适合诵读困难者的乳胶包装

Question

我的物理课上有个学生患有阅读障碍，我需要为他们调整一些内容。我一直在研究这个问题，但找不到很多相关资料。

我已经找到了打开 Dyslexia 字体这可能会有所帮助，但我还被要求在单词之间设置水平间距像这样更宽。

有没有办法可以全局执行此操作而不影响文档的对齐？

Answer 1

您可以影响单词间（和句子间）的间距，如下所示，可能需要增加基线间距来补偿，所以我也展示了这一点。

\documentclass[12pt]{article}

\begin{document}

\sffamily

One two three four five six seven eight nine ten eleven twelve.
One two three four five six seven eight nine ten eleven twelve.
One two three four five six seven eight nine ten eleven twelve.

\bigskip

\setlength\spaceskip{.75cm plus .5cm minus .25cm}
\setlength\xspaceskip{1cm plus .75cm minus .25cm}
\renewcommand\baselinestretch{1.2}\selectfont


One two three four five six seven eight nine ten eleven twelve.
One two three four five six seven eight nine ten eleven twelve.
One two three four five six seven eight nine ten eleven twelve.

\end{document}

Answer 2

正如评论中提到的那样，英国阅读障碍协会有一个时尚指南这为创作适合阅读障碍人士的材料提供了一个良好的起点。

• 正如评论中提到的，患有阅读障碍的学生会有各种各样的偏好，这只是一个起点。
• 我觉得有趣的是，没有提到针对阅读障碍的特定字体。

这是一份用来fontspec与样式指南对齐的文档。它使用无衬线字体（Deja Vu Sans）和匹配的数学字体（TeX Gyre DejaVu Math）。 fontspec用于调整字间距，并setspace调整行间距。

\documentclass{article}
%\url{https://tex.stackexchange.com/q/715510/86}
\usepackage[scale=.7]{geometry}

\usepackage{amsmath}
\usepackage{fontspec}
\usepackage{unicode-math}

\setmainfont[LetterSpace=2, Ligatures=NoCommon, WordSpace={3.5}]{Deja Vu Sans}
\setmathfont{TeX Gyre DejaVu Math}

\parskip=2\baselineskip

\usepackage{setspace}

\begin{document}

\onehalfspacing

  Pythagoras' theorem is often recited as \(a^2 + b^2 = c^2\) and is commonly proven by looking at squares drawn on the sides of a triangle.
  This is problematic.
  Firstly, the \(a\), \(b\), and \(c\) in the formula have \emph{meaning}.
  They are not arbitrary but are the sides of a right-angled triangle.
  Moreover, the side \(c\) has to represent the hypotenuse of this triangle.
  Secondly, the theorem is not actually related to the concept of area.
  It is actually about how similar triangles behave.
  In fact, I prefer to rearrange it as follows.
  Starting with \(a^2 = c^2 - b^2 = (c + b)(c - b)\), then divide through to get \(\frac{a}{c + b} = \frac{c - b}{a}\).
  Or written in ratio form as \(a : c + b = c - b : a\).

  \begin{gather*}
  a^2 + b^2 = c^2 \\
  \int_0^1 \log(x) \mathrm{d} x \\
  \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} \\
  u_n = a + (n - 1) d
  \end{gather*}

  
\end{document}