I looked at this some more. The first time I didn't notice this phrase in the right light... "line drawn perpendicular to the gradient-line through the center" While I prefer the simple "here's the math" approach, I think this might work. Brad, can take a look and see if it works for you? Behnam should probably take a look as well since he opened the original "Gradient Magic" thread. -Brian > -----Original Message----- > From: www-style-request@w3.org [mailto:www-style-request@w3.org] On > Behalf Of Brian Manthos > Sent: Monday, August 08, 2011 6:42 PM > To: Tab Atkins Jr. > Cc: Brad Kemper; fantasai; www-style list > Subject: RE: magic corner gradient revisited > > # 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. > > If you don't like my slope description (from Tue 7/26/2011 12:17 PM > PST), I can provide an alternative involving around-center rotation of > the box by 90 degrees.Received on Tuesday, 9 August 2011 04:25:25 UTC
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