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

[fourfathom]

Well, obviously you can compress a spring by pressing on only one end of it, while leaving the other end free... 

The other end of the spring is being pushed by the wind itself.  Obviously.  Or something.

[gnuarm]

Actually, if you look at the car/wheel/belt system carefully, and make the following assumptions:

   1. The system is in steady state (no acceleration)
   2. The system is ideal (no friction losses from bearings or air resistance)

Then from these assumptions the power everywhere is zero, as power is only required to overcome losses or produce acceleration.

Since the power is everywhere zero, it is not possible to analyze this system in terms of power flows, and no conclusions can be drawn using that approach.

To analyze the system correctly, a different approach is required.

Yes, don't assume zero losses.  Assume some losses from the air resistance and rolling resistance.  It doesn't need to be complex.  Just consider that you can't move through the air with zero loss. 

[gnuarm]

I have never claimed blackbird is not going faster than the wind.
What I was saying is that it can only do so using energy storage and in case of blackbird that energy storage is the pressure differential created by the propeller.
A gear can not amplify power thus output power will always be lower than the input.

What does energy storage accomplish?  The vehicle moves faster than the wind continuously.  In the examples where you talk about microslipping or some such nonsense, the vehicle is clearly moving faster that the relative wind continuously because your microslipping is such a short time duration.

[Brumby]

Yes, don't assume zero losses.  Assume some losses from the air resistance and rolling resistance.  It doesn't need to be complex.  Just consider that you can't move through the air with zero loss. 

I suggest it is better to not complicate the key discussion with real world minutiae.  It is far more practical to work out the basic principles using ideal elements.  That is, to work out if the desired result is possible. 
Once that is established, then add in the real world losses to see what sort of experimental result you can expect.

If it can't be found in the ideal situation, then there's no point in even considering losses.

[IanB]

Actually, the analysis does not change if you introduce friction losses. All that happens is that any losses due to friction are overcome by the power of the motors driving the belts. You just have to assume that the motors are powerful enough to achieve the stated belt speeds, and that the wheels of the cart do not slip on the belts.

[electrodacus]

Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.

Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.

You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released.  Include any necessary description and formulae that would be required for it to stand up to examination.

I will make an attempt to explain pressure differential.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.
Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.
If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.
All this is possible because air is a compressible fluid and it is much heavier than you actually imagine.
This is also the reason I insist that none of the wheels only vehicles can represent direct down wind faster than wind and all of them represent direct upwind faster than wind as there pressure differential energy storage is not used.

I saw some comments about my example with vehicle drag.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
I was making a point of the fact that the equation I insist on is one of the most important equations in anything that has to deal with air from wind turbines to any type of wind powered vehicles and to any aerodynamic drag calculation.
And yes using power to define drag is absolutely the right type of unit. When you design a vehicle you need to know the drag power as that is the most significant part affecting vehicle consumption.
I see people comparing Tesla model 3 a very small super aerodynamic vehicle with a large SUV in therms of consumption at highway driving speeds and they think that the SUV was just badly designed with a very inefficient drive-train where the reality is that the larger air drag is by far the largest factor in an EV consumption.     

[bdunham7]

If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.

Nope.  It will need 1/6 of the power.

[electrodacus]

If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.

Nope.  It will need 1/6 of the power.

You have no idea what air is.

[IanB]

That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.

Why do you keep going back to old things that we have proved to you are not true?

A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.

You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage

If you keep going back and re-stating wrong things that have been corrected earlier, then this thread is going round in circles and cannot make progress. That is why we think you are trolling.

Insisting on something doesn't make you right, no matter how often you repeat it. In order to demonstrate the correctness of what you are saying, you have to be able to prove it with appropriate equations and logic, which is something you consistently fail to do. Moreover, you keep ignoring inconvenient facts when they go against your preconceptions.

[IanB]

Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.
I will no longer post the graph but you can check that here https://en.wikipedia.org/wiki/Axial_fan_design
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.

If air is compressible, why does your power formula only have a single, constant density in it?

If air is compressed like a spring, then the density will vary with pressure and your equation for power cannot be correct.

[Brumby]

I will make an attempt to explain pressure differential.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.

This is pretty straightforward.  You didn't need to explain this as we are all well aware of this.

Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.

Are you saying that energy is stored across the entire swept area of the propeller?

[Kleinstein]

If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.

The analogy to the inductor is good and it includes 2 not so convenient effects:
1) With a normal conductor the energy storrage is for limited time only. If you short out the inductor, the current will go down with time - usually one the oder of less than a second. A similar thing happens with the pressure near the fan: one the fan is no longer creating a new pressure difference, the pressure will drop, even if the ernergy is not used to accelerate the vehicle. So in about a second or less the erngy in the air is gone, if the is no container around it. By coincidence the time scales for the usual inductors and the blackbird vehicle are comparable.

2) It is very difficult to use the erngy storred in the inductor to pump more current through the inductor. You only get the energy out from the inductor by lowering the current. Similar with the pressure behind the fan. The pressure is linked to the speed of the fan. So when making the fan spin faster, it would also cause an increase in the pressure difference. So with the fan propelling the vehicle, you can not speed up the fan from the pressure difference.
In the vehicle on the treadmil the fan speed is directly linked to the wheels.  So without some magic control for the fan pitch you can only use the energy from the pressure when slowing down.

[gnuarm]

Yes, don't assume zero losses.  Assume some losses from the air resistance and rolling resistance.  It doesn't need to be complex.  Just consider that you can't move through the air with zero loss. 

I suggest it is better to not complicate the key discussion with real world minutiae.  It is far more practical to work out the basic principles using ideal elements.  That is, to work out if the desired result is possible. 
Once that is established, then add in the real world losses to see what sort of experimental result you can expect.

If it can't be found in the ideal situation, then there's no point in even considering losses.

Sounds great, but if "ideal" conditions prevent the analysis as was done here, then there's no way to draw any conclusions.  Did you see the part where under no losses there's no power flow and so no analysis?  That's not a key discussion.

It's all moot.  The thing was built, tested under many conditions and it works.  Anyone who wishes to say it doesn't work is a denialist without reason.

[gnuarm]

Actually, the analysis does not change if you introduce friction losses. All that happens is that any losses due to friction are overcome by the power of the motors driving the belts. You just have to assume that the motors are powerful enough to achieve the stated belt speeds, and that the wheels of the cart do not slip on the belts.

I believe you said you could not analyze the power flow because it was non-existent in the ideal case.  So I'm saying don't assume a perfectly ideal case and you can analyze the power flow. 

[gnuarm]

Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.

Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.

You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released.  Include any necessary description and formulae that would be required for it to stand up to examination.

I will make an attempt to explain pressure differential.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.
Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.
If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.
All this is possible because air is a compressible fluid and it is much heavier than you actually imagine.
This is also the reason I insist that none of the wheels only vehicles can represent direct down wind faster than wind and all of them represent direct upwind faster than wind as there pressure differential energy storage is not used.

I saw some comments about my example with vehicle drag.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
I was making a point of the fact that the equation I insist on is one of the most important equations in anything that has to deal with air from wind turbines to any type of wind powered vehicles and to any aerodynamic drag calculation.
And yes using power to define drag is absolutely the right type of unit. When you design a vehicle you need to know the drag power as that is the most significant part affecting vehicle consumption.
I see people comparing Tesla model 3 a very small super aerodynamic vehicle with a large SUV in therms of consumption at highway driving speeds and they think that the SUV was just badly designed with a very inefficient drive-train where the reality is that the larger air drag is by far the largest factor in an EV consumption.   

That was a massive fail.  He started to explain the pressure differential and how it relates to the downwind faster than wind case, then shifted gears and started talking about other stuff.  I guess that says it all. 

[gnuarm]

That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.

Why do you keep going back to old things that we have proved to you are not true?

A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.

If you are talking about having no other losses than air drag and ignore ground effects of the air stream, you are completely right except that you are wrong.

If you are including other losses such as tire losses, then yes, there is an element of lower resistance since the ground speed is lower.  The calculator you show below includes tire losses and has an input to select the type of tire.

You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage

I think he is talking about only the wind effects, so he is right. 

[IanB]

I believe you said you could not analyze the power flow because it was non-existent in the ideal case.  So I'm saying don't assume a perfectly ideal case and you can analyze the power flow.

I think you misunderstand. I said you cannot perform an analysis of that system using power flows as the basis, because that is not the correct analysis. The correct analysis is to use belt speeds, wheel speeds, and gear ratios, as I did. That analysis is the same regardless of any friction losses in the system.

Power requirements are an outcome of the calculations, not an input to the calculations.

[IanB]

I think he is talking about only the wind effects, so he is right.

No. He said the power required to overcome wind resistance is the same when cycling at 60 km/h as it is when cycling at 10 km/h against a 50 km/h headwind. That is wrong. It takes much less power to go at 10 km/h.

(This has nothing to do with rolling resistance. This is only about wind effects.)

[Kleinstein]

One can do the analysis based on power flow also with losses. It is also relatively easy. However one should not look at the power taken from the wind and the power transferred to the vehicle mass. The easy way is looking at the power to drive the prop and the power you can get from the wheels. The movement is possible if the wheels can generate more power than needed for the prop + the power needed to overcome friction.
The calculation is relatively easy for the vehicle going at the wind speed and the question is than of the wheels can generate more power than actually needed. With excess power available the vehicle could go faster than the wind too.

There is no need to care about energy storage - all is steady state.
There is no need to care about wind resistance / aerodynamics as the relative wind speed is zero in the calculation. Just take a given working point of the prop (a given thrust and the power needed for that).
There is no need to calculate the maximum available power (it is enough to show a way to get enough).

Diretly trying to calculate the maximum available power from the wind is tricky, as there are different ways to used the wind and not all have a well defined area and the like. Chances are that for a vehicle going faster than the wind this may be actually near infinite (e.g. getting to the speed of sound were some of the approximations fail), as more and more air mass is available with an ever faster speed.

Prooving that something is impossible can be quite tricky and would usually be of the type showing a violation of the laws or equavalence to something known to be impossible (e.g. a perpetu-mobile). The other way around is often much easier, as it only has to show 1 possible implementation and this does not even have to be a good one.

[electrodacus]

A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.

You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage

If you keep going back and re-stating wrong things that have been corrected earlier, then this thread is going round in circles and cannot make progress. That is why we think you are trolling.

Insisting on something doesn't make you right, no matter how often you repeat it. In order to demonstrate the correctness of what you are saying, you have to be able to prove it with appropriate equations and logic, which is something you consistently fail to do. Moreover, you keep ignoring inconvenient facts when they go against your preconceptions.

Yes that bicycle calculator is incorrect (you are not the only one to get this wrong so that includes those people that did the calculator)
That calculator will say you can drive at 1km/h in 230km/h headwind with just 300W (easy for a cyclist)
This is maybe 80km/h no where near 230km/h and here is what happens

[fourfathom]

No. He said the power required to overcome wind resistance is the same when cycling at 60 km/h as it is when cycling at 10 km/h against a 50 km/h headwind. That is wrong. It takes much less power to go at 10 km/h.

(This has nothing to do with rolling resistance. This is only about wind effects.)

Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.

[Labrat101]

Hello guys . He is playing with you and you are falling for it .
He is side tracking you .
We are talking about can blackbird can go faster than tail wind .
He has mentioned SO Far :: Inductors . because they store energy    .Yes, Not that energy that blows a ship across the water.
Turbine on a car . Yes .. Wind turbine blade for a small blade is about 25 mtr dia .. does the blade stand still & the car rotate?
Now push bikes .
     This is even Funnier than   Monty Python's Flying Circus
Now he has realized that he has  ed up . And drawing you All into his warped web of confusion . 

And Now for something completely Different .. Reality

 key parameters in propeller design, the main ones being the power to drive the propeller,
 and the thrust that the propeller delivers.
    • The angle between blade chord and propeller plane is the geometric pitch (blade angle β
    • )
    • Angle of attack of the blade element α=β−Φ
    • The effective pitch angle is Φ=arctan(V/ω∗r)
. This is often given relative to rotational speed n = rounds per second of the propeller, and the propeller diameter R:
Φ=arctanVn⋅D∗1π⋅r/R
The advance ratio J of the propeller is
J=Vn⋅D
Induced velocity should be constant over the blade, implying that β
decreases linearly with increasing r: the propeller blade twist. Because the twist changes linearly, one point on the blade can be taken as the representative blade β
, and this is usually taken at either 70% or 75% of the radial distance.
It can be shown that for a given blade geometry, the power and thrust coefficients CP
and CT of the propeller are determined only by J and β0.75. If tip speed is below speed of sound and the blades are not stalled, Mach and Reynolds number effects are negligible.
CP=Pρ⋅n3⋅D5
CT=Tρ⋅n2⋅D4
    now look up CP
as function of J in for instance NACA reports from the NACA 

Regarding the profiles: propeller design is pretty specialist but old fashioned NACA profiles are still being used in helicopter blades for both main rotor and tail rotor. (Example) For a symmetrical profile NACA 0012 would be a representative choice, again available from the NASA .
 A reference book on this is Theory Of Wing Sections by Abbott & Von Doenhoff.

 Sorry if I am a bit rusty the last design stuff was over 45 years ago .

[bdunham7]

Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.

Yes, that determines the wind force.  The power used to move against that force is the product of the force and the speed with which you move against it, not the speed (apparent or otherwise)  of the medium. 

[IanB]

Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.

Yes, as bdunham7 said, the power required is the product of the vehicle speed and the resistance to motion. So you have:

  (power) = (speed) x (resistance)

In SI units, that is:

  [W] = [m/s] x [N]

Since 1 W = 1 Nm/s, it becomes this:

  [Nm/s] = [m/s] x [N]

So, if you are moving at 60 km/h in still air, the speed is 60 km/h, and the drag force corresponds to a wind velocity of 60 km/h.

On the other hand, if you are moving at 10 km/h against a 50 km/h headwind, the speed is 10 km/h, and the drag force corresponds to an apparent wind velocity of 10 + 50 = 60 km/h (same).

The drag force is the same in both cases, but in the second case the speed is 1/6 of the first case, so the power required is also 1/6.

[Domagoj T]

Full disclosure, I skipped quite a few pages of this discussion, for obvious reasons.
In any case, how I interpret the downwind faster than wind is just as a simple lever. The wind has arbitrarily large amount of energy (size of the propeller is not specified), it's just a matter of devising a mechanism to push against the huge energy source that is the wind.
A simple demonstration is a yo-yo (yes, the spiny toy on a string).
 

Put the yo-yo on the table with the string coming from the underside of the bobbin part and pull. The yo-yo will catch up with you.
If you don't have a yo-yo at hand, substitute with a spool of solder wire.