Re: magic corner gradient revisited

Yes, that is very similar to how I was originally saying it, but with quadrant to make sure the direction along the path is correct. This wording should be fine. 

Sent from my iPad

On Aug 8, 2011, at 9:24 PM, Brian Manthos <> wrote:

> 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: [] 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 06:25:32 UTC