# delimiters with exact size

I'm looking to get a delimiter at an exact and user specified (point) size, rather than the default "a discrete size big enough to hold the contents" of \left( etc.

Is this possible? I've tried futzing around with the definitions of \big*,

\newlength{\vbraceheight}
\setlength{\vbraceheight}{50pt}
\def\vbig#1{{\hbox{$\left#1\vbox to\vbraceheight{}\right.\n@space$}}}
\def\vbigl{\mathopen\vbig}
\def\vbigr{\mathclose\vbig}


but these still have jumps as one varies \vbraceheight continuously (and I'm not really sure why).

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Does this answer help? –  Ian Thompson Aug 20 '12 at 16:04
@Ian, thanks, but is seems not -- as the final comment on the answer says: the size goes up in discrete jumps depending on the font size (I am not sure why this is the case) –  n00b Aug 20 '12 at 16:20
@n00b -- the reason for discrete jumps is that large braces are constructed from pieces, and the size of the extension pieces is a fixed vertical length. for a brace, two pieces are used, one in the top half and one in the bottom half. the glyph is located in font cmex10 at position hex 3E (octal 76). –  barbara beeton Aug 20 '12 at 18:35

I would capture the contents of \vbig and resize vertically (keeping the aspect ratio) to \vbraceheight:

\documentclass{article}
\usepackage{graphicx}% http://ctan.org/pkg/graphicx
\makeatletter
\newlength{\vbraceheight}
\setlength{\vbraceheight}{50pt}
\def\vbig#1{{\resizebox{!}{\vbraceheight}{$\left#1\vbox to\vbraceheight{}\right.\n@space$}}}
\def\vbigl{\mathopen\vbig}
\def\vbigr{\mathclose\vbig}
\makeatother
\begin{document}
$\setbox0=\hbox{\displaystyle\vbigl\{\frac{1}{2}\vbigr\}}\usebox0\ \mbox{has height \the\ht0}$
\end{document}

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[Answer written by OP, moved from question]

Thanks for the comments and reply. I think I've got my head around why this is a difficult thing to do. I posted a similar question on comp.text.tex and received the following "manual construction" from Dan Luecking, I thought it worth sharing (I have made some minor spacing modification and include the code only for the left brace):

\DeclareMathSymbol{\bracetr}{\mathord}{largesymbols}{"38}
\DeclareMathSymbol{\bracetl}{\mathord}{largesymbols}{"39}
\DeclareMathSymbol{\bracebr}{\mathord}{largesymbols}{"3A}
\DeclareMathSymbol{\bracebl}{\mathord}{largesymbols}{"3B}
\makeatletter
\def\leftbracefill{%
\setbox\z@\hbox{$\m@th\bracetr$}%
\@tempdima=.5\wd\z@
\offinterlineskip
\hbox{\kern-\@tempdima$\m@th\bracetr$}
\hbox{\kern-\@tempdima$\m@th\bracebl$}
\kern-.5\p@
\hbox{\kern-\@tempdima$\m@th\bracetl$}
\hbox{\kern-\@tempdima$\m@th\bracebr$}
}
\newcommand\leftbrace[1]{%
\setbox\z@\hbox{$\m@th\bracetl$}%
\@tempdima=.5\wd\z@
\hbox{$\m@th\displaystyle{#1}$}\crcr