In SQL 2005, the data type geography does not exist. If you take into account the real form of the earth, this becomes complicated (in addition, there are many reference systems in which you may want to interpret the coordinates.

Simplifying everything to the fact that the earth was a sphere (which it is not, and therefore the result if you implement this response will be only an approximation of reality), you could do something like this, since in the end the calculation is not more than a little trigonometry:

```
create function deg2rad(@deg float) returns float
as
begin
return @deg * 3.1415926536 / 180;
end
go
create function DistanceFromLatLonInKm (@lon1 float, @lat1 float, @lon2 float, @lat2 float) returns float
as
begin
declare @R float = 6371; --radio de una _tierra_ esférica en Km.
declare @dLat float = dbo.deg2rad(@lat2 - @lat1);
declare @dLon float = dbo.deg2rad(@lon2 - @lon1);
declare @a float;
set @a = sin(@dLat / 2) * sin(@dLat / 2)
+ cos(dbo.deg2rad(@lat2)) * cos(dbo.deg2rad(@lat2)) * sin(@dlon / 2) * sin(@dlon / 2);
declare @c float = 2 * atn2(sqrt(@a), sqrt([email protected]));
declare @d float = @R * @c;
return @d;
end
go
```

With these defined functions, you would get this result:

```
select dbo.DistanceFromLatLonInKm(0, 51.5, -3, 56);
534.04509287162
```