From: Chris Lilley <chris@w3.org>

Date: Thu, 5 Aug 2004 19:43:17 +0200

Message-ID: <37538405.20040805194317@w3.org>

To: David Turner <novalis@fsf.org>

Cc: www-svg@w3.org

Date: Thu, 5 Aug 2004 19:43:17 +0200

Message-ID: <37538405.20040805194317@w3.org>

To: David Turner <novalis@fsf.org>

Cc: www-svg@w3.org

On Thursday, August 5, 2004, 6:34:27 PM, David wrote: DT> I'm trying to implement proper direcionality for path DT> markers in Inkscape (the current version is entirely DT> nonconforming). DT> Here's what 11.6.2 tells me: DT> "(a) if there is a path segment coming into the vertex and DT> another path segment going out of the vertex, the marker's DT> positive x-axis should point toward the angle bisector for DT> the angle at the given vertex, where that angle has one side DT> consisting of tangent vector for the path segment going into DT> the vertex and the other side the tangent vector for the DT> path segment going out of the vertex ..." DT> Question: What is the tangent vector to the initial point of DT> a curve, when the first control point is identical to the DT> initial control point? You are correct that it doesn't have one, its a degenerate case. There are various ways of dealing with it. The spec should clarify this case. If its a vector sum of two vectors and there is only one vector, a useful result is to just take the second vector. DT> My guess (based on eyeballing the DT> curves onscreen) would be to devolve to the direction from DT> the initial point to the second control point, then to the DT> end point, then to the rules for zero-length path segments DT> (in the implementation notes). DT> Is this correct? I would be glad of other opinions but it sounds correct to me. DT> I'm not subscribed to the mailing list, so please CC me on DT> replies. -- Chris Lilley mailto:chris@w3.org Chair, W3C SVG Working Group Member, W3C Technical Architecture GroupReceived on Thursday, 5 August 2004 13:43:19 UTC

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