Great Circle Distance Calculation
Oct 15, 2007
Great Circle distance calculation
Is there any stored procedure or application that implements Great Circle distance calculation
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Great Circle Distance Calculation
Oct 15, 2007Great Circle distance calculation
Is there any stored procedure or application that implements Great Circle distance calculation
Great Circle Distance Function - Haversine Formula
Mar 28, 2007This 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
Help W/ Distance Calculation Query
Mar 28, 2007I'm trying to run a dyncamic query that returns all records within a specific distance of a certain point. The longitude and latitude of each record is stored in the database. The query is constructed from two dynamic variables $StartLatitude and $StartLongitude with represent the starting point.
SELECT UserID, ACOS(SIN($StartLatitude * PI() / 180) * SIN(Latitude * PI() / 180) + COS($StartLatitude * PI() / 180) * COS(Latitude * PI() / 180) * COS(($StartLongitude - Longitude) * PI() / 180)) * 180 / PI() * 60 * 1.1515 AS Distance
FROM HPN_Painters
HAVING (Distance <= 150)
It runs fine until I add the 'HAVING (Distance <= 150)' clause, in which I recieve the error: Invalid column name 'Distance' It seems that Distance cannot be referenced in the HAVING clause.
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Jan 12, 2008I have a user defined function, I want to determine the distance between the 2 points. I have it working but i'm having a problem getting to print.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Code Snippetcreate function dbo.Distance( @lat1 float , @long1 float , @lat2 float , @long2 float)
returns float
as
begin
declare @DegToRad as float
declare @Ans as float
declare @Miles as float
set @DegToRad = 57.29577951
set @Ans = 0
set @Miles = 0
if @lat1 is null or @lat1 = 0 or @long1 is null or @long1 = 0 or @lat2 is
null or @lat2 = 0 or @long2 is null or @long2 = 0
begin
return ( @Miles )
end
set @Ans = SIN(@lat1 / @DegToRad) * SIN(@lat2 / @DegToRad) + COS(@lat1 / @DegToRad ) * COS( @lat2 / @DegToRad ) * COS(ABS(@long2 - @long1 )/@DegToRad)
set @Miles = 3959 * ATAN(SQRT(1 - SQUARE(@Ans)) / @Ans)
set @Miles = CEILING(@Miles)
return ( @Miles )
end
DECLARE @RC float
EXEC Distance '39.943762', '-78.122265', '32.334709', '-96.633546'
PRINT @RC /* in miles */
Help With Great Plains Query
May 31, 2006Greetings,
We currently use GP 8.00 with the SQL Server 8.0.
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The calculation runs fine, we multiply end quanity x componet quanity found on the BOM. But we are not getting the correct componet item numbers to display.
Here is the syntax we are using:
SELECT DISTINCT
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We have tried several different types of joins but still no luck. The upshot is we need to know that when a finished good is placed on hold, - how many of its componet parts are placed on hold.
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DECLARE @MyLatitude NUMERIC(9, 6), @MyLongitude NUMERIC(9, 6)
Set @Latitude = 42.329596;
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Jun 24, 2005See here www.merriampark.com/ld.htm for information about the algorithm. This page has a link (http://www.merriampark.com/ldtsql.htm) to a T-SQL implementation by Joseph Gama: unfortunately, that function doesn't work. There is a debugged version in the also-referenced package of TSQL functions (http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=502&lngWId=5), but this still has the fundamental problem that it only works on pairs of strings up to 49 characters.
CREATE FUNCTION edit_distance(@s1 nvarchar(3999), @s2 nvarchar(3999))
RETURNS int
AS
BEGIN
DECLARE @s1_len int, @s2_len int, @i int, @j int, @s1_char nchar, @c int, @c_temp int,
@cv0 varbinary(8000), @cv1 varbinary(8000)
SELECT @s1_len = LEN(@s1), @s2_len = LEN(@s2), @cv1 = 0x0000, @j = 1, @i = 1, @c = 0
WHILE @j <= @s2_len
SELECT @cv1 = @cv1 + CAST(@j AS binary(2)), @j = @j + 1
WHILE @i <= @s1_len
BEGIN
SELECT @s1_char = SUBSTRING(@s1, @i, 1), @c = @i, @cv0 = CAST(@i AS binary(2)), @j = 1
WHILE @j <= @s2_len
BEGIN
SET @c = @c + 1
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j-1, 2) AS int) +
CASE WHEN @s1_char = SUBSTRING(@s2, @j, 1) THEN 0 ELSE 1 END
IF @c > @c_temp SET @c = @c_temp
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j+1, 2) AS int)+1
IF @c > @c_temp SET @c = @c_temp
SELECT @cv0 = @cv0 + CAST(@c AS binary(2)), @j = @j + 1
END
SELECT @cv1 = @cv0, @i = @i + 1
END
RETURN @c
END
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