Re: Tangent to Beziers (Tangiers?)

Chris Lilley wrote:

> 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> 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.

   I think if you do some basic algebra you will see that it has a
useful result (the one you suggest):

   The derivative of a Bezier curve is:

   f'(t) = -3(1-t)(1-t)P0 + 3(1-3t)(1-t)P1 + 3t(2-3t)P2 + 3t*tP3

   If P0 == P1:
   f'(t) = -3(2t)(1-t)P0 + 3(2t)(1-(3/2)t)P2 + 3t*tP3
         = 6t((1-(3/2)t)P2 - (1-t)P0) + 3t*tP3

   As t approaches zero we can ignore 3t*tP3 as it's contribution is
insignificant.  This leaves:

         = 6t((1-(3/2)t)P2 - (1-t)P0)

    The 6t term is simply a scaling factor (which makes our vector
zero length at t=0) but if you look at the rest of the eqn:

         (1-(3/2)t)P2 - (1-t)P0

    You can see that this becomes:  P2-P0 as t goes to zero

    Thus your hunch below is proven out.  The directionality of
the derivative is P2-P0 but the length of the vector is zero.

    A similar argument can be followed for the case P0=P1=P2 to
find out that the directionality is P3-P1:

          6t((1-(3/2)t)P0 - (1-t)P0) + 3t*tP3
         = -3t*tP0 + 3t*tP3
         = 3t*t(P3-P0)

> 
> 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.
> 
> 
> 
>

Received on Tuesday, 10 August 2004 15:17:03 UTC