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

[erobz]

Just to be clear, I meant $0° < \theta < 45°$. For the $v_x = w$ case you can keep your 90° deflection vane, just rotate it by 90° clockwise, so it deflects the relative flow from -y to -x.

It should be obvious right away that this slows down the air relative to the ground, so it removes energy from the ground-air system.

This is what I thought you meant?

 

[A.T.]

This is what I thought you meant?

The way this is drawn, it would not even go downwind, but rather upwind (opposite to the v arrow). The force on the vane is along the vector from the apex of $\beta$ to the middle of the vane. And this force must have a positive component along v.

For the case $v_x = w$ you can make $\beta = 90°$ by extending the right end of the vane anti-clockwise, if that makes your life simpler. It's not efficient but sufficient to show you can still accelerate at $v_x = w$, which is the whole point.

Most importantly: The flow direction you have drawn applies only initially for $v = 0$. As soon it starts moving the incoming flow relative to the vane starts rotating. The 2D vector math is:

flow_relative_to_vane = flow_relative_to_ground - vane_velocity_relative_to_ground

 

[erobz]

The way this is drawn, it would not even go downwind, but rather upwind (opposite to the v arrow). The force on the vane is along the vector from the apex of $\beta$ to the middle of the vane. And this force must have a positive component along v.

For the case $v_x = w$ you can make $\beta = 90°$ by extending the right end of the vane anti-clockwise, if that makes your life simpler. It's not efficient but sufficient to show you can still accelerate at $v_x = w$, which is the whole point.

Most importantly: The flow direction you have drawn applies only initially for $v = 0$. As soon it starts moving the incoming flow relative to the vane starts rotating. The 2D vector math is:

flow_relative_to_vane = flow_relative_to_ground - vane_velocity_relative_to_ground

You've told me that turning the flow by $\beta = 90^{\circ}$ was horribly inefficient (personally I don't see the relevance since I'm only looking at the steady state solutions - but whatever). I'm just going to solve it in general for an arbitrary angle $\beta$.

Most importantly: The flow direction you have drawn applies only initially for �=0. As soon it starts moving the incoming flow relative to the vane starts rotating. The 2D vector math is:

What do you mean? That is not the case I'm examining. I'm imagining a vertical ($y$ direction) array of jets. The incoming momentum (momentum entering the control volume) is always $\dot m w ~\mathbf{ \hat{x}}$, relative to the inertial frame.

 

[A.T.]

I'm just going to solve it in general for an arbitrary angle $\beta$.

It's more important that you solve it for every orientation of the vane, because the one you have assumed in the sketch is obviously wrong for downwind travel.

I'm imagining a vertical ($y$ direction) array of jets.

Then why individual jets at all? Why not simply a continuous moving body of air, like the actual wind?

The incoming momentum (momentum entering the control volume) is always $\dot m w ~\mathbf{ \hat{x}}$, relative to the inertial frame.

So even if the control volume moves at windspeed parallel to the wind ($v_x = w, v_y = 0$), you still have non-zero momentum flow into the volume? Since no air enters the volume in this case, how does this work?

 

[erobz]

It's more important that you solve it for every orientation of the vane, because the one you have assumed in the sketch is obviously wrong for downwind travel.

You asked me to rotate the vane 90 degrees clockwise for what was considered in post 20.

Then why individual jets at all? Why not simply a continuous moving body of air, like the actual wind?

The array of jets is "the wind" in the limit. This has to be analyzed in a framework that makes some sense.

So even if the control volume moves at windspeed parallel to the wind ($v_x = w, v_y = 0$), you still have non-zero momentum flow into the volume? Since no air enters the volume in this case, how does this work?

$\dot m \to 0$ as $v_x \to w$ in this analysis that must be the case. The steady state is the steady state, the cv does not accelerate in the limit. It has some component of velocity $v_x$ and $v_y = v_x \tan \theta$ in the limit.

 

[A.T.]

You asked me to rotate the vane 90 degrees clockwise for what was considered in post 20.

Yes, but you also removed half of it. The wrong half for downwind travel.

$\dot m \to 0$ as $v_x \to w$

So what is the formula for $\dot m$ in the general case when $v_y > 0$ ?

 

[erobz]

Yes, but you also removed half of it. The wrong half.

You wish to have that?

Is the flow coming in tangentially or horizontally?

 

[A.T.]

You wish to have that?

Yes, that would work better for downwind travel at $v_x = w$.

Is the flow coming in tangentially or horizontally?

The 2D vector math is:

flow_relative_to_vane = flow_relative_to_ground - vane_velocity_relative_to_ground

If you want the incoming flow_relative_to_vane to be exactly tangential to the vane, then you to have adjust the vane orientation for every vane_velocity_relative_to_ground.

But for $v_x = w$ the above configuration would have tangential relative incoming flow along negative y.

 

[erobz]

like this?

 

[A.T.]

like this?

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 can still accelerate.

 

[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 $v_y = v_x \tan \theta$