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

[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.

 

[erobz]

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.

View attachment 322944

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?

Yeah, scalars. I don't know why you are crossing out $F_{nc}$? That is the force I am asking you to find. It is the force being applied to the cart as to resist acceleration.

 

[Gleb1964]

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

I think the magnitude of $F_{nc}$ should be opposite to the drag force, which components are (according to your drawing):

$F_x = \dot m(v_j-v_c)+\dot m(v_j-v_c)cos(\theta )$
$F_y = \dot m(v_j-v_c)sin(\theta )$

 

[Gleb1964]

I don't know why you are crossing out $F_{nc}$? That is the force I am asking you to find. It is the force being applied to the cart as to resist acceleration.

Because the flow is going sideway, that mean the drag force has x and y components, as mention at formulas above. To keep boat at the constant speed, the balancing reaction force should be opposite to the drag force.

 

[A.T.]

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

If the frame you use is fixed to the cart, then you can assume (as an idealized case) outflow speed = inflow speed.

 

[erobz]

Because the flow is going sideway, that mean the drag force has x and y components, as mention at formulas above. To keep boat at the constant speed, the balancing reaction force should be opposite to the drag force.

There is no shear on the vane. This is under the assumption of inviscid flow. $F_{nc}$ is not a drag force from flow over the vane, its a force being applied by something external.

 

[erobz]

I think the magnitude of $F_{nc}$ should be opposite to the drag force, which components are (according to your drawing):

$F_x = \dot m(v_j-v_c)+\dot m(v_j-v_c)cos(\theta )$
$F_y = \dot m(v_j-v_c)sin(\theta )$

$F_x$ is the momentum outflow - momentum inflow. I think you have some signs flipped

$$F_x = \dot m ( v_j -v_c ) \cos \theta - \dot m ( v_j -v_c )$$

$F_{nc}$ is as you say the opposite $F_{nc} = -F_x$

 

[A.T.]

There is no shear on the vane. This is under the assumption of inviscid flow. $F_{nc}$ is not a drag force from flow over the vane, its a force being applied by something external.

Drag is the component of the aerodynamic force parallel to the incoming relative flow, so the horizontal component here, which equals $-F_{nc}$.

 

[erobz]

Drag is the component of the aerodynamic force parallel to the incoming flow, so the horizontal component here, which equals $-F_{nc}$.

There are no shear forces from the flow over the vane. The flow is assumed inviscid. The drag force ( if the flow wasn't inviscid) from the flow over the vane would be opposite of $F_{nc}$, with a component in the vertical direction as well. $F_{nc}$ is not a drag force in the traditional sense. It is a force I am applying to the cart.

 

[Gleb1964]

There are no shear forces from the flow over the vane. The flow is assumed inviscid.

That is what I assume with the formulas I wrote. I confess that I am not accurate with the sign. But x and y components are in accordance with picture you made.

 

[erobz]

That is what I assume with the formulas I wrote. I confess that I am not accurate with the sign. But x and y components are in accordance with picture you made.

The blue arrows aren't forces. That’s partially my fault. I have hybridized what is traditionally two separate diagrams ( momentum diagram, and force diagram) for brevity without stating that.