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Re: Tangent to Beziers (Tangiers?)

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

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 Group
Received on Thursday, 5 August 2004 13:43:19 UTC

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