Crosswind problem (pgs. 34-35, Thinking Physics, 3rd edition)

[erobz]

Yes, this is the relative flow at $v_x = w$, and the force on the vane has now a component in the positive direction of $v$ so it will still accelerate.

So what is the hypothesis for the limits of $v_x,v_y$? I ask because this completely re-orients the problem. $v_x$ is now orthogonal to the incoming wind! The whole question appears to be flipped on its head. $v_x$ is no longer in the direction of the wind.

 

[A.T.]

So what is the hypothesis for the limits of $v_x,v_y$? I ask because this completely re-orients the problem.

The limit for $v$ is fully determined by the maximal lift/drag ratio of the vane and $\theta$ (because the track is frictionless). More generally it depends on both lift/drag ratios: at the air and at the surface.

This is explained here:
https://www.onemetre.net//design/CourseTheorem/CourseTheorem.htm

And visualized here (see also references in the description):
https://www.geogebra.org/m/tj5qf3w2

In short the lift/drag ratios determine the apparent wind angle AWA (usually called $\beta$).

AWA = atan(1 / LD_surface) + atan(1 / LD_air)

This AWA-angle then determines the size of the polar circle, that limits the speed for any given course (true wind is to north and the scale is true wind multiples):

The above example assumes AWA = 6°, which is based on GPS measurements with iceboats:
https://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

This allows a downwind velocity component of more than 5 times true windspeed.

 

[erobz]

I get the following system of equations:

$y$ direction
$$ N \cos \theta = M \frac{dv_y}{dt} + \dot m v_y - \dot m w \cos \beta + \dot m w \tag{1} $$

$x$ direction
$$ -N \sin \theta = M \frac{dv_x}{dt} + \dot m v_x - \dot m w \sin \beta \tag{2}$$

Using the constraint and it derivative:

$$\frac{v_y}{v_x} = \tan \theta \tag{3} $$

$$ \implies \frac{dv_y}{dt} = \frac{dv_x}{dt} \tan \theta \tag{4} $$

The final ( sub 2 $\to$ 1, then 3,4, $\to$ 1 ) resulting ODE for $x$ direction:

$$ M \frac{dv_x}{dt} ( 1 + \tan^2 \theta ) + \dot m v_x ( 1 + \tan^2 \theta ) + \dot m w ( \tan \theta - \cos \beta \tan \theta - \sin \beta) = 0 \tag{5}$$

From which the steady state response is given by:

$$\lim_{\dot{v}_x \to 0 }(5) \implies v_x \to -w\frac{( \tan \theta - \cos \beta \tan \theta - \sin \beta)}{( 1 + \tan^2 \theta )}$$

My position is that this result does not support the claim that $v_x$ can exceed $w$.

 

[A.T.]

My position is that this result does not support the claim that $v_x$ can exceed $w$.

You should do some reality-checks with simple cases, like the one shown in your current diagram ($v_x = w$). It's easy to see what the direction of force F on the vane would be, given the flow deflection. And since F has a component in the direction of v, the cart will accelerate further.

Not sure where the error in your general solution is, but you have no parameter for the vane orientation, so I assume it's fixed as shown in the image. In this case the incoming relative flow along the vane will not be tangential most of the time, and somewhere between $v_x = 0$ and $v_x = w$ the tangential component of the relative flow flips orientation. There is obviously still a radial flow component onto the vane which pushes it at this point. However, if your math only considers tangential flow, then you might have found this false limit.

 

[erobz]

You should do some reality-checks with simple cases, like the one shown in your current diagram ($v_x = w$). It's easy to see what the direction of force F on the vane would be, given the flow deflection. And since F has a component in the direction of v, the cart will accelerate further.

I don't think its easy to see anything of the sort. If things like this were easy to see (trivial as you put it), the pioneers of fluid mechanics wouldn't have developed an analytic approach to solve such problems.

To me what seems trivial is that if you can only approach $w$ by taking an impinging jet and completely reversing its momentum (post #29), then anything else is at most second best.

I'm clearly not an expert, so if I'm interpreting the mechanics incorrectly, I want to know...but I need some clear - "right there is the issue" specific gripes with what the math is actually alluding to, or how it is being mishandled, and most importantly what to do to remedy it within the accepted framework of Newtonian Mechanics...

 

[A.T.]

I don't think its easy to see anything of the sort.

The force on the vane F is in the opposite direction of the momentum change of the fluid. Take the vector difference of your incoming and outgoing relative flow (blue arrows). That gives you the direction of F.

To me what seems trivial is that if you can only approach $w$ by taking an impinging jet and completely reversing its momentum (post #29),...

Post #29 is about going directly downwind, and is consistent with the limits on conventional sailcraft on that course.

... then anything else is at most second best.

Wrong. The configuration in post #29 maximizes $a$ for $v = 0$ (initial acceleration). But the goal is to maximize $v$ for $a = 0$ (terminal velocity).

Your argument is like saying: "To maximize bike speed, nothing can be better than the lowest gear, because it gives you maximal acceleration from rest."

I need some clear - "right there is the issue"

What do you assume for the vane orientation at different speeds? To explore the limits you would have to assume the optimal vane orientation for each speed: The one that maximizes the component of F in the positive direction of v. Is that included in your general approach?

 

[erobz]

What do you assume for the vane orientation at different speeds? To explore the limits you would have to assume the optimal vane orientation for each speed: The one that maximizes the component of F in the positive direction of v. Is that included in your general approach?

The vane orientation is fixed relative to the cart in the position shown. Just like the sail in all the animations that supposedly "described how this is achieved"...

 

[A.T.]

The vane orientation is fixed relative to the cart in the position shown.

You cannot determine the maximal possible speed using a fixed vane orientation, because the optimal vane orientation is a function of the speed itself.

But you can use the vane orientation as shown for $v_x = w$, and check if acceleration is possible at this condition. If yes, then this is enough to demonstrate that $v_x = w$ is not a limiting speed.

 

[erobz]

You cannot determine the maximal possible speed using a fixed vane orientation, because the optimal vane orientation is a function of the speed itself.

But you can use the vane orientation as shown for $v_x = w$, and check if acceleration is possible at this condition. If yes, then this is enough to demonstrate that $v_x = w$ is not a limiting speed.

$v_x$ is not in the direction of the wind though. We are talking about downwind travel faster than the wind in the direction of the wind.

And you have completely ignored the comment that every one of the animations you have shown to describe how this is possible has the vane fixed relative to the cart...please explain the cognitive dissonance?

 

[erobz]

But you can use the vane orientation as shown for $v_x = w$, and check if acceleration is possible at this condition. If yes, then this is enough to demonstrate that $v_x = w$ is not a limiting speed.

This is false. You cannot simply "plug in" $v_x = w$, if you can't properly show that $w$ is in the domain of possible values for $v_x$. Its contradictory.

 

[A.T.]

$v_x$ is not in the direction of the wind though. We are talking about downwind travel faster than the wind in the direction of the wind.

We are talking about a velocity component in the direction of the true wind, that is greater than the true wind.

If the true wind (relative to the ground) is in the same direction as $v_x$, then at $v_x = w$ the apparent wind (relative to the cart) is as shown in the image (towards negative y).

And you have completely ignored the comment that every one of the animations you have shown to describe how this is possible has the vane fixed relative to the cart...please explain?

Since you didn't provide an example I can only guess what your confusion is about:

1) If an animation shows a constant speed, then there is no need to change the sail setting,

2) There is a difference between "possible" and "optimal". If you seek a general solution for the limit, then you must use the optimal sail setting.

 

[A.T.]

But you can use the vane orientation as shown for $v_x = w$, and check if acceleration is possible at this condition. If yes, then this is enough to demonstrate that $v_x = w$ is not a limiting speed.

This is false. You cannot simply "plug in" $v_x = w$, if you can't properly show that $w$ is in the domain of possible values for $v_x$. Its contradictory.

It shows that if you push the cart to $v_x = w$ and release it at that speed, then it can accelerate further on its own to $v_x > w$.

If you have doubts that it can also get from $v_x = 0$ to $v_x = w$ on its own, then you have to solve it generally for every speed in between, with the optimal vane setting.

 

[erobz]

This analysis I suffering from scope creep. Before it was enough to change the sail orientation, then change the wind direction, and then vane angle. Now we are optimizing the sail angle as a function of velocity… “it’s trivial” -your words. What’s that function then?

 

[A.T.]

Now we are optimizing the sail angle as a function of velocity…

That's what every racing sailor does. The analysis is much simpler in terms of lift/drag ratios and vector algebra. See post #102.

What’s that function then?

Every airfoil has an angle of attack that maximizes lift/drag. That's how you set the sail to the apparent wind.

 

[erobz]

That's what every racing sailor does. The analysis is much simpler in terms of lift/drag ratios and vector algebra. See post #102.Every airfoil has an angle of attack that maximizes lift/drag. That's how you set the sail to the apparent wind.

I’m going to try and figure out how to optimize…

Most likely an exercise in futility.

 

[A.T.]

I’m going to try and figure out how to optimize…

Most likely an exercise in futility.

Before you put effort into optimizing the vane orientation, please consider my comment below as to why your fixed vane analysis might have failed. It's quite possible that even the fixed vane as shown, while not optimal, sill allows acceleration at any $v_x <= w$.

Not sure where the error in your general solution is, but you have no parameter for the vane orientation, so I assume it's fixed as shown in the image. In this case the incoming relative flow along the vane will not be tangential most of the time, and somewhere between $v_x = 0$ and $v_x = w$ the tangential component of the relative flow flips orientation. There is obviously still a radial flow component onto the vane which pushes it at this point. However, if your math only considers tangential flow, then you might have found this false limit.

 

[erobz]

Before you put effort into optimizing the vane orientation, please consider my comment below as to why your fixed vane analysis might have failed. It's quite possible that even the fixed vane as shown, while not optimal, sill allows acceleration at any $v_x <= w$.

I don't understand it. If it's going to be on the vane the flow is going to be tangential to the vane. You might be saying that because of $v_x > 0$, its actually just taking a glancing blow. I would say that is detrimental to the argument, because the momentum change in that scenario will be one of reduced capacity in comparison.

 

[erobz]

So, I think the objective is to maximize the component of velocity $v_x$ of the cart by maximizing the force $F_x$. A maximal $v_x$ maximizes $v_y$ by constraint (3) in post #103.

To achieve this, I introduce the parameter $\varphi$ for the angle of rotation of the vane relative to the cart. The intention is to find $\varphi$ as a function of $v_x$ that maximizes $F_x$ for a fixed angle of turn $\beta$.

From the diagram above we find that:

$$ F_x = \dot m \left( v_x - w \sin( \beta + \varphi )\right) \tag{1}$$

Next, attempt to optimize $F_x$ w.r.t. $v_x$

$$ \frac{dF_x}{dv_x} = \frac{ d \dot m }{d v_x} \left( v_x - w \sin( \beta + \varphi ) \right) + \dot m \left( 1 - w \cos ( \beta + \varphi ) \frac{ d \varphi}{d v_x} \right) \tag{2}$$

Under continuity the mass flowrate entering-exiting the control volume is given by:

$$ \dot m = \rho A ( w + v_y ) = \rho A ( w + v_x \tan \theta ) \tag{3}$$

This implies that:

$$ \frac{ d \dot m }{d v_x} = \rho A \tan \theta = constant. \tag{4} $$

Making the substitutions $(3)$ and $(4) \to (2)$ becomes:

$$ \frac{dF_x}{dv_x} = \rho A \tan \theta \left( v_x - w \sin( \beta + \varphi ) \right) + \rho A ( w + v_x \tan \theta ) \left( 1 - w \cos ( \beta + \varphi ) \frac{ d \varphi}{d v_x} \right) \tag{2'}$$

Optimization implies looking for the solution to:

$$ 0 = \frac{dF_x}{dv_x} = \rho A \tan \theta \left( v_x - w \sin( \beta + \varphi ) \right) + \rho A ( w + v_x \tan \theta ) \left( 1 - w \cos ( \beta + \varphi ) \frac{ d \varphi}{d v_x} \right) \tag{5}$$

Solving (5) for $\varphi '$ yields the following first order nonlinear ODE:

$$ \frac{ d \varphi }{ d v_x} = \frac{w - w \tan \theta \sin ( \beta + \varphi)+ 2 \tan \theta v_x}{ \cos ( \beta + \varphi ) \left( w^2 +w v_x \tan \theta \right) } \tag{6}$$

How am I doing so far?

 

[A.T.]

by maximizing the force $F_x$.

No, as already said:

....the optimal vane orientation for each speed: The one that maximizes the component of F in the positive direction of v.

Also shown by the dotted blue arrow:

 

[erobz]

Maximizing $v_x$ maximizes $v$. The motion is constrained by $v_y = v_x \tan \theta$

 

[A.T.]

Maximizing $v_x$ maximizes $v$. The motion is constrained by $v_y = v_x \tan \theta$

I was replying this part:

... by maximizing the force $F_x$.

Maximizing $F_x$ doesn't maximize $v$ or $v_x$. For example: If $F$ is perpendicular to the track, then no matter how large you make $F$ and $F_x$ it won't start moving.

 

[erobz]

I was replying this part:

Maximizing $F_x$ doesn't maximize $v$ or $v_x$. For example: If $F$ is perpendicular to the track, then no matter how large you make $F$ and $F_x$ it won't start moving.

Ok, I see the issue.

 

[A.T.]

I don't understand it.

Just plug in some numbers into your solution to see if it makes any sense.

$$\lim_{\dot{v}_x \to 0 }(5) \implies v_x \to -w\frac{( \tan \theta - \cos \beta \tan \theta - \sin \beta)}{( 1 + \tan^2 \theta )}$$

My position is that this result does not support the claim that $v_x$ can exceed $w$.

The angles in the image below are roughly $\theta=33°$ and $\beta = 43°$. When you plug these values into your solution you get a limit for $v_x$ of about $0.36 w$.

If the track is completely frictionless, why would the below cart (even with the fixed vane) stop accelerating at $v_x = 0.36w$?

The incoming relative flow w' would be ~20° clockwise from the x-axis, blowing right onto the vane, creating a force F roughly as shown, which only needs to have some non-zero component in the direction of v for acceleration.

 

[erobz]

Just plug in some numbers into your solution to see if it makes any sense.The angles in the image below are roughly $\theta=33°$ and $\beta = 43°$. When you plug these values into your solution you get a limit for $v_x$ of about $0.36 w$.

If the track is completely frictionless, why would the below cart (even with the fixed vane) stop accelerating at $v_x = 0.36w$?

The incoming relative flow w' would be ~20° clockwise from the x-axis, blowing right onto the vane, creating a force F roughly as shown, which only needs to have some non-zero component in the direction of v for acceleration.

View attachment 320802

$w$ is not in the direction you show?

 

[A.T.]

$w$ is not in the direction you show?

$w$ : true wind, air motion relative to the ground, is along positive x direction
$w'$ : apparent wind, air motion relative to cart, is given by the vector equation:

$w' = w - v$ (see also vector diagram top-left)

 

[erobz]

$w$ : true wind, air motion relative to the ground, is along positive x direction
$w'$ : apparant wind, ari motion relative to cart, is given by the vector equation:

$w' = w - v$ (see also vector diagram top-left)

View attachment 320808

No the true wind $w$ is in $-y$ direction for the analysis you are quoting.

 

[A.T.]

No the true wind $w$ is in $-y$ direction for the analysis you are quoting.

That doesn't make any sense. Why are you solving for the limit of $v_x$ if true wind is along $-y$. The whole dispute was about the limit of velocity component parallel to true wind.

 

[erobz]

That doesn't make any sense. Why are you solving for the limit of $v_x$ if true wind is along $-y$. The whole dispute was about the limit of velocity component parallel to true wind.

Because you were asking me to flip flop all the directions of things, the sail, the angle of the vane and the wind! I had said this doesn't make sense for the question a long time ago. see post #109, and prior to that post #101. I brought up this point several times now.

The fluid jet is the true wind. What configuration do you want me to solve?

 

[A.T.]

Because you were asking me to flip flop all the directions of things, the sail, the angle of the vane and the wind!

I never told you to change the true wind from positive x. I told you to compute the wind relative to the vane, which for $v_x = w$ is indeed along negative y.

The fluid jet is the true wind.

There is no jet. The moving airmass is continuous. Your jet is just a source of confusion you have introduced for yourself. Please learn how to add vectors and transform velocities between frames. I have posted the correct vector math many times.

 

[erobz]

I never told you to change the true wind from positive x. I told you to compute the wind relative to the vane, which for $v_x = w$ is indeed along negative y.There is no jet. The moving airmass is continuous. Your jet is just a source of confusion you have introduced for yourself. Please learn how to add vectors and transform velocities between frames. I have posted the correct vector math many times.

No, you're not understanding the intentions. The jet(s) - an array of what is shown are the are the wind!

We think about the wind flow as lamina (the individual jets impacting the sail as the cart moves along the track), in the limit that the distance between the jets goes to zero.

 

[A.T.]

the individual jets impacting the sail

The fluid jet is the true wind.

The sail experiences the apparent wind (relative to the boat), not the true wind. If your jets are the true wind (relative to the ground) then the interaction between jet and sail in your image is nonsense. This is how the interaction between sail and apparent wind looks like.

Forget your jets. Just do the proper vector math.

 

[erobz]

The sail experiences the apparent wind (relative to the boat), not the true wind. If your jets are the true wind (relative to the ground) then the interaction between jet and sail in your image is nonsense. This is how the interaction between sail and apparent wind looks like.

Forget your jets. Just do the proper vector math.

Well, correct it within the appropriate framework then i.e. Newtonian Mechanics. That's how we do Classical Mechanics. We don't add randomly drawn vectors and say, "look at the way they point, that proves it..."

OR Just forget about it and go enjoy your faster than wind down wind sailing. I'm through wasting any more time on this back and forth, there is no ground to gain.

Take Care!

 

[A.T.]

Well, correct it within the appropriate framework then i.e. Newtonian Mechanics.

I did correct your errors, as far I could understand what you are trying to do. But if you can't even keep true and relative wind apart, because you insist on thinking in terms of "jets", then there is not much I can do.

In the end, the much simpler vector approach agrees with empirical evidence. Your "jets" don't. And that's all that matters in physics.

 

[A.T.]

We don't add randomly drawn vectors and say, "look at the way they point, that proves it..."

If those vectors look "random" to you, then I can recommend some reading that explains them further:

High-speed sailing, Wolfgang Püschl 2018 Eur. J. Phys. 39 044002
https://iopscience.iop.org/article/10.1088/1361-6404/aab982

Physics of Sailing, John Kimball
https://books.google.de/books?id=Xe_i23UL4sAC&lpg=PP1&hl=de&pg=PA49#v=onepage&q&f=false

Course Theorem, Lester Gilbert
https://www.onemetre.net//design/CourseTheorem/CourseTheorem.htm

 

[A.T.]

What configuration do you want me to solve?

This is the situation at $v_x = w$:

$w$: true wind (relative to the ground) along positive x
$v$: velocity of the cart relative to the ground
$w'_{in}$: incoming apparent wind (relative to the vane) along negative y (vector equation: $w'_{in} = w - v$)
$w'_{out}$: deflected apparent wind (relative to the vane)
$F$: force on the vane (vector equation: $F = \dot{m}w'_{in} - \dot{m}w'_{out}$)
$F_v$: component of $F$ parallel to $v$ (vector equation: $F_v = F \cdot \hat{v}$)

As long as $F_v$ is positive, we can accelerate further, and $v_x = w$ is not a limit.

 

[Gleb1964]

There is no jet. The moving airmass is continuous.

I would agree even with unrealistic "jet wind" concept, asking consider infinity long sail instead.
Your explanations are good.
Once in past disputing the same subject (if its possible to descent faster than the wind speed), I did a picture with the boat on a flow. Using a lever connected to the ground, the boat is going faster than the flow. The boat is using the speed difference between flow and ground, exchanging momentum both with flow and ground.

 

[A.T.]

I would agree even with unrealistic "jet wind" concept, asking consider infinity long sail instead.

The problem with @erobz 's jet model is that causes him to confuse true and relative wind. He makes his jets parallel to the true wind, and then seems to falsely assume that these are the stream-lines of the relative wind that the airfoil interacts with. But the relative wind has a completely different direction when you move across the true wind.

Once in past disputing the same subject (if its possible to descent faster than the wind speed), I did a picture with the boat on a flow. Using a lever connected to the ground, the boat is going faster than the flow. The boat is using the speed difference between flow and ground, exchanging momentum both with flow and ground.

Your image is about going directly downwind faster than the wind, which we even didn't get into here. Some similar animations:

 

[Gleb1964]

Your image is about going directly downwind faster than the wind, which we even didn't get into here.

Descending directly downwind faster than wind would need lateral speed component for sail, like making spinning propeller sail, where propeller rotation forced by motion relative to water.

The problem with @erobz 's jet model is that causes him to confuse true and relative wind. He makes his jets parallel to the true wind, and then seems to falsely assume that these are the stream-lines of the relative wind that the airfoil interacts with.

Perhaps, the sailing boat can be considered in any frame with the same outcome.
I would take the model of boat going downwind with the speed many times exceed the wind speed to get it the obvious that the apparent wind would be filling like counter wind moving in any directions. But nevertheless, in the most directions it is possible to find a vane position when sail is getting a positive force, keeping boat moving. The boat can descend faster than wind, but the sail surface can descend slower than wind, taking wind momentum into the sail.

 

[A.T.]

Perhaps, the sailing boat can be considered in any frame with the same outcome.

The change in the air's momentum (and thus the force) is frame independent. But the direction of the incoming and outgoing flows are frame dependent. In the image below the blue/purple line is the stream line in the ground frame (similar to @erobz jets). But the stream relative to the sail is indicated by the red dots.

Here it is animated:

 

[erobz]

Same as for a boat with a fixed sail going directly downwind. But the boats that achieve downwind components greater than windspeed are not going directly downwind. Your original model was more relevant, just your sail model was bad.

That was in reply to the following post #29. My emphasis.

What is the maximum velocity of the cart in the following image?

View attachment 320367

The original configuration that you refer to was in post #20, so lets move back to that and fix it...I will analyze on any angle a jet(s) impinging on a vane. How do you want me to orient\select geometry for the vane such that the craft will outrun the jet(s) in the direction of the jet for some track angle $\theta$?

Assumptions:

1) The vane geometry must be fixed ( it's not changing from a quarter turn to a half turn etc...)
2) The direction of the impinging fluid jet(s) is fixed to the right $\rightarrow^+$
3) The velocity $w$ of the incoming fluid jet(s) relative to the ground is constant magnitude.
4) The frame which we consider the forces is the inertial frame.

If you have a further stipulation that we need to optimize the vane orientation w.r.t. the craft, as a function of cart velocity, I'll work on that after those conditions\assumptions are agreed upon and I complete a fixed vane orientation result.

 

[erobz]

I would agree even with unrealistic "jet wind" concept, asking consider infinity long sail instead.

I can begin with that. So you want a single jet, infinitely long planar sail oriented relative to the cart such that the force is maximized on the cart in the direction of motion?

 

[A.T.]

How do you want me to orient\select geometry for the vane such that the craft will outrun the jet(s) in the direction of the jet for some track angle $\theta$?

You can orient the sail like in the image below:

w : wind relative to the ground
w' : wind relative to the boat
v: boat velocity relative to the ground

2) The direction of the impinging fluid jet(s) is fixed to the right $\rightarrow^+$

This sounds wrong. The direction of fluid inflow onto the sail (w') is not fixed. It changes with v:

w' = w - v

See blue arrows above.

4) The frame which we consider the forces is the inertial frame.

The force by the air on the sail is frame invariant.

 

[erobz]

You can orient the sail like in the image below:

https://www.physicsforums.com/attachments/322853
w : wind relative to the ground
w' : wind relative to the boat
v: boat velocity relative to the groundThis sounds wrong. The direction of fluid inflow onto the sail (w') is not fixed. It changes with v:

w' = w - v

See blue arrows above.The force by the air on the sail is frame invariant.

So it looks like we can't agree...can't say I'm surprised. The wind caries momentum into the sail,and is carried out in the direction the sail deflects it. This relative wind you keep trying to invoke for the analysis doesn't exist in the inertial frame. You are making up momentum that doesn't exist in the inertial frame. I'm doing the analysis in the inertial frame (a frame fixed to the ground)

 

[A.T.]

So it looks like we can't agree...can't say I'm surprised.

Fortunately real world evidence agrees with me.

The wind caries momentum into the sail,and is carried out in the direction the sail deflects it. This relative wind you keep trying to invoke for the analysis doesn't exist in the inertial frame. You are making up momentum that doesn't exist in the inertial frame.

1) We have empirical data on lift/drag ratios, which are all based on the relative wind. Those are way more realistic than the simplifying assumptions about the air deflection that you have to make.

2) The way you yourself keep drawing the flow (as if the vane was fixed) can only refer to the relative wind, The deflection in the ground frame, where both: the air and the vane are moving looks different.

I'm doing the analysis in the inertial frame (a frame fixed to the ground)

You can do that, but keep in mind: The mass flow rate into your moving control volume depends on the wind relative to the control volume. No matter how much you try to ignore it, the varaying relative wind still comes in.

 

[erobz]

2) The way you yourself keep drawing the flow (as if the vane was fixed) can only refer to the relative wind, The deflection in the ground frame, where both: the air and the vane are moving looks different.

How I draw it, or how it looks is not important. What is important is properly accounting for the change in momentum of the flow w.r.t the stationary inertial frame.

You can do that, but keep in mind: The mass flow rate into your moving control volume depends on the wind relative to the control volume. No matter how much you try to ignore it, the varaying relative wind still comes in.

Yes. Thank you.

 

[Gleb1964]

.. This relative wind you keep trying to invoke for the analysis doesn't exist in the inertial frame. You are making up momentum that doesn't exist in the inertial frame. I'm doing the analysis in the inertial frame (a frame fixed to the ground)

Not quite agree. You can consider inertial frame with the boat at any speed. Been sailing a bit, the apparent (relative) wind is a very real thing, especially when you are sitting in the boat.

 

[erobz]

Not quite agree. You can consider inertial frame with the boat at any speed. Been sailing a bit, the apparent (relative) wind is a very real thing, especially when you are sitting in the boat.

You are in a non-inertial frame when you are on the boat accelerating. Let's just talk about how we do Newtonian Mechanics. If you see something is off with the physics I will present, say something. That is what should be the main focus here.

 

[Gleb1964]

At any moment of acceleration it is possible to induce an instant inertial frame to the boat. That inertial frame is good for understand the correct change of the air momentum.
I would even not complicate the task, but just assume the boat moving with constant speed. Make a solve with the constant boat speed. It is always possible to assume that boat has some resistance, keeping it speed constant. Forgot about how boat is getting to the speed. Make it simple.

 

[erobz]

This is the diagram for your suggested infinite planar sail. My goal is to optimize the angle of the sail w.r.t. the cart $\beta$ such that the component of the force $F$ ( shown in purple ) in the direction of motion is maximized. I.e. I'm hoping to maximize $F \cos \gamma$. Seeing the equations I'm developing I don't suspect to have a prayer of getting there in general, but I'll see what I can come up with.

If you don't like it, shoot it down before I go further.

 

[erobz]

At any moment of acceleration it is possible to induce an instant inertial frame to the boat. That inertial frame is good for understand the correct change of the air momentum.
I would even not complicate the task, but just assume the boat moving with constant speed. Make a solve with the constant boat speed. It is always possible to assume that boat has some resistance, keeping it speed constant. Forgot about how boat is getting to the speed. Make it simple.

Unfortunately it can't work like that. I must form some first order differential equation and look at the limit of said equation as $\frac{dv}{dt} \to 0$.