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1238pt = 1157dd 是 Tex 中 Didot 点的准确定义吗？

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1238pt = 1157dd 是 Tex 中 Didot 点的准确定义吗？

阅读完以下答案后以 mm 表示的各种单位 (ex、em、in、pt、bp、dd、pc) 是什么？和一些相关问题，我意识到我对 Tex 的一个基本假设是错误的：我认为由于缩放点sp是基本内部基本单位（约 5.36 nm），所以所有其他（绝对）单位都将以整数表示。这应该是下面顶级项目符号中的值，粗体值由源代码报告，如下所示1pt = \number\dimexpr 1pt\relax sp：

  • 1pt=65536sp（精确定义，2^16）
  • 1pc= 12pt=786432sp
  • 1bp= 65781.76sp⇒65781sp至65782sp（72.27/72 = 803/800 pt/bp）
    • 来自 int in： 相同的 int 值
    • 100bp = \number\dimexpr 100bp\relax sp⇒6578176sp（不是 6578100sp！）
  • 1in= 72.27pt = 4736286.72sp ⇒4736286sp到4736287sp
    • 从 int bp: 72bp=4736232sp到4736304sp
    • 100in = \number\dimexpr 100in\relax sp⇒473628672sp（不是 473628600sp！）

好的，点和英寸的实现很准确，但是毫米和厘米呢？

  • 1mm= 186467.9811…sp⇒186467sp到186468sp
    • 来自inint pt：相同的 int 值
    • 从intin到bp:186465sp或186466sp到186468sp186469sp
    • 1000mm = \number\dimexpr 1000mm\relax sp⇒ 186467981sp
    • 2540mm = \number\dimexpr 2540mm\relax sp⇒ 473628672sp（= 100in）
  • 1cm= 1864679.811…sp⇒1864679sp到1864680sp
    • 来自inint pt：相同的 int 值
    • 从intin到bp:186465sp或186466sp到186468sp
    • 来自 int mm：10mm= 1864650sp，1864660sp，1864670sp，1864680sp或1864690sp
    • 100cm = \number\dimexpr 100cm\relax sp⇒ 186467981sp
    • 254cm = \number\dimexpr 254cm\relax sp⇒ 473628672sp（= 100in）

似乎有理由预期实际的内部定义实际上是1in= 1cm= 10mm，但由于值在转换为 时会被截断sp，因此当然会出现舍入误差，如上面链接的答案所示。

但是，Tex 中狄多点（和西塞罗点）的实际定义是什么？它试图近似哪个历史定义？我见过多次提到 1238/1157 pt/dd 或 1.07 pt/dd，但没有实际的权威参考。

  • 1dd=sp或sp(70124sp）
    • ⅜ 月/日：0.375mm=
    • Tschichold 100/266 毫米/日：0.37594mm=
    • 迪多约：0.37597mm=
    • 伯特霍尔德：0.376mm=
    • 16/15 分/日：69905sp
    • 1.07pt=
  • 1cc= 12dd= `` 或sp(841489sp， 不是841488sp！）
    • 4.5mm=
    • Tschichold 1200/266 mm/cc：约。4.5112782mm
    • 伯特霍尔德：4.512mm=
    • 192/15 点/件：12.8pt=
    • 12.84pt=
\noindent1pt = \number\dimexpr 1pt\relax sp

\noindent1pc = \number\dimexpr 1pc\relax sp

\noindent1bp = \number\dimexpr 1bp\relax sp

\noindent1in = \number\dimexpr 1in\relax sp

800bp = 803pt = \number\dimexpr 800bp\relax sp = \number\dimexpr 803pt\relax sp %= 52625408sp

\noindent1mm = \number\dimexpr 1mm\relax sp

\noindent1cm = \number\dimexpr 1cm\relax sp

2540mm = 254cm = 100in = 7200bp = 7227pt \hfill\break %= 473628672sp
= \number\dimexpr 2540mm\relax sp = \number\dimexpr 254cm\relax sp = \number\dimexpr 100in\relax sp = \number\dimexpr 7200bp\relax sp = \number\dimexpr 7227pt\relax sp

\noindent1dd = \number\dimexpr 1dd\relax sp %= 70124sp

1.07pt = \number\dimexpr 1.07pt\relax sp %= 70124sp

Berthold: \number\dimexpr 0.376mm\relax sp %= 70113sp

Didot approx.: \number\dimexpr 0.37597mm\relax sp %= 70107sp

Tschichold: \number\dimexpr 0.37593985mm\relax sp (100/266 mm/dd) %= 70101sp

3/8 mm/dd: \number\dimexpr 0.375mm\relax sp %= 69925sp

16/15 pt/dd: 69905sp

1238pt = 1157dd = \number\dimexpr 1238pt\relax sp = \number\dimexpr 1157dd\relax sp %= 81133568sp

\noindent1cc = 12dd = \number\dimexpr 1cc\relax sp = \number\dimexpr 12dd\relax sp  %= 841489sp != 841488sp

12.84pt = \number\dimexpr 12.84pt\relax sp %= 841482sp != 841488sp

Berthold: \number\dimexpr 4.512mm\relax sp %= 841342sp != 841356sp

Tschichold:  \number\dimexpr 4.5112782mm\relax sp (1200/266 mm/cc) %= 841208sp != 841212sp

4.5mm = \number\dimexpr 4.5mm\relax sp %= 839105sp != 839100sp

192/15 pt/pc: \number\dimexpr 12.8pt\relax sp %= 838861sp != 838860sp

1238pc = 1157cc = \number\dimexpr 1238pc\relax sp = \number\dimexpr 1157cc\relax sp %= 973602816sp
\vskip1em

10000pt = \number\dimexpr 10000pt\relax sp %= 655360000sp

10000bp = \number\dimexpr 10000bp\relax sp %= 657817600sp

10000dd = \number\dimexpr 10000dd\relax sp %= 701240864sp
\bye

答案1

哎哟，在我发布问题几分钟后，我想我自己找到了答案：

13773 行及后续行。

The next two parameters, |num| and |den|, are positive integers that define
the units of measurement; they are the numerator and denominator of a
fraction by which all dimensions in the \.{DVI} file could be
multiplied in order to get lengths in units of $10^{-7}$ meters. Since
$\rm 7227{pt} = 254{cm}$, and since \TeX\ works with scaled points
where there are $2^{16}$ sp in a point, \TeX\ sets
$|num|/|den|=(254\cdot10^5)/(7227\cdot2^{16})=25400000/473628672$.
@^sp@>

行 #10443ff。

@ The necessary conversion factors can all be specified exactly as
fractions whose numerator and denominator sum to 32768 or less.
According to the definitions here, $\rm2660\,dd\approx1000.33297\,mm$;
this agrees well with the value $\rm1000.333\,mm$ cited by Bosshard
@^Bosshard, Hans Rudolf@> in {\sl Technische Grundlagen zur
Satzherstellung\/} (Bern, 1980). The Didot point has been newly
standardized in 1978; it's now exactly $\rm 1\,nd=0.375\,mm$.
Conversion uses the equation $0.375=21681/20320/72.27\cdot25.4$. The
new Cicero follows the new Didot point; $\rm 1\,nc=12\,nd$. These
would lead to the ratios $21681/20320$ and $65043/5080$, respectively.
The closest approximations supported by the algorithm would be
$11183/10481$ and $1370/107$.  In order to maintain the relation $\rm
1\,nc=12\,nd$, we pick the ratio $685/642$ for $\rm nd$, however.

行 #10460ff。

@d set_conversion_end(#)== denom:=#; end
@d set_conversion(#)==@+begin num:=#; set_conversion_end

@<Scan for \(a)all other units and adjust |cur_val| and |f|...@>=
if scan_keyword("in") then set_conversion(7227)(100)
@.in@>
else if scan_keyword("pc") then set_conversion(12)(1)
@.pc@>
else if scan_keyword("cm") then set_conversion(7227)(254)
@.cm@>
else if scan_keyword("mm") then set_conversion(7227)(2540)
@.mm@>
else if scan_keyword("bp") then set_conversion(7227)(7200)
@.bp@>
else if scan_keyword("dd") then set_conversion(1238)(1157)
@.dd@>
else if scan_keyword("cc") then set_conversion(14856)(1157)
@.cc@>
else if scan_keyword("nd") then set_conversion(685)(642)
@.nd@>
else if scan_keyword("nc") then set_conversion(1370)(107)
@.nc@>
else if scan_keyword("sp") then goto done
@.sp@>

因此 1238/1157 pt/dd 或 pc/cc 确实是精确的，并且 375 µm 的“公制狄多点”及其西塞罗实际上有两个单位：nd和nc：

  • 1nd= ⌊685/642 pt⌋ = 69925sp ≈ 0.375mm
  • 1nc= ⌊1370/107 pt⌋ = 839105sp ≈ 4.5mm

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