Horizontal braces with XeTeX take too much space

Question

After some experiments and tests, I managed to create the following code for \upbracefill and \downbracefill in Plain XeTeX, which it seems is that used for \underbrace and \overbrace:

\def\midshift#1{
\setbox0=\hbox{#1}\dimen0=\ht0\advance\dimen0by+\dp0\advance\dimen0by-1ex
\lower.5\dimen0\box0 }

\def\rotatebrace#1{%
\leavevmode\setbox0=\hbox{#1}\rlap{%
\kern.5\wd0\dimen0=\ht0\advance\dimen0by-\dp0%\advance\dimen0by+1ex%
\raise.5\dimen0\hbox{\special{x:gsave}\special{x:rotate 90}}}%
\box0\special{x:grestore}}

[...]

\XeTeXmathchardef\bracelu = 0 3 `\⎧
\XeTeXmathchardef\bracemu = 0 3 `\⎨
\XeTeXmathchardef\braceru = 0 3 `\⎩
\XeTeXmathchardef\bracebar = 0 3 `\⎪
\XeTeXmathchardef\braceld = 0 3 `\⎫
\XeTeXmathchardef\bracemd = 0 3 `\⎬
\XeTeXmathchardef\bracerd = 0 3 `\⎭

\def\upbracefill{\rotatebrace{\midshift{$\bracelu$}}%
    \cleaders\hbox{\rotatebrace{\midshift{$\bracebar$}}}\hfill
    \rotatebrace{\midshift{$\bracemu$}}%
    \cleaders\hbox{\rotatebrace{\midshift{$\bracebar$}}}\hfill
    \rotatebrace{\midshift{$\braceru$}}}%
\def\downbracefill{\rotatebrace{\midshift{$\braceld$}}%
    \cleaders\hbox{\rotatebrace{\midshift{$\bracebar$}}}\hfill
    \rotatebrace{\midshift{$\bracemd$}}%
    \cleaders\hbox{\rotatebrace{\midshift{$\bracebar$}}}\hfill
    \rotatebrace{\midshift{$\bracerd$}}}%

Where Cambria Math is loaded for the 3. family.
However, the height of these braces are bigger because of the middle part of the brace, and I can't find a way to shrink/limit the height of the box, without which the brace would fall apart.

I originally used the :vertical parameter for the font, which I figured has no dimension in horizontal text.

So how can I make this horizontal brace height smaller? (preferably to the character width, which is now the height.)

Answer 1

After I saw that \upbracefill is used for \underbrace, I felt the need to correct this problem, so I take some time to it and look at it to see what choices do I have. Soon I realized that the only good solution is if I shrink the bounding box of each character, with the help of \setbox. (The only problem was to do this in the characters right state.) Since the most biggest character is the middle one, I arrange their height to that. (The rest of that characters height is aligned to that, however 0pt height is also ok.)
So here is the result using the above \rotatebrace and \midshift macros:

\def\upbracefill{%
    \setbox0=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\bracemu$}}}}\ht0=.1\wd0\dp0=0pt%
    \setbox1=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\bracelu$}}\kern-.2em}}\ht1=.1\wd0\dp1=0pt%
    \setbox2=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\bracebar$}}}}\ht2=.1\wd0\dp2=0pt%
    \setbox3=\hbox{\lower.64ex\hbox{\kern-.2em\rotatebrace{\midshift{$\braceru$}}}}\ht3=.1\wd0\dp3=0pt%
    \box1\cleaders\copy2\hfill\box0\cleaders\box2\hfill\box3}

\def\downbracefill{%
    \setbox0=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\bracemd$}}}}\ht0=.1\wd0\dp0=0pt%
    \setbox1=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\braceld$}}\kern-.2em}}\ht1=.1\wd0\dp1=0pt%
    \setbox2=\hbox{\lower.64ex\hbox{\rotatebrace{\midshift{$\bracebar$}}}}\ht2=.1\wd0\dp2=0pt%
    \setbox3=\hbox{\lower.64ex\hbox{\kern-.2em\rotatebrace{\midshift{$\bracerd$}}}}\ht3=.1\wd0\dp3=0pt%
    \box1\cleaders\copy2\hfill\box0\cleaders\box2\hfill\box3}%

Here is an example: The example above uses Calibri for normal text and Cambria for mathematical equations, (just as Office 2007 does) as well as for the braces, as in a Hungarian notes of a class of mine.
Note: In some viewer in certain zoom it appears as if it would fallen apart, and the join of the characters seems thicker than they should be. This will be OK in print.