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

[A.T.]

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.

I don't think the orientation of the jet changes between the frames, just the velocity of the air particles, that the jet consists of.

The jet is animated here in different frames:

 

[A.T.]

What you are proposing is not parallel to the agreed upon analysis technique shown in the video?

In your video the cart moves parallel to the jet, so even in the cart frame the incoming air in the jet still moves parallel to the jet.

But this is not the case if the cart moves diagonally, because here in the cart frame the air velocity is different from the jet orientation.

When you are trying to generalize from a simpler 1D-transformation (your video) to a more complex 2D-transformation, you have to be more careful.

 

[erobz]

It certainly can. Just like the stick in the video below can push the cart in the direction of the stick faster than the stick is going in the direction of the stick.

That velocity vector is neither constant in magnitude nor direction over the duration of the push. It wobbles all over the place. There is a clear difference in direction in which the push begins and which the push ends. The labeled velocity vector at the end that was supposed to describe it perfectly over the duration of motion is a gross distortion of reality...as evidenced by the need for "velocity made good". Furthermore, if you pause the video an take some screenshots you'll see that cart has deviated from the assumed trajectory over the push as well. That is not a controlled experiment in the least bit. It's simply shoddy analysis, that leads to an absurd conclusion that goes against everything we learn in physics. It's a con artist tool.

This is no different conceptually than the little Pytagorean Puzzle, where it gets cut up and a square is suddenly missing. You are allowing yourself to be fooled.

This is precisely why vectors both magnitude and directions are so important. So when we get a result, we see that gross simplifications lead to deviation from perfect. We never say, lets adjust perfect!

 

[A.T.]

That velocity vector is neither constant in magnitude nor direction over the duration of the push. It wobbles all over the place.

Do you seriously think it would stop working if the push was "cleaner". It is a kinematically constrained system, which can only move the way it does.

It's simply shoddy analysis, that leads to an absurd conclusion that goes against everything we learn in physics.

Which law of physics specifically do you see violated by this simple toy?

 

[erobz]

Do you seriously think it would stop working if the push was "cleaner". It is a kinematically constrained system, which can only move the way it does.Which law of physics specifically do you see violated by this simple toy?

None of the vectors that supposedly describe it are the reality! Everything is changing! The cart is on a curved trajectory! The pusher has neither controlled velocity nor constant direction. Everything is somewhat wrong.

You are being fooled!

https://en.wikipedia.org/wiki/Missing_square_puzzle#/media/File:Missing_Square_Animation.gif

This puzzle is proof positive of the very issue.

"velocity made good" is absolute madness.

 

[A.T.]

Everything is changing! from what it is upposed to be over the duration of the push.

It is a kinematically constrained system. The relative displacements will be the same, regardless if the speed is constant or not.

 

[erobz]

It is a kinematically constrained system. The relative displacements will be the same, regardless if the speed is constant or not.

All the directions are changing too.

 

[A.T.]

All the directions are changing too.

It would work even better, if it would wobble less. So what is your point?

 

[erobz]

It would work even better, if it would wobble less. So what is your point?

Bull. Its a simple perfect geomety problem. That "demonstration" is utter nonsense.

 

[A.T.]

bull.

Can we be grown ups now?

 

[erobz]

Can we be grown ups now?

That "demonstration" is utter nonsense. I am going to be a grown up and let you with you delusions.

 

[A.T.]

Its a simple perfect geomety problem.

Absolutely correct. The relative displacements are fully determined by simple geometry.

That "demonstration" is utter nonsense.

How does it contradict geometry?

 

[Gleb1964]

I don't think the orientation of the jet changes between the frames, just the velocity of the air particles, that the jet consists of.

View attachment 5_AT.png.webp

Ok, thank you for correction. You are right. I would need to think about it before posting.

 

[erobz]

How does it contradict geometry?

Its a rigid body on a track. Every point in the upper position is fully constrained and mapped by the right angle triangle to its lower position. If you end up with something else, things were bouncing sliding, bending, etc... Nothing was as it appeared to be, they are telling you a subtle lie, you are gobbling it up...asking for more...and worst of all...sharing it with others as "Physics"!

 

[A.T.]

Its a rigid body on a track. Every point in the upper position is fully constrained and mapped by the right angle triangle to its lower position.

Good start. Now add the pushing stick for both positions.

I would also recommend rotating the vane slightly clockwise, so its orientation bisects the angle between vertical and track. This will make the illustration clearer.

 

[erobz]

Good start. Now add the pushing stick for both positions.

I would also recommend rotating the vane slightly clockwise, so its orientation bisects the angle between vertical and track. This will make the illustration clearer.

That red arrow is the change in length of the pushing stick. Bye. This time its for real.

 

[A.T.]

That red arrow is the change in length of the pushing stick.

Wrong. Draw the stick in both positions. It must touch the vane for both positions, and be only displaced horizontally.

 

[erobz]

Wrong. Draw the stick in both positions. It must touch the vane for both positions, and be only displaced horizontally.

Ahh. S.o.b. There is some kind of mechanical advantage here. Are you breaking me like a wild stallion...maybe a pony?

 

[A.T.]

There is some kind of mechanical advantage here.

Absolutely, and so is in sailing.

 

[erobz]

Absolutely, and so is in sailing.

So then the analysis I was trying to get to before this will reveal this mech advantage too...if I do it properly?

 

[A.T.]

So then the analysis I was trying to get to before this will reveal this mech advantage too...if I do it properly?

If I understand your aim correctly, it was about checking momentum conservation. Mechanical advantage roughly means that you can trade force for speed without violating energy conservation.

 

[erobz]

If I understand your aim correctly, it was about checking momentum conservation. Mechanical advantage roughly means that you can trade force for speed without violating energy conservation.

Don't worry, you won't have to worry about correcting me anymore today. I'm already going to be in trouble for this one.

Thanks for your help, I let my emotions get the best of me. I'm sorry.

 

[A.T.]

Thanks for your help, I let my emotions get the best of me. I'm sorry.

No problem. Here is another mechanical analogy: A thin wedge squeezed between oblique surfaces:

 

[erobz]

The frame is fixed to the origin (not the control surface), $\boldsymbol{i}$ parallel to the slope.

I'm trying to find the velocity of the jet "$v_j$ relative to the cart:

$$ v_{j/O} \boldsymbol{i} = v_{j/c} \boldsymbol{i} + v_{c/O} \boldsymbol{i} \implies v_{j/c} \boldsymbol{i} = \left( v_j \cos \theta - v_c \right) \boldsymbol{i} $$

$$ v_{j/c} \boldsymbol{j} = -v_j \sin \theta \boldsymbol{j}$$

That means the magnitude of $\vec{v}_{j/c}$ is given by:

$$ \left| v_{j/c} \right| = \sqrt{ \left( v_j^2 - 2 v_j v_c \cos \theta + v_c^2 \right) }$$

Letting $\cos \theta \to 0$ looks as expected:

$$\lim_{\theta \to 0 } \left| v_{j/c}\right| = \sqrt{v_j^2 - 2 v_j v_c + v_c^2} = \sqrt{\left( v_j - v_c\right)^2} = v_j- v_c$$

However, the limit as $\theta \to 90°$:

$$\lim_{\theta \to 90° } \left| v_{j/c}\right| = \sqrt{v_j^2 + v_c^2} $$

That seems unlikely. What have I goofed on already?

 

[Gleb1964]

I'm trying to find the velocity of the jet "$v_j$ relative to the cart:

There are two relative velocities of jet, relative to cart and relative to the sail. Jet is crossing the planar sail, the velocity regarding to this crossing point is not the same, as velocity regarding to the cart.

 

[erobz]

I'm trying to find the velocity of the jet entering the control volume relative to the control volume. I'm not picking up on what you are saying.

 

[Gleb1964]

The drawing has 1) cart speed, $v_c$ 2) down wind cart speed $v_x$ 3) the speed of the jet line crossing sail. The crossing sail speed 3 is a point of the interest. The jet speed should be taking relative the speed of point where jet is crossing sail.

 

[erobz]

The drawing has 1) cart speed, $v_c$ 2) down wind cart speed $v_x$ 3) the speed of the jet line crossing sail. The crossing sail speed 3 is a point of the interest. The jet speed should be taking relative the speed of point where jet is crossing sail.

What's the answer for the mass flow rate entering the control volume? Assuming a stream with density $\rho$ and cross-section $A$.

 

[Gleb1964]

Cart velocity $v_c$

$$ v_c = \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} v_c \cdot cos( \theta ) \\ v_c \cdot sin( \theta ) \end{pmatrix} \;\;\;\;\; (1) $$​

Down wind cart velocity

$$ \begin{pmatrix} v_c \cdot cos( \theta ) \\ 0 \end{pmatrix} \;\;\;\;\; (2)$$​

The velocity of jet-crossing-sail point ($/beta$ is angle between sail and cart line)

$$ \begin{pmatrix} v_c \cdot cos( \theta ) - v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) ) \\ 0 \end{pmatrix} \;\;\;\;\; (3) $$
​

Jet velocity

$$ \begin{pmatrix} w \\ 0 \end{pmatrix} \;\;\;\;\; (4) $$
​

Jet velocity regarding to the sail cross point:

$$ \begin{pmatrix} w \\ 0 \end{pmatrix} \; - \; \begin{pmatrix} v_c \cdot cos( \theta ) - v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) ) \\ 0 \end{pmatrix} \;\;\;\;\; (5) $$

 

[erobz]

Cart velocity $v_c$

$$ v_c = \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} v_c \cdot cos( \theta ) \\ v_c \cdot sin( \theta ) \end{pmatrix} \;\;\;\;\; (1) $$​

$$\vdots$$

The control volume is a black box. We only care about the momentum efflux across its surface. Nothing about that has changed from the configuration shown in the video. You are doing "funny business" inside of the control volume, there is no need to be mucking about inside of it. What goes in must come out. The flow rate entering the control volume establishes the velocity of the flow relative to the vane. The vane establishes the direction of the flow on the way out...nothing more.

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.

 

[Gleb1964]

Jet is contacting only with sail and only on a line of jet, which is along $x+$. The relative jet-sail velocity at jet line is done on (5). The ingoing momentum in the control volume is
$$ \dot m \cdot \begin{pmatrix} w - v_c \cdot cos( \theta ) - v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) ) \\ 0 \end{pmatrix} \;\;\;\;\; $$If we not agree at this point, not possible to look at flow outgoing from the control volume.

 

[erobz]

Jet is contacting only with sail and only on a line of jet, which is along $x+$. The relative jet-sail velocity at jet line is done on (5). The ingoing momentum in the control volume is
$$ \dot m \cdot \begin{pmatrix} w - v_c \cdot cos( \theta ) - v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) ) \\ 0 \end{pmatrix} \;\;\;\;\; $$If we not agree at this point, not possible to look at flow outgoing from the control volume.

Sorry, I'm not able to see how the incoming flowrate is dependent on $\beta$. You have "un-black boxed" the control volume.

 

[A.T.]

What have I goofed on already?

As I explained in post #247, using a 1D jet for the 2D case doesn't make much sense. It just complicates the analysis. For the 2D case the wind is best modeled as a 2D velocity field, which can be transformed into the rest frame of the control volume, to get the mass flow rate into it.

 

[Gleb1964]

Sorry, I'm not able to see how the incoming flowrate is dependent on $\beta$. You have "un-black boxed" the control volume.

At the point, where jet line is crossing the sail, here are several velocity components.
$v_c \cdot cos( \theta )$ is projection of cart velocity $v_c$ to jet line $x+$

$- v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) )$ is additional component, because cart has $v_y$ component and sail plane has angle $\beta$ to the cart motion direction. Depending on angle $\beta$ the acting point on sail can go faster or slower than $v_x$ component.

 

[erobz]

Jet is contacting only with sail and only on a line of jet, which is along $x+$. The relative jet-sail velocity at jet line is done on (5). The ingoing momentum in the control volume is
$$ \dot m \cdot \begin{pmatrix} w - v_c \cdot cos( \theta ) - v_c \cdot sin( \theta ) \cdot tan( \pi/2 - (\theta+ \beta) ) \\ 0 \end{pmatrix} \;\;\;\;\; $$If we not agree at this point, not possible to look at flow outgoing from the control volume.

After looking at the diagram in #270, I can see how this relates to the pushcart video now. Also, your relationship passes the sanity check at $\beta = 90°, \theta = 0$ ( you've defined $\beta$ as the supplementarily angle of how I had defined it). That can't be coincidental... trying to check $\beta = 90°, \theta = 90°$ creates some issues, but that should be expected too.