- From: Tab Atkins Jr. <jackalmage@gmail.com>
- Date: Wed, 8 Sep 2010 00:06:42 -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 11:13 PM, fantasai <fantasai.lists@inkedblade.net> wrote: > 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. ...right. That's what I said. That definition is the *opposite* of what I'm using for determining the ending point, where I just project the perpendicular line forwards rather than backwards. ~TJ
Received on Wednesday, 8 September 2010 07:07:35 UTC