Author 
Topic 
JamesTT
Starting Member
2 Posts 
Posted  20070313 : 18:41:27

I have a database that holds a postcode/zipcode the town and then the longitude and latitude, what I wanted to know is it possible to write a query to say If I wanted to get a list of postcodes that were in a radius of 4 miles or kilometers from say Postcode 4612 or the town of Abbeywood. Some thing tells me I will need to get my old trigonometry maths books out but if any one could give me some pointers that would be great.
4612 Abbeywood 151.61991 25.9974 2176 Abbotsbury 150.86645 33.87245 2046 Abbotsford 151.12888 33.84981


jezemine
Master Smack Fu Yak Hacker
2886 Posts 
Posted  20070313 : 19:44:40

if a bounding box was good enough (min/max lat/long), that would be trivial to implement in sql.
a true radius is more complicated.
www.elsasoft.org 


JamesTT
Starting Member
2 Posts 
Posted  20070313 : 19:50:43

Hi jezemine thank you for the reply I will go back and find out if a bounding box would be good enough but if not would I have to create a another table holding holding a list of all the post codes and there distrances between one another as I can see that would create a massive table. 


SwePeso
Patron Saint of Lost Yaks
30421 Posts 
Posted  20070314 : 01:54:41

If not correcting for Earth spherical (which only makes a difference of 0.515 % anyway) , use pythagoras theorem
SQRT(POWER(Lat1  Lat2, 2) + Power(Long1  Long2, 2))  This the distance in coordinates. Just multiply with some constant to get real distance in either mile or km.
gives the "real" distance to the other coordinates. But since you almost certainly want the nearest, get rid of the SQRT part (saves a lot a calculations) and use only
POWER(Lat1  Lat2, 2) + Power(Long1  Long2, 2)
to get the shortest "pseudodistance".
Peter Larsson Helsingborg, Sweden 


Kristen
Test
22859 Posts 
Posted  20070314 : 04:58:59

here's what we do:
1) Find the Min/Max Lat/Long of a bounding box. Select rows "within" that first. Use an index for Lat/Long. That will be quick!
2) For rows that are within the bounding box then use a Great Circle algorithm, or similar, to filter the bounding "Box" to a "Circle".
Assuming you are looking for proximity of Miles, rather than fractions of an inch!, sacrifice trigonometric accuracy for speed.
See also:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=40295  Great Circle Formula http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=57242  Accuracy v. Speed
Kristen 


Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)
7020 Posts 
Posted  20070314 : 07:26:16

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, @Logitude1 float, @Latitude2 float, @Logitude2 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: @Logitude1  Logitude in degrees of point 1 @Latitude1  Latitude in degrees of point 1 @Logitude2  Logitude 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( @Logitude1 ) set @lon2 = radians( @Logitude2 ) 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
CODO ERGO SUM 


codezilla94
Starting Member
6 Posts 
Posted  20071112 : 20:09:25

(spam removed graz) 


SwePeso
Patron Saint of Lost Yaks
30421 Posts 

