What limits the speed of a sail boat?

[Phrak]

What limits the speed of a sail boat?

Give a sail boat that, for whatever reason, doesn't heal and dump wind, what would limit it's speed?

For the sail, the lift and drag are proportional the the square of the velocity. The same goes for the keel. So as the force of drag increases with speed to slow the boat, the lift increases proportionally to pull it along.

 

[HallsofIvy]

When the wind strikes the sail, you can separate the winds velocity vector into two components: perpendicular to the sail and parallel to the sail. The component parallel to the just "slides" off and does not contribute any push to the sail. Further, unless you are going directly down wind, your sail will be at an angle to the motion of the boat. You now separate that "perpendicular to the sail" component into components that are perpendicular to the boat's motion and parallel to the boat's motion. The force perpendicular to the motion is blocked by the boats body and keel. It is, finally, the force parallel to the boats motion that accelerates the boat. As long as there is any non-zero component in the direction of the boat, the boat will accelerate. If, for example, the boat were sailing directly downwind, with the sail completely perpendicular to the boat, the entire force of the sail wind would be pushing the boat. As long as the boat is going slower than the wind there will be some non-zero component of wind accelerating the boat: the limit on the sail boat speed is the wind speed.

If the sail is set at all off the direction of the wind, the component of the wind perpendicular to the sail will be smaller that the full speed of the wind. Further, if the sail is not perpendicular to the boat, the component of that in the direction of the boat will be smaller yet- so the boat cannot go as fast as the wind. Contrary to what some people believe, a boat's fastest run is downwind and a sail boat cannot go faster than the wind. (On a tack, it may feel faster because there will be more wind crossing the boat than running with the wind where you feel very little wind at all.)

 

[cesiumfrog]

Contrary to what some people believe, a boat's fastest run is downwind and a sail boat cannot go faster than the wind.

Good grief, why are the mentors here still perpetuating this myth?

 

[cesiumfrog]

HallsofIvy, here is a https://www.physicsforums.com/showthread.php?t=274996" to which I'd like to draw your attention. The second post both explains the speed limit of sail boats and links a very well presented and qualified source for further detail. (If you would reopen those discussions then I think it is a place where we could resume addressing valid misconceptions.)

 

[schroder]

I think it would do everyone some good to reopen the thread in question, which was “locked pending moderation”. The locking was done even though there was no hostility or animosity in the thread, and some key questions were finally being addressed. NO evidence at all has ever been presented for DDWFTTW, none whatsoever! And yet, this myth continues to be perpetuated again and again all over the Internet, including this forum. A cart can go down wind, driven by the wind, and a cart can advance on a treadmill, driven by the treadmill. Nobody is seriously challenging that! But, a cart cannot go directly down wind, driven by the wind, at a faster velocity than the wind which is driving it! If it ever got close, it would be find itself pushing against that same wind as a headwind, in front of the cart! Can a tailwind push faster than its own headwind! Can a dog run faster than his own tail? It is high time to show that this DDWFTTW is nonsense, and I am glad to see that one of Physics Forums most qualified mathematicians, (HallsofIvy) has taken a position on this. Let the discussion begin!

 

[mgb_phys]

I think that's a different question.
A sail cannot do downwind faster than the wind is true - but that does not prove that downwind is the fastest a sail can go.

I'm not sure there is an obvious speed limit to a boat with a side wind.
The wind causes a pressure difference across the sail so a forward force, the force depends on the area of the sail and the sideways velocity of the wind. The only limit I can think of is when the speed is high enough that the air pushed along in front of the sail causes a high pressure that cancels out the low pressure are caused by the curved sail.

 

[schroder]

I think that's a different question.
A sail cannot do downwind faster than the wind is true - but that does not prove that downwind is the fastest a sail can go.

I'm not sure there is an obvious speed limit to a boat with a side wind.
The wind causes a pressure difference across the sail so a forward force, the force depends on the area of the sail and the sideways velocity of the wind. The only limit I can think of is when the speed is high enough that the air pushed along in front of the sail causes a high pressure that cancels out the low pressure are caused by the curved sail.

Yes, you are right. The OP is a different question altogether. I was responding in a knee jerk reaction to the reference to the other (closed) thread, and that old saw about DDWFTTW. It is no secret that as long as the sailing skiff is angling to the wind, it can achieve a speed which is quite a bit faster than wind speed. But of course, this speed does have a limit! As I understand it, the drag of the hull, as well as the pattern of the waves created by the hull, determine what the limit is, and it is about twice the wind speed. Notice I use speed, not velocity, since speed is an absolute value and not a vector. Whether or not this speed can be translated into a velocity made good, along the original line of wind direction, which exceeds the wind velocity along that same line, is still a subject of debate. I have not seen any convincing evidence that it has ever been done. The skiff would need to accelerate at an angle away from the direct line of the wind, then decelerate and tack back at an angle towards the true line, and accelerate again to the finish. Even with a maximum burst speed of 2 times wind velocity within the legs, the skiff is covering a much greater distance, and it does not seem realistic that the overall velocity will ever exceed constant wind velocity along the true wind direction. Also, constant wind velocities and constant wind directions are very hard to find, and so the issue is hard to prove in any case. Sailing directly down wind, the skiff cannot ever achieve wind velocity, only reaching about 70% on average. As for those carts, the less said the better!

 

[Carid]

In a sailing boat the important wind is the apparent wind.
If you sail at right angles to the true wind the apparent wind is at a lesser angle due to the boat's own motion. (Sum of vectors).
For high performance catamarans this is the eventual limiting factor. The wind goes too far aft.

 

[mgb_phys]

So the fastest sail would have to have a large area (more force) but be very narrow so the cross wind travel across it before there is significant forward motion - so something like a glider (sailplane) wing mounted vertically.
Presumably for a pure sidewind you wouldn't need a keel and so less drag - the boat would capsize immediately if you turned but that's somebody else's problem.

For a land yacht (lower friction than a hull) the speed record is 116mph
Cool looking attempt to beat it - http://news.bbc.co.uk/2/hi/technology/7610786.stm

 

[rcgldr]

This site has some numbers:
http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

For a sailcraft, the apparent crosswind on a sailcraft depends only on the sailcrafts heading and the true wind speed, and is independent of the sailcrafts forward speed, since the apparent crosswind is perpendicular to the direction of travel (and therefore independent of sailcraft speed).

From page 4 of the pdf linked to above: An iceboat heading 30 degrees offset to an 18mph wind, total speed 70mph and apparent wind speed 55mph, which I adjusted to 55.15mph (so heading was 30 degrees). Note that Beta is 9 degress, not 8 as shown in the diagram so there's at least one mistake there. Back to the math:

The apparent crosswind is 9mph (sin(30) x 18mph), and the iceboat could achieve an apparent wind speed of 55.15mph, composed of a 9mph apparent crosswind and a 54.4mph apparent headwind for a L/D of about 6:1. Downwind, the iceboats total speed was given as 70mph and it's VMG downwind would be 60.6mph (cos(30) x 70mph). Upwind, the iceboats apparent wind speed remains the same at 55.15mph. It's total speed would be > 38.8mph, and VMG upwind would be > 33.6mph, assuming the drag from the ice would be a bit less at the slower true speed.

In the downwind case, the sail diverts part of the apparent headwind directly upwind, so that the true wind is slowed down by interaction of the sail. In the above case where the VMG was 60.6 mph with an 18 mph wind, the component of 70 mph apparent headwind deflected directly upwind would need to be > (60.6 mph - 18 mph =) 42.6 mph.

I'm curious to find out if there's a way to approximate the air flow velocities in the vicinity of a moving sail (or wing), or a better mathematical model or explanation of what's going on.

 

[Phrak]

In a sailing boat the important wind is the apparent wind.
If you sail at right angles to the true wind the apparent wind is at a lesser angle due to the boat's own motion. (Sum of vectors).
For high performance catamarans this is the eventual limiting factor. The wind goes too far aft.

Hi Carid. This is about what I expected, but why?

 

[Phrak]

When the wind strikes the sail, you can separate the winds velocity vector into two components: perpendicular to the sail and parallel to the sail.

In aeronautics, compontents are usually given in the lift-drag system of coordinates. It makes things simpler. Perpendicular vs. parallel velocitites WRT the sail would be obtained using angle of attack. Perhaps nautical people uses different methods?

Lift is defined as the force perpendicular to the relative free-air stream.
Drag is defined as the force oppositly directed from the free-air stream.

 

[Phrak]

So the fastest sail would have to have a large area (more force) but be very narrow so the cross wind travel across it before there is significant forward motion - so something like a glider (sailplane) wing mounted vertically.
Presumably for a pure sidewind you wouldn't need a keel and so less drag - the boat would capsize immediately if you turned but that's somebody else's problem.

The keel is a water foil. It does in the water what the sail does in the air. When tacking, both are needed. Exceptions might be a cat, where the hulls serve as (rather lousy aspect ratio) foils, I guess. Does a cat have a keel at all?

 

[Phrak]

Jeff, I looked over the "Downwind Angles, Skeeter" drawing. If boat speed means speed over the ice, then the vectors are wrong. V_a and V_b are swapped.

To keep the vector labels unchanged on the drawing, then:

V_t = velocity of air with respect to the ice. (or just "wind velocity")
V_a = velocty of boat with resect to the ice. (or just "boat velocity")
V_b = velocity of air with respect to boat. (or "apparent wind")

 

[rcgldr]

In aeronautics, components are usually given in the lift-drag system of coordinates.

The problem with a sailcraft is that it interacts with the air and the ground (or water), so you have two sets of components, one set relative to the wind, the other set relative to the ground.

The keel is a water foil. It does in the water what the sail does in the air. When tacking, both are needed. Exceptions might be a cat, where the hulls serve as (rather lousy aspect ratio) foils, I guess. Does a cat have a keel at all?

Some cats have deployable "daggerboards", that can be raised or lowered. I've read that the daggerboards are mostly used for upwind tachs, but I also read that top speeds are faster with the daggerbords. The keel in a sailboat uses weight below the boat to counter the torque from the wind on the sail. Cat's don't require the weight, which is why they are faster than monohull sailboats.

I looked over the "Downwind Angles, Skeeter" drawing. If boat speed means speed over the ice, then the vectors are wrong. V_a and V_b are swapped.

Looks correct to me, and it's what I based my example numbers from. Ice boat speed 70 mph, apparent wind speed 55.15 mph, true wind speed 18 mph. This corresponds to the iceboats heading being 30 degrees offset from the true wind. The angle of the apparent wind (relative to iceboat) is arctan (9 mph / 54.4 mph) ~= 9.4 degrees (Beta). The angle of the sail (relative to sailcraft) would have to be smaller still.

 

[Phrak]

Looks correct to me, and it's what I based my example numbers from. Ice boat speed 70 mph, apparent wind speed 55.15 mph, true wind speed 18 mph. This corresponds to the iceboats heading being 30 degrees offset from the true wind. The angle of the apparent wind (relative to iceboat) is arctan (9 mph / 54.4 mph) ~= 9.4 degrees (Beta). The angle of the sail (relative to sailcraft) would have to be smaller still.

My Mistake! I read the drawing wrong. The craft is sailing downwind. I presumed it was sailing upwind.

 

[rcgldr]

I wasn't questioning your numbers

Not my numbers, I'm just using the ones from that pdf file, filling in some digits to get the heading angle 30 degrees (adjusting 55 mph to 55.15 mph).

"Apparent wind" = air velocity with respect to the boat
"True wind" = velocity of air with respect to the water

agreeed

From this, you can see, that if the wind is comming from the north, and your velocity over the ice is northward, the apparent wind speed better be greater than your speed over the ice.

The way I see the diagram, you have 3 vectors. A true wind vector, heading 180 (south) at 18 mph, the boat velocity vector, heading 150 (30 degrees east of south) at 70 mph, and the apparent wind, heading 320.6 (39.4 degrees west of north), 55.15 mph. Apparent wind vector = true wind vector - iceboat velocity vector.

 

[Phrak]

The way I see the diagram, you have 3 vectors. A true wind vector, heading 180 (south) at 18 mph, the boat velocity vector, heading 150 (30 degrees east of south) at 70 mph, and the apparent wind, heading 320.6 (39.4 degrees west of north), 55.15 mph. Apparent wind vector = true wind vector - iceboat velocity vector.

Jeff. I hastily corrected my last post, when I discovered my mistake (I assumed the craft was sailing upwind.). But I was way late. Sorry about that.

 

[Phrak]

I see this had roots in the 'DDWFTTW' business. (which is very much possible according to Newton, but not the way most people have in mind.)

 

[rcgldr]

I see this had roots in the 'DDWFTTW' business. (which is very much possible according to Newton, but not the way most people have in mind.)

The basic principle of any wind powered device is that it has to "slow" down the wind in order to use the wind as a power source. A sailcraft diverts enough of the apparent wind upwind that the net effect of the sailcraft's interaction with the air is to slow it down with respect to the ground the sailcraft travels on, so that even though the sailcraft's component of downwind speed may be greater than the true wind, the true wind is slowed down by interaction with the induced upwind wash off the sailcraft's sail.

The same principle applies to a DDWFTTW cart, the cart generates an upwind thrust speed at a fraction (< 1) of it's forward ground speed, allowing the cart to go DDWFTTW, although it's upper limit will be lower than a sailcraft, in my opinion (one issue is that the "operating" ground force opposes motion for the cart, but is perpendicular to the motion for a sailcraft).

 

[Phrak]

In a sailing boat the important wind is the apparent wind.
If you sail at right angles to the true wind the apparent wind is at a lesser angle due to the boat's own motion. (Sum of vectors).
For high performance catamarans this is the eventual limiting factor. The wind goes too far aft.

In the attachment there's a force-velocity diagram that seems to confirm your statement.

The difficult part has been to determine what should remain constant, and what should vary.

The terminal speed of the boat through the water is reached when the sum of the forces of lift and drag sum to zero—the acceleration is equal to zero.

Apparently, the best speed of the boat through the water, for any given tack, is obtained when the speed of the boat through the water is equal to the speed of the boat through the air. This is the scenario examined to analyse terminal velocity. (Does it hold true for any tack?)

There are a few assumptions that are made, such as the angle of attack of sail and keel remain constant—so that the L/D of both sail and keel are each constant.

I've concluded that when the magnitude of V_{B-W} is held equal to the magnitude of V_{B-A} , then V_{B-W} will reach its terminal velocity when alpha+beta=gamma. (see attached file)

 

[rcgldr]

Apparently, the best speed of the boat through the water, for any given tack, is obtained when the speed of the boat through the water is equal to the speed of the boat through the air.

Do you mean the overall speed or only the component of speed parallel to the wind? Note that in the example from the link to the pdf file the iceboats speed was 70mph, with an apparent wind of 55.15 mph, not the same. The wiki article includes an unverified claim that catamarans can go 1.5 times wind speed. Assuming this is a near perpendicular tack, then the water speed is 1.5 times the true wind speed, while the apparent wind would be 1.8 times the true wind speed.

Ignoring drag, the best tack for maximum total speed, is Beta (the smallest angle between apparent wind and sailcraft direction that the sailcraft can maintain) degrees downwind of perpendicular. One of the other threads included the best tack for maximum downwind speed:

Hi Jeff. I made some mathematical calculations based on the "sailing" vector diagrams and making the over-simplifying assumption that the craft can always achieve a speed such that the apparent wind reaches a fixed minimum angle relative to the crafts direction. I know this assumption is not really justified in *practice but it's a useful starting point.

The results were interesting in their simplicity and symmetry between the headwind and tailwind cases. (BTW, here I'm considering simple sailing craft such as a sailboat or iceboat, w = wind velocity magnitude and $\phi$ is the closest angle that the craft can make to the apparent wind.)

Upwind. The upwind VMG (upwind velocity component) is maximized when the craft direction is $\pi/4 + \phi/2$ off the headwind and the resulting maximum upwind VMG is $w (\mbox{\rm cosec}(\phi) - 1) / 2$

Downwind. The Downwind VMG is maximized when the craft direction is $\pi/4 - \phi/2$ off the tailwind and the resulting maximum upwind VMG is $w (\mbox{\rm cosec}(\phi) + 1) / 2$

Note that with these results the difference between the maximum downwind VMG and the maximum upwind VMG is precisely equal to the windspeed W, exactly as mender hypothesised in the start of this thread. :)

* Though these calculations keep the closest the angle to the apparent wind as a constant, the relative magnitudes of craft speed and wind speed vary greatly depending sailing direction. So this assumption cannot be true in practice.

** cosec(x) = 1/sin(x)

 

[Phrak]

Do you mean the overall speed or only the component of speed parallel to the wind? Note that in the example from the link to the pdf file the iceboats speed was 70mph, with an apparent wind of 55.15 mph, not the same. The wiki article includes an unverified claim that catamarans can go 1.5 times wind speed. Assuming this is a near perpendicular tack, then the water speed is 1.5 times the true wind speed, while the apparent wind would be 1.8 times the true wind speed.

Ignoring drag, the best tack for maximum total speed, is Beta (the smallest angle between apparent wind and sailcraft direction that the sailcraft can maintain) degrees downwind of perpendicular. One of the other threads included the best tack for maximum downwind speed:

One of the assumptions I've made is that the L/D is the same in air and water. D includes parasitic drag in both cases. For instance the drag of the hull through the water would be included in the force I've labeled K_D. For an ice craft the L/D of the skates will be a great deal better than the sail--but it's not zero.

As an example, in my attachment where the two L/D's are equal, I used L/D=12/5, true wind was 10, speed of the boat through the water was 13, and the apparent wind speed was also 13. With the assumption of equal L/D's, maximum speed of the boat through the water is obtained when the boat progresses atan(5/12) degrees downwind from directly across the wind.

I don't follow uarts terminology, so I can't comment.

I fixed-up the drawing slightly to make these things more apparent.

 

[rcgldr]

I don't follow uarts terminology, so I can't comment.

There's a smallest angle between the direction a sailcraft can sustain versus an apparent near headwind, and this is called Beta. Beta = tan(apparent crosswind / apparent headwind). Uart called it $\phi$. Does this help?

 

[Phrak]

There's a smallest angle between the direction a sailcraft can sustain versus an apparent near headwind, and this is called Beta. Beta = tan(apparent crosswind / apparent headwind). Uart called it $\phi$. Does this help?

Great. I got it. By best convention, the wind comes from the north. Best speed according to uart is at 90 degrees plus (1/2)phi. This is what I got, eventually, so I'm happy.

The only cavet is that we assume the L/D of sail and keel don't change substancially around this critical direction. Ideally, if speed is your thing, you want to the sail in the low drag pocket, if there is such a thing for two sided sails. You'd want the pocket to cover the region where you expect beta will fall.

Where you've said the best direction is beta degrees downwind from perpendicular to the wind, you should have said half of beta. No big thing.

This beta angle is really quite a useful value. I hadn't taken it seriously enough. Beta and and wind speed will tell you all you need for any given heading, given you keep in mind that beta is a function of the L/D's of both sail and keel.

Given all this, the maximum speed of the boat over the water is about

$$\frac{V_{true.wind}}{2} = V_{boat} \ tan{\left( \frac{\beta}{2} \right)$$

$$\beta = arctan(LD.sail) + arctan(LD.keel)$$

(The reality is something of a disappointment, but expected. I guess I can't fly a sail over a keel arrangment and expect to approch the sound barrier anytime soon. But in a hurrican and a beta=8 degrees...)

 

[rcgldr]

Where you've said the best direction is beta degrees downwind from perpendicular to the wind, you should have said half of beta.

I meant best direction for maximum speed, not maximum VMG made downwind. Uart's formula differs from what I've read on some sailing web sites, so now I'm not sure which formulas are correct.

 

[Phrak]

I meant best direction for maximum speed, not maximum VMG made downwind. Uart's formula differs from what I've read on some sailing web sites, so now I'm not sure which formulas are correct.

So many new terms. I don't sail, so I don't use them. I took uart to mean the maxium velocity of the boat through the water since his equation for downwind gave the answer for it.

Given a constant beta and a constant wind, sovling this sailing formula geometrically is pretty simple. Solving it mathematically requires the Pythagorean theorem. From trigonometry, you need three values of eihter sides or angles, to get the triangle's remaining sides and angles. We're given two values: the wind speed and beta. This means that as one of the remaining sides of the triangle vary in length the triangle changes shape.

It turns out that an isosceles trangle give you the greatest speed of the boat through the water, when you don't care about heading. The wind speed is the base of the triangle.

 

[rcgldr]

So many new terms. I don't sail, so I don't use them. I took uart to mean the maxium velocity of the boat through the water since his equation for downwind gave the answer for it.

VMG means velocity made good. Rather that use the boat actual speed, VMG is the average velocity to reach a specifc target, which may be a point or a line. In UART's case the target is a line directly downwind or upwind from the boat. The VMG downwind speed is cos(relative heading) times boat speed. I was looking for maximum boat speed regardless of heading.

The second diagram from the sailcraft link before shows an ice boat with a heading 30 degrees offset from downwind, achieving a speed of 70 mph with an 18 mph wind (apparent headwind of 55.15 mph), with a VMG downwind speed of 70 mph x cos(30) = 60.6 mph. Beta in this case is 9.4. Using his formula, the heading should be (pi/4 - beta/2) or 40.3 degrees offset from downwind, which gives a boat speed of 80 mph and downwind speed of cos(40.3) x 80 mph = 64.1 mph.

Maximum boat speed seems to occur at (90 - beta) = 80.6 degrees from downwind, for a theoretical boat speed of 110.2 mph (this is assumes that a beta of 9.4 would still hold at this high speed). The apparent crosswind in this case is 18 sin(80.6) = 17.76 mph. The apparent headwind would be 17.76 / tan(9.4) = 107.27 mph. The apparent wind would be 108.73 mph. The VMG downwind equals the true wind = 18 mph.

 

[Phrak]

Jeff, I was so sure you were wrong about the maximum speed of a sail boat occurring at $$\ 90 + \beta$$ degrees from the direction of the wind, rather than $$\ 90 + \beta / 2$$.

The equation of vectors is

$$\boldf{V_{a}} + \boldf{V_{b}} = \boldf{V_{t}}$$

$$\ \boldf{V_{b}}$$ – velocity of boat (measured as the velocity of the boat over the water)
$$\ \boldf{V_{t}}$$ – true wind (the velocity of the air over the water)
$$\ \boldf{V_{a}}$$ – apparent wind (velocity of wind as measured by the boat)

I couldn't find an elegant mathematical solution. A geometric method came to mind. So I stuck a couple straight pins in a board in vertical line. I called the distance between the pins $$\ V_t$$. The true wind blows from the top pin down.

The particular $$\ \beta$$ wasn't important, so I stuck the sharpest end of a 30-60-90 draftsman's triangle between the pins, from the left, touching each pin. This gave me a $$\ \beta$$ of 30 degree. $$\ \boldf{V_b}$$ points from the top pin to the 30 degree vertex . $$\ \boldf{V_a}$$ points from the 30 degree vertex to the bottom pin .

Sliding the triangle around, always touching the pins, the point of the triangle describes a circle. I don’t know why; but a nice development. You should try it. $$\ V_b$$, the speed of the boat doing its fastest, regardless of direction, turns out to be equal to the diameter of the circle. Knowing the diameter simplifies every other problem about best heading for upwind and downwind speed.

So you were right, hurrah for Euclid, and I’ll be ding-donged.

 

[rcgldr]

maximum speed of a sail boat occurring at (90 - beta) degrees from the direction of the wind

I couldn't find an elegant mathematical solution.

I just used a spreadsheet and noticed where the peak values were occurring. However I've since figured out the math for an ideal ice boat (where beta is independent of alpha and wind):

alpha = angle between heading and true wind
beta = angle between heading and apparent wind
beta = atan(Vac / Vah)

Vb = velocity of boat
Vw = velocity of true wind
Vac = velocity of apparent crosswind
Vah = velocity of apparent headwind

Vac = Vw sin(alpha)

beta = atan(Vac / Vah)
tan(beta) = Vac / Vah
Vah = Vac / tan(beta)

Vb = Vah + (Vw cos(alpha))
Vb = (Vac / tan(beta)) + (Vw cos(alpha))
Vb = ((Vw sin(alpha)) / tan(beta)) + (Vw cos(alpha))
Vb = Vw ((sin(alpha) / tan(beta)) + cos(alpha))
Vb = Vw ((sin(alpha) cos(beta) / sin(beta)) + cos(alpha) sin(beta)/sin(beta))
Vb = Vw ((sin(alpha) cos(beta) + cos(alpha) sin(beta))/sin(beta))
Vb = Vw ((sin(alpha + beta) / sin(beta))

setting the derivative = 0

d(Vb)/d(alpha) = Vw ((cos(alpha + beta) / sin(beta)) = 0
Vb is a max when cos(alpha + beta) = 0
Vb is a max when alpha = 90 - beta

Also it's clear that sin(...) is a max when sin(...) = 1
So Vb is a max when sin(alpha + beta) = 1
or Vb is a max when alpha = 90 - beta

 

[Phrak]

I couldn't sort-out your math.

Our terminology is getting in the way of communicating, which is why I could say 90+beta, and it means the same when you say 90-beta. One of the problems in defining angles. This seems to originate where the angle between two lines or vectors could be taken to mean a particular angle, or its complimentry angle.

I've found Microsoft Word is pretty easy at drawing simple diagrams that aren't too crude, once you get started. It relys heavy on right mouse clicks to select operations.

On a second note, all of these models we've come-up with are flawed, as the sail and keel don't stall. Upwind, stalling is called 'going into 'irons', I'm sure you know. I don't know what the downwind problem is called. Geometrically, you can see what happens. For sailing upwind, for instance, the sail is doing nicely in a fast apparent wind, but the keel is moving through the water relatively slowly. To balance the force from the sail, the keel is at a large angle of attack. Beta, as defined in this link

http://www.nalsa.org/Articles/Cetus/Iceboat%20Sailing%20Performance-Cetus.pdf"

grows large, and fairly quickly thereafter the keel stalls. It's the sail that stalls going downwind. This is why beta isn't actually constant, but varies for a sail boat and an ice boat as well. You should run that boat program I sent you, if you haven't. It isn't actually a 'toy' program--all the vector are there, and some work went into accurately modeling the LD curves of keel and sail. So it does model going into irons.

 

[rcgldr]

I couldn't sort-out your math.

Assume the wind is traveling left to right on a X Y graph, then alpha is the same as theta in polar coordinates.

Alpha is the angle between the ice boat heading and the true wind. If the ice boat is heading directly downwind, then alpha = 0. Increasing alpha means counter clockwise, decreasing alpha means clockwise. Positive alpha corresponds to a "positve" apparent crosswind (left to right), while negative alpha corresponds to a "negative" apparent crosswind (right to left).

To get the simplified form, I took advantage of the fact that

sin(a+b) = sin(a) cos(b) + cos(a) sin(b)