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

[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.

 

[A.T.]

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.

You requested mathematical rigor, but when the math contradicts your intuition, you are making excuses why you won't accept it. I'm just asking you to be consistent for both axes x & y.

In still air you cannot push something with a sail and expect it to accelerate away from you.

Correct, but we are not talking about no wind. We are talking about your own scenario with wind $w$ along the positive x-direction. Post #57 explains why it doesn't work without wind relative to the ground.

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.

That is exactly how it works. That's why there is no fixed limit on the windspeed multiple a sailcraft can achieve, and this applies to the downwind component as well.

But there are practical limits. As you accelerate, the relative flow comes more and more from the front, so you need increasing high lift/drag ratios to still produce a forward force component (see post #57). The achievable lift/drag ratios for the interactions with the air and the surface are limiting the speed of a sailcraft.

 

[A.T.]

...It is effectively a fluid jet impinging on a sail in the vacuum of space...

Fortunately the real wind is a large moving airmass that extends in all directions, so it doesn't have the limitations of such a thin jet model. Just use a very wide jet instead.

 

[erobz]

That is exactly how it works. That's why there is no fixed limit on the windspeed multiple a sailcraft can achieve, and this applies to the downwind component as well.

Thats exactly how it works my foot. You cannot give a sail craft an initial shove in still air and have it accelerate away. That is absolutely ludicrous!

 

[A.T.]

,,,in still air ,,,

Why do you misrepresent what I am saying. I stated many times, that it doesn't work without wind relative to the ground and explained why:

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.

 

[erobz]

Why do you misrepresent what I am saying.

That is exactly how it works.

in reply to;

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.

 

[A.T.]

Yes, that is exactly how it works, if there is wind relative to the ground.

 

[erobz]

You say this is so trivial early on, but then give no foundational proof. Your idea of mathematical rigor is " just look at the (arbitrarily drawn) vectors on the diagram"!

That doesn't cut it. Color me unconvinced. Good day to you.

 

[A.T.]

You say this is so trivial early on, but then give no foundational proof. Your idea of mathematical rigor is " just look at the arbitrarily drawn vectors on the diagram"!

What about the vectors is wrong? The only assumption there is a lift/drag ratio of about 5, which is very conservative. Airfoils reach 20 and more.

 

[erobz]

If I gave the boat a shove in still air in the direction of the "true wind" in your diagram, that would produce relative flow over the air foil in the direction of the "apparent wind", that in turn would provide a force that accelerated the boat in the direction of the lift force (increasing its velocity), which would increase the sail force by way of increasing the apparent wind! $\circlearrowright$. So in that case (absence of a drag force on the air foil - a function of the apparent flow velocity) the craft would accelerate away from me indefinitely. Now...add in the drag force. A drag force ( since you are talking about drag/lift ratio ) would grow (from 0) such that its component in the direction of the boats velocity matched the component of the lift force, thus accelerating away from me for a while, approaching moving away from me at a constant velocity.

 

[A.T.]

If I gave the boat a shove in still air in the direction of the "true wind" in your diagram,

In "still air" the "true wind" vector is zero. It doesn't have a direction. Can you draw vector diagram of what you mean?

that would produce relative flow over the air foil in the direction of the "apparent wind",

If true_wind = 0 then apparent_wind = -boat_velocity. Since sail_force cannot be more than 90° from the apparent_wind, it cannot have a positive component in the direction of boat_velocity.

So if true_wind = 0 there cannot be a propulsive forward_force from the sail, no matter how much you push initially. There is nothing in the diagram that suggests it would accelerate without true_wind relative to the ground.

 

[erobz]

Your diagram is just arbitrarily drawn vectors that have scant applicability to a proper momentum transfer analysis. The wind applies a force to the sail. In Newtonian Mechanics we talk about forces, the wind force on the sail is not fully described by the wind vector in your diagram. The force on the sail depends on how it deflects the incident momentum from the wind. That is what determines which way and how much the boat accelerates.

The "apparent wind" in still air in your diagram would be the component of the boats velocity in the direction of the sail. That was your whole argument about scooping momentum from the $y$ direction by turning my vane clockwise 90 degrees. If there is apparent wind from me giving it initial velocity the airfoil produces lift. If there is lift $\circlearrowright$. Perpetual motion ensues.

 

[A.T.]

The wind applies a force to the sail.

Yes, as shown in both bottom diagrams as sail_force.

... the wind force on the sail is not fully described by the wind vector in your diagram.

The sail_force direction is fully determined by the apparent_wind direction and the lift/drag ratio (assumed to be ~5 here).

The "apparent wind" in still air in your diagram would be the component of the boats velocity in the direction of the sail.

Wrong. The vector math is:

apparent_wind = true_wind - boat_velocity

For true_wind = 0 you have apparent_wind = -boat_velocity. And since sail_force cannot be more than 90° from the apparent_wind, it cannot have a positive component in the direction of boat_velocity. So no acceleration is possible without true_wind.

 

[erobz]

Sorry, that's not how the dot product works.

If there were no wind and I push the boat in the direction shown in your diagrams, the "apparent wind" (from the perspective of the sail) would have magnitude $|v| |s| \cos \theta$ and direction opposite the direction of motion.

This is exactly the fact that allows planes to fly in still air.

 

[A.T.]

Sorry, that's not how the dot product works.

There is no dot product here. It's just vector subtraction:

apparent_wind = true_wind - boat_velocity

If there were no wind and I push the boat in the direction shown in your diagrams, the "apparent wind" (from the perspective of the sail) would have magnitude $|v| |s| \cos \theta$ and direction opposite the direction of motion.

Your magnitude computation is confused, but the key is that you got the direction right:

If true_wind = 0 then apparent_wind is exactly opposite to the direction of motion.

And since sail_force cannot be more than 90° from the apparent_wind, it cannot have a positive component in the direction in the direction of motion. So no acceleration is possible without true_wind.

 

[erobz]

There is no dot product here. It's just vector subtraction.

The apparent wind in still air is the projection of $\vec v$ onto $\vec s$. That is not debatable. ( I should clarify $\vec s$ is a unit vector)

 

[A.T.]

The apparent wind in still air is the projection of $\vec v$ onto $\vec s$. That is not debatable.

Maybe you misunderstand the sailing terms, so let me translate into in physics terms:

boat_velocity = velocity of boat relative to the surface
true_wind = velocity of airmass relative to the surface
apparent_wind = velocity of airmass relative to the boat

Thus per Galilean transformation the vector relationship is:

apparent_wind = true_wind - boat_velocity

There is no projection involved.

 

[erobz]

Maybe you misunderstand the sailing terms, so let me translate into in physics terms:

boat_velocity = velocity of boat relative to the surface
true_wind = velocity of airmass relative to the surface
apparent_wind = velocity of airmass relative to the boat

Thus per Galilean transformation the vector relationship is:

apparent_wind = true_wind - boat_velocity

There is no projection involved.

$\vec v$ has components parallel and orthogonal, to $\vec s$. It is the component of $\vec v$, parallel to the chord $\vec s$ that is responsible for lift, hence the dot product.

 

[A.T.]

$\vec v$ has components parallel and orthogonal, to $\vec s$. It is the component of $\vec v$, parallel to the chord $\vec s$ that is responsible for lift, hence the dot product.

You are confusing lift with normal force. Lift and drag are defined based on the relative flow direction (apparent wind), not based on the chord:

Difference between lift (L) and drag (D) versus normal force (N) and axial force (A)

From: https://aerospaceweb.org/question/aerodynamics/q0194.shtmlThe direction of the apparent_wind is not affected by how you orient the sail, just by true_wind and boat_velocity per Galilean transformation:

apparent_wind = true_wind - boat_velocity

 

[erobz]

You are confusing lift with normal force. Lift and drag are defined based on the relative flow direction (apparent wind), not based on the chord:

View attachment 320519
Difference between lift (L) and drag (D) versus normal force (N) and axial force (A)

From: https://aerospaceweb.org/question/aerodynamics/q0194.shtmlThe direction of the apparent_wind is not affected by how you orient the sail, just by true_wind and boat_velocity per Galilean transformation:

apparent_wind = true_wind - boat_velocity

the apparent wind is coming in the direction of $\vec A$. In the case of no wind it is equal in magnitude opposite in direction to the component of $\vec v$ in the direction of $\vec A$ ( $\vec s$ in my diagram ).

 

[A.T.]

the apparent wind is coming in the direction of $\vec A$. In the case of no wind it is equal in magnitude opposite in direction to the component of $\vec v$ in the direction of $\vec A$ ( $\vec s$ in my diagram ).

No, your $\vec s$ (sail orientation) is completely irrelevant to the apparent wind direction. Only the relative velocities matter in the Galilean transformation:

boat_velocity = velocity of boat relative to the surface
true_wind = velocity of airmass relative to the surface
apparent_wind = velocity of airmass relative to the boat

Galilean transformation:

apparent_wind = true_wind - boat_velocity

Note that these are free stream velocities, not the air motion deflected by the sail. But that is also how lift is defined: as perpendicular to the free stream relative flow.

 

[erobz]

No, your $\vec s$ (sail orientation) is completely irrelevant to the apparent wind direction. Only the relative velocities matter in the Galilean transformation:

boat_velocity = velocity of boat relative to the surface
true_wind = velocity of airmass relative to the surface
apparent_wind = velocity of airmass relative to the boat

Galilean transformation:

apparent_wind = true_wind - boat_velocity

Note that these are free stream velocities, not the air motion deflected by the sail. But that is also how lift is defined: as perpendicular to the free stream relative flow.

Oh, so you are saying that the vectors on your diagram don't have any relevance to the forces acting on the sail?

 

[A.T.]

Oh, so you are saying that the vectors on your diagram don't have any relevance to the forces acting on the sail?

On the contrary. The apparent_wind vector in my diagram is very important for the sail_force, as already explained:

The sail_force direction is fully determined by the apparent_wind direction and the lift/drag ratio (assumed to be ~5 here).

But it seems you fail to understand that apparent_wind is not defined based on the sail orientation, but on the motion of the freestream airmass in the rest-frame of the boat. It's a function of boat_velocity and true_wind only.

The sail in the diagram is just for informal reasons, shown roughly how you would orient it to maximize lift/drag ratio. But the motion freestream airmass is not affected by the sail setting.

 

[erobz]

On the contrary. The apparent_wind vector in my diagram is very important for the sail_force, as already explained:But it seems you fail to understand that apparent_wind is not defined based on the sail orientation, but on the motion of the freestream airmass in the rest-frame of the boat. It's a function of boat_velocity and true_wind only.

The sail in the diagram is just for informal reasons, shown roughly how you would orient it to maximize lift/drag ratio. But the motion freestream airmass is not affected by the sail setting.

In the absence of wind, the apparent wind is the boat velocity (opposite it). The component of the free stream velocity parallel to the airfoil is flowing over the air foil. It is generating lift.

 

[A.T.]

In the absence of wind, the apparent wind is the boat velocity (opposite it).

Correct

The component of the free stream velocity parallel to the airfoil is flowing over the air foil, that is generating lift.

What "generates" lift is topic of lengthy discussions you can read on this and other forums, but irrelevant here.

How lift is defined is key: Lift is the force component perpendicular to the free-stream relative flow (apparent_wind). And that's what the force vectors are based on in my diagram, not on the sail orientation.

 

[erobz]

Correct

What "generates" lift is topic of lengthy discussions you can read on this and other forums, but irrelevant here.

How lift is defined is key: Lift is the force component perpendicular to the free-stream relative flow (apparent_wind). And that's what the force vectors are based on in my diagram, not on the sail orientation.

Well, that's a deceptive diagram in my opinion.

 

[A.T.]

Well, that's a deceptive diagram in my opinion.

I agree that one could make the sail orientation and apparent_wind more distinct. But then some will complain, that with such a large angle of attack you could never achieve the shown lift/drag ratio. The other alternative is not to show the boat and sail at all, just the vectors.

So now that this is clarified, do you understand how it works?

 

[erobz]

So now that this is clarified, do you understand how it works?

No. I can't get from where I started to there.

I've shown (I believe correctly) in the vacuum of space a cart on a frictionless track can't outrun the jet of fluid pushing it with momentum change. Obviously, the models are absent the force of lift. Turning the vane around clockwise 90 deg. ( as you suggest) is detrimental to your argument. The momentum change goes from applying a force $\nearrow$ to applying a $\searrow$ to the cart.

If lift is somehow the answer, it's a long way from where I'm at.

Oh, and to one of your earlier posts about modeling the thickness of the jet ( you said something like the thin jet was the problem and I needed a thick one) The "wind" would be analogous to an array of jets, which the cart was intersecting as it passed by, taken in the limit as the distance between each jet goes to zero, we get a reasonable approximation for wind in my opinion.

 

[A.T.]

No. I can't get from where I started to there.

Ok, but do you understand that if lift/drag ratio of 5 is empirically possible, then based on the diagram, you still can have a propulsive force, when the downwind velocity component is 1.5 x true wind-speed?

I 've shown ( I believe correctly) in the vacuum of space a cart on a frictionless track can't outrun the jet of fluid pushing it with momentum change.

If you insist on using a thin jet, then you have to make your sail very large instead. This would look something like this animation

Turning the vane around clockwise 90 deg. ( as you suggest) is detrimental to your argument. The momentum change goes from applying a force $\nearrow$ to applying a $\searrow$ to the cart.

But a $\searrow$ force will propel your cart forward, if $\theta < 45°$. And this will also happen when $v_x = w$, where the relative flow into the vane is along negative y direction. Of course you need a wide jet here, like the actual wind. To go even faster you have to turn the vane further clockwise, and make it less curved. As already explained, deflecting the air by 90° is very inefficient.

 

[erobz]

Ok, but do you understand that if lift/drag ratio of 5 is empirically possible, then based on the diagram, you still can have a propulsive force, when the downwind velocity component is 1.5 x true wind-speed?If you insist on using a thin jet, then you have to make your sail very large instead. This would look something like this animation

But a $\searrow$ force will propel your cart forward, if $\theta < 45°$. And this will also happen when $v_x = w$, where the relative flow into the vane is along negative y direction. Of course you need a wide jet here, like the actual wind. To go even faster you have to turn the vane further clockwise, and make it less curved. As already explained, deflecting the air by 90° is very inefficient.

View attachment 320527

I'll solve for 45 degree turn tomorrow. Im burned out.

 

[A.T.]

I'll solve for 45 degree turn tomorrow. Im burned out.

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.

 

[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.