Question

Get Latitude/Longitude for a Known Distance From Another Latitude/Longitude

Asked by: wwarby

I need a pair of T-SQL functions that look something like the attached code snippet. Essentially, my aim is to create a bounding box 200m around a known lat/lng so I need to get the co-ordinates 200m due east and due north from the known point. I have a function for calculating the distance between two known Latitudes/Longitudes (that was easy to find on the web) but not being a mathematician, I don't know how to refactor the equation for this purpose.

The reason I need this is a little convoluted but I'll gladly explain if asked. Ideally, the solution would use the same mathematics as my existing distance calculator function, just re-organised to get the lat/lng values instead of the distance, but I'll take any solution that works ;)

CREATE FUNCTION dbo.LatOffset(@Lat FLOAT, @Lng FLOAT, @MetresOffset) 
	RETURNS FLOAT
	AS
	BEGIN
		--do some magic and return the Latitude @MetresOffset due north of the point (@Lat, @Lng)
	END
 
CREATE FUNCTION dbo.LngOffset(@Lat FLOAT, @Lng FLOAT, @MetresOffset) 
	RETURNS FLOAT
	AS
	BEGIN
		--do some magic and return the Longitude @MetresOffset due east of the point (@Lat, @Lng)
	END
 
 
 
--my existing distance calculation function
CREATE FUNCTION dbo.Distance(@Lat1 FLOAT, @Lng1 FLOAT, @Lat2 FLOAT, @Lng2 FLOAT) 
	RETURNS FLOAT
	AS
	BEGIN
		RETURN(6378.14 * ACOS((COS(RADIANS(@Lat1)) * COS(RADIANS(@Lng1))) * (COS(RADIANS(@Lat2)) * COS(RADIANS(@Lng2))) + (COS(RADIANS(@Lat1)) * SIN(RADIANS(@Lng1))) * (COS(RADIANS(@Lat2)) * SIN(RADIANS(@Lng2))) + SIN(RADIANS(@Lat1)) * SIN(RADIANS(@Lat2))))
	END

                                  
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Asked On
2008-11-25 at 05:52:50ID23933874
Tags

T-SQL

Topics

GIS & GPS Programming

,

Math & Science

,

MS SQL Server

Participating Experts
3
Points
100
Comments
5

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Answers

 

by: deightonPosted on 2008-11-25 at 08:33:19ID: 23035783

assuming circumference of earth is 40030km

then 200m north is increase in latitude of

.2 / 40030 * 360 = .001799 degrees

for longitude it depends on latitude you're at

.2 * 360 / (cos(latitude) * 40030)

that's what I think, there might be refinements due to the fact earth is not a perfect sphere

I'm working in degrees there, not radians

 

by: ozoPosted on 2008-11-25 at 15:14:32ID: 23038510

lat2: =ASIN(SIN(lat1)*COS(d/ER) + COS(lat1)*SIN(d/ER)*COS(brng))
lon2: =lon1 + ATAN2(COS(d/ER)-SIN(lat1)*SIN(lat2), SIN(brng)*SIN(d/ER)*COS(lat1))

 

by: ozoPosted on 2008-11-25 at 15:21:59ID: 23038547

For due East and West, following a line of latitude instead of a great circle,
long2 = long1 +/- dist/(radius*cos(latitude)) * 180/pi

 

by: wwarbyPosted on 2008-11-26 at 00:41:33ID: 23040409

Thanks guys - for some reason my email notification of these responses got delayed and I found a solution in the meantime. For anyone else trying to solve the same problem I've attached my completed functions. I'll split the point between both people who answered by way of thanks ;)

CREATE FUNCTION dbo.LatitudeOffset(@Lat FLOAT, @Offset DECIMAL(6,4)) 
	RETURNS FLOAT
	AS
	BEGIN
		DECLARE @Rads FLOAT
		SET @Rads = RADIANS(@Lat)
		RETURN (@Lat + ((@Offset * 1000)  / (111132.92 + (-559.82 * COS(2 * @Rads)) + (1.175 * COS(4 * @Rads)) + (-0.0023 * COS(6 * @Rads)))))
	END
 
CREATE FUNCTION dbo.LongitudeOffset(@Lat FLOAT, @Lng FLOAT, @Offset DECIMAL(6,4)) 
	RETURNS FLOAT
	AS
	BEGIN
		DECLARE @Rads FLOAT
		SET @Rads = RADIANS(@Lat)
		RETURN (@Lng + ((@Offset * 1000)  / ((111412.84 * COS(@Rads)) + (-93.5 * COS(3 * @Rads)) + (0.118 * COS(5 * @Rads)))))
	END

                                              
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by: Infinity08Posted on 2008-11-26 at 01:52:21ID: 23040698

>> lon2: =lon1 + ATAN2(COS(d/ER)-SIN(lat1)*SIN(lat2), SIN(brng)*SIN(d/ER)*COS(lat1))

shouldn't that be :

lon2: =lon1 + ATAN2(SIN(brng)*SIN(d/ER)*COS(lat1), COS(d/ER)-SIN(lat1)*SIN(lat2))

since atan2 takes the y coordinate first, and then the x coordinate (some applications decided to do it the other way around against common convention though).

20120131-EE-VQP-002

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