- From: Tab Atkins Jr. <jackalmage@gmail.com>
- Date: Tue, 7 Sep 2010 20:17:02 -0700
- To: fantasai <fantasai.lists@inkedblade.net>
- Cc: David Singer <singer@apple.com>, Simon Fraser <smfr@me.com>, "www-style@w3.org" <www-style@w3.org>
On Tue, Sep 7, 2010 at 8:04 PM, fantasai <fantasai.lists@inkedblade.net> wrote: > On 09/07/2010 06:09 PM, Tab Atkins Jr. wrote: >> On Tue, Sep 7, 2010 at 5:47 PM, David Singer<singer@apple.com> wrote: >>> why isn't the starting point for a radial linear gradient >>> "those point(s) on a line perpendicular to the gradient angle, >>> in the most negative position possible, that still just fit(s) >>> within the shape to be filled"? >> >> I can't understand what you're trying to say here. > > He's saying what you're saying, just generalizing it so you don't need > different rules depending on the angle's value. > > Here's the idea: You have an angle. You want a gradient. > > 1. Draw a directed line at that angle. > 2. Place it over the centerpoint of the box. Call this point zero. > 3. Draw a line perpendicular to the line. > 4. Shift the line backwards (i.e. negatively) until the line's > intersection with the directed line is as negative as possible > while still intersecting the box. > > The intersection point in step 4 is the start point of the gradient. Oh, got it. Yeah, that's just the opposite of how I'm currently defining the ending point (which will soon just be the point from which you measure percentages in the <length> part of the angle gradient). ~TJ
Received on Wednesday, 8 September 2010 03:17:54 UTC