Philippe Wittenbergh wrote:
> In a perfect world, 'Arial Black' shouldn't get bolded, as it already
> is a bold face with "weight=900".
That would be the natural interpretation, but then "Arial Black" should
not be a font-family value at all, just a particular font with
font-family: Arial and font-weight: 900. However, many browsers probably
treat it as a font family of its own.
> You can't get any bolder than that.
In CSS terms you cannot, but in typography, there might be extra black.
For some odd reason, CSS font-weight has theoretically 9 values, whereas
typographic weight scale may have 10 values.
> And indeed, testing on OS X: WebKit, Gecko1.9 (fx3), Opera do not
> render it bolder than it already is. Gecko 1.8 (fx2) does make it
> bolder - at least on OS X 10.4 & 10.5. But that is a bug.
IE 7 bolds it, too.
Whether it's strictly a bug is a matter of definition. If "Arial Black"
is taken as a font family of its own, with only one true weight
available, then it might arguably be acceptable to generated other
weights true algorithmic transformations. Actually that's what IE does
with Arial Unicode MS, for example; if specify font-weight: bold for it,
IE will produce something bold, even though Arial Unicode MS has only
one weight available. This is comparable to "fake italics".
Anyway, there is no way in CSS to say "bold this if the font family is
not xxx" or anything like that.
Jukka K. Korpela ("Yucca")
http://www.cs.tut.fi/~jkorpela/
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