Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147194 times)

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Offline gnuarm

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This has long since stopped being a technical discussion and has become a psychological interview.  I'm very interested in understanding ED's motivation. 

It could just be troll baiting, trying to keep this thread going as long as possible. 

It could be that he really does think he is the only person to understand this issue.  Clearly he is wrong and so either is not capable of understanding the real issues or is just playing with us. 

Orrrr... he might know he is wrong, but figures as long as he is not forced to admit it, he isn't technically wrong. 

Hmmm.... I wonder which it is.
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Offline fourfathom

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

Are you sure about that?  When wind speed = vehicle speed, that equation still reduces to zero.
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Offline Labrat101

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This has long since stopped being a technical discussion and has become a psychological interview.  I'm very interested in understanding ED's motivation. 

It could just be troll baiting, trying to keep this thread going as long as possible. 

YES this exactly what he is doing . I mentioned that he was doing this about 10 pages back .
 Every one is getting sucked into some sort of weird elusions of trying to convince him.
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Offline gnuarm

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I don't care about the wind power.  This equation was being tossed about when there seemed to be a discrepancy of whether the power to move a vehicle falls to zero when the vehicle approaches a zero velocity.  To determine that, the force is required which is

0.5 * air density * area * (wind speed - vehicle speed)²

This is the force provided by the wind which must then be applied by the wheels to move the vehicle at a speed in the face of the wind.  To find the power that must be applied to the vehicle you multiply by the relative speed of the vehicle and the point applying the force, the ground, or "vehicle speed".  The equation for the applied power then becomes

0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)

There's no point in shifting the conversation to rivers or moats or spacecraft.  We are talking about a ground vehicle in the wind.  Your equation is for power in the wind and the wind alone and has nothing to do with the power required to move the vehicle. 

Until you understand your mistake here, you have no possibility of understanding the blackbird.  You really need to listen to those who are trying to help you understand.  This is not a battle, this is your friends who want to see you succeed in learning the correct way to understand this issue. 
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Offline electrodacus

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This has long since stopped being a technical discussion and has become a psychological interview.  I'm very interested in understanding ED's motivation. 

It could just be troll baiting, trying to keep this thread going as long as possible. 

It could be that he really does think he is the only person to understand this issue.  Clearly he is wrong and so either is not capable of understanding the real issues or is just playing with us. 

Orrrr... he might know he is wrong, but figures as long as he is not forced to admit it, he isn't technically wrong. 

Hmmm.... I wonder which it is.

I have no intention for this to continue in fact my goal is to get this done as soon as possible.
I'm sure not the only person understanding this as many product designs depend on using the correct equation.
I work in the field where understanding power and energy is super important (I designed my own wind turbine) and my business is in the renewable energy storage thus I investigated all energy storage sources including but not limited to eletrochemical energy storage, kinetic energy storage and thermal energy storage.
There will always be people that do not understand some parts of physics and I will not get bothered by this particular problem if it was not such a wide spread misinformation involving science communicators and university professors. 

Offline gnuarm

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

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

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

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I don't care about the wind power.  This equation was being tossed about when there seemed to be a discrepancy of whether the power to move a vehicle falls to zero when the vehicle approaches a zero velocity.  To determine that, the force is required which is

0.5 * air density * area * (wind speed - vehicle speed)²

This is the force provided by the wind which must then be applied by the wheels to move the vehicle at a speed in the face of the wind.  To find the power that must be applied to the vehicle you multiply by the relative speed of the vehicle and the point applying the force, the ground, or "vehicle speed".  The equation for the applied power then becomes

0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)

There's no point in shifting the conversation to rivers or moats or spacecraft.  We are talking about a ground vehicle in the wind.  Your equation is for power in the wind and the wind alone and has nothing to do with the power required to move the vehicle. 

Until you understand your mistake here, you have no possibility of understanding the blackbird.  You really need to listen to those who are trying to help you understand.  This is not a battle, this is your friends who want to see you succeed in learning the correct way to understand this issue.

The equation for force is correct and includes wind speed - vehicle speed  the equation for power will also need to include both.
Just imagine air as a solid hitting the vehicle. Can you see why there power available and not just force will include the speed of that solid ?
This is just wrong equation  0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)  the correct one is this for all applications
0.5 * air density * area * (wind speed - vehicle speed)² * (wind speed - vehicle speed)

Online Kleinstein

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electrodacus starts to repeat of old nonsense over an over again. Chance are he not at all interested in learning something. Given his inablity to see the contradictions his claims cause, I don't think it makes much sense to argue any more. Chances are all other have got the point by now.


 

Offline gnuarm

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I have no intention for this to continue in fact my goal is to get this done as soon as possible.

Lol!  Your first post was Reply #45 and your last was 1280!!!  If you didn't intend to keep this going so long you sure seem impotent to end it.


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I'm sure not the only person understanding this as many product designs depend on using the correct equation.

Yes, they do, just not the way you think they do.


Quote
I work in the field where understanding power and energy is super important (I designed my own wind turbine) and my business is in the renewable energy storage thus I investigated all energy storage sources including but not limited to eletrochemical energy storage, kinetic energy storage and thermal energy storage.

I pity anyone buying your products.  Really!


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

The real issue is that both upwind and downwind motion is possible being powered by the wind.  You seem to want to complicate the issue by dragging in imaginary storage of energy and refusing to discuss details that would show you wrong.  Even after we point to the spots in the conversation where you stop discussing an idea that would prove you wrong, you continue to refuse to discuss those issues.

I'm voting for number 3.  He does now understand that he is wrong (it would be so nearly impossible for him to not see this) and is continually back peddling so he doesn't have to admit it. 

The funny thing is the only thing worse than admitting you are wrong, is refusing to admit it in the face of incontrovertible proof.  So he applies the wrong equation to the right situation (or is it the other way around?), changes the discussion to extraneous topics and poor analogies and refuses to continue a thread of discussion when it is clear it will prove him wrong.  This has to be knowing evasion of the truth.  Even a troll eventually gets tired of the game.
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Online IanB

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

It is supposed to reduce to zero. When the vehicle is moving at the same speed as the wind, the power transfer from the wind to the vehicle is zero.

See here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830
« Last Edit: December 28, 2021, 09:10:51 pm by IanB »
 

Offline electrodacus

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Black bird is hit by an air particle. What will happen there ?
Stop thinking that a vehicle has a brake.

Offline gnuarm

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The equation for force is correct and includes wind speed - vehicle speed  the equation for power will also need to include both.
Just imagine air as a solid hitting the vehicle. Can you see why there power available and not just force will include the speed of that solid ?
This is just wrong equation  0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)  the correct one is this for all applications
0.5 * air density * area * (wind speed - vehicle speed)² * (wind speed - vehicle speed)

Again, not trying to calculate the power in the wind.  We are calculating the power to move the object in the wind.  That power is the force times the vehicle velocity.  You can dance around all you want, but the fact remains you are applying the wrong equation to this.

Imagine the sail is stationary and the wind blows on it.  The force on the sail is applied through linkages to the vehicle.  Now the vehicle is not in the wind, only the sail.  Clearly in this situation the force that must be applied to the vehicle to maintain a constant velocity is the force on the sail from the wind and the ONLY velocity that is relevant to the vehicle is the vehicle speed, so

0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)

I've explained it, I've derived it.  You are not capable of understanding it or you are just refusing to acknowledge you understand it in public. 
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Offline fourfathom

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

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

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

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

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Again, not trying to calculate the power in the wind.  We are calculating the power to move the object in the wind.  That power is the force times the vehicle velocity.  You can dance around all you want, but the fact remains you are applying the wrong equation to this.

Imagine the sail is stationary and the wind blows on it.  The force on the sail is applied through linkages to the vehicle.  Now the vehicle is not in the wind, only the sail.  Clearly in this situation the force that must be applied to the vehicle to maintain a constant velocity is the force on the sail from the wind and the ONLY velocity that is relevant to the vehicle is the vehicle speed, so

0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)

I've explained it, I've derived it.  You are not capable of understanding it or you are just refusing to acknowledge you understand it in public.

We are trying to calculate the available wind power to the vehicle as that is all the vehicle can use since it is a wind only powered vehicle thus the most important equation to get right.

The vehicle will not be stationary if there is no friction including no brake. The lowest energy state for that vehicle will be vehicle speed = wind speed.
If vehicle without brakes wants to maintain zero speed relative to ground it will need to provide this power
0.5 * air density * area * (wind speed - vehicle speed)² * (wind speed - vehicle speed)
And since you want vehicle speed to be zero relative to ground you have 0.5 * air density * area * (wind speed)3

This shows that even an ideal vehicle able to get all wind power can at most just use that to stay in the same place as if it will want to move upwind it will need that plus something extra.
As for direct down wind the ideal vehicle can get to same speed as wind speed but not above that without having some stored energy or an external energy source other than wind.

You all seems to agree that is correct for a sail vehicle but refuse to believe that is valid for any other type of vehicle including blackbird.
Blackbird is not using wind power when above wind speed but it is using earlier stored wind energy and the difference is stored energy is a limited resource so a real vehicle with friction at some point will start to slow down.

Online IanB

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

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

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

It breaks down into three parts:

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

Hence:

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

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

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

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

Offline electrodacus

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


A real vehicle that has friction will never be able to get to wind speed it will stay at some lower than wind speed forever. The only way to even get to wind speed or exceed that is to use an external energy source or an energy storage device to store wind energy when it is available as it is the case for direct downwind blackbird.

Offline electrodacus

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

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

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

It breaks down into three parts:

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

Hence:

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

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

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

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

Actually a sail on ideal wheels friction less will be 100% efficient in using that available wind power.

Air is a fluid so trillion of small particles that have mass and an average speed in a particular direction.
So each time a particle hits that area of the sail it will lose half of the kinetic energy since the other half will be transferred to sail.
Stop involving brakes or an anchor. It is like a wind turbine had the electrical brakes enabled of course there will be no output power.   

Online IanB

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

The same general theory still applies. The golden equation, which is always true, is this one:

(power applied to vehicle) = (force applied to vehicle) x (speed of vehicle)

You can apply this equation twice. You can apply it first to the wind pushing the vehicle along (power supplied), and you can apply it second to the rolling resistance and drag on the vehicle as it moves (power consumed).

When (power supplied) = (power consumed) then everything is balanced and the vehicle reaches a steady speed. For a simple sail, this steady speed will be lower than the wind speed.

For Blackbird, the spinning prop adds some extra thrust, allowing the (force applied to vehicle) to be positive even when exceeding the wind speed. Therefore, Blackbird can reach a steady speed faster than the wind where it is still true that (power supplied) = (power consumed).

For Blackbird to do this requires careful design of gear ratios on the propeller shaft and pitch angle on the propeller blades.
 

Online bdunham7

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So each time a particle hits that area of the sail it will lose half of the kinetic energy since the other half will be transferred to sail.

Say what?  Is this another gem that 'university professors' don't understand?
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.
 

Online IanB

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Actually a sail on ideal wheels friction less will be 100% efficient in using that available wind power.

Newton's first law of motion says that an object will not change its motion unless a force acts on it. If the sail boat is moving at the same speed as the wind there is no wind force: (w-v) = 0. If there is no friction, there is no friction force. Therefore no force at all. If there is no friction, the boat requires no power at all to continue moving at the same speed, so it neither needs nor gets any power from the wind. The wind power in this situation is zero, so the efficiency must be 0%.
 

Offline electrodacus

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For Blackbird, the spinning prop adds some extra thrust, allowing the (force applied to vehicle) to be positive even when exceeding the wind speed. Therefore, Blackbird can reach a steady speed faster than the wind where it is still true that (power supplied) = (power consumed).


Do you read your own comments ?
"extra thrust" ?  You get that vehicle is only powered by wind ? There is no extra thrust and all the thrust as in the case of a sail is provided by the wind.
Wind power supplied is zero when vehicle speed equal wind speed and even with your wrong equation you will get the same thing as force will be zero.
 

Offline electrodacus

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So each time a particle hits that area of the sail it will lose half of the kinetic energy since the other half will be transferred to sail.

Say what?  Is this another gem that 'university professors' don't understand?

Alex the professor what lost 10K knows this and has never said anything wrong as far as I remember. He just failed to see that there is energy storage involved thus was unable to defend his position when Derek showed him the working treadmill model.

Offline electrodacus

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Newton's first law of motion says that an object will not change its motion unless a force acts on it. If the sail boat is moving at the same speed as the wind there is no wind force: (w-v) = 0. If there is no friction, there is no friction force. Therefore no force at all. If there is no friction, the boat requires no power at all to continue moving at the same speed, so it neither needs nor gets any power from the wind. The wind power in this situation is zero, so the efficiency must be 0%.


Wind power drops as vehicle speed approaches wind speed thus efficiency is still 100% there is just no more wind power available when speeds are equal in an ideal system.

Offline PlainName

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Quote
As for direct down wind the ideal vehicle can get to same speed as wind speed but not above that without having some stored energy or an external energy source other than wind.

You are wrong. I was leading you gently to the realisation (and others are trying the more direct way), but you can't let yourself suffer the possibility that you might be wrong, so you keep diverting to this rubbish.

Quote
Blackbird is not using wind power when above wind speed

And I was on the way to showing you it could. But rather than learn you prefer to stick your head in the sand.
 

Offline electrodacus

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OK lets compare the output of the two equations

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

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

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

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


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