Fluid mechanics, why does the air flow faster over the wing?

[boneh3ad]

The ocean example would not be called lift by anyone I know in the fluids community. Instead that is a wave dynamics problem.

The definition of lift is an upward aerodynamic force on a body as a result of the movement of the body through that fluid. In other words, it has to do with the flow of that fluid around a body. In your wave example, it is an entirely different fluid below the "bump" and any upward force is as a result of the dynamics of the Kelvin-Helmholtz instability rather than lift.

It may be easier to think about in terms of other descriptions of lift. Lift requires a net circulation about the body moving through the fluid. This can't happen unless the body has the same fluid on all sides.

 

[SDEric]

It seems to me that requiring the hill to have the working fluid underneath it to generate lift is unnecessary. I understand that there is no such thing as "negative pressure", only less pressure than somewhere else. We can work around that requirement, however. The question is not whether the net forces on the hill are in any particular direction (I know the hill is not able to move), but whether the net pressure on the hill is less than ambient.

Suppose the hill were like an iceberg, floating in water. If there were lift on the hill, it is free to move up out of the water. With no wind, the air pressure on the hill is balanced by the air pressure on the water (and, of course, the mass of the hill displacing mass of water, etc.). There is enough of the hill below the water that the curved section begins exactly at the water surface.

Then, when the wind is blowing, is the net force on the hill up? Down? Parallel to the wind and surface?

The air encounters the hill and is turned up to perpendicular to its original direction. Then it turns and follows the same radius of curvature back down until it is again parallel, then again perpendicular, and then turns back the other direction until it is again parallel to the original flow.

I think the air, generally speaking, slows and turns, then speeds up and turns in the other direction, then slows and turns again in the first direction.

I can vaguely see that there might be net force upward because the speed (and hence acceleration) is lower at the concave curvature sections than at the convex curvature section at the top. Thus the integral of the pressure over the hill is less than ambient.

(Nothing about that construction explains drag, however.)

 

[boneh3ad]

There would be a net upward force in that case... as a result of buoyancy. Of course then the berg would rise again, the forces would all be equal and there would be no net force.

Despite how you seem to be interpreting things, lift is a force on a body arising due to its relative motion through a fluid. Forit to exist, it has to be immersed in that fluid, not partially immersed. That is just the definition of lift.

 

[SDEric]

There would be a net upward force in that case... as a result of buoyancy. Of course then the berg would rise again, the forces would all be equal and there would be no net force.

Despite how you seem to be interpreting things, lift is a force on a body arising due to its relative motion through a fluid. Forit to exist, it has to be immersed in that fluid, not partially immersed. That is just the definition of lift.

Understand. However, you can measure lift on an airplane wing in a wind tunnel, supported by thin rods or wires, and measure the force on those rods or wires. The wing does not deflect in the wind, yet you can measure the force imposed by the lift and drag. So it is definitely not necessary that the body be freely moving through the air. It can be rigidly supported.

So what is the integral of the pressure over the hill ("integral P dHill")? Is it less when the air is moving? Specifically, at each point on the curve, the pressure is orthogonal to the curve. At each point that orthogonal pressure has an x (horizontal) component and a y (vertical) component. Is the y component of the pressure less when the air is moving than when it is still?


A related thought experiment: If the hill is simply mated to its mirror image, the net force would obviously be zero (or, more to the point, straight back due to drag). If you attached the two halves by a force meter, would the meter measure tension when the wind blew? Would the two halves pull apart?


Thanks for lending your expertise. I am amazed that after more than a hundred years, explanations of lift do not simply take apart the various aspects of lift and discuss them independently.

 

[boneh3ad]

I never said it couldn't be rigidly mounted, only that it had to be immersed in such a way that a net circulation in the flus can form around the body.

The y component of the integral of the pressure should be less when the wind is moving.

If the hill was mated to a mirror image by a force sensor, it would measure zero force assuming a perfect mate with no air seeping into it.

 

[olivermsun]

The ocean example would not be called lift by anyone I know in the fluids community. Instead that is a wave dynamics problem.

While the development of a pressure gradient over a disturbance on a surface is a subject of interest for waves people, the dynamics are very similar (at least at the outset) to flow around an airfoil (and not surprisingly so).

The definition of lift is an upward aerodynamic force on a body as a result of the movement of the body through that fluid. In other words, it has to do with the flow of that fluid around a body.

To take the Kutta-Joukowsky theorem literally, it seems there would be a "lift" associated with circulation at an interface. However, if your feeling is that the proper usage is only associated with circulation around a solid object (an airfoil) then I can certainly accept that convention.

In your wave example, it is an entirely different fluid below the "bump" and any upward force is as a result of the dynamics of the Kelvin-Helmholtz instability rather than lift.

Look closer at the dynamics of the K-H instability -- similarities between that and flow around a thin wing are quite interesting. In the first case, you have vortices bound to a wing. In the second, the vortices are bound to the interface, with the net pressure distribution contributing to growth of the instability (raising the bumps and pushing down the troughs).

It may be easier to think about in terms of other descriptions of lift. Lift requires a net circulation about the body moving through the fluid. This can't happen unless the body has the same fluid on all sides.

Are you sure that the same fluid on all sides is a requirement? What about lift in a parallel stratified flow?

 

[boneh3ad]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities. It is not a vorticity-based phenomenon or circulation-based phenomenon.

Kutta-Joukowski was developed for solid objects in a single fluid. While there may be analogous phenomenon in other situations, there are no corollaries to Kutta-Joukowski that apply to those situation. The concept of lift applies to a solid object immersed in a fluid.

Stratified flows are still all the same gas.

 

[olivermsun]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities.

I find there to be some resemblance between the mathematics of the growth term in K-H instability and the lift theorem. There is also similar phenomenology between the flow around a wing and the flow around the K-H disturbances (which are often seen to develop into a roll of vortices). Again, I just find it to be an interesting parallel (no pun intended). If you maintain that there is no similarity whatsoever, well, you are certainly entitled to your view and I accept that.

As an aside: we observe K-H instabilities between miscible fluids all the time. In fact, K-H instabilities are believed to be quite important for mixing in a variety of environmental flows.

It is not a vorticity-based phenomenon or circulation-based phenomenon.

There is a velocity discontinuity (in the idealized limit) but there is no vorticity at the surface?

Stratified flows are still all the same gas.

Perhaps in the stratified flows you have looked at.

 

[boneh3ad]

I find there to be some resemblance between the mathematics of the growth term in K-H instability and the lift theorem. There is also similar phenomenology between the flow around a wing and the flow around the K-H disturbances (which are often seen to develop into a roll of vortices). Again, I just find it to be an interesting parallel (no pun intended). If you maintain that there is no similarity whatsoever, well, you are certainly entitled to your view and I accept that.

As an aside: we observe K-H instabilities between miscible fluids all the time. In fact, K-H instabilities are believed to be quite important for mixing in a variety of environmental flows.

There is a velocity discontinuity (in the idealized limit) but there is no vorticity at the surface?

Sure there is vorticity, but that isn't the nature of the growth of the instability. If you had a situation where the Kelvin-Helmholtz instability would ordinarily occur that was in equilibrium, you could see a completely lack of any waves regardless of the vorticity inherent in the shear layer. You would still need some kind of initial perturbation to cause the instability to manifest.

The Kelvin-Helmholtz instability early in its life can be fully described without any viscosity terms and need not roll over to still be an example of this instability. It is an inherently irrotational problem. Treating the governing equations linearly (by dropping viscous terms that are the only source of vorticity) still results in observation of the instability. Therefore it is not a vortical phenomenon. It starts out as linear waves that grow according to the dynamics of the linearized Euler equations. Upon reaching a certain amplitude, only then to they start exhibiting any sort of nonlinear (and vortical) component. At that point they create the familiar rolls that are highly nonlinear and fractal in nature.

Lift on an airfoil, however, is an inherently viscous phenomenon and depends on the effect of viscosity in generating circulation about a solid body. I suppose I fail to see the parallel here because lift on an airfoil is intimately related to viscosity and vorticity and cannot exist without it. The Kelvin-Helmholtz instability is still present in the inviscid limit.

You can certainly find Kelvin-Helmholtz phenomena occurring in miscible fluids, so I overstepped with that description. In that you are correct. All you really need is a velocity discontinuity.

Perhaps in the stratified flows you have looked at.

I read it (and clearly typed it) wrong. In my mind, I was thinking rarefied. You are, of course, correct. My mistake.

 

[RandomGuy88]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities. It is not a vorticity-based phenomenon or circulation-based phenomenon.

There are examples of KH instabilities without two immiscible fluids. Jets issuing into a motionless fluid are an example. It can occur in shear layers with an inflection point. There were a few other requirements but I don't remember them.

 

[boneh3ad]

There are examples of KH instabilities without two immiscible fluids. Jets issuing into a motionless fluid are an example. It can occur in shear layers with an inflection point. There were a few other requirements but I don't remember them.

Already covered in my previous post. I erred when stating the immiscible part. All that is necessary is a discontinuity in velocity in a flow. Most examples that are visible in nature involve two different fluids (or the same fluid of different properties such that the two parts stand out from one another). It is the velocity discontinuity that is important. If the velocities were the same, it would either be stable or unstable under the Rayleigh-Taylor instability, not Kelvin-Helmholtz.

The inflection point is an entirely different phenomenon. An inflection point in the shear layer profile is the criterion for inviscid instability, and I am not 100% sure that it is valid for shear layers in general, but definitely for boundary layers. It is known as the Rayleigh criterion.

 

[Stan Butchart]

Just found this thread and thought I would throw out a bananna peel just for fun.

The flow (ideal) next to a cylinder is accelerated to the velocity of the cylinder at the fwd stag point. It remains at constant 1V speed until the aft stag point.

The flow (ideal) next to a common wing is accelerated to the velocity of the wing at the fwd stag point. It SLOWS down over the top before speeding up again tothe TE.

 

[boneh3ad]

Just found this thread and thought I would throw out a bananna peel just for fun.

The flow (ideal) next to a cylinder is accelerated to the velocity of the cylinder at the fwd stag point. It remains at constant 1V speed until the aft stag point.

The flow (ideal) next to a common wing is accelerated to the velocity of the wing at the fwd stag point. It SLOWS down over the top before speeding up again tothe TE.

That isn't true at all. Maybe you have your reference frames mixed up or something. If you use the same frame of reference for both objects, say, stationary object with moving fluid for simplicity, you see qualitatively the same behavior. For example, if you place a cylinder in an inviscid flow with free stream velocity $U_{\infty}$ moving left to right in $x$, the maximum velocity will be $\vec{V}=2U_{\infty}\hat{\imath}$ at the top and bottom of the cylinders.

If you similarly place a NACA 0012 airfoil (for simplicity) into the same flow, the maximum flow speed is roughly $\vec{V}=1.2U_{\infty}\hat{\imath}$.

Both flows are accelerated over the wing.

If you then want to look at it in the other frame of reference with the object moving with velocity $U_{\infty}$ through a stagnant medium, you need only add $\vec{V} = -U_{\infty}\hat{\imath}$ to all quantities. In other words, in both cases, the air is still moving the same direction and is still accelerated over the object, accelerating at it reaches a max (whose location varies depending on the object) before decelerating back to the stagnation point. I think you must have looked at two different reference frames. The cylinder and the airfoil are qualitatively similar. The cylinder just accelerates the air to a greater degree. This is, as previously mentioned, all inviscid.

 

[Stan Butchart]

T the maximum velocity will be $\vec{V}=2U_{\infty}\hat{\imath}$ at the top and bottom of the cylinders.QUOTE]

I'm sorry but I must stick to my guns.
For a moving cylinder, there is a movingsurface and a moveing fluid.
IF i wish to identify the inertial velocity of the fluid following the surface, as if I were to calculate Bernoulli, I need to relate path and relative to its original position (or better the remote still field)
A particle located on the fwd stagnation is accelerated to1V right to left by a cylinder moveing 1V right to left.
Using the the accepted v=2sinA*V, a little high school calculating will allow graphing of its path and velocity.

At the top and bottom the particle velocity is 1V left to right. The cylinder surface velocity is 1V righ to left.

by the same token , over the0012,the particle velocity is .2V left to right while while the wing surface is 1V right to left.

A particle is used since a following particle is on a different inertial flowpath.

Maybe we are just talking the importance of reference.

 

[boneh3ad]

Either way, you said that the particle slows down in the case of an airfoil. That isn't true. What you just said in this post is exactly what my previous post said.

 

[Stan Butchart]

Either way, you said that the particle slows down in the case of an airfoil. That isn't true. What you just said in this post is exactly what my previous post said.

Yah, I realized tha halfway through my post but it was too late to stop!

But tell me, it is moving at 1V at the stag point and .2V at the top ( all relative to the remote air.) How is that not slowing down?

 

[boneh3ad]

Well, -0.2V. It is slowing down in that frame and the speeding back up until it reaches that -0.2V, but in your earlier post you said it sped up over te cylinder, which does not happen in that frame.

 

[Stan Butchart]

in your earlier post you said it sped up over te cylinder, which does not happen in that frame.

I certainly hope that it did not come across that way. If it did I apologise.

The cylinder is constant speed, stag point to stag point.

 

[boneh3ad]

No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.

 

[Stan Butchart]

No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.

Oohh yes it is! This is all in friendship but communication is difficult.

I am looking at the inertial path relative to the remote still air only. The flowpath is as a lower case script "e".
Look up the source/sink pattern for the cylinder. While it is an instantaneous picture, all of the surface intercepts are at 1V. The "e" is an integration of those intercepts.

 

[boneh3ad]

Listen, the potential flow around a cylinder shows that the the velocity can be described by the potential function of a doublet and a free stream, which give you
$$\phi(r,\theta) = U_{\infty}\left(r+\frac{R^2}{r}\right)\cos\theta$$
Differentiating to get velocities
$$v_{r} = U_{\infty}\left(1 - \frac{R^2}{r^2}\right)\cos\theta$$
$$v_{\theta} = -U_{\infty}\left(1 + \frac{R^2}{r^2}\right)\sin\theta$$

At the surface where $r = R$,

$v_r = 0$
$$v_{\theta} = -2U_{\infty}\sin\theta$$
Clearly, that is not uniform for all $\theta$. The velocity is symmetric about the $x$-axis, but it is not uniform along the surface. This is in the frame of reference of the cylinder, by the way. If you look at it, that means at the stagnation points, the velocity in this frame is zero and at the top and bottom, it is $2U_{\infty}\hat{\imath}$ in Cartesian coordinates. Between these points along the surface the velocity varies smoothly. It is not a constant on the surface. You can find that canonical example in any aerodynamics book or other fluids book that includes potential flow.

 

[Stan Butchart]

Clearly, that is not uniform for all $\theta$. The velocity is symmetric about the $x$-axis, but it is not uniform along the surface. This is in the frame of reference of the cylinder, by the way. If you look at it, that means at the stagnation points, the velocity in this frame is zero and at the top and bottom, it is $2U_{\infty}\hat{\imath}$ in Cartesian coordinates. Between these points along the surface the velocity varies smoothly. It is not a constant on the surface.

Absolutly right! However, I have been trying to get across the velocity relative to the remote field. The reason was to show that the Bernoulli Principle did not apply tangent to the surface. Converting the flow relative to the surface to relative to the field is a pretty simple process.

(As a uneducated tech designer, I configurated missles and tactical aircraft using the most basic equations. I never had time for real "rocket science".)

 

[boneh3ad]

I have absolutely no idea what you are trying to get at then. Based on this quote

While it is an instantaneous picture, all of the surface intercepts are at 1V.

It leads me to believe you are saying the velocity is 1V everywhere on the surface. It isn't.

 

[Stan Butchart]

It leads me to believe you are saying the velocity is 1V everywhere on the surface. It isn't.

Right, relative to the surface it is not. However, while it moves along the surface it is moving thru space at 1V because the cylinder is moving.

My reason to continue to pursue this is to try to understand better how to convey concepts. (Which I am failing!)

You might give http://svbutchart.com a shot

 

[RandomGuy88]

The reason was to show that the Bernoulli Principle did not apply tangent to the surface.

I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface?

In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.

 

[boneh3ad]

Right, relative to the surface it is not. However, while it moves along the surface it is moving thru space at 1V because the cylinder is moving.

My reason to continue to pursue this is to try to understand better how to convey concepts. (Which I am failing!)

You might give http://svbutchart.com a shot

But this just isn't true. In the frame where the object is moving, the only places where the flow is moving with speed $U_{\infty}$ are the two stagnation points and the points on top and bottom where $\theta = \pm 90^{\circ}$.

Keep in mind I said speed. If you look at it, at the stagnation points, $V = -U_{\infty}\hat{\imath}$, but at the top and bottom points, $V = U_{\infty}\hat{\imath}$. These are not the same quantities. Also note that in between these points along the circumference, the velocity varies between these two, meaning there are points where $V = 0$, points where $V = \frac{1}{2}U_{\infty}\hat{\imath}$ and any other concoction $V = nU_{\infty}\hat{\imath}$ where $-1\le n\le1$

Velocity is absolutely not constant along the circumference in any frame or reference.

I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface?

In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.

Expanding on this, Stan originally specified an ideal fluid, meaning zero viscosity.

 

[Stan Butchart]

I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface?

In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.

The point is to present an understanding to the hundreds that search these pages.

Bernoulli is based upon differences in the speed at different points in the inertial flow. Speed calculated related to different points on the surface does not reflect the actuual inertial flow. Bernoulli Princple does not fit.
However, the Bernoulli Equation with velocity relative to the curved surface does fit when paired with the normal acceleration of turning flow.

If we take the accepted tangential velocity and reconcile it with the cylinder velocity, the inertial velocity will be 1V at all points.

 

[boneh3ad]

If we take the accepted tangential velocity and reconcile it with the cylinder velocity, the inertial velocity will be 1V at all points.

No it won't. This is an objective fact.

 

[Stan Butchart]

at the top and bottom points, $V = U_{\infty}\hat{\imath}$. These are not the same quantities. QUOTE]

Could you expand on that statement?

Another way using the accepted equation for velocity relative to the surface. v=2sinA*V
i run a radius line from the center out past the surface at angle A.
At the intercept I place a 1V vector running fwd.
I reflect the 1V vector across the radius line. It defines 2 A from horizontal.
The base of the 2A triangle is the velocity, tangent and relative to the cylinder.

The reflected vector is the inertial vector of the flow relative to far field. It is always 1V.

I am not sure why I am finding this concept so difficult!

 

[boneh3ad]

Because your concept is wrong. There is no frame of reference where the speed is 1V along the entire surface.

And to expand on what you referenced, the speed at those points I mentioned is the same but the velocity is not. I was noting the difference.

 

[Stan Butchart]

There is no frame of reference where the speed is 1V along the entire surface.

One last shot.
we have a fluid that is moving relative to the remote air. We have a body that is moving relative to the remote air. The body and moving air are moving relative to each other.

I mark the lip of a glass and place it upside down on the table and rotate it. (I should accelerate and decclerate it but that is a detaile.) Now as i rotate it as I move it across the table. What has the mark traced on the table?

I have resolved it from the source/sink pattern. I have resolved it from the tangental velocity relative tothe surface and i have resolved it from the cylinder surface velocity equation.
I have graphed its path as described in http://svbutchart.com
I have never ever suggested that the constant 1V was relative to the cylinder.

 

[boneh3ad]

One last shot.
we have a fluid that is moving relative to the remote air. We have a body that is moving relative to the remote air. The body and moving air are moving relative to each other.

I mark the lip of a glass and place it upside down on the table and rotate it. (I should accelerate and decclerate it but that is a detaile.) Now as i rotate it as I move it across the table. What has the mark traced on the table?

I have resolved it from the source/sink pattern. I have resolved it from the tangental velocity relative tothe surface and i have resolved it from the cylinder surface velocity equation.
I have graphed its path as described in http://svbutchart.com
I have never ever suggested that the constant 1V was relative to the cylinder.

You have to be careful with your choice of words. Velocity and speed are not the same thing. Going through the math, in the frame you are describing, the speed of a parcel of fluid is, in fact, going to be $U_{\infty}$ along the circumference. The velocity isn't, and is a vector quantity. Additionally, your site is quote difficult to follow. For anyone else on here that has been reading along and thought Stan's explanations were somewhat opaque, I go through the math showing this below.

Earlier, I talked about potentials. The equation Stan keeps quoting is the same equation as mine for $v_{\theta}$, only you have the angle defined differently. My angle was defined in a typical coordinate system with 0 being on the right at the x-axis and increasing counter-clockwise. Otherwise, the two are identical. I am starting from my equation since it will make things easier to change to a Cartesian coordinate system and change frames.

Now, if we want to switch into a Cartesian system, we take my results for $v_{\theta}$ and $v_{r}$ and plug them into
$$u = v_r\cos\theta - v_{\theta}\sin\theta$$
$$u = v_r\sin\theta + v_{\theta}\cos\theta$$

Since we already determined that on the surface, $v_{r}=0$, this simplifies to
$$u = 2U_{\infty}\sin^2\theta$$
$$v = -2U_{\infty}\sin\theta\cos\theta$$

This defines the vector field on the surface of the cylinder in Cartesian coordinates in the frame where we are riding along with the moving cylinder (so it has no velocity and the free stream is moving). Now to change reference frames such that we are viewing the cylinder moving through the stagnant fluid, we just have to add the following:
$$(-U_{\infty},0)$$

Leaving us with
$$u = 2U_{\infty}\sin^2\theta-U_{\infty}$$
$$v = -2U_{\infty}\sin\theta\cos\theta$$

This is the velocity vector along the surface in a Cartesian coordinate system in the inertial frame. The cylinder is moving through stagnant air. We can transform this back into cylindrical coordinates for simplicity using
$$v_r = u\cos\theta + v\sin\theta$$
$$v_{\theta} = -u\sin\theta + v\cos\theta$$

This leaves us with (and you can do the algebra if you want)
$$v_r = -U_{\infty}\cos\theta$$
$$v_{\theta} = -U_{\infty}\sin\theta$$

Taking the magnitude of this, it is easy to then show that
$$|\bar{V}|=U_{\infty}$$

So yes, indeed, the fluid parcel is moving with a constant speed of $U_{\infty}$ in an inertial frame. The problem is, this doesn't tell us a lot about the forces on the cylinder. The important quantity is the speed with respect to the cylinder.

 

[Stan Butchart]

So yes, indeed, the fluid parcel is moving with a constant speed of $U_{\infty}$ in an inertial frame. The problem is, this doesn't tell us a lot about the forces on the cylinder. The important quantity is the speed with respect to the cylinder.

The only point here was that the Bernoulli Principle was not at work along the surface but that the Bernoulli Equation was, based on normal acceleration.

 

[boneh3ad]

The only point here was that the Bernoulli Principle was not at work along the surface but that the Bernoulli Equation was, based on normal acceleration.

That doesn't make any sense. Bernoulli's equation makes perfect sense in this situation, especially since we are doing this ideally. Just because the result of something is zero in a certain frame or reference doesn't mean it doesn't apply. Regardless, the important frame is that which follows the body.

 

[Stan Butchart]

Bernoulli's equation makes perfect sense in this situation,

In the long run you might be right.If we only want numbers the Bernoulli Equation dos just fine. However, most of the thousands that search the web and books are only looking to uderstand the lifting process.

The largest percentage of the hundreds of articles and sites firmly define the concept that pressure changes are created by linear variations in the flow tangent and relative to the surface. (in the sense of the venturi)

There is additional group that explains that the velocity changes are in response to the pressure variations.

The primary flow is from displacement. The source/sink pattern describes the motion effects throughout the entire flowfield. These instantaneous velocity vectors are relative to the remote (still) fluid. When we add the fwd motion of the cylinder we can find the tangent velocity relative to the surface. This tangent velocity, in combination with the surface radius produces the normal acceleration, v^2/R, creating a pressure gradient that must be integrated across the entire flowfield. It is almost luck that all of this can be reduced again to the Bernoulli Equation.

We find that we can still apply the Bernoulli Equation to the tangent velocity, relative to the surface. But is this the Bernoulli Principle??