Mess with your minds: A wind powered craft going faster than a tail wind speed.

[electrodacus]

How would that be violated by a sail dissipating all or most of the wind energy and only transferring a small portion of that energy to the vehicle?

I was talking about the wrong explanation of how blackbird works and that will be indefinitely above wind speed (not possible).
Under wind speed you can drive as much as you want as you have access to wind power.
If a sail vehicle is less efficient due to friction losses then it will just accelerate a bit slower but you will need to have horrible wheel friction to take 1.66 seconds to get to 1m/s in 10m/s winds with 1m^2 sail and 100kg mass.

The formula used earlier that seems to be widely accepted as correct (the drag equation) is also exponential and doesn't diverge to infinity as the vehicle speed approaches zero.  In fact, it doesn't matter whether the vehicle is stationary or moving beyond how that changes the relative speed of the wind and vehicle. 

So now you are positing some new drag or sail formula that would give you an extremely high pressure and thus an extremely high initial acceleration for a 10m/s wind and a stationary vehicle.  So what would the pressure and acceleration be if you had an 11m/s wind and a vehicle moving at 1m/s?  Would that be different?

What experimental tests are you referring to?

I'm very confident that online calculator that I liked is accurate in describing pressure on either a wall or a sail directly downwind.
It is also similar to how pressure differential looks on a axial fan the graph from wikipedia.
As mentioned no matter the method used if implemented correctly will provide the same result and using the results from that calculator for pressure behind the sail (pressure is variable decreasing as you get away from the sail/wall and it makes perfect sense for a compressible fluid).
If you use that variable pressure then how fast the vehicle accelerate will look exactly the same as the one predicted by my formula where wind power and kinetic energy where used.
Both formula for wind power and kinetic energy where the correct ones so the result not only look correct they match reality if you were to add the friction.
The thing is even in that ideal setup a direct downwind vehicle can not exceed wind speed without energy storage.
Also as an extra result no vehicle can also travel at any speed direct upwind unless it also uses energy storage and a trigger mechanism like stick slip hysteresis.  Not only that is the conclusion for correct use of equations but also I demonstrated that in practice.

[fourfathom]

That energy storage device will be charged during the initial acceleration phase

Where is the acceleration phase on the treadmill?

I think I explained that before here but it is when you put the vehicle on the treadmill.

When you put the vehicle on the treadmill and hold it in place, this is equivalent to the vehicle traveling downwind exactly at windspeed [  (wind speed - vehicle speed) = 0 in your equation ]  So according to you, there's no energy to be stored -- and you're right, at least about that.    But when you let go, the vehicle instantly rolls forward, faster than the wind.

[electrodacus]

When you put the vehicle on the treadmill and hold it in place, this is equivalent to the vehicle traveling downwind exactly at windspeed [  (wind speed - vehicle speed) = 0 in your equation ]  So according to you, there's no energy to be stored -- and you're right, at least about that.    But when you let go, the vehicle instantly rolls forward, faster than the wind.

There is maybe a second or two when you touch the vehicle to treadmill until all energy storage devices are fully charged up that means all wheels get to same speed as the treadmill the propeller gets to nominal speed and the pressure differential is created.
For that small treadmill model we are talking about 0.25 to maybe 1Ws (1J) of total stored energy.
The kinetic energy needed for that vehicle to accelerate from zero (when you release) up to 1m/s (never got that fast in the video) is 0.5 * 0.5kg * 1m/s^2 = 0.25Ws (0.25J) basically nothing is needed to be stored in pressure differential to be able to accelerate that small vehicle to 1m/s even to 2m/s will not take much and vehicle was only demonstrated for way lower speed.  The treadmill is just to short to show the vehicle accelerating to whatever max speed it is designed for and then show also how it will slow down (that will take again as much as to get there).

[fourfathom]

When you put the vehicle on the treadmill and hold it in place, this is equivalent to the vehicle traveling downwind exactly at windspeed [  (wind speed - vehicle speed) = 0 in your equation ]  So according to you, there's no energy to be stored -- and you're right, at least about that.    But when you let go, the vehicle instantly rolls forward, faster than the wind.

There is maybe a second or two when you touch the vehicle to treadmill until all energy storage devices are fully charged up that means all wheels get to same speed as the treadmill the propeller gets to nominal speed and the pressure differential is created.
For that small treadmill model we are talking about 0.25 to maybe 1Ws (1J) of total stored energy.
The kinetic energy needed for that vehicle to accelerate from zero (when you release) up to 1m/s (never got that fast in the video) is 0.5 * 0.5kg * 1m/s^2 = 0.25Ws (0.25J) basically nothing is needed to be stored in pressure differential to be able to accelerate that small vehicle to 1m/s even to 2m/s will not take much and vehicle was only demonstrated for way lower speed.  The treadmill is just to short to show the vehicle accelerating to whatever max speed it is designed for and then show also how it will slow down (that will take again as much as to get there).

What stored energy is there when the vehicle is stationary?  You claim that there is zero available energy when the relative windspeed is zero, so there's nothing to store.  By your thinking, all the storage occurs when the vehicle is traveling slower than windspeed.

So, how long would we need to hold that vehicle stationary for all that "stored energy" to dissipate?  And why do we care about max speed?  Isn't any speed beyond wind speed proof that it works?

Your model, and your understanding, is flawed.

[electrodacus]

What stored energy is there when the vehicle is stationary?  You claim that there is zero available energy when the relative windspeed is zero, so there's nothing to store.  By your thinking, all the storage occurs when the vehicle is traveling slower than windspeed.

So, how long would we need to hold that vehicle stationary for all that "stored energy" to dissipate?  And why do we care about max speed?  Isn't any speed beyond wind speed proof that it works?

Your model, and your understanding, is flawed.

What do you mean by stationary vehicle ? Vehicle before it touches the treadmill ? if yes then of course there is no stored energy.
There is zero wind power when vehicle speed equals wind speed but blackbird and by extension the treadmill model have store energy that they can use to accelerate past wind speed.
The stored energy will not be used unless you release the vehicle so same amount of store energy will be present no matter how long you keep the vehicle in place. Once you let go it will start to use that stored energy and you do not want to interact with vehicle in any way after that just let it run on that stored energy and you will be able to see how acceleration rate drops to the point that it will start to decelerate down below wind speed.

Vehicle works as shown. I never claimed that blackbird either the large one or the treadmill model is not working as shown in tests. What I insist on is that current explanation of why is works is completely wrong.
Vehicle can not be powered as claimed by wind when above wind speed directly down wind but it is powered by store energy and since this stored energy is limited and there is friction the vehicle will be able to get to some peak speed and then start to slow down as there is no longer any stored energy and it will slow down way below wind speed.

It is just clear to me that energy storage is not something people think to much about but it is as prevalent and important as friction in real world.

[fourfathom]

What do you mean by stationary vehicle ? Vehicle before it touches the treadmill ? if yes then of course there is no stored energy.

No.  By "stationary" I mean the vehicle is being held stationary on the treadmill.  The wheels are in contact with the treadmill, turning, and the propeller is spinning.  The vehicle speed and the windspeed are equal.  This is exactly equivalent to a vehicle traveling directly downwind at the speed of the wind.  And no matter how long you hold the vehicle stationary, when released the vehicle will accelerate.

You claim there is no available energy when the vehicle is stationary, since "windspeed - vehicle speed = 0".   Even if we ignore your equation, and at windspeed there is stored energy in a pressure differential (which I dispute -- I claim there is no stored energy adequate for even a brief burst of acceleration), how long do you think this stored energy will persist before dispersing into the surrounding environment?

There is measurable stored kinetic energy in the spinning propeller, and rotating wheels, and the mass of the vehicle itself, but unless you have some sort of gearshift or prop-feathering (which the treadmill vehicle does not) none of these will cause the vehicle to accelerate, they will only slow the rate of friction-caused deceleration.  And yet, the vehicle accelerates.

[electrodacus]

What do you mean by stationary vehicle ? Vehicle before it touches the treadmill ? if yes then of course there is no stored energy.

No.  By "stationary" I mean the vehicle is being held stationary on the treadmill.  The wheels are in contact with the treadmill, turning, and the propeller is spinning.  The vehicle speed and the windspeed are equal.  This is exactly equivalent to a vehicle traveling directly downwind at the speed of the wind.  And no matter how long you hold the vehicle stationary, when released the vehicle will accelerate.

You claim there is no available energy when the vehicle is stationary, since "windspeed - vehicle speed = 0".   Even if we ignore your equation, and at windspeed there is stored energy in a pressure differential (which I dispute -- I claim there is no stored energy adequate for even a brief burst of acceleration), how long do you think this stored energy will persist before dispersing into the surrounding environment?

There is measurable stored kinetic energy in the spinning propeller, and rotating wheels, and the mass of the vehicle itself, but unless you have some sort of gearshift or prop-feathering (which the treadmill vehicle does not) none of these will cause the vehicle to accelerate, they will only slow the rate of friction-caused deceleration.  And yet, the vehicle accelerates.

You are confusing power with energy.
What I say is that a vehicle power only by wind and that is at same speed as the wind directly downwind has no wind power available.
What that vehicle has is stored energy in pressure differential created by the propeller and that energy can allow the vehicle to exceed wind speed but only for a limited amount of time depending on amount of stored energy and vehicle friction losses.

The treadmill model is not quite the same thing until you release the vehicle as by changing the frame of reference the vehicle has no kinetic energy at that starting point so all friction losses on the vehicle will be provided by the treadmill and what you have there is just an inefficient fan powered by treadmill motor.
As soon as you release the vehicle from your hand the vehicle will simulate fairly well what happens with direct down wind blackbird from the wind speed forward. 
At release there will be energy stored in pressure differential (propeller will have a constant speed until release and thus stored energy will stay the same no matter how long you keep the vehicle there).
When vehicle is released the pressure differential will start to drop as it will supply both vehicle losses as well as the energy needed to increase speed basically increasing kinetic energy of the vehicle.
The treadmill no longer helps the vehicle in any way once released from the hand and as soon as all the pressure differential is used up the vehicle will start to slow down but because the treadmill is so short you will not be able to see that.
What you can see if you take a video from the side is the decrease acceleration rate and that will not correspond to Derke's or anyone else's prediction.
Wind speed will need to be higher than vehicle speed in order to be able to power the vehicle and since that is not the case vehicle can only accelerate if it has an energy storage device.
A sail vehicle while more efficient in converting wind power to kinetic energy has no energy storage device thus can never exceed wind speed but Blackbird due to the use of a propeller fan has the pressure differential energy storage and that is what it allows it to exceed for a limited amount of time (few minutes depending on design and wind speed) the wind speed.

The crazy explanation that vehicle when above wind speed directly down wind will take power from the wheel to supply the propeller is exactly like saying a motor and generator connected together with a belt are a free energy generator.
That is because when you take energy from the wheel that is directly taken out of the vehicle kinetic energy Ws or Joules and since propeller is maybe at best 70% efficient can only put back 70% of that back in to kinetic energy and so vehicle will have lower speed. Luckily there is that pressure differential to cover all losses and accelerate the vehicle until it will be used up.   

[fourfathom]

Pressure differential energy storage?  No.  But I am slowly backing away from this apparently futile discussion for a while. 

[Kleinstein]

Also my point is that no vehicle can exceed wind speed directly downwind unless it has an energy storage device.

And that is wrong and there is no sense in repeating this all over !. 

I'm very confident that online calculator that I liked is accurate in describing pressure on either a wall or a sail directly downwind.

Very confident is has to be seen with a grain of salt: The calculator is for a different problem: the force a wall has to withstand, when close to an edge (kind of clif if you want). The Zones are not the place where the pressure in the air is measured, but the distance of the wall from the edge. Close to the edge there is some concentration of the wind - so the real value to compare is the one far away from the edge.
The calculated force is not that impressive to support the claimed high power from a sail.

So I am glad you are confident in a calculator that contradicts your claims.

Better have more confidence in the calculator for the bicycle power that showed that it is possible to drive with relatively low power against an head wind.

[Labrat101]

@electrodacus
A Bumblebee can't fly .!!
Same as electrodacus can't  believe a simple
Solution.  That if everyone says the 🌎 is round .
He would argue its flat because my spirit level says so ..
So if everyone here has proven it does work .
Plus there has been a video showing it works
. There for Bumblebee can fly also faster the tail wind .
Your ideas of what is true and false , was is, and what can . 
  Maybe beyond your comprehension .
  It wouldn't matter if Albert Einstein could tell you that your wrong .
You are living in some sort of self denial.
Get over it and except that it can and does work.

[Brumby]

The crazy explanation that vehicle when above wind speed directly down wind will take power from the wheel to supply the propeller is exactly like saying a motor and generator connected together with a belt are a free energy generator.

No, it's not.  Your motor and generator are operating at the same speed - but the the two sides of the Blackbird mechanism are not.

This is clearly apparent in the following statement:

That is because when you take energy from the wheel that is directly taken out of the vehicle kinetic energy Ws or Joules and since propeller is maybe at best 70% efficient can only put back 70% of that back in to kinetic energy and so vehicle will have lower speed.

Yes, there will be a reduced efficiency in the power extracted from the wheels through to the propeller - but the propeller is pushing against the wind.  With the blackbird travelling at wind speed, the wind will appear as a stationary wall of air - and THIS is what the propeller is pushing against.  <== IMPORTANT POINT!

Warning: The following uses terms like power and energy in a conversational style - NOT an engineering one!  Change the words as you see fit - but follow the intent.

 - Let's say wind speed is 20 km/h
 - Power is extracted from the wheels which are rotating at a tangential speed of 20km/h
 - This power is transferred through a gearbox to rotate a propeller.  (The ratio of the gearing is important here - which includes propeller characteristics.)
 - The propeller then pushes against a wall of air that appears to be stationary ( 0 km/h ) from the perspective of the vehicle.

In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference.

The extracted energy will cause a drop in speed, but the additional thrust will cause an increase.  Finding the equilibrium point is the trick.  Is is below wind speed?  Is it above?  Is it equal to wind speed?

[Brumby]

What you can see if you take a video from the side is the decrease acceleration rate

It doesn't matter even if the acceleration rate drops to zero.  If the resulting "terminal velocity" is higher than wind speed, then we have found the equilibrium point which supports Blackbird's achievement.

and that will not correspond to Derke's or anyone else's prediction.

You say "will not".  That sounds like an intuitive belief rather than a mathematical proof.

[Brumby]

After re-reading, I felt this statement was worth emphasising.

In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference.

[fourfathom]

[I just can't help myself!]

One more way to consider the situation (and this has been said here before, by myself and others) is that the vehicle is located between and mechanically coupled to the interface of the ground and the moving air.  It really doesn't matter if the vehicle is moving or stationary, the difference in velocity is always there and can be used to generate power.

But you need to have an appropriate mechanism to extract this power and drive the vehicle DDWFTTW.  A rock can't do it.  The Blackbird can.

[electrodacus]

And that is wrong and there is no sense in repeating this all over !. 

You need to explain it not repeat something you memorised.

Very confident is has to be seen with a grain of salt: The calculator is for a different problem: the force a wall has to withstand, when close to an edge (kind of clif if you want). The Zones are not the place where the pressure in the air is measured, but the distance of the wall from the edge. Close to the edge there is some concentration of the wind - so the real value to compare is the one far away from the edge.
The calculated force is not that impressive to support the claimed high power from a sail.

So I am glad you are confident in a calculator that contradicts your claims.

Better have more confidence in the calculator for the bicycle power that showed that it is possible to drive with relatively low power against an head wind.

The calculator is for exactly this type of problem. You may not like it as it supports my theory and calculations for a compressible fluid are much more complex.
Look at the type of teren you can select and no there is no close to the edge of a clif. Please look more closely at that calculator and input some data. The calculation will perfectly apply to a sail.
The pressure is what is shown there and is the average for a certain volume.

Use the following data
Terrain category 0
Wind speed 10m/s
Wall Length 10m
Wall height 2m
Distance from base say 1m (so wall is not touching the ground)
length return corner 0
Solidity ratio 1
Ortogonal factor 1
Air density 1.2

Then you will see that average pressure for 0.3*h so 0.6m behind the wall/sail  you have 407N/m^2
Then from that point 60cm to 4m the average pressure is 253N/m^2
You can see how average pressure increases as you get closer to the wall/sail and it makes sense as this is a compressible fluid so way more air molecules will be squeezed just right next to the wall so much higher air density in those regions.
Same was shown for a propeller with the difference that propeller can increase this pressure differential even more while vehicle is below wind speed and it has access to wind power.

[electrodacus]

@electrodacus
A Bumblebee can't fly .!!
Same as electrodacus can't  believe a simple
Solution.  That if everyone says the 🌎 is round .
He would argue its flat because my spirit level says so ..
So if everyone here has proven it does work .
Plus there has been a video showing it works
. There for Bumblebee can fly also faster the tail wind .
Your ideas of what is true and false , was is, and what can . 
  Maybe beyond your comprehension .
  It wouldn't matter if Albert Einstein could tell you that your wrong .
You are living in some sort of self denial.
Get over it and except that it can and does work.

Solution is not simple and I never believe anything I need to understand.
All works as shown in tests and is fully described by my theory.
On the other hand the equations you proposed do not predict what happens in real tests.

[electrodacus]

No, it's not.  Your motor and generator are operating at the same speed - but the the two sides of the Blackbird mechanism are not.

Most of those have a different gear ratio so no they do not operate at the same speed and they will still not work.

Yes, there will be a reduced efficiency in the power extracted from the wheels through to the propeller - but the propeller is pushing against the wind.  With the blackbird travelling at wind speed, the wind will appear as a stationary wall of air - and THIS is what the propeller is pushing against.  <== IMPORTANT POINT!

Warning: The following uses terms like power and energy in a conversational style - NOT an engineering one!  Change the words as you see fit - but follow the intent.

 - Let's say wind speed is 20 km/h
 - Power is extracted from the wheels which are rotating at a tangential speed of 20km/h
 - This power is transferred through a gearbox to rotate a propeller.  (The ratio of the gearing is important here - which includes propeller characteristics.)
 - The propeller then pushes against a wall of air that appears to be stationary ( 0 km/h ) from the perspective of the vehicle.

In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference.

The extracted energy will cause a drop in speed, but the additional thrust will cause an increase.  Finding the equilibrium point is the trick.  Is is below wind speed?  Is it above?  Is it equal to wind speed?

Clearly shows you do not understand power and energy. 20km/h ~ 5.5m/s
If you take 5.5W form the wheel (1N * 5.5m/s) and put all that in to propeller (ideal case) the most propeller can output is 5.5W thus no acceleration possible even in ideal case.
The gear ratio is irrelevant since you can still not output more power than you input.

This is exactly what others including Derek will try to say it is so silly I do not even know what to say.  Luckily there are people that understand all this they just seems to be unwilling to waste time with you (all).

If you think with a 5.5W input the propeller can output more then I have a free energy generator for sale.

[Kleinstein]

Clearly shows you do not understand power and energy. 20km/h ~ 5.5m/s
If you take 5.5W form the wheel (1N * 5.5m/s) and put all that in to propeller (ideal case) the most propeller can output is 5.5W thus no acceleration possible even in ideal case.
The gear ratio is irrelevant since you can still not output more power than you input.

This is exactly what others including Derek will try to say it is so silly I do not even know what to say.  Luckily there are people that understand all this they just seems to be unwilling to waste time with you (all).

If you think with a 5.5W input the propeller can output more then I have a free energy generator for sale.

The gear ratio is very relevant, as it determines how much thrust / force the propeller will produce. With a good gear ratio propeller can create more than 1 N of thrust from the given 5.5 W. This is possible as the prop sees essentially standing air. 1 N at zero relative speed is zero ouput power from the prop.

What is a "free" energy generator good for that can create 0 W when the wind is blowing. There is quite some energy taken from the wind.

[electrodacus]

The gear ratio is very relevant, as it determines how much thrust / force the propeller will produce. With a good gear ratio propeller can create more than 1 N of thrust from the given 5.5 W. This is possible as the prop sees essentially standing air. 1 N at zero relative speed is zero ouput power from the prop.

What is a "free" energy generator good for that can create 0 W when the wind is blowing. There is quite some energy taken from the wind.

When you take 5.5W say for 1 second then that 5.5Ws (5.5J) energy will be subtracted from the vehicle kinetic energy and since kinetic energy is proportional to speed speed will also be reduced.
Now all you have is this 5.5Ws (5.5J) and no matter what gear ratio you have you can not deliver more than the same 5.5J to the vehicle kinetic energy assuming ideal case.

So yes you can have a gearbox with say 2:1 ratio so input is 5.5m/s 1N and output can be 2.75m/s at 2N it will still only put back 5.5Ws (5.5J) in the vehicle kinetic energy since gear box is not magic and can not output higher power than it has available at the input.
So you just get confused as you do not use the correct speeds in your equations like Derek subtracting wind speed from vehicle speed (ridiculous).
 

[Kleinstein]

The gear ratio is very relevant, as it determines how much thrust / force the propeller will produce. With a good gear ratio propeller can create more than 1 N of thrust from the given 5.5 W. This is possible as the prop sees essentially standing air. 1 N at zero relative speed is zero ouput power from the prop.

What is a "free" energy generator good for that can create 0 W when the wind is blowing. There is quite some energy taken from the wind.

When you take 5.5W say for 1 second then that 5.5Ws (5.5J) energy will be subtracted from the vehicle kinetic energy and since kinetic energy is proportional to speed speed will also be reduced.
Now all you have is this 5.5Ws (5.5J) and no matter what gear ratio you have you can not deliver more than the same 5.5J to the vehicle kinetic energy assuming ideal case.

So yes you can have a gearbox with say 2:1 ratio so input is 5.5m/s 1N and output can be 2.75m/s at 2N it will still only put back 5.5Ws (5.5J) in the vehicle kinetic energy since gear box is not magic and can not output higher power than it has available at the input.
So you just get confused as you do not use the correct speeds in your equations like Derek subtracting wind speed from vehicle speed (ridiculous).

The gear box does not increase the power, but the prop or wheels on the treadmill work relative to a different speed. Because of the lower speed the gear box can reduce the speed so much to generate more than 1 N ( e.g. 2 N). With 2 N of force to drive the vehicle  forward and 1 N of force to work against it, there would be still 1 N going forward and thus an increase in speed. It is the force that decides which way the vehicle moves, not the power.

The gain in power is from slowing down the wind with those 2 N. 2 N pushing against the 5.5 m/s wind is a power source of 11 W and thus plenty of power. It is not free energy, but energy taken from the wind.

[electrodacus]

The gear box does not increase the power, but the prop or wheels on the treadmill work relative to a different speed. Because of the lower speed the gear box can reduce the speed so much to generate more than 1 N ( e.g. 2 N). With 2 N of force to drive the vehicle  forward and 1 N of force to work against it, there would be still 1 N going forward and thus an increase in speed. It is the force that decides which way the vehicle moves, not the power.

The gain in power is from slowing down the wind with those 2 N. 2 N pushing against the 5.5 m/s wind is a power source of 11 W and thus plenty of power. It is not free energy, but energy taken from the wind.

There is no option to get 11W out of a wheel or propeller that is powered with 5.5W.
If there are 11W at the propeller then it is not because you got that from the ground wheel but because either vehicle is well below wind speed and air pushes the propeller or if above wind speed the stored pressure differential.
You will never (I mean never) get 11W with an input of 5.5W. Thinking like this is that resulted in this wild explanations.
For wind to power the vehicle air molecules needs to move faster and in the same direction of travel.

[bdunham7]

You will never (I mean never) get 11W with an input of 5.5W. Thinking like this is that resulted in this wild explanations.

Sure you will, and the Blackbird does just that.  There's no big mystery how, either.  You simply don't understand the energy conservation laws that you continually cite.

[electrodacus]

Sure you will, and the Blackbird does just that.  There's no big mystery how, either.  You simply don't understand the energy conservation laws that you continually cite.

 
You are to funny.

What if I say I have a DC-DC converter for sale that you can input say 5.5V at 1A thus 5.5W and that it can output 5.5V but at 2A thus 11W ?
The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.
That is what happens here the pressure differential stored energy is what supplies the vehicle and allows it to accelerate for some limited amount of time.

[bdunham7]

The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.

Yes, that's the principle.  The propeller and the wind are 'in series', so to speak.  If you like bad analogies, you can compare the various 'speed' of the players with voltage.  If the wind and the vehicle are at the same speed you can compare that to a power source and a sink--say a charger and battery--that are both at the same voltage so that no power flows.  But if you add an isolated DC-DC converter across the source and then put its output in series between the source and sink, now power will flow.

So the wheel-to-propeller link can be considered as a sort of boost converter that allows power to go from an original source that is the same or even lower voltage than the sink. 

[Kleinstein]

If the generator is sufficiently large and needs a strong wind to work, there is no problem with this.

There is not just the 5.5 W from driving the prop, but also 11 W of wind power. 16.5 W in and 11 W out sounds feasable, though maybe demanding on the prop quality. With less extra energy the demands on the prop get lower, so that even the relatively simply toy model on the treadmill worked.
Using some energy input to later get out more power is common: The combustion engines usually work this way, by first compressing air to than get back more energy from the expansion of the hotter gas.

The calculation does not include an extra form of energy storrage. The prop driven from the wheel mechanism works without it.

For the experiment we have not found any good hint that energy storrage is relevant:
There is no extra motor - so no battery to use.

Kinetic energy can only be used when going slower.
The slightly more tricky to analyse pressure differential idea also does not work,  with two strong aguments found against it:
  a) the fast decay
  b) similar to the kinetic energy the pressure amplitude is linked to the speed.
Eleastic energy (e.g. in a chain) would not last very long and there is no visible change in the vehicle configuration (e.g. change in the length)  while going at the speed of the wind.
The rubber band example showed elastic energy storrage, but with the energy storrage still accumulationg while at the speed of the simulated wind.

The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.

Yes, that's the principle.  The propeller and the wind are 'in series', so to speak.  If you like bad analogies, you can compare the various 'speed' of the players with voltage.  If the wind and the vehicle are at the same speed you can compare that to a power source and a sink--say a charger and battery--that are both at the same voltage so that no power flows.  But if you add an isolated DC-DC converter across the source and then put its output in series between the source and sink, now power will flow.

So the wheel-to-propeller link can be considered as a sort of boost converter that allows power to go from an original source that is the same or even lower voltage than the sink. 

I like that analogy, but the DCDC converter would be more like a step down one to get more current, though at a lower voltage.
Going at the speed of wind would be equivalent to charing one 12 V battery with another 12 V battery. It does not work directly, but it works if an extra DCDC converter is used to add some (e.g. 2 V) to the voltage.