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Using the same ensatz, and the revised wind velocity conditions, First I invoked Bernoulli's equation using$\ v_{x}$ values for the trough on one side and valuse for$\ v_{x}$ at the crest on the other. This analysis gave me $\gamma$($\ h$) = (-$\ gh$)/$\ v_{o}$. By approximating the pressure differences in the air at the crest and trough to be negligible, the continuity equation for air becomes $\grad$($\ v$) = 0. Plugging in $\ v_{x}$,$\ v_{z}$, and$\ v_{y}$ (which is a constant and therefore does not effect $\grad$($\ v$)) we obtain a relationship between the product ($\alpha$)($\beta$) = ($\ kgh$)/$\ v_{o}$. In order to determine $\alpha$ and $\beta$, I substituted $\beta$ = $\ kgh$/(($\ v_{o}$)($\alpha$)) into the Bernoulli relation between the trough ($\ x$ =$\ 0$,$\ z$ = -$\ h$) and equilibrium ($\ x$ = $\pi$/$\ 2k$,$\ z$ = $\ 0$) using the full wind velocity, given by$\ v^2$ = ($\ v^2_{x}$ + $\ v^2_{z}$). From this relation we obtain the expression $\alpha$ = (1/$\ k$)($\ gh$/$\ v_{o}$ -$\ 2v_{o}$). It is important to notice here that, in requiring that $\alpha$ is positive we can see that the ensatz will hold only when$\ v_{o}$ $\leq$ $\sqrt((gh)/2)$. This restraint implies that my ensatz set of solutions approximates the air flow over waves for low wind speeds. Pelicans often compression surf in glassy conditions, therefore this is a reasonable regime to analyze. Plugging back into the product relation we can see that $\beta$ = $\ ghk^2$/($\ gh - 2v_{o}^2$). This equation adds another constraint to the inequality above: if the two sides equal then $\beta$ becomes an undefined, nonphysical quantity and alpha decay drops out, which also does not make sense physically. Therefore the domain expression should actually read$\ 0$ $\leq$ $\ v_{o}$ < $\sqrt((gh)/2)$. Now we have full expressions and their functional domain for the still picture of these waves. To add in time dependence we simply tack on -($\omega$)$\ t$, where $\omega$ is the angular frequency of the swell. Now there are two defining cases for $\omega$ I will consider: the deep and shallow water dispersion relations. In deep water we have$\ kH$ >>$\ 1$ where$\ H$ is the depth of the sea floor. In this case $\omega$ = $\sqrt(kg)$ (for a wave propagating to positive $\hat x$; for a wave propagating to negative $\hat x$, $\omega$ acquires a minus sign). the wave velocity in deep water is then given by$\ v_{ph}$ = ($\omega$)/$\ k$ = $\sqrt(g/k)$. In shallow water we have$\ kH$ <<$\ 1$, yeilding $\omega$ = ($\ k$)$\sqrt(gH)$ and$\ v_{ph}$ = $\sqrt(gH)$. Now if we substitute the expressions obtained for $\alpha$, $\beta$, $\gamma$, and the $\omega$ relevant to the depth, we have full equations for the x and z components of wind velocity, given by:  $\ v_{x}$($\ x$,$\ z$,$\ h$) = (-$\ gh$)/$\ v_{o}$($\ h$)cos($\ kx$ - ($\omega$)$\ t$)[exponential] +$\ v_{o}$  $\ v_{z}$($\ x$,$\ z$,$\ h$) = $\ ghk^2$/($\ gh - 2v_{o}^2$)($\ h$)sin($\ kx$ - ($\omega$)$\ t$)[exponential]    An ancient creature, estimates have it that this majestic bird has been cruising the skies for at least thirty million years. Historical geologists would call this period of time the "Oligocene epoch" of the "Paleogene period," but to the rest of us these implications can be put in much simpler terms--Pelicans are dinosaurs! Perhaps it is to this that the Brown Pelican owes its aerial expertise. Thirty million years has allowed evolution to take its course, and over the generations the pelican has been able to develop the artform of "compression surfing."