適合唸讀困難者的乳膠包裝

Question

我的物理課上有一名患有誦讀困難的學生，我需要為他們調整一些內容。我一直在研究這個問題，但找不到很多相關內容。

我已經找到了打開誦讀困難字體這可能會有所幫助，但我還被要求在單字之間設定水平間距像這樣更寬。

有沒有一種方法可以在全域範圍內進行而不影響文件的合理性？

Answer 1

您可以影響字間（和句子間）空間，如下所示，可能您需要增加基線間距來補償，所以我也展示了這一點。

\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

正如評論中提到的，英國閱讀障礙協會有一個時尚指南這為創建適合閱讀障礙者的材料提供了一個良好的起點。

• 正如評論中所提到的，患有閱讀障礙的學生會有各種偏好，這只是一個起點。
• 我發現有趣的是沒有提到閱讀障礙特定的字體。

fontspec這是一個用於與樣式指南保持一致的文件。它使用無襯線字體 (Deja Vu Sans) 和匹配的數學字體 (TeX Gyre DejaVu Math)。 fontspec用於調整字間距，以及setspace調整行間距。

\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}