deletions | additions       diff --git a/k.tex b/k.tex  index 0bef170..ea6a89d 100644  --- a/k.tex  +++ b/k.tex 
...
$\ C_{L} = 2mg$/($\rho_{air}A_{wings}v^2$) which I will call the coefficient of lift equation.  The solution to the second problem is beyond the scope of simple geometric arguments. To solve for the air flow over ocean swells, a good method is to introduce an ensatz, or in other words, a guess. As a first step, I approximated an ocean swell as a sinusoidal wave of amplitude$\ h$. From there I required that at a large height above the wave the wind would feel no effects from the wave, and would travel with$\ v_{x}$ = constant =$\ v_{o}$, the undisturbed wind speed. Both$\ v_{x}$ and$\ v_{z}$ decay at equal rates with additional altitude$\ z$ and therefore obey the same exponential decay factor, exp[-($\alpha$)($\ z$)] Since at this point I do not have any intuition for what this factor is, I introduced the undetermined constant $\alpha$. Initially I thought that due to radial accelleration along the trough, combined with the force of gravity pushing air down the backside of an arbitrary swell, that the maximum wind velocity would occur at the deepest point of the trough. On the flipside the wind would be least on top of the wave as it has to fight gravity in traversing up the face. One may think the wave would block the wind, but I am considering waves under the constraint that wavelength $\lambda$ >>$\ h$ to negate that factor. For purposes of analysis I picked a segment of wave to work with such that the maximum$\ v_{z}$ value occurs at$\ x$ = $\pi$/$\ 2k$, water displacements from equilibrium given by $\xi$($\ x$) = -$\ h$cos($\ kx$) I picked this zone because when I put the pelican in the picture it will center its body directly over the line in the$\ y$ axis where$\ z$ = $\ 0$. This will make the third independent problem of solving for $\phi$ much easier, let alone the solution for$\ W_{w}$, the work done by the wind. Following from this condition, the minimum $\ v_{z}$ occurs at $\ x$ =$\ 3pi$/$\ 2k$, maximum$\ v_{x}$ occurs at $\ x$ = $\ 0$ and minimum $\ v_{x}$ occurs at $\ x$ = $\pi$/$\ k$. The velocity equations that correspond to these conditions are given by  Anyways, the velocity equations that correspond to these conditions are given by$\ $\  v_{x}$($\ x$,$\ z$,$\ h$) = $\gamma$($\ h$)cos($\ kx$)exp[-($\alpha$)($\ z$] +$\ v_{o}$ and$\ $\  v_{z}$($\ x$,$\ z$,$\ h$) = $\beta$($\ h$)sin($\ kx$)exp[-($\alpha$)($\ z$). In each of these$\ k$ is the wave number. $\gamma$ and $\beta$ represent how much the x and z velocities actually change due to the wave. After solving the whole problem under these conditions, a serious contradiction presented itself. After I set $\alpha$, $\gamma$, and $\beta$ such that they solved Bernoulli's relation and the continuity equation everywhere, I plugged the conditions that I initially imposed on$\ v_{x}$ and$\ v_{z}$ back into my equations to see if they made sense. The conditions on$\ v_{z}$ were not met; in fact my equation stated that the minimum z-velocity would occur at $\ x$ = $\pi$/$\ 2k$. This was exactly where I required that the maximum velocity to occur. In formulating the contradictory scenario, I had $\gamma$ come out to the positive value ($\ gh$)/($\ v_{o}$), and the product ($\beta$)($\alpha$) equal the negative quantity, (-$\ kgh$)/($\ v_{o}$). Clearly as the magnitude of the disturbances due to the wave die off with additional altitude, $\alpha$ must be set positive such that the exponential term decays in this manner. This implies $\beta$ would be negative. Reviewing my calculations, it became apparent that both the continuity equation and the condition that$\ v_{z}$(max) happens at$\ x$ = $\pi$/$\ 2k$ could not be satisfied simultaneously with a negative $\beta$ value. Clearly it would be unreasonable to change the continuity equation. However, if we change the requirements on velocity so that now the minimum value of$\ v_{x}$ occurs at the trough, where$\ x$ =$\ 0$, $\xi$,$\ v_{z}$,$\ v_{x}$, bernoulli's equation, the continuity equation, can all be solved in harmony. Therefore I will carry on working under the revised conditions imposed on the wind velocity components. 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 tells us that the gradient of $\ v$ = 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.$\ v_{o}$ cannot equal zero either as this results in an undefined term in the exponent. Therefore the domain expression should actually read$\ 0$ <$\ v_{o}$ < $\sqrt((gh)/2)$. 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: