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

[electrodacus]

It's not outputting any more.  It's extracting power from a wheel system that is travelling at one speed (frame of reference is the ground) and then applying that power to a propeller system travelling at a different speed (frame of reference is the wind).

This is the concept you continually dismiss, bypass or ignore.  It is THE heart of the Blackbird doing what it can - and has - done.  By refusing to even consider this concept as a possibility you are condemned to remain ignorant.

Do you understand that power you take from the wheel will brake the vehicle ? Then putting that power even all (ideal case) in propeller will not do more than just compensate (ideal case).
That is the part that you all get wrong power is not force so you can not play in the same way.  You are talking about different speed at the wheel and at the propeller but when you use power that is already included there is nothing to talk about.

That university professor that lost the bet (his fault for gambling) said nothing wrong in Derek's video but since it was unable to explain how the treadmill prototype worked it was forced to pay the money not to be seen as not honoring his obligations.  He should have asked for help and try to understand why it works then make that information public.

Poor, dumb professor.

He is not dumb just did not think about the obvious energy storage and without that he will not be able to explain how this vehicle works as there is no other alternative solution.

I do suggest you spend more time thinking about this (valid for all people here).
I spent enough time trying to help and I feel I repeated my self more than I needed to so it will not be helpful for me to help if you do not want to put the brain to work.
Try to learn what power and energy is as that should help.

[IanB]

You did not comment on the vector diagram of the wind and the boat, so I assume you agree that a boat can indeed sail faster than the wind in the direction of the wind?

[Brumby]

I do suggest you spend more time thinking about this (valid for all people here).
I spent enough time trying to help and I feel I repeated my self more than I needed to so it will not be helpful for me to help if you do not want to put the brain to work.
Try to learn what power and energy is as that should help.

"There Are None So Blind As Those Who Will Not See."

*sigh*

Let's try once more....

Do you understand that power you take from the wheel will brake the vehicle ?

I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.

Do you realise that while that extraction of power will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed?  This results in an increase in total power applied to the vehicle.  Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power.  Even if that additional power is just 2W or 1W or even 1mW, the overall power delivered to the vehicle by the wind has increased.

You have never mentioned this increased power, which indicates your understanding is less than comprehensive and your analysis certainly incomplete.

And we haven't even looked at the harvested power component.  Apply that harvested power into driving the propeller and now the backward thrust will be trying to slow the wind, but the wind (having a practically infinite energy store) will just push against this - applying even more power into the system.

The extra power required for the Blackbird to do its thing, comes precisely from the wind as the wind pushes against those forces which come from the elements that are trying to slow the vehicle down down.

At this point, even your intuition must be curious as to where this total system will settle - and what parameters will affect that point.

I would really encourage you to do a proper job of investigating this - rather than do your typical handwaving dismissal.  The total power has increased, so you cannot just say the additional power is of no significance without providing real numbers.

[Kleinstein]

Please read again my formula as you forgot a very important part and that is the area that wind pushes against.
0.5 * air density * area * windspeed³
This is the formula anyone designing a wind turbine will use except that is is ideal 100% of available wind power and to this they need to add the turbine efficiency usually around 40% and the generator efficiency that can be over 90%
For a vehicle driving directly down wind ideal case all you need is to subtract vehicle speed from wind speed since wind speed relative to vehicle is what can power the vehicle.
This formula
0.5 * air density * area * (wind speed - vehicle speed)³ is what will apply to any wind only powered vehicle driving directly downwind.
This is also the ideal case so absolutely max that is available to the vehicle meaning that if you do not add an energy storage device to this vehicle or some external energy source like gasoline and an engine then your vehicle can not exceed wind speed.
For direct downwind blackbird is clear to me that energy is stored in pressure differential generated by the propeller and that is what allows it to exceed wind speed even if it may be for just a few minutes depending on the design, how fast it will use that stored energy.

The point is not in reading the euqations, the point is understand what they are actually calculating. The first one is the power theortically available from the wind as the kinetic energy in the moving air.

The second equation is giving the energy available at the reduces wind speed, e.g. for a wind-turbine on the moving vehicle. This is not the maximum energy available to the vehicle.

The wind turbine also creates some drag force, comparable to a sail and this drag force also creates power to a moving vehicle. This power contribution is force times vehicle speed. For a sail this is the only way it creates power.

The wind turbine (=porp) on the vehicle could create some extra power directly. For the sail the drag force is proportional to (wind_speed-vehicle_speed)². For the prop the drag force (called thrust when caused by the driven prop) also depends on the speed the prop rotates.  depending on the direction and speed this can be less, but also more than just a sail.  With the driven prop there can be drag/thrust even with zero wind speed. 

[electrodacus]

I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.

Do you realise that while that extraction of power will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed?  This results in an increase in total power applied to the vehicle.  Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power.  Even if that additional power is just 2W or 1W or even 1mW, the overall power delivered to the vehicle by the wind has increased.

You have never mentioned this increased power, which indicates your understanding is less than comprehensive and your analysis certainly incomplete.

And we haven't even looked at the harvested power component.  Apply that harvested power into driving the propeller and now the backward thrust will be trying to slow the wind, but the wind (having a practically infinite energy store) will just push against this - applying even more power into the system.

The extra power required for the Blackbird to do its thing, comes precisely from the wind as the wind pushes against those forces which come from the elements that are trying to slow the vehicle down down.

At this point, even your intuition must be curious as to where this total system will settle - and what parameters will affect that point.

I would really encourage you to do a proper job of investigating this - rather than do your typical handwaving dismissal.  The total power has increased, so you cannot just say the additional power is of no significance without providing real numbers.

Again do we discus about power or energy.  You seems to confuse the two way to much.
If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension.
So if your conditions are vehicle exactly at same speed as wind speed and you are talking in therms of power you take 5.5W from wheel speed will be the same and then apply those 5.5W to propeller (ideal case) and since there is no imbalance of power between input and output nothing will change thus no deceleration but also no acceleration.
 

Not sure if I can help you. Try to understand the difference between power and energy then you will not even need my help in understanding this problem. 

[IanB]

power is instantaneous contains no time dimension

Power is defined as "the rate of doing work" and has dimensions of energy/time. So it certainly does contain a time dimension.

[IanB]

Here are the correct equations for wind power available to a vehicle traveling directly down wind
0.5 * air density * area * (wind speed - vehicle speed)^3    yes this is the correct equation. Notice how you have highest power at low vehicle speed and power drops to zero by the time vehicle speed equals wind speed.

But why is this the correct equation?

Please derive this equation step by step from first principles, stating any assumptions made at each step.

Then you will understand where and how this equation can be applied, and what limitations exist in its usage.

[electrodacus]

The point is not in reading the euqations, the point is understand what they are actually calculating. The first one is the power theortically available from the wind as the kinetic energy in the moving air.

The second equation is giving the energy available at the reduces wind speed, e.g. for a wind-turbine on the moving vehicle. This is not the maximum energy available to the vehicle.

The wind turbine also creates some drag force, comparable to a sail and this drag force also creates power to a moving vehicle. This power contribution is force times vehicle speed. For a sail this is the only way it creates power.

The wind turbine (=porp) on the vehicle could create some extra power directly. For the sail the drag force is proportional to (wind_speed-vehicle_speed)². For the prop the drag force (called thrust when caused by the driven prop) also depends on the speed the prop rotates.  depending on the direction and speed this can be less, but also more than just a sail.  With the driven prop there can be drag/thrust even with zero wind speed.

As I said many times before please learn the difference between power and energy. Both equation refer to power only nothing to do with energy.
The second equation outputs the ideal case wind power available to any direct down wind vehicle (any means is true for blackbird and for a sail vehicle).

The difference is that sail vehicle (an ideal one) takes all this available power and stores it as kinetic energy thus not able to exceed wind speed.
Blackbird will split this available wind power in to two. One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential  by increasing the difference in pressure between upwind side of the propeller and down wind side.

Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.

[electrodacus]

But why is this the correct equation?

Please derive this equation step by step from first principles, stating any assumptions made at each step.

Then you will understand where and how this equation can be applied, and what limitations exist in its usage.

There are multiple resources here is one
http://web.mit.edu/wepa/WindPowerFundamentals.A.Kalmikov.2017.pdf
The equation can be applied anywhere starting with wind turbine design, power available to any wind powered vehicle, drag power needed for vehicle design .....

[IanB]

One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential  by increasing the difference in pressure between upwind side of the propeller and down wind side.

Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.

But the pressure after the propeller is lower than the pressure in front of the propeller: https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan

I have shown you a video illustrating this. Do you think that video was manipulated or faked in some way? Every engineer knows that video shows the truth. You can do the same experiment yourself. Just use any household fan. You will find the pressure in the moving airstream is lower than the surroundings.

You should stop looking at that Wikipedia article. It does not say what you think it says.

[IanB]

There are multiple resources here is one
http://web.mit.edu/wepa/WindPowerFundamentals.A.Kalmikov.2017.pdf
The equation can be applied anywhere starting with wind turbine design, power available to any wind powered vehicle, drag power needed for vehicle design .....

No, I want you to derive it yourself, and give your assumptions in the derivation.

If you cannot do that, you are merely quoting facts and equations from others, which in your own words shows a lack of understanding.

Unfortunately school is mostly for those that can memorize facts and equations an understanding is completely irrelevant.   

[electrodacus]

One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential  by increasing the difference in pressure between upwind side of the propeller and down wind side.

Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.

But the pressure after the propeller is lower than the pressure in front of the propeller: https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan

I have shown you a video illustrating this. Do you think that video was manipulated or faked in some way? Every engineer knows that video shows the truth. You can do the same experiment yourself. Just use any household fan. You will find the pressure in the moving airstream is lower than the surroundings.

You should stop looking at that Wikipedia article. It does not say what you think it says.

Not sure what you mean by after and front. Here is the graph for pressure difference.
Never claimed that anything was faked. It works exactly as shown from the equations and theory I presented.

[electrodacus]

No, I want you to derive it yourself, and give your assumptions in the derivation.

If you cannot do that, you are merely quoting facts and equations from others, which in your own words shows a lack of understanding.

Unfortunately school is mostly for those that can memorize facts and equations an understanding is completely irrelevant.   

Do not be afraid just open the pdf and look at page 4.

[IanB]

Not sure what you mean by after and front. Here is the graph for pressure difference.
Never claimed that anything was faked. It works exactly as shown from the equations and theory I presented.

I mean, if you do the experiment, yourself, with a real fan, you will find that P₂ < P_A.

Never claimed that anything was faked.

But you did. I showed you a video where Matthias Wandel did the experiment, and you said you didn't believe the video. If you disbelieve a video of an actual, real world experiment, showing what actually happens, then you must think the video was somehow faked?

[fourfathom]

The difference is that sail vehicle (an ideal one) takes all this available power and stores it as kinetic energy thus not able to exceed wind speed.

So why is there no pressure differential energy stored behind the sail?

Blackbird will split this available wind power in to two. One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential  by increasing the difference in pressure between upwind side of the propeller and down wind side.
Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.

Stored rotational kinetic energy can not cause a fixed-geometry propeller to increase rotational speed.  It can at best reduce the rate that the propeller rotation will slow down.
In your models and equations, stored pressure differential energy will not cause the vehicle to accelerate to above windspeed, since any pressure differential will diminish to zero at windspeed, and go negative when the vehicle is above windspeed.

Your errors in analysis have been explained many times.

[IanB]

Do not be afraid just open the pdf and look at page 4.

Do you think the teachers do not know how to do the exercise?

But apparently, you cannot do it. You are just quoting facts and equations, which by your own words shows you do not understand.

[electrodacus]

So why is there no pressure differential energy stored behind the sail?

There is but that pressure drops to zero by the time sail vehicle gets to wind speed so it is not useful to exceed wind speed
With propeller that already existing pressure differential created by wind is increased by the propeller that pushes air in the opposite direction while vehicle is below wind speed. It can no longer increase that pressure while it is above wind speed so at some point that pressure differential drops to nothing and then is when blackbird will start to slow down.

Stored rotational kinetic energy can not cause a fixed-geometry propeller to increase rotational speed.  It can at best reduce the rate that the propeller rotation will slow down.
In your models and equations, stored pressure differential energy will not cause the vehicle to accelerate to above windspeed, since any pressure differential will diminish to zero at windspeed, and go negative when the vehicle is above windspeed.

Your errors in analysis have been explained many times.

Yes that is true the rotational kinetic energy can not do that but the pressure differential can. And yes that stored rotational energy will help sow down the deceleration after the pressure differential stored energy is used up.
My models (see in my video) perfectly describe how wind speed is exceeded using stored pressure differential energy.

[IanB]

Do not be afraid just open the pdf and look at page 4.

Yes, and pages 3 and 4 of that pdf contain proof that your equation is wrong, which is why I am asking you to derive it for yourself, so you can understand why you have the wrong equation.

[Brumby]

Do you understand that power you take from the wheel will brake the vehicle ?

If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension.

Which is it?  You can't have it both ways.

[Brumby]

as power is instantaneous contains no time dimension.

Now that's just funny ... in a sad, sad way.


You're tripping over yourself.

[Brumby]

Let's revisit:

Do you understand that power you take from the wheel will brake the vehicle ?

I presume you mean energy.

I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.

Do you realise that while that extraction of energy will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed?  This results in an increase in total power applied to the vehicle.  Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power and, thus, more energy.  Even if that additional power is just 2W or 1W or even 1mW, the overall energy delivered to the vehicle by the wind has increased.


The above is perfectly reasonable and reads pretty clearly, even before throwing in the word "energy".  There is no conflict in using the terms power or energy.  Each describes a particular quantity and those terms can be mixed quite happily.  It is for the reader to understand what information each quantity provides.  The only difference is time and the context gives enough structure for the appropriate consideration of time.

[electrodacus]

Let's revisit:

Do you understand that power you take from the wheel will brake the vehicle ?

I presume you mean energy.

I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.

Do you realise that while that extraction of energy will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed?  This results in an increase in total power applied to the vehicle.  Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power and, thus, more energy.  Even if that additional power is just 2W or 1W or even 1mW, the overall energy delivered to the vehicle by the wind has increased.


The above is perfectly reasonable and reads pretty clearly, even before throwing in the word "energy".  There is no conflict in using the terms power or energy.  Each describes a particular quantity and those terms can be mixed quite happily.  It is for the reader to understand what information each quantity provides.  The only difference is time and the context gives enough structure for the appropriate consideration of time.

I do not want to discredit you I'm just pointing the fact that you do not know what power is and also do not know the difference between power and energy.
Yes taking power from the wheel will mean braking the vehicle but if you do not mention the amount of time you took the power then you have no idea how much the vehicle has slowed down.
If you mention a time period say 1ns, 1ms or 1s then you also mentioned the amount of energy you extracted meaning you can calculate exactly by how much the vehicle slowed down.
If you just mention power then you look at power input (what you take from the wheel) and the output power the one that propeller will output and if the output is equal or lower that the input (the only way it can be) then there is no acceleration.
This example I used in my video uses power to calculate if there is a net power meaning acceleration or if that value is negative deceleration.
Keep in mind this is the equivalent vehicle with wheels only so there is no air involved and no possibility to store energy in pressure differential.
Also keep in mind that there is also no energy storage consider in the internal mechanism so no option to store energy inside the vehicle like it is the case with direct down wind version's including the toy vehicle with rubber band I showed in the video.
Below are the correct equations and shows that if there is no energy storage available that will be the result.

[IanB]

Below are the correct equations

If you would stop making assertions like this, and instead ask questions like, "Are these the correct equations?" or "What is the correct way to analyze this system?", then you would become far more enlightened.

As it is, you are unable to make any progress in your understanding.

[electrodacus]

Below are the correct equations

If you would stop making assertions like this, and instead ask questions like, "Are these the correct equations?" or "What is the correct way to analyze this system?", then you would become far more enlightened.

As it is, you are unable to make any progress in your understanding.

If you disagree post what you think are the correct ones.  I do not make a video without checking multiple times and making sure what I say is correct. I can of course still make mistakes as anyone else.
Looking forward to your results for that problem.

[Kleinstein]

The case with the wheels is a bit different from the prop case, but I think it contains the same difficulty in understanding. As an advantage the basics of mechanics are a bit simpler and less of an approximation. So it may be OK to look again at this simpler picture for the start.

To judge if the equations are correct or not the first thing is to make clear what is meant with P_out and P_net mean and the condition for the vehicle.
For the condition of the vehicle the situation to look at, the most useful case would be the vehicle standing still in the picture and with F_G = F_M to have a stationary case with no acceleration.

So what do P_out and P_net stand for ?