I've got a dyslexic student in my physics class and I need to adapt some of my content for them. I've been researching about this but I couldn't find a lot of stuff about it.

I already found the Open Dyslexia Font which will probably help, but I was also asked to make the horizontal spacing between words wider  just   like   this.

Is there a way to do it globally without affecting the justification of the document?

  • 2
    If you are using xelatex or lualatex with fontspec you can use the WordSpace property \setmainfont[WordSpace=1.4]{Arial} (change Arial for your OTF font).
    – Mane32
    Apr 13 at 13:10
  • 3
    in classic tex you can use \setlength\spaceskip{1cm plus 1cm minus .5cm} or whatever you need Apr 13 at 14:25
  • Can somebody post a little example with these suggestions? Thank you
    – MS-SPO
    Apr 13 at 15:57
  • 4
    There are different kinds of dyslexia, and what helps one may not do much for another. I’d make small sample documents with different typefaces, leading, etc., and ask which my students preferred. You may like bigelowandholmes.typepad.com/bigelow-holmes/2014/11/… and bdadyslexia.org.uk/advice/employers/….
    – Thérèse
    Apr 13 at 16:58
  • 1
    @Thérèse As I understand the question, these are specific requests for a specific student, so the things the OP is trying to do are already adapted to that student's needs. (At least, this fits my experience of how it usually works in both the US and UK. Somebody else figures out what does/doesn't help and then instructors get requests specific to the students they're teaching. So the recommendations are figured out once and distributed to all the instructors who need to know. For lifelong learning students it is often different, but then I may be the one suggesting screening.)
    – cfr
    Apr 14 at 6:11

2 Answers 2


You can affect the interword (and inter sentence) space as shown below, probably you need to increase baseline spacing to compensate so I showed that as well.

enter image description here




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.


\setlength\spaceskip{.75cm plus .5cm minus .25cm}
\setlength\xspaceskip{1cm plus .75cm minus .25cm}

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.


As mentioned in the comments, the British Dyslexia Association has a style guide which provides a good starting point for creating material that is suitable for people with dyslexia.

  • As also mentioned in the comments, students with dyslexia will have a variety of preferences and this is intended as a starting point.
  • I find it intriguing that there is no mention of dyslexia-specific fonts.

Here's a document using fontspec to align to the style guide. It uses a sans serif font (Deja Vu Sans) with matched math font (TeX Gyre DejaVu Math). fontspec is used to adjust the word spacing, and setspace to adjust the line spacing.



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





  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 \(a : c + b = c - b : a\).

  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


A document showing the result of the above code where the font is sans serif, the word spacing is 3.5, and the line spacing is 1.5

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .