Re: magic corner gradient revisited

On Aug 4, 2011, at 1:35 PM, "Tab Atkins Jr." <jackalmage@gmail.com> wrote:

> On Thu, Aug 4, 2011 at 1:23 PM, Brian Manthos <brianman@microsoft.com> wrote:
>> Tab Atkins Jr.:
>>> As well, there are two distinct and very different ways
>>> of achieving the magic corner gradient that I know of
>>> (adjust the angle, or do the entire thing in
>>> objectBoundingBox coords scale).
>> 
>> I think by "adjust the angle" you're referring to my algorithm.
> 
> Yes.
> 
> 
>> I'm unclear on what you mean by the "scale" approach.
>> 
>> Can you elaborate on this?  Or maybe offer some renderings?
> 
> Whoops, didn't mean to leave the word "scale" in there.
> 
> I'm referring to how SVG does linear gradients when
> gradientUnits="objectBoundingBox".  For example:
> 
> <!doctype html>
> <svg>
> <defs>
>  <linearGradient id=foo x1=0 y1=0 x2=1 y2=1 gradientUnits="objectBoundingBox">
>   <stop offset=0% stop-color=red />
>   <stop offset=50% stop-color=white />
>   <stop offset=100% stop-color=blue />
>  </linearGradient>
> </defs>
> <rect x=0 y=0 width=300 height=100 fill=url(#foo) />
> </svg>
> 
> In this coordinate space, the units are such that 1 unit horizontally
> is the width of the box it's used in, and 1 unit vertically is the
> height of the box it's used in.  In other words, you draw the gradient
> into a square, and then scale it to fill the box.  This gives you the
> "magic corner" behavior, in addition to a few other interesting bits.
> You lose, however, the ability to use absolute lengths when placing
> stops, unless some extra magic goes on.

That would be a pretty huge loss to get to an end result that is otherwise in no way better than Brian's formula. We should not be considering such a drastic reinvention at this point, since what we had already was extremely workable, and the small change proposed in this thread previously didn't change that except to make it render much more suitably.

Received on Thursday, 4 August 2011 21:00:42 UTC