[texhax] 1em after scaling/rounding

[Barbara Beeton]

what's going on here is based on the philosophy that
typefaces in different sizes should have different
design sizes.  in this, computer modern follows the
conventions of generations of metal type, where the
width of a particular letter is proportionally wider
in small sizes, and proportionally narrower in large
sizes.  (you've probably noticed that newspaper
headlines have quite narrow type.)  this enhances
readability and legibility, especially in math,
where individual letters are often used in isolation.

there's a nice example in the texbook, on p.16, second
paragraph from the bottom, that shows the difference
between 10pt and 5pt cmr.  in fact, much of that
chapter is instructive in regard to design sizes.

anyhow, in tex terminology, 1em is a multiple of the
unit width, which is defined as part of the design;
this is covered more thoroughly in volume e (the
computer modern fonts) of computers & typesetting.

one could dispute whether the definition of em in
tex is traditional, but that's possibly subjective.
one of my "back-burner" projects is to determine
from actual historical printed examples whether
in fonts of different sizes the widths of the
em-dash and en-dash are absolute (based on nominal
font size) or proportional in the same manner as
the letters.
 							-- bb

---------- Original message ----------
Date: Wed, 09 Nov 2005 11:17:30 +0100
From: "Uwe [iso-8859-1] Lück" <uwe.lueck at web.de>
To: texhax at tug.org
Cc: Alexander Rozhenko <roj at front.ru>
Subject: [texhax] 1em after scaling/rounding

Some rather academic/numerical question
(to font experts like Karl Berry or experts in numerics like Alex Rozhenko):

According to TeXbook p. 433, 1em is 10pt with unscaled cmr10.
I am measuring by   \normalsize\setbox0\hbox to...{\hfil}\showthe\wd0.
This yields 1em = 1.00002 pt -- OK, a rounding error
that doesn't matter at all.

Now I load \documentclass[12pt]{article}.
I get 1em = 11.74988pt, i.e.: 11.75pt
-- to me this appears very different from which one would expect
(12pt).
By contrast:   \normalsize %% standard 10pt
   \@tempdima = 1em
   \multiply\@tempdima by 6
   \divide\@tempdima by 5
   \showthe\@tempdima
The final line yields 12pt -- when ignoring
the same absolute rounding error that 10pt produces.

... any ideas about 12.00002pt vs. 11.74988pt, i.e.,
12pt vs. 11.75pt!?

Best,
   Uwe.