Re: magic corner gradient revisited

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