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Re: [css3-images] [css3-values] Inconsistent Angles

From: fantasai <fantasai.lists@inkedblade.net>
Date: Tue, 07 Sep 2010 23:13:49 -0700
Message-ID: <4C87299D.5020405@inkedblade.net>
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.

Received on Wednesday, 8 September 2010 06:14:26 UTC

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