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Re: Is there a formula for converting a linear gradient to a box-shadow?

From: Behrang Saeedzadeh <behrangsa@gmail.com>
Date: Tue, 5 Feb 2013 10:26:34 +1100
Message-ID: <CAERAJ+8Oof579X-HurDSLAN3eg=C8FkcmzyFqMXxTDTYbqn96A@mail.gmail.com>
To: Simon Sapin <simon.sapin@kozea.fr>
Cc: François REMY <francois.remy.dev@outlook.com>, W3C CSS Mailing List <www-style@w3.org>
Hi Simon,

Thanks for the reply. I think this now proves that it makes sense to have
box-shadow-[top, right, bottom left]. Please see this thread for more info:
http://lists.w3.org/Archives/Public/www-style/2012Dec/0029.html

Cheers,
Behrang Saeedzadeh
http://www.behrang.org


On Thu, Jan 31, 2013 at 7:45 PM, Simon Sapin <simon.sapin@kozea.fr> wrote:

> Le 31/01/2013 00:50, Behrang Saeedzadeh a écrit :
>
>  - And the question is: is there a formula for creating a box shadow that
>> renders identical to a given linear gradient?
>>
>
> I think not, because there is no syntax to specify a different blur radius
> in the horizontal and vertical direction.
>
> And even if you had that, a linear gradient to transparency and a box
> shadow that look similar are not quite the same. Their respective
> "profiles" would be a triangle (linear interpolation) and a bell curve (a
> Gaussian function.)
>
> --
> Simon Sapin
>
Received on Monday, 4 February 2013 23:27:01 GMT

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