- From: fantasai <fantasai.lists@inkedblade.net>
- Date: Tue, 07 Sep 2010 23:13:49 -0700
- To: "Tab Atkins Jr." <jackalmage@gmail.com>
- CC: David Singer <singer@apple.com>, Simon Fraser <smfr@me.com>, "www-style@w3.org" <www-style@w3.org>
On 09/07/2010 08:17 PM, Tab Atkins Jr. wrote: > 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 Starting point, Tab. He's using this to define the starting point. To find the ending point, you'd take step 4 to the most negative possible intersection. ~fantasai
Received on Wednesday, 8 September 2010 06:14:26 UTC