Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147314 times)

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Offline electrodacus

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As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.

Indeed, and for reference, here is an analysis of the cart/wheel/belt system.

Refer to the diagram attached below.



Assume that it is an ideal system, no friction losses, and both wheels are able to spin freely on their bearings.

There are five variables: \$v_W\$, \$v_R\$, \$v_M\$, \$v_G\$ and \$v\$

\$v_M\$ and \$v_G\$ are the rim speeds of the wheels, in m/s.

\$v_W\$ and \$v_R\$ are given, leaving three degrees of freedom.

Two equations can be written relating the belt velocities, the rim speeds of the wheels, and the cart speed:
$$v=v_M-v_W$$ $$v=v_G-v_R$$
These two equations take up two degrees of freedom, leaving one degree of freedom remaining.

With no additional constraints this system has an infinite number of solutions. The cart velocity, \$v\$ can have any value we choose.

This makes sense, intuitively, since the wheels are free running, and without friction losses any initial cart velocity will remain the same forever.

To remove the remaining degree of freedom in the system, the wheels can be linked with a connecting gear. If we let the gear ratio be \$\alpha\$, then we can write a linking equation:
$$v_G=\alpha\,v_M$$
For each revolution of wheel M, wheel G makes \$\alpha\$ revolutions.

(Note that the gearing could be mechanical, or it could be electrical. Mechanical is simpler.)

By making a suitable choice of \$\alpha\$, we can arrange for a desired cart speed, \$v\$.

For instance, suppose we wish the cart to move at 1 m/s to the right.

We set \$v\$ = 1 m/s, which gives \$v_M=v+v_W\$ = 5 m/s, and \$v_G=v+v_R\$ = 11 m/s.

From this, we have: \$\alpha = v_G / v_M\$ = 11 / 5 = 2.2

Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.

I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.

You just looked at speed here no forces ? How will that even make sense for this particular problem ?
I wish I knew how to help you from here but you are just way off thus I'm sorry but I cant help.

Offline fourfathom

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[...]
Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.

Thank you for this!  It's also a good model for the "plank over rollers on floor" demonstration.  Harnessing the speed difference of two media, zero energy storage required, moving "DWFTTW".  Couldn't be much simpler.

Friction, drag, coupling efficiency, don't change the fundamentals either, they just set some requirements or limits for windspeed, transducer design, acceleration, and upper-bound on vehicle speed.
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Offline electrodacus

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If you remove the brakes those 600W will be available to vehicle instead of earth.

And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'?  The vehicle is still pushing on the earth...

As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).

Say you want to maintain a 1m/s  then brakes will need to deal with 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W
If your brakes where not designed to be able to get rid of those 437.4W then they will overheat and get damaged.

We can setup a scale down model of this if you do not believe that will be the case.
« Last Edit: December 21, 2021, 08:28:20 pm by electrodacus »
 

Offline fourfathom

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I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.

Look again.  Actually think about it.
We'll search out every place a sick, twisted, solitary misfit might run to! -- I'll start with Radio Shack.
 

Offline electrodacus

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I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.

Look again.  Actually think about it.

I did look and I just hope IanB has no engineering degree.

He just looked at what gear ratio he will need if he wanted to push the vehicle with 1m/s (by push I mean apply external power to the system as that is the only way that vehicle can move from left to right).
« Last Edit: December 21, 2021, 08:31:46 pm by electrodacus »
 

Online IanB

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He just looked at what gear ratio he will need if he wanted to push the vehicle with 1m/s (by push I mean apply external power to the system as that is the only way that vehicle can move from left to right).

The system does, of course, have external power. The two belts are not moving by themselves, so we must assume there is some kind of motor driving them, and if we make any attempt to slow down the belts then the motors must supply more power to maintain the belt speed. So the cart can pick up any power it needs from the difference in speed of the belts.

But leaving that aside, power is only required to overcome friction or to accelerate the cart. In an ideal system, in steady conditions, with no friction, then the power requirement is zero. So no external power is actually needed for the cart to be moving at any steady velocity.
 

Online Kleinstein

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If you remove the brakes those 600W will be available to vehicle instead of earth.

And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'?  The vehicle is still pushing on the earth...

As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).

Say you want to maintain a 1m/s  then brakes will need to deal with 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W
If your brakes where not designed to be able to get rid of those 437.4W then they will overheat and get damaged.

We can setup a scale down model of this if you do not believe that will be the case.
No need to build a model, just have think: The mechanical power from a sail is transferered as force times vehicle speed. So the force would be power divided by speed.   Does it really make sense to get an even increasing force from the sail, the slower you make the vehicle. Thas is obviously not going to happen.  The wind power of some 600 W is theoretically (that is with an ideallized wind turbine, ignoring the Beetz limit and similar details) available at zero speed, but a simple sail can not harvest it. A sail is not 100% efficient.  So doing the math with a sail you can not exclude anything for a vehichle used other means than a simple sail. Using the wrong equations comes one top of this.
 

Offline electrodacus

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No need to build a model, just have think: The mechanical power from a sail is transferered as force times vehicle speed. So the force would be power divided by speed.   Does it really make sense to get an even increasing force from the sail, the slower you make the vehicle. Thas is obviously not going to happen.  The wind power of some 600 W is theoretically (that is with an ideallized wind turbine, ignoring the Beetz limit and similar details) available at zero speed, but a simple sail can not harvest it. A sail is not 100% efficient.  So doing the math with a sail you can not exclude anything for a vehichle used other means than a simple sail. Using the wrong equations comes one top of this.

Yes it makes sense and that is exactly what you will get when a compressible fluid like air is involved.
Yes 600W is the ideal case no friction losses so a theoretical 100% efficient wind turbine will get 600W Benz limit is somewhere around 59% if I remember correctly and a real wind turbine can extract about 40% so about 240W
A sail even a real one will be very close to having access to 100% of that an ideal sail as discussed here will be exactly 100% and if that 100% ideal sail can no help vehicle exceed wind speed then is clear no vehicle no matter the design can do that unless energy storage is used.
That equation will apply to all cases where wind power is involved that means wind turbines and any type of wind powered direct down wind vehicle.
Also same equation will be used to calculate the power needed to overcome drag on any vehicle.

Offline bdunham7

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As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).

Is the car (and the wind through the car) not still pushing on the earth?
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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He just looked at what gear ratio he will need if he wanted to push the vehicle with 1m/s (by push I mean apply external power to the system as that is the only way that vehicle can move from left to right).

The system does, of course, have external power. The two belts are not moving by themselves, so we must assume there is some kind of motor driving them, and if we make any attempt to slow down the belts then the motors must supply more power to maintain the belt speed. So the cart can pick up any power it needs from the difference in speed of the belts.

But leaving that aside, power is only required to overcome friction or to accelerate the cart. In an ideal system, in steady conditions, with no friction, then the power requirement is zero. So no external power is actually needed for the cart to be moving at any steady velocity.

When I mentioned external power I was excluding the two belts / treadmills.
You need to be able to demonstrate that using the energy from road treadmill (generator wheel) and applying that to Motor wheel you are able to advance forward (meaning accelerate from left to right).
You will not be able to do that without adding energy storage.
But your equations involved just speeds like thus are completely unable to demonstrate anything.

I do not think I can continue to argue here as your level of understanding is way below what is required. Not sure if this is a failing of the education system or a limitation of most humans brain (or maybe a combination of both).
Even if I setup an experiment and show conclusively that my theory is correct all you will do is defend the new theory without understanding what you are defending.

Offline electrodacus

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As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).

Is the car (and the wind through the car) not still pushing on the earth?

Think about the wind turbine (normal fixed to the ground) and then think that the more efficient that turbine is the less the tower and ground needs to deal with.
In the case of a cart with a sail you can see the stationary part as the vehicle having the same mass as the entire earth as they move together.
If there are no brakes then vehicle will accelerate based on the vehicle weight minus the friction losses that can include a friction brake.
If there is no friction all power will be available to cart to accelerate so nothing will be transferred to the ground.

A sail vehicle is just the most efficient way to convert wind power to kinetic energy. If you take the ideal case with no friction then it 100% of available power will be converted to kinetic energy. Still even this ideal vehicle can not exceed wind speed not because is not as efficient as it gets but because it has no energy storage thus there is no way for it to accelerate above the speed of the thing that is pushing it.

Online IanB

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When I mentioned external power I was excluding the two belts / treadmills.
OK, no belts or treadmills. They are excluded.

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You need to be able to demonstrate that using the energy from road treadmill (generator wheel) and applying that to Motor wheel you are able to advance forward (meaning accelerate from left to right).
It seems that won't be possible, as the treadmills are excluded. If the road treadmill is excluded it cannot supply energy.

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You will not be able to do that without adding energy storage.
Energy storage from where? The treadmills are excluded, so they cannot supply any energy to store. There is no other source of energy in the system.

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I do not think I can continue to argue here as your level of understanding is way below what is required. Not sure if this is a failing of the education system or a limitation of most humans brain (or maybe a combination of both).
Even if I setup an experiment and show conclusively that my theory is correct all you will do is defend the new theory without understanding what you are defending.
It's certainly true that your brain is different from most humans. You are exceptional. Congratulations!
 

Online IanB

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The mechanical power from a sail is transferered as force times vehicle speed.

The sail situation is actually rather interesting.

Take a square sail, directly downwind. We can analyze the system like this:

Bernoulli's principle in stationary coordinates:
$$P_a + \frac{1}{2} \rho v_w^2 = P_s + \frac{1}{2} \rho v_s^2$$
Shifting to the reference frame of the sail:
$$P_a + \frac{1}{2} \rho (v_w - v_s)^2 = P_s$$
Therefore, the pressure difference across the sail is:
$$P_s - P_a = \frac{1}{2} \rho (v_w - v_s)^2$$
And the power transferred to the sail is:
$$W_s = (P_s - P_a) A v_s = \frac{1}{2} \rho (v_w - v_s)^2 A v_s$$
If we plot this, putting the relative vehicle speed on the horizontal axis as a fraction of wind speed, and the fraction of available wind energy captured by the sail on the vertical axis, we get a graph like this:



If we do the analysis by differentiating and finding the maximum, we find that the maximum power transfer occurs when:
$$v_s = \frac{1}{3} v_w$$
And under these conditions, the power transferred to the sail is:
$$v_s = \frac{2}{27}\rho Av_w^3$$
Therefore the sail only captures at most 15% of the available power from the wind, which is far worse than the Betz limit for a wind turbine of about 59%.
« Last Edit: December 21, 2021, 11:46:06 pm by IanB »
 

Offline Brumby

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I do not see how else you can call the pressure differential created by the propeller other than a form of energy storage.
I would not call it energy storage - but I see why you are saying that.

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There is loop but it is as follow (proportions are just an example).
Wind power available to vehicle will be split say in two equal parts.  One half accelerates the vehicle directly thus ends up as kinetic energy the other half will be taken from wheels and sent to propeller witch then increases the pressure differential
I'm not sure I agree with the proportions but, as you say, they are just to illustrate the example.  The important part here is they represent two non-zero values.

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(you can look at this as increasing the apparent wind speed relative to vehicle).
You have just given the answer as to why the Blackbird could be capable of exceeding wind speed.

While ever the Blackbird is moving, the propeller will always produce this pressure (also known as thrust) which means the "apparent wind speed" (as you have called it) which the Blackbird experiences will always be greater than the actual wind speed.

All your claims about the Blackbird's abilities (or lack thereof) are 100% correct if you were to use this "apparent wind speed".  It could never exceed the "apparent wind speed" for any length of time, if at all.

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... ... ...
For blackbird since it takes in this example half of the available power and stores that as pressure differential basically increasing the available potential energy it allows the blackbird to exceed wind speed for some limited amount of time as when above wind speed direct downwind there is no longer any wind power available to vehicle and it is starting to use the energy stored as pressure differential but it will continue to use just half of the power provided by the pressure differential to accelerate (increase kinetic energy) and then the other half it will put back in to increasing the pressure differential.
Again, I'm not enthusiastic about some of the terminology and I'd be cautious about the proportions used - but as a qualitative description, that's not too bad.

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Obviously since only half is put back the overall pressure differential will drop and vehicle acceleration rate will continue to drop
(Again, proportions) - But this is not unexpected.  Acceleration will tend to zero.

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until there is not enough
As in not enough "pressure differential" (as you call it) to provide acceleration?  I can agree with that.  There will be a point where everything balances out.  This will be the "steady state" situation I was asking about earlier.

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and it will need to start to slow down.
Now this statement is where (as I see it) you have made your error.

When acceleration drops to zero, it does not mean the Blackbird slows down.  It means the Blackbird's velocity does not change.  As long as the wind blows the same, it will continue at that same speed.

As long as the Blackbird is moving, it is continually producing (not storing) this "pressure differential" as you call it and the "apparent wind speed" will be greater than the actual wind speed.

How this translates into the Blackbird's final velocity is very dependent on the proportions you mentioned.  These proportions and the mechanisms that determine them need to be properly worked out.
« Last Edit: December 22, 2021, 12:30:59 am by Brumby »
 

Offline eti

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“Mess with your minds”? 🤨

No, not really, in fact not at all. It’s pretty obvious that the thing which everyone expects to work one way, is gonna work the opposite way. Let’s face it - one could spend months going down countless rabbit holes watching YouTube videos, but there’s only 24hrs in the day, and so what how a land sail works!
 

Offline electrodacus

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OK, no belts or treadmills. They are excluded.

I must be very bad at explaining myself. The two treadmills are the only thing interacting with vehicle. So they need to power the vehicle not some external source if you do not want to accept that there is a energy storage device then you need to explain how it will move from left to right using just the two treadmills.


Quote
You need to be able to demonstrate that using the energy from road treadmill (generator wheel) and applying that to Motor wheel you are able to advance forward (meaning accelerate from left to right).
It seems that won't be possible, as the treadmills are excluded. If the road treadmill is excluded it cannot supply energy.

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You will not be able to do that without adding energy storage.
Energy storage from where? The treadmills are excluded, so they cannot supply any energy to store. There is no other source of energy in the system.

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I do not think I can continue to argue here as your level of understanding is way below what is required. Not sure if this is a failing of the education system or a limitation of most humans brain (or maybe a combination of both).
Even if I setup an experiment and show conclusively that my theory is correct all you will do is defend the new theory without understanding what you are defending.
It's certainly true that your brain is different from most humans. You are exceptional. Congratulations!

In any case do not waste your time as you will also waste mine and you have no clue of how this works and you are not even anywhere close.

Offline electrodacus

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Now this statement is where (as I see it) you have made your error.

When acceleration drops to zero, it does not mean the Blackbird slows down.  It means the Blackbird's velocity does not change.  As long as the wind blows the same, it will continue at that same speed.

As long as the Blackbird is moving, it is continually producing (not storing) this "pressure differential" as you call it and the "apparent wind speed" will be greater than the actual wind speed.

How this translates into the Blackbird's final velocity is very dependent on the proportions you mentioned.  These proportions and the mechanisms that determine them need to be properly worked out.

Not understanding that pressure differential is energy storage will make you come to wrong conclusions like the fact that vehicle can always accelerate (as Derek and Rick claim at ever higher rate) or that it can maintain higher than wind speed indefinitely (not even close).
After vehicle is above wind speed wind can not offer anything to the vehicle so vehicle is on his own and need to use earlier stored energy that will eventually be all used up.
Should be relatively easy to test if that is the only way you can be convinced. I fully understand how it works but answering the question exactly how many minutes it will take to start to slow down is not possible as wind speed is not constant in real world and in case of blackbird pilot has access to propeller pitch control same as if you where allowed to change the gear ratio.

Online IanB

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I must be very bad at explaining myself. The two treadmills are the only thing interacting with vehicle. So they need to power the vehicle not some external source if you do not want to accept that there is a energy storage device then you need to explain how it will move from left to right using just the two treadmills.

That's exactly what I did.

If you do not like the result, you need to tell me which step in the logic contains a mistake.
 

Online IanB

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I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.
When you make this statement, you are starting with a conclusion, "No vehicle can...", and working backwards to a justification. But that is not how reasoning works. You need to start with some definitions and a statement of the problem, and work forward from there using logical reasoning to reach a conclusion.

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You just looked at speed here no forces ? How will that even make sense for this particular problem ?
I wish I knew how to help you from here but you are just way off thus I'm sorry but I cant help.
There are no forces because this is a steady state analysis of an ideal system with no friction. There are no forces, and there is no power involved. This is a pure problem in mechanics.

If we introduce friction, then there will be some forces, and there will be some power input to the system from the belts, but there still will be no storage of energy because the conditions are steady, with nothing changing over time.
 

Offline Brumby

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After vehicle is above wind speed wind can not offer anything to the vehicle so vehicle is on his own
At wind speed, the wind itself is not directly driving the vehicle.  It's like a wall of air that never makes contact with the vehicle - that part is true.  However, we still have a rotating propeller which is producing thrust.  It is this thrust which is pushing against that wall of air, giving the "apparent wind speed" phenomenon YOU mentioned.
 

Offline Brumby

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I need to bring back the discussion to what is relay important.

What is the available wind power to direct down wind vehicle.

I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

If anyone disagree with that please provide the correct equation so we can compare the predictions.

For a passive sail, then you may be right, but for the Blackbird, I believe you need something extra:

0.5 * air density * area * (wind speed - vehicle speed + speed of propeller thrust)^3
 

Offline bdunham7

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+ speed of propeller thrust)^3

Don't go there!  Thrust is force and force doesn't have a speed, although I can understand what you are getting at.  I think what you want is the effective pitch of the propeller blades multiplied by its angular velocity.  So if the pitch, which can be stated as distance per revolution, is 0.1m/rev and the propeller is spinning at 100rpm, then it's pitch speed (probably the wrong term) is 10m/s. 
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline Brumby

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+ speed of propeller thrust)^3

Don't go there!  Thrust is force and force doesn't have a speed, although I can understand what you are getting at.  I think what you want is the effective pitch of the propeller blades multiplied by its angular velocity.  So if the pitch, which can be stated as distance per revolution, is 0.1m/rev and the propeller is spinning at 100rpm, then it's pitch speed (probably the wrong term) is 10m/s.

WHOA!  I'm just looking at the OUTPUT of the propeller - which is air moving at a non zero speed.

I'm not talking about how that is calculated.  That is not important for this equation, just that it is the missing part of the formula which was presented.  Our friend asked where that formula was wrong and I have obliged.
« Last Edit: December 22, 2021, 06:35:54 am by Brumby »
 

Offline Brumby

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Besides, I use the word "thrust" here simply as a descriptive term.  If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
 

Offline electrodacus

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Besides, I use the word "thrust" here simply as a descriptive term.  If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3

There is nothing you need to do to that equation.  The correct one is the one I presented.
The extra bit that is missing the the stored energy. The propeller was powered by wind power when well below wind speed then as the wind power decreased the stored energy started to provide most of the power to vehicle.

The way that stored energy is calculated is a bit complicated but it can be done as I already did in the spreadsheet calculator.
Say 60% of wind power is sent to propeller (from the wheel) then the other 40% gets to accelerate the vehicle increasing the vehicle kinetic energy and obviously speed.
The 60% that gets to propeller will be contributing to increased pressure differential (of course propeller may be 50% efficient so half of that 60% ends up as stored energy) You can see this increase in pressure differential about the same way as if natural wind speed has increased.
In the calculator all that energy put in the propeller increases the potential wind energy and with such a mechanism 2x even 3x wind speed is not a problem but since this is stored energy as soon as it is all used up vehicle will slow down below wind speed.
All that it will need to be done is to run the experiment fully not stop the experiment just before vehicle will start to slow down.

So if anyone has a treadmill and one of those small blackbird models you can test by increasing the vehicle weight. If vehicle weight is increased enough so that vehicle starts to slow down before vehicle gets to the end of the treadmill then you can show the entire cycle of charge peak speed and then slow down.
Likely some people that did this test and saw the vehicle slow down very soon after release where thinking their vehicle is just not build well enough.

As for direct upwind that can be tested with wheels only no propeller or air is needed to demonstrate that but a relatively high speed camera at least 120fps will be good to see any direct upwind vehicle and the fact that there is a charge discharge cycle multiple times per second no matter if it uses a belt or gears.
I showed those in my video but for those that think my test setup was not done properly do your own setup and take a high speed video.
You may never understand the theory or equations but you can at least see the real test.

It is a shame that this problem is wrongly presented in some schools and I wish I could do something about it.  I was much more optimistic about being able to explain this but it seems so far I'm failing to do so.


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