Square root in fraction space problem

Question

I'm having some troubles with the square root symbol within a fraction. This code

\documentclass{book}

\begin{document}
\begin{equation}
h_4 = \frac{\sqrt{1.5 \gamma_4 -3.5}-1}{18}
 \end{equation}
\begin{equation}
h_3 = \frac{\gamma_3}{4+ \sqrt{6\gamma_4-14}}
\end{equation}
\end{document}

returns this

Now, is it just me, or the horizontal line of the square root is really close both to the numbers below it and to the fraction line above it?

I find this a little clumsy.

Is there a solution? Or this is just my paranoia? :P

Thank you very much

Answer 1

If it is not too large, \strut may be a solution (an original one left for comparision). According Mico's sugestion, \mathstrut gives a value between them. You can also adjust vertical spacing exactly to expected values, using e.g. \rule.

\documentclass{book}
\newcommand\uprule{\rule{0mm}{1.9ex}} %shortcut macro
\begin{document}
\begin{equation}
h_4 = \frac{\sqrt{1.5 \gamma_4 -3.5}-1}{18}
 \end{equation}
\begin{equation}
h_3 = \frac{\gamma_3}{4+ \sqrt{6\gamma_4-14}}
\end{equation}

strut

\begin{equation}
h_4 = \frac{\sqrt{\strut1.5 \gamma_4 -3.5}-1}{18}
 \end{equation}
\begin{equation}
h_3 = \frac{\gamma_3}{4+ \sqrt{\strut6\gamma_4-14}}
\end{equation}

mathstrut

\begin{equation}
h_4 = \frac{\sqrt{\mathstrut1.5 \gamma_4 -3.5}-1}{18}
 \end{equation}
\begin{equation}
h_3 = \frac{\gamma_3}{4+ \sqrt{\mathstrut6\gamma_4-14}}
\end{equation}

rule

\begin{equation}
h_4 = \frac{\sqrt{\uprule1.5 \gamma_4 -3.5}-1}{18}
 \end{equation}
\begin{equation}
h_3 = \frac{\gamma_3}{4+ \sqrt{\uprule 6\gamma_4-14}}
\end{equation}
\end{document}

Answer 2

With \mathstrut the size of the square root sign would increase. In these cases, masking the descender is probably the best way around:

\documentclass{book}
\usepackage{amsmath}

\begin{document}
\begin{align}
h_4 &= \frac{\sqrt{1.5 \smash[b]{\gamma^{}_4} -3.5}-1}{18}
\\[1ex]
h_3 &= \frac{\gamma_3}{4+ \sqrt{6\smash[b]{\gamma^{}_4}-14}}
\end{align}
\end{document}

Here is the same with \mathstrut:

\begin{align}
h_4 &= \frac{\sqrt{\mathstrut 1.5 \gamma^{}_4 -3.5}-1}{18}
\\[1ex]
h_3 &= \frac{\gamma_3}{4+ \sqrt{\mathstrut 6 \gamma^{}_4-14}}
\end{align}