Ian Stokes edited k.tex  over 9 years ago

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$\tan$( $\theta$ ) =$\ C_{D}$/$\ C_{L}$. I will refer to this as the glide angle 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, $\ e^{(\verb|-$\alpha$|)(\verb|x|)]}$ $\exp${-($\alpha$)($\ z$)  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 $\ x$ = $\pi$/$\ k$. Anyways, the velocity equations that correspond to these conditions are given by$\ v_{x}$($\ x$,$\ z$,$\ h$) = $\gamma$($\ h$)cos($\ kx$)[exponential] +$\ v_{o}$ and$\ v_{z}$($\ x$,$\ z$,$\ h$) = $\beta$($\ h$)sin($\ kx$)[exponential]. 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$\ x$ =$\ 0$, In can easily be verified that $\xi$,$\ v_{z}$,$\ v_{x}$, bernoulli's equation, the continuity equation, can all be solved in harmony under the revised conditions imposed on the wind velocity components.