Recent PostsRecent Posts Popular TopicsPopular Topics
 Home Search Members Calendar Who's On

 Geometry - Parallel lines Rate Topic Display Mode Topic Options
Author
 Message
 Posted Wednesday, March 19, 2014 4:31 AM
 SSCommitted Group: General Forum Members Last Login: Tuesday, November 10, 2015 4:58 AM Points: 1,596, Visits: 1,786
 Hi,Assuming I have a line, is there a function I can call to create a parallel line at a given distance away.i.e - with the below I would want to draw a parallel line to the one output.DECLARE @line geometry = 'LINESTRING(1 1, 2 2, 3 3, 4 4)' SELECT @lineAny help would be appreciated.Thanks,Nic ------------------------------------------------------------Check out my bloghttp://www.sqlservercentral.com/articles/Best+Practices/61537/
Post #1552530
 Posted Wednesday, March 19, 2014 1:20 PM
 Keeper of the Duck Group: Moderators Last Login: Tuesday, November 24, 2015 12:26 PM Points: 6,634, Visits: 1,898
 I don't think you're going to find a function with just distance. Maybe with a point. The reason I say this is in 2D space there would be two lines parallel to a give line. In 3D space you'd form a cylinder. So there wouldn't just be a line output in either situation. K. Brian Kelley, CISA, MCSE, Security+, MVP - SQL ServerRegular Columnist (Security), SQLServerCentral.comAuthor of Introduction to SQL Server: Basic Skills for Any SQL Server User| Professional Development blog | Technical Blog | LinkedIn | Twitter
Post #1552776
 Posted Saturday, April 5, 2014 12:48 AM This worked for the OP
 SSCarpal Tunnel Group: General Forum Members Last Login: Today @ 12:18 AM Points: 4,816, Visits: 12,634
 This can be done by shifting each point on the line by x and y (@OFFSET_X/Y in the code)`DECLARE @line geometry = 'LINESTRING(1 1, 2 2, 3 3, 4 4, 4 5, 5 7)' ;DECLARE @pp_line geometry;DECLARE @OFFSET_X FLOAT = 2;DECLARE @OFFSET_Y FLOAT = 0;;WITH NUMBERS(N) AS (SELECT NM.N FROM (VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10)) AS NM(N))SELECT @pp_line = geometry::Parse ( CONCAT ( 'LINESTRING(' ,STUFF((SELECT CONCAT ( CHAR(44) ,CHAR(32) ,CAST(@line.STPointN(NM.N).STX + @OFFSET_X AS VARCHAR(12)) ,CHAR(32) ,CAST(@line.STPointN(NM.N).STY + @OFFSET_Y AS VARCHAR(12)) ) AS [text()] FROM NUMBERS NM WHERE NM.N <= @line.STNumPoints() FOR XML PATH(''), TYPE).value('.[1]','VARCHAR(8000)'),1,1,'') ,CHAR(41) ));SELECT @pp_lineUNION ALL SELECT @line;`
Post #1558699
 Posted Sunday, April 6, 2014 2:42 PM This worked for the OP
 Ten Centuries Group: General Forum Members Last Login: Monday, October 26, 2015 8:19 PM Points: 1,207, Visits: 3,247
 HiSorry about the late post, but parallel is always a tricky one. First decision to make is which side of the line do you want to parallel to? The next problem you hit is how do you handle angles greater than 180 degrees in a line string? Do you single point or multiple points at the distance?Here's some code to parallel to a simple line (2 point)`DECLARE @simpleLineString Geometry = Geometry::STGeomFromText('LINESTRING (0 0, 10 3)',0);DECLARE @sideMod FLOAT = -1; -- Right = -1, Left = 1DECLARE @offset FLOAT = .5;WITH linePoints AS ( SELECT X1 = @simpleLineString.STPointN(1).STX ,Y1 = @simpleLineString.STPointN(1).STY ,X2 = @simpleLineString.STPointN(2).STX ,Y2 = @simpleLineString.STPointN(2).STY ,L = @simpleLineString.STLength() ) ,calcOffset AS ( SELECT xOffSet = (((Y2 - Y1) * (1 - (L - @offset) / L)) * @sideMod) * -1, yOffset = ((X2 - X1) * (1 - (L - @offset) / L)) * @sideMod FROM linePoints ) ,buildParallel AS ( SELECT parallelLine = Geometry::STGeomFromText( CONCAT('LINESTRING (', X1 + xOffset,' ',Y1 + yOffset,', ', X2 + xOffset,' ',Y2 + yOffset,')'), 0) FROM linePoints l CROSS APPLY (SELECT * FROM calcOffset) o )SELECT 'Original' Name, @simpleLineString GeomUNION ALLSELECT 'Parallel' Name, parallelLine GeomFROM buildParallel;`If you want to do a multi-point line strings you will need to start working out half angles etc.
Post #1558861
 Posted Sunday, April 6, 2014 8:03 PM
 Ten Centuries Group: General Forum Members Last Login: Monday, October 26, 2015 8:19 PM Points: 1,207, Visits: 3,247
Hi

Had a bit of time to play around with a multi point line. The following will do a parallel for both sides of an input geometry. I've left all the calculations exploded out to try and make it a bit easier to follow. There is probably better math for this though

`DECLARE @LineString Geometry = Geometry::STGeomFromText('LINESTRING (7 5, 10 3, 11 4, 13 4, 13 -2, 7 1, 5 -2)',0);DECLARE @offset FLOAT = .5;WITH cteTally AS (	SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) N	FROM (VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) E1 (N)		,(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) E2 (N)		,(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) E3 (N)	)	,linePoint AS (	SELECT TOP(@LineString.STNumPoints()) N		,X = @LineString.STPointN(N).STX 		,Y = @LineString.STPointN(N).STY		,B2Next = CASE WHEN N < @LineString.STNumPoints() THEN 			CAST(90 -				DEGREES(					ATN2(					@LineString.STPointN(N + 1).STY - @LineString.STPointN(N).STY,					@LineString.STPointN(N + 1).STX - @LineString.STPointN(N).STX					)				) + 360			AS DECIMAL(38,19)) % 360		END 	FROM cteTally	)	,offsetBearings AS (		SELECT b1.N, b1.X, b1.Y, b1.B2Next,			offsetAngleLeft = CASE 				WHEN b1.B2Next is NULL THEN b2.B2Next - 90				WHEN b2.B2Next is NULL THEN b1.B2Next - 90				ELSE (360 + b1.B2Next - ((360 - ((b2.B2Next + 180) - b1.B2Next)) / 2)) % 360				END,			offsetAngleRight = CASE 				WHEN b1.B2Next is NULL THEN b2.B2Next + 90				WHEN b2.B2Next is NULL THEN b1.B2Next + 90				ELSE (b1.B2Next + ((((b2.B2Next + 180) - b1.B2Next)) / 2)) % 360				END		FROM linePoint b1			LEFT OUTER JOIN linePoint b2 ON b1.N  = b2.N + 1	)	,offsetDistance AS (	SELECT *,		offsetDist = CASE 			WHEN N = 1 or B2Next is null THEN @offset			ELSE @offset / (SIN(RADIANS(((b2Next - offsetAngleLeft) + 360) % 360)))			END	FROM offsetBearings	)	, parallelCoords AS (	SELECT *		, XL = X + (offsetDist * COS(RADIANS(90 - offsetAngleLeft)))		, YL = Y + (offsetDist * SIN(RADIANS(90 - offsetAngleLeft)))		, XR = X + (offsetDist * COS(RADIANS(90 - offsetAngleRight)))		, YR = Y + (offsetDist * SIN(RADIANS(90 - offsetAngleRight)))	FROM offsetDistance	)SELECT 'Left' Name, ParallelLineLeft = geometry::Parse (    CONCAT        (            'LINESTRING('            ,STUFF((SELECT                   CONCAT                    (                      CHAR(44)                     ,CHAR(32)                     ,CAST(XL AS VARCHAR(12))                     ,CHAR(32)                     ,CAST(YL AS VARCHAR(12))                       ) AS  [text()]                FROM parallelCoords NM               ORDER BY N               FOR XML PATH(''), TYPE).value('.[1]','VARCHAR(8000)'),1,1,'')            ,CHAR(41)        ))UNION ALLSELECT 'Right' Name, ParallelLineRight = geometry::Parse (    CONCAT        (            'LINESTRING('            ,STUFF((SELECT                   CONCAT                    (                      CHAR(44)                     ,CHAR(32)                     ,CAST(XR AS VARCHAR(12))                     ,CHAR(32)                     ,CAST(YR AS VARCHAR(12))                       ) AS  [text()]                FROM parallelCoords NM               ORDER BY N                FOR XML PATH(''), TYPE).value('.[1]','VARCHAR(8000)'),1,1,'')            ,CHAR(41)        ))UNION ALLSELECT 'Orig' Name, @LineString`

Edit Added some ORDER BY clauses in to ensure line vertexes are added in the correct order

Looks like this

Post Attachments
 parallel.png (148 views, 11.56 KB)
Post #1558884
 Posted Wednesday, April 9, 2014 7:48 AM
 Old Hand Group: General Forum Members Last Login: Wednesday, September 30, 2015 7:28 AM Points: 321, Visits: 1,308
 K. Brian Kelley (3/19/2014)I don't think you're going to find a function with just distance. Maybe with a point. The reason I say this is in 2D space there would be two lines parallel to a give line. In 3D space you'd form a cylinder. So there wouldn't just be a line output in either situation.That is only the case where the lines are similar as well as parallel otherwise even in 2D there will still be an infinite number of parallel lines - the length of the line does not affect whether or not it is parallel.Any straight line in 2d can be defined by the equation y=mx+c bounded by its min and max ordinates. By changing the bounds or the value of c you will generate a parallel line.
Post #1559948
 Posted Wednesday, April 9, 2014 7:56 AM
 SSCommitted Group: General Forum Members Last Login: Tuesday, November 10, 2015 4:58 AM Points: 1,596, Visits: 1,786
 Hi all,Thanks for your responses, since the lines in our case are to be the same length we achieved this without to much complexity.Thank you all for your help.Nic ------------------------------------------------------------Check out my bloghttp://www.sqlservercentral.com/articles/Best+Practices/61537/
Post #1559957
 Posted Wednesday, September 2, 2015 12:43 PM
 Forum Newbie Group: General Forum Members Last Login: Friday, November 13, 2015 11:52 AM Points: 1, Visits: 16
 @Ten CenturiesThis is nice, however the resulting left and right side lines have the incorrect order of vertices. Do you know how to fix?Thanks!Note - the query produces proper results when running the left side, or the right side, but not when displaying all together. I solved my need by just looking at one side as needed.
Post #1716564
 Posted Wednesday, September 2, 2015 7:16 PM
 Ten Centuries Group: General Forum Members Last Login: Monday, October 26, 2015 8:19 PM Points: 1,207, Visits: 3,247
 roy jackson (9/2/2015)@Ten CenturiesThis is nice, however the resulting left and right side lines have the incorrect order of vertices. Do you know how to fix?Thanks!Note - the query produces proper results when running the left side, or the right side, but not when displaying all together. I solved my need by just looking at one side as needed.I'm going to assume that you are aiming this at me:)I see what you mean about my original query. It must have worked back when I did, but I just tried it on my current server and can see the problem.I will edit my original post to fix the issue. It is all around taking the order for granted, which I shouldn't have done.
Post #1716630
 Posted Thursday, September 3, 2015 1:43 AM
 Old Hand Group: General Forum Members Last Login: Wednesday, September 30, 2015 7:28 AM Points: 321, Visits: 1,308
 NicHopper (4/9/2014)Hi all,Thanks for your responses, since the lines in our case are to be the same length we achieved this without to much complexity.Thank you all for your help.NicIf you are requiring the lines to be the same length as the original line then you are likely to experience a number of cases where the polylines will cross and some of the line segments may overlap when the offset is forced to be along the length of the original segment.
Post #1716663

 Permissions