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

[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.

 

[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...