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

[erobz]

This makes sense, if I interpret your definitions correctly: $\beta$ is the flow deflection in the ground frame, right?

I treated $\beta$ as the angle the flow leaves w.r.t. the control volume. i.e the flow is coming out of the boat at a constant angle $\beta$ in the frame of the boat. My momentum equations for the angle of that outflow in the ground frame would not be constant. If this is an issue, I'm not seeing it...

 

[A.T.]

I treated $\beta$ as the angle the flow leaves w.r.t. the control volume. i.e the flow is coming out of the boat at a constant angle $\beta$ in the frame of the boat. My momentum equations for the angle of that outflow in the ground frame would not be constant. If this is an issue, I'm not seeing it...

Any velocity that is parallel to the trajectory in the ground frame, is also parallel to the trajectory in the boat frame.

So for v -> infinity, the outflow velocity is parallel to the trajectory in both frames.

But the inflow is not:
- In the ground frame the inflow velocity is always vertical (true wind).
- In the boat frame (with v -> infinity) the inflow velocity is parallel to the trajectory (just like the outflow)

So the deflections (change in velocity direction) for v -> infinity are:
- In the ground frame: $\beta$
- In the boat frame: 0

 

[erobz]

Any velocity that is parallel to the trajectory in the ground frame, is also parallel to the trajectory in the boat frame.

So for v -> infinity, the outflow velocity is parallel to the trajectory in both frames.

But the inflow is not:
- In the ground frame the inflow velocity is always vertical (true wind).
- In the boat frame (with v -> infinity) the inflow velocity is parallel to the trajectory (just like the outflow)

So the deflections (change in velocity direction) for v -> infinity are:
- In the ground frame: $\beta$
- In the boat frame: 0

I don’t follow what you are trying to tell me, but I suspect it is that you believe the analysis is flawed?

 

[A.T.]

I don’t follow what you are trying to tell me, but I suspect it is that you believe the analysis is flawed?

I'm not sure, that's why I wanted to confirm if it is consistent with the result of vector analysis, which I described in post #302. Which part of it is unclear? I'm just listing the directions of the air's motion before and after the interaction for both frames of reference, in the limiting case of v -> infinity.

 

[erobz]

I'm not sure, that's why I wanted to confirm if it is consistent with the result of vector analysis, which I described in post #302. Which part of it is unclear? I'm just listing the directions of the air's motion before and after the interaction for both frames of reference, in the limiting case of v -> infinity.

It doesn’t sound like it’s consistent with what you are saying. For the analysis $\beta$ is a fixed angle. It’s dictated by the vane geometry inside the control volume that’s is turning the flow.

 

[A.T.]

For the analysis $\beta$ is a fixed angle.

Okay, but you have solved it for different values of $\beta$.

It’s dictated by the vane geometry inside the control volume that’s is turning the flow.

The vane geometry is determing how the flow relative to the vane is turned. When v -> infinite that turning angle of the relative flow should go to zero (corresponding to lift/drag -> infinity), not to 45° as indicated in your plot.

I suspect you are assuming the wrong inflow direction relative to the vane. It's not vertically down, like the true wind, but depends on v. And the vane has to be aligned with the inflow relative to the vane, so the vane orientation also depends on v. The outflow relative to the vane is then offset from the inflow relative to the vane by the vane geometry angle.

I think it would help to avoid confusion if you use different symbols for the flow in different frames, like $w$ for flow relative to the ground, and $u$ for flow relative to the vane.

 

[erobz]

Okay, but you have solved it for different values of $\beta$.

Correct. The vane is fixed relative to the cart. Just like it is in your animations... I'm saying the wind is always pointing down, what is the top speed along the track if $\beta$ is always $a,b,c \cdots$

I think it would help to avoid confusion if you use different symbols for the flow in different frames, like $w$ for flow relative to the ground, and $u$ for flow relative to the vane.

$w$ is the wind in the ground frame.

$w_{out}$ is the speed of the exiting flow in the carts frame. In the ground frame it's velocity is: $$\boldsymbol{w_{out}} =\left( v_x - \sqrt{(v_y+w)^2 + v_x^2 }\sin \beta \right) \boldsymbol{u_x} + \left( v_y - \sqrt{(v_y+w)^2 + v_x^2 } \cos \beta \right) \boldsymbol{u_y} $$

 

[A.T.]

Correct. The vane is fixed relative to the cart. Just like it is in your animations.

In my animation there is no vane, just a flat sail, which allows for elastic collision at different relative inflow directions, and thus different amounts of relative flow turning. A vane is a different model, because it fixes the relative flow turning angle.

I'm saying the wind is always pointing down, what is the top speed along the track if $\beta$ is always $a,b,c \cdots$

The true wind is always pointing down, but not the relative wind.

$w$ is the wind in the ground frame.

OK

$w_{out}$ is the exiting flow in the carts frame.

Why not call it $u_{out}$ to avoid confusion with the ground frame ?

In the ground frame it's: $$\boldsymbol{w_{out}} =\left( v_x - \sqrt{(v_y+w)^2 + v_x^2 }\sin \beta \right) \boldsymbol{u_x} + \left( v_y - \sqrt{(v_y+w)^2 + v_x^2 } \cos \beta \right) \boldsymbol{u_y} $$

This seems to still assume fixed vane? For a aligned vane $\beta$ is the angle between $\boldsymbol{u_{in}}$ and $\boldsymbol{u_{out}}$. Then $\boldsymbol{w_{out}} = \boldsymbol{u_{out}} + \boldsymbol{v}$

 

[erobz]

In my animation there is no vane, just a flat sail, which allows for elastic collision at different relative inflow directions, and thus different amounts of relative flow turning. A vane is a different model, because it fixes the relative flow turning angle.The true wind is always pointing down, but not the relative wind.OK

Why not call it $u_{out}$ to avoid confusion with the ground frame ?This seems to still assume fixed vane? For a aligned vane $\beta$ is the angle between $\boldsymbol{u_{in}}$ and $\boldsymbol{u_{out}}$. Then $\boldsymbol{w_{out}} = \boldsymbol{u_{out}} + \boldsymbol{v}$

The outflow always comes out of the control volume at a fixed angle $\beta$ relative to the control volume.

The flow field surrounding the control volume always looks something like this in the inertial frame.

You've made me "un black box" the control volume to explain the momentum exchange. It is not necessary to do so using Newton Laws.

 

[A.T.]

The flow field surrounding the control volume always looks like this in the inertial frame.

View attachment 333667

The indicated flow directions are fine for the ground frame, not for the vane frame.

The indicated vane orientation is wrong, because it must be aligned with the flow relative to the vane, not relative to the ground.

You've made me "un black box" the control volume to explain the momentum exchange. It is not necessary to do so using Newton Laws.

Then don't unmask it. But then also don't define $\beta$ in terms of vane geometry. Define it in terms of relative flow turning, as the angle between $\boldsymbol{u_{in}}$ and $\boldsymbol{u_{out}}$.

 

[erobz]

The indicated flow directions are fine for the ground frame, not for the vane frame.

The indicated vane orientation is wrong, because it must be aligned with the flow relative to the vane, not relative to the ground.Then don't unmask it. But then also don't define $\beta$ in terms of vane geometry. Define it in terms of relative flow turning, as the angle between $\boldsymbol{u_{in}}$ and $\boldsymbol{u_{out}}$.

There is something in the cv that makes the air come out of the box at a constant angle relative to the box. Are you telling me this is not allowed by the laws of physics?

 

[A.T.]

There is something in the cv that makes the air come out of the box at a constant angle relative to the box.

What do you mean by that exactly? Between which two vectors is the angle constant? Specify the frame of reference for every frame dependent vector you are using to define the angle.

 

[erobz]

What do you mean by that exactly? Between which two vectors is the angle constant? Specify the frame of reference for every frame dependent vector you are using to define the angle.

Between the wind in the ground frame and the outflow in the c.v. frame. Is there no possible entity inside the control which could make it true?

 

[A.T.]

Between the wind in the ground frame and the outflow in the c.v. frame.

And why are you defining the angle $\beta$ based on velocities from two different frames?

Is there no possible entity inside the control which could make it true?

Not a vane with fixed orientation and geometry. But if you allow those two parameters to change for different v (there is no reason why you shouldn't), then you can keep that $\beta$ you defined constant across different v. But it's not clear why you decided to keep it constant.

I also think that with this definition of $\beta$ your result from post #299 makes sense. For v -> inf the angle $\beta$ as you defined will be 45°. It's just a weird parameter, defined across different frames.

 

[Gleb1964]

With the optimal vane orientation (bisecting apparent wind angle in the moving frame), the deflected wind is moving in opposite direction to the boat speed (parallel to the boat speed, but in opposite direction). The air leaving the control volume in direction opposite to the boat speed.
Assuming that the vane angle is adjusted to the optimal at any speed, that means that at any speed the air is deflected in direction opposite to the boat moving direction. And that is valid both in the ground frame and in the boat frame.

 

[erobz]

I still can't figure out what all the confusion is surrounding $\beta$, perhaps I'm miss interpreting what it "looks like", but I think all of that not an issue. If the outflow is exiting at a constant angle $\beta$ (imagine there is a pipe sticking out of the control volume at angle $\beta$ relative to vertical in the diagram, it checks out; No bounds on velocity for some angle in this idealism. Any real body faces significant opposition to that proposition via drag. In theoretically idealistic cases I have to cast out my prior doubt of this.

 

[Gleb1964]

One question, may be I missed before. Do you have control volume stationary at ground frame or it is moving with the carriage? If it is moving, the pipe sticking out moving volume would not blow the flow in direction of the pipe pointing.

 

[A.T.]

I still can't figure out what all the confusion is surrounding $\beta$,

It's not clear why $\beta$ is defined between two velocity vectors taken from two different frames of reference. It's not wrong, just odd.

If the outflow is exiting at a constant angle $\beta$ (imagine there is a pipe sticking out of the control volume at angle $\beta$ relative to vertical in the diagram,

You didn't specify the frame in which you measure the outflow direction. If the thing in your diagram is a pipe, then the outflow in the pipe-frame will indeed be $\beta$ relative to vertical. But in the ground-frame the outflow in will not will be $\beta$ relative to vertical (unless v = 0 or $\beta$ = 45°).

More importantly, the inflow side of you pipe is not aligned with the inflow in the pipe-frame (as it should be for efficient operation), instead it is aligned with the inflow in the ground-frame (which seems like an arbitrary choice that makes no sense).

 

[erobz]

One question, may be I missed before. Do you have control volume stationary at ground frame or it is moving with the carriage? If it is moving, the pipe sticking out moving volume would not blow the flow in direction of the pipe pointing.

I don't have it blowing the flow in the direction of the pipe in the inertial frame. This should be apparent from the mathematics surrounding the momentum transfers. The Reynolds Transport Theorem is valid in the inertial frame only. So every momentum flow is converted to the inertial frame in the equations.

For instance, I don't say the momentum outflow in the $\boldsymbol{u_x}$ is simply:
$$ - \dot m \sqrt{ ( w+y)^2 + v_x^2 }\sin \beta $$

I say it is in the inertial frame:

$$ \dot m \left( v_x - \sqrt{ ( w+y)^2 + v_x^2 }\sin \beta \right) $$

I'm not drawing the "true outflow flow momentum vector" in the inertial frame...It's just a diagram.

 

[A.T.]

With the optimal vane orientation (bisecting apparent wind angle in the moving frame), the deflected wind is moving in opposite direction to the boat speed (parallel to the boat speed, but in opposite direction). The air leaving the control volume in direction opposite to the boat speed.
Assuming that the vane angle is adjusted to the optimal at any speed, that means that at any speed the air is deflected in direction opposite to the boat moving direction. And that is valid both in the ground frame and in the boat frame.

Are you referring to the tacking upwind case here? Because compared to that, in the case of tacking downwind with an VMG > windspeed, the roles of water and air are swapped. So what you wrote about air deflection by the sail here, applies to the water deflection by the keel there.

 

[erobz]

More importantly, the inflow side of you pipe is not aligned with the inflow in the pipe-frame (as it should be for efficient operation), instead it is aligned with the inflow in the ground-frame (which seems like an arbitrary choice that makes no sense).

This is irrelevant, infinite velocity is still obtained even with some theoretical inefficiency you claim exists. Its like you are trying to talk me out of it now! I was of the opinion that you were blowing smoke before, and now that I derive the EOM's it see that its not quackery. Take your win, otherwise write your own equations and show me, because I'm tired of playing these games.

 

[A.T.]

This is irrelevant, infinite velocity is still obtained even with some theoretical inefficiency you claim exists.

In this upwind case you considered here, the air will enter the inlet of the pipe for any v, even if you keep it vertical for all v. So you can get away with this.

But for the downwind case (which was the actual starting point of the debate) the range of relative inflow direction if far greater (for v = 0 to inf). So such a fixed pipe model is problematic, because it might have the inlet pointing the wrong way (more than 90° off the relative inflow). This could create a false-limit, while an actual sail can always be oriented in the optimal direction.

 

[erobz]

In this upwind case you considered here, the air will enter the inlet of the pipe for any v, even if you keep it vertical for all v. So you can get away with this.

But for the downwind case (which was the actual starting point of the debate) the range of relative inflow direction if far greater (for v = 0 to inf). So such a fixed pipe model is problematic, because it might have the inlet pointing the wrong way (more than 90° off the relative inflow). This could create a false-limit, while an actual sail can always be oriented in the optimal direction.

We aren't discussing a sail at this point, we are discussing a "black box" control volume, and even without examining the actual mechanics of the sail, it still works. There is nothing to argue about anymore IMO, unless you think the mathematics is truly abusive.

In reality all of this infinite velocity talk is a ridiculous notion anyhow, even with an always perfectly oriented sail! The question was "can one theoretically sail faster than the wind". I would say "in a world without friction or drag it's not theoretically prohibited nor limited". Calculating the real limit in the real world is not something I'm trying to tackle, that's for certain.

 

[A.T.]

In reality all of this infinite velocity talk is a ridiculous notion anyhow, even with an always perfectly oriented sail! The question was "can one theoretically sail faster than the wind". I would say "in a world without friction or drag it's not theoretically prohibited nor limited".

You are conflating different things here:
- moving at infinite multiples of true wind-speed requires drag to be zero
- moving at more than 1 multiple of true wind-speed, including downwind or upwind component > true wind, works fine with real-world drag

The relationship of velocity limits to drag is described here:
https://www.physicsforums.com/threa...king-physics-3rd-edition.1048870/post-6861400

 

[erobz]

You are conflating different things here:
- moving at infinite multiples of true wind-speed requires drag to be zero
- moving at more than 1 multiple of true wind-speed, including downwind or upwind component > true wind, works fine with real-world drag

The relationship of velocity limits to drag is described here:
https://www.physicsforums.com/threa...king-physics-3rd-edition.1048870/post-6861400

I'm not conflating anything. I made no such claim in that statement about the possibility of moving at more than 1 multiple of wind speed in the real world. I said I'm not trying to calculate the limit in the real world. i.e. the model (without drag - nor general relativity) is not bounded, in reality (with drag - and relativity) its certainly is bounded nowhere "near" infinity . Thats the only claim I confidently make without updating the model.

 

[A.T.]

I said I'm not trying to calculate the limit in the real world. i.e. the model (without drag - nor general relativity) is not bounded, in reality (with drag - and relativity) its certainly is bounded nowhere "near" infinity .

Fair enough. But note that relativity puts a limit on the speed terms of distance / time, not on the ratio of boat-speed to wind-speed. That ratio can go to infinity for a fixed boat-speed, as the wind-speed needed to achieve it goes to zero, because efficiency goes to one.

 

[Gleb1964]

Are you referring to the tacking upwind case here? Because compared to that, in the case of tacking downwind with an VMG > windspeed, the roles of water and air are swapped. So what you wrote about air deflection by the sail here, applies to the water deflection by the keel there.

First I was looking the upwind case only. But I have checked the downwind case as well and my conclusion is still keep holding.
In the case of the optimal vane orientation (meant bisecting the direction of motion and the apparent wind in the frame of carriage) the out flow from control volume pointed ether in direction of motion (downwind) or opposite to motion (upwind). That is valid in both the ground frame and the carriage frame.
For me that is interesting finding, I haven't been aware about it before.

 

[A.T.]

First I was looking the upwind case only. But I have checked the downwind case as well and my conclusion is still keep holding.
In the case of the optimal vane orientation (meant bisecting the direction of motion and the apparent wind in the frame of carriage) the out flow from control volume pointed ether in direction of motion (downwind) or opposite to motion (upwind). That is valid in both the ground frame and the carriage frame.
For me that is interesting finding, I haven't been aware about it before.

I think I now understand what you mean by the optimal vane orientation. If we just take a flat sail, that bisects the angle between negated boat velocity and apparent wind, and assume a perfectly elastic collision with it, then after the collision the flow will be along the negated boat velocity. Yes that's always true.