\subsection{Layered media}\inputdir{vrms}\parFrom the assumption that experimental datacan be fit to hyperbolas(each with a different velocity and each with a different apex $\tau$)let us next see howwe can fit an earth model of layers,each with a constant velocity.Consider the  horizontal reflectoroverlain by a stratified \bx{interval velocity} $v(z)$ shown in Figure~\ref{fig:stratrms}.%\sideplot{stratrms}{width=3.00in}{  Raypath diagram for normal moveout in a stratified earth.} %The separation between the source and geophone,also called the offset, is $2h$ and the total travel time is $t$.Travel times are not be precisely hyperbolic,but it is common practice to find the best fitting hyperbolas,thus finding the function $V^2(\tau)$.\begin{equation}t^2 \eq \tau^2 + \frac{4h^2}{V^2(\tau)}\label{eqn:vrmshyp}\end{equation}where $\tau$ is the zero-offset two-way traveltime.\inputdir{vscan}\parAn example of using equation~(\ref{eqn:vrmshyp})to stretch $t$ into $\tau$is shown in Figure~\ref{fig:nmogath}.(The programs thatfind the required $V(\tau )$ and do the stretching are coming up inchapter~\ref{vela/paper:vela}.)\plot{nmogath}{width=6.00in,height=3.6in}{  If you are lucky and get a good velocity,  when you do NMO, everything turns out flat.  Shown with and without mute.}\parEquation (\ref{eqn:vrmsdefine}) shows that$V(\tau)$isthe ``root-mean-square'' or``\RMS'' velocity defined byan average of $v^2$ over the layers.Expressing it for a small number of layers we get\begin{equation}V^2(\tau) \eq \frac{1}{\tau}\  \sum_i v^2_i \Delta\tau_i\label{eqn:vrmsdefn}\end{equation}where the zero-offset traveltime $\tau$ is a sum over the layers:\begin{equation}\tau \eq \sum_i \ \Delta\tau_i\label{eqn:onetaudefn}\end{equation}The two-way vertical travel time $\tau_i$in the $i$th layer is related to thethickness $\Delta z_i$ and the velocity $v_i$  by\begin{equation}\Delta\tau_i \eq  \frac{2\ \Delta z_i}{v_i}  \ \ .\label{eqn:twotaudefn}\end{equation}\parNext we examine an important practical calculation,getting interval velocities from measured RMS velocities:Definein layer $i$,the interval velocity $v_i$and the two-way vertical travel time $\Delta\tau_i$.Define the RMS velocityof a reflectionfrom the bottom of the $i$-th layerto be $V_i$.Equation (\ref{eqn:vrmsdefn}) tells us that forreflections from the bottom of the first, second, and third layers we have\begin{eqnarray}V_1^2 &=& {v_1^2\Delta\tau_1                                          \over \Delta\tau_1                 }\\\label{eqn:bot2}V_2^2 &=& {v_1^2\Delta\tau_1+ v_2^2\Delta\tau_2                                \over \Delta\tau_1 + \Delta\tau_2        }\\\label{eqn:bot3}V_3^2 &=& {v_1^2\Delta\tau_1+ v_2^2\Delta\tau_2 +v_3^2\Delta\tau_3                                 \over \Delta\tau_1 + \Delta\tau_2 +\Delta\tau_3}\end{eqnarray}Normally it is easy to measure the times of the three hyperbola tops,$\Delta\tau_1$, $\Delta\tau_1 + \Delta\tau_2$and$\Delta\tau_1 + \Delta\tau_2 +\Delta\tau_3$.Using methods in chapter \ref{vela/paper:vela}we can measure the RMS velocities $V_2$ and $V_3$.With these we can solve for the interval velocity $v_3$ in the third layer.Rearrange (\ref{eqn:bot3}) and (\ref{eqn:bot2}) to get\begin{eqnarray}\label{eqn:next3}                                    (\Delta\tau_1 + \Delta\tau_2 +\Delta\tau_3)V_3^2 &=& v_1^2\Delta\tau_1+ v_2^2\Delta\tau_2 +v_3^2\Delta\tau_3  \\\label{eqn:next2}                                    (\Delta\tau_1 + \Delta\tau_2)V_2^2 &=& v_1^2\Delta\tau_1+ v_2^2\Delta\tau_2 \end{eqnarray}and subtract getting the squared interval velocity $v_3^2$\begin{equation}v_3^2 \eq {        (\Delta\tau_1 + \Delta\tau_2 +\Delta\tau_3) V_3^2  -        (\Delta\tau_1 + \Delta\tau_2              ) V_2^2        \over        \Delta\tau_3}\label{eqn:estint}\end{equation}For any real earth model we would not like an imaginary velocitywhich is what could happen if the squared velocity in (\ref{eqn:estint})happened to be negative.You see that this means that the RMS velocity we estimatefor the third layer cannot be too much smaller than the one weestimate for the second layer.\par%Experimentalists like equations~\EQN{vrmshyp} and \EQN{vrmsdefn}.%Suppose data contains two hyperboloids%(curves that look like hyperbolas)%one with a top at time $\tau_1$ and the other with a top at $\tau_2$.%By finding the best fitting hyperbolas,%they get $V(\tau_1)$ and $V(\tau_2)$.%If they presume the earth has three constant-velocity layers%with interfaces between them at $\tau_1$ and $\tau_2$%then%they can determine the velocity of each of the top two layers.\subsection{Nonhyperbolic curves}\sx{hyperbolic, non}\sx{nonhyperbolic}Occasionally data does not fit a hyperbolic curve very well.Two other simple fitting functions are\begin{eqnarray}\label{eqn:fourthorder}t^2              &=& \tau^2 \ +\ { x^2   \over  v^2 } \                                                 +\ x^4 \times {\rm parameter}\\\label{eqn:datum}(t-t_0)^2        &=& (\tau-t_0)^2 \ +\ { x^2   \over  v^2 } \end{eqnarray}Equation~(\ref{eqn:fourthorder}) has an extra adjustable parameterof no simple interpretation other than the beginning of a power series in $x^2$.I prefer Equation~(\ref{eqn:datum}) where the extra adjustable parameteris a time shift $t_0$ which has a simple interpretation,namely, a time shiftsuch as would result from a near-surface low velocity layer.In other words, a datum correction.\todo{ XXXXXX \subsection{Can I abandon the material in this section?}\parWe could work out the mathematical problemof finding an analytic solution forthe travel timeas a function of distance in an earth with stratified $v(z)$,but the more difficult problem isthe practical one which is the reverse,finding $v(\tau)$ from the travel time curves.Mathematically we canexpress the travel time (squared)as a power series in distance $h$.Since everything is symmetric in $h$,we have only even powers.The practitioner's approach is to look at small offsetsand thus ignore $h^4$ and higher powers.Velocity then enters only as the coefficient of $h^2$.Let us why it is the \RMS\ velocity,equation~(\ref{eqn:vrmsdefn}),that enters this coefficient.%% above is new%\parThe hyperbolic form of equation~(\ref{eqn:vrmshyp}) will generally not be exactwhen $h$ is very large.For ``sufficiently'' small $h$,the derivation of the hyperbolic shape followsfrom application of Snell's law at each interface.Snell's law implies that the Snell parameter $p$, defined by\begin{equation}p \eq \frac{\sin\theta_i}{v_i}\label{eqn:pdefine}\end{equation}is a constant along both raypaths in Figure~\ref{fig:stratrms}.%This law implies that $\cos\theta_i$%can be written in terms of $p$ as%\begin{equation}%\cos\theta_i \eq \sqrt{1-(pv_i)^2} \ \ \ .%\end{equation}Inspection of Figure~\ref{fig:stratrms} shows that in the $i$th layerthe raypath horizontal distance $\Delta x_i$ and travel time $\Delta t_i$are given on the left below by\begin{eqnarray}\label{eqn:xrms}\Delta x_i &=& \Delta z_i \tan\theta_i           \eq%          \frac{p}{2}%          \frac{\Delta\tau_i v_i^2}           \frac{v_i\Delta\tau}{2} \            \frac{p v_i}           {\sqrt{1-p^2v_i^2}}           \eq \frac{p}{2} \Delta\tau_i v_i^2 + O(p^3)    \\\label{eqn:trms}\Delta t_i &=& \frac{2\,\Delta z_i}{v_i \cos\theta_i}            \eq \frac{\Delta\tau_i}{\sqrt{1-p^2v_i^2}}           \eq \Delta\tau_i \left(1+                {\tiny 1 \over 2}  p^2 v_i^2 \right) + O(p^4) \ \ .\end{eqnarray}The center terms above arise by using equation~(\ref{eqn:pdefine})to represent $\tan\theta$ and $\cos\theta$as a function of $\sin\theta$ hence $p$,and the right sides above come from expanding in powers of $p$.Any terms of order $p^3$ or higher will be discarded,since these become important only at large values of $h$.Summing equation~(\ref{eqn:xrms}) and~(\ref{eqn:trms}) over all layers yields the half-offset $h$ separating the midpointfrom the geophone location and the total travel time $t$.\begin{eqnarray}\label{eqn:hrms}h &=&\frac{p}{2}\ \tau \  V^2(\tau) + O(p^3) \\\label{eqn:310}t &=&\tau \left( 1 + \frac{1}{2} p^2 V^2(\tau) \right) + O(p^4) \ \ \ .\end{eqnarray}Solving equation~(\ref{eqn:hrms}) for $p$ gives $p=2h/(\tau V^2)$,justifying the neglect of the $O(p^3)$ terms when $h$ is small.Substituting this value of $p$ into equation~(\ref{eqn:310}) yields\begin{equation}t \eq \tau \left( 1 + \frac{2h^2}{\tau^2 V^2(\tau)} \right) + O(p^4) \ \ \ .\end{equation}Squaring both sides and discarding terms of order $h^4$ and $p^4$yields the advertised result, equation~(\ref{eqn:vrmshyp}). XXXXXXXX }\subsection{Velocity increasing linearly with depth}Theoreticians are delighted by velocity increasing linearly with depthbecause it happens that many equations work out in closed form.For example, rays travel in circles.We will need convenient expressions for velocityas a function of traveltime depthand \RMS\ velocity as a function of traveltime depth.Let us get them.We take the \bx{interval velocity} $v(z)$ increasing linearly with depth:\begin{equation}v(z) \eq v_0 + \alpha z\end{equation}%This presumption can also be written as a differential equation:\begin{equation}\frac{dv}{dz} \eq \alpha  .\end{equation}%The relationship between $z$ and vertical two-way traveltime $\tau(z)$(see equation~(\ref{eqn:twotaudefn})) is also given by a differential equation:\begin{equation}\frac{d \tau}{dz} \eq \frac{2}{v(z)}.\end{equation}%Letting $v(\tau)=v(z(\tau))$and applying the chain rulegives the differential equation for $v(\tau)$:\begin{equation}\frac{dv}{dz}\frac{dz}{d \tau}\eq\frac{dv}{d \tau}\eq\frac{v \alpha}{2},\end{equation}whose solution gives us the desired expression for \bx{interval velocity}as a function of traveltime depth.\begin{equation}v(\tau) \eq v_0 \ e^{\alpha \tau / 2 }  .\label{eqn:priorint}\end{equation}%\subsection{Prior RMS velocity}Substituting the theoretical interval velocity $v(\tau)$from equation~(\ref{eqn:priorint})into the definition of\RMS\ velocity $V(\tau)$(equation~(\ref{eqn:vrmsdefn}))yields:\begin{eqnarray}\tau \ V^2(\tau) &=& \int_{0}^{\tau} v^2(\tau') \ d \tau'\\               &=& v_0^2 \ \frac {e^{\alpha \tau} - 1} {\alpha} .\end{eqnarray}Thus the desired expression for \RMS\ velocityas a function of traveltime depth is:\begin{equation}V(\tau) \eq v_0 \         \sqrt{        \frac{e^{\alpha \tau} - 1 }{\alpha \tau}        }\label{eqn:Vrms}\end{equation}For small values of $\alpha \tau$,this can be approximated as\begin{equation}V(\tau) \quad\approx \quad v_0\ \sqrt{1 + \alpha \tau / 2}  .\end{equation}%\iex{Exer/Intro}{assignment}%\iex{Exer/Shift}{assignment}%\iex{Exer/2Dft}{assignment}%\iex{Exer/Phasemod}{assignment}%\iex{Exer/Phasedown}{assignment}%\iex{Exer/Phasemig}{assignment}%\iex{Exer/Stolt1}{assignment}%\iex{Exer/Stolt2}{assignment}