- From: Tab Atkins Jr. <jackalmage@gmail.com>
- Date: Mon, 8 Aug 2011 18:54:15 -0700
- To: Brian Manthos <brianman@microsoft.com>
- Cc: Brad Kemper <brad.kemper@gmail.com>, fantasai <fantasai.lists@inkedblade.net>, www-style list <www-style@w3.org>
On Mon, Aug 8, 2011 at 6:41 PM, Brian Manthos <brianman@microsoft.com> wrote: > # If the argument specifies a corner to angle towards, the gradient must be rendered identically to an angle-based gradient with an angle chosen such that the endpoint of the gradient is in the same quadrant as the indicated corner, and a line drawn perpendicular to the gradient-line through the center of the box intersects the two neighboring corners. > > This phrasing doesn't narrow down the angle that must be used. In fact, pretty much any angle in the quadrant satisfies the specified criteria. No, there is a unique line that both passes through the center of the box and intersects the appropriate corners. (It's just the line between the two corners.) There is thus a unique gradient angle that places the endpoint in the right quadrant and produces a gradient line perpendicular to the aforementioned line. ~TJ
Received on Tuesday, 9 August 2011 01:55:10 UTC