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 01:42:09 UTC