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

[Gleb1964]

The orientation of the sail is not correct at #149.

 

[erobz]

The orientation of the sail is not correct at #149.

The sail has no orientation in that diagram. Its orientation relative to the cart is characterized by the parameter $\beta$ which I hope to first optimize as a function of $v_x$. I.e the expectation is that there is a little man on the boat that is turning the sail to maximize the component of the applied force $F$ in the direction of motion over the entire duration of the analysis.

 

[Gleb1964]

The optimal sail would be bisecting between apparent wind direction and boat direction, the force on the sail is perpendicular to the sail plane. That mean the optimal force direction is known for every boat speed.

 

[erobz]

The optimal sail would be bisecting between apparent wind direction and boat direction, the force on the sail is perpendicular to the sail plane. That mean the optimal force direction is known for every boat speed.

If the sail were fixed at the angle $\beta$(making it a constant), that you say it is ( whatever it is), with the force $F$ perpendicular to the sail that implies that $\frac{F_y}{F_x}$ or $\frac{F_x}{F_y}$ is a constant w.r.t. to velocity.

The problem with that is:

$$ F_x = \dot m \left( w \sin ( \beta + \theta ) + v_x - w \right) $$

$$F_y = \dot m \left( v_y - w \cos ( \beta + \theta ) \right) $$

If you sub $v_y = v_x \tan \theta$ it should be apparent by inspection that isn't the case as:

$$ \frac{F_x}{F_y} = \frac{w \sin ( \beta + \theta ) + v_x - w}{ v_x \tan \theta - w \cos ( \beta + \theta ) } = f(v_x)$$

There is no one that wants it to be a constant more than I do right now, as the solution to that problem is straight forward...but its not. @A.T. already made a point of this in an earlier post.

 

[Gleb1964]

Here is illustration of sail boat diagram.
The boat is descending downwind faster than wind, but you can see, that the sail line is moving slower than wind. That makes possible for wind to make a force on the sail, which has a component (not illustrated) keeping or accelerating the boat speed.

Do you have questions to this diagram?

 

[erobz]

Here is illustration of sail boat diagram.
The boat is descending downwind faster than wind, but you can see, that the sail line is moving slower than wind. That makes possible for wind to make a force on the sail, which has a component (not illustrated) keeping or accelerating the boat speed.

Do you have questions to this diagram?

View attachment 322863

Its not analysis. Quit patronizing me with "vectors" that supposedly reduce everything I just went over to nothing. I am applying Newtons Laws of motion, as instructed to do so by the Fluid Mechanics textbook sitting in front me. I don't accept..."but look at randomly drawn vectors", so stop wasting your time. If you are a physicist, you should be much more inclined to hear what Newton has to say about, or figure out how what I'm saying is incorrect in an agreed upon framework of Classical Physics i.e. Newtonian Mechanics and tell me about it.

 

[Gleb1964]

Your flow analysis suppose to come to the same. What way do you defining the outgoing flow?

 

[erobz]

Your flow analysis suppose to come to the same. What way do you defining the outgoing flow?

Look at the diagram. It shows everything I have planned.

This equation represents Newtons Second in Fluid Mechanics under the assumption of constant cross-sectional properties:

$$ \sum F = \frac{d}{dt} \int_{cv} \rho v ~d V\llap{-} + \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i $$

 

[A.T.]

Quit patronizing me with "vectors" that supposedly reduce everything I just went over to nothing.

Vectors indeed simplify the analysis substantially.

I am applying Newtons Laws of motion, as instructed to do so by the Fluid Mechanics textbook sitting in front me.

Just because you have a hammer, doesn't mean every problem is a nail. We don't have to go into the complex details of fluid dynamics, because we have empirical data on achievable lift/drag ratios, which leads to the much simpler vector analysis.

I don't accept..."but look at randomly drawn vectors", so stop wasting your time.

What have vectors done to you, that you hate them so much?

If you are a physicist, you should be much more inclined to hear what Newton has to say about,

Here is a paper by a physics professor in a physics journal, analyzing the same thing and using vectors of course:

High-speed sailing, Wolfgang Püschl 2018 Eur. J. Phys. 39 044002
https://iopscience.iop.org/article/10.1088/1361-6404/aab982

 

[A.T.]

The sail has no orientation in that diagram. Its orientation relative to the cart is characterized by the parameter $\beta$ which I hope to first optimize as a function of $v_x$. I.e the expectation is that there is a little man on the boat that is turning the sail to maximize the component of the applied force $F$ in the direction of motion over the entire duration of the analysis.

OK, but note that:
- the example $\beta$ you have drawn is wrong for maximizing $v$. See post #142.
- you have drawn the relative outgoing flow again, because it is parallel to the sail. This is not how the outgoing flow looks like in the ground frame where the sail is moving.

 

[A.T.]

What way do you defining the outgoing flow?

Look at the diagram. It shows everything I have planned.

This equation represents Newtons Second in Fluid Mechanics under the assumption of constant cross-sectional properties:

$$ \sum F = \frac{d}{dt} \int_{cv} \rho v ~d V\llap{-} + \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i $$

But your diagram doesn't have $\dot m_o v_o$ in it. The outgoing flow is labeled $\dot m w$, which is of course wrong, because the flowspeed $w$ (in the ground frame) will change after the interaction with the sail.

So how do you plan to determine that $v_o$ for all possible speeds of the boat?

 

[erobz]

But your diagram doesn't have $\dot m_o v_o$ in it. The outgoing flow is labeled $\dot m w$, which is of course wrong, because the flowspeed $w$ (in the ground frame) will change after the interaction with the sail.

The problem is that you don't understand that I'm adding vectors ( well, technically scalars that are associated with the component vectors)! I wrote out the forces acting on the sail in post #154. For example: The momentum of the incoming flow for $F_x$ has three terms; the first two combined describe the momentum of the outflow (notice the dependency of the outflow momentum on component of cart velocity w.r.t the ground frame). The last term describes the momentum inflow relative to inertial frame.

So how do you plan to determine that $v_o$ for all possible speeds of the boat?

The subscripts $i,o$ refer to inflow, outflow respectively. The problem here, given this statement and the fact that I have done this analysis 5 times over with this notation in this thread means that you haven't actually tried to digest any of it until just now.

 

[erobz]

Vectors indeed simplify the analysis substantially.Just because you have a hammer, doesn't mean every problem is a nail. We don't have to go into the complex details of fluid dynamics, because we have empirical data on achievable lift/drag ratios, which leads to the much simpler vector analysis.What have vectors done to you, that you hate them so much?Here is a paper by a physics professor in a physics journal, analyzing the same thing and using vectors of course:

High-speed sailing, Wolfgang Püschl 2018 Eur. J. Phys. 39 044002
https://iopscience.iop.org/article/10.1088/1361-6404/aab982

Vectors are not physics! They are a small part of the process. They do not in and of themselves encapsulate it, and or complete an analysis.

If we are talking fluid mechanics, we are talking Newtons Second, whether it be through the simplified model im proposing, or the more detailed Navier-Stokes Equations.

 

[erobz]

View attachment 322880

OK, but note that:
- the example $\beta$ you have drawn is wrong for maximizing $v$. See post #142.

I said $\beta$ is a parameter I was proposing to optimize as a function of $v_x$, to maximize $F \cos \gamma$.

- you have drawn the relative outgoing flow again, because it is parallel to the sail. This is not how the outgoing flow looks like in the ground frame where the sail is moving.

This is not a diagram of the flow throughout all time and space. It is a snapshot capturing the momentum change of the impinging jet immediately before and after impacting the sail at some specific time, location on the infinite planar sail.

 

[A.T.]

If we are talking fluid mechanics, we are talking Newtons Second,...

Sure, but you don't have to go back to Newton's Laws for every fluid mechanics problem. In particular for the aerodynamic forces on airfoils we have tons of experimental data, which is more reliable, than what you get by simplifying the situation, in order to solve it analytically.

 

[erobz]

Sure, but you don't have to go back to Newton's Laws for every fluid mechanics problem. In particular for the aerodynamic forces on airfoils we have tons of experimental data, which is more reliable, than what you get by simplifying the situation, in order to solve it analytically.

I can tell you right now I don't have a prayer of solving it analytically. Numerically...maybe. But that has been my whole point. We have this analysis (which is not the analysis of an airfoil, but is a step in that direction) which appears to be virtually humanly unsolvable in general, and you were telling me to just look at some randomly drawn vectors and animations, while critizing my "vectors" that are used as part of an analysis in a traditional way.

If you are able to site a CFD analysis that specifically proves downwind travel faster than the wind in the idealized case presented here, post it. I mean, if this is such a hot button issue I'm finding it hard to believe there aren't any experts that have created the rather theoretically simple CFD model to demonstrate it.

 

[A.T.]

...you were telling me to just look at some randomly drawn vectors...

No, I'm telling you to look at vectors, which are consistent with real world lift/drag-data for airfoils. This removes the whole messy fluid dynamics from the problem. Even for a numerical approach you can use empirical lift/drag-data, instead of CFD.

If you are able to site a CFD analysis that specifically proves downwind travel faster than the wind in the idealized case presented here, post it

"Sure, it works in reality, but can you show it in CFD?"

 

[erobz]

"Sure, it works in reality, but can you show it in CFD?"

"Reality" is messy and an uncontrolled environment. Show it in the laboratory, or simulation where experimental parameters are mostly controlled\fully controlled.

 

[A.T.]

"Reality" is messy and an uncontrolled environment.

Ice boats can do 4 x windspeed in the downwind direction. This is way more than the variability of the wind.

Show it in the laboratory,

The experimental lift/drag-data for airfoils comes from the lab (wind-tunnel). And and it implies that downwind faster than the wind is possible. Which is consistent with outdoor experimental data.

 

[erobz]

Ice boats can do 4 x windspeed in the downwind direction. This is way more than the variability of the wind.The experimental lift/drag-data for airfoils comes from the lab (wind-tunnel). And and it implies that downwind faster than the wind is possible. Which is consistent with outdoor experimental data.

Color me skeptical, and prove it with CFD.

 

[A.T.]

prove it with CFD.

That is not how physics works. It's an empirical science based on experiments. We know lift/drag-data for airfoils from experiments, so we don't need CFD to "prove" that it is possible.

 

[erobz]

That is not how physics works. It's an empirical science based on experiments. We know lift/drag-data for airfoils from experiments, so we don't need CFD to "prove" that it is possible.

This wasn't about a wing, this is about a vane. There is a difference. You are telling me that the solution for a simple vane should be faster than the wind in the direction of the wind...given the complexity of the equations I will derive I say, prove that with CFD. A thin straight (or curved) sheet of canvas, does not an airfoil make.

 

[A.T.]

This wasn't about a wing, this is about a vane.

It doesn't mater how you call it. All that matters is that you have something that can create much more lift than drag. The rest is trigonometry and vector math.

 

[erobz]

It doesn't mater how you call it. All that matters is that you have something that can create much more lift than drag. The rest is trigonometry and vector math.

I just showed you that is substantially more complicated than that! You are ignoring it and defaulting to your "just add the vectors" non-analysis "analysis".

 

[A.T.]

I just showed you that is substantially more complicated than that!

No. You just showed that one can make it more complicated than it needs to be.

There is no need to go back to Newton's Laws or use CFD here, because the question is not how lift is generated on the fundamental level. The question is how fast the boat can go downwind, given the empirically known performance of airfoils.

 

[erobz]

Why have you goaded me back into this conversation? I walked away 50 posts ago, last month to allow you your delusions of simplicity. You called me back for a fight when you thought you had reinforcements.

I'm walking away again. I won't coming back to this, so feel free to trash what I've done all you want.

Take Care.

 

[A.T.]

... your delusions of simplicity....

I'm proposing to keep the fluid dynamics details of lift generation out, because I know it's not simple. But we also know empirically that it works, so it can be taken as given.

 

[erobz]

I'm proposing to keep the fluid dynamics details of lift generation out, because I know it's not simple. But we also know empirically that it works, so it can be taken as given.

I was keeping fluid mechanics of lift generation out of it. There is no lift generated on the vane. It is not an airfoil.

 

[A.T.]

There is no lift generated on the vane.

Lift is the force component perpendicular to the relative flow. Your vane will certainly produce some lift, at most angles of attack (all except +/- 90°). Even a brick can produce some lift. But key to performance is the ratio of lift to drag, which determines the angle between relative flow and aerodynamic force. The rest follows from those angles.

 

[erobz]

Lift is the force component perpendicular to the relative flow. Your vane will certainly produce some lift, at most angles of attack (all except +/- 90°). Even a brick can produce some lift. But key to performance is the ratio of lift to drag, which determines the angle between relative flow and aerodynamic force. The rest follows from those angles.

There is no flow over the vane. There is a force from the impulse of the jet changing it’s momentum. Is that what you are calling lift, because that’s not what I would call it. My analysis is happening in a vacuum. There is no lift, there is no drag….

 

[Gleb1964]

Look at the diagram. It shows everything I have planned.

This equation represents Newtons Second in Fluid Mechanics under the assumption of constant cross-sectional properties:

$$ \sum F = \frac{d}{dt} \int_{cv} \rho v ~d V\llap{-} + \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i $$

I confess I haven't penetrated deep into your fluid analysis. Can you elaborate about it. I am not an expert on it.
Can we consider we are shooting balls into infinity planar sail. Every ball interact with sail, one ball at the time. Every ball jump out and deliver a portion of momentum to the sail, incrementing its speed. That should look like a sum of momentum translated to sail. Like your formula for fluid.

 

[erobz]

I confess I haven't penetrated deep into your fluid analysis. Can you elaborate about it. I am not an expert on it.
Can we consider we are shooting balls into infinity planar sail. Every ball interact with sail, one ball at the time. Every ball jump out and deliver a portion of momentum to the sail, incrementing its speed. That should look like a sum of momentum translated to sail. Like your formula for fluid.

I'm not an expert either. Your probably better off googling "Reynolds Transport Theorem: Momentum Equation" first to familiarize with its full derivation. I've expressed an already simplified version. If you have some specific questions, I'll do my best to tell you what little I know.

 

[A.T.]

There is no flow over the vane. There is a force from the impulse of the jet changing it’s momentum. Is that what you are calling lift, because that’s not what I would call it. My analysis is happening in a vacuum. There is no lift, there is no drag….

Lift and drag are defined based on the velocity of the air relative to the vane before the interaction. Lift is the force on the vane perpendicular to that velocity, while drag is parallel to it. Even if you shoot a jet of particles in a vacuum at the vane, you can still apply this definition and compute lift and drag.

But note that the velocity of the air relative to the vane before the interaction is not parallel to your jets (true wind direction) when the vane is moving, because it has a y-component too.

 

[A.T.]

This relative wind you keep trying to invoke for the analysis doesn't exist in the inertial frame. You are making up momentum that doesn't exist in the inertial frame. I'm doing the analysis in the inertial frame (a frame fixed to the ground)

What you seem to be missing is this: The force by the air on the sail is frame invariant. So it is perfectly valid to compute that force in the rest frame of the sail, and then use that force in any other frame (like the ground frame).

This greatly simplifies the estimation of the air deflection and and its momentum change of the air, because in the rest frame of the sail you have only the air moving, while the sail is static. So for a super simple model you could assume an elastic collision with a flat sail (angle incidence = angle of rebound). You just have to transform the air velocity from the ground frame (true wind) into the rest frame of the sail (relative wind), including the y-component (as vectors: w' = w - v, see image below).

If you are worried that the rest frame of the sail might not be inertial, this is not a problem. As long as we are dealing with instantaneous quantities (force is rate of momentum transfer), we can use the instantaneous inertial rest frame of the sail.

 

[Gleb1964]

The diagram above showing the optimal vane only for the last case, v_x>w.

May be I can extend the above diagram in steps of boat speed v relative to true wind w , illustrating the optimal vane orientation at different speeds.

Here you can see, how the sail orientation changed during increasing the boat speed v (the boat is sailing down wind). At the very high speed the sail would be tending very close to direction of the boat motion.

 

[erobz]

In an attempt to find out when exactly opinions diverge:

Lets say the cart below is moving at a constant velocity $v_c$ with application of $F_{nc}$. With respect to an inertial frame fixed to the ground the jet has velocity $v_j$

Assumptions:
Constant flow properties across the jet
Steady Flow
Inviscid Flow

The diagram is indicating the rate of momentum entering\exiting the cart w.r.t the inertial frame per the assumptions.

What is the force $F_{nc}$ acting on the cart?

 

[A.T.]

Why do you again propose a scenario where the vehicle moves parallel to the wind? It doesn't work like this. Ýou have the correct setup in the two posts just above.

 

[Gleb1964]

I assume that you mean v_c is a vector that can be pointed in any direction, not only along x axis, right?

 

[erobz]

I assume that you mean v_c is a vector that can be pointed in any direction, not only along x axis, right?

No, this exact set up. What is $F_{nc}$. This a plant a flag in the ground problem to find out where our opinions diverge. It sets up the base model from which analysis can be gradually expanded.

 

[Gleb1964]

If the boat speed direction is collinear with the true wind, that is not possible for boat to exceed the wind speed.

 

[erobz]

If the boat speed direction is collinear with the true wind, that is not possible for boat to exceed the wind speed.

Ok, so what is the force acting on the cart?

 

[Gleb1964]

As I see on your picture, you are at the boat inertial frame, the incoming and outcoming flow has the same speed (v_j-v_c). I have illustrated the result force direction and sail on your picture.

 

[A.T.]

In an attempt to find out when exactly opinions diverge:

Lets say the cart below is moving at a constant velocity $v_c$ with application of $F_{nc}$. With respect to an inertial frame fixed to the ground the jet has velocity $v_j$

Assumptions:
Constant flow properties across the jet
Steady Flow

The diagram is indicating the rate of momentum entering\exiting the cart w.r.t the inertial frame per the assumptions.

View attachment 322941
What is the force $F_{nc}$ acting on the cart?

You cannot assume that outflow speed equals inflow speed ($v_j - v_c$) in the ground frame, if the vehicle is moving at constant speed against the force $F_{nc}$. To maintain constant speed the air must be doing work on the vehicle, so it must loose kinetic energy (slow down). Only in the rest frame of the vehicle you can assume (in the ideal case) that inflow/outflow speeds are equal, because no work is done on the static vehicle there.

Once you corrected that: $F_{nc}$ is the rate of horizontal momentum change of the air.

 

[erobz]

As I see on your picture, you are at the boat inertial frame, the incoming and outcoming flow has the same speed (v_j-v_c). I have illustrated the result force direction and sail on your picture.

View attachment 322943

What is the magnitude of the force labled $F_{nc}$ in the diagram?

 

[erobz]

You cannot assume that outflow speed equals inflow speed ($v_j - v_c$) in the ground frame, if the vehicle is moving at constant speed against the force $F_{nc}$. To maintain constant speed the air must be doing work on the vehicle, so it must loose kinetic energy (slow down). Only in the rest frame of the vehicle you can assume (in the ideal case) that inflow/outflow speeds are equal, because no work is done on the static vehicle there.

Once you corrected that: $F_{nc}$ is the rate of horizontal momentum change of the air.

Yea I most certainly can. The control volume is not accelerating in this example.

A frame fixed to the ground has to give me the same force as a frame fixed to the cart in this example.

 

[A.T.]

You cannot assume that outflow speed equals inflow speed ($v_j - v_c$) in the ground frame, if the vehicle is moving at constant speed against the force $F_{nc}$. To maintain constant speed the air must be doing work on the vehicle, so it must loose kinetic energy (slow down). Only in the rest frame of the vehicle you can assume (in the ideal case) that inflow/outflow speeds are equal, because no work is done on the static vehicle there.

Yea I most certainly can. The control volume is not accelerating in this example.

Acceleration is not relevant to work done. If the air exerts a horizontal force on the moving vehicle, then it is doing work on it, and thus must loose energy.

 

[erobz]

Acceleration is not relevant to work done. If the air exerts a horizontal force on the moving vehicle, then it is doing work on it, and thus must loose energy.

A frame fixed to the ground must give me the same force as a frame fixed to the cart for this example. The cart is not accelerating.

 

[A.T.]

A frame fixed to the ground must give me the same force as a frame fixed to the cart for this example.

Yes, but the inflow/outflow speeds are different across frames. And outflow speed cannot be equal to inflow speed in the ground frame.

The cart is not accelerating.

Doesn't matter, The braking force $F_{nc}$ could be used to drive a generator and charge a battery. Where is that energy comming from, if the air doesn't slow down?

 

[Gleb1964]

The control volume is not accelerating in this example.

I have change the resistance force on your drawing in accordance with your statement.
Now the boat is not accelerating. The picture is at the boat reference frame.

And I am confused with ($v_j$) and ($v_c$) - is it vectors? Probably not, because you use the same in outgoing flow, but outgoing flow has a different direction. It is scalar values?

 

[erobz]

Yes, but the inflow/outflow speeds are different across frames. And outflow speed cannot be equal to inflow speed in the ground frame.Doesn't matter, The braking force $F_{nc}$ could be used to drive a generator and charge a battery. Where is that energy comming from, if the air doesn't slow down?

I see your point. However, if the frame is fixed to the cart what I have diagrammed is accurate.