# Talk:Haversine formula

## Different results[edit]

I find it interesting that there are two 'clusters' of results around 2886.4 and 2887.26 with very few 'significantly' differing ones- Now: who is 'correct' here ??? and why are so many others 'wrong'??? --Walterpachl 19:08, 16 August 2012 (UTC)

2886 km (1793 miles) 2886.326609413624 2886.327 2886.4 km 2886.44 2886.44 kilometers (or 1793.55 miles) 2886.44444099822 2886.444442837984 2886.44444 2886.4 2887 km. 2887.2599 km 2887.2599506071097 2887.25995060711 2887.2599506071106 2887.2599506071106 2887.2599506071106 2887.259950607113 2887.25995060711 2887.26 km. 2887.26 km. 2887.260 km. 2887.2600 km 2887.26 2887.3 km (1794.1 mi.) 2887.3 km 2889.68

- This might be due to the different values given for the rough radius of the Earth, in kilometers, for different examples. --Paddy3118 19:22, 16 August 2012 (UTC)

Some reasons probably are:

- using a different (mean) radius of the earth
- using a derived radius from the earth's circumference
- use of difference precisions
- use of different formulas for trigonometric functions
- use of smaller numbers via radian (or degree) reduction/normalization.

-- Gerard Schildberger 19:25, 16 August 2012 (UTC)

Apparently it's the two different values used for the eartH's radius: 6372.8 and 6371.0, respectively- --Walterpachl 20:14, 16 August 2012 (UTC)

- Right, “6371.0” is the authalic radius based on/extracted from surface area, while “6372.8” is an approximation of the radius of the average circumference (i.e., the average “great-elliptic” or “great-circle radius”), where the boundaries are the meridian (≈ 6367.45 km) and the equator (≈ 6378.14 km).
- See
*Ellipsoidal quadratic mean radius*. ~Kaimbridge~ 17:44, 19 August 2012 (UTC)

- Thank you for the excellent explanation. I'd still want that explanation in the task description
- and would advise programmers to use one specific value in order to get comparable results
- Maybe the one yielding a result of about 2887.26 km as this is the majority. --Walterpachl 18:11, 19 August 2012 (UTC)

- It's probably too late to ask for an update to the tasks. We know what causes the differences and the two results that are obtained from using the two radii, and have noted it here. --Paddy3118 20:57, 19 August 2012 (UTC)

- It's never too late:-) I suggest to MENTION these radii in the task and RECOMMEND one for future implementers
- Was it you who said that it's difficult to create watertight task descriptions:-)
- Thanks anyway --Walterpachl 05:07, 20 August 2012 (UTC)
*Recommend a radius for future implementors*. Mention both and then recommend one sounds good as it would leave the examples already there alone or allow them to be updated. Or would we want to force all the examples to use the one radius? (Is it that much of a problem? --Paddy3118 07:26, 20 August 2012 (UTC)

- I have a sneaking suspicion that the fortran code c = 2*atan2(sqrt(a),sqrt(1-a)) is wrong. I can't prove it. Does anyone have a Fortran compiler handy?

I've change the recommended radius to the mean earth radius. This choice minimizes the RMS relative error and is consistent with the choice which conserves the volume and the area of the earth (in the limit of small flattening). The previous recommendation was the "quadratic mean radius". Unfortunately, its derivation contained a flaw in the way azimuths were sampled, so it does not minimize any reasonable error metric. For an extended discussion see this archived talk page for Great-circle distances on Wikipedia. cffk (talk) 22:48, 10 February 2014 (UTC)

- With 53 examples, it is too late to make most radical changes. Try appending a recommendation instead. --Paddy3118 (talk) 07:55, 11 February 2014 (UTC)

If you want to use a high-accuracy function for the *actual* distance on earth's geoid, you should probably look at the Frink entry. Frink's navigation library contains high-accuracy calculations of distances on earth's ellipsoid. These calculations are due to:

"Direct and Inverse Solutions of Geodesics on the Ellipsoid with Application
of Nested Equations", T.Vincenty, *Survey Review XXII*, 176, April 1975.
http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf

There is also a slightly higher-accuracy algorithm (if you want nanometer accuracy instead of sub-millimeter accuracy):

"Algorithms for geodesics", Charles F. F. Karney, *Journal of Geodesy*, January 2013, Volume 87, Issue 1, pp 43-55
http://link.springer.com/article/10.1007%2Fs00190-012-0578-z

In short, a most accurate distance on the Earth's geoid, given the WGS84 geoid, is:

2892.776957 km --Eliasen (talk) 07:35, 24 April 2022 (UTC)

- But at that point you're no longer using the Haversine formula. The task is not to compute the most accurate distance possible between two points on the surface of the Earth. -- Markjreed (talk) 22:38, 24 April 2022 (UTC)

- Additionally, these implementations are very close. The bc implementation with WGS84 ellipsoid¹ is guaranteed to have less than ¼% of error (0.21877% from your given number). This is “good enough” for something this easy to port to various programming languages that might not have enough functionality to support the “better” maths. 【¹) Though I switched to the Astronomical Almanac 2021 radius of i=6378136.600, x stays the same, in my “current” version, which is supposedly more precise. It’s very very close to WGS84, still.】 mirabilos (talk) 01:00, 25 April 2022 (UTC)