Re: magic corner gradient revisited

Brian,

Thanks, but I lost the track of the discussion and ain't sure what your
phrase meant to express.

But please let me note that I'm in favour of not changing the current
behaviour of corner gradients, because of two reasons.

First, the current spec is already implemented in many platforms and this
feature might be already in use by the platforms (like Mozilla) internally
and/or externally. Changing this behaviour may cause either a big
incompatibility between versions of these platforms, or introduce a lot of
complexity to the platform just to keep the new ones compatible.

Second, we should respect that "connecting corners" has a well-defined
meaning for (professional) designers, thus changing the current behaviour
will be confusing for them in some degree, and probably more important,
inconsistent with some graphic applications.

The later is why I proposed introducing a new keyword for this matter in the
first place.

-Behnam



On Tue, Aug 9, 2011 at 12:24 AM, Brian Manthos <brianman@microsoft.com>wrote:

> I looked at this some more.  The first time I didn't notice this phrase in
> the right light...
>
> "line drawn perpendicular to the gradient-line through the center"
>
> While I prefer the simple "here's the math" approach, I think this might
> work.
>
> Brad, can take a look and see if it works for you?
>
> Behnam should probably take a look as well since he opened the original
> "Gradient Magic" thread.
>
> -Brian
>
>
>
> > -----Original Message-----
> > From: www-style-request@w3.org [mailto:www-style-request@w3.org] On
> > Behalf Of Brian Manthos
> > Sent: Monday, August 08, 2011 6:42 PM
> > To: Tab Atkins Jr.
> > Cc: Brad Kemper; fantasai; www-style list
> > Subject: RE: magic corner gradient revisited
> >
> > # If the argument specifies a corner to angle towards, the gradient
> > must be rendered identically to an angle-based gradient with an angle
> > chosen such that the endpoint of the gradient is in the same quadrant
> > as the indicated corner, and a line drawn perpendicular to the
> > gradient-line through the center of the box intersects the two
> > neighboring corners.
> >
> > This phrasing doesn't narrow down the angle that must be used.  In
> > fact, pretty much any angle in the quadrant satisfies the specified
> > criteria.
> >
> > If you don't like my slope description (from Tue 7/26/2011 12:17 PM
> > PST), I can provide an alternative involving around-center rotation of
> > the box by 90 degrees.
>



-- 
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    '     Behnam Esfahbod
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