# Pelican Compression Surfing

Few of our planet’s residents have managed to adapt to their environmental niche as gracefully as the Brown Pelican. Imagine for a moment your typical river salmon, who spends their entire existence struggling against raging water currents, slowly hopping up rapids until they get eaten by a bear. Think of a seagull, now, or a crow. Surely you have been in a storm and seen the foolish foul flying straight into the wind, making either no progress or even getting blown backwards and landing with negative displacement from their starting point. Rather than fighting with the forces of nature as many animals do, the Brown Pelican has managed to develop ways of harnessing these dynamic forces of nature –that is, moving air and water, or “wind” and “waves”–in ways that actually make their lives easier.

One of the most fascinating ways that the species accomplishes this goal is by gliding just in front of ocean swells, such that they are being donated additional lift by updrafts that naturally occur in this region. The pelicans essentially “surf” waves by funneling, or “compressing” these updrafts between their wings and the water–a phenomenon known as “compression surfing.” This goes hand in hand with the so-called “ground effect.” Ground-effect refers to the observation that birds such as skimmers and pelicans are able to maintain gliding at a constant altitude for longer periods of time without additional energy input (through flapping) when near the water’s surface than when farther away (Hainsworth 431-444). Ground effect in relation to compression surfing can be thought of as the work done (by updrafts associated with the frontside of moving ocean swells) in allowing a pelican to fly at a constant altitude without flapping.

Pelicans spend most of their time in the air, looking for sardines and anchovies. To support this exhausting habit, they eat roughly four pounds of fish per day. (U.S. Fish and Wildlife service). The cycle becomes circular in that Pelicans have to fly to feed, and must be well fed in order to fly. In order to preserve the food they eat in case they cannot find a fish for a while, pelicans take to compression surfing. This allows them to travel more distance with respect to the portion of their diet they burn off. If the bird is unable to find a fish before using all of their stores, they wont make it to the next day–in this sense the artform of compression surfing is necessary to the species survival.

Furthermore, pelicans travel enourmous distances throughout their lifetime. To avoid predators, the brown pelican only nests on specific offshore islands, the majority of which lie off the coast of Baja California. On the Pacific coast of the United States, the channel islands national park makes up the only brown pelican breeding zone. (National Park Service). During the summertime these birds make it considerably far north before returning to their home nesting sites. Compression surfing is critical in these massive continental traverses the pelican embarks on each year. My intention in writing this paper is to model a pelican “compression surfing” and make quantitative conclusions about the prescence of “ground-effect” in this phenomenon.

Before looking at the picture of a pelican compression surfing, four independent physical situations must be considered:

• First, for a pelican flying in absence of ocean swells, the angle (below horizontal) at which it naturally descends while gliding at a constant velocity must be determined. I will refer to this angle as $$\theta$$. Theta is a function of the coefficients of drag and lift,$$\ C_{D}$$ and$$\ C_{L}$$, respectively. Therefore $$\theta$$ = $$\theta(C_{D}, C_{L})$$.

• Second, I must model the air currents, or wind velocities associated with the region immediately above ocean swells. Broken into components$$\ v_{x}$$ and $$\ v_{z}$$,$$\ v_{x}$$ represents the horizontal component of the wind in the direction of the wave’s propagation and$$\ v_{z}$$ represents the vertical component of the wind which is, on the frontside of the wave, the updraft factor that donates the added lift to the pelican which in turn keeps her afloat. I am assuming symmetry in $$\hat y$$ for simplicity of calculation, such that$$\ v_{y}$$ is constant. It turns out in nature most ocean swells are relatively symmetric in $$\hat y$$, which is the direction perpindicular to the wave’s propagation.

• Third, I will introduce the angle $$\phi$$ which represents the direction that the pelican must fly in order to stay on the wave. That is, if she flies too parallel to the wave, the wave will pass her up and she will be on the backside of the swell where$$\ v_{z}$$ is negative, corresponding to a downdraft. If this happened the bird has two choices: to either crash into the water or bail to a higher altitude by flapping. In either case she is no longer compression surfing. On the contrary if the bird flies too much in the$$\ x$$ direction it will be going to slow to generate the threshold lift required for flight and will fall out of the sky. In order to compression surf for extended periods of time and avoid these miserable outcomes the bird must match$$\ v_{x}$$ with$$\ v_{ph}$$ where$$\ v_{ph}$$ is the velocity of the wave. Therefore, $$\phi$$ is a function of the pelican’s velocity (call it$$\ v_{b}$$ for bird’s velocity) and the wave’s velocity$$\ v_{ph}$$. Furthermore if there is a breeze$$\ v_{o}$$ blowing in the opposite direction of the wave’s propagation the bird must match this windspeed as well as the wavespeed. Thus, $$\phi$$ is a function of$$\ v_{b}, v_{ph}$$ and$$\ v_{o}$$. So we have $$\phi$$ = $$\phi(v_{b}, v_{ph}, v_{o})$$.

• The proficient compression surfing pelican inherently will strive to situate itself over the zone of the wave with the greatest updraft factor, thereby gaining the most lift it can from the wave. From observation, I have noticed that often times when pelicans compression surf they will actually drag the tip of their wing in the water. From the angle $$\phi$$, the wingspan$$\ l$$ of the bird, and the fact that the bird will center itself over the point of concavity change of the wave, we can determine how high the pelican must be during compression surfing (assuming the bird is level with the horizontal plane) and therefore, as$$\ v_{z}$$ is a funtion of height off the wave, we can determine the updraft factor that the pelican experiences.

Each of these situations is equally crucial in solving for the work done by the wind. At this point I will name this quantity$$\ W_{w}(d_{t},{h})$$ where$$\ d_{t}$$ is the distance travelled while sustaining constant altitude compression surfing and$$\ h$$ is the amplitude height of the wave.

The first situation is the simplest and thus I will hash out this scenario before the others. The magnitude of the lift force on an object in flight is given by the density of the fluid the object is traveling through times the velocity of the object times the circulation, where circulation is the closed path integral of the fluid velocity around a wing’s cross section. For a bird in flight, the circulation ( $$\Gamma$$ ) can be approximated as $$\Gamma$$ = $$\ (C_{L}/2)(v_{o}A_{p})$$ where$$\ A_{p}$$ is the planiform area of the wing: that is, the area of the wing projected onto the relevant xy plane. Therefore the magnitude of the lift force is given by (1/2)($$\rho_{o}$$)($$\ A_{p})(v^2)$$. The drag force follows the same relation, but$$\ C_{L}$$ is replaced by$$\ C_{D}$$. Choosing the xy plane to be parallel with the plane of the bird’s flight in a standard, right handed cartesian coordinate system, we can set $$\Sigma(F_{x})$$ and $$\Sigma(F_{z})$$ both equal to zero as we are solving for the angle where the bird flies at a constant altitude. Solving for this angle, $$\theta$$, we find that

$$\tan$$( $$\theta$$ ) =$$\ C_{D}$$/$$\ C_{L}$$.

I will refer to this as the glide angle equation. From this expression we can easily formulate an equation that gives the work done by a pelican flying at a constant altitude$$\ W_{p}$$ as a function of distance using the definition of potential energy,$$\ U = mga$$. If we label the altitude lost while gliding under constant velocity descent as$$\ a_{l}$$ and the distance travelled in the horizontal xy plane as$$\ d_{trav}$$, we obtain the work equation for a pelican, mass M_p, flying independent of ocean swell.

$$\ W_{p} = M_{p}a_{l}g = M_{p}g(d_{trav})[(C_{D}/C_{L})]$$

Work done by the pelican in maintaing constant altitude flight, which can be interpretted as the energy expended by the bird, in joules is given by the expression above. We can also solve for$$\ C_{L}$$ using the glide angle equation. From the relationship between weight and lift, we can determine that the minimum gliding speed$$\ v$$ = $$\sqrt(2mg$$/($$\rho_{air}A_{wings}C_{L}))$$. From this we have

$$\ C_{L} = 2mg$$/($$\rho_{air}A_{wings}v^2$$) which I will call the coefficient of lift equation. From Schnell and Hellack, we have a minimum gliding speed of $$\approx$$ $$\ 6 m/s$$. From the “Brown Pelican Fact Sheet” we have an average mass of $$\approx 4kg$$. This value gives $$\ C_{L} \approx 1.54$$.

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

$$\ v_{x}$$($$\ x$$,$$\ z$$,$$\ h$$) = $$\gamma$$($$\ h$$)cos($$\ kx$$)exp[-($$\alpha$$)($$\ z$$] +$$\ v_{o}$$

$$\ v_{z}$$($$\ x$$,$$\ z$$,$$\ h$$) = $$\beta$$($$\ h$$)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 this contradictory scenario, $$\gamma$$ became the positive value ($$\ gh$$)/($$\ v_{o}$$), and the product ($$\beta$$)($$\alpha$$) was shown equal to 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 $$\beta$$ individually, 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

$$\alpha$$ = ($$\ k$$)($$\ v_{o}$$/($$\ gh$$ -$$\ 2v^2_{o}$$).

Notice here that, in requiring that $$\alpha$$ is defined we can see that the domain expression should actually read$$\ 0 < v_{o}$$ < $$\sqrt((gh)/2)$$. 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 $$\beta$$ =$$\ [(gh)^2 - 2ghv^2_{o}]$$/($$v_{o}^2$$). This equation adds another constraint to the inequality above: if the two sides equal then $$\alpha$$ becomes an undefined, nonphysical quantity and alpha decay drops out, which does not make sense physically.$$\ v_{o}$$ cannot equal zero either as this results in an undefined term $$\beta$$. 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 t