Hi Troels,
One more LaTeX tip here ;) If you change:
+ & - \frac{1}{T_{\textrm{rel}}}\ln{\left(
\frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D
)\right)} \\
to:
+ & \qquad - \frac{1}{T_{\textrm{rel}}}\ln{\left(
\frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D
)\right)} \\
see how that looks for you. Or there is the phantom trick I added to
http://wiki.nmr-relax.com/CR72 and http://wiki.nmr-relax.com/CR72_full
pages
(http://wiki.nmr-relax.com/index.php?title=CR72&curid=300&diff=2403&oldid=2402
and
http://wiki.nmr-relax.com/index.php?title=CR72_full&curid=317&diff=2404&oldid=2399).
In this case:
+ & \phantom{=} -
\frac{1}{T_{\textrm{rel}}}\ln{\left( \frac{1+y}{2} +
\frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D )\right)} \\
The formatting idea is that the multi-line maths should align to the
right of the equal sign.
Regards,
Edward
On 7 May 2014 10:14, <tlin...@nmr-relax.com> wrote:
> Author: tlinnet
> Date: Wed May 7 10:14:09 2014
> New Revision: 23030
>
> URL: http://svn.gna.org/viewcvs/relax?rev=23030&view=rev
> Log:
> Used LaTeX subequations instead, and using R2eff parameter is defined in the
> relax.tex
>
> Using the defined \Rtwoeff, \RtwozeroA, \RtwozeroB, \kAB, \kBA, \kex.
>
> sr #3154: (https://gna.org/support/?3154) Implementation of Baldwin (2014)
> B14 model - 2-site exact solution model for all time scales.
>
> This follows the tutorial for adding relaxation dispersion models at:
> http://wiki.nmr-relax.com/Tutorial_for_adding_relaxation_dispersion_models_to_relax#The_relax_manual
>
> Modified:
> trunk/docs/latex/dispersion.tex
>
> Modified: trunk/docs/latex/dispersion.tex
> URL:
> http://svn.gna.org/viewcvs/relax/trunk/docs/latex/dispersion.tex?rev=23030&r1=23029&r2=23030&view=diff
> ==============================================================================
> --- trunk/docs/latex/dispersion.tex (original)
> +++ trunk/docs/latex/dispersion.tex Wed May 7 10:14:09 2014
> @@ -565,21 +565,26 @@
> This is the model for 2-site exchange exact analytical derivation on all
> time scales (with the constraint that $\pA > \pB$), named after
> \citet{Baldwin2014}.
> It is selected by setting the model to `B14 full'.
> The equation is
> -\begin{eqnarray}
> - R_{2,\textrm{eff}} & = &
> \frac{R_2^A+R_2^B+k_{\textrm{EX}}}{2}-\frac{N_{\textrm{CYC}}}{T_{\textrm{rel}}}\cosh{}^{-1}(v_{1c})
> \nonumber \\
> - & - & \frac{1}{T_{\textrm{rel}}}\ln{\left(
> \frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2k_{\textrm{AB}}p_D
> )\right)} \nonumber \\
> - & = & R_{2,\textrm{eff}}^{\textrm{CR72}} -
> \frac{1}{T_{\textrm{rel}}}\ln{\left( \frac{1+y}{2} +
> \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2k_{\textrm{AB}}p_D )\right)} ,
> -\end{eqnarray}
> +\begin{subequations}
> +\begin{align}
> + \Rtwoeff & = \frac{\RtwozeroA + \RtwozeroB + \kex }{2}-\frac{
> N_{\textrm{CYC}} }{ T_{\textrm{rel}} } \cosh{}^{-1}(v_{1c}) \\
> + & - \frac{1}{T_{\textrm{rel}}}\ln{\left( \frac{1+y}{2}
> + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D )\right)} \\
> + & = \Rtwoeff^{\textrm{CR72}} - \frac{1}{T_{\textrm{rel}}}\ln{\left(
> \frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2\kAB p_D )\right)} ,
> +\end{align}
> +\end{subequations}
> +
>
> where
> -\begin{eqnarray}
> - v_{1c} & = &
> F_0\cosh{\left(\tau_{\textrm{CP}}E_0\right)}-F_2\cosh{\left(\tau_{\textrm{CP}}E_2\right)}
> \nonumber \\
> - v_{1s} & = &
> F_0\sinh{\left(\tau_{\textrm{CP}}E_0\right)}-F_2\sinh{\left(\tau_{\textrm{CP}}E_2\right)}
> \nonumber \\
> - v_{2}N & = & v_{1s}\left(O_B-O_A\right)+4O_B F_1^a
> \sinh{\left(\tau_{\textrm{CP}}E_1\right)} \nonumber \\
> - p_D N & = & v_{1s} +
> \left(F_1^a+F_1^b\right)\sinh{\left(\tau_{\textrm{CP}}E_1\right)} \nonumber \\
> - v_3 & = & \left( v_2^2 + 4 k_{\textrm{BA}} k_{\textrm{AB}} p_D^2
> \right)^{1/2} \nonumber \\
> - y & = & \left( \frac{v_{1c}-v_3}{v_{1c}+v_3} \right)^{N_{\textrm{CYC}}}
> -\end{eqnarray}
> +\begin{subequations}
> +\begin{align}
> + v_{1c} & =
> F_0\cosh{\left(\tau_{\textrm{CP}}E_0\right)}-F_2\cosh{\left(\tau_{\textrm{CP}}E_2\right)}
> \\
> + v_{1s} & =
> F_0\sinh{\left(\tau_{\textrm{CP}}E_0\right)}-F_2\sinh{\left(\tau_{\textrm{CP}}E_2\right)}
> \\
> + v_{2}N & = v_{1s}\left(O_B-O_A\right)+4O_B F_1^a
> \sinh{\left(\tau_{\textrm{CP}}E_1\right)} \nonumber \\
> + p_D N & = v_{1s} +
> \left(F_1^a+F_1^b\right)\sinh{\left(\tau_{\textrm{CP}}E_1\right)} \\
> + v_3 & = \left( v_2^2 + 4 \kBA \kAB p_D^2 \right)^{1/2} \\
> + y & = \left( \frac{v_{1c}-v_3}{v_{1c}+v_3} \right)^{N_{\textrm{CYC}}}
> +\end{align}
> +\end{subequations}
>
> The advantage of this code will be that you will always get the right answer
> provided you got 2-site exchange, in-phase magnetisation and on-resonance
> pulses.
>
>
>
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