Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147190 times)

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Offline Alex Eisenhut

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Offline electrodacus

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https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf

Good find. Thank you!

That will take some time to digest.

Do not waste your time it is the same rubbish from another person that has no understanding of power and energy.

Offline gnuarm

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So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed

Are you sure about that?  When wind speed = vehicle speed, that equation still reduces to zero.

Yes, that is correct.  When the wind speed and the vehicle speed are the same, there is no wind force on the vehicle and so no force required to maintain the vehicle speed... other than friction which we are not factoring in here.  This is just about the wind forces.

OK.  I will shut up and wait until friction is re-introduced.  I want to see the situation when the vehicle speed = windspeed, and force has to be found to overcome friction.  I know this works, but how the math applies to the actual real-world blades, gears, and wheels still isn't clear to me.

I assume you are talking downwind?  That's not complex, but figuring out the math is not so simple as it depends on a lot of details regarding not just the gearing but the propeller design and adjustment.  The details really aren't important.  When the vehicle is moving the same speed as the wind, the wheels are being turned by the motion of the vehicle.  The wheels in turn work through the gearing to turn the prop.  The prop pushes against the wind, in essence adding to the wind speed. 

I'm not going to try to figure out the details of how much force or power is happening where.  This is remotely reminiscent of an old regenerative radio I built as a kid.  It used feedback to amplify the signal.  This is analogous to that.  The wind blows pushing the car initially, which in turn spins the prop.  The prop makes its own wind blowing against the oncoming wind making the "virtual wind" higher.  By the time the vehicle is moving at the speed of the wind this virtual wind is significantly higher.  The only limitation is from the losses, friction and propeller drag, ect.  It takes work to spin the prop and work the machinery, etc.  That has to come from the wind itself.  So when all the losses add up to the power from the wind, that's the top speed.

Earlier someone drew an imaginary vehicle that used sails like oars to move the vehicle.  Clearly that vehicle could move faster than the wind as the sails moved backwards when the vehicle moved forward.  The prop is the same thing.
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Offline gnuarm

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The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.

There is an equation, 0.5 x (air density) x (area) x (wind speed)^3

This equation gives the flow of wind kinetic energy through the swept area of a turbine.

It breaks down into three parts:

1) kinetic energy of wind per unit mass of air
2) mass flow of air per unit area
3) swept area of turbine

Hence:

(flow of kinetic energy) = (energy per unit mass of air) x (mass flow per unit area) x (area)

If any one of those terms is zero, the available energy will be zero.

If you try to apply this to a sail, you find that the mass flow of air though the sail is zero (because the sail is a wind barrier). So you cannot use this equation in this form for sails.

When I brought up this point with ED and asked him to explain it, he quickly changed the subject and avoided giving an answer.

I'm not going to get into all the details of how the propeller or a sail interacts with the wind.  That is very complex and I expect the equation above is actually a limit or something else fundamental rather than a practical equation for a given propeller. 

Doesn't matter.  Either way it is for the power in the wind.  The issue is how much power does it take to move the vehicle??  That is set by the force applied to the vehicle by the wheels which are propelling it against the wind.  That force is the same force as generated by the wind, but the speed is not the relative wind speed (the difference in wind and vehicle speed) but rather is just the vehicle speed because that's what the wheels see pushing against the ground. 

So the power applied to move the vehicle against the wind is the force the wind exerts (which the wheels must push against) times the speed the wheels are turning against, not the relative speed of the vehicle and the wind. 

The equation with the cubed factor is incorrect because it is the wrong power being calculated.  As you say, it is the kinetic power in the wind rather than the power applied to the car to maintain a speed.

Is that clear? 
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Offline gnuarm

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OK lets compare the output of the two equations

air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b)  10m/s, c) 20m/s

0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a)  4800W
b)  600W
c)  0W

0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W  (how will your wind powered vehicle ever start)
b) 600W
c) 0W

Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.

The first equation is the power available in the wind.  Not relevant to the discussion of how much power it takes to move the vehicle. 

The second equation is the power required to move the vehicle at that speed against the wind.  So when the vehicle is not moving, it takes no power since there is no motion. 

P = F * V, if V = 0, P = 0. 

See ma?  No hands!!!

You even admitted this in a previous message saying the bike rider can simply keep his foot on the pedal to hold the bike against the wind with no power being exerted.  Now you wish to retract that and claim a vehicle at "anchor" is a special case somehow. 

Whatever.  Your arguments get lamer every time you post.
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Offline gnuarm

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https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf

Good find. Thank you!

That will take some time to digest.

Do not waste your time it is the same rubbish from another person that has no understanding of power and energy.

LOL!!!!  By virtue of the fact that the author comes to a conclusion that does not agree with ED the author must not understand power or energy. 

If that is the case, clearly ED is not qualified.  He has already shown he doesn't understand the nature of a "rate" being related to time such as power.  He claims it is energy that is related to time.  His arguments remind me of an "engineer" who couldn't apply basic math.  They were doing a survey which required calculating the area of a room.  He wanted to divide length by width rather than multiply.  When someone asked him to explain he said, "How many quarters in a dollar?  How many quarters in two dollars?  See???" 

That's ED.  Nobody understands quarters like ED does.
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Online IanB

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Is that clear?

Yes, all clear and correct.

In the case of a wind turbine, the equation with wind speed cubed gives the total kinetic energy of the wind flowing through the swept area of the turbine. The turbine blades are able to capture some, not all, of this energy. This capture ratio can be expressed as an efficiency factor, which has a theoretical maximum at the Betz limit of 59.3%.

For simplified design purposes when installing a wind turbine, you can just make use of a quoted efficiency factor as given by the designer of the turbine, which will typically be presented in a chart looking something like this one:

 

Offline electrodacus

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OK lets compare the output of the two equations

air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b)  10m/s, c) 20m/s

0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a)  4800W
b)  600W
c)  0W

0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W  (how will your wind powered vehicle ever start)
b) 600W
c) 0W

Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.

The first equation is the power available in the wind.  Not relevant to the discussion of how much power it takes to move the vehicle. 

The second equation is the power required to move the vehicle at that speed against the wind.  So when the vehicle is not moving, it takes no power since there is no motion. 


The first equation is the wind power available to any wind powered vehicle. Second equation is not representing anything as it is just the wrong version of the first one.

The first equation (the correct one is valid for all cases).
When vehicle wants to travel direct upwind the vehicle speed will be negative so equation will look like this
 0.5 * air density * area * (wind speed - (-vehicle speed))2 * (wind speed - (-vehicle speed)) = 0.5 * air density * area * (wind speed + vehicle speed)3
And yes for any vehicle speed direct down wind you need the wind power available plus some more meaning it is impossible for any wind powered vehicle to drive directly upwind without energy storage or external energy source.

Offline gnuarm

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OK lets compare the output of the two equations

air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b)  10m/s, c) 20m/s

0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a)  4800W
b)  600W
c)  0W

0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W  (how will your wind powered vehicle ever start)
b) 600W
c) 0W

Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.

The first equation is the power available in the wind.  Not relevant to the discussion of how much power it takes to move the vehicle. 

The second equation is the power required to move the vehicle at that speed against the wind.  So when the vehicle is not moving, it takes no power since there is no motion. 


The first equation is the wind power available to any wind powered vehicle. Second equation is not representing anything as it is just the wrong version of the first one.

The first equation (the correct one is valid for all cases).
When vehicle wants to travel direct upwind the vehicle speed will be negative so equation will look like this
 0.5 * air density * area * (wind speed - (-vehicle speed))2 * (wind speed - (-vehicle speed)) = 0.5 * air density * area * (wind speed + vehicle speed)3
And yes for any vehicle speed direct down wind you need the wind power available plus some more meaning it is impossible for any wind powered vehicle to drive directly upwind without energy storage or external energy source.

Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind?  Lets simplify the messy bits and assume the force generated by the wind on the vehicle is
 
Fw = Kv * (wind speed + vehicle speed)2

Where Kv is an accumulation of all the factors that relate the force to the relative wind speed. 

So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed?

Now, forget wind.  Lets say there is a drag on the vehicle defined by

Fdrag = Kdrag * (vehicle speed)2

What is the power required to maintain the speed of this vehicle?

I bet $50 he won't answer this and ducks the question.  I suppose there's a chance he comes up with something amazingly convoluted.  Probably starts talking about driving the vehicle on a river bottom. 

I really can't believe this guy designs anything.  He must be pulling our legs. 
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Offline electrodacus

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Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind?  Lets simplify the messy bits and assume the force generated by the wind on the vehicle is
 
Fw = Kv * (wind speed + vehicle speed)2

Where Kv is an accumulation of all the factors that relate the force to the relative wind speed. 

So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed?

Now, forget wind.  Lets say there is a drag on the vehicle defined by

Fdrag = Kdrag * (vehicle speed)2

What is the power required to maintain the speed of this vehicle?

I bet $50 he won't answer this and ducks the question.  I suppose there's a chance he comes up with something amazingly convoluted.  Probably starts talking about driving the vehicle on a river bottom. 

I really can't believe this guy designs anything.  He must be pulling our legs.

It is quite simple. Power needed to counter drag will be

Fw * (wind speed + vehicle speed).

If there is no headwind then

Fdrag * (0 + vehicle speed)

So to give real numbers.
If vehicle speed is 20m/s it will require the same power to maintain speed as if vehicle is at 10m/s with a headwind of 10m/s (ignoring the difference in friction loss inside the vehicle and roiling resistance).

There is only one formula for all cases
0.5 * air density * area * (wind speed - vehicle speed)3

When you go against wind direction the vehicle speed is negative so (wind speed - (-vehicle speed)) thus basically (wind speed + vehicle speed)

Offline gnuarm

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Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind?  Lets simplify the messy bits and assume the force generated by the wind on the vehicle is
 
Fw = Kv * (wind speed + vehicle speed)2

Where Kv is an accumulation of all the factors that relate the force to the relative wind speed. 

So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed?

Now, forget wind.  Lets say there is a drag on the vehicle defined by

Fdrag = Kdrag * (vehicle speed)2

What is the power required to maintain the speed of this vehicle?

I bet $50 he won't answer this and ducks the question.  I suppose there's a chance he comes up with something amazingly convoluted.  Probably starts talking about driving the vehicle on a river bottom. 

I really can't believe this guy designs anything.  He must be pulling our legs.

It is quite simple. Power needed to counter drag will be

Fw * (wind speed + vehicle speed).

If there is no headwind then

Fdrag * (0 + vehicle speed)

So to give real numbers.
If vehicle speed is 20m/s it will require the same power to maintain speed as if vehicle is at 10m/s with a headwind of 10m/s (ignoring the difference in friction loss inside the vehicle and roiling resistance).

There is only one formula for all cases
0.5 * air density * area * (wind speed - vehicle speed)3

When you go against wind direction the vehicle speed is negative so (wind speed - (-vehicle speed)) thus basically (wind speed + vehicle speed)

You ignored the rest of my post just as I expected you would.  if you looked at the calculations you would see that the force is only multiplied by the vehicle speed to get the power required to maintain speed.  But you don't want to admit that, so you ignored it.  That was my first prediction, that you would duck the question.

You are hopeless.  You know you are wrong, but refuse to discuss it in any meaningful way.  Silly rabbit.  Trix are for kids. 

Yeah, I do pity anyone depending on you for sound engineering.  I don't believe you even have a technical job.  How may quarters???
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Offline electrodacus

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You ignored the rest of my post just as I expected you would.  if you looked at the calculations you would see that the force is only multiplied by the vehicle speed to get the power required to maintain speed.  But you don't want to admit that, so you ignored it.  That was my first prediction, that you would duck the question.

You are hopeless.  You know you are wrong, but refuse to discuss it in any meaningful way.  Silly rabbit.  Trix are for kids. 

Yeah, I do pity anyone depending on you for sound engineering.  I don't believe you even have a technical job.  How may quarters???

:) So I correct your mistake and you are saying that I ignored your post ?
You should do a test and see that your assumption is wrong.
You just believe the same absurd thing that those people that designed those online bicycle calculators.
Like 300W is plenty to bike at 1km/h against a 230km/h head wind.  People that say that is a possibility are just clueless.   

Online IanB

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OK, we are making progress here. We have established that university professors are getting it wrong. We have established that Wikipedia is getting it wrong. We have established that the bicycle power calculators are getting it wrong.

What now remains is to figure out where the physics textbooks are getting it wrong, then we can have them recalled and pulped, and replaced with corrected versions. It is astonishing how the whole world has been getting this wrong for so long.

Unfortunately there are enough people that do not understand what conservation of energy really means and so with that there is a bad understanding of how the world works.
It seems the education system is failing in this regards.  I say this seeing university level physics professors getting this wrong.

See this link and scroll down to air drag https://en.wikipedia.org/wiki/Bicycle_performance
You will find two equations

This is the same that I use for max wind power available.
And this

That is incorrect (you can not always expect much from wikipedia).

Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics)


They add vehicle speed and wind speed for force part of the equation but not for the power part ?
How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.

This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.

There will always be people that do not understand some parts of physics and I will not get bothered by this particular problem if it was not such a wide spread misinformation involving science communicators and university professors.

You just believe the same absurd thing that those people that designed those online bicycle calculators.
Like 300W is plenty to bike at 1km/h against a 230km/h head wind.  People that say that is a possibility are just clueless.
 
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Offline electrodacus

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OK, we are making progress here. We have established that university professors are getting it wrong. We have established that Wikipedia is getting it wrong. We have established that the bicycle power calculators are getting it wrong.

What now remains is to figure out where the physics textbooks are getting it wrong, then we can have them recalled and pulped, and replaced with corrected versions. It is astonishing how the whole world has been getting this wrong for so long.

I had not seen any university professor getting this exact equation wrong but it looks like some just do not know the equation at all or think it is not relevant here while it clearly is.
Wikipedia and the few online bicycle calculators I checked (at least 3) all got this wrong.

What other proof do you want other than the correct equation next to wrong equation on Wikipedia presented as being the same thing but provide different results.
And also for the bicycle calculators what else will you want as proof other than to get the result that 300W is enough to bike at 1km/h against a 230km/h head wind.
I mean most any person can put out 300W at least for a few seconds and all you need is get some strong wind 80 to 100km/h is enough (no chance to ever see 230km/h) and try to pedal against that wind even for a minute at 1km/h
Even the 80km/h will require 3kW and not sure if anyone in the world can manage that.

I already posted this as evidence for those that never experienced high wind speeds and those are professionals in way better shape than most of us and they where not able to bike against 80km/h headwind that is about 60x milder (lower power) than a 230km/h wind


So how can anyone seeing this video say the online bike calculator is providing an accurate result ?

Offline Domagoj T

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I'm still waiting for you to respond to my wind powered toy design from page 45.
It has no storage, operates in steady state continuously and has infinite range. At no point do extended sails travel faster than the wind, but the vehicle as a whole does.
I'm waiting.

There are two very different versions of blackbird. There is the direct downwind that works based on pressure differential energy storage and there is the direct upwind version that uses small capacity internal storage and stick slip hysteresis and that version since it always have access to wind power it can work continuously as long as there is wind.
All the wheels only toys where air is not involved are the equivalent of that direct upwind version of blackbird using small capacity energy storage device maybe charging in a few ms and then discharging triggered by stick slip hysteresis either internally in the mechanism or externally at the wheels.

The wrong equation for power either drag or generation is what made the math possible without energy storage. If the correct equation for power is used then is clear from the math that neither direct downwind faster than wind nor direct up wind at any speed is not possible without energy storage.
And again you manage to deflect and avoid answering.
My design is different than Blackbird, so let's stick to mine, please.
I'm not talking about upwind, only downwind.
There is no stick slip hysteresis in my design (substitute string with chain, and wheels and ground with rack and pinion, if you so choose).
There is no energy storage in my design, unless you consider sails themselves to be storage, in which case sure, whatever, doesn't matter, let's not mention it ever again since it's irrelevant.
The steady state of this design is that the vehicle as a whole will continuously travel above wind speed, directly downwind.
 

Offline electrodacus

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And again you manage to deflect and avoid answering.
My design is different than Blackbird, so let's stick to mine, please.
I'm not talking about upwind, only downwind.
There is no stick slip hysteresis in my design (substitute string with chain, and wheels and ground with rack and pinion, if you so choose).
There is no energy storage in my design, unless you consider sails themselves to be storage, in which case sure, whatever, doesn't matter, let's not mention it ever again since it's irrelevant.
The steady state of this design is that the vehicle as a whole will continuously travel above wind speed, directly downwind.

You can be offended if you want but you do not even deserve an answer as it will be a long and involved one from me and you will have no clue of what I just said.
My answer was generic and included all possible vehicle types powered only by wind directly down wind and no vehicle can exceed wind speed unless it uses an energy storage device or an external energy source.
Not only the correct equation for wind power will show that but even the wrong equation will predict the same as I just shown in an example earlier today.

Offline lordium

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@electrodacus,

Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.
 
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Offline PlainName

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Quote
and no vehicle can exceed wind speed unless it uses an energy storage device or an external energy source.

The authoratitive source for that statement of fact?

Well, seems to be just you. Surprised you aren't the nobel laureate yet.
 
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Online Kleinstein

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...

It is quite simple. Power needed to counter drag will be

Fw * (wind speed + vehicle speed).
This only applies if you use the power to drive the vehicle with a prop.

The drag force is not different from other forces and the mechanical power is force times speed between the 2 part that the force is acting between (ground and the vehicle).


OK lets compare the output of the two equations

air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b)  10m/s, c) 20m/s

0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a)  4800W
b)  600W
c)  0W

0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W  (how will your wind powered vehicle ever start)
b) 600W
c) 0W

Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.
The 2nd expression (it is not even an equation, as there is no "=" sign ) never claimed to calculate the available wind power, but the power available from a sail (which is not the most power efficient way) or the power needed to drive against the wind, like a sail in reverse.
The 1 st expression is the power available to a wind turbine on a muving vehicle, not including the power needed or available from moving the vehicle. The problem is using that same expression for different things. So why not use Ohms law instead ? This give a wrong result too, and arguably maybe sometimes even a correct if the result is 0.
 

Offline Labrat101

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@ electrodacus

Your Forum is also full of stupid answers .. Now you have plagued us with your stupidity of Not seeing what is true life Working .
 If it works don't brake it . Your Maths is no better than a 1st grader . Just picking up numbers @ random will never solve this .
 Nor will you ever learn . and worst still your not willing to learn .
As hard as it is for you to except you are making a terrible mistake .
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Offline electrodacus

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Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.

I have made a video but is maybe a bit long and boring

In any case I will try to make a summary.
The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.
That can be proved with the correct equation for wind power available to any wind powered vehicle

Pw = 0.5 * air density * area * (wind speed - vehicle speed)3

It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
The air density can be considered a constant so not very relevant and then there are just two other therms the area of the vehicle that in the particular case of blackbird will increase with speed up to a max of swept area of the propeller but while that increase in area helps make more power available to vehicle it will only be valid as long as vehicle speed is smaller than wind speed since as it can be seen in the equation if vehicle speed equals wind speed the wind power will be zero.

This formula is all that is needed in order to demonstrate that any wind powered only vehicle will need an energy storage device in order to exceed wind speed directly down wind.
It seems that many people do not work with power and prefer to work with force and speed separately and this is how a wrong equation ended everywhere.
The equation for force is correct
Fw = 0.5 * air density * area * (wind speed - vehicle speed)2

But then when they want to calculate wind power they just multiply by vehicle speed and leave the wind speed out of the equation.  The most likely reason they do that is because someone made this mistake first and they just copy paste the same wrong equation everywhere without thinking to much or testing to see if it is true.

They just think that if vehicle speed is zero the wind power available to that vehicle is zero and they think that since they imagine zero speed as a vehicle with brakes engaged.  That seems as a super silly mistake to make is like saying that a sail boat has zero wind power available when boat speed is zero because the boat is anchored to the ground.

Let me know if you agree with the above and then I can continue with how the Blackbird can actually exceed wind speed for a limited amount of time using energy storage in pressure differential created by the propeller with part of the wind power.

Offline fourfathom

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That can be proved with the correct equation for wind power available to any wind powered vehicle

Pw = 0.5 * air density * area * (wind speed - vehicle speed)3

It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).

So do the wheels in contact with the ground make no difference?  Your equation might be appropriate for a hot-air balloon.
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Offline electrodacus

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    • electrodacus
That can be proved with the correct equation for wind power available to any wind powered vehicle

Pw = 0.5 * air density * area * (wind speed - vehicle speed)3

It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).

So do the wheels in contact with the ground make no difference?  Your equation might be appropriate for a hot-air balloon.

They are ideal wheels so yes they make no difference as there is no friction thus best case scenario.
If you use real wheels with friction then vehicle acceleration rate will be lower and max speed for a direct down wind vehicle will be lower than wind speed.

Online Kleinstein

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The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.

Let me know if you agree with the above and then I can continue with how the Blackbird can actually exceed wind speed for a limited amount of time using energy storage in pressure differential created by the propeller with part of the wind power.
The calculation this way is absolutely not easy, as there is no accepted formula for the maximum power available to a vehicle. So the difficulty is  in deriving (not just propose a solution and claim it must but ture as an act of god). As there are multiple possible ways to harness the wind power this is a really difficult task. To make it a proof it needs a really good explaination that most people would agree with.

Just presenting an expression that most people think is wrong is far from supporting the claim. It is more like showing poor understanding of logic and science in general.

The V_w * F_w form was never claimed to be the available wind power, but the power needed by the vehicle. Just like for another force to push against the power is force times speed (by definition and not by mistake). Thinking the power would stay constant, essentially independent of the speed is just a rediculous idea, that causes obvious contradictions.  One such contraticion would be that the power when going in the same direction as the wind would be the same as the maximum avialable power and thus all wind genrator would be 100% efficient. I don't think that sounds plausible.
 


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