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 Great Circle Distance Function - Haversine Formula

Author  Topic 

Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2007-03-28 : 22:43:31
This function computes the great circle distance in Kilometers using the Haversine formula distance calculation.

If you want it in miles, change the average radius of Earth to miles in the function.

create function dbo.F_GREAT_CIRCLE_DISTANCE
(
@Latitude1 float,
@Longitude1 float,
@Latitude2 float,
@Longitude2 float
)
returns float
as
/*
fUNCTION: F_GREAT_CIRCLE_DISTANCE

Computes the Great Circle distance in kilometers
between two points on the Earth using the
Haversine formula distance calculation.

Input Parameters:
@Longitude1 - Longitude in degrees of point 1
@Latitude1 - Latitude in degrees of point 1
@Longitude2 - Longitude in degrees of point 2
@Latitude2 - Latitude in degrees of point 2

*/
begin
declare @radius float

declare @lon1 float
declare @lon2 float
declare @lat1 float
declare @lat2 float

declare @a float
declare @distance float

-- Sets average radius of Earth in Kilometers
set @radius = 6371.0E

-- Convert degrees to radians
set @lon1 = radians( @Longitude1 )
set @lon2 = radians( @Longitude2 )
set @lat1 = radians( @Latitude1 )
set @lat2 = radians( @Latitude2 )

set @a = sqrt(square(sin((@lat2-@lat1)/2.0E)) +
(cos(@lat1) * cos(@lat2) * square(sin((@lon2-@lon1)/2.0E))) )

set @distance =
@radius * ( 2.0E *asin(case when 1.0E < @a then 1.0E else @a end ))

return @distance

end


Edit: corrected spelling


CODO ERGO SUM

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-03-29 : 08:44:40
Well done Michael!

Here is some more information about Haversine formula and also the average radius in miles (3956)
http://www.movable-type.co.uk/scripts/GIS-FAQ-5.1.html

It's interesting to see that the old Pythagorean Theorem is only a few percent wrong
quote:
less than 30 meters (100 ft) for latitudes less than 70 degrees
less than 20 meters ( 66 ft) for latitudes less than 50 degrees
less than 9 meters ( 30 ft) for latitudes less than 30 degrees


Haversine formula can error as much as 2 km (1 mi). But only under the context "half around the world", 20000 km / 12000 miles.


Peter Larsson
Helsingborg, Sweden
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SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-03-29 : 08:49:15
Just for the curiosity, here is Vincenty's formula (which is accurate to the millimeter)
http://www.movable-type.co.uk/scripts/LatLongVincenty.html

Haversine is much faster and accurate enough
quote:
Vincentyfs 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).



Peter Larsson
Helsingborg, Sweden
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Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2007-03-29 : 18:15:55
quote:
Originally posted by Peso

Just for the curiosity, here is Vincenty's formula (which is accurate to the millimeter)
http://www.movable-type.co.uk/scripts/LatLongVincenty.html

Haversine is much faster and accurate enough
quote:
Vincentyfs 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).



Peter Larsson
Helsingborg, Sweden



I looked into that before, and my head started hurting when I saw the formula.

In any case, there are so many other factors that introduce distance inaccuracy, such as difference in altitude between locations, that it seemed unlikely that it would even produce a more accurate result. Basically, the input data is likely to have so much inaccuracy built into it that anything more complex than the Haversine distance calculation is unlikely to produce better results. Garbage in, garbage out.


When I get to it, I want to post a follow-up on the method of calculating a square around the circle in order to limit the number of locations to be searched with an index lookup on longitude and latitude. For database applications, this is likely to have the most benefit. Basically you calculate a maximum and minimum longitude and latitude, and use that to limit the locations to be searched. Using this method, you eliminate most locations because they are outside the square, and use the Haversine function on the remaining locations. By chance, about 75% on the locations inside the square will be inside the search circle.

I did some work on this before, and found that calculating the east/west limits was much more complex than it may seem at first glance. The kilometers per degree along the latitude circle vary according to latitude. There is also an error that increases the further you are from the equator, because the distance east/west is shorter along the great circle distance, than just calculating the distance along the latitude circle. The magnitude of this error also increases as the size of the search circle increases. This means that you can leave out locations by making the square too small. More complications happen if your search circle is so large that one of the earth’s poles is inside the circle, or if the circle crosses the International Date Line.





CODO ERGO SUM
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SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-03-29 : 18:59:20
I did a test measuring the distance between the two cities Stockholm and Kiruna in Sweden.

The Haversine formula said the distance is 954.2 kilometers (593 miles).
Pythagorean Theorem said the distance is 957.7 kilometers (595 miles).

The error is only 0.35% between these two methods!
A mere 3.5 kilometer (2 mile) stray when travelling more than 950 kilometers (590 miles)!


Peter Larsson
Helsingborg, Sweden
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Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2007-03-29 : 20:38:36
Well, what I really want is a simple, accurate method to compute the min/max longitude to get the limits for the query. For performance, I think it is far more important to limit the number of points for which you do a computation, than to use a less CPU intensive calculation.

Instead of "How far away is this point?", I would like to find, "What are the longitudes that are exactly X number of distance units east or west of a given latitude/longitude using the great circle distance?"

I found the following formula in Wikipedia that may be what I need. Unfortunately, they failed to post a TSQL implementation, so I guess I have to do that myself.

http://en.wikipedia.org/wiki/Longitude
"As opposed to a degree of latitude, which always corresponds almost exactly to sixty nautical miles or about 111 km (69 statute miles, each of 5280 feet), a degree of longitude corresponds to a distance that varies from 0 to 111 km: it is 111 km times the cosine of the latitude, when the distance is laid out on a circle of constant latitude; if the shortest distance, on a great circle were used, the distance would be even a little less. More precisely, one degree of longitude = (111.320 + 0.373sin²ö)cosö km, where ö is latitude)."








CODO ERGO SUM
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jezemine
Master Smack Fu Yak Hacker

2886 Posts

Posted - 2007-03-31 : 16:13:37
minor typo in your function: @Logitude ought to be @Longitude.

that is unless you did it that way on purpose so it would line up nicely with @Latitude

also I agree Haversine will be plenty accurate enough for most all applications. You might notice a small discrepancy going from the equator north 90 degrees, as opposed to east or west 90 degrees. but even then the difference would small.

if the earth's period were shorter, like spinning at 100Hz or so, there would be more of a difference because it would be much more ellipsoidal. of course then we'd be too dizzy to care...



www.elsasoft.org
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Kristen
Test

22859 Posts

Posted - 2007-04-02 : 03:32:04
I've got these two links in my file of "Ready to add to FAQ", included here for completeness. The discussion is mostly about Accuracy v. Speed

http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=40295
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=57242

Kristen
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7695af9b
Starting Member

1 Post

Posted - 2007-10-24 : 04:03:28
What do you mean?
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AndrewMurphy
Master Smack Fu Yak Hacker

2916 Posts

Posted - 2007-10-24 : 07:26:38
"What do you mean?"....This must qualify as one of the more obscure questions ever posted here. To whom is this latest question posed? And to what are you referring?
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Kristen
Test

22859 Posts

Posted - 2007-10-24 : 07:50:53
not to mention the 6.5 elapsed months
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Kristen
Test

22859 Posts

Posted - 2007-11-02 : 09:15:08
"( 2.0E *asin(case when 1.0E < @a then 1.0E else @a end ))"

I'm wondering if this is a robust alternative to

2 * atan2(sqrt(a), sqrt(1-a))

because in the tests I've been doing this seems to be where I'm getting some error creeping in (i.e. comparing SQL's calculation above to a hand-calculation using atan2.

Obviously SQL doesn't have atan2(x, y) ...

Kristen
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Zoroaster
Aged Yak Warrior

702 Posts

Posted - 2007-11-02 : 10:08:15
quote:
Originally posted by 7695af9b

What do you mean?



42



Future guru in the making.
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SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-11-06 : 04:42:01
quote:
Originally posted by Kristen

Obviously SQL doesn't have atan2(x, y) ...

http://msdn2.microsoft.com/en-us/library/ms173854.aspx

ATN2



E 12°55'05.25"
N 56°04'39.16"
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Kristen
Test

22859 Posts

Posted - 2007-11-06 : 10:32:05
Would you Adam and Eve it? Its even in SQL2000 ... I wonder how I missed that :(

I'll retro fit that and see if it helps my accuracy problem. Thanks.

Kristen
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Van
Constraint Violating Yak Guru

462 Posts

Posted - 2007-11-06 : 10:50:18
In other news .9999999 repeating = 1
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SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-11-06 : 11:45:27
Try these constants for earth radius

km - 6366.70701949371
mi - 3956.0883313286096695299



E 12°55'05.25"
N 56°04'39.16"
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jezemine
Master Smack Fu Yak Hacker

2886 Posts

Posted - 2007-11-06 : 15:17:39
quote:
Originally posted by Van

In other news .9999999 repeating = 1



http://qntm.org/pointnine


elsasoft.org
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jezemine
Master Smack Fu Yak Hacker

2886 Posts

Posted - 2007-11-06 : 15:22:16
quote:
Originally posted by Peso

Try these constants for earth radius

km - 6366.70701949371
mi - 3956.0883313286096695299



E 12°55'05.25"
N 56°04'39.16"




those are some very accurate measurements! the last one has the radius of the earth to within a millionth of the size of a hydrogen atom!




elsasoft.org
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Kristen
Test

22859 Posts

Posted - 2007-11-06 : 17:17:16
Yeah, but by the time you've sum'd the squares of the other two sides to get the hypotenuse you're up to nearly a 100th of an electron ...
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Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2007-11-06 : 17:36:34
quote:
Originally posted by jezemine

quote:
Originally posted by Peso

Try these constants for earth radius

km - 6366.70701949371
mi - 3956.0883313286096695299



E 12°55'05.25"
N 56°04'39.16"




those are some very accurate measurements! the last one has the radius of the earth to within a millionth of the size of a hydrogen atom!




elsasoft.org




Yes, but that's gigantic compared to the size of the strings a hydrogen atom is (allegedly) made from.

CODO ERGO SUM
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