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

[erobz]

This is the problem fully worked. as well as some other surrounding questions.

 

[erobz]

@A.T. What are the incoming/outgoing flow velocities w.r.t a frame fixed to the ground. If I ever hope to get to accelerating control volumes I need to get a better handle on this. The equation is only valid for inertial reference frames.

 

[A.T.]

$F_{nc}$ is not a drag force in the traditional sense.

But it is equal and opposite to the drag on the vane, as usually defined in aerodynamics: component of the aerodynamic force parallel to the incoming relative flow.

 

[erobz]

There is no drag on the vane. The flow is inviscid. Drag is a viscous force. I am holding the cart with my finger.

 

[A.T.]

There is no drag on the vane. The flow is inviscid.

Doesn't matter. Look at the definition: drag is the component of the aerodynamic force parallel to the incoming relative flow. There is nothing in there about inviscid vs viscid.

 

[erobz]

Doesn't matter. Look at the definition: drag is the component of the aerodynamic force parallel to the incoming relative flow. There is nothing in there about inviscid vs viscid.

You seem to be arguing about nothing to avoid getting to anything of substance? You are setting up strawman arguments to knock down. It's really annoying. Can we be grown ups now?

 

[A.T.]

You seem to be arguing about nothing to avoid getting to anything of substance?

I'm clarifying the terms so we can communicate.

 

[erobz]

I'm clarifying the terms so we can communicate.

There is absolutely zero need to talk about drag forces and lift forces here. Force $F_x$ and $F_y$ are more than sufficient.

 

[A.T.]

@A.T. What are the incoming/outgoing flow velocities w.r.t a frame fixed to the ground.

If we assume that in the frame of the cart: outflow speed = inflow speed = $v_j -v_c$ (idealized case), then you can transform the vectors into the ground frame:

$$Vin_{ground} = \begin{pmatrix} v_j \\ 0 \end{pmatrix}$$
$$Vout_{ground} = \begin{pmatrix} cos \theta (v_j - v_c) + v_c \\ sin \theta (v_j - v_c) \end{pmatrix}$$

 

[erobz]

If we assume that in the frame of the cart: outflow speed = inflow speed = $v_j -v_c$ (idealized case), then you can transform the vectors into the ground frame:

$$Vin_{ground} = \begin{pmatrix} v_j \\ 0 \end{pmatrix}$$
$$Vout_{ground} = \begin{pmatrix} cos \theta (v_j - v_c) - v_c \\ sin \theta (v_j - v_c) \end{pmatrix}$$

Something isn't sitting right with me.

If we let $\theta = 90$ ( a quarter circle vane) we have a change in momentum in the $x$ direction of

$-\dot m( v_j+v_c)$

according to your transformation

if the jet is traveling in with velocity $v_j \boldsymbol {\hat i}$, it should have momentum $\dot m v_c \boldsymbol {\hat i}$ leaving the control volume in the $\boldsymbol {\hat i}$ direction?

Thats a change in momentum of $-\dot m ( v_j- v_c ) \boldsymbol {\hat i}$

There seems to be a discrepancy?

 

[Gleb1964]

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.

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

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.

That's clear. You have mentioned all in a citation above. It is a steady inviscid flow. Flow is transferred momentum into control volume. That's all.
No need mention any air dynamic drag, the control volume is a black box, and a change of transferred momentum enough to calculate force components.
I only make remark, that from the question of "sailing down wind faster than wind" this configuration is irrelevant, because carriage in this case will never exceed the speed of flow. But I hope we will come to the relevant configuration.

 

[erobz]

But I hope we will come to the relevant configuration.

Well, I would like to get this ironed out in the straight forward case first. Do you have any input on what I'm asking in post #220?

 

[Gleb1964]

If we assume that in the frame of the cart: outflow speed = inflow speed = $v_j -v_c$ (idealized case), then you can transform the vectors into the ground frame:

$$Vin_{ground} = \begin{pmatrix} v_j \\ 0 \end{pmatrix}$$
$$Vout_{ground} = \begin{pmatrix} cos \theta (v_j - v_c) - v_c \\ sin \theta (v_j - v_c) \end{pmatrix}$$

I think, if it should be $+v_c$ for x component for back transformation.

 

[erobz]

I think, if it should be $+v_c$ for x component for back transformation.

I like that:

$$\vec{v_{o/O} } = \vec{v_{j/c}}+\vec{v_{c/O}}$$

$$v_{o/O} \boldsymbol{i} = ( v_j - v_c ) \cos \theta \boldsymbol{i} + v_c \boldsymbol{i}$$

$$ v_{o/O} \boldsymbol{j} = ( v_j - v_c ) \sin \theta \boldsymbol{j}$$

$O$ is the origin. Seems ok to me.

 

[A.T.]

I think, if it should be $+v_c$ for x component for back transformation.

Yes, fixed it.

 

[A.T.]

I like that:

$$\vec{v_{o/O} } = \vec{v_{j/c}}+\vec{v_{c/O}}$$

Note that in the ground frame the air slows down and loses kinetic energy. So if for example $F_{nc}$ is a frictional force, that results in heat dissipation, this is where the energy for it comes from.

But what about the cart frame, where the air doesn't slow down, but the same heat is generated? Here the ground is moving and can provide energy. The reaction frictional force $-F_{nc}$ by the cart on the ground is doing negative work on the ground in the cart frame.

Another interesting inertial frame is the rest frame of the incoming air, where the air initially doesn't have any kinetic energy, but ends up with some. Here again the energy comes from the ground and goes into heat and air's KE.

 

[erobz]

Note that in the ground frame the air slows down and loses kinetic energy. So if for example $F_{nc}$ is a frictional force, that results in heat dissipation, this is where the energy for it comes from.

But what about the cart frame, where the air doesn't slow down, but the same heat is generated? Here the ground is moving and can provide energy. The reaction frictional force $-F_{nc}$ by the cart on the ground is doing negative work on the ground in the cart frame.

Another interesting inertial frame is the rest frame of the incoming air, where the air initially doesn't have any kinetic energy, but ends up with some. Here again the energy comes from the ground and goes into heat and air's KE.

We’ll, I just need to be careful to have an inertial frame of reference for the governing equations to be valid. The cart frame is the obvious choice when the control volume is not accelerating, but soon I hope it will be and I needed to be careful about referencing all momentum correctly from the fixed ground frame. I got this wrong in some of those earlier analysis and hope to correct that moving forward.

 

[A.T.]

The cart frame is the obvious choice when the control volume is not accelerating, but soon I hope it will be and I needed to be careful about referencing all momentum correctly from the fixed ground frame.

You could keep the cart frame inertial, and show that the force $F_{nc}$ needed to keep the cart at constant speed points backwards. This proves that acceleration is possible.

Or you could use the instantaneous inertial rest frame of an accelerating cart, for analysis of instantaneous quantities.

But even if you want to use the ground frame, it is preferable to compute the frame invariant rate of momentum change of the air (force) in the cart frame.

 

[erobz]

You could keep the cart frame inertial, and show that the force $F_{nc}$ needed to keep the cart at constant speed points backwards. This proves that acceleration is possible.

I was thinking about this approach too. I need to show that the magnitude of the force $F_{nc}$ is non-zero at $v_x = w$.

 

[Gleb1964]

I think we are talking that projection of force $F_{nc}$ on a speed vector $v$ is non-zero.

 

[erobz]

I think we are talking that projection of force $F_{nc}$ on a speed vector $v$ is non-zero.

https://www.physicsforums.com/attachments/322992

That was not my intent. Just as in the last example there is going to non-conservative force acting on the control volume opposite $v$ that prevents the control volume from accelerating...thus, keeping a frame attached to the cart an inertial reference frame. If there is a resultant non-zero net force being applied to the cart from the jet at $v_x = w$, meaning $F_{nc} >0,$ given $v_x=w$ then if said force were removed, acceleration beyond $v_x$ would be possible.

 

[Gleb1964]

We are at the frame moving with speed $v_x=w$, no wind.
Because $v_x=w$, $w-v_x = 0$
But the boat speed $v$ has $v_x$ and $v_y$ components. Because is $v_y$ component, the "control volume" is moving up with $v_y = v sin( \theta )$, the sail is "smashing" the air flow upwards and back.

 

[erobz]

We are at the frame moving with speed $v_x=w$, no wind.
Because $v_x=w$, $w-v_x = 0$
But the boat speed $v$ has $v_x$ and $v_y$ components. Because is $v_y$ component, the "control volume" is moving up with $v_y = v sin( \theta )$ smashing the air flow upwards and back.

What you are saying is that you don't agree with what the analysis is going to show. That was the whole point of agreeing on the analysis in the video! This is a complete waste of my time!

Me: look at this video...here are all the assumptions
You: Ok looks reasonable.
You: That's what I get too, but...the car needs to be moving at some angle relative to the jet, blah, blah,blah
Me: turn it by some angle, and fiddle with the sail to maximize that force?
You: That will never work out the way I want.
You: lets skip the formalities we just agreed upon and just say I'm correct!

This conversation has been going on like this for three cycles now. I'm just waiting for "just look at the vectors" at this point; As if that not what I've been doing the entire time. Any minute now...

 

[Gleb1964]

I just showed my analysis. Have you seen I have mention that agree or disagree with something?

 

[erobz]

I just showed my analysis. Have you seen I have mention that agree or disagree with something?

What you are proposing is not parallel to the agreed upon analysis technique shown in the video? Suddenly you are telling me the jet will rebound off of the vane.

 

[Gleb1964]

Because there is outcoming flow, there should be incoming flow.
May be need to go to the reference frame of control volume? I remind, the reference volume is not fixed, its moving up. That mean I have wrong illustration of incoming flux. It is non-zero.

 

[erobz]

Because there is outcoming flow, there should be incoming flow.
May be need to go to the reference frame of control volume? I remind, the reference volume is not fixed, its moving up. That mean I have wrong illustration of incoming flux. It is non-zero.

The cv is the reference frame. All the same assumptions about flow characteristics apply. We can do this so long as the cv in not accelerating, it is not accelerating because it being acted on by an external force opposite the direction of motion...just as it was in the video. If said force is non-zero at $v_x= w$ you've won the debate.

 

[Gleb1964]

I should been looking on a diagram of apparent wind to not confuse myself.
Now is corrected drawing. The reference frame is fixed on the cart, control volume is fixed.
There is no wind in $+x$ -direction, $v_x=w$, but there is a wind in $-y$ -direction.
From the point of control volume there is incoming flow from top with speed $w'$ and outcoming flow with the same speed, showed as out-flow on the picture.

 

[erobz]

I should been looking on a diagram of apparent wind to not confuse myself.
Now is corrected drawing. The reference frame is fixed on the cart, control volume is fixed.
There is no wind in $x$ -direction, $v_x=w$, but there is a wind in $-y$ -direction.
From the point of control volume there is incoming flow from top with speed $w'$ and outcoming flow with the same speed, showed as out-flow on the picture.

View attachment 322994

Stop with the apparent jet! Was there an "apparent jet" in the analysis? You are adding it in because you believe you can fool me. I'm not amused...

It is apparent you are desperately grasping for straws now because you already realize what the "agreed upon" analysis is going to say...just as I do!

The jet CANNOT push the cart in the direction of the jet FASTER than the jet! No matter what you do to the sail orientation momentum CANNOT enter ( nor exit) the cart at $v_x = w$. The mass flow rate $\dot m$ is a true "mathematical factor" in the sense that it is multiplied by anything you can come up with for the velocities- any configuration your heart desires. Short of DeVine intervention by the hand of GOD no matter what you do to the sail to maximize the acceleration of the cart there is a hard limit.

$$ \lim_{v_x \to v_j} \dot m = 0 \implies F_{nc} \to 0 $$

Period!

 

[Gleb1964]

Stop with the apparent jet! Was there an "apparent jet" in the analysis? You are adding it in because you believe you can fool me. I'm not amused...

It is apparent you are desperately grasping for straws now because you already realize what the "agreed upon" analysis is going to say...just as I do!

Here is your jet, as you wish. One in the ground frame and another in the cart frame. I have to split it in the volumes to illustrate somehow the transformation.