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

[erobz]

The proof that it works is neither the animation nor math, but racing boats which do it for years (ice boats for a century). The animation just shows how trivial it is.

Thats not a good argument in Physics which seeks to understand the physical phenomenon by way of mathematical model. If everyone said, "well, there it is...I don't need to model it". What would we know?

 

[A.T.]

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

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.

Mathematical rigor is useless, if in order to formulate / solve your math, you have to make assumptions which completely change the scenario. I recommend using vectors in this case (see post #21).

 

[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. You original model was more relevant, just your sail model was bad.

Mathematical rigor is useless, if in order to formulate / solve your math, you have to make assumptions which completely change the scenario. I recommend using vectors in this case (see post #21).

I was trying to simplify in order to discuss maximum possible impulse.

There is no "real" sail that can completely reverse the momentum of the incoming flow, as shown in #29. It is the theoretical maximum possible impulse that can be applied to the boat in the direction of motion. There is no possible alteration of a vane which could impart any greater momentum to the cart in the $x$ direction than what is shown in #29. For the $x$ component of velocity $w$ is a hard limit. I've shown it with Newtons laws, Eulers, and Reynolds laws twice now for two independent vane configurations.

Real wind gusts, swirls, changes direction., any "true" measurement of downwind speed greater than the wind is going to be obscured by this reality. Conservation of Momentum is not.

 

[A.T.]

I've shown it with Newtons laws, Eulers, and Reynolds laws twice now for two independent sail configurations.

But neither one of your configurations corresponds to a boat with downwind component greater than windspeed.

Go back to your first one, but rotate the sail by 90° clockwise. What happens when Vx = W? Can you still accelerate?

 

[erobz]

But neither one of your configurations corresponds to a boat with downwind component greater than windspeed.

Go back to your first one, but rotate the sail by 90° clockwise. What happens when Vx = W? Can you still accelerate?

No it can not accelerate. The mass flowrate $\dot m = \rho A (w-v_x)$ entering the control volume always goes to $0$ as $v_x \to w$. The fluid jet can only just keep up with the cart, it can no longer impart any further momentum to it. If your hope was that the momentum outflow changes to $\dot m (v_y +w)$, sorry $\dot m \to 0$.

 

[A.T.]

No it can not accelerate. The mass flowrate entering the control volume always goes to $0$ as $v_x \to w$.

You think there is no relative wind at the boat when Vx = W? Think again, what about Vy?

 

[erobz]

You think there is no relative wind at the boat when Vx = W? Think again, what about Vy?

What $v_y$, the wind is coming in horizontal (purely in the $x$ direction). Are you trying to say that $w$ now has a vertical component $w_y$ as well?

 

[A.T.]

What $v_y$, the wind is coming in horizontal (purely in the $x$ direction). Are you trying to say that $w$ now has a vertical component $w_y$ as well?

You have a moving control volume. The mass flowrate entering the control volume depends on the motion of the airmass relative to the control volume, not relative to the ground.

 

[erobz]

You have a moving control volume. The mass flowrate entering the control volume depends on the motion of the airmass relative to the control volume, not relative to the ground.

The amount of mass entering and exiting the control volume per unit time is frame independent.

 

[erobz]

If the cart has velocity $v_x = w$, none of the flow mass can enter the control volume. it is chasing it at the same speed! You are telling me that I can run at the same speed as a car in front of me and end up inside it. This is just absurd.

 

[A.T.]

The amount of mass entering and exiting the control volume per unit time is frame independent.

Yes, but you compute it based on the velocity of the air relative to the volume, not relative to some arbitrary object like the ground, as you did.

 

[A.T.]

If the cart has velocity $v_x = w$, none of the flow mass can enter the control volume.

See post #36.

 

[erobz]

Yes, but you compute it based on the velocity of the air relative to the volume, not relative to some arbitrary object like the ground, as you did.

What are you talking about? They are the same, what part of frame independent are you not understanding?

The mass flow rate entering the control volume is $\dot m = \rho A ( w - v_x )$. Period.

 

[A.T.]

The mass flow rate entering the control volume is $\dot m = \rho A ( w - v )$.

It's a 2D situation (your first model in post #20), so W and V are 2D vectors.
Is the (W - V) vector zero when Vx = Wx?

 

[erobz]

If I am running with speed $w$ behind a car at speed $w$ holding a bag of rocks, I can't simply smash out the rear window by simply letting a rock go...I must throw the rock with a speed $v$ relative to me. The only way the rock can get into the car is if it has speed $w+v$ relative to the ground.

 

[erobz]

It's a 2D situation (your first model in post #20), so W and V are 2D vectors.
Is the (W - V) vector zero when Vx = Wx?

$w$ is a vector in the $x$ direction. The wind is blowing in the $x$ direction...only in the $x$ direction. Just like it is in #29. You pick a direction of the wind then you try to move at some angle $\theta$ relative to it and examine the "Momentum Equation" in each direction (component basis)

 

[A.T.]

$w$ is a vector in the $x$ direction. The wind is blowing in the $x$ direction...only in the $x$ direction. Just like it is in #29.

You didn't answer my question:
Is the vector (W - V) zero when Vx = Wx?

 

[erobz]

Is the vector (W - V) zero when Vx = Wx?

Firstly $w_x$ makes it seem like $w$ has multiple directions! It doesn't its just plain $w$ in post#20

But Yes. By definition the $w - v_x = 0$, when $w = v_x$.

 

[A.T.]

But Yes. by definition the $w - v_x = 0$, when $w = v_x$.

That wasn't my question. I'm asking about the 2D vector (W - V) with W = (w, 0) and V = (v_x, v_y).

 

[erobz]

Well, its not clear what you are asking. Are you asking if $v_y$ is 0 when $v_x = w$?

The answer to that is no. $v_y = v_x \tan \theta = w \tan \theta$

 

[A.T.]

The mass flow rate entering the control volume is $\dot m = \rho A ( w - v_x )$

You have edited your post and made it wrong. Why would the mass flow rate entering the control volume depend on the motion of the control volume along X only, if the control volume moves along Y as well.

 

[erobz]

You have edited your post and made it wrong. Why would the mass flow rate entering the control volume depend on the motion of the control volume along X only, if the control volume moves along Y as well.

because that is how its getting in! It can only enter from the direction of the wind.

 

[A.T.]

because that is how its getting in! It can only enter from the direction of the wind.

Think again. If there was no wind at all, but the control volume was moving along X and Y, would there be air entering the control volume? From which direction?

 

[erobz]

Think again. If there was no wind at all, but the control volume was moving along X and Y, would there be air entering the control volume? From which direction?

Thats counterproductive to your argument. Scooping mass from in front of $y$direction would impart momentum on the cart opposite its direction of motion! You are trying to use the circular reasoning that the wind can drive the cart into the wind and harness more forward momentum.

 

[A.T.]

You are trying to use the circular reasoning that the wind can drive the cart into the wind and harness more forward momentum.

Nope. The no wind case is just there to demonstrate that your approach for mass flow rate into a moving control volume makes no sense.

So again: If there was no wind at all, but the control volume was moving along X and Y, would there be air entering the control volume? From which direction?

 

[erobz]

Nope. The no wind case is just there to demonstrate that your approach for mass flow rate into a moving control volume makes no sense.

So again: If there was no wind at all, but the control volume was moving along X and Y, would there be air entering the control volume? From which direction?

This is where we are at... If there is no wind, and you give a cart with a sail an initial push it will accelerate away from you!!! Forget about it. You win.

 

[A.T.]

If there is no wind, and you give a cart with a sail an initial push it will accelerate away from you!!!

Nope. Look at the vectors. If true_wind = 0 then apparent_wind = -boat_velocity. So even a infinite lift/drag ratio won't produce a forward_force. The sail_force can be at most 90° from the apparent_wind.

 

[A.T.]

If I am running with speed $w$ behind a car at speed $w$ holding a bag of rocks, I can't simply smash out the rear window by simply letting a rock go..

If the car additionally has a velocity component perpendicular to w (like the boat in your model below) then you can smash a side window by just letting the rocks go.

Or alternatively the car passengers can use a surface to deflect your rocks, and get some propulsion from that (for this the sail in your model needs to be turned by 90° clockwise)

 

[A.T.]

The amount of mass entering and exiting the control volume per unit time is frame independent.

It is frame independent exactly because, no matter which frame you use for analysis, the mass flow rate entering the control volume is computed based on the velocity of the air relative to the control volume. Not relative to your picked frame, like the ground.

You must know this, because you do it for the x-component:

The mass flow rate entering the control volume is $\dot m = \rho A ( w - v_x )$.

The $( w - v_x )$ term in your formula is the x-component: of the flow relative to the control volume. Now you just have to do the same for y-component of the relative flow, which is $( 0 - v_y )$.

So when $v_x = w$ the relative flow is $\begin{pmatrix} 0 \\ -v_y \end{pmatrix}$, therefore you have to rotate the vane in your sketch by 90° clockwise to create propulsion in this situation.

 

[erobz]

It is frame independent exactly because, no matter which frame you use for analysis, the mass flow rate entering the control volume is computed based on the velocity of the air relative to the control volume. Not relative to your picked frame, like the ground.

You must know this, because you do it for the x-component:

The $( w - v_x )$ term in your formula is the x-component: of the flow relative to the control volume. Now you just have to do the same for y-component of the relative flow, which is $( 0 - v_y )$.

So when $v_x = w$ the relative flow is $\begin{pmatrix} 0 \\ -v_y \end{pmatrix}$, therefore you have to rotate the vane in your sketch by 90° clockwise to create propulsion in this situation.

You cannot propel something initially and expect a chain reaction to occur. Momentum entering the vane as $-v_y$ would counter the momentum it was gaining from changing the direction of $w$. The very momentum that is driving it.

In still air you cannot push something with a sail and expect it to accelerate away from you. If I turned the vane as you wish, having it scoop air, and gave it a shove, it would scoop mass from the $y$ direction and sent it to $-x$ direction, thus providing an impulse, which would scoop more mass, and give another impulse, ad infinitum. It just doesn't work like that.

Another thing...my analysis has no mass to "scoop" in the $y$ direction. It is effectively a fluid jet impinging on a sail in the vacuum of space. The analysis provides the highest possible rate of momentum transfer imaginable for a fluid transferring momentum to a sail.