RE: magic corner gradient revisited

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.


> -----Original Message-----
> From: [] 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