Understanding DDWFTTW: Exploring Its Principles and Addressing Common Questions

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In summary, the propeller can apply more force at faster-than-wind speeds because it does not travel as far through the air. This allows the cart to extract more power from the wind.
  • #36
rorix_bw said:
@thinairdesign:

Are you saying Blackbird - if it floated - would qualify for Americas Cup? I would like a citation on that if possible.

If that's a serious question on your part I'm going to have to conclude you aren't serious about the exchange.

You have asserted is that the BB is not equipped with sails. Several people have asked you for your definition of a "sail" and you have yet to respond. I only pointed out that *if* your definition includes the 'hard sail' wings such as used in the America's Cup, then the BB does indeed use sails.

Is there a separate Americas Cup for land?

Actually there is.

It's not just a rigid wingsail: it's a set of them on a rotating mount, like windmill.

And has been pointed out to you, both the AC wing sails and the BB sails are rotating -- only the diameter of the circle is different -- hardly a sound distinction

Further, some of those designs link them to wheels or propellers.

The connection between the sails and the wheels on the BB is the exact same kinematic constraint as the link between the sail and the keel on the AC boats. In both cases, this constraint ensures that for every the sail travels downwind, it must also travel some distance cross wind. Without this constraint, both the AC boat and the BB would simply drift downwind at a speed less than the wind.

I don't know to call it.

If you learn how they work you will know what to call them -- "sails". It's what they do and what they are.

JB
 
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  • #37
jduffy77 said:
I read it, I guess I did not understand it. Could you elaborate on why you think the sail to prop video is a flawed analogy? I agree that in scenario 1 we see a sail tacking faster than the wind but not directly down wind. It is in scenario 2 that we seem to disagree. If we tack around a cylinder we are clearly going directly downwind, faster than the wind.

For tacking to allow velocity made good to exceed windspeed, the angle between sails and the apparent wind cannot be held fixed with time as has been depicted in those videos including the gear videos and the sail to prop video. Sailcraft slow down when facing headwind so it cannot be steady state on wind power alone.
 
  • #38
kmarinas86 said:
For tacking to allow velocity made good to exceed windspeed, the angle between sails and the apparent wind cannot be held fixed with time as has been depicted in those videos including the gear videos and the sail to prop videos. Sailcraft slows down when facing headwind so it cannot be steady state on wind power alone.

Conventional sail craft can't. The prop driven cart can. That is the whole point of the sail to prop simulation video. It is explaining how it is accomplished. Perhaps if you point out exactly what you think is wrong with it I could help you understand. Assuming I have not misunderstood you and you do think something is wrong.
 
  • #39
kmarinas86 said:
For tacking to allow velocity made good to exceed windspeed, the angle between sails and the apparent wind cannot be held fixed with time.
An ice boat can maintain a steady speed with velocity made good down wind exceeding wind speed indefinitely (as long as there's room) with the sail at a fixed angle. There is a vector chart in this article:

ice_boat_performance.pdf

that shows a true wind speed of 18 mph, ice boat speed of 70 mph, and apparent wind of 55 mph. Cleaning up the numbers a bit, change apparent wind to 55.15 mph. This corresponds to a true wind heading of 0° at 18 mph, and an ice boat heading of 30° at 70 mph, with an apparent wind heading of -140.61° at 55.15 mph (= 39.39° at -55.15 mph). The apparent wind can be separated into components perpendicular and in the direction of the ice boat. The apparent crosswind is 18 mph x sin(30) = 9 mph, regardless of the ice boats speed. The apparent headwind is ice boat speed - 18 x cos(30) = sqrt(apparent_wind2 - apparent_crosswind2) = 54.41 mph.

The key factor here is that apparent crosswind is independent of the ice boat's speed, so the maximum speed is related to the maximum ratio of apparent headwind versus apparent crosswind that the ice boat can achieve for a given true wind and ice boat heading relative to true wind.

The other requirement is that the ice boat's sail must divert the apparent wind upwind by some amount, so that the diverted apparent wind slows some of the true wind air (wrt ground), since slowing the air is the source of power for an ice boat.

I assume that land boats can also achieve vmg downwind greater than wind speed. I don't know if a catamaran in water could accomplish this.
 
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  • #40
jduffy77 said:
Conventional sail craft can't. The prop driven cart can. That is the whole point of the sail to prop simulation video. It is explaining how it is accomplished. Perhaps if you point out exactly what you think is wrong with it I could help you understand. Assuming I have not misunderstood you and you do think something is wrong.

Semantics have confused some of the discussion here.

I don't refer to the prop driven cart as a "sailcraft".

The videos I am referring to are those uploaded to the YouTube channels of xelerat3d and eyytee:

http://www.youtube.com/user/axelerat3d/videos
* (This video is mistitled as "DDWFTTW" when it is only "DWFTTW")
* (This video presents tacking as sufficient for exceeding wind speed in the downwind direction. It's not. You must change the tack back and forth (i.e. jibe) as was explained in comments in the "DDWFTTW" video).

http://www.youtube.com/user/eyytee/videos (All the following videos are completely flawed as they treat the air mass as pushing the craft despite the craft clearly having a headwind which should by Newton's laws help to slow it down.)
*
*
*

These three videos display the same constant angle assumption applied to a sail, which is not at all analogous to the propeller+transmission+wheel apparatus of the DDWFTTW land yacht, in an incorrect attempt to explain the DDWFTTW land yacht, in a similar manner to one of axelerat3d's videos:

 
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  • #41
rcgldr said:
An ice boat can maintain a steady speed with velocity made good exceeding wind speed indefinitely (as long as there's room) with the sail at a fixed angle. There is a vector chart in this article:

ice_boat_performance.pdf

that shows a true wind speed of 18 mph, ice boat speed of 70 mph, and apparent wind of 55 mph. Cleaning up the numbers a bit, change apparent wind to 55.15 mph. This corresponds to a true wind heading of 0° at 18 mph, and an ice boat heading of 30° at 70 mph, with an apparent wind heading of -140.61° at 55.15 mph (= 39.39° at -55.15 mph). The apparent wind can be separated into components perpendicular and in the direction of the ice boat. The apparent crosswind is 18 mph x sin(30) = 9 mph, regardless of the ice boats speed. The apparent headwind is ice boat speed - 18 x cos(30) = sqrt(apparent_wind2 - apparent_crosswind2) = 54.41 mph.

The key factor here is that apparent crosswind is independent of the ice boat's speed, so the maximum speed is related to the maximum ratio of apparent headwind versus apparent crosswind that the ice boat can achieve for a given true wind and ice boat heading relative to true wind.

The other requirement is that the ice boat's sail must divert the apparent wind upwind by some amount, so that the diverted apparent wind slows some of the true wind air (wrt ground), since slowing the air is the source of power for an ice boat.

I assume that land boats can also achieve vmg greater than wind speed. I don't know if a catamaran in water could accomplish this.

The document you cite (http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf) has a chart on page 2 of 6 that proves that the ice boats in question were not in a steady state (i.e. not having both constant speed and direction simultaneously).

In a few past posts of mine here, I made the mistake of using the term "velocity made good" where I have actually meant "velocity made good, assuming that the next mark is directly downwind as is (effectively) the case of the tests of the Blackbird DD(T)WFTTW land yacht."

Directly (or dead) downwind means alpha=180 degrees (See http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind#Speed_made_good and http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind#Sailing_dead_downwind_faster_than_the_wind).
 
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  • #42
kmarinas86 said:
Semantics have confused some of the discussion here.

I don't refer to the prop driven cart as a "sailcraft".

The videos I am referring to are those uploaded to the YouTube channels of xelerat3d and eyytee:

http://www.youtube.com/user/axelerat3d/videos
* (This video is mistitled as "DDWFTTW" when it is only "DWFTTW")
* (This video presents tacking as sufficient for exceeding wind speed in the downward direction. It's not. You must change the tacking (jibing) as was explained in comments in the "DDWFTTW" video).


The prop blade is doing just that. It is turning around an axis so the blade surface is presenting a constant angle to the wind just like the tacking sail.

kmarinas86 said:
http://www.youtube.com/user/eyytee/videos (All the following videos are completely flawed as they treat the air mass as pushing the craft despite the craft clearly having a headwind which should by Newton's laws help to slow it down.)
*
*
*

These three videos display the same constant angle assumption applied to a sail, which is not at all analogous to the propeller+transmission+wheel apparatus of the DDWFTTW land yacht, in an incorrect attempt to explain the DDWFTTW land yacht, in a similar manner to one of axelerat3d's videos:



Unfortunately it is clear that you do not understand the cart mechanism. These three videos are not incorrect. The first one especially describes the cart system exactly and completely.
 
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  • #43
kmarinas86 said:
The document you cite has a chart on page 2 of 6 that proves that ice boats in question were not in a steady state.
Yes, in real world circumstances, the wind speed was changing as well as the heading, but during the downwind tacks (near constant heading) the ice boat's average vmg downwind exceeded wind speed. I was trying to eliminate the concept that the sail was somehow being pumped to create a momentary burst in speed.
 
  • #44
jduffy77 said:
The prop blade is doing just that. It is turning around an axis so the blade surface is presenting a constant angle to the wind just like the tacking sail.
Unfortunately it is clear that you do not understand the cart mechanism. These three videos are not incorrect. The first one especially describes the cart system exactly and completely.

Install a second identical propeller on the same DDTWFTTW vehicle but facing the other way. Attach this second propeller to another set of four wheels through an identical transmission placed just as the first was (of course in the same vehicle!). You kept the same (absolute magnitude of the) angles (with respect to the vehicle and the wind) and the rotational rates and thus the tacking, but then watch the vehicle not be able to go Dead-DTWFTTW as the two props will cancel each others' thrust!

The correct way:
attachment.php?attachmentid=42345&d=1325349305.gif
 
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  • #45
kmarinas86 said:
(All the following videos are completely flawed as they treat the air mass as pushing the craft despite the craft clearly having a headwind which should by Newton's laws help to slow it down.)

It is this statement which illustrates your confusion most clearly. The cart is not being pushed by the air mass. The wheels spin the prop to create thrust. This slows the air mass with respect to the ground. In this way it is taking kinetic energy from the air and using it to power the cart. If the thrust generated by the prop is greater than the losses due to aero drag and rolling resistance the fact that there is a headwind relative to the craft is irrelevant.
 
  • #46
jduffy77 said:
It is this statement which illustrates your confusion most clearly. The cart is not being pushed by the air mass. The wheels spin the prop to create thrust. This slows the air mass with respect to the ground. In this way it is taking kinetic energy from the air and using it to power the cart. If the thrust generated by the prop is greater than the losses due to aero drag and rolling resistance the fact that there is a headwind relative to the craft is irrelevant.

The cart, when it begins operating in the Blackbird test, it is clear that the tailwind consisting of air mass pushes this car. This is represented by the left side of the line graph in the following picture.

attachment.php?attachmentid=42345&d=1325349305.gif


The wheels are spun because the tailwind pushes the frame of the vehicle which holds the axles. So the axles are pushed, dragging the wheels over the ground, causing them to rotate (due to ground friction) which, by connection with the transmission, spins the prop in the direction opposite that the tailwind wants to turn it. In other words, the power of the tailwind that is diverted to turn propeller to generate a "head" wind component that passes from the propeller in the thrustward direction (opposite of the tailwind). Two "winds" blowing at each other (opposite directions toward one another) increases the pressure, and such increases the potential energy stored in the air (at the expense of kinetic energy of the winds). So energy is stored entirely outside the vehicle, if we ignore the phonons traveling through the vehicle's structure. The latter is of little importance as its energy density is very small (acoustic resonance is not an important feature of the Blackbird vehicle).

The faster the vehicle moves, the faster the prop turns. When the vehicle velocity matches the wind velocity, the props will still produce thrust. Thus, steady state for a successful DDTWFTTW vehicle is not at wind velocity (with respect to initial frame of reference) but faster than it (as per the above diagram).
 
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  • #47
kmarinas86 said:
Install a second identical propeller on the same DDTWFTTW vehicle but facing the other way. Attach this second propeller to another set of four wheels through an identical transmission placed just as the first was (of course in the same vehicle!). You kept the same (absolute magnitude of the) angles (with respect to the vehicle and the wind) and the rotational rates and thus the tacking, but then watch the vehicle not be able to go Dead-DTWFTTW as the two props will cancel each others' thrust!

I do not know what you mean by this or what it has to do with our topic. Do you mean install a turbine? A propeller facing the other way through an identical transmission would not function.
 
  • #48
kmarinas86 said:
The cart, when it begins operating in the dry lake bed test (i.e. the Blackbird test), it is clear that the tailwind consisting of air mass pushes this car. This is represented by the left side of the line graph in the following picture.

attachment.php?attachmentid=42345&d=1325349305.gif


The wheels are spun because the tailwind pushes the frame of the vehicle which holds the axles. So the axles are pushed, dragging the wheels over the ground, causing them to rotate to spin the prop in the direction opposite that the tailwind wants to turn it. In other words, the power of the tailwind that is diverted to turn propeller to generate a "head" wind component that passes from the propeller in the thrustward direction (opposite of the tailwind).

The faster the vehicles move, the faster this prop turns. When the vehicle velocity matches the wind velocity, the props will still produce thrust. Thus, steady state for a successful DDTWFTTW vehicle is not at wind velocity (with respect to initial frame of reference) but faster than it.

Right. I agree. About 3x faster in the case of the current cart. What are we arguing about again? The applicability of the tacking analogy?
 
  • #49
jduffy77 said:
I do not know what you mean by this or what it has to do with our topic. Do you mean install a turbine? A propeller facing the other way through an identical transmission would not function.

It would provide thrust in the other direction.

This invalidates the argument that all you need for the craft to work is for angled sails to revolve around a "cylindrical earth" (so to speak). Counter-thrusting the props (matched-angled sails ("matched-angled" due to the 180-degree rotation) moving in opposite directions around this "cylindrical earth" (so to speak)) would eliminate the very thing that the Blackbird requires to operate DDTWFTTW: the propeller thrust as advantaged by the interaction between the ground, wheels, transmission, frame, propellers, and the wind.

Note: The notion of a "cylindrical earth" is found in Scenario 2 of the video that has been referenced in this thread several times.



In both scenarios, the sail (seen here as not jibbing) should only be able to beat the balloons drifting with the wind if the wind is not parallel to the left-to-right axis. However, in DDTWFTTW, velocity made good must be parallel to downwind (and anti-parallel to upwind). The video therefore relies upon two contradicting assumptions about the direction of the wind even though it speaks of only one of them.
 
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  • #50
kmarinas86 said:
In other words, the power of the tailwind that is diverted to turn propeller to generate a "head" wind component that passes from the propeller in the thrustward direction (opposite of the tailwind). Two "winds" blowing at each other (opposite directions toward one another) increases the pressure, and such increases the potential energy stored in the air (at the expense of kinetic energy of the winds). So energy is stored entirely outside the vehicle, if we ignore the phonons traveling through the vehicle's structure.

I do not see how an increase in pressure has anything to do with it. The prop is slowing down the wind. There is no increase in potential energy stored in the air at any point. The prop together with the carts drive mechanism is acting as an air brake.
 
  • #51
kmarinas86 said:
It would provide thrust in the other direction.

It would not do so using an identical transmission.

kmarinas86 said:
This invalidates the argument that all you need for the craft to work is for angled sails to revolve around a "cylindrical earth" (so to speak). Counter-thrusting the props (matched-angled sails ("matched-angled" due to the 180-degree rotation) moving in opposite directions around this "cylindrical earth" (so to speak)) would eliminate the very thing that the Blackbird requires to operate DDTWFTTW: the propeller thrust as advantaged by the interaction between the ground, wheels, transmission, frame, propellers, and the wind.

Note: The notion of a "cylindrical earth" is found in Scenario 2 of the video that has been referenced in this thread several times.



It is not an "argument" to say that angled sails are revolving around a "cylindrical earth". It is a fact. It is called the prop. I do not see how your second prop scenario would invalidate anything even if it were true. You have yet to say exactly what is wrong with the scenario in the video. Do you not believe it could be built?
 
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  • #52
jduffy77 said:
I do not see how an increase in pressure has anything to do with it. The prop is slowing down the wind. There is no increase in potential energy stored in the air at any point. The prop together with the carts drive mechanism is acting as an air brake.

Two winds blowing on each other is just like two like-charges repelling each other, so it is wrong to say that opposing winds cannot store potential energy. Heat does not form immediately. Energy in opposing winds produces pressure (both dynamic pressure and static pressure) before the stored energy can be released as heat.
 
  • #53
jduffy77 said:
A propeller facing the other way through an identical transmission would not function.
The end result would be a directly downwind slower than the wind vehicle. Look at the graph posted by A.T., if the propeller is "facing the wrong way", the vehicle advance ratio is negative, downwind but slower than wind speed. An advance ratio greater than zero but less than 1 resutls in DDWFTTW, an advance ratio greater than 1 results in an upwind vehicle (between 1 and 2, upwind faster than wind on an idealized vehicle (no losses), greater than 2, upwind slower than wind).

jduffy77 said:
I do not see how an increase in pressure has anything to do with it. The prop is slowing down the wind.
The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop. Using a ground based frame of reference, the total energy of the affected air is decreased and consumed by all the losses in a DDWFTTW vehicle, aerodynamic drag (due to relative headwind), drivetrain losses, rolling resistance, ..., and consumed by the increase in kinetic energy of the vehicle during acceleration.
 
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  • #54
kmarinas86 said:
A sail boat whose velocity made good exceeds wind speed uses water as a reaction mass to attain this excess speed (for all one must do with that is to turn the vehicle, after it exceeds wind speed, toward direct downwind),
No. A sailcraft can achieve and hold a steady state downwind VMG > true wind at a constant course in constant true wind. You don't have to turn the vehicle at all.

http://img253.imageshack.us/img253/6694/downwindvectorsen3.png

kmarinas86 said:
On top of this, many of your videos (the ones showing the gears) apparently depict the tailwind as providing a steady thrust when actually that tailwind is replaced by an unmistakable headwind.
Propellers can produce thrust in apparent head wind.
kmarinas86 said:
The sailboats simply cannot exceed windspeed downwind without changing the angle of the apparent wind...
For a sailboat the angle of the apparent wind changes, even when you accelerate at a constant course in constant true wind. The angle of attack at the propeller blades changes in the same way during acceleration.
kmarinas86 said:
...either by turning the boat (requiring a reaction mass) and/or the wind itself.
Nope. Constant course. Constant true wind. Steady state downwind VMG > true wind. See vectors above.
kmarinas86 said:
Thus, your analogy with sail boats is flawed.
Once you understand sailboats, you will see that the analogy is perfectly valid.
 
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  • #55
kmarinas86 said:


In both scenarios, the sail (seen here as not jibbing) should only be able to beat the balloons drifting with the wind if the wind is not parallel to the left-to-right axis.


The true wind is perpendicular to the blue stripes on the surface. The balloon goes directly downwind at windspeed. The boat goes at an angle to the true wind, with a downwind VMG > true wind. So the boat outruns the balloon along the true wind direction. This is a physically perfectly valid steady state situation (see vectors in previous post).

Jibbing is irrelevant to achieve this. In fact, changing tacks only slows down the boat as it goes through DDW. The continuous helical tack of the propeller blades eliminates that problem.
 
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  • #56
rcgldr said:
The end result would be a directly downwind slower than the wind vehicle. Look at the graph posted by A.T., if the propeller is "facing the wrong way", the vehicle advance ratio is negative, downwind but slower than wind speed. An advance ratio greater than zero but less than 1 resutls in DDWFTTW, an advance ratio greater than 1 results in an upwind vehicle (between 1 and 2, upwind faster than wind on an idealized vehicle (no losses), greater than 2, upwind slower than wind).

A propeller facing the other way using the same transmission would not function. As I mentioned if you changed the angle of attack to the point where you had a turbine you would be able to generate an upwind force vector. As I mentioned though I do not think this has any bearing at all on the applicability of the tacking prop analogy, least of all invalidating it.

rcgldr said:
The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop. Using a ground based frame of reference, the total energy of the affected air is decreased and consumed by all the losses in a DDWFTTW vehicle, aerodynamic drag (due to relative headwind), drivetrain losses, rolling resistance, ..., and consumed by the increase in kinetic energy of the vehicle during acceleration.

I agree with all of the above as long as you mean the area of decreased pressure is in front of the prop, downwind relative to the cart. The cart is functioning exactly as a prop plane would except instead of engines the prop cart is powered by the wheels. As long as that is what is meant by the following explanation then ok.

kmarinas86 said:
The wheels are spun because the tailwind pushes the frame of the vehicle which holds the axles. So the axles are pushed, dragging the wheels over the ground, causing them to rotate (due to ground friction) which, by connection with the transmission, spins the prop in the direction opposite that the tailwind wants to turn it. In other words, the power of the tailwind that is diverted to turn propeller to generate a "head" wind component that passes from the propeller in the thrustward direction (opposite of the tailwind). Two "winds" blowing at each other (opposite directions toward one another) increases the pressure, and such increases the potential energy stored in the air (at the expense of kinetic energy of the winds). So energy is stored entirely outside the vehicle, if we ignore the phonons traveling through the vehicle's structure. The latter is of little importance as its energy density is very small (acoustic resonance is not an important feature of the Blackbird vehicle).

I do not think it is a very accurate or illuminating explanation of what is happening with the cart. I would be curious to get your opinion on that.
 
  • #57
@kmarinas86:

It appears that you do not believe that traditionally rigged sailing vessels (be they on land, water or ice) do not and cannot achieve VMGs of greater than wind speed while on a steady course heading. This is simply not true. Until you correct this aspect of your physics understanding, you will be fighting a steep uphill battle in understanding the DDWFTTW vehicles.

Perhaps this link would help with your understanding of the topic. It contains all the maths, vector diagrams and real world examples (along with plenty of enlightening exchanges on the discussion pages) that show your position on this to be flawed.

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

Traditionally rigged sailing vessels easily and regularly exceed 1.0 vmg on steady course -- often by a very large margin.

JB
 
  • #58
kmarinas86 said:
Two winds blowing on each other is just like two like-charges repelling each other, so it is wrong to say that opposing winds cannot store potential energy. Heat does not form immediately. Energy in opposing winds produces pressure (both dynamic pressure and static pressure) before the stored energy can be released as heat.

Whether or not it makes sense to talk about winds "storing potential energy" I will leave to others more knowledgeable. I will say it does not make make sense to talk about in the context of the cart. No one is saying the cart is 100% efficient. Obviously there is energy being converted to heat at the propeller air interface as well as all over the drive train. What does it have to do with what we are talking about? The cart takes kinetic energy from the air by slowing it with respect to the Earth and uses it to accelerate ddwfttw.
 
  • #59
jduffy77 said:
A propeller facing the other way using the same transmission would not function.
This is how the Bauer DDWFTTW cart started off, it had a variable pitch prop and started off in turbine mode, where the wind drove the propeller which drove the wheels (DDW slower TTW). Once pitch of the propeller is increased from negative to positive but less than 1:1 ratio, the setup is in DDWFTTW mode and the wheels drive the propeller that produces thrust against the wind. The chart that A.T. created on the first page of this thread shows this.

jduffy77 said:
As I mentioned if you changed the angle of attack to the point where you had a turbine you would be able to generate an upwind force vector.
But that angle of attack corresponds to a vehicle advance ratio greater than 1. Note that in the upwind case, both the propeller and the wheels turn the opposite direction as the downwind case, so the transmission direction doesn't change, only the ratio.

rcgldr said:
The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop.

jduffy77 said:
I agree with all of the above as long as you mean the area of decreased pressure is in front of the prop, downwind relative to the cart.
It's what I meant, although the wording was a bit awkward. The main point was that relative to the ground, the prop extracts energy from the air.

jduffy77 said:
As I mentioned though I do not think this has any bearing at all on the applicability of the tacking prop analogy, least of all invalidating it.
A propeller may be more efficient than other means of generating thrust, but the tacking part of the thrust generation isn't a requrement. Ignoring efficiency issues, any method that uses a force from the wheels to generate thrust (greater force, slower speed) would work for a DDWFTTW cart, and no component of movment perpendicular to the true wind is required for a DDWFTTW cart. Note that for an under the ruler faster than the ruler vehicle, tacking is not used:



The point here is that tacking surfaces aren't required for a DDWFTTW cart, but since a propeller is more efficient at producing thrust than a paddle wheel type device, a propeller is used.
 
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  • #60
ThinAirDesign said:
@kmarinas86:

It appears that you do not believe that traditionally rigged sailing vessels (be they on land, water or ice) do not and cannot achieve VMGs of greater than wind speed while on a steady course heading. This is simply not true.

It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!). If you are able to travel a straight line and "beat the wind" with a sail, then the wind has to be changing direction over time. In reality, wind does exactly that, but the changing of the wind's direction is supposed to be negligible for the very scenario tested by DDWFTTW Blackbird sand yacht. So for the test, you must rely upon a steady wind of unchanging direction and travel parallel only with respect to that wind.



The blades travel faster than wind speed, but they do not move directly downwind.

ThinAirDesign said:
Perhaps this link would help with your understanding of the topic. It contains all the maths, vector diagrams and real world examples (along with plenty of enlightening exchanges on the discussion pages) that show your position on this to be flawed.

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

Traditionally rigged sailing vessels easily and regularly exceed 1.0 vmg on steady course -- often by a very large margin.

JB

Not so if vmg is parallel to steady wind of unchanging course! I've been to that page several times, and it still doesn't refute my point. Take for example the picture for boat speed downwind:

http://upload.wikimedia.org/wikipedia/en/4/4f/Wiki_sailing_vector_downwind.png

Note that alpha+beta=180-(90-alpha)-(90-beta). So the component of the apparent wind applied to the boat in the direction opposite of the true wind is equal to apparent wind speed*cos(alpha+beta).

Given that alpha and beta are positive angles, when the boat is going downwind faster than true wind (0 degrees < 180-(90-alpha)+(90-beta) < 90 degrees), cos(alpha+beta) > 0. Therefore, as soon as the boat overtakes the true wind, the apparent wind applies a force component on the boat against the direction with respect to the true wind. As alpha+beta approach 0 (parallel condition), the apparent wind speed on the boat against the direction of the true wind approaches boat speed minus true wind speed.

kmarinas86 said:


In both scenarios, the sail (seen here as not jibbing) should only be able to beat the balloons drifting with the wind if the wind is not parallel to the left-to-right axis. However, in DDTWFTTW, velocity made good must be parallel to downwind (and anti-parallel to upwind). The video therefore relies upon two contradicting assumptions about the direction of the wind even though it speaks of only one of them.


The above was the further elaboration of my point about this not having to do with tacking.
 
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  • #61
kmarinas86 said:
alpha ... beta
Can you swap the name alpha and beta in your post and diagram(s)? This would allow beta in your diagram to correspond to the term "beta" as used by the sailing world (where beta is the angle of apparent wind relative to boat's heading).
 
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  • #62
rcgldr said:
Can you swap the name alpha and beta in your post and diagram(s)? This would allow beta in your diagram to correspond to the term "beta" as used by the sailing world (where beta is the angle of apparent wind relative to boat's heading).

It would be helpful them for sure, but this is not really my diagram, and I have no interest in swapping the alpha and beta terms in the image. One can replace alpha and beta with x and y if they like. They will just have to think a little bit harder when observing the diagram, that's all.
 
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  • #63
kmarinas86 said:
Therefore, as soon as the boat overtakes the true wind, the apparent wind applies a force component on the boat against the direction with respect to the true wind.
Split up the apparent wind into components perpendicular and parallel to the boat's heading. An apparent headwind applies an aerodynamic drag to the boat, while an apparent crosswind is diverted to produce thrust by the sail. Note that apparent crosswind is independent of the boats speed: apparent crosswind = true wind speed x sin(boat heading - true wind heading). As long as the thrust to drag ratio is high enough, a boat can achieve vmg downwind greater than true wind speed while tacking at some angle (around 30° to 40°, depending on circumstances).
 
  • #64
rcgldr said:
Split up the apparent wind into components perpendicular and parallel to the boat's heading. An apparent headwind applies an aerodynamic drag to the boat, while an apparent crosswind is diverted to produce thrust by the sail. Note that apparent crosswind is independent of the boats speed: apparent crosswind = true wind speed x sin(boat heading - true wind heading). As long as the thrust to drag ratio is high enough, a boat can achieve vmg downwind greater than true wind speed while tacking at some angle (around 30° to 40°, depending on circumstances).

The components you mentioned were already in the original picture and had nothing to do with the force directed parallel to the direction of the wind.

Instead of projecting the apparent wind onto axes parallel and perpendicular to the boat speed, I split up the apparent wind into components perpendicular and parallel to the wind's heading in lime green.

attachment.php?attachmentid=42379&stc=1&d=1325531529.gif


Your components:
* x = Boat speed - cos(beta)
* sin(beta)

My components:
* apparent wind speed*cos(alpha+beta)
* apparent wind speed*sin(alpha+beta)

In the second image below, I split up your two components into four, shown in red.

attachment.php?attachmentid=42380&stc=1&d=1326137078.gif
 

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  • #65
kmarinas86 said:
It is possible for them to exceed VMG, but not if VMG is dead downwind (and stays that way!).
They don't "exceed VMG". They achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind.

downwind VMG = boat velocity component pointing directly downwind.

attachment.php?attachmentid=42384&stc=1&d=1325536019.png
 

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  • #66
A.T. said:
They don't "exceed VMG". They achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind.

I didn't fix the typo in time. I meant this:

kmarinas86 said:
It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!). If you are able to travel a straight line and "beat the wind" with a sail, then the wind has to be changing direction over time.

A.T. said:
downwind VMG : boat velocity component pointing directly downwind.

attachment.php?attachmentid=42384&stc=1&d=1325536019.png

Project the apparent wind speed vector over the boat speed vector component parallel to the true wind speed vector:

attachment.php?attachmentid=42386&stc=1&d=1325539649.gif


That is not steady state. It is obvious that the craft will slow down.
 

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  • #67
kmarinas86 said:
I didn't fix the typo in time. I meant this:
It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!).
Well, that is wrong. They achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind.

downwind VMG = boat velocity component pointing directly downwind.

kmarinas86 said:
Project the apparent wind speed vector over the boat speed vector component parallel to the true wind speed vector:
Why should I do this? It is utter nonsense.
kmarinas86 said:
Nope, this makes no sense at all. Acceleration is determined by forces, not velocities. Where are the forces in your diagram?
kmarinas86 said:
That is not steady state. It is obvious that the craft will slow down.

It is steady state. And it is obvious that the craft will accelerate if the L/D ratio of the sail is good enough. L/D must be at least tan(90°-alpha), but since we have hull drag it must be better than that.

attachment.php?attachmentid=42391&stc=1&d=1325542203.png
 

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  • #68
attachment.php?attachmentid=42386&stc=1&d=1325539649.gif


A.T. said:
Nope, this makes no sense at all. Acceleration is determined by forces, not velocities. Where are the forces in your diagram?

None, as this was not a force diagram. It is a velocity diagram. The red text I added was not identify the red vectors as forces. The vectors are obviously velocities as none of them are forces. The text "This affects d'' / t' of the boat[,] and it causes d' t' to change." is a description of the consequences of the red vectors in question, not of the red vectors themselves. Two of three red vectors are components of the apparent wind velocity. The third and longest one represents the apparent wind velocity itself. The consequences (not depicted graphically here) are the forces which vary by the square of the vector magnitudes here depicted and the power which varies by the cube of the vector magnitudes here depicted, so taking that into consideration, it is clear that decomposing them in the same way as the velocity vectors would result in a diagram where the angles have no physical relevance to path of the boat other than its derivative with respect to time, and indeed I have not done so with the above modified diagram, as it is only a velocity diagram applicable for an instant of time. The force on the sail is not even balanced here without incorporating friction. Once you do that, then you would have explain how friction would act against the apparent wind, when it is obvious that the apparent wind is able to push as a result of the fact that it is itself a source of friction providing a force on the craft.
 
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  • #69
kmarinas86 said:
None, as this was not a force diagram. It is a velocity diagram.
Then it cannot tell us if the boat will accelerate. Acceleration is determined by forces, not velocities. And the force vectors say it can accelerate:

attachment.php?attachmentid=42391&stc=1&d=1325542203.png
 
  • #70
A.T. said:
Then it cannot tell us if the boat will accelerate. Acceleration is determined by forces, not velocities. And the force vectors say it can accelerate:

attachment.php?attachmentid=42391&stc=1&d=1325542203.png

The deflection of the apparent wind off the sail means that the apparent wind does not maintain the same direction when approaching versus leaving the sail. Any deflection at an exact right angle to the sail's momentum direction does not add energy to the boat. It simply redirects the direction of the boat. Any force component at a right angle to a momentum direction only deflects it, leaving its norm unaffected. Does this deflection add to the kinetic energy of the boat? No. It can only deflect whatever kinetic energy is already there. Only forces parallel or anti-parallel to the motion in question will allow for the kinetic energy of the boat to change.
 

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