Newton's third law problem.

[electrodacus]

He is getting confused by the two equations and how they apply.

Nope.  They are perfectly in agreement with the experiments that you claim are misconstructed or exhibit some special slip-stick hysteresis or magical energy storage pixies.

They are also perfectly in alignment with my old University Physics textbooks on Classical Mechanics.  I have done the work and passed those courses without problems.  (Modern/quantum mechanics too, but that's not relevant here.)

The only thing there that can be discussed – among physicists – is how balancing static forces against the ground means the relative wind speed is irrelevant.  To laymen, the analog to keels and hydrofoils in sailing vessels provides the concept and the practical example.

Nothing can affect the misconceptions your kind of people have.  You refuse to acknowledge or even perceive anything that is contrary to your preconceptions, and instead disregard them as being "wrong" somehow.  That really reminds me of the people who insist that despite hundreds of millions of murders, communism is still a viable political system; it's just that because they themselves have not been in the lead, nobody has implemented it truly correctly yet.  Their argument, too, is that until you can prove it would not work when they are in charge, it is proven to work and everybody else is wrong.  Invalid logic, irrational thinking, and evasive random argumentation, just to prop up your ego and misplaced beliefs of your own 'understanding'.

It is horrible, and yet interesting, in the pathological sense.  There are those who interview monsters, so why not engage with delusional lunatics pushing irrational concepts?

You know why engineers exist ?  It is so that physicist have someone to look-up to

Is sad but you will be only convinced by doing the test and seeing the measurement results.  They will match what this calculator provides https://www.electromotive.eu/?page_id=12&lang=en

You do not have a good understanding of what power and energy is and always use just forces that clearly are not properly understood also.
Forces come in pairs always equal and opposite unless they accelerate an object.

Your first equation is a general case for a stationary object relative to ground like a wind turbine fixed to ground.
Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.
The above linked calculator implements this equation correctly.

I need to remind you that all of you not even just on this forum used this equation that is completely wrong meaning it has no applications.
Same speed will be used for calculating the Force to overcome drag as it to calculate power to overcome drag.

[Nominal Animal]

Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.

That equation describes the power needed to overcome drag.  It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift.  Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow.  In most useful airfoils, lift is much, much greater than drag.

Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency.  There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces.  All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed.  For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.

Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind.  None.  It only describes aerodynamic losses, and certain very poor turbine models, nothing more.

As an example, consider the Gorlov helical turbine I described some time back.  It has three airfoils, narrow wing-like elements, that "coil" around the common axis.  In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$).  More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel.  Moreover, it typically rotates much faster than the wind speed relative to its axis.

(In water, according to Bachant and Wosnik's experiments in 2011, Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)

I don't know how more clearly that can be stated.  You claim I don't know this or that, only because you live in your own private little world with your private little custom physics.  I do not know those, that's for sure; nobody can.

[electrodacus]

Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.

That equation describes the power needed to overcome drag.  It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift.  Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow.  In most useful airfoils, lift is much, much greater than drag.

Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency.  There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces.  All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed.  For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.

Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind.  None.  It only describes aerodynamic losses, and certain very poor turbine models, nothing more.

As an example, consider the Gorlov helical turbine I described some time back.  It has three airfoils, narrow wing-like elements, that "coil" around the common axis.  In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$).  More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel.  Moreover, it typically rotates much faster than the wind speed relative to its axis.

(In water, according to Bachant and Wosnik's experiments in 2011, Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)

I don't know how more clearly that can be stated.  You claim I don't know this or that, only because you live in your own private little world with your private little custom physics.  I do not know those, that's for sure; nobody can.

There is only one way that air interacts with an object and that is trough elastic collisions between object and air molecules.

So the same equation applies to any type of wind turbine as well as to any wind powered vehicle and also to any other vehicle when you want to know the power needed to overcome drag.

This will be that equation

P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)³

So if it is a propeller type wind turbine you use the swept area of the propeller thus there is no drag coefficient involved or you can consider that as 1 and the vehicle speed is zero if the wind turbine is mounted on the ground.

So for wind turbine that same equation will look like (still the same equation)

P = 0.5 * air density * swept area * (wind speed)³ * efficiency  (basically the first equation you mentioned in an earlier comment).

If the wind turbine is installed on a vehicle that drives say upwind same equation will look like this

P = 0.5 * air density * swept area * (wind speed-(-vehicle speed))³ * efficiency

If it is a vehicle driving direct downwind same equation will look like

P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)³ this is the wind power available to vehicle ideal case

If vehicle drives upwind the same equation shows the power needed to overcome drag

P = 0.5 * air density * area * coefficient of drag * (wind speed-(-vehicle speed))³
 
So there is no other equation it just looks a bit different depending on what it applies to but is the exact same one in all cases.


Look at energy balance for example.

You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.

KE = F_d * d

Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s  so 1.39m

KE_vehicle = (0.5*1.2*0.827*(35/3.6)²) * 1.39 = 65.2Ws

Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period

KE_air = (0.5*1.2*0.827*(35/3.6)²) * 8.33 = 391Ws

Then you will add this two 391 + 65.2 = 456Ws

So total energy that should be provided by vehicle propulsion for one second is 456Ws so basically the same as what average power needs to be.

Also say you want to be powered by the wind so you do not use the motor just let the wind push the bike your acceleration power will be 391W and will drop as the bicycle speed increases so you will need to integrate.
So say over 1ms that 391W will not change then bicycle gained  0.391Ws of kinetic energy and from that and mass of the bike + rider you get the speed.

[Kleinstein]

The interaction of the wind with the vehicle is no way close to 100% efficient - with a standing obstacle it is more like 0%. Even a good wind turbine is not close to 100% efficient. So it does not make sense to base the calculation on the power that the wind could provide with 100% efficiency.

So it is not about an energy or power balance, but about the forces.

You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.

KE = F_d * d

Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s  so 1.39m

KE_vehicle = (0.5*1.2*0.827*(35/3.6)²) * 1.39 = 65.2Ws

Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period

KE_air = (0.5*1.2*0.827*(35/3.6)²) * 8.33 = 391Ws

The power you calculate for  the air drag is coming from the wind and not form vehicle to drive. It gets obvious for a stationary obstacle - not power needed by the obstacle to keep standing. There is no sudden change from stationaly objects and very slow moving objects. The air drag power is taken from wind and converted to heat.   A wind turbine could use part of that power for other purposes, like driving the vehicle.

[PlainName]

If it is a vehicle driving direct downwind same equation will look like

P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)3 this is the wind power available to vehicle ideal case

This is where you are going wrong. Let's try and fix it...
Once again I will say that previously I would have given you one step and let you agree on it before moving to the next, because otherwise you just jump to the chase and insist it can't be done, skipping everything about how we get there. But life's too short and it's your loss.

So, what we want to know is how much power is available to make the vehicle move (because, ultimately, we need power to make it move faster than the wind is blowing). Your Someone's copied equation can be made simpler thus:

Terms:
  Pu - Useful power. This is what we want to use to move the vehicle
  Vw - Wind velocity. Importantly, this is relative to the ground and in the same direction as the vehicle moves.
  Vv - Vehicle velocity. Again, relative to the ground and in the same direction as the wind

The details (right now) of how Pu is derived from Vw and/or Vv are unimportant - it is sufficient to say that Pu is proportional to Vw - Vv. That is:

    Pu ∝ (Vw - Vv)

Your premise is that when the vehicle is at wind speed there is no power available. That is, Pu = 0. And that's correct for this simple case. What we're interested in is a case where Vw - Vv = 0 and Pu > 0.

Moving on, there's the small problem of the propeller, which is driven by the wheels. The propeller is sucking power, via the wheels, so let's add another term:

   Pv - Vehicle power. This is the power the wheels soak up when driving the propeller.

That changes things thus:

    (Pu + Pv) ∝ (Vw - Vv)

or

    Pu ∝ (Vw - Vv) - Pv

Two things to note here: first that now when Pu = 0, Vw - Vv > 0. Second, Pw is proportional to the vehicle speed since it depends on how fast the wheels turn:

    Pv ∝ Vv

Next, another term is needed for the effect of the propeller:

    Vt = Thrust velocity. This is the speed of the air being pushed backwards relative to the vehicle.

We know from seeing jets take off that the vehicle speed is wind speed plus thrust, and you even agreed earlier in this thread.

    Vv = Vw + Vt

What makes this complicated is that Vt is also proportional to the vehicle speed since the power to turn the prop is derived from the speed of the wheels:

    Vt ∝ Vv

OK, so to fill in the blanks, the usable power is proportional to the wind speed less the vehicle speed less the thrust velocity, minus the power to produce the thrust:

    Pu + Pv ∝ Vw - (Vv - Vt)

or

    Pu ∝ (Vw - (Vv - Vt)) - Pv

Essentially, this means that Vw = Vv - Vt can't occur because of the power Pv used up by the prop. In words, there isn't enough power from the wind to have the vehicle speed less thrust get up to wind speed. But, by the same token, if the thrust Vt is high enough then the vehicle speed Vv can be higher than wind speed whilst the same constraint exists. Pv, thus Vt, is proportional to wheel speed, so the faster the wind the higher these are and at some value (left for someone clever to work out if they can be arsed) we  will find that Vt > Vw when Pu != 0.

Edit: noted that Vt is relative to the vehicle whereas Vw and Vv are reltive to the ground.

[electrodacus]

The interaction of the wind with the vehicle is no way close to 100% efficient - with a standing obstacle it is more like 0%. Even a good wind turbine is not close to 100% efficient. So it does not make sense to base the calculation on the power that the wind could provide with 100% efficiency.

So it is not about an energy or power balance, but about the forces.

You are correct that a wind turbine is no more than 40 to 45% efficient but a sail is way more efficient approaching ideal.
In any case the reason for using ideal numbers is so that there can be no discussion about improving efficiency as even with ideal case no vehicle can drive upwind without using energy storage and a trigger mechanism.

No sail based vehicle can drive directly upwind not because the wind power is zero at that point is actually maximum is because that power can only be used to accelerate the vehicle direct downwind not direct upwind.
So if a stationary vehicle has 300W of wind power available when stationary it can accelerate down wind using all that but it can not move upwind at all unless it has >300W
Changing the sail for anything else like a less efficient wind turbine will not help things as you still need more than wind turbine power in order to accelerate upwind so what happens is that a very small energy storage is involved and that will be charged then used to move upwind even for a super small distance then energy storage can be recharged and things will repeat many times a second but human brain will translate that as continues motion so you will in most cases need a slow motion video to observe what happens.
And I showed that video many times and it was dismissed. 

The power you calculate for  the air drag is coming from the wind and not form vehicle to drive. It gets obvious for a stationary obstacle - not power needed by the obstacle to keep standing. There is no sudden change from stationary objects and very slow moving objects. The air drag power is taken from wind and converted to heat.   A wind turbine could use part of that power for other purposes, like driving the vehicle.

Actually the stationary obstacle is moving is just locked to earth so it is moving together with earth. The earth is accelerated by that wind hitting the obstacle but since on the planet wind is moving in multiple direction it all basically cancels out  not to mention the earth mass vs air density.
So you will need to ignore the case where brakes are engaged.
Also thinking in terms of forces gets you much higher chances of making mistakes but in both cases you will get the same result.

A sail vehicle can also drive upwind but in order to do so it will need first to charge the only type of energy storage it has (kinetic energy storage) and so drive at an angle maybe even perpendicular to wind direction and when it has enough stored energy it can just change direction to directly upwind and yes it will slow down as stored energy is used up but it will move significant distance upwind.
This things happen much slower so anyone can observe but the propeller based moves directly upwind at all times is just that it accelerates and stops accelerating multiple times per second as the very small energy storage is charged and discharged at many Hz so human brain perceive/interpret that as smooth motion.

[electrodacus]

    Pu ∝ (Vw - Vv)

Your premise is that when the vehicle is at wind speed there is no power available. That is, Pu = 0. And that's correct for this simple case. What we're interested in is a case where Vw - Vv = 0 and Pu > 0.

That is the same thing with what I say is just that is correct for any case unless it involves energy storage. You may call that stored wind power.

Moving on, there's the small problem of the propeller, which is driven by the wheels. The propeller is sucking power, via the wheels, so let's add another term:

   Pv - Vehicle power. This is the power the wheels soak up when driving the propeller.

That changes things thus:

    (Pu + Pv) ∝ (Vw - Vv)

All this and what followed is incorrect.

I will try to explain so please read:

This Pu is all that it is so when you take what you call Pv from the wheel you are actually subtracting that from Pu
If that Pu is 1000W that is what is available to accelerate the vehicle in the wind direction and so if you take 700W at the wheel you are left with only 300W to accelerate the vehicle so the vehicle acceleration is reduced.
Now you can take those 700W and send it to propeller that will be used for propulsion and so if this was 100% efficient and air was not a compressible fluid then all this 700W will end up accelerating the vehicle adding to the existing 300W so 1000W in total thus you did nothing compared to original where all 1000W where directly provided by wind.

The reality is that air is a compressible fluid so very little of those 700W you put in propeller will result as net thrust as it pushes against a compressible fluid the equivalent of a spring so most of that is stored so pressure differential is increased.
The more energy you put there the more pressure differential will increase and sometime around when vehicle speed is about 0.3x wind speed direct down wind the power provided by the stored energy matches the power available from wind directly and from that point the stored energy is the main force pushing the vehicle and when vehicle is at wind speed all power comes from this created pressure differential that is also maintained moved along with the vehicle but is more complex to visualize.

This is also what pushes the treadmill vehicle that is indoor with zero wind speed.  The treadmill provide vehicle with power as long as human keeps the vehicle constrained so all energy is used to create the pressure differential then when it is full it will be maintained and so when human releases the vehicle the force pushing the vehicle comes from this stored pressure differential and there is no longer any power coming from the treadmill. 

[PlainName]

This Pu is all that it is so when you take what you call Pv from the wheel you are actually subtracting that from Pu
If that Pu is 1000W that is what is available to accelerate the vehicle in the wind direction and so if you take 700W at the wheel you are left with only 300W to accelerate the vehicle so the vehicle acceleration is reduced.

You are missing it still.

Yes, Pv is subtracted from Pu (that's what the equation says after all!) and, no doubt to your vast delight, when Vw-(Vv-Vt) is 0, Pu is 0 so there is nothing to extract from, hence no power to turn the prop. That's why Vv-Vt can never reach wind speed.

So, Vv-Vt must be slower than wind speed because around the point Pu is 0, Pv is still being sucked out leading to an actual negative Pu. The whole thing will slow (although, in reality accelerate¹) until Pu = Pv.

If that Pu is 1000W

Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.

---
[1] Kind of. As we see from the experiments, it needs a helping had (some models, literally) to get up to speed whereupon the model will take over on its own. This is really just saying that it would slow down if for some reason (wind gust, kick up the pants) it did go overspeed.

[electrodacus]

Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.

I use values as people do not see a limit.

There is only wind power as input. There is absolutely nothing else.
There is a single equation describing the amount of power available to any wind powered vehicle and as you started to realize it is proportional with wind speed minus vehicle speed.

This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
And also same equation shows that no vehicle can drive directly upwind because in order to move upwind you need more than wind power.

So there is nothing to discuss other than if this equation is the correct one or not. Once you agree with that equation the conclusion is clear.

Now if you want to know how those vehicle work because they work then that is a separate discussion and involves energy storage.

[PlainName]

This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.

You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".

[james_s]

This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.

You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".

That's because he's a troll, he's succeeding in wasting your time and frustrating you which is exactly what trolls seek to do. We already know that he's wrong because multiple people have built vehicles that do exactly what he says is not possible, there's no point in arguing.

[PlainName]

You're right, of course. I thought it useful to me to crystalise my thinking in that post, so not completely wasted.

[electrodacus]

You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".

Propeller is powered by wind so it will not add up.
All you have as input is wind power.

Equation for available wind power cares only about the vehicle shape (coefficient of drag) and projected frontal area.

[electrodacus]

That's because he's a troll, he's succeeding in wasting your time and frustrating you which is exactly what trolls seek to do. We already know that he's wrong because multiple people have built vehicles that do exactly what he says is not possible, there's no point in arguing.

Where I ever claimed any of those vehicles do things that are not possible?
They all work as shown but not for the reasons you think they do.

They will also not work as you will predict they will based on flawed theory.

For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.

[Kleinstein]

Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.

I use values as people do not see a limit.

Doing much of the math with numbers can be confusing, but it can sometimes also help. Getting obviously wrong resucts (e.g. reduculous numbers that contradict experiance) is one way to show that a claimed equation is obviously wrong. However this still does not work with electrodacus: he refuses to see how rediculous his claims are (e.g. walking against a 30 km/h head wind would need some 400 W of power and thus be a kind of heavy sports).

There is only wind power as input. There is absolutely nothing else.
There is a single equation describing the amount of power available to any wind powered vehicle and as you started to realize it is proportional with wind speed minus vehicle speed.

This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
And also same equation shows that no vehicle can drive directly upwind because in order to move upwind you need more than wind power.

So there is nothing to discuss other than if this equation is the correct one or not. Once you agree with that equation the conclusion is clear.

Now if you want to know how those vehicle work because they work then that is a separate discussion and involves energy storage.

The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power.  At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausibile explaination. 
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.
The equation claimed by electrodacus is known to be wrong. AFAIR his reasoning kind of works for a simple sail, but it does not apply for the fan/ porp case. There are cases when a prop can be more efficient than a sail.  The calculations for the prop drivne vehicle show that one can get more power and one can go faster than the wind. So the equation claimed by electrodacus is proven wrong - it is just that he refuses to accept (and maybe understand - though it is relatively easy) the proof.

Beside the calculation there are also videos to show that it is possible to go faster than the wind / or the equivalent case of moving against the moving treadmil, driven by a prop. With the plausible forms of ernergy storage the times shown are well long enough to consider them to show a stationary case - so the energy storrage argument / read herring is dead.

[fourfathom]

For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.

Would you mind providing more information on this energy stored in the pressure differential behind the propeller?  The amount, and duration of this storage would be helpful.  There are equations for the dissipation of a pressurized area -- the one I've seen apply to bomb blast overpressure, and the rate of dissipation seems quite high, the pressure dissipates much too quickly to provide power beyond a few milliseconds.  Of course there is also kinetic energy stored in the spinning propeller blades, and in the momentum of the vehicle.  Do you have numbers for those as well?

You have obviously studied this in some detail, and I'm sure we would all be interested in what you have learned.

[electrodacus]

Doing much of the math with numbers can be confusing, but it can sometimes also help. Getting obviously wrong resucts (e.g. reduculous numbers that contradict experiance) is one way to show that a claimed equation is obviously wrong. However this still does not work with electrodacus: he refuses to see how rediculous his claims are (e.g. walking against a 30 km/h head wind would need some 400 W of power and thus be a kind of heavy sports).

You may not get strong winds at your location so your intuition may be wrong. Also keep in mind that wind speed close to ground is lower and this is a theoretical calculation that asume same wind speed on the entire surface.

The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power.  At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausibile explaination. 
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.
The equation claimed by electrodacus is known to be wrong. AFAIR his reasoning kind of works for a simple sail, but it does not apply for the fan/ porp case. There are cases when a prop can be more efficient than a sail.  The calculations for the prop drivne vehicle show that one can get more power and one can go faster than the wind. So the equation claimed by electrodacus is proven wrong - it is just that he refuses to accept (and maybe understand - though it is relatively easy) the proof.

Beside the calculation there are also videos to show that it is possible to go faster than the wind / or the equivalent case of moving against the moving treadmil, driven by a prop. With the plausible forms of ernergy storage the times shown are well long enough to consider them to show a stationary case - so the energy storrage argument / read herring is dead.

It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.

Equation is universal and the fan/prop is powered by the wind power provided by that equation.

I can build a vehicle with sail that exceeds wind speed and will not involve any propeller.
All you do is take wind energy at the wheel and store it in a battery. The same thing is done by the propeller version but instead of a battery it uses pressure differential to store energy.

If you could prove that vehicle (direct downwind blackbird or the analog treadmill model) can maintain above wind speed indefinitely then you will be right but nobody has proved that because they can not.
All those examples are incomplete and people rushed to get the wrong conclusion about how they work. Going as far as to modify correct equations and posible one of the reasons all those wrong equations are everywhere.

The large heavy blackbird needed about 6Wh less than half the energy in a cell phone battery to achieve that speed record that with pressure differential on each side of a 20m² swept area propeller.
The small lightweight treadmill model will need orders of magnitude less energy storage.

I showed that calculating using that equation and alternative kinetic energy equation will provide the same result.

You will need to prove the same that calculating in this two different ways you get the same result.
The only alternative I seen to my equation is that incorrect one where for drag force you have (w-v) correctly but for power you want to use just vehicle speed.


All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed and show that power needed for vehicle is lower than the equation I claim to be correct predicts.

And why do you not contact this guys https://www.electromotive.eu/?page_id=12&lang=en  to let them know they have a defective calculator and explanation on their website ?

[electrodacus]

For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.

Would you mind providing more information on this energy stored in the pressure differential behind the propeller?  The amount, and duration of this storage would be helpful.  There are equations for the dissipation of a pressurized area -- the one I've seen apply to bomb blast overpressure, and the rate of dissipation seems quite high, the pressure dissipates much too quickly to provide power beyond a few milliseconds.  Of course there is also kinetic energy stored in the spinning propeller blades, and in the momentum of the vehicle.  Do you have numbers for those as well?

You have obviously studied this in some detail, and I'm sure we would all be interested in what you have learned.

Yes you can watch my video starting at  10:30     calculations there do not include friction and that is important if you want to know how long it will take until vehicle will slow down below wind speed.
And starting at 15:35 you can see the details about pressure differential

[PlainName]

All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed

The previously posted video shows the electric motor version. It DOES NOT WORK AT LOW SPEED as we keep pointing out - it  has to get up to speed somehow and then it works fine.

I also stressed that in the explanation of why your love child of an equation is inappropriate.

[electrodacus]

All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed

The previously posted video shows the electric motor version. It DOES NOT WORK AT LOW SPEED as we keep pointing out - it  has to get up to speed somehow and then it works fine.

I also stressed that in the explanation of why your love child of an equation is inappropriate.

I remember some video with an electric wind turbine probably battery and motor. Is that what you refer to or something else ?
The minimum wind speed is related to wind turbine. Wind speed needs to be above a minimum threshold 3 to 4m/s in order to be enough wind power to cover the turbine friction losses. That is valid for any wind turbine no matter if it is connected to an electric generator or directly to wheels.

The equation is both correct and appropriate as it predicts exactly what it is observed.  You did not provide an alternative equation that works (can do correct predictions).
 

[Kleinstein]

It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.

This will be that equation
P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)³

Luckyly the equation is mainly used by fools and not be engineers, because it is flawed.
The part sometimes used is the special case the vehicle speed = 0 and with a slightly different coeffcient - that it get right.  Adding the vehicle speed is a point making a flawed.

It would actually your part to prove the equation, not my part to proof it is false.

I can still give you the proof by showing how the backbird vehichle works and gets power out from the wind, when driving at the speed of the wind. So giving an example that poofs the equation wrong.

A prop running in still air needs a  given power P to provide a thrust force of F. We actually don't care very much how good the prop is, it just has to work in the ideal case with no extra friction loss.
The force from the prob can than be used to have equal force on the wheels to counter-act the movement. This way the wheels provide a power of  vehichel speed times force-. This power can than be used to drive the Fan and excess power could be used for what ever other task. That excess power would be generated from the wind, as the wind is the only power soruce to the system.
To make this there needs to be excess power available. So one need   F*V_vehicle > P_prop  or  V_vehicle > P_prop / F. So given enough speed for the vehicle = wind this condition can be met.
For only somewhat realistic prop quality it does not even need a high speed, nowhere near the speed of sound where you can expect the equations to break down.

So this gives a proof that one can get power at the speed of the wind and thus that the equation with  (wind speed - vehicle speed)³ is wrong.

[PlainName]

The minimum wind speed is related to wind turbine. Wind speed needs to be above a minimum threshold 3 to 4m/s in order to be enough wind power to cover the turbine friction losses.

It is not a wind turbine, it's a propeller. And the effect it more pronounced the fast the wind is because the propller thrust is proportional to the vehicle speed (not the wind speed). Thus the faster the vehicle goes over the ground, the greater the power sucked up  by the wheels and the greater the thrust.

Think about it (I know, you need someone else to do if for you, but ask them nicely) - at zero ground speed there is no propeller power regardless of how fast the wind is blowing, or how many megawatts your favoured equation says there is.

[electrodacus]

It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.

This will be that equation
P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)³

Luckyly the equation is mainly used by fools and not be engineers, because it is flawed.

That is not good enough.
If you want to say that equation is wrong you need to provide the correct one so we can compare the prediction each of them makes.
The other equation that floats around is wrong and is so easy to show it is as it will predict zero power available for a stationary vehicle also the values for available power will be way off what is measured in real world.


That calculator comes from "electromotive engineering & consulting" they use the same equation and they must have at least one engineer based on the company name so equation is used by engineers.

You can not move a vehicle with the brakes engaged so when you want to start moving you need to disengage the brakes thus you need to deal with the power applied by the headwind if you want to move in that direction.

[IanB]

The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power.  At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausible explanation. 
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.

There are some ways of making estimates, however.

It doesn't help Electrodacus that he makes assertions that are demonstrably untrue and then relies on them to derive conclusions. For example, saying that a sail is nearly 100% efficient at extracting wind energy:

You are correct that a wind turbine is no more than 40 to 45% efficient but a sail is way more efficient approaching ideal.

We covered this in another thread, and it appears that a simple square sail is not very efficient at all. A sail self-evidently has zero efficiency when stationary, and zero efficiency when moving at the same speed as the wind. It seems the maximum efficiency of a simple square sail occurs somewhere in between, and is not much more than 15% at best. See this post for an explanation:

https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830

Yacht sails are clearly much more efficient, but that is because they use angles and aerodynamics to do a better job.

[electrodacus]

It is not a wind turbine, it's a propeller. And the effect it more pronounced the fast the wind is because the propller thrust is proportional to the vehicle speed (not the wind speed). Thus the faster the vehicle goes over the ground, the greater the power sucked up  by the wheels and the greater the thrust.

Think about it (I know, you need someone else to do if for you, but ask them nicely) - at zero ground speed there is no propeller power regardless of how fast the wind is blowing, or how many megawatts your favoured equation says there is.

This is getting old. How many times I need to mention that there are two very different vehicle being discussed.

Direct downwind uses a propeller for propulsion
Direct upwind uses a propeller as a generator so wind turbine.

There is power available for a vehicle at 0 speed while wind is blowing and in fact that is when the max wind power is available for a direct down wind vehicle.
By applying the brakes you transfer all this wind power directly to earth.