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

[nish95]

Hi everyone,

I was looking to develop my physical insight when I encountered this book by Lewis Caroll Epstein. For the crosswind problem, I couldn't understand the author's explanation; in particular, his concept of "artificial wind," and the force being larger in this case than the previous one. I understand that the sail will sag when the speed of the boat in the direction of normal to the sail is equal to that of wind in the same direction. I have attached the relevant pages of the book below.

 

[malawi_glenn]

Which book? You just state the authors name.
What is your question really?

Dirac was notoriously precise about his lectures. Victor Weisskopf used to tell the famous joke about Dirac’s answering questions after he’d given a lecture. One student said, “I don’t understand that second equation, Professor Dirac.” Dirac remained silent. “Aren’t you going to answer the question?” asked Weisskopf. “That was not a question, that was a statement,” said Dirac.
https://physicstoday.scitation.org/doi/10.1063/1.4796250

 

[Lnewqban]

Could you show us "the previous one"?

 

[nish95]

In the previous one, the boat is sailing vertically in the same direction as the wind.

 

[malawi_glenn]

What is your question?
How is the sail oriented (as in the previous question)?

 

[nish95]

What is your question?
How is the sail oriented (as in the previous question)?

Relative to the boat, the sail is oriented the same way in this and the previous question.
One would think that the answer then, is that the boat sails at the same speed in both cases but that is not what the author says. I'm trying to understand why this is so. More details are in the original post.

 

[malawi_glenn]

How is the sail oriented in the previous question?

 

[Baluncore]

Which book? You just state the authors name.

I can answer that difficult question.

Title: Thinking Physics: Practical Lessons in Critical Thinking.
Author: Lewis Carroll Epstein.
Publisher: Insight Press. Year: 2002.
2nd Edition. ISBN: 9780935218060

 

[malawi_glenn]

I can answer that difficult question.

It was also written in the thread-title :)

 

[andrewkirk]

When the boat is sailing with the wind directly behind it ('previous case'), the force depends on the excess of the wind speed over the boat's speed. As the boat speed increases, that excess reduces so the force reduces. When the two speeds match, that excess is zero, so there is no wind force on the sail and it sags. From the point of view of the sail, there is no wind.

When sailing across the wind (case of the diagram above), the sail will get a force from the wind as long as the wind is hitting the inside of the sail (the port side in the diagram), not the outside (starboard side), and not directly in line with the sail. That force will have direction perpendicular to the direction of the wind relative to the sail, pointing outwards to starboard in the diagram.
We then decompose that force into one perpendicular to the boat's direction, which is offset by the keel (so it doesn't affect the boat's motion), and one parallel to the direction, that pushes the boat forward.
So in this case, the force doesn't go to zero until the relative wind is blowing in line with the sail. The sail looks to be at an angle of about 30 degrees to the diagram's horizontal, so that will happen when the boat speed equals the wind speed divided by tan(30 degrees), which equals boat speed times about 1.73. So unlike the with-the-wind case, the boat can go up to 1.73 times wind speed before losing its wind propulsion. Whether the boat can actually reach that speed will depend on other things like the boat's sail size and drag in the water.

"Artificial wind" means the component of the windspeed relative to the sail, that arises from the boat's motion relative to the water. Think about riding a bike. If there's no wind, as soon as you start, you have a headwind. The faster you go, the stronger the headwind. That's all 'artificial wind'. The wind perceived by the rider - the 'relative wind' is a combination of the actual wind (wind speed relative to ground) and artificial wind (rider speed relative to ground). The combination is a 'vector sum'.
If there's a sidewind, perpendicular to the direction you will ride, the artificial wind starts at zero and increases speed as you speed up, making the relative wind swing from the side towards the front, getting closer to the front the faster you go

 

[A.T.]

Relative to the boat, the sail is oriented the same way in this and the previous question.
One would think that the answer then, is that the boat sails at the same speed in both cases but that is not what the author says. I'm trying to understand why this is so. More details are in the original post.

The force on the sail depends on the relative airflow (apparent wind)

When you sail directly downwind that relative airflow is going to zero, as you approach windspeed.

When you sail perpendicular across the wind, the downwind comportment of the relative airflow (vertical in the book) stays constant, while the induced component (horizontal in the book) increases with your speed. So the total relative airflow increases.

In the book the sail is fixed, so the speed is limited by the fact that the relative airflow also changes direction. But in reality you can set the sail freely so the speed perpendicular across the wind is limited only by the efficiency of the boat. It can be much higher than the windspeed relative to the ground.

You can even outrun the moving airmass along the downwind direction:

 

[Arjan82]

Actually, I think the question is a bit misguided... The question is: "will you sail faster or slower than before". Regardless of the valid points made earlier that the relative wind speed will go up with cross-wind instead of down as is the case when you sail downwind, the angle of the lift force is also rotated. This means that a smaller component of the lift force is in forward direction, which is the only one contributing to the speed of the ship. So you have two effects which work in opposite ways. While it is clear that with cross-wind you have at least the potential to go faster than the wind, it is not clear at all if you will be actually faster than going downwind.

Having said that, it is actually true that you are faster with cross-wind. And if you bear-away* from the wind up to when you're sailing 'broad reach'*, when the wind is coming in at an angle greater than 90º, say 135º relative to in-the-wind, you will sail even faster still...

But I would say that the main reason for that is that your sail is acting like a blunt body when sailing down-wind and actually like a wing when sailing cross-wind. A sail works much more efficient when acting like a wing rather than like a blunt body. In other words, with the same relative wind speed a sail will generate much more lift when acting like a wing rather than like a blunt body.

The reason that you are even faster when sailing 'broad reached'* is that at that point the lift vector of the sail is still in forward direction but the sail is now acting as a wing, generating more lift. However, this is still not the whole story. Because the apparent wind, besides becoming stronger, will also change angle and will come progressively more from the forward direction as you increase speed. Since the sails only feel the apparent wind, you'll have to haul-in* your sails to adjust, making the lift vector point away again.

This takes on extreme forms with the extremely fast America's cup racing sailboats where they have the sail pretty much constantly in closed-hauled* condition, also when sailing 'broad reached'*. The America's cup sailboats also actually go well above wind-speed, often more than twice that. Even when sailing close-hauled! They will therefore never go directly downwind, since that limits their speed to the actual wind speed and for them it is much faster to tac* (or zig-zag) downwind.

But for most leisure type sailboats this effect is much less strong. They actually pretty much never sail faster than the wind (also because these ships are generally not designed to go faster than their hull-speed).

*) I'm not very familiar with English sailing terms (I'm Dutch) and how they are used, but Google said it is called this way... :)

 

[A.T.]

But I would say that the main reason for that is that your sail is acting like a blunt body when sailing down-wind and actually like a wing when sailing cross-wind. A sail works much more efficient when acting like a wing rather than like a blunt body. In other words, with the same relative wind speed a sail will generate much more lift when acting like a wing rather than like a blunt body.

Yes, but this is a very non-technical book for laymen. Lift (L) and drag (D) is one possible to decompose the resultant force (R). Normal (N) and axial force (A) is another:

Source: https://www.aerospaceweb.org/question/aerodynamics/q0194.shtml

The simple argument in the book is that when sailing across-wind at windspeed or even faster, you can still have N pointing in the right direction that propels the boat, while this is not possible when sailing directly downwind.

 

[Arjan82]

The simple argument in the book is that when sailing across-wind at windspeed or even faster, you can still have N pointing in the right direction that propels the boat, while this is not possible when sailing directly downwind.

True as it might, but that still doesn't explain why, if at all, you will actually go faster...

Oversimplifying to an extent that the physics changes is never helpful...

 

[anorlunda]

I am fond of a related fun question. "What is the maximum theoretical speed of a sailing vessel?" Note the word theoretical rather than practical.

The answer is that it is undefined. Speed is limited only by drag and friction. A very low friction vessel (like an iceboat) pointing up into the wind can go much faster to the wind, and there is no defined limit to how close to the wind it can get.

http://www.carbondragon.us/iceboat/speed_record.htm

That source claimed 150 mph (241 kph), but it is not official. Sorry, I don't have the time right now to look up the official record.

By the way, if sailplanes also count, you might be interested in this thread. In that case, the sailplane accelerates going downwind.

https://www.physicsforums.com/threads/the-835kph-sailplane-and-dynamic-soaring.999852/

 

[A.T.]

True as it might, but that still doesn't explain why, if at all, you will actually go faster...

It explains the ability to produce a forward force at higher speeds, which is the key geometric concept here.

Oversimplifying to an extent that the physics changes is never helpful...

The basic physics concept of how it's possible is not misrepresented here. Details like hull resistance and aerodynamic efficiency are the engineering part.

 

[A.T.]

http://www.carbondragon.us/iceboat/speed_record.htm

That source claimed 150 mph (241 kph), but it is not official. Sorry, I don't have the time right now to look up the official record.

Those old ice-boat record claims are not very reliable. Modern GPS tracked attempts are much slower. I don't have the current record, but a decade ago it was at 135 km/h.

However, recently the land sailing record was improved from 202.9 to 222.4 km/h:
https://www.abc.net.au/news/2022-12...d-speed-record-set-on-lake-gairdner/101761048

Here is a nice documentary about a previous record breaking land yacht:

 

[tech99]

Yes, but this is a very non-technical book for laymen. Lift (L) and drag (D) is one possible to decompose the resultant force (R). Normal (N) and axial force (A) is another:

View attachment 320175
Source: https://www.aerospaceweb.org/question/aerodynamics/q0194.shtml

The simple argument in the book is that when sailing across-wind at windspeed or even faster, you can still have N pointing in the right direction that propels the boat, while this is not possible when sailing directly downwind.

In this diagram I believe the boat is propelled by the lift force L, which acts at 90 degrees to the (apparent) wind, and not by the normal force N. The drag force D is counteracted by the keel/centreboard/hull which themselves provide lift in the water.

 

[A.T.]

In this diagram I believe the boat is propelled by the lift force L, which acts at 90 degrees to the (apparent) wind, and not by the normal force N.

If you mean the diagram on page 35 of the book in the OP, then your statement makes no sense. At the low angle of attack shown there, lift and normal force are very similar vectors, and they have a very similar component in the direction of travel, that propels the boat.

 

[erobz]

You can even outrun the moving airmass along the downwind direction:

I can't arrive at that. Imagine a cart on a frictionless track moving in a horizontal plane (top view) inclined at some angle $\theta$, and a $90^{\circ}$ vane (sail).

Begin by applying the "Momentum Equation" for flow with constant cross-sectional properties:

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

$cv$ is the control volume
$cs$ is the control surface - dashed box
the subscripts $o,i$on the momentum terms refer to the outflow, inflow respectively.

This becomes two equations in the direction $x,y$:

Force

$$ \sum F_x = -N \sin \theta $$

Accumulation

$$ \frac{d}{dt}\int_{cv} \rho v ~d V\llap{-} = \frac{d}{dt} \left( v_x\int_{cv} \rho ~ d V\llap{-}\right) = M\frac{d}{dt}( v_x )$$

$M$ is the total mass of the control volume ( cart + flow ) and it is constant assuming incompressible flow (continuity)

Momentum

$$ \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i = \dot m_{o_x} v_x - \dot m_{i_x} w $$

under continuity $\dot m_{o_x} = \dot m_{i_x} = \dot m$

and w.r.t. the inertial frame the flow momentum outflow is non-zero $\dot m v_{x}$ ( the flow is following the cart in the $x$ direction).

Putting it all together for the $x$direction:

$$ -N \sin \theta = M \frac{d v_x}{dt} + \dot m v_x - \dot m w \tag{1}$$

Repeat for the $y$ direction ( final result shown):

$$ N \cos \theta = M \frac{dv_y}{dt} - \dot m ( w - v_y ) \tag{2}$$

The objective is to solve the resulting system of linear first order ODE's by substitution using the track constraint:

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

$$(3) \implies \frac{d v_y}{dt} = \frac{d v_x}{dt} \tan \theta \tag{4} $$

I solved (2) for $N$ and sub into (1), then sub (3) and (4) ( eleminating $v_y$ and its derivative) to get the following ODE:

$$ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta ) + \dot m ( 1 + \tan \theta ) ( v_x - w) = 0 \tag{5}$$

Correction to (5)
$$ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta ) + \dot m ( 1 + \tan \theta )v_x - \dot m w ( 1+ \tan \theta) = 0 \tag{5} $$

Edit: The solution is not valid as mass flow rate has dependecy on $v_x$ as well.
With the substitution:

$$ u = \dot m ( 1 + \tan \theta ) ( v_x - w) \tag{6}$$

Implies

$$ \frac{du}{dt} = \dot m ( 1 + \tan \theta ) \frac{d v_x}{dt} \tag{7} $$

Sub (6) and (7) into (5):

$$\frac{M ( 1 + \tan^2 \theta)}{\dot m ( 1+ \tan \theta) } \frac{du}{dt} + u = 0 \tag{8}$$You get a solution of the form:

$$ u_f = u_o e^{-\beta t} \tag{9} $$

The conclusion of all this is in the limit as $t \to \infty$ the RHS $\to 0$, hence:

$$ \lim_{t \to \infty} u_f = 0 \implies \dot m ( 1 + \tan \theta ) ( v_x - w ) = 0 $$

In other words $v_x \to w$ as $t \to \infty$.

We don't have to find a solution to (5), just observe the steady state response:$$ \cancel{ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta )}^0 + \dot m ( 1 + \tan^2 \theta )v_x - \dot m w ( 1+ \tan \theta) = 0 \tag{6} $$

$$ \lim_{\frac{dv_x}{dt} \to 0} \implies v_x \to w\frac{1 + \tan \theta}{1 + \tan^2 \theta} \tag{7}$$

the limits of (7) in terms of $\theta$:

$$\theta \to 0 , v_x \to w$$

$$ \theta \to 90^{\circ} , v_x \to 0 $$

So, as far as I can see it's never going to outrun the wind $w$ in the direction of the wind.

$v$(the magnitude of the cart velocity) is given by:

$$v = \frac{w ( 1 + \tan \theta)}{ \cos \theta (1 + \tan^2 \theta)} \tag{8}$$
Where have I blundered that allows the cart to go faster than the wind in the direction of the wind?

 

[A.T.]

I can't arrive at that. Imagine a cart on a frictionless track moving in a horizontal plane (top view) inclined at some angle $\theta$, and a 90^{\circ} vane (sail).

....

Where have I blundered that allows the cart to go faster than the wind in the direction of the wind?

Not sure, but my quick guess is right at the start with the assumptions about the sail.
- Is the orientation of the sail fixed? Because the way it is drawn is good for intial acceleration but wrong for going with Vx > W, when the relative wind comes from north-east in your picture.
- Diverting the air by 90° corresponds to a lift/drag ratio of 1 at best, which is horribly inefficient.

The vectors for the situation look like below:

 

[erobz]

Not sure, but my quick guess is right at the start with the assumptions about the sail.
- Is the orientation of the sail fixed? Because the way it is drawn is good for intial acceleration but wrong for going with Vx > W, when the relative wind comes from north-east in your picture.
- Diverting the air by 90° corresponds to a lift/drag ratio of 1 at best, which is horribly inefficient.

The vectors for the situation look like below:

View attachment 320332

This thing about "lift". Where does that come from. The sail is a sheet of canvas?

 

[A.T.]

This thing about "lift". Where does that come from. The sail is a sheet of canvas?

Doesn't really matter. Even canvas sails can achieve a lift/drag ~= 5 as shown in my diagram. Rigid wing sails can do much better.

 

[anorlunda]

This thing about "lift". Where does that come from. The sail is a sheet of canvas?

But the backside has roughly the same shape as an airfoil.

Note that some modern racing sailboats have rigid sails. So the idea of putting up an airplane wing in place of the mast and cloth sails is not crazy. In the picture below, it looks like they took an entire airplane.

 

[erobz]

How something outruns what is pushing it is not an easy pill to swallow. I suppose lift is the hidden variable here for this to work because it’s not 100% relying on changing the momentum of the air impacting the sail…I guess.

 

[A.T.]

How something outruns what is pushing it is not an easy pill to swallow.

You don't really outrun the airmass in the sense of leaving it behind, because you stay embedded in it.

I suppose lift is the hidden variable here ...

The key is rather that you have contact with two objects (like air and water) which are moving relative to each other. The concept works beyond lift and fluids.

 

[A.T.]

... it’s not 100% relying on changing the momentum of the air impacting the sail…

Regarding this part: It is 100% relying on changing the momentum of the air. There is no magic here and even the simplest model with individual particles hitting the sail (very low density gas) can explain how it works:

 

[erobz]

Regarding this part: It is 100% relying on changing the momentum of the air. There is no magic here and even the simplest model with individual particles hitting the sail (very low density gas) can explain how it works:

This is all very nice animation but it’s severely lacking any mathematical rigor. In the cartoon, Wiley coyote runs in mid air chasing the Roadrunner…until he realizes it.

 

[erobz]

What is the maximum velocity of the cart in the following image?

 

[A.T.]

This is all very nice animation but it’s severely lacking any mathematical rigor.

The proof that it works is neither the animation nor math, but racing boats which do it for years (ice boats for a century). The animation just shows how trivial it is.

 

[erobz]

The proof that it works is neither the animation nor math, but racing boats which do it for years (ice boats for a century). The animation just shows how trivial it is.

Thats not a good argument in Physics which seeks to understand the physical phenomenon by way of mathematical model. If everyone said, "well, there it is...I don't need to model it". What would we know?

 

[A.T.]

What is the maximum velocity of the cart in the following image?

Same as for a boat with a fixed sail going directly downwind. But the boats that achieve downwind components greater than windspeed are not going directly downwind. Your original model was more relevant, just your sail model was bad.

Mathematical rigor is useless, if in order to formulate / solve your math, you have to make assumptions which completely change the scenario. I recommend using vectors in this case (see post #21).

 

[erobz]

Same as for a boat with a fixed sail going directly downwind. But the boats that achieve downwind components greater than windspeed are not going directly downwind. You original model was more relevant, just your sail model was bad.

Mathematical rigor is useless, if in order to formulate / solve your math, you have to make assumptions which completely change the scenario. I recommend using vectors in this case (see post #21).

I was trying to simplify in order to discuss maximum possible impulse.

There is no "real" sail that can completely reverse the momentum of the incoming flow, as shown in #29. It is the theoretical maximum possible impulse that can be applied to the boat in the direction of motion. There is no possible alteration of a vane which could impart any greater momentum to the cart in the $x$ direction than what is shown in #29. For the $x$ component of velocity $w$ is a hard limit. I've shown it with Newtons laws, Eulers, and Reynolds laws twice now for two independent vane configurations.

Real wind gusts, swirls, changes direction., any "true" measurement of downwind speed greater than the wind is going to be obscured by this reality. Conservation of Momentum is not.

 

[A.T.]

I've shown it with Newtons laws, Eulers, and Reynolds laws twice now for two independent sail configurations.

But neither one of your configurations corresponds to a boat with downwind component greater than windspeed.

Go back to your first one, but rotate the sail by 90° clockwise. What happens when Vx = W? Can you still accelerate?

 

[erobz]

But neither one of your configurations corresponds to a boat with downwind component greater than windspeed.

Go back to your first one, but rotate the sail by 90° clockwise. What happens when Vx = W? Can you still accelerate?

No it can not accelerate. The mass flowrate $\dot m = \rho A (w-v_x)$ entering the control volume always goes to $0$ as $v_x \to w$. The fluid jet can only just keep up with the cart, it can no longer impart any further momentum to it. If your hope was that the momentum outflow changes to $\dot m (v_y +w)$, sorry $\dot m \to 0$.