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

[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.

 

[A.T.]

So what is the hypothesis for the limits of $v_x,v_y$?

Here the speed limits plotted as function of the minimal apparent wind angle (AWA), normalized to multiples of true wind speed (TWS):

And here the speeds for AWA = 6° (ice boats) and AWA = 20° (racing boats), as function of the true wind angle (TWA), or the course relative to the true wind:

For the AWA = 20°, here the vectors for maximal speed:

Maximal downwind component:

Maximal upwind component:

The math:

LD_air = lift/drag at the air
LD_surface = lift/drag at the surface

AWA : apparent wind angle
TWA : true wind angle
TWS : true wind speed

AWA = atan(1 / LD_surface) + atan(1 / LD_air)

MAX_SPEED = TWS * 1 / sin(AWA)
MAX_DOWNWIND_COMPONENT = TWS * (1 / sin(AWA) + 1) / 2
MAX_UPWIND_COMPONENT = TWS * (1 / sin(AWA) - 1) / 2

SPEED = TWS * sin(TWA - AWA) / sin(AWA)
DOWNWIND (+) UPWIND (-) COMPONENT= - SPEED * COS(TWA)

For more information see:
http://www.onemetre.net/design/CourseTheorem/CourseTheorem.htm
https://books.google.de/books?id=Xe_i23UL4sAC&lpg=PP1&hl=de&pg=PA49#v=onepage&q&f=false

 

[erobz]

I believe you’ve already posted this, I didn’t understand it then, and I don’t now. The links have almost no supporting analysis. Surely someone has completed a full analysis?

 

[A.T.]

I believe you’ve already posted this, I didn’t understand it then, and I don’t now.

It's just basic geometry. Are you familiar with inscribed angles?
https://en.wikipedia.org/wiki/Inscribed_angle

The apparent wind angle (AWA) is the inscribed angle and determines the size of the polar circle, which constrains the velocity vector. The greater the lift/drag ratios, the smaller the AWA, and the larger the circle of possible velocities.

Just play around with this interactive diagram, move the sliders around and try different situations:
https://www.geogebra.org/m/tj5qf3w2

 

[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.

Here is an animation to visualize the kinematics with respect to the 2D air-mass, rather than a single jet.

The relative flow is along the bottom/left or top/right edge of the red area. And relative to the flow we have empirical data on lift(L)/drag(D) we can achieve. Shown below is a conservative L/D=8 which still results in the aerodynamics force (F) having a positive component along the direction of motion (P).

 

[erobz]

Here is an animation to visualize the kinematics with respect to the 2D air-mass, rather than a single jet.

View attachment 333400

The relative flow is along the bottom/left or top/right edge of the red area. And relative to the flow we have empirical data on lift(L)/drag(D) we can achieve. Shown below is a conservative L/D=8 which still results in the aerodynamics force (F) having a positive component along the direction of motion (P).

View attachment 333399

My interest was theoretical (I can't even say I have interest anymore - this was started almost a year ago). If you are insistent in dragging us back in, I would like to see a "Newtons 2nd Law" approach, or figure out how to model it myself using Newton 2nd Law. I'm not really interested in further discussion unless its analytical. If the result is empirical(as you say), then there must be some set of stepping stone models to get to the empirical evidence.

 

[A.T.]

My interest was theoretical (I can't even say I have interest anymore - this was started almost a year ago). If you are insistent in dragging us back in, I would like to see a "Newtons 2nd Law" approach, or figure out how to model it myself using Newton 2nd Law. I'm not really interested in further discussion unless its analytical. If the result is empirical(as you say), then there must be some set of stepping stone models to get to the empirical evidence.

If you are interested in finding a theoretical limit on the ratio between a boat's downwind velocity component and windspeed, then that is doomed to fail, as there is no theoretical limit on mechanical advantage / gear ratio. All limits here are practical and stem from our limitations in achieving higher efficiency while preserving sufficient structural stability. For airfoils / hydrofoils those efficiencies are expressed as lift/drag ratios, and lead to velocity limits via the course theorem as explained in post #281. Those lift/drag ratios can be derived empirically or numerically (CFD).

If you are interested in a purely analytical derivation from Newton 2nd Law, then you have to simplify a lot. For example you could adopt an idealized elastic collision with a plane model like shown in the animation below. Of course, if you make it frictionless and perfectly elastic, you won't find any limits on the velocity. If you introduce some friction, then that will determine the velocity limits. While those limits will likely differ from realistic limits, they can give you an idea would would happen in an artificial rare gas scenario.

 

[erobz]

$$ N \cos \theta = M \frac{dv_y}{dt} + \dot m ( v_y - w \cos \beta ) + \dot m w \tag{1}$$

$$ -N \sin \theta = M \frac{dv_x}{dt} + \dot m ( v_x - w \sin \beta ) \tag{2} $$

$$ \frac{v_y}{v_x} = \tan \theta \tag{3} $$

$$ \frac{dv_x}{dt}= \frac{dv_y}{dt} \frac{1}{\tan \theta} \tag{4} $$

• The frame is inertial fixed to the frictionless track.
• The wind $w$ speed is assumed constant via Bernoulli's ($P_{out} = P_{in} = P_{atm} \implies w _{out} = w_{in} = w)$
• The control volume is a black box. All we see is momentum inflows/outflows and external forces

This is standard analysis. Your claim is that there is no limit to which component of the control volumes velocity in this scenario?

 

[A.T.]

$w _{out} = w_{in}$

Do you mean the flow speed relative to control volume, or relative to a frame where the control volume moves?

 

[erobz]

Do you mean the flow speed relative to control volume, or relative to a frame where the control volume moves?

Relative to the control volume. Applying conservation of energy.

 

[A.T.]

Relative to the control volume.

If your $w$ is relative to the control volume, then the direction of $w_{in}$ is not constant, but changes as function of $v$.

What is the wind direction relative to the ground here?

 

[erobz]

If your $w$ is relative to the control volume, then the direction of $w_{in}$ is not constant, but changes as function of $v$.

What is the wind direction relative to the ground here?

It is the direction of the blue arrow at the top of the diagram pointing "down". Are you saying the momentum inflow needs to be:

$$\dot m ( v_y - w ) $$

And if I choose $w$ to be measured in the ground frame, then what? I think it is actually what I have in mind. The frames of reference are definitely a point of confusion. I would rather reference everything from ground frame to be sure. $w$ is the wind relative to the ground frame. On the way in it is fixed, and it doesn't care about the cart. On the way out, I'm not certain about conservation of energy statement now, but I want to figure that out.

 

[A.T.]

It is the direction of the blue arrow at the top of the diagram pointing "down".

If the wind relative to ground is in the negative y direction, then you are modelling upwind tacking, not downwind tacking. That's OK, as there is no theoretical limit on that either. Just making sure this is what you want.

And if I choose $w$ to be measured in the ground frame, then what?

Then $w _{out} = w_{in}$ makes no sense. Of course you have to slow down the air in the ground frame to accelerate the vehicle.

 

[erobz]

If the wind relative to ground is in the negative y direction, then you are modelling upwind tacking, not downwind tacking. That's OK, as there is no theoretical limit on that either. Just making sure this is what you want.

Thats fine by me.

Then $w _{out} = w_{in}$ makes no sense. Of course you have to slow down the air in the ground frame to accelerate the vehicle.

Then this has to be worked on. Let me think about it.

 

[erobz]

I don't see an issue with trying to work out the outgoing flow velocity using mass flowrate constraint. Mass is not accumulating in the control volume, that must be true in every frame? I would probably switch to the frame of the cart for a moment... it seems to be trickier than I originally suspected. I will probably try again tomorrow.

 

[A.T.]

I don't see an issue with trying to work out the outgoing flow velocity using mass flowrate constraint. Mass is not accumulating in the control volume, that must be true in every frame? I would probably switch to the frame of the cart for a moment... it seems to be trickier than I originally suspected. I will probably try again tomorrow.

The in/outflow rates are always based on the flow relative to the boundary, no matter what reference frame you use otherwise.

You have to distinguish ground relative and boat relative velocities clearly, and account for x & y components.

 

[erobz]

So if we switch to the frame of the cart, the flowrate entering the cart is:

$$ \dot m_{in} = \rho A ( v_y + w ) $$

And that must equal the outgoing mass flowrate in the frame of the cart ( with no mass accumulating in the control volume )

$$ \dot m_{out} = \rho A w_{out}$$

Which I believe implies:

$$w_{out} = w+v_y $$

Before switching to the inertial frame is that ok? Or does the cross-sectional area have to change? It seems a bit unsettling.

 

[A.T.]

So if we switch to the frame of the cart, the flowrate entering the cart is:

$$ \dot m_{in} = \rho A ( v_y + w ) $$

I think you have forgotten $v_x$. The inflow velocity in the cart frame is not purely vertical and the flowrate depends on the magnitude of the whole velocity vector.

 

[erobz]

I think you have forgotten $v_x$. The inflow velocity in the cart frame is not purely vertical and the flowrate depends on the magnitude of the whole velocity vector.

I think you are saying its this:

$$ w_{out} = \sqrt{( v_y + w )^2 + v_x^2 } $$

?

 

[erobz]

Updating the diagram and the system of equations to reflect the changes:$$ N \cos \theta = M \dot v_y + \dot m \left( v_y + w - \sqrt{(v_y + w )^2 + v_x^2} \cos \beta \right) \tag{1} $$

$$ -N \sin \theta = M \dot v_x + \dot m \left( v_x - \sqrt{(v_y + w )^2 + v_x^2} \sin \beta \right) \tag{2} $$

$$ v_y = v_x \tan \theta \tag{3} $$

$$ \dot v_y = \dot v_x \tan \theta \tag{4} $$

Solving the system for the limiting velocity component $v_y$ as a function of $\beta$ ( $\dot v_y = 0$ ) is get the quadratic:

$$ \left( \left( \frac{ \varphi }{ \lambda }\right)^2 - \varphi \right) v_y^2 + 2 w \left( \frac{\varphi}{ \lambda^2} - 1 \right) v_y + w^2\left( \frac{1}{\lambda^2}-1\right) = 0 $$

Here is a plot of the solution over $0^{\circ} \leq \beta \leq 44^{\circ}$ for $\theta= 45^{\circ}$:

As the outflow becomes parallel with the trajectory, the steady state velocity increases without bound. @A.T. can now rest their case, unless I've committed an error in the mechanics...I have been converted.

 

[A.T.]

As the outflow becomes parallel with the trajectory, the steady state velocity increases without bound. @A.T. can now rest their case, unless I've committed an error in the mechanics...I have been converted.

This makes sense, if I interpret your definitions correctly: $\beta$ is the flow deflection in the ground frame, right? Because in the boat frame, the flow deflection should tend towards zero in the limiting case of velocity towards infinity. This corresponds to lift/drag ratio tending towards infinity, or the aero dynamic force tending towards 90° off the relative inflow (which tends towards parallel with the trajectory too).

So now that you see that there is no theoretical limit on the upwind component, you could check the same for downwind component, for example for $\theta = 135°$. Here the relative flow has a greater range, as it changes direction by more than 90° as v goes from 0 to infinity.