probably is the coordinate reference system your coordinates are related (from GPS on any (smart-)device).

Althought the posted formula was considered best in the post (Greater Circle), it is not (always). Typical application calculate small distances, like your 5km. For that the formula posted is not well suited, because it uses cos() of a small value which is numerically ill-conditioned.

For small distances the haversine is better suited, it avoids the cos() operation. But it is a bit slower, because it uses a sqrt() which is very slow.

About accuracy:
In typical applications there is no true real distance (e.g a road goes up, and down, on a 3d earth surface). So this has more influence than the accuracy of the said formulas.

So, yes for most application the accuracy is acceptable.

If you want it more precise, you can use Vicency's formula. It treats the earth as an ellipsoid and not as a sphere.

Answer 3

For an explanation of the problem on an ellipsoid, see the Wikipedia article Geodesics on an ellipsoid. Java code to compute distances accurately (within 15 nanometers) on the ellipsoid is available as part of GeographicLib.