Vincenty formula for distance between two Latitude/Longitude points

 

For the benefit of the terminally obsessive (as well as the genuinely needy), Thaddeus Vincenty (‘TV’) devised formulae for calculating geodesic distances between a pair of latitude/longitude points on the earth’s surface, using an accurate ellipsoidal model of the earth.

When I looked up the references (‘Direct and Inverse Solutions of Geodesics on the Ellipsoid with application of nested equations’), I discovered to my surprise that while the mathematics is utterly beyond me, it is actually quite simple to program.

Vincenty’s formula is accurate to within 0.5mm, or 0.000015″ (!), on the ellipsoid being used. Calculations based on a spherical model, such as the (much simpler) Haversine, are accurate to around 0.3% (which is still good enough for most purposes, of course).

Note: the accuracy quoted by Vincenty applies to the theoretical ellipsoid being used, which will differ (to varying degree) from the real earth geoid. If you happen to be located in Colorado, 2km above msl, distances will be 0.03% greater. In the UK, if you measure the distance from Land’s End to John O’ Groats using WGS-84, it will be 28m – 0.003% – less than using the Airy ellipsoid, which provides a better fit for the UK. (Since of course no one can travel on the surface of the theoretical ellipsoid, these differences are generally of no relevance at all).

Enter the co-ordinates into the text boxes to try it out (using deg-min-sec suffixed with N/S/E/W, or signed decimal degrees):

Lat 1: Long 1:

Lat 2: Long 2:

Note: nearly-antipodal points may fail to give a solution, in which case NaN is returned.

Vincenty wrote this at a time when computing resources were very expensive, and made it very computationally efficient. The JavaScript should be quite straightforward to translate into other languages, if required.

See the ‘Direct’ formula to calculate the destination point given the start point, bearing and distance.

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Vincenty’s formula as it is used in the script:

a, b = major & minor semiaxes of the ellipsoid
f = flattening (a−b)/a
φ1, φ2 = geodetic latitude
L = difference in longitude
U1 = atan((1−f).tanφ1) (U is ‘reduced latitude’)
U2 = atan((1−f).tanφ2)
λ = L (first approximation)
iterate until change in λ is negligible (e.g. 10-12 ≈ 0.06mm) {
	sinσ = √[ (cosU2.sinλ)² + (cosU1.sinU2 − sinU1.cosU2.cosλ)² ]	(14)
	cosσ = sinU1.sinU2 + cosU1.cosU2.cosλ	(15)
	σ = atan2(sinσ, cosσ)	(16)
	sinα = cosU1.cosU2.sinλ / sinσ	(17)
	cos²α = 1 − sin²α (trig identity; §6)
	cos2σm = cosσ − 2.sinU1.sinU2/cos²α	(18)
	C = f/16.cos²α.[4+f.(4−3.cos²α)]	(10)
	λ′ = L + (1−C).f.sinα.{σ+C.sinσ.[cos2σm+C.cosσ.(−1+2.cos²2σm)]}	(11)
}
u² = cos²α.(a²−b²)/b²
A = 1+u²/16384.{4096+u².[−768+u².(320−175.u²)]}		(3)
B = u²/1024.{256+u².[−128+u².(74−47.u²)]}		(4)
Δσ = B.sinσ.{cos2σm+B/4.[cosσ.(−1+2.cos²2σm) − B/6.cos2σm.(−3+4.sin²σ).(−3+4.cos²2σm)]}		(6)
s = b.A.(σ−Δσ)		(19)
α1 = atan2(cosU2.sinλ, cosU1.sinU2 − sinU1.cosU2.cosλ)		(20)
α2 = atan2(cosU1.sinλ, −sinU1.cosU2 + cosU1.sinU2.cosλ)		(21)

Where:

• s is the distance (in the same units as a & b)
• α₁ is the initial bearing, or forward azimuth
• α₂ is the final bearing (in direction p₁→p₂)

Note: Vincenty observes that eqn. (18) becomes indeterminate over equatorial lines (since cos²α → 0); in this case, set cos2σ_m to 0 and the result is computed correctly. He also points out that the formula may have no solution between two nearly antipodal points; an iteration limit traps this case (Vincenty says “this will occur when λ, as computed by eqn. (11), is greater than π in absolute value”, but this is not always a reliable test).
Note: some implementations of Vincenty’s formula inefficiently use a large number of trig functions; Vincenty devised this solution with an eye for efficiency in implementation, and this one uses just one each of sin, cos, sqrt, and atan2 for each iteration – only 3 or 4 iterations are generally required. [Formulation updated Dec 05 to make it closer to Vincenty’s original and computationally more efficient.]

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The most accurate and widely used globally-applicable model for the earth ellipsoid is WGS-84, used in this script. Other ellipsoids offering a better fit to the local geoid include Airy (1830) in the UK, International 1924 in much of Europe, Clarke (1880) in Africa, GRS-67 in South America, and many others. America (NAD83) and Australia (GDA) use GRS-80, functionally equivalent to the WGS-84 ellipsoid.

	WGS-84	a = 6 378 137 m (±2 m)	b ≈ 6 356 752.314245 m	f ≈ 1 / 298.257223563
	GRS-80	a = 6 378 137 m	b ≈ 6 356 752.314140 m	f = 1 / 298.257222101
	Airy 1830	a = 6 377 563.396 m	b = 6 356 256.910 m	f ≈ 1 / 299.3249646
	Internat’l 1924	a = 6 378 388 m	b ≈ 6 356 911.946 m	f = 1 / 297
	Clarke mod.1880	a = 6 378 249.145 m	b ≈ 6 356 514.86955 m	f = 1 / 293.465
	GRS-67	a = 6 378 160 m	b ≈ 6 356 774.719 m	f = 1 / 298.247167

Note that to locate latitude/longitude points on these ellipses, they are associated with specific datums: for instance, OSGB36 for Airy in the UK, ED50 for Int’l 1924 in Europe; WGS-84 defines a datum as well as an ellipse. See my convert coordinates page for converting points between different datums.

 

One eagle-eyed reader asked why a degree of latitude is longer at the pole than at the equator (111,694m vs. 110,574m), despite the radius being less.

The answer is that (geodetic) latitude is determined by the angle between the normal of the spheroid and the plane of the equator, as you can see on this exaggerated ellipse:

If the angles are taken from the centre of the Earth, the (geocentric) latitudes would get squashed up toward the poles – try drawing a 45º angle from the centre of this ellipse…

 

Aside for fellow geeks: it is often said that WGS-84 and GRS-80 are ‘functionally equivalent’. Well, I would concur that a 0.1 mm difference in Earth radius is pretty equivalent, but why any difference? Apparently, when defining WGS-84, the normalized second degree zonal harmonic gravitational coefficient C_2,0, derived from the GRS-80 value for the dynamic form factor J₂, was truncated to 8 significant digits in the normalization process!

Some of the terms involved are explained in Ed Williams’ notes on Spheroid Geometry.

Test case (from Geoscience Australia), using WGS-84:
	Flinders Peak	37°57′03.72030″S, 144°25′29.52440″E
	Buninyong	37°39′10.15610″S, 143°55′35.38390″E
	s	54 972.271 m
	α1	306°52′05.37″
	α2	127°10′25.07″ (≡ 307°10′25.07″ p1→p2)

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Notes:

• Trig functions take arguments in radians, so latitude, longitude, and bearings in degrees (either decimal or degrees/minutes/seconds) need to be converted to radians, rad = π.deg/180. When converting radians back to degrees (deg = 180.rad/π), West is negative if using signed decimal degrees. For bearings, values in the range -π to +π [-180° to +180°] need to be converted to 0 to +2π [0°–360°]; this can be done by (brng+2.π)%2.π [brng+360)%360] where % is the modulo operator.
• The atan2() function used here takes two arguments, atan2(y, x), and computes the arc tangent of the ratio y/x. It is more flexible than atan(y/x), since it handles x=0, and it also returns values in all 4 quadrants -π to +π (the atan function returns values in the range -π/2 to +π/2).
• If you implement any formula involving atan2 in Microsoft Excel, you will need to reverse the arguments, as Excel has them the opposite way around from JavaScript – conventional order is atan2(y, x), but Excel uses atan2(x, y). To use atan2 in a (VBA) macro, you can use WorksheetFunction.Atan2(). You will also have to rename the variables A, B / a, b, as VBA is case-insensitive.
• All bearings are with respect to true north, 0°=N, 90°=E, etc; if you are working from a compass, magnetic north varies from true north in a complex way around the earth, and the difference has to be compensated for by variances indicated on local maps.
• For miles, divide km by 1.609344
• For nautical miles, divide km by 1.852

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See below for the source code of the JavaScript implementation. These functions should be simple to translate into other languages if required.

I offer these formulæ & scripts for free use and adaptation as my contribution to the open-source info-sphere from which I have received so much. You are welcome to re-use these scripts [under a simple attribution license, without any warranty express or implied] provided solely that you retain my copyright notice and a link to this page.

If you would like to show your appreciation and support continued development of these scripts, I would most gratefully accept donations.

If you have any queries or find any problems, contact me at ku.oc.epyt-elbavom@oeg-stpircs.

© 2002-2010 Chris Veness

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/* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -  */
/* Vincenty Inverse Solution of Geodesics on the Ellipsoid (c) Chris Veness 2002-2010             */
/*                                                                                                */
/* from: Vincenty inverse formula - T Vincenty, "Direct and Inverse Solutions of Geodesics on the */
/*       Ellipsoid with application of nested equations", Survey Review, vol XXII no 176, 1975    */
/*       http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf                                             */
/* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -  */

/**
 * Calculates geodetic distance between two points specified by latitude/longitude using 
 * Vincenty inverse formula for ellipsoids
 *
 * @param   {Number} lat1, lon1: first point in decimal degrees
 * @param   {Number} lat2, lon2: second point in decimal degrees
 * @returns (Number} distance in metres between points
 */
function distVincenty(lat1, lon1, lat2, lon2) {
  var a = 6378137, b = 6356752.314245,  f = 1/298.257223563;  // WGS-84 ellipsoid params
  var L = (lon2-lon1).toRad();
  var U1 = Math.atan((1-f) * Math.tan(lat1.toRad()));
  var U2 = Math.atan((1-f) * Math.tan(lat2.toRad()));
  var sinU1 = Math.sin(U1), cosU1 = Math.cos(U1);
  var sinU2 = Math.sin(U2), cosU2 = Math.cos(U2);
  
  var lambda = L, lambdaP, iterLimit = 100;
  do {
    var sinLambda = Math.sin(lambda), cosLambda = Math.cos(lambda);
    var sinSigma = Math.sqrt((cosU2*sinLambda) * (cosU2*sinLambda) + 
      (cosU1*sinU2-sinU1*cosU2*cosLambda) * (cosU1*sinU2-sinU1*cosU2*cosLambda));
    if (sinSigma==0) return 0;  // co-incident points
    var cosSigma = sinU1*sinU2 + cosU1*cosU2*cosLambda;
    var sigma = Math.atan2(sinSigma, cosSigma);
    var sinAlpha = cosU1 * cosU2 * sinLambda / sinSigma;
    var cosSqAlpha = 1 - sinAlpha*sinAlpha;
    var cos2SigmaM = cosSigma - 2*sinU1*sinU2/cosSqAlpha;
    if (isNaN(cos2SigmaM)) cos2SigmaM = 0;  // equatorial line: cosSqAlpha=0 (§6)
    var C = f/16*cosSqAlpha*(4+f*(4-3*cosSqAlpha));
    lambdaP = lambda;
    lambda = L + (1-C) * f * sinAlpha *
      (sigma + C*sinSigma*(cos2SigmaM+C*cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)));
  } while (Math.abs(lambda-lambdaP) > 1e-12 && --iterLimit>0);

  if (iterLimit==0) return NaN  // formula failed to converge

  var uSq = cosSqAlpha * (a*a - b*b) / (b*b);
  var A = 1 + uSq/16384*(4096+uSq*(-768+uSq*(320-175*uSq)));
  var B = uSq/1024 * (256+uSq*(-128+uSq*(74-47*uSq)));
  var deltaSigma = B*sinSigma*(cos2SigmaM+B/4*(cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)-
    B/6*cos2SigmaM*(-3+4*sinSigma*sinSigma)*(-3+4*cos2SigmaM*cos2SigmaM)));
  var s = b*A*(sigma-deltaSigma);
  
  s = s.toFixed(3); // round to 1mm precision
  return s;
  
  // note: to return initial/final bearings in addition to distance, use something like:
  var fwdAz = Math.atan2(cosU2*sinLambda,  cosU1*sinU2-sinU1*cosU2*cosLambda);
  var revAz = Math.atan2(cosU1*sinLambda, -sinU1*cosU2+cosU1*sinU2*cosLambda);
  return { distance: s, initialBearing: fwdAz.toDeg(), finalBearing: revAz.toDeg() };
}

/* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -  */

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Usage:

<p>Lat 1: <input name="lat1" value="53 09 02N"> Long 1: <input name="long1" value="001 50 40W"></p>
<p>Lat 2: <input name="lat2" value="52 12 19N"> Long 2: <input name="long2" value="000 08 33W"></p>

<input type="button" value="calculate distance" 
    onClick="result.value = distVincenty(Geo.parseDMS(lat1.value), Geo.parseDMS(long1.value), 
                                         Geo.parseDMS(lat2.value), Geo.parseDMS(long2.value)) + ' m'">
<input name="result" value="">

Reference
http://www.movable-type.co.uk/scripts/latlong-vincenty.html