Talk:High-performance sailing/Archive 3

Simple Diagram
Many (including some editors) seem to have a problem understanding how achieving a downwind-VMG greater than true-wind-speed is possible. I propose to include a introductory vector diagram, that shows how the sail can produce a force with a positive forward-component that propels the boat, under this conditions. I cannot upload files yet, so I put the proposed diagram here:. If the diagram needs improvement, I can change it. Eyytee (talk) 22:13, 7 June 2010 (UTC)


 * Your diagram is very good! Also the two force diagrams at the bottom are illuminating. Maybe you could add the lateral force of the water acting on the keel/hull/dagger board balancing the large lateral component of the sail force, which is essential for the whole argument (without it the hot-air baloon argument would be correct). In addition I much agree with your first sentence. MrBeanBob (talk) 23:38, 7 June 2010 (UTC)

Good start, but the force diagram needs a drag vector, so the true boat velocity can be derived, not just a given...--Paul (talk) 00:49, 8 June 2010 (UTC)


 * There is no need for that (nice as it would be). All they need show is that there is a forward component of the force on the sail in conditions where the free floating balloon is being beaten to downwind.


 * This is exactly what the proposed vector diagram shows. A forward component of the force on the sail while going with a downwind VMG = 1.5 true wind speed.


 * You don't mean VMG (i.e. velocity made good to the desired [directly downwind] point), you mean speed over the ground.


 * This is factually not true. The proposed vector diagram clearly shows a downwind VMG = 1.5 true wind speed.


 * I never said that speeds faster than true wind velocity were impossible. And that isn't a proper vector diagram showing proper decomposition of forces.  It's just a sketch supposedly showing what you want it to.  Paul Beardsell (talk) 15:13, 8 June 2010 (UTC)


 * There is none. The forward force on the vessel disappears when (speed over the ground)/cos(angle to the true wind)=(speed of the true wind).  That's the argument in a nut shell.  Impossible.  And any CORRECTLY drawn vector diagram makes this plain.  Paul Beardsell (talk) 08:18, 8 June 2010 (UTC)


 * Mr. Beardsell: if I understand that statement as written, then what you are saying is equivalent to saying that the maximum (speed over the ground) = cos(angle to the true wind) * (speed of the true wind).  Am I correct that that is indeed what you are saying?  At what points of sail do these statements hold true?  At what points of sail do they not hold true?  Could you check your original formulation to see if you meant to multiply by the cosine instead of divide? Sirclicksalot (talk) 10:33, 14 June 2010 (UTC)


 * Paul Beardsell seems to suggest that the proposed vector diagram is NOT CORRECTLY drawn. Yet he fails to explain what exactly is not correct in the diagram. He merely makes assertions without providing proof or references.


 * It's like when something is incorrectly spelt in English. The literate ones don't need to be told the correct spelling.  Those mathematically and physically illiterate cannot be explained to, and those that are don't need to be told.  But, if you need a few pointers (ha!), start by properly labelling everything, say what is being decomposed into what, make the lengths proportionate to the speeds, if that is what you're drawing, or forces, if that is what your drawing.  You're strumming on a guitar without having even tuned it.  Paul Beardsell (talk) 15:13, 8 June 2010 (UTC)

what a properly drawn diagram shows
At the point where (speed over the ground)/cos(angle to the true wind)=(speed of the true wind) the sail provides no forward force (i.e. in the direction of the keel / wheels / skates). The slightest friction brings the vessel gradually to a halt. But, if all friction is eliminated the downwind point is reached at EXACTLY the same time as the free floating balloon. Paul Beardsell (talk) 08:33, 8 June 2010 (UTC)


 * This statement by Paul Beardsell is Original Research. It is not supported by any citation.  It is also incorrect, as shown by the citations that Paul deleted from the original article.  I will post more on this later.--Gautier lebon (talk) 08:46, 8 June 2010 (UTC)


 * It isn't OR. It's Physics 101.  And even if it were OR, we are on a Talk page, and we can write here what is NOT allowed in the encyclopedia pages.  I will not allow OR there.  And it is not me who wants to put WP:OR and WP:SYN material there but others such as GL.  Anything there has to be properly cited as per WP:V.  Here, we can talk and argue but EVEN IF WE REACH A CONSENSUS what goes on the page itself has to be WP:V.  Paul Beardsell (talk) 08:58, 8 June 2010 (UTC)


 * It is not Physics 101, and you have not provided a citation to show that it is. Worse, you have added to the main article the statement “Nevertheless this remains slower than the time taken for a free floating balloon to travel the downwind track directly”.  This is Original Research, not supported by any citation, in the article itself.--Gautier lebon (talk) 09:11, 8 June 2010 (UTC)


 * I will continue to remove material from the actual article which does not satisfy WP:V. Paul Beardsell (talk) 08:58, 8 June 2010 (UTC)


 * It seems many of the rest of you cannot do force diagrams properly, and seem to want to disregard 2nd law of thermodynamics, as if science is not a matter of fact but merely what you would like to believe. Paul Beardsell (talk) 08:58, 8 June 2010 (UTC)


 * It might be the other way around, given that all text books support the force diagrams that Paul Beardsell is criticizing.--Gautier lebon (talk) 09:11, 8 June 2010 (UTC)


 * No they don't or there's a revolution in physics happening and it all started in Yachting Monthly. Paul Beardsell (talk) 15:43, 8 June 2010 (UTC)

VMG greater than windspeed section does not conform to Wikipedia specifications
First and foremost, there really don't seem to be any RELIABLE, VERIFIABLE sources. Secondly, what sources there are appear to be reports of original research, none of which has made it through a peer review process since they first appeared in November of 2008.

Wikipedia's recommendations are: Exceptional claims require exceptional sources

Certain red flags should prompt editors to examine the sources for a given claim:

* surprising or apparently important claims not covered by mainstream sources;

* claims that are contradicted by the prevailing view within the relevant community, or that would significantly alter mainstream assumptions, especially in science, medicine, history, politics, and biographies of living persons. This is especially true when proponents say there is a conspiracy to silence them.

Exceptional claims in Wikipedia require high-quality sources.

If such sources are not available, the material should not be included.

The major "sources" in the section are to self-published videos demonstrating original research.

This "issue" has been publicized since at least 2008 with no peer-reviewed publications of such a spectacularly counter-intuitive notion. It is VERY hard to believe that such a scientific breakthrough could remain an orphan with no better explanation than a very dubious "thought experiment".

As indicated in the Wikipedia recommendations for such unauthenticated exceptional claims, this section should not be included.

Possibly a supporting site - but still not up to what I suspect Wikipedia needs - ((http://www.idniyra.org/articles/IceboatSailingPerformance.html)). Polars at ((http://www.west.net/~lpm/hobie/archives/v1-i3/feature3.htm)) show an Aussie 18 almost making downwing VMG of windspeed.

In fact, several sections in the article - The last 2 paragraphs and table in Sailing perpendicular to the wind are erroneous and speculative. Sailing on a broad reach, velocity made good are similarly speculative and unsupported. Aerodynamic drag on the sail is completely ignored. Boat speed is NOT a simple multiple of apparent wind.
 * Sailing Perpendicular to the Wind from paragraph 5, beginning "If hull speed is not a limiting factor, and if the strength of the wind is sufficient to overcome the surface resistance....." should be removed - erroneous and speculative.
 * Sailing on a Broad Reach from paragraph 4, beginning "Suppose that a boat ...." should be removed - erroneous and speculative.
 * Vector diagrams and formulas should be removed - erroneous and speculative. Once again makes the assumption that boat speed is a simple vector of apparent wind speed and ignores aerodynamic drag.
 * Speed made good from the 5th sentence of paragraph 5, beginning "Again, simple trigonometry can be used to calculate the velocity made good." should be removed. I didn't find any references to AC course times in any of the AC references (32, 34 - 38). Some of those pages are now gone - therefore no VERIFIABLE sources - do you have others? Ref 31 repeats one of the refs I cite above - still only 1 possible verification. I disagree with your interpretation of ref 41 and point to the same author illustrating actual data in the second citation I make above. Another sentince in this section - "Of course real boats cannot equal those performances,[30] although iceboats can come close to them." indicates the section is speculation, not reality. If tables are acceptable, they should be based upon real data, not hypothetical calculations based on a poor understanding of aerodynamics.


 * You appear to be starting from the assumption that VMG greater than wind speed cannot be achieved, and thus propose to delete various bits. How about starting from the assumption that VMG greater than wind speed can be achieved, and then proposing ways to clarify the article so that it is easier to understand?  A team of people worked for months to do that, but further improvement is surely possible.  How can you interpret the polar chart in Ref 41 in any other way than showing downwind VMG greater than wind speed?  You are correct to state that the USA 17 data is no longer readily available on the 2010 Cup site.  I had not realized that it had been removed.  I will restore it.  You will find the partial timings in older versions, for example [].  The "no hull speed" limitation is not speculative, it is the actual situation of an iceboat.  It is easier to understand this stuff if you think of an iceboat.  Then to transpose that to a high-performance catamaran or planing hull.--Gautier lebon (talk) 11:41, 14 June 2010 (UTC)


 * Sailing dead downwind faster than the wind for reasons already stated needs removing.

I don't currently have library access to the book, "High Performance Sailing". These citations need to be checked and perhaps verified against second sources - even book authors can be wrong!


 * They are supported by numerous other sources, please see below.--Gautier lebon (talk) 11:41, 14 June 2010 (UTC)

No problems with sailing faster than the wind - just with a downwind velocity made good greater than windspeed.--QuietJohn (talk) 06:17, 14 June 2010 (UTC)


 * The cited book by Bethwaithe is a perfectly reliable, high-quality source, that summarizes years of data and developments. The NALSA web site contains many references and data that show that iceboats routinely go downwind at VMG greater than wind speed, and it also explains why this is possible.  The Gautier lebon (talk) 09:09, 14 June 2010 (UTC)

ALmost all of the citations I have checked have been dubious or inaccessible to me. If the NALSA site (which I have looked at and searched for relevant data) has the supporting data, please CITE the web pages that support your claims. I'm quite neutral about the DDWFTTW phenomenon. It doesn't make sense to me, and I haven't found any data to convince me otherwise, so it's either wrong, or the page isn't providing the necessary information. Guide me to the polars- or other corroborating data and I'll be happy to concede. Until then, much of this article isn't appropriate for a Wikipedia entry.--QuietJohn (talk) 09:43, 14 June 2010 (UTC)


 * The Bethwaite book is easily available. The fact that you do not have it does not mean that it cannot be used as a citation in Wikipedia. The polar chart at  shows downwind VMG greater than windspeed and it is consistent with the polar charts published by Bethwaite.  As stated many times on this talk page, and in the article, the published data from the first regatta of the 2010 America's Cup shown that USA 17's downwind VMG was far in excess of wind speed.  As stated in the article, the sandyacht data at  also show that.  As does the iceboat data at .  And it is all explained at the bottom of the page at .  So you have multiple sources providing data that confirm the physics explained in the article.  And that physics is not Original Research, it is simply a summary of the physics found in most text books on sailing.--Gautier lebon (talk) 11:28, 14 June 2010 (UTC)


 * I disagree that the material in this article is in conflict with wiki guidelines. The material discussed here is standard, well-known to any expert on the physics of sailing or wind power and to many informed amateurs.  The fact that some people find it non-intuitive or have trouble accessing the references isn't relevant.  There are many published sources, although I think the article is sufficiently referenced as it stands now.  If you want a published book reference for ddwfttw, this one is very good http://www.amazon.com/Symmetry-Sailing-Physics-Yachtsman/dp/1574090003 (and it also discusses sailing with a dw vmg>ws).  Peer review is going to be hard to find, because no one would bother (or be able to) publish an article in a scientific journal on something this basic.  It would be rejected by the editors (before the peer review stage) as too obvious and not of sufficient scientific interest. Waleswatcher (talk) 11:57, 14 June 2010 (UTC)

Sailing downwind faster than the wind
I know that this section seems hard to believe. Please read the citations and study them carefully before posting comments to the effect that this section is obviously wrong. I too did not believe that this was possible, until I did the research and found the citations given in this section. Others (including a physicist) were also skeptical, but then became convinced, see and .--Gautier lebon (talk) 08:39, 10 December 2009 (UTC)

I've posted some responses in talk:sailing. Another interesting topic is sailing directly into the wind. I've never seen a turbine / prop craft "work" directly upwind in what I consider to be a controlled test - the so-called proof has been less than convincing. The "spork33" device can be simply adapted to demonstrate that this is possible on land. It's just inefficiency of props and turbines that make it hard to get something to work on water, as there's no fundamental law to say it can't work. In a thought experiment, this can be extended to make something work on water by say dropping and raising large sea anchors, and using energy from the wind through a gearing mechanism to winch the boat forward against the sea anchor, but necessitating the use of a complex device to drop sea anchors "upwind" and raise them with minimal energy loss if sustained progress is to be made. There are other possible variations of the same idea (really not so different from a turbine-prop in principle). Using a sea anchor to hold a boat in position with a turbine harnessing wind energy, storing the energy (ie in a battery) and then using it to propel the craft forward so that progress is made is fine, but this isn't continuous uninterrupted progress, and is somewhat unremarkable. —Preceding unsigned comment added by 202.180.87.174 (talk) 23:56, 14 December 2009 (UTC)


 * I'm not sure what you mean by "continous uninterrupted progress". Hardly any boat progresses at uniform speed.  Variations in the strength of the wind and waves cause any boat to advance at varying speed.  As you say, the wind-and-anchor driven device that you describe is not particuarly surprising and demonstrates that it is possible to advance directly into the wind while using only energy captured from the wind.  What seems to puzzle people is the claim that it is possible to progress downwind faster than the wind.  But the trick is to think in terms of apparent wind.  High-performance boats go so fast, with respect to the true wind, that they are sailing upwind with respect to the apparent wind, even though they are going downwind with respect to the true wind.  Seems paradoxical, but think of a boat that is on a 90-degree reach and has accelerated to the point where it is close-hauled with respect to the apparent wind.  If that boat bears off just a little, would its speed start to drop off until it equalled the speed of the wind?  No: why should there be a sharp discontinuity in the speed/course curve?  In reality, the boat will accelate even more (unless it is already at hull speed), until it is again close-hauled with respect to the apparent wind.--Gautier lebon (talk) 12:30, 18 December 2009 (UTC)

Indeed the discussion in Talk:Sailing should be read by anybody interested in this topic.--Gautier lebon (talk) 12:16, 18 December 2009 (UTC)
 * In partcular the summary.--Gautier lebon (talk) 11:26, 13 January 2010 (UTC)


 * I just removed "Sailing dead downwind faster than the wind" section as there is no agreement on that in the discussion Talk:Sailing as Gautier lebon is well aware of. The section should not be added unless a reliable source is referred to (not likely as it is impossible to sail dead downwind faster than the wind). Prillen (talk) 10:23, 19 January 2010 (UTC)


 * Thank you for your edits. Regarding the section in question, please specify what exactly you are objecting to.  There was NO disagreement regarding the theoretical possibility.  There was skepticism regarding whether or not the propeller-driven cart was was a hoax or not.  You (and others) state that it is impossible to sail dead downwind faster than the wind, but I have not yet seen anybody explain why this is theoretically impossible, nor have I seen anybody explain why the thought-experiments that I provided are not correct.  And please note that what I wrote is not original research, it is merely a summary of the various sources that I cited.  In my view, it would have been more appropriate to flag the section and to discuss the matter further before deleting it.  Regarding clean-up, I am all in favor of improvement.  What specific items would you suggest?--Gautier lebon (talk) 13:25, 19 January 2010 (UTC)
 * What to be cleaned? Here are some suggestions:
 * The introduction is too much about speed records.
 * The drawings are nice but have a too large font size and the arrow indicator on the wind vectors are pointing the wrong direction.
 * The headings "Sailing on a broad reach" and "Sailing perpendicular to the wind" should use the same terminology (angle -vs- reach/broad reach).
 * "Normal cruising boats yachts can sail at a about 45 degrees off the apparent wind (50 to 60 degrees off the true wind)." is (still) not representative for a modern yacht (40-50 degrees off true wind is more the case).
 * The tables should use a more standard layout (i.e. without border).
 * "Further Reading" should be cleaned up.
 * The language should be clarified/shortened in general. Prillen (talk) 10:43, 20 January 2010 (UTC)


 * OK, that seems sensible to me. I'm busy with other things this week, but I will try to work on it next week.--Gautier lebon (talk) 11:56, 21 January 2010 (UTC)

I have corrected the introduction and the terminology. Regarding the drawings, the arrows point in the correct directions for the vector algebra to work: just think of the case when the boat is moving directly upwind or directly downwind; I wear thick glasses, so I like the large font, and I presume that others will too, since it does not make the drawings any larger. Regarding performance of normal cruising boats, I just took the figure given in the sailing article; if you don't agree with it, then please update both articles so that they are consistent, and provide a reference. I like the borders on the tables, they are easier to read that way. I don't know what should be cleaned up in "Further Reading", maybe somebody can take care of that. Same regarding general improvement of the language: I've many revisions to the article, incorporating many good suggestions that were made on Talk:Sailing. I don't know what more to do.--Gautier lebon (talk) 10:20, 29 January 2010 (UTC)

Common fallacies
If you think that anything stated in the article is impossible, please consider that the following two reasons have already been stated, and shown not to apply:--Gautier lebon (talk) 11:22, 11 June 2010 (UTC)

1) Velocity made good downwind cannot exceed the speed of the wind because of the laws of conservation of energy

2) Velocity made good downwind cannot exceed the speed of the wind because there is no force on the sails when the boat reaches wind speed.

3) If it were possible to go faster downwind by tacking, then why do people bother with spinnakers?

4) There isn't any easily accessible published data that supports the statements that boats can achieved downwind VMG greater than wind speed.

5) The cited sources are not sufficiently reliable.

Conservation of energy
A balloon drifts downwind at the speed of the wind. It uses no energy to do that. Similarly, an iceboat can drift dead downwind at very close to the speed of the wind, because the friction of its runners on the ice is negligle.

That is, to proceed downwind at the speed of the wind requires essentially no energy.

If a boat can capture some energy from the wind, then it can use that energy to propell itself downwind faster than the wind. There is no violation of the law of conservation of energy because the energy captured from the wind is used to overcome the resistance of the surface (insignificant in the case of an iceboat) and the resistance of the apparent headwind induced by the boat's progress. As explained in the main article, when a boat sails at an angle to the wind, it can capture energy from the wind, even if its downwind progress is faster than the wind itself. This is because of the apparent wind shift, see below.

Also, it is easy to see that a device can be constructed that can capture the energy from the wind even when moving dead downwind, see below Talk:Sailing faster than the wind.

No force on the sails
As explained in the main article, the forward motion of the boat induces a wind that must be added to the true wind in order to find that apparent wind that strikes the sails.
 * Sorry, the apparent wind created by the downwind motion of the boat must be vectorially subtracted. At some point,the downwind vector of the boat velocity must be equal to the wind velocity, at which point. The only remaining wind vector is perpendicular to the wind direction and is entirely due to motion of the boat. At this point and beyond, the "wind" is created entirely by the motion of the boat and thus becomes a true perpetual motion hype.--QuietJohn (talk) 05:20, 14 June 2010 (UTC)
 * Sorry, you are confusing velocities with energy. To make a "perpetual motion" claim, you have to show that energy would not be conserved. It is not sufficient just to state, that something about the velocity vectors seems counter-intuitive to you.Eyytee (talk) 09:12, 14 June 2010 (UTC)


 * You are correct, it is vectorially subtracted. The drawings in the article show this.  You are correct that, at some point, the downwind vector of the boat velocity is equal to the wind velocity: for a specific example, see the drawings for 135 degrees below.  You are correct that the apparent wind is created by the motion of the boat, but the motion of the boat is created by the TRUE wind, thus there is no perpetual motion.  Think of a motor boat: it uses energy from its engine to create an apparent wind.  That apparent wind could be used to drive a propeller-driven power generator. Similarly, a sailboat uses energy from the wind to create an apparent wind, which allows it to obtain even more energy from the wind.  The point is that the energy required to drive the boat downwind always comes from the true wind.--Gautier lebon (talk) 09:03, 14 June 2010 (UTC)


 * Gautier lebon: this - "a sailboat uses energy from the wind to create an apparent wind, which allows it to obtain even more energy from the wind." - is misleading in general and is incorrect even after the apparent wind direction goes forward of abeam.  See discussion below about the perpendicular component of the wind.Sirclicksalot (talk) 10:39, 14 June 2010 (UTC)


 * QuietJohn: It was obvious from the next sentence below, and from the grammar of the sentence you tried to correct, that the original statement refers to adding the true and induced winds vectorially; the original statement is correct:  the induced wind and the true wind are to be added as vectors to give the apparent wind vector.  Your statement confuses induced wind and apparent wind.  The induced wind (vector) is the opposite (negative) of the boat's motion (vector) over the water, and that induced wind vector is to be added to the true wind (vector) over the water to give the apparent wind relative to the boat.  Since in this case they have opposite direction, adding true and induced winds as *vectors* is equivalent to the difference (i.e. subtraction) of their *magnitudes*.  In an arbitrary reference frame, there are three velocity vectors:  True Wind=Vtrue; Water=Vwater; Boat=Vboat.  An observer (not the boat in question) floating in, and moving with, the water sees the true wind as [Vtrue - Vwater] [1].  The boat in question moving downwind is moving as [Vboat - Vwater] [2].  The induced wind of the boat is the opposite (negative) of the boat i.e. [Vwater - Vboat] [3].  Adding (not subtracting) the true wind (1) and the induced wind (3) vectors:  [Vtrue - Vwater] + [Vwater - Vboat]  =  [Vtrue - Vboat] [4] yields the apparent wind i.e. the wind that the boat experiences i.e. the wind that determines where the telltales on the boat point (direction) and what speed (magnitude) the anemometer on the boat measures.  Also, if the boat is going straight downwind, there can never be a non-zero "remaining wind vector [which] is perpendicular to the [true] wind  direction."Sirclicksalot (talk) 09:27, 14 June 2010 (UTC)


 * It is correct that, if the boat is going straight downwind, then the apparent wind will be nearly zero. But that is not the point.  The situation that is of interest is when the boat is NOT going straight downwind: for example when it is going at 135 degrees off the true wind.  In that situation, as shown from the various charts in the article (all derived from textbooks), there will be an apparent wind that can be forward of the beam.  You cannot simply subtract the magnitude of the boat's speed from the magnitude of the true wind.  You need to do a vector addition (or subtraction) of the two speeds, and that gives you the magnitude and direction of the apparent wind.  Again, this is explained, with drawing, in most basic manuals on sailing.--Gautier lebon (talk) 10:56, 14 June 2010 (UTC)


 * This is getting silly. All three of us are saying to add vectors.  Where is the disagreement?  Nobody is saying that subtracting magnitudes is correct for the general case, or is correct for any case but straight downwind.  The current case, which QuickJohn mistakenly tried to correct but which I and others responded to, *IS* for sailing straight downwind; look at the next sentence"  "If an iceboat sails dead downwind, ..."  I agree the interesting case is later and not straight downwind, but that is not the case where QuickJohn made his comment above.Sirclicksalot (talk) 11:42, 14 June 2010 (UTC)


 * Yes, I am probably confused about who is saying what about what. Maybe it would be better if you started a new section?  I'm no longer sure about what QuickJohn is arguing, apart from the fact that, according to him, the citations are insufficient or insufficiently reliable: he has made that clear.--Gautier lebon (talk) 11:53, 14 June 2010 (UTC)

If an iceboat sails dead downwind, then it will soon reach a speed close to the speed of the wind and the apparent wind on the sails will be nearly zero.

But if an iceboat sails downwind at an angle to the wind, there will be an apparent wind shift: the apparent wind will move forward. As explained in the main article, the iceboat will eventually find itself on a broad reach with respect to the apparent wind (apparent wind at 90 degrees to the boat's course). The apparent wind will still generate a forward force component, so the boat will continue to increase in speed and the apparent wind will shift even further forward. As explained in the main article, this can result in velocity made good (progress in the direction of the wind) that is greater than the velocity of the wind.


 * The conclusion here is correct but the explanation is faulty: the apparent wind shifting forward must *reduce* the forward force component.  It cannot be otherwise.  It is possible that is not what is meant here; if so it should be written more clearly.  See discussion below about component perpendicular to boat's motion which never changes.  —Preceding unsigned comment added by Sirclicksalot (talk • contribs) 10:00, 14 June 2010 (UTC)


 * Well, it must be true that it should be explained more clearly: I don't know how to do that, help would be appreciated. You are correct that the apparent wind shifting forward reduces the forward force component.  BUT the apparent wind become stronger as it shifts forward, so the resulting forwar force component is still signficiant: see the calculations below for the iceboat sailing at 135 degrees off the true wind.  I will work out the actual forward force component and post that on this discussion page, but it will take me a couple of days.--Gautier lebon (talk) 10:56, 14 June 2010 (UTC)


 * There is no increase in forward force as the apparent wind goes forward of abeam. The apparent wind getting stronger as the boatspeed increases cannot not make more forward force available (for fixed-wing sails).  Neither is the concept needed to explain DDWFTTW.  Claiming that the increasing strength of the apparent wind, once it goes forward of abeam, provides more forward force (and therefore power; force x speed is power) is the same as claiming perpetual motion and we know that does not work.  Once the apparent wind goes forward of abeam, it is the component of the wind perpendicular to the direction of motion, which is constant, that provides all forward force and is more than adequate to explain DDWFTTW.Sirclicksalot (talk) 11:57, 14 June 2010 (UTC)


 * I agree that whether or not the forward force increases is not particularly important for this discussion. What matters is whether the forward force (net of the drag of the sail itself) is sufficient to overcome the resistance of the surface and the resistance caused by the induced headwind.  But I'm not convinced that the forward force component cannot increase.  You are correct that this would imply more power, but why can't the sail capture more energy from the wind at a tighter angle than it did at a broader angle? That would not violate conservation of energy and would not constitute perpetual motion. It seems to me that the force on the sail is proportional to the square of the speed of the apparent wind, so it could well be that the forward force component increases as the boat accelerates.  As stated before, I will work out some numbers and post them at the bottom of the page, for further discussion.--Gautier lebon (talk) 12:35, 14 June 2010 (UTC)

Note the key point: what drives the boat is the apparent wind, that is what the sails "see" and what the sails react to and what propelles the boat.
 * The apparent wind is created by the boat. So what this sentence says is that the motion of the boat creates the apparent wind which then speeds up the boat. Once again, perpetual motion.--QuietJohn (talk) 05:20, 14 June 2010 (UTC)
 * It's motion that lasts as long as the wind does, just like sailing slower than the wind. If the wind dies, it stops.  So it's as "perpetual" as the wind, or a windmill, or any other wind-powered device.
 * As for the boat "creating the apparent wind", not really. The apparent wind is just the average speed of air molecules in a reference frame in which the boat is at rest (just as the "true" wind is the average air speed in a frame in which the land is at rest).  The boat obviously didn't "create" that wind, because it didn't have any effect on the air far away from it.  Anyway, Newton's laws apply in any frame.  So all one has to do is apply them, and they tell us there can be a net force that acts to accelerate the boat even when the downwind vmg of the boat is 1 or greater.  See my example below.Waleswatcher (talk) 11:46, 14 June 2010 (UTC)


 * The apparent wind is not just resulting from the motion of the boat but also from the true wind. Apparent wind is the air velocity relative to the airfoil, and it is the standard approach to use that relative velocity to calculate the force on an airfoil based on Angle_of_attack and Lift-to-drag_ratio. If you suggest, that we should compute the force on the airfoil WITHOUT using the air velocity relative to the airfoil (apparent wind), then I want to see your maths.Eyytee (talk) 09:12, 14 June 2010 (UTC)
 * QuietJohn is confusing induced wind (created by the boat) with apparent wind (sum of true wind and induced wind). However, original article is also a bit fuzzy/careless.  First:  the boat is on a beam, not a broad, reach, wrt the apparent wind when that wind is perpendicular to the boat's motion.  Second, once the apparent wind direction goes forward of abeam, the only "wind" the boat has available from which to extract energy for forward motion is the vector component of the apparent wind which is perpendicular to the boat's direction of motion (the component parallel to the boat's direction of motion is from ahead and can therefore only produce drag unless the boat has a wind turbine driving a sub-surface propeller - but that is another article).  Note, however, that since the induced wind is always parallel to the direction of the boats motion, that perpendicular component is the same as that available from the true wind, and therefore that component never changes.  The only thing that changes is the direction and strength of the apparent wind, which in turn affects how efficiently the sails operate:  it is easier to direct aft a wind coming from abeam than one coming from in front of abeam.  The increasing strength (magnitude) of the apparent wind does not make available more energy from the wind, neither is it necessary, so there is no perpetual motion claim made here.  The only issue is the perpendicular component of the wind from which energy is extracted, and the only limit to the speed of the boat is how efficiently it can convert that energy into a forward motion.Sirclicksalot (talk) 09:54, 14 June 2010 (UTC)


 * It is not perpetual motion, please see the explanation above.--Gautier lebon (talk) 09:03, 14 June 2010 (UTC)

Why use spinnakers
Most sailboats cannot accelerate enough when they sail downwind so that the apparent wind shifts to come forward of the beam, and this because the resistance of the water is so large. That is, the resistance from the hull prevents the boat from moving fast enough so that the apparent wind shifts forward of the beam. Therefore, the apparent wind will always be aft of the beam and a spinnaker will increase the speed of the boat. However, downwind progress might still be faster if the boat gybes downwind, even with a spinnaker, and indeed most books on racing tactics say that it will usually be faster to avoid a dead downwind course and to gybe back and forth, even if the sailboat is using a spinnaker.

Published data
The article provides citations to various observed data. The most accessible data are those from the first regatta of the 2010 America's Cup. The winning yacht, USA 17, completed the 40 nautical mile course (20 miles dead upwind, 20 miles dead downwind) in 2h32, in 5-10 knots of wind, see for example. If the wind had been 10 knots all the time (which was not the case) and if the downwind VMG had been equal to the wind speed, then USA 17 would have needed 2 hours to complete the downwind leg. That means that it would have completed the 20 mile upwind leg in 32 minutes, that is, at about 40 knots made good, meaning well over 50 knots over the ground. Of course USA 17 did not achieve such speeds upwind: anybody can watch the video and see the displayed upwind speeds, which were just over 20 knots. In fact, it took USA 17 63 minutes to complete the 20 mile downwind leg. Even if winds had been 10 knots (and they were less), the downwing VMG was nearly 2 times wind speed.--Gautier lebon (talk) 10:16, 15 June 2010 (UTC)

Insufficiently reliable citations
The citations in the article include many web sites, which are easily accessible, and a published book (by Bethwaite) which is still in print and easily available. One of the contributors to this discussion has provided a second book, also still in print and easily available: The Symmetry of Sailing: The Physics of Sailing for Yachtsman, by Ross Garret.--Gautier lebon (talk) 10:16, 15 June 2010 (UTC)

As explained below, the material in the article is not sufficiently original or suprising to be published in a peer-reviewed scientific journal. So it is not clear what additional citations should be provided.--Gautier lebon (talk) 10:16, 15 June 2010 (UTC)

Sailing at 135 degrees off the true wind
As promised, I've developed force diagrams that are meant to make the situation easier to understand. But I'm not convinced that what I have done can be understood easily. At first, I thought of making three diagrams, showing velocities and forces for three situations: (1) standstill (2) apparent wind at 90 degrees (3) apparent wind at 45 degrees. But then I decided to make separate diagrams for the velocities and the forces, with each diagram showing the three situations.--Gautier lebon (talk) 12:27, 11 June 2010 (UTC)

I would much appreciate comments on how to make the text below, and the diagrams, easy enough to understand so that they could be incorporated into the main article.--Gautier lebon (talk) 12:27, 11 June 2010 (UTC)

Consider an iceboat that is sailing at a course that is 135 degrees off the true wind. At the beginning, the boat is stationary. We will use the value "1" for the speed of the true wind. The apparent wind is equal to the true wind and comes from the same direction as the true wind. The velocity made good (VMG) downwind is zero. This is shown by the value "V0" in the chart below.

The boat will accelerate and will eventually reach a speed equal to to .707 times the speed of the wind. At this point, the apparent wind will come from 90 degrees (broad reach) and the speed of the apparent wind will also be .707. VMG is .5 This is shown by the value "V1" in the chart below.

'''At this point, the only resistance to forward motion is the friction of the iceboat's runners on the ice. But that is negligible.'''
 * Unfortunately, not true. Wings (and sails) come with a penalty called drag in addition to lift. Lift would typically be perpendicular to airflow and drag in the same direction as airflow. So, in addition to runner resistance, you have aerodynamic drag which acts to slow the boat down. --QuietJohn (talk) 06:43, 14 June 2010 (UTC)


 * That is correct. I should have referred to the lift, which is the force on the sails net of drag. According to Bethwaite, lift is about 2/3 of the total force. But this does not change the argument, because lift and drag increase proportionally, with the square of wind speed.  So there is an excess of lift over drag, and there is still a forward force which will accelerate the boat past a beam reach.--Gautier lebon (talk) 09:12, 14 June 2010 (UTC)

So the boat will continue to accelerate and will reach a speed equal to 1.41 times the speed of the wind. At this point, the apparent wind will come from 45 degrees (close hauled) and the speed of the apparent wind will be 1. VMG is 1. This is shown by the value V2 in the chart below.



If the forward component of the force on the sails is sufficient, the boat can continue to accelerate, the apparent wind will shift further forward, and VMG will be greater than 1, that is, greater than the speed of the true wind. That is, the boat will progress downwind faster than the speed of the wind. The text and charts below show the force on the sails and the forward component of that force. It is not possible to determine theoretically whether the forward component at V2 would be sufficient to accelerate the boat further; but measurements of actual iceboat performances show that indeed it is. Iceboats accelerate until they are sailing at about 10 degrees (or less) off the apparent wind, and thus achieve speeds of 5 or more times the speed of the wind. This is the case even if the boat is sailing at 135 degrees off the true wind. So the VMG (progress in the dead downwind direction) is far greater than the speed of the wind.

The chart below shows the apparent wind and the sails on the iceboat when it is at a standstill (V0), on a broad reach (V1) and close hauled (V2). The angle of the sail is half of the angle of the apparent wind, which is the optimal angle in terms of generating a forward driving force. The force generated by the apparent wind on the sail is proportional to the square of the speed of the apparent wind.



The chart below shows the forward force resulting from the apparent wind when the iceboat is at a standstill (V0), on a broad reach (V1) and close hauled (V2). The force generated by the wind is perpendicular to the sails. It can be seen from elementary trigonometry that the forward component of that force is sin(sail angle).



Thus, the forward force when the iceboat is close hauled will be 38% of the total force on the sail. Again, whether or not that is sufficient to accelerate the boat further will depend on the resistance caused by the apparent wind (the headwind induced by the speed of the boat). As stated earlier, in practice it has been found that the force is indeed sufficient to further accelerate the iceboat (and in fact also other high-performance boats, including sailboats, as explained elsewhere in the article).

Force and speed table
Here is the table that I've come up with for a course that is 135 degrees off the true wind. Please check the formulas and the calculations.--Gautier lebon (talk) 12:41, 15 June 2010 (UTC)

The formulas are:

V: boat speed

Alpha: angle of apparent wind

AW: Apparent wind speed


 * Angle of apparent wind: 180-(ATAN(SQRT(2)/(SQRT(2)-2*V))
 * Apparent wind speed: SQRT(V^2-V*SQRT(2)+1)
 * Force of apparent wind: AW^2
 * Forward force component (assuming lift coefficient is 1): SIN(Alpha/2)*force of apparent wind
 * Drag force component (drag from the induced headwind): COS(ALPHA)*force of apparent wind
 * Ratio of drag to forward force: drag/forward

If an iceboat has a sail area of 6 square meters, and if it is sailing at 5 times the speed of the wind, the the foward force will be about 6*1.54 = 9, assuming the lift cofficient is 1. But the lift cofficient is about .66 (according to Bethwaite), so the forward force will be about 6.

The drag component will be about 18. So the speed of 5 times the speed of the wind can be achieved if the frontal surface that creates the drag (the surface of the mast, hull, runners, etc.) is less than 0.33 square meters (18*.33 = 6).

The popular International DN iceboat has a sail area of 6.5 square meters and can achieve 5 times the speed of the wind. This is perfectly consistent with the calculations above, given that the aerodynamically equivalent frontal surface of the iceboat is indeed probably about .33 square meters.

I think you should be more explicit here in step 2
Why are we settling on English when this can be settled in Math? Both Gautier has posted diagrams above, and I have posted diagrams showing exactly how this works, given drag sufficiently small (doesn't need to be zero). What is appealing about these types of disputes over religious or political, is that they should be able to be resolved very quickly around a chalkboard. I suggest we attempt to do that.Generacy (talk) 13:31, 14 June 2010 (UTC)
 * The first problem I have with the diagrams is that force (lift or drag) are not directly proportional to airspeed vectors. There are coefficients involved, one for lift and a different one for drag. A consideration of lift coefficient, drag coefficient, lift-induced drag, lift-to-drag ratio etc. would provide a more convincing treatment. Given that wing angles of attack can be as low as 5 degrees and lift/drag ratios can be as much as 10, the ideas presented here seem feasible. However, the analysis presented seems dubious. I also remember my math teacher proving that 1 = 2. What I've seen so far is similarly unrigorous.--QuietJohn (talk) 17:15, 14 June 2010 (UTC)


 * As stated below, it doesn't really matter what the lift/drag ratio is, so long as lift is greater than drag. According to Bethwaite, lift is about 2/3 of the total force on the sail.  What matters is not the drag induced by the sail, because that is always overcome by the lift induced by the sail.  What matters is the drag induced by (1) the surface (insignificant for an iceboat) and (2) the windage on the mast, hull, etc.  It is that drag (the drag from the induced headwind) that limits the ultimate speed.--Gautier lebon (talk) 09:52, 15 June 2010 (UTC)


 * Failing to see a division by zero, and acknowledging the possibility of a lift/drag ratio of 10, I'm not sure what the problem is. As long as the airspeed shown in the diagram is in the correct direction, and the proportionality of airspeed to force is not negative (the coefficients are > 0), then the accelerating force is pointing in the right direction.  Finally, if it is possible that the accelerating force is greater than the drag force, which it is, we have a boat that is accelerating past VMG.Generacy (talk) 17:44, 14 June 2010 (UTC)
 * Additionally, one needn't even get into particular L/D ratios to mathmatically demonstrate that no physical laws are violated -- one merely need to do a simple energy balance. Using the simplest of 'force * distance' formulas one can easily see that in any tailwind, when the vehicle (any of them) has a downwind VMG of 2x the speed of the wind, there is twice as much energy available from the surface as is required to do the needed work on the air.  If anyone needs such a simple analysis to be convinced, I'm happy to produce it ... meanwhile here is a link to a very comprehensive one produced by the legendary MIT Professor of Aerodynamics Mark Drela:  http://www.boatdesign.net/forums/attachments/propulsion/28168d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddwe.pdf ThinAirDesigns (talk) 18:35, 14 June 2010 (UTC)

Also, if folk are looking for for more math, here is a link to another extensive treatment by MIT Professor Mark Drela on the topic. His treatment is specifically aimed at a DDWFTTW water craft, but near the end he addresses a ground based wheeled vehicle and makes the following statement "This confirms that the DDWFTTW condition V/W > 1 is achievable with a wheeled vehicle without too much difficulty". http://www.boatdesign.net/forums/attachments/propulsion/28167d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddw2.pdf ThinAirDesigns (talk) 18:41, 14 June 2010 (UTC)


 * Dear ThinAir, thank you for these. I have incorporated these citations into the main article.--Gautier lebon (talk) 10:21, 15 June 2010 (UTC)


 * You have included Drelas power analysis twice. Please also include the Drelas general DDWFTTW analysis. That is where this quote is from "This confirms that the DDWFTTW condition V/W > 1 is achievable with a wheeled vehicle without too much difficulty" —Preceding unsigned comment added by Eyytee (talk • contribs) 11:11, 15 June 2010 (UTC)


 * Fixed, thank you.--Gautier lebon (talk) 11:42, 15 June 2010 (UTC)


 * The following references might be usfull as well: A lecture about up-wind-carts & ddwfttw-carts from the Technical University of Denmark, Andrew Bauer's paper from 1969 about his DDWFTTW-vechicle —Preceding unsigned comment added by Eyytee (talk • contribs) 11:25, 15 June 2010 (UTC)


 * Very helpful, than you. I have added them.--Gautier lebon (talk) 11:42, 15 June 2010 (UTC)

Basic concepts
I have the impression that some people are, at times, confusing some basic concepts. So here is a reminder of what we are talking about.--Gautier lebon (talk) 10:43, 15 June 2010 (UTC)


 * F=m*a, meaning that a force applied to a mass will produce an acceleration. If there is no force, then there is no accelaration.
 * V=a*t, meaning that an acceleration a applied for time t will increase velocity by V. If there is no acceleration, then the velocity does not change.
 * W=F*d, meaning that work W (also called energy) results from force F applied over a distance d.
 * P=W/t, meaning that power P is work W applied over time t.
 * Momentum = V*m, and it is conserved (unless a force is applied).
 * Kinetic energy = 0.5*V^2*m, and it is conserved (unless a force is applied).

The energy that drives a boat comes from the transfer of kinetic energy from the wind to the boat's sails: the molecules of air strike the sail, a force is induced, and momentum and energy are transferred from the air to the sail.

Billiard ball model
I don't know whether the following explanation will be helpful, but I'm tossing it in for comment.--Gautier lebon (talk) 11:32, 15 June 2010 (UTC)

One can think of the molecules of air as billiard balls, and the sail as a flat panel that is struck by the billiard balls. There is a transfer of kinetic energy and momentum from the balls to the flat panel. If there is no contsraint on the panel, it will wind up moving in the same direction and the same speed as the billiard balls: this is drifting dead downwind at nearly the speed of the wind.

At that point the boat has kinetic energy, but it's velocity does not change.

Now imagine that the flat panel (the sails) and the boat are constrained to travel at an angle (say 135 degrees) to the direction from which the billiard balls are coming. for example because it is mounted on a rail. When the balls hit the panel, there will be a transfer of kinetic energy and momentum, but now the panel does not wind up moving dead downwind, it winds up moving along the rail, that is at an angle with respect to the direction from which the billiard balls are coming.

Consequently, the direction from which the billiard balls are coming changes: they appear to come from further forward as the boat increases in speed. If you adjust the angle of the flat panel, the incoming balls will still create a forward force on the panel (in addition to the sideways force that has to be resisted by the rail on which the panel is mounted). There is still a transfer of energy from the balls to the panel, and the boat may or may not accelerate, depending on the resistance to its motion along the rail.

Now imagine that the billiard balls are relatively spaced out, so that one hits the panel only every second. In between billiard balls, the boat will be slowed down by the resistance of its motion along the rail (assuming that the resistance from the air is negligle, for example because this experiment takes place in a vacuum). If that resistance from its motion along the rail is very small, then the boat will continue to advance along the rail (conservation of momentum and kinetic energy), slowing down only slightly. Then the panel is hit by a new billiard ball: this will result in the transfer of some energy and in an acceleration of the boat.

So the boat will continue to accelerate until the energy lost due to the resistance of its motion along the rail is greater than the energy transferred to it when a billiard ball hits the flat panel.

The boat's ultimate speed depends on the friction from the surface, on how efficiently it captures the energy from the billiard balls, and on how much energy there is in the billiard balls.

Billiard balls II
Sirclicksalot (talk) 19:17, 15 June 2010 (UTC)

(updated description of forces' contribution to conservation of momentum Sirclicksalot (talk) 19:27, 15 June 2010 (UTC))

I have alway preferred this model, it is easier to understand and makes much more sense than the reams of bogus verbiage I have seen over the years invoking Bernoulli and phantom venturis. I used it yesterday to explain all points of sail to a neophyte in about ten minutes. I will try here to extend this description.

As noted above, the billiard billiard ball model is a momentum- and energy-transfer model. That is, generally speaking energy and momentum are transferred between the two media - air and water (or air and land) - and anything those media touch. To be more accurate, it is the differential momentum and differential energy between the air and water that provides the capability to sail. If there is no difference in velocities (vector quantities) between the air and the water, a sailboat cannot move through the water in any direction without an external energy source (paddle, motor, etc.). But if there is no wind over the ground, a sailboat can move on a river with current because of the differential momentum and energy between the still air and the moving water. And now that DDWFTTW has been demonstrated, it is theoretically possible even to sail directly upstream on a calm day.

Ultimately, the existence of differential momentum and energy between the two media are the reason it is possible to move in any direction relative to the true wind, while using that same true wind as the sole motive force.

One thing to note in Gabon's example above is that the flat panel (sail) is perpendicular to the true wind direction. In that case the sail and boat VMGDW (Velocity Made Good DownWind i.e. parallel to the true wind) cannot exceed the true windspeed. Mr. Beardsell's analysis (Physical mechanics, above) is an excellent description of why. The reason I bring this up is to make the point that when momentum transfer occurs, momentum must be conserved, but it is the change in momentum between the wind (billiard ball) before and after it encounters the sail that is transferred to the sail. There is another point to be made here: momentum is a vector so by definition it has a magnitude and a direction; if the direction of the momentum of a billiard ball changes but the magnitude of the momentum does not, that still constitutes a change in momentum transferred from the billiard ball to the sail which changed it. Counter to intuition, the direction and magnitude of the momentum of the incoming medium (true wind, billiard balls, etc) does not matter, only the change in momentum between the incoming and outgoing wind determines the actual momentum transferred. Intuition may say the best performance has to do with making use of the momentum of the wind in the direction it is blowing, but that is not always the case. If the sail can be configured to push (change the momentum of) the wind in a given direction, the sail will experience a force in the opposite direction (Newton's first law plus the concept that a change in momentum per unit time is the same as a force). So, if instead of keeping the sail perpendicular to the true wind it is angled such that the back of the sail is further upwind, it is possible for the boat to be making VMGDW greater than the true windspeed and simultaneously to have the sail's intersection with a true wind streamline moving downwind more slowly than the true wind which means the sails are pushing against i.e. upwind the true wind and receiving a push downwind ([IMG]VMGdwfttw_2.png[/IMG]).

Other important concepts are the conservation of momentum and energy in any inertial reference frame. If a quantity of momentum is extracted (subtracted in a vector sense) from the wind, that same quantity of momentum must be added (again, in a vector sense) to the boat or to the water; note also that a force applied over time is the same as momentum and may be an alternative trade to maintain conservation of momentum. The same goes for energy, which is a scalar (or magnitude only, non-vector) quantity. This is why the boat-fixed reference frame is often the easiest in which to make vector diagrams for analyzing sailing performance: the energy of the apparent wind in the boat-fixed frame does not change because the boat and sail are static and passive and have no external energy sources or sinks with which to give or take energy from the wind or water (ignoring frictional losses for the moment). The advantage that comes from this is simple: the magnitude of the momentum of the wind and the water in this frame do not change. Here is why:

The momentum (M) of a billiard ball (our analog for the wind) is the product of its mass (m) a scalar, and its velocity (V, a vector):

M = m * V

The kinetic energy (KE) of a billiard ball is half the product of its mass and the square of its speed (S = the magnitude of its velocity):

KE = M * S^2 / 2

In a typical apparent wind diagram of a sail, the billiard ball (apparent wind) comes into the sail at some angle of attack off the centerline of the boat and leaves the sail going parallel to the centerline. Because the sail is not moving in this reference frame it can neither give energy to nor receive energy from the billiard ball, so the KE of the ball cannot change (again, ignoring frictional losses for now). Solving for the speeds of the ball coming in and out of its interaction with the sail:

Sin = SQRT(2 * KEin  / Min ) Sout = SQRT(2 * KEout / Mout)

Neither the mass of the ball nor its KE change as it goes over the sail (Min=Mout and KEin=KEout) in this reference frame, so the speed of the ball must remain constant (Sin=Sout) in this reference frame. This makes the calculation of the change in momentum of the ball across the sail a simple trigonometric exercise involving the speed of the ball and the angle it is turned by the sail. The change in momentum (a vector) is the same as the force (also a vector) exerted by the ball on the sail.

Sirclicksalot (talk) 19:17, 15 June 2010 (UTC)

Physical mechanics
I had started by saying this is WP:OR but no, that's giving me airs and graces. This is plain classical mechanics, unadorned. It's fact, plain and simple. There is no need to put this in the article, this below is the reason other stuff must not go in the article. This is high school physics only. This could easily all be substantiated from WP articles, but there is no need, as it so uncontroversial. If you disagree with it then you are not literate in the most basic physics. Sorry. Leave aside your beliefs that sailing downwind with a VMG towards the directly downwind mark is possible. Leave aside what you've read the skipper of Oracle in the AC as having said. Just stay with the physics. I write this just to try and shake your convictions ever so slightly.

In order for one object to propel another in a given direction (consider two moving balls, or whatever) the first must have a component of it's motion in that direction relative to the motion of the second object. If I am travelling south at 10mph and you collide with me and I end up travelling south at more than 10mph then you must have had a southerly component to your velocity of more than 10mph. Otherwise the collision would have slowed me down in that particular direction. Uncontroversial, and of course we can decompose any particular motion so that it consists of the sum of one NS vector and one WE vector. When considering changes in NS velocity the WE velocity of the colliding objects van be disregarded. That is what vector decomposition lets us do, and it's done all the time in high school and undergraduate physics. It's not controversial and it's not rocket science. (Well, actually it is rocket science, but you get my drift, no pun intended).

So when a particle of the air collides with my yacht and the yacht is accelerated in a particular direction, the particle of the air must have had a velocity component in that direction faster greater in magnitude than the speed velocity component of the yacht in that particular direction.

The question then is, and you should all have seen this coming, if the yacht is going faster than the air in a southerly direction, how can collision with the air accelerate it in a southerly direction? And, sorry to disappoint, that's a rhetorical question, there is no satisfactory answer. Therefore sailing downwind VMG faster than the air is impossible. (I removed Mr. Beardsell's QED hereSirclicksalot (talk) 03:19, 15 June 2010 (UTC)).


 * Yes, we do need a diagram, the one showing how the true wind is shifted forward and becomes an APPARENT wind that allows the boat to accelerate past the speed of the true wind. Any sailor knows about apparent wind shift and the resulting decrease or increase in apparent wind speed.  You deleted the diagram that explains this.--Gautier lebon (talk) 07:44, 9 June 2010 (UTC)

You don't need a diagram, do you?

Paul Beardsell (talk) 14:52, 8 June 2010 (UTC)

Then someone says oh yes you can! And they're the skipper of an America's Cup yacht. Or they have endless qualifications, and they've read every book on yachting, and they've maybe even written one. Or someone has a PhD in Physics and they disagree? Well, then they're wrong, they forget the basics, if they ever knew them. Paul Beardsell (talk) 14:57, 8 June 2010 (UTC)


 * As shown below, this is not true: they are not wrong, and they do know the basics.--Gautier lebon (talk) 12:46, 15 June 2010 (UTC)

Criticism zero
The only thing wrong in what was written above here is the claim that there is, and can be, no satisfactory answer to this question: if the yacht is going faster than the air in a southerly direction, how can collision with the air accelerate it in a southerly direction?. It is difficult to explain a satisfactory answer when using a reference frame fixed to the water, but here it is (using the reference frame moving with the boat makes the answer obvious). (removed some snarkiness Sirclicksalot (talk) 01:24, 16 June 2010 (UTC))



Simply put, the satisfactory answer is that even if the yacht is going faster - in the VMG sense - than the wind in the southerly direction, the portion of the sail, on which the true wind acts and which is not perpendicular to the true wind, can indeed be going more slowly - again in the VMG sense - than the true wind in the southerly direction.Sirclicksalot (talk) 03:19, 15 June 2010 (UTC)

The actual steady-state numbers will depend on the actual configuration (sail, hull type - iceboat, land-yacht, etc), but the diagram above clearly demonstrates that there can be a southward force of a southerly true wind on the fixed sail of a craft moving across that wind with a downwind (i.e. southward) VMG greater than the true wind. The angles chosen are not at all optimal but were selected to make understanding the concept as simple as possible. This model has a theoretical limit to the maximum VMG downwind so thermodynamic principles are not violated, which an important criterion in assessing any such model. In the case above the VMG limit is 1/(1-sin(22.5deg) ~ 1.62 * Vt.Sirclicksalot (talk) 03:19, 15 June 2010 (UTC)

The part of the sail on which the wind acts is indeed, relative to the true wind, moving upwind (again, relative to the true wind - see Note below), and therefore "pushes" northward on the wind, from which Newton's Third Law plainly grants an ongoing thrust southward onto the sail, and, through the rigging, from the sail onto the yacht, more than adequate to counter all frictional forces for iceboats, land yachts, catamarans and other high-performance craft.Sirclicksalot (talk) 03:19, 15 June 2010 (UTC) (Corrected which of Newton's laws was applicable 74.79.24.116 (talk) 14:55, 17 June 2010 (UTC))

Note: I am not saying the sail is moving upwind relative to the water, only that the part of the sail on which true wind (i.e. a streamline) acts is moving downwind more slowly than the true wind.Sirclicksalot (talk) 03:19, 15 June 2010 (UTC)

If the sail were perpendicular to the true wind, then Mr. Beardsell's analysis above would of course hold (you can't beat simple physics). However, in the actual case it is obvious that the sail is not perpendicular to the true wind direction, but is instead angled between the boat's velocity vector and the vector perpendicular to the true wind. That angle means that relative to the streamlines of the wind moving south at the true windspeed, the point at which the sail intersects any given streamline is moving more slowly than the true wind directly south along that streamline. You can't beat simple physics, especially when they are done correctly. QEDSirclicksalot (talk) 03:19, 15 June 2010 (UTC)

Streamlines are not some sort of hand-waving black magic invention; please stop and attempt to understand them. The vector diagram above is equivalent to moving the sail sideways and slightly down in calm air (less than 22.5 degrees South of West in the diagram, assuming South is down. In that case the angle of the sail is such that the sail would push North on the calm air which means the air would push South on the sail.  Anyone who claims to understand the equivalence of different reference frames but thinks this diagram, and the calm-air equivalent description in this paragraph, are not physically realistic needs to show the error(s) in the diagram above and provide a corrected diagram.  Continuing with baseless claims that it is not possible with neither correct nor valid supporting arguments may still provide humor for many but is otherwise getting a bit old.Sirclicksalot (talk) 03:19, 15 June 2010 (UTC)

criticism 1
"So when a particle of the air collides with my yacht and the yacht is accelerated in a particular direction, the particle of the air must have had a velocity component in that direction faster greater in magnitude than the speed velocity component of the yacht in that particular direction."

If the above paragraph was true boats could not tack UP wind, which is an observed fact for centuries.(Eyytee (talk) 15:52, 8 June 2010 (UTC))


 * The wind force experienced by a sail is perpendicular to the set of the sail. That force is most conveniently decomposed into two forces at right angles to one another (the vector sum of which is the force on the sail).  The first force is in the axis of the vessel i.e. in the rolling, ski-ing, keel direction.  The other force is at right angles to that, the one that tends to tip the vessel to the side.  When the set of the sail is such that the axial force is forward that drives the boat forward and vv.  If on an upwind tack the keel/whatever is at 40 degs to the wind and the sail is at 20 degs then a proportion of the force experienced by the sail will be in a vessel-forward direction.  If the keel is at 40 degs to the wind but the sail at 60 then there will be a backwards force.  None of this contradicts the classical mechanics.  We have found the component of the force necessary to explain the upwind sailing.  Paul Beardsell (talk) 16:09, 8 June 2010 (UTC)


 * This is correct. And if you work out the force diagrams for an iceyacht sailing at 135 degrees of the true wind, you will see why it can accelerate to 6 times the speed of the wind, and thus achieve VMG far greater than wind speed.  I will work out those diagrams and publish them here, but it will take me a few days.--Gautier lebon (talk) 07:47, 9 June 2010 (UTC)


 * It'll take longer than that. Paul Beardsell (talk) 08:07, 9 June 2010 (UTC)


 * No, it won't, because I have already done them by hand. What will take time is to make them pretty.--Gautier lebon (talk) 15:35, 10 June 2010 (UTC)


 * Nobody here claimed that tacking upwind contradicts classical mechanics. It just contradicts your initial statement on the top of this section. On an upwind tack the air doesn't have a velocity component in the direction of the boat's acceleration which is greater than the boat's velocity component in that direction. Therefore your statement, on which you base your entire argument, is wrong. Eyytee (talk) 22:22, 8 June 2010 (UTC)


 * We're comparing components of [vector] velocities in a particular direction, not [scalar] speeds (talk) 06:26, 9 June 2010 (UTC)


 * Then please explain what you mean by "faster" without invoking scalar speeds, when you say "velocity component in that direction faster than the speed of the yacht in that particular direction"? What do you compare and how? Eyytee (talk) 07:57, 9 June 2010 (UTC)


 * Give me just a little slack. It would have been technically more correct to say "greater in magnitude" than "faster", but I suspect you did know that's what I meant. I will make that adjustment to my original text, thanks.  Paul Beardsell (talk) 08:07, 9 June 2010 (UTC)


 * I gave a new counter example to your claim based on your clarification. Find it below. Eyytee (talk) 11:14, 9 June 2010 (UTC)


 * and I am preparing example(s) to illustrate. Note also that 5 > -6 (if you find that patronising I apologise but there are lots of things that simple which I am having to explain), which is why sailing towards the wind presents no fundamental problem in my argument. It is DOWNWIND TO A POINT QUICKER THAN A BALLOON which is impossible.  Examples in prep.  Paul Beardsell (talk) 06:26, 9 June 2010 (UTC)


 * Don't spend too much time on it, because sail-powered vehicles CAN get to a point downwind faster than a balloon. Are you saying you don't believe yachts like USA-17 and almost any iceboat or land-yacht can't achieve VMG downwind at a multiple of the true wind speed????--Paul (talk) 06:45, 9 June 2010 (UTC)


 * You misuse the term VMG. By VMG we mean the velocity (not speed) towards the desired point, which in our case, is the point directly downwind, the point towards which the balloon is drifting.  As to having travelled on a downwind tack faster than the speed of the wind I have done so myself!  But the ice yacht travelling at 3x windspeed over the ice does so at such a shallow angle that it does not get to the downwind point before the balloon.  The ice/land/sea yacht is constrained thus, downwind:  (speed over the ground)/cos(angle to the true wind) < (speed of the true wind)  (speed over the ground) * cos(angle to the true wind) < (speed of the true wind) The term on the left of the inequality is component of the yacht's velocity directly downwind, the term to the right is that of the balloon.  If the inequality is not true then there is no possible impulse of the wind on the yacht in the windward direction.  Paul Beardsell (talk) 07:18, 9 June 2010 (UTC)


 * There is progress here! You (nearly) talk about the same thing: VMG = (speed over ground)*cos(angle to the true wind) (I guess we all agree it should be * and not / : VMG < speed over ground and cos ≤ 1). The "only" remaining difference now is that VMG > speed of the true wind (parallel vectors pointing in the same direction) has been achieved by many crafts, not least USA-17! That's an empirical fact, and there are models ('explanations' if you want) in the framework of classical Newtonian mechanics. MrBeanBob (talk) 07:39, 9 June 2010 (UTC)


 * thanks for spotting and fixing my typo Paul Beardsell (talk) 07:51, 9 June 2010 (UTC)


 * You and others allude to this evidence and then fail to produce it. This does not surprise me: One can talk of the impossible, one cannot do the impossible.  Paul Beardsell (talk) 07:47, 9 June 2010 (UTC)


 * No, the issue is that you are refusing to accept multiple citations that disprove your assertion.--Gautier lebon (talk) 15:35, 10 June 2010 (UTC)

Velocity = speed. Downwind VMG = progress over the water (or ground) towards the downwind mark = cos(ang) time velocity. USA 17 had VMG of 19 knots in 5-10 knots of wind. That is evidence. The models and explanations are found in the material that you deleted, which was all supported by citations, in particular to Bentwaithe's book. Benthwaithe provides polar charts showing VMG greater than wind speed.--Gautier lebon (talk) 07:49, 9 June 2010 (UTC)


 * No USA17 did not! It had a VMG to the mark but the mark was NOT directly downwind! This point has been made to you already by me and on another article's talk page by someone else. You did say you were sending me extract's from B's book.  I have not received them.  But no matter, I am sure.  Paul Beardsell (talk) 07:54, 9 June 2010 (UTC)


 * I have sent you the extracts by E-Mail on the morning of 9 June. If you have not received them, then please send me an E-Mail to that effect.  Re USA 17, the mark was dead downwind when the race started, and the wind shifts were not significant.  So, for all practical purposes, the VMG to the mark was downwind VMG.--Gautier lebon (talk) 15:35, 10 June 2010 (UTC)


 * VMG <= boat-speed. Where the desired point is directly downwind, VMG <= windspeed.  Note we talking about the component of the boats velocity directly downwind, not its speed over the ground.  Have I said that before?  Paul Beardsell (talk) 07:56, 9 June 2010 (UTC)


 * Yes, we understand what you are saying. The problem is that published polar charts show the opposite, as does data from USA 17, despite your skepticism (the wind shifts were nowhere near big enough to counter the fact that USA 17 had a downwind VMG of 19 knots in 5-10 knots of wind).--Gautier lebon (talk) 15:35, 10 June 2010 (UTC)

I have a very simple counter example that disproves PB's statement at the top of this section (in its current form) : Boat is moving perpendicularly to the true wind, and accelerates forward. The magnitude of the velocity component of the air in the direction of the boat's acceleration is zero. The magnitude of the velocity component of the boat in the direction of the boat's acceleration is greater than zero. This is exactly the opposite of PBs claim, that the magnitude of the air's velocity component (in boat's acceleration direction) must be greater than the magnitude of the boat's velocity component (in boat's acceleration direction) Eyytee (talk) 11:14, 9 June 2010 (UTC)

criticism 2
Paul Beardsell is making a fundamental error in the above analysis. He assumes that the boat interacts only with the air mass. He neglects the interaction with the water. While in reality the resulting acceleration of a boat depends on both forces(keel and sail), and vector diagrams show that the acceleration can be partially opposed to the apparent wind direction(Eyytee (talk) 15:52, 8 June 2010 (UTC))


 * Apparent wind direction is usually a red herring in this type of discussion. It is useful if you want to explain what is happening from the perspective of someone IN the boat. No, as physics allows me, I will continue to look at this from the stationary (in comparison to the ground) spectator viewpoint.  (Otherwise that the ground is moving relative to the observer makes it more complicated - our target point would be moving too!)  Having thus being allowed to discard apparent wind, where is the component of the wind pushing the vessel faster downwind than the wind itself!?!?  There cannot be one.  Paul Beardsell (talk) 16:14, 8 June 2010 (UTC)


 * Apparent wind is NOT a red herring. The sail sees only apparent wind and can react only to the apparent wind. This is elementary Newtonian physics.--Gautier lebon (talk) 07:59, 9 June 2010 (UTC)


 * What is also elementary is that the analysis can be done from a spectator's POV, one who is not in the boat and one who does not experience apparent wind. That is what is meant by the equivalence of frames of reference.  Paul Beardsell (talk) 08:29, 9 June 2010 (UTC)


 * Apparent wind direction and velocity is not a red herring. It is part of the physics of what allows a sail powered craft to travel faster than the wind. The sail is an airfoil, not just a flat obstruction to the wind. An airplane does not fly because the wing is tilted at an angle such that the wind pushes the plane up, it flies because of a pressure difference between the top and bottom of the wing caused by Bernoulli's principle.  For sailing craft that travel faster than the wind, the force generated on the airfoil by the apparent wind is an important part of the force driving the craft.  It isn't a perpetual motion machine, first, because it won't work without wind, and second, it does have a limiting case where drag prevents further acceleration. I have been sailing for close to 50 years, and when I first heard that modern mult-hulls sail downwind with apparent wind forward, I didn't believe it. But I did some research, and read up on vector analysis, and saw the truth. Everyone has a bit of difficulty understanding how this can be, but it is. Did you read the Physics Today article I linked to, above?--Paul (talk) 16:37, 8 June 2010 (UTC)


 * Very much a side issue but your understanding of aircraft flight is also limited! Whatever the mechanism, Bernoulli pressure differnces, whatever, those are one step removed from the basics. Newton still applies.  To exert a force upwards a force force downwards is required (Newton 3: every force etc).  Whatever the mechanism, whatever the meta-explanation, airfoil lift RESULTS from the directing of the airflow downwards.  Flight is not a special case:  Newton applies.  [Bernoulli simply helps explain the most effective shape of the airfoil.]  Paul Beardsell (talk) 19:51, 8 June 2010 (UTC)


 * Paul H has it right. An aircraft's wing sees only the apparent wind.  That is why you don't want to take off with a tail wind: it would take too long to accelerate enough so that the apparent wind allows you to take off.  All pilots are well aware of the fact that only the apparent wind counts, please read any elementary manual on flying.--Gautier lebon (talk) 07:59, 9 June 2010 (UTC)


 * Standing at the side of the runway I see the true wind, I see the speed of the a/c relative me. I add/subtract the two.  So what?  Both POVs are equivalent, the pilot's and the spectator's.  What happens in reality is the same in both frames.  some things are easier to measure and calculate in one or the other frame.  The pilot has it easier deciding his airspeed, the spectator has it easier calculating ground speed.  As for the boat I want to monitor progress to a point fixed my frame of reference which is not fixed in the sailors.  I'll present / I am presenting the maths.  Easier for this particular purpose in my shore-based frame.  Sail trimming? Easier if you're in the boat.  Paul Beardsell (talk) 08:26, 9 June 2010 (UTC)


 * Well, if you actually do work it out, you will see that it is far simpler in the frame of the boat. The same hold for planes: pilots look at the relative wind speed, not speed over the ground plus or minus true wind speed.--Gautier lebon (talk) 15:49, 10 June 2010 (UTC)


 * Maybe the term red herring is not the most helpful BUT one has to decide what one's coordinate system is and then stick with it. By changing one's frame of reference one can become disoriented and start believing three impossible things before breakfast. If one chooses one's frame of reference to be the vessel then analysis becomes difficult.  The ground is moving, not the boat!  A course change requires everything else to swivel including wind direction, the movement of the ground etc etc.  Paul Beardsell (talk) 19:34, 8 June 2010 (UTC)


 * This is not difficult at all. It is perfectly valid to use a reference frame where where the sail is at rest in order to compute the force on the sail. The air velocity in this frame is called "apparent wind". Since the coordinate transformation doesn't involve rotation, the force vector on the sail is the same as in the ground frame. Eyytee (talk) 05:55, 9 June 2010 (UTC)


 * It is perfectly valid but I will force you to do the maths if you insist on using that frame and you will find the trig more complicatedPaul Beardsell (talk) 07:44, 9 June 2010 (UTC)


 * Just to clarify: I'm not suggesting using the sail's frame for the entire analysis, just for the calculation of the force on the sail. This is the standard approach to calculate the force on an airfoil based on Angle_of_attack and Lift-to-drag_ratio. And this is not confusing because as I said: The coordinate transformation doesn't involve rotation, so the force vector in the ground frame is the same as in the sail's frame.


 * You on the other hand suggest, that we should compute the force on the airfoil WITHOUT using the air velocity relative to the airfoil(apparent wind). I will force you to do the maths if you insist on this non-standard approach.Eyytee (talk) 08:14, 9 June 2010 (UTC)


 * and I believe it will be difficult for readers to follow the maths. I am simply using another equally valid frame, as you acknowledge.  And, as I am sure you must know, the RESULTS calculated in one frame are the same as those calculated in every other frame - that is what equivalence means in this context.  If the boat does not beat the balloon in my "spectator on the shore" frame it does not do so in your "sailor on the boat" frame.  Paul Beardsell (talk) 07:44, 9 June 2010 (UTC)


 * Easier is to choose one's frame of reference as the ground. Then the boat moves and everything else remains constant!  Wind relative to the motion of the boat remains important to those IN THE BOAT but unimportant to those in OUR frame of reference.  Physics (in particular certain conservation principles) allows us to analyse the system from any frame of reference guaranteeing the same outcome.  That is why I suggest that the perspective of those in the boat be disregarded.  We will look at this from the shore.  The maths and the arguments become easier.  In particular we can disregard APPARENT WIND as the only wind we will witness is TRUE WIND, i.e. wind relative to the earth.  Those who do not follow this argument are disqualified from further comment.  [If you want to discuss Shakespeare you must at least read the plays!]  Paul Beardsell (talk) 19:34, 8 June 2010 (UTC)


 * Therefore I disregard any argument that depends upon apparent wind. Good bye!  Paul Beardsell (talk) 19:34, 8 June 2010 (UTC)


 * This means that you fail to understand the basic physical principles involved. Which makes it difficult to see how to continue the discussion.--Gautier lebon (talk) 07:59, 9 June 2010 (UTC)


 * Paul Beardsell seems to deny basic physical principles like the equivalence of inertial reference frames, by suggesting the Earth's frame is the only valid one, and generally denying the validity of analysis which uses a different frame. Eyytee (talk) 05:55, 9 June 2010 (UTC)


 * No, I do not. My argument depends on the equivalence of inertial frames. I am grateful to you acknowledging this in advance as it avoids argument later.  As every inertial frame is equivalent to every other I am doing the maths in the frame where the maths is easiest.  No very complicated trigonometry is required im my "spectator on the shore" frame of reference: I do not even have to calculate apparent wind. Much easier. But I am glad you acknowledge the frames' equivalence as then you will not be able to come back to me and say "but, the apparent wind is different" as you will acknowledge, in my equivalent frame, there is no apparent wind, only true wind.  Paul Beardsell (talk) 07:38, 9 June 2010 (UTC)


 * "In my equivalent frame, there is no apparent wind, only true wind". That is your fundamental fallacy.  The sails see the apparent wind, not the true wind.  If that were not the case, a boat sailing dead downwind with square sails would keep accelerating, becuase the true wind does not change.  But you know perfectly well that a boat with square sails sailing dead downwind cannot sail faster than the wind: and this is precisely because the apparent wind drops to zero, so there is no more any force that can accelerate the boat.  So the ONLY wind that matters is the apparent wind.  Again, this is not OR, it is basic physics of sailing, explained in all textbooks, including those referenced in the material that you deleted.--Gautier lebon (talk) 08:13, 9 June 2010 (UTC)


 * This is fundamentally nonsense. I ask you to think again. I have explained why below where at criticism 4. You need to review your physics.  In your sailing books they will be talking about sail trim, they talk about the POV of the sailor.  I use an alternative easier but equivalent frame of ref.  Equivalent.  Please think again.  08:19, 9 June 2010 (UTC)


 * Sails are trimmed with respect to the apparent wind in order to optimize the force generated by the APPARENT wind. The sails generate a force, which drives the boat (technically, the force causes an acceleration, which results in an increase in velocity).  The friction from the surface and the resistance of the APPARENT headwind generate an opposite force, so eventually the boat stops accelerating. The point is that the driving force, the one from the sails, is generated by the APPARENT wind, and that is what you have to use when doing the physics.  Again, this is explained in all basic textbooks, which disuss not only trim, but also the forces and the speed vectors.--Gautier lebon (talk) 15:53, 10 June 2010 (UTC)


 * Hi Paul H (this is going to be fun - two Pauls!). If you look at Lift (force) you will see that the Bernoulli explanation of lift is now discredited. And anyway, thin cloth sails are so easy to understand, we do not need it. Paul B is nearly correct saying that, "The wind force experienced by a sail is perpendicular to the set of the sail". The force due to the wind on every square inch of sail-cloth has a component at a right-angle to the square inch, and another (hopefully much smaller one) roughly backwards along the cloth due to drag. The curve or 'belly' near the front of a close-hauled sail (the luff) contains lots of square inches facing almost forwards, and it is largely held full and pressurised by the existence of the rest of the sail creating a nice trailing edge for, and maintaining laminar flow over, this driving front part near the luff. This is what allows sailboats to do the apparently magical thing of making progress upwind. Overtaking the wind downwind is no more special than this. You are right that the key to it is to go so fast on a broad reach that you end up close hauled, but once you do, I'm sure, looking at that luff and imagining all the force-vectors sticking out of the forward-facing side of the sailcloth, it will make as much sense as when imagining them on a beat to windward. You can't overtake the wind in the same direction as the wind, but you can at an angle to it, provided you can sail very efficiently very close-hauled. So, my point is, there is no problem with the physics, we just need reliable sources that say people have done it, and hopefully to explain the physics as well (since we must not reference our own explanations). What I'm hoping to do is to convince Paul B that the physics is not screwy, so that he starts to help us find these references, rather than saying they can't exist. --Nigelj (talk) 18:15, 8 June 2010 (UTC)


 * Nigel: I already provided the references, but Paul Beardsell deleted them. Bentwaithe's book explains all this (and more) in excruciating detail.  And explains how high VMG is achieved. Again, all I did was summarize what is in Bentwaithe and the other cited sources.  But Paul Beardsell has deleted all of that because he thinks that he knows better.  It this any way to run Wikipedia?--Gautier lebon (talk) 07:59, 9 June 2010 (UTC)


 * No, too difficult, I'm on the shore, you want to invoke relative (or apparent) wind without doing the very complex maths that requires. You're out of order!  (Different argument:  I use the "set of the sail" as a device to average out the sail's curvature and drag.  There is no important consequence of this.)  Paul Beardsell (talk) 19:34, 8 June 2010 (UTC)


 * Actually the math is not at all complicated in the frame of the boat (not the frame of the wind): it is elementary trigonometry and you will find it in the article. Again, you will also find it in basic texbooks on sailing.  And I would appreciate it if you stopped using expressions such as "you are out of order".--Gautier lebon (talk) 15:56, 10 June 2010 (UTC)

Paul Beardsell is correct in saying that the frame does not matter. However, the vector algebra is easier in some frames than in others. The deleted material used the frames that are adopted by all textbooks, precisely because the presentation is simpler. I still cannot understand why Paul refuses to look at Bentwaithe's book, which explains it all.--Gautier lebon (talk) 07:59, 9 June 2010 (UTC)


 * I do not have the book. GL said he would send me extracts. Not received. Paul Beardsell (talk) 08:21, 9 June 2010 (UTC)


 * I have sent you the extracts by E-Mail on 9 June at 8:38 European time. If you did not receive them, then please send me an E-Mail to that effect.--Gautier lebon (talk) 15:28, 10 June 2010 (UTC)

criticism 3
Your criticism here [where? this used to be part of a different thread until moved Nigelj (talk) 20:04, 8 June 2010 (UTC)] is based on the 'conservation of velocity', which is not a recognised principle. 'Conservation of momentum' is, but as Eyytee says below [well, it's above now, but who cares anymore? Nigelj (talk) 20:04, 8 June 2010 (UTC)], in order to 'see' momentum being conserved as a sailboat goes by, you would have to 'see' all the disturbances left behind in the air and the water after the sails and the keel have done their thing. Racing sailors are very familiar with 'dirty air' downwind of another boat, but to me that is all too complicated to visualise in such detail. In answer to your rhetorical question, the yacht is going faster than the south-going air, but in a south-westerly direction. By re-using the same magic that allows it to sail upwind, say 30 deg off the wind, it has now overtaken the southerly vector of the wind and is using it's 'upwind' ability to sail 30 deg off an apparent wind that is made up mostly of the wind due to its own movement, plus a small reduction and an angle in that wind due to the angle between its movement and the real wind. Just like in the vector diagrams in the article. One thing that's weird is that, unusually, the wind due to the boat's movement is much larger than the real wind. The real wind mostly serves to add a small angle to the 'movement wind', an angle that this very efficient boat can continue to claw itself along by, close hauled, getting just enough drive to overcome its minimal hull-drag. --Nigelj (talk) 18:39, 8 June 2010 (UTC)


 * No, I never said "conservation of velocity". Yet you are wrong to seemingly ridicule the term, it is Newton One and is thus PROFOUNDLY TRUE.  Nevertheless I did not invoke it.  And you cannot exert a force in particular direction in collision with a 2nd object unless one is travelling faster in that directon than the 2nd object.  That is why the wind cannot propel you faster than itself IN THE DIRECTION of the wind.  I.e., to the physics-illiterate: that is why downwind sailing to the directly downwind point quicker than the free floating balloon is impossible.  Paul Beardsell (talk) 19:16, 8 June 2010 (UTC)


 * Once again, you fail to undertand that the only relevant wind is the APPARENT wind, not the true wind. Have you actually ever sailed a boat?  All sailors know that sails are trimmed with respect to the apparent wind, not with respect to the true wind. And that it is the force of the apparent wind that propels the boat.  That's why it is so slow to go dead downwind with square sails: the apparent wind is pretty weak.--Gautier lebon (talk) 08:07, 9 June 2010 (UTC)


 * Sirclicksalot (talk) 16:26, 17 June 2010 (UTC): Mr. Beardsell, it is your application of physics that is not up to the task. I am not insulting your physics literacy:  I believe you understand the basic principles well enough to overcome your incorrect first impression of the physics involved.  However, you do need to be more careful when you analyze the physics and not be limited by your first intuitive response (as so many of us have).  My response was the same as yours and I had put out one foolish statement (elsewhere in one of the DDWFTTW threads) before I drew the simple diagram that convinced me.  I have shown (above) in your preferred water-fixed reference frame - and can show in any reference frame - that the hull of a boat can be making VMG downwind faster than the true wind (i.e. hullspeed * cosine(angle between course and downwind) > [true windspeed]) while still having air particles moving at the true wind velocity colliding with the windward side of the sail of that same boat.  Before you reflexively post "impossible" consider the implications of the following:


 * the primary force of the true wind on the boat is against the sail, and
 * the sail's orientation is not perpendicular to the true wind,


 * so you can draw your diagrams carefully and correctly this time; personally I look forward to those diagrams because we all learn from seeing different perspectives on any topic. Once that is done, note that there are recorded on this page a very many wrong things you have stated not only with a great deal of misplaced certainty but also in an ungracious manner.  You have acted not so much, as you perhaps imagined, as the misunderstood prophet truthfully declaring the Holiness of God to a lost and sinful people, but more as the drunk in the middle of the street directing traffic.  I note that you are a consultant and therefore expect you may wish to be careful about your web-public image; what you do next determines how you will be remembered for the future.  If you apologize, all is forgiven (it will stroke our egos so much, that we actually taught someone else something, that we will forget what you said and how you said it), and welcome to the fold of those who have misapplied Sir Newton but realized their error (we're having jackets made up and there's a twelve-step program;-).  If instead you are acting the part of a troll, then know that the game is up (see below), but looking at all the time others have spent responding to your posts you should consider it a job well done and know that you have perhaps helped improved their explanations on this topic.  Either way, best regards.  Your servant, Sirclicksalot (talk) 16:26, 17 June 2010 (UTC)

OK, I'm outa here. If you are going to start rearranging other people's helpful comments so that they make no sense any more, and flooding the page with your own disordered comments, and getting abusive ("physics-illiterate" indeed!) and SHOUTING, then you can continue to argue without me. After you're done, we'll restore the page to a consensus version, but this is not helpful to improving the article at the moment. --Nigelj (talk) 20:04, 8 June 2010 (UTC)


 * Play the ball. I moved your critique to a subsection to allow BETTER addressing of your points. Sorry about Newton 1, but that was your error not mine.  Heat, kitchen etc Paul Beardsell (talk) 20:12, 8 June 2010 (UTC)


 * I am leaning towards agreeing with Nigel. As all of you know, I am very patient and more than willing to try to explain admittedly counter-intuitive concepts.  But, given my two years of physics at MIT (followed by a BA in Math and a PhD from Harvard), I am beginning to be a bit tired of being told that I don't understand physics 101, especially when the physics in question can be found in any basic book on sailing.  During the week end, I will try to produce the force diagrams that were mentioned earlier, and that show how an iceboat sailing at 135 degrees off the true wind will easily accelerate to a speed that results in VMG far greater than the wind.  Those force diagrams will also show why there is no issue of conservation of energy.  If that still does not convince Paul Beardsell, then we might have to envisage another approach.--Gautier lebon (talk) 08:04, 9 June 2010 (UTC)

criticism 4
Correct. But what counts is the APPARENT wind, not the true wind. That was explained in the material that you deleted.--Gautier lebon (talk) 07:44, 9 June 2010 (UTC)


 * Anyone with any physics (I know you have some) will tell you that all inertial frames of reference are equivalent. If you do the analysis from the POV of the sailor in the boat you will need to cope with his measurements of the wind, speed and direction, the apparent wind.  If you do the analysis from the shore then the wind is the true wind.  We all are assuming the true wind remains constant during our stopwatch-speed and balloon tests.  In the shore-frame-of-ref the wind is always the same.  In the boat it varies for every point of sail and every boat speed.  The boat-frame-ref is difficult to analyse.  The shore-frame-ref is EASY.  But they are equivalent (=equally valid i.e. give the same results).  That's basic physics.  I choose the easy frame.  If you choose the other frame I bet you cannot do the maths - no one here has done it so far.  I can do the maths in the easy frame and I think you can follow.  If you believe the assertion of the equivalence of the frames of ref.  Paul Beardsell (talk) 08:16, 9 June 2010 (UTC)


 * Of course the frames are equivalent, but the easy frame is the boat's frame: that is why all textbooks use that frame. The shore frame is hard to get right, and that may explain why you are having trouble seeing this.  Please try the following: imagine an iceboat that is sailing at 135 degrees off the true wind, and work out the speed and force diagrams relative to the boat as it accelerates.  Then, if you wish, convert them to the stationary frame of the earth.  The mathematics of the boat frame are simple, they are presented in the article.  The mathematics of the shore frame are more complicated, which is why you won't find them in textbooks.--Gautier lebon (talk) 15:26, 10 June 2010 (UTC)