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Thread: Vertical Spacing

  1. #1
    Rai
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    Vertical Spacing

    Hello, had a quick search but couldn't find any other threads about this.

    I'm using the align environment to typeset a series of equations. They are fairly large equations with lots of fractions and as such are looking rather squashed together with the standard spacing. I was wondering if there's a way of stretching the vertical spacing within this environment? I did have a play at using \smallskip etc, but it doesn't seem to make any difference - I'm not sure if I'm just putting it in the wrong part of the line though.

    Cheers,
    Rai
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Can you show a picture or something?
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  3. #3
    Rai
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    Hmmmm, for some reason it's telling me there's a Syntax error when I try to post it using [ math] [ /math] to show you. The code copied straight from my document is

    \begin{align*}
    g'_{real}&=C_g-2.5\log(ADU_g)_{instr}-b_g[-2.5\log(ADU_g)_{instr}+2.5\log(ADU_r)_{instr}]\\
    &=2.5\log(10^{C_g/2.5})-2.5\log(ADU_g)_{instr}-b_g[2.5\log(ADU_r)_{instr}-2.5\log(ADU_g)_{instr}]\\
    &=2.5\log\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)-b_g\left[2.5\log\left(\frac{ADU_r}{ADU_g}\right)_{instr}\ri ght]\\
    &=2.5\log\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)-\left[2.5\log\left(\frac{ADU_r}{ADU_g}\right)_{instr}^{b _g}\right]\\
    &=2.5\log\left[\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)\left/\left(\frac{ADU_r}{ADU_g}\right)_{instr}^{b_g}\rig ht.\right]
    \end{align*}

    if you want to compile it to see for yourself.

    Cheers
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    It looks pretty cool to me.

    But if you want to stretch vertical spacing just add another \\ when you finish a line.

    For example

    \begin{align*}
    g'_{real}&=C_g-2.5\log(ADU_g)_{instr}-b_g[-2.5\log(ADU_g)_{instr}+2.5\log(ADU_r)_{instr}]\\\\
    &=2.5\log(10^{C_g/2.5})-2.5\log(ADU_g)_{instr}-b_g[2.5\log(ADU_r)_{instr}-2.5\log(ADU_g)_{instr}]\\\\
    &=2.5\log\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)-b_g\left[2.5\log\left(\frac{ADU_r}{ADU_g}\right)_{instr}\ri ght]\\\\
    &=2.5\log\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)-\left[2.5\log\left(\frac{ADU_r}{ADU_g}\right)_{instr}^{b _g}\right]\\\\
    &=2.5\log\left[\left(\frac{10^{C_g/2.5}}{(ADU_g)_{instr}}\right)\left/\left(\frac{ADU_r}{ADU_g}\right)_{instr}^{b_g}\rig ht.\right]
    \end{align*}

    You can also use the eqnarray environment.
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  5. #5
    Rai
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    Thanks, that looks much clearer now.
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