# Do Ocean Waves Experience Lower Downwards Air Pressure?

• moatilliatta
In summary, the conversation discusses the concept of airfoil lift and its relation to water waves. The general explanation of airfoil lift is that the top-side curve of the airfoil causes air on the topside to move faster than the bottom-side, resulting in a reduction of air pressure and lift. However, there is some disagreement about the role of downward air pressure in airfoil lift. The conversation then delves into the specifics of how a wave interacts with air, with the conclusion that the effect on air pressure is similar to that of wind blowing over a curved object. Overall, the conversation concludes that a wave does experience reduced downward air pressure, but the effect is minimal compared to the pressure changes caused by wind blowing over a wave.
moatilliatta
Of the few occasions I've seen Airfoil Lift described to general audience (on documentaries), it has focused on the top-side curve of the Airfoil causing air on the topside to move faster than the bottom-side. This is said to reduce air pressure on the topside and result in lift.

If a wave is rolling through still air, it will accelerate the air over the wave, much like an Airfoil does. Will this acceleration of air reduce the downwards air pressure on the wave?

stedwards
aerofoil dynamics also involve an 'angle of attack' in order to produce the lift.
In the case of water waves that angle is zero so I think it unlikely that any significant pressure drop would occur.
You might get some small amount of pressure increase ahead of the wave and an equally small pressure drop behind the wave crest, so this would have a small drag effect on the wave rather than anything corresponding to 'lift';.

If I remember correctly, the water at the surface of the waves moves mainly up and down, rather than horizontally. It only appears that the water is moving horizontally. It's like "the wave" at a baseball game. The people are not moving around the stadium. So you wouldn't expect any lift at the top of a water wave.

Chet

I've always thought of an Airfoil's lift is a result of air being forced downwards, which to me differs from the general explanation (see 1.18 of the below video)

An unbroken wave can't force air below itself, so lowered downwards air pressure would add credence to this general explanation of how an Airfoil works.

moatilliatta said:
I've always thought of an Airfoil's lift is a result of air being forced downwards, which to me differs from the general explanation (see 1.18 of the below video)

An unbroken wave can't force air below itself, so lowered downwards air pressure would add credence to this general explanation of how an Airfoil works.

Did you not read what I wrote?

Chet

The air will be accelerated by the front face of the moving wave, causing it to accelerate away from and over the top of the wave. For the air that flows over the top of the wave, some of it curves to follow the top of the wave, so part of that acceleration is perpendicular, and the pressure of the air at the top and back of the wave is reduced since the wave effectively sweeps out a volume of air, leaving what would otherwise be a void if the air didn't accelerate to follow the back of the wave. Beyond the back of a wave, the water returns from below average height back to average height, so that would involve upwards acceleration of air coexistent with an increase in pressure.

I'm not sure of the overall effect on air pressure by a wave traveling through the air. The interaction involves some small generation of heat, but I'm not sure if that small amount of heat energy coexists with a reduction in pressure of the air, since pressure of the air is due to the weight of all the air above any area at the surface of the water or the earth.

Chestermiller said:
Did you not read what I wrote?

Chet
Yes, It just didn't come across that you are very sure of your answer. If the air was stationary relative to the water when a wave passes, I would expect that there be no effect.

moatilliatta said:
Yes, It just didn't come across that you are very sure of your answer. If the air was stationary relative to the water when a wave passes, I would expect that there be no effect.
On the ocean, when a wave passes, objects on the water bob up and down, they don't travel along with the wave. They act as tracer particles for what the water is doing at the surface.

EDIT: After further consideration, I take back what I was saying. I don't know what I was thinking. You are absolutely right.

As reckoned from an inertial frame of reference traveling along with the wave, the wave is simply a bump on the surface of the water, with the air blowing over it. This is very much the same kind of situation as air flow over a wing, particularly if you treat the flow as potential (so that the non-slip boundary condition does not have to be satisfied). Sorry for my confusion.

Chet

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stedwards and Ophiolite
Chestermiller said:
On the ocean, when a wave passes, objects on the water bob up and down, they don't travel along with the wave.
Just because the water does not traverse the ocean and merely bobs up and down, doesn't mean the air above the wave isn't in motion.

moatilliatta said:
Just because the water does not traverse the ocean and merely bobs up and down, doesn't mean the air above the wave isn't in motion.
Thanks. See my EDIT in the previous post.

Chet

Chestermiller said:
the wave is simply a bump on the surface of the water, with the air blowing over it. This is very much the same kind of situation as air flow over a wing, particularly if you treat the flow as potential (so that the non-slip boundary condition does not have to be satisfied). Sorry for my confusion.
Chet

So, a wave will experience reduced downwards air pressure?

moatilliatta said:
So, a wave will experience reduced downwards air pressure?
An ocean wave traveling left to right in still air is not the same as wind moving right to left over a curved object.

A series of waves could be thought of as a bunch of very small pistons moving up and down in the still air.

256bits said:
An ocean wave traveling left to right in still air is not the same as wind moving right to left over a curved object.
Actually, it is. I was incorrect when I said this same thing in my first post. From an inertial frame of reference moving along with the velocity of the wave (or waves), the wave is stationary and the air is blowing past. We know from Newton's first law that any inertial frame of reference is valid for analyzing the problem.
A series of waves could be thought of as a bunch of very small pistons moving up and down in the still air.
Yes, but the waves are also traveling, and this is the same as the wave being still, and the air blowing by.

Chet

Chestermiller said:
Actually, it is. I was incorrect when I said this same thing in my first post. From an inertial frame of reference moving along with the velocity of the wave (or waves), the wave is stationary and the air is blowing past. We know from Newton's first law that any inertial frame of reference is valid for analyzing the problem.

Chet
Yes. It is a repetition of your first post.
A reference following the wave, has a Δm association, where as the moving solid contoured shape does not.
Thus I have been trying to get my head around on how a reference following the wave so it can be considered stationary can be called inertial.

In the same vein, the air should also be considered the exact same way, in which case we have 2 waves, one air and one water, for example, moving at the same velocity ( and direction, which is redundant ) but 180 degrees out of phase.

CWatters said:
Diagram shows pressure fluctuations as wind blows over waves..

Taken from.
http://en.wikipedia.org/wiki/Wind_wave
Nice picture, which invalidates the simple piston model.

I did fail to mention that the surface of the pistion(s) has to have a variable surface orientation depending upon its phase angle. Thus, there would be variable horizontal forces on the pistion surface during its travel.

As said before, the question is would that qualify as lift, by rootone

That diagram shows wave amplification where the state of motion of waves, water and air are all in relative motion.

If the pressure were reduced on wave crests in still air, this would amplify the wave amplitude. The effect would be to increase the energy. However, even with inviscid fluids there would be an induced drag reducing the energy, so I don't know what gives.

stedwards said:
That diagram shows wave amplification where the state of motion of waves, water and air are all in relative motion.

If the pressure were reduced on wave crests in still air, this would amplify the wave amplitude. The effect would be to increase the energy. However, even with inviscid fluids there would be an induced drag reducing the energy, so I don't know what gives.
This is not a very straightforward air dynamics problem. Even if one knew the shape of the wave and its velocity, one would still need to solve the potential flow equations for the pressure distribution and the velocity distribution in the air (relative to the wave).

Chet

Chestermiller said:
This is not a very straightforward air dynamics problem. Even if one knew the shape of the wave and its velocity, one would still need to solve the potential flow equations for the pressure distribution and the velocity distribution in the air (relative to the wave).

Chet

It is a very interesting problem. I had a thought. Assume a flow velocity of an ideal fluid over a planar surface then conformally transform the surface into a wave shape. Will this produce pressure information over the surface?

1. The flow field is symmetrical about crests and troughs, so maybe there isn't an induced drag as I had thought. 2. I think that the forces on the wave are allowed to change the shape of the waves as long as energy is neither removed nor added to the waves, which would violate conservation of energy.

It is indeed highly complex and nonlinear. The relative motion of air versus water will not give the entire picture.

As an ocean sailor I can tell you that the shape of waves is radically different when the wind acts against current as opposed to with it. Consider a 5 knot north flowing current and a 20 knot north flowing wind. 15 knots relative motion. Then consider a 5 knot north flowing current and a 10 knot south flowing wind. Same 15 knot relative speed.

With wind and current in the same direction, waves are nearly sinusoidal. With wind and current in opposite directions, the shape approximates a square wave facing the wind with a sinusoidal trailing edge.Therefore, the horizontal momentum of the water in the wave plays a role.

stedwards said:
It is a very interesting problem. I had a thought. Assume a flow velocity of an ideal fluid over a planar surface then conformally transform the surface into a wave shape. Will this produce pressure information over the surface?
I would do it numerically.
1. The flow field is symmetrical about crests and troughs, so maybe there isn't an induced drag as I had thought. 2. I think that the forces on the wave are allowed to change the shape of the waves as long as energy is neither removed nor added to the waves, which would violate conservation of energy.
Just solving the problem of wave motion without interaction with the air is a complicated enough problem. As anorlunda said, solving the coupled problem would be much or complicated. Of course, it definitely can be done.

Chet

Thanks to fzero, I got some mathematical help finding the velocity field of the air. https://www.physicsforums.com/threads/conformal-transformation-fluid-flow-over-surface-waves.818636/The shape of the wave is described by a trochoid.

The last curve, but inverted about the x-axis, describes the wave.

$u_0 = a \theta – b \sin \theta$
$v_0 = – a + b \cos \theta$,
parameric in $\theta$.
a and b are constants where a>b.

$u$ is the horizontal coordinate.
$v$ is the verticle coordinate.

The wave velocity is $c= \sqrt{\frac{g\lambda}{2 \pi}}$ where g is the acceleration due to gravity and $\lambda$ is the wavelength.

The streamlines of air, in a coordinate system where the wave is stationary are
$u = a \theta – b \sin \theta e^{-y}$
$v = ay – a + b \cos \theta e^{-y}$
where y is held constant. The so called velocity potentials are found by holding $\theta$ constant and varying y.

At $y=0$, the wave surface,
$u = a \theta – b \sin \theta$
$v = – a + b \cos \theta$.
These are the first two equations.

Over constant intervals of $[\theta]$, say every 5 degrees, we have a system of points where the velocity potential intersects the wave surface. The velocity, c is inversely proportional to spacing between points.

$c^2 =(\frac{du}{d \theta})^{-2} + (\frac{dv}{d \theta})^{-2}$
$\frac{du}{d\theta} = a-b cos(\theta)$
$\frac{dv}{d\theta} = -b sin(\theta)$

A wave peak occurs where v is maximal; $\theta= 0$.
A trough is found at $\theta=\pi$.

At the peak
$\frac{du}{d\theta}=a-b$
$\frac{dv}{d\theta}=0$
and $c^2=(a-b)^{-2}$.

At the trough
$\frac{du}{d\theta}=a+b$
$\frac{dv}{d\theta}=0$
and $c^2=(a+b)^{-2}$.

The free stream velocity, high above the waves is $c=1/a$.

We can now apply Bernoulli, $V^2+2p=H$ to find the gage pressure at points along the wave. H is a constant.
For the free steam, we establish the value of H. $H=c^2=\frac{1}{a^2}$.

At a peak, $(a-b)^{-2}+2p= a^{-2}$
The pressure, $p=(1/2)[(a-b)^{-2}-a^{-2}]$.

At a trough, $(a+b)^{-2}+2p= a^{-2}$
The pressure, $p=(1/2)[(a+b)^{-2}-a^{-2}]$.

These are difficult to evaluate without some numbers. For a=1 and b=1/2, peak pressure is less than gauge over the peaks and greater than gauge over the troughs, as expected.

The pressure acts symmetrically about the peaks and troughs. This seems to mean that the pressure profile is in phase with the waves so that no energy is transferred in this idealized analysis and no violation of the conservation of energy.

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moatilliatta

## 1. How do ocean waves experience lower downwards air pressure?

Ocean waves experience lower downwards air pressure due to the Bernoulli principle. As the waves move, they create a low-pressure zone behind them, causing the air above the ocean to move downwards and create a high-pressure zone. This high-pressure zone pushes down on the ocean's surface, creating the characteristic "dip" in the water's surface behind the wave.

## 2. What causes the change in air pressure above ocean waves?

The change in air pressure above ocean waves is caused by the movement of the waves. As the waves move, they create a disturbance in the air above them, causing the air molecules to move and creating a change in air pressure.

## 3. Does the size of ocean waves affect the downwards air pressure?

Yes, the size of ocean waves does affect the downwards air pressure. Larger waves will create a greater disturbance in the air above them, resulting in a larger change in air pressure. This can also contribute to stronger winds and more turbulent weather patterns.

## 4. Is the downwards air pressure consistent across all types of ocean waves?

No, the downwards air pressure is not consistent across all types of ocean waves. The shape and size of the waves, as well as the strength and direction of the wind, can all affect the downwards air pressure experienced by the waves. For example, stormy seas with larger, choppier waves will experience greater changes in air pressure compared to calm, small waves.

## 5. How does the downwards air pressure impact marine life?

The downwards air pressure can impact marine life in various ways. Strong, turbulent waves can cause changes in water pressure, which can affect the behavior and movement of marine animals. Additionally, fluctuations in air pressure can also affect the oxygen levels in the water, which can have detrimental effects on certain marine species.

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