# [OpenJDK 2D-Dev] X11 uniform scaled wide lines and dashed lines; STROKE_CONTROL in Pisces

Denis Lila dlila at redhat.com
Wed Oct 20 10:48:36 PDT 2010

```> Also, how is A(t) and B(t) are parallel not the same as "the curves A
> and B are parallel at t"?

Well, suppose A and B are lines with endpoints (0,0), (2,0) for A
and (0,1),(2,1) for B. Obviously, for all t, A and B are parallel at t.
However let t = 0.5. Then A(t) = (1,0) and B(t) = (1, 1). The vectors
(1,0) and (1,1) are not parallel, so saying A(t) || B(t) is the same
as saying that there exists c such that (1,0) = c*(1,1), which isn't true.

However, A'(t)=(2,0) and B'(t)=(2,0), and the vectors
(2,0) and (2,0) are parallel.

Does this make more sense?

Regards,
Denis.

----- "Jim Graham" <james.graham at oracle.com> wrote:

> On 10/20/10 7:54 AM, Denis Lila wrote:
> >> In #2, you have a bunch of "I'() || B'()" which I read as "the
> slope
> >> of the derivative (i.e. acceleration) is equal", don't you really
> mean
> >> "I() || B()" which would mean the original curves should be
> parallel?
> >> Otherwise you could say "I'() == B'()", but I think you want to
> show
> >> parallels because that shows how you can use the dxy1,dxy4 values
> as
> >> the parallel equivalents.
> >
> > Not really. I've updated the comment explaining what || does, and
> > it should be clearer now. Basically, A(t) || B(t) means that
> vectors
> > A(t) and B(t) are parallel (i.e. A(t) = c*B(t), for some nonzero
> t),
> > not that curves A and B are parallel at t.
>
> I'm not sure we are on the same page here.
>
> I'() is usually the symbol indicating the "derivative" of I().  My
> issue
> is not with the || operator, but with the fact that you are applying
> it
> to the I'() instead of I().
>
> Also, A(t) = c*B(t) is always true for all A and B and all t if you
> take
> a sample in isolation.  Parallel means something like "A(t) = c*B(t)
> with the same value of c for some interval around t", not that the
> values at t can be expressed as a multiple.
>
> Again, I'() || B'() says to me that the derivative curves are
> parallel,
> not that the original curves are parallel...
>
> 			...jim

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