sql基于坐标计算距离的优化

如题，下面是最初版的sql,取与目标坐标150千米范围内的记录

select top 10 * from T_Temple a with(nolock) where dbo.[fnGetDistance](37.356976,116.8425,lat,lon)<150 order by a.Id asc

因为我们知道，如果要与目标坐标在150公里，所以经纬度应该在目标坐标点的正负2度范围内（粗略这么认为啊），所以加上一个前提条件，如下：
select top 10 * from T_Temple a with(nolock) where lat between 37.356976-2 and 37.356976+2 and lon between 116.8425-2 and 116.8425+2 and dbo.[fnGetDistance](37.356976,116.8425,lat,lon)<150 order by a.Id asc

这样的话，相比于方案1，方案2还是有性能上的提升，特别是T_Temple表记录比较多时，可在lat,lon上可以建立坐标。

附上ms sql 版的fnGetDistance

CREATE FUNCTION [dbo].[fnGetDistance](@LatBegin REAL, @LngBegin REAL, @LatEnd REAL, @LngEnd REAL) RETURNS FLOAT 
   
  AS 
 BEGIN 
   --距离(千米) 
  DECLARE @Distance REAL 
  DECLARE @EARTH_RADIUS REAL 
  SET @EARTH_RADIUS = 6378.137   
  DECLARE @RadLatBegin REAL,@RadLatEnd REAL,@RadLatDiff REAL,@RadLngDiff REAL 
   SET @RadLatBegin = @LatBegin *PI()/180.0   
  SET @RadLatEnd = @LatEnd *PI()/180.0   
  SET @RadLatDiff = @RadLatBegin - @RadLatEnd   
  SET @RadLngDiff = @LngBegin *PI()/180.0 - @LngEnd *PI()/180.0  
  SET @Distance = 2 *ASIN(SQRT(POWER(SIN(@RadLatDiff/2), 2)+COS(@RadLatBegin)*COS(@RadLatEnd)*POWER(SIN(@RadLngDiff/2), 2))) 
  SET @Distance = @Distance * @EARTH_RADIUS   
  SET @Distance = Round(@Distance * 10000,4) / 10000   

  RETURN @Distance 
 END