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

[Jim Graham]

OK, I can see how your terminology works now, but it seems odd to me.  I 
never consider re-expressing the coordinates on a curve as a vector and 
basing geometric properties on those constructed vectors.  I either 
consider the points on the curve, or its tangent or its normal - none of 
which is the value you are expressing.  You are, essentially, operating 
on tangent vectors, but you aren't calling them that, you are calling 
them something like "the vector of the derivative" which is a relative 
(direction only) version of the tangent vector (which has location and 
direction).  When one talks about curves and being parallel, my mind 
tends to think of the tangents of the curves being parallel and tangents 
are directed by the first derivative.

Also, if you are going to use your definition of "vector" then parallel 
is an odd term to use for values that originate from the same point 
(points considered as a vector are taken to originate from 0,0) - really 
you want those "vectors" to be collinear, not (just) parallel.

So, either || means "the coordinates of the curves expressed as vectors 
are collinear" or it means "the curves (i.e. the tangents of the curve 
at the indicated point) are parallel".  Saying "vector I() is parallel 
to vector B()" didn't really have meaning to me based on the above biases.

So, I get your comment now and all of the math makes sense, but the 
terminology seemed foreign to me...

			...jim

On 10/20/10 10:48 AM, Denis Lila wrote:
>> 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