1

The effect of \strut on the vertical spacing under a square root is very severe. I would like the effect to be about 20% of what it is. Otherwise, strut solves a lot of problems.

% without \strut
\noindent \hspace{18pt}\large$\bigg[\hspace{1.4pt}x\hspace{1.4pt}'\hspace{-0.6pt}+\dfrac{\big(\hspace{0.4pt}9\hspace{-0.6pt}-\hspace{-0.6pt}5\hspace{0.4pt}\surd{\hspace{1.0pt}2}\hspace{0.8pt}\big) \hspace{1.2pt}\sqrt{  \hspace{1.2pt}4\hspace{-0.6pt}+\hspace{-0.6pt}2\hspace{0.4pt}\surd{\hspace{1.0pt}2}}}{62}\hspace{0.8pt}\bigg]^{\hspace{0.8pt}2}$

% with \strut
\vspace{6pt}
\noindent \hspace{18pt}$\bigg[\hspace{1.4pt}x\hspace{1.4pt}'\hspace{-0.6pt}+\dfrac{\big(\hspace{0.4pt}9\hspace{-0.6pt}-\hspace{-0.6pt}5\hspace{0.4pt}\surd{\hspace{1.0pt}2}\hspace{0.8pt}\big) \hspace{1.2pt}\sqrt{\strut\hspace{1.2pt}4\hspace{-0.6pt}+\hspace{-0.6pt}2\hspace{0.4pt}\surd{\hspace{1.0pt}2}}}{62}\hspace{0.8pt}\bigg]^{\hspace{0.8pt}2}$
TeXnician
  • 33,589
keith77777
  • 424

1 Answers1

2

You will have to try to use a bit smaller \strut. As you can see in the following example, a difference of 0.1pt can lead to quite a jump in the size:

\documentclass{article}
\usepackage{amsmath}

\begin{document}
 $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\ht\strutbox}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.1pt}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.2pt}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.3pt}(x_2-x_1)^2+(y_2-y_1)^2}$.
\end{document}

enter image description here

Ulrike Fischer
  • 327,261
  • Ulrike, Yes, the effect is very much non-linear. The effect is hardly perceptible, then dramatically too much. – keith77777 Aug 17 '17 at 17:55
  • Barbara, I've tried \mathstrut, but is appears to do nothing. Whatever I try, appears to give a non-linear result: i.e. very little space, then a huge change. It is as if there is an either/or in the code. – keith77777 Aug 17 '17 at 18:09
  • Barbara Beeton/ Ulrike Fischer I think this is a common problem for many writers with no satisfactory solution; could someone look into it, please. – keith77777 Aug 21 '17 at 19:20
  • There is not much anyone can do. It is the way the root is implemented: At some point it switches to the next symbol with a more vertical base. So it is never linear. – Ulrike Fischer Aug 21 '17 at 19:53
  • I agree, that's the way it currently is, but that's not to say it cannot be fixed so as to be perfectly progressive, is it? – keith77777 Aug 22 '17 at 21:41
  • perfectly progressive would be a bit much -- you can't put an infinity of glyphs in a font. But if you want more intermediary root symbols you will have to contact the various font designers. I would concentrate on the open type math fonts. – Ulrike Fischer Aug 22 '17 at 22:07