## Geometry - Parallel lines

 Author Message NicHopper SSCommitted Group: General Forum Members Points: 1963 Visits: 1900 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/ K. Brian Kelley Keeper of the Duck Group: Moderators Points: 10450 Visits: 1917 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@‌kbriankelley Eirikur Eiriksson SSCoach Group: General Forum Members Points: 15328 Visits: 18612 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:arse ( 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;` mickyT SSCommitted Group: General Forum Members Points: 1594 Visits: 3317 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. mickyT SSCommitted Group: General Forum Members Points: 1594 Visits: 3317 HiHad 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:-D`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:arse ( 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:arse ( 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 orderLooks like this Attachments parallel.png (221 views, 11.00 KB) crmitchell SSChasing Mays Group: General Forum Members Points: 659 Visits: 1761 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. NicHopper SSCommitted Group: General Forum Members Points: 1963 Visits: 1900 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/ roy jackson Forum Newbie Group: General Forum Members Points: 7 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. mickyT SSCommitted Group: General Forum Members Points: 1594 Visits: 3317 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 meI 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. crmitchell SSChasing Mays Group: General Forum Members Points: 659 Visits: 1761 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.