【发布时间】:2015-12-16 10:10:34
【问题描述】:
考虑以下函数来计算two points之间的距离
CREATE FUNCTION CoordinateDistanceMiles(
@Latitude1 float,
@Longitude1 float,
@Latitude2 float,
@Longitude2 float
)
RETURNS float
AS
BEGIN
-- CONSTANTS
DECLARE @EarthRadiusInMiles float;
SET @EarthRadiusInMiles = 3963.1
DECLARE @PI float;
SET @PI = PI();
-- RADIANS conversion
DECLARE @lat1Radians float;
DECLARE @long1Radians float;
DECLARE @lat2Radians float;
DECLARE @long2Radians float;
SET @lat1Radians = @Latitude1 * @PI / 180;
SET @long1Radians = @Longitude1 * @PI / 180;
SET @lat2Radians = @Latitude2 * @PI / 180;
SET @long2Radians = @Longitude2 * @PI / 180;
RETURN Acos(
Cos(@lat1Radians) * Cos(@long1Radians) * Cos(@lat2Radians) * Cos(@long2Radians) +
Cos(@lat1Radians) * Sin(@long1Radians) * Cos(@lat2Radians) * Sin(@long2Radians) +
Sin(@lat1Radians) * Sin(@lat2Radians)
) * @EarthRadiusInMiles;
END
以及以下使用地理类型的简化版本:
CREATE FUNCTION [dbo].[GetDistanceInMiles]( @lat1 FLOAT , @lon1 FLOAT , @lat2 FLOAT , @lon2 FLOAT)
RETURNS FLOAT
AS
BEGIN
DECLARE @result FLOAT;
DECLARE @source GEOGRAPHY = GEOGRAPHY::Point(@lat1, @lon1, 4326)
DECLARE @target GEOGRAPHY = GEOGRAPHY::Point(@lat2, @lon2, 4326)
SELECT @result = @source.STDistance(@target) / 1609.344
RETURN @result
END
当我跑步时
SELECT dbo.CoordinateDistanceMiles(50.73521,-1.96958,50.75822,-2.07768)
它返回 4.99171837612563
然而
SELECT dbo.GetDistanceInMiles(50.73521,-1.96958,50.75822,-2.07768)
返回 5.0005149496216
我得到的结果彼此略有不同。谁能解释一下
- 以上哪个函数更准确?
- 我怎样才能让它们返回相同的结果?
【问题讨论】:
-
数字非常接近。也许sqlgeorraphy没有假设地球是一个完美的球体?
标签: sql sql-server tsql sql-server-2012 sqlgeography