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Re: Gradients by angle (was Re: [CSSWG] Minutes and Resolutions 2009-08-12)

From: Brad Kemper <brad.kemper@gmail.com>
Date: Thu, 13 Aug 2009 16:15:55 -0700
Message-Id: <FF3AD07B-6B56-4BAD-8033-0EA57CBD8905@gmail.com>
To: "robert@ocallahan.org" <robert@ocallahan.org>
Cc: "L. David Baron" <dbaron@dbaron.org>, "www-style@w3.org" <www-style@w3.org>
No. 100% should mean corner to corner.

Sent from my iPhone

On Aug 13, 2009, at 3:27 PM, "Robert O'Callahan"  
<robert@ocallahan.org> wrote:

> On Fri, Aug 14, 2009 at 10:14 AM, L. David Baron <dbaron@dbaron.org>  
> wrote:
> Consider the image here:
> http://lists.w3.org/Archives/Public/www-archive/2009Aug/att-0041/gradient.png
> In this image:
>  * the dark blue box is the boundary of the box
>  * the white->black area is the area covered by the gradient
>  * the red area is the area outside the gradient
>  * the line blue line is the ray given by your proposal
>
> Here, I use "top left" as the first position.  But unless the angle
> given is 45deg (or that plus some multiple of 90deg), a gradient in
> this square won't actually have the gradient area touch both
> corners.  In this example, the red area extends well into the box at
> the bottom right corner.
>
> I don't see the problem here. This is what you asked for, and I  
> think people will ask for.
>
> You can choose the following behaviours for the red region
> -- fill with white --- specify color stops "black, white" and  
> background-repeat:no-repeat
> -- fill with transparent (or any other color) --- specify color  
> stops "black, 100% white, rgba(0,0,0,0)" and background-repeat:no- 
> repeat
> -- repeat with black to white --- color stops "black, white" and  
> background-repeat:repeat (default)
>
> Rob
> -- 
> "He was pierced for our transgressions, he was crushed for our  
> iniquities; the punishment that brought us peace was upon him, and  
> by his wounds we are healed. We all, like sheep, have gone astray,  
> each of us has turned to his own way; and the LORD has laid on him  
> the iniquity of us all." [Isaiah 53:5-6]
Received on Thursday, 13 August 2009 23:16:43 GMT

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