Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147242 times)

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Offline Kleinstein

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Also if you double the speed drag power increases 8x not 2x or even 4x

It is darg force, not power.


No, they got it exactly right here and we've all been telling you the exact same thing....
OK then why that second formula that should represent the same thing provide a different value ?
The problem is not with the 2nd formular, but with the 1st.
If to claculations don't agree one (at least one) has to be wrong (or using different approximations). If you get one of the formulas from text books and many sources chances are that that's the good one and the other is the bad one.
For the second formular there is a simle derivation:  The drag force is proportional to relative wind speed squared and the mechanical power is force times relative speed (in this case bicycle relative to ground).
 
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Offline gnuarm

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Except my car has no battery and neither does the blackbird.  Also, the blackbird has both a wheel and a propeller while my car ONLY has a wheel. 

It was my assumption that you own a Tesla due to referral program mentioned in your signature.

I'm talking about my imaginary car for this example. 


Quote
Why do you think a propeller that is less efficient than a wheel can help ? The propeller itself has no advantages over wheels other than it is to travel trough air and it can store energy as pressure differential.

The issue is the blackbird has wheels that pick up power from the motion and provide it to the propeller.  This additional energy turns the propeller so it pushes against the wind so the car can run faster than the wind.  There is no need to invoke imaginary storage. 


Ok, now you are going into pointless details because this is not how the blackbird works.

That is exactly how blackbird or any wind powered vehicle works.  This is your fallacy.  Until you can get past this, you will never understand why the blackbird works.


Nothing, literally NOTHING you have talked about here has to do with the blackbird, so just stop the insanity! 

It is clear that you will never understand anything anyone tells you that doesn't agree with your idea because you reject it out of hand without understanding it. 

I keep coming to the conclusion that you can't be taught anything, but I get suckered in when I see a point that is so crystal clear to anyone else that it must be possible to show it to you.  However, as others have pointed out, instead of understanding what they tell you, you simply duck the issue and start talking about something different. 

So no more replying to your nonsense.

The fact that you do not understand how blackbird works will not make all that I mentioned very relevant.
That is about what I think about you but I'm still hopeful.
[/quote]

The blackbird is actually quite simple to understand.  It is when you start imagining energy storage being a part of the matter that you go astray.

I know nothing I'm posting will have any impact on your understanding.  I don't know where this comes from.  Every time someone gives you a clear explanation you seem compelled to invent something new that explains how it doesn't work, in spite of the fact that there are videos showing the blackbird working both downwind and upwind.  The only limitation seems to be the length of the track they run on.  So your storage idea is bogus.  If there was a track that circled the earth they probably could circumnavigate the globe easily... without storing any energy.
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Offline gnuarm

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No, you really don't and that's where you are failing hard.  You do not understand or correctly apply the basic tenets of junior high-school level physics, namely:

1.  The principles of Archimedes
2.  Newton's laws of motion
3.  The principle of conservation of energy--the one you keep bringing up but clearly have no clue as to how to apply it.

OK answer this.

Max wind power available to a direct down wind powered vehicle is when vehicle just starts moving so low speed (Fact 1).
equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)

What powers the vehicle when vehicle speed is above wind speed ?

This has been explained to you many times.  The propeller pushes against the wind, so the car gets energy from the combined speed of the wind and the exhaust of the propeller.   This forward motion is from the combined forces of the wind and the propeller.  The wheels pick up energy from this combined motion and uses it to turn the prop.  Since the prop only has to supply part of the combined energy, the resistance of the wheels is less than the total of the wind and the prop. 
 

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Clearly not wind power (at least not directly but stored wind power).
I'm starting to get very tiered by this level of stupidity.
I asked for the equation showing the available wind power for Blackbird and nobody presented one that will match the observed results in both blackbird and treadmill model.

The observed results being that the blackbird travels faster than the wind or upwind?   Yes, I get tired of the insanity as well.
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Offline gnuarm

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equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)

Except:
a) You cannot derive this equation from first principles when asked to do so,
b) You have not pointed to any authoritative source that gives this equation (online or textbook).

Therefore you cannot take this as fact, and if you rely on it all your arguments are flawed.

It is a fact.  Please provide an equation that you think is correct.
This equation is used everywhere from wind turbine design to vehicle drag so it is one of the most used equations and you can find it literally everywhere.
Your problem seems to be not understanding power and working with forces only not understanding what speed will correspond to get the correct result.

If you claim this equation is universal, why don't you provide a link showing this?  If you can't provide a link, it must not be very universal.
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Offline gnuarm

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OK then why that second formula that should represent the same thing provide a different value ?

Because it doesn't represent the same thing.  You are 'intuiting' (assuming wrongly) what the result should be and then assuming the formula is wrong when it doesn't match  your preconceived notion.  The result of the formula, which is correct,  is that it takes more power to go 20km/h in still air than it does to go 10km/h against a 10km/h headwind.

I don't want to go through the entire history of this.  Why would that be true?  Are they taking into account the other losses like tire drag?  The air drag would only be a matter of bike vs. air, no?
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Offline gnuarm

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This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.

I don't know what you should and shouldn't learn in school, but I do agree that the question is extremely simple.  And you have simply gotten it wrong.  The drag is the same, the power is not.  Drag is a force and is the same in each case.  Thus when I pedal the bicycle, I will have to apply a certain amount of force to the pedals to counter that drag.  If I'm going 20km/h in still air however, I have to pedal twice as fast with the same force as I do pedaling an identical bicycle at 10km/h into a 10km/h wind.  Same force, twice as fast.  Twice the power.  Extremely simple.  No calculus.  Everybody understands it except you.

Ah, just saw this part that emphasized "power" vs. force.  Got it.
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Offline IanB

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I don't want to go through the entire history of this.  Why would that be true?  Are they taking into account the other losses like tire drag?  The air drag would only be a matter of bike vs. air, no?

This is not about rolling resistance or any other extraneous factors.

There is a simple formula for the mechanical power of a moving vehicle:

  power = force ("volts") x speed ("amps")

In SI units you get force [N] x speed [m/s] = power [Nm/s = J/s = W]

The force in this case comes from resistance to movement of the vehicle, such as aerodynamic drag. The speed is how fast the vehicle is moving (no movement, no power).

The aerodynamic drag is proportional to the square of the effective wind velocity as seen by the vehicle. The effective wind velocity is the sum of the vehicle speed and the wind speed.
 

Offline gnuarm

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Thus this sort of wrong understanding is fairly common as it ended up in Wikipedia.

But every physicist in the world, every engineer in the world, every textbook in the world, agrees with that formula in Wikipedia. So are you going to tell the whole world they have been getting it wrong for the past 200 years and re-write all the physics textbooks?

This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.

This is interesting. That formula that appears in Wikipedia is not derived from logic. It is derived from experiment. Many experiments. Over the years, scientists and experimenters measured the amount of power required in different situations, and they found that all of their experiments match the formula given by Wikipedia. What is more, you can repeat those experiments yourself, and if you do so, you will also obtain results matching that formula.

Yeah, but it doesn't take into account the stored energy...  ;-)
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Offline gnuarm

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Also if you double the speed drag power increases 8x not 2x or even 4x

It is darg force, not power.


No, they got it exactly right here and we've all been telling you the exact same thing....
OK then why that second formula that should represent the same thing provide a different value ?
The problem is not with the 2nd formular, but with the 1st.
If to claculations don't agree one (at least one) has to be wrong (or using different approximations). If you get one of the formulas from text books and many sources chances are that that's the good one and the other is the bad one.
For the second formular there is a simle derivation:  The drag force is proportional to relative wind speed squared and the mechanical power is force times relative speed (in this case bicycle relative to ground).

Why is anyone concerned about drag?  I'm being serious.  On the blackbird the speeds are low enough that the drag has to be pretty minimal.  The issue is more one of showing the forces to create acceleration once you reach wind speed.  No?  The forces from the propeller are going to be much greater than wind drag at 15 mph. 
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Offline IanB

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Why is anyone concerned about drag?  I'm being serious.  On the blackbird the speeds are low enough that the drag has to be pretty minimal.  The issue is more one of showing the forces to create acceleration once you reach wind speed.  No?  The forces from the propeller are going to be much greater than wind drag at 15 mph.

You are right to ask this. When examining a wind powered vehicle, it is necessary to consider the balance of forces on the vehicle. The resulting net force will show what direction it moves and how fast it will eventually go. The limiting speed is where drag comes in, later on.
 

Offline electrodacus

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But every physicist in the world, every engineer in the world, every textbook in the world, agrees with that formula in Wikipedia. So are you going to tell the whole world they have been getting it wrong for the past 200 years and re-write all the physics textbooks?
That is your opinion and not a fact. Anyone that designs any sort of vehicle will know that second formula is not correct.
Also I made the mathematical proof that it is not correct since it will not provide the same result as the first equation and they should provide the same result if both where correct equation calculating the same thing.


This is interesting. That formula that appears in Wikipedia is not derived from logic. It is derived from experiment. Many experiments. Over the years, scientists and experimenters measured the amount of power required in different situations, and they found that all of their experiments match the formula given by Wikipedia. What is more, you can repeat those experiments yourself, and if you do so, you will also obtain results matching that formula.

Witch one ?  The first one is correct while the second one is not as I already demonstrated.

Double the wind speed and drag power increases 8x (that is a fact and anyone can test to see that is the case).
Why do you think vehicle traveling trough air is any diff rent in therms of drag than air traveling at the same speed (headwind).
As far as drag surface is concerned it will not be able to differentiate.
Air is a fluid so imagine the drag on a boat traveling on a lake at some speed v and then same boat in a river that flows at speed v (there will be the same drag in both cases).
Air is no different from water as is still a fluid the only difference is lower density and important for blackbird it is a compressible fluid.

Online bdunham7

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Anyone that designs any sort of vehicle will know that second formula is not correct.

Can you cite any actual person that 'designs any kind of vehicle?  Is not the designer of the Blackbird such a person?  Do you think he agrees with you?

Quote
therms of drag

 |O :-DD

Why not 'pints of drag'?  Or perhaps 'acres of drag'.
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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Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics)


They add vehicle speed and wind speed for force part of the equation but not for the power part ?
How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.

This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.

Check this https://www.omnicalculator.com/sports/cycling-wattage  use 50kg cyclist weight 20kg bike weight 1km/h (0.277m/s) bike speed and 230km/h (63.88m/s) headwind.
The Cd*A for tops is 0.408
So 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2 * 0.277 = 284.9  less than they get 300W with no elevation but they add the other less significant parts as riling resistance
So yes you think 300W is enough to deal with 230km/h head wind ? If so you never experienced strong winds (I'm sure nobody here experienced 230km/h).
The most I have ever experienced is 110km/h gusts (not even continues wind speed) and this is 8x less damaging than 230km/h

Correct answer for that is 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2*(63.88 + 0.277) = 66kW
Unless you are a super hero with super human strength you will not going to pedal against a 230km/h wind at any speed not even 1km/h

Offline Kleinstein

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There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.


The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part.  It would still need a good explaination if the 2 parts together are actually the maximum avialable power.

So both formulas can be correct, but are made for different problems.  For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.
 
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Offline gnuarm

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Anyone that designs any sort of vehicle will know that second formula is not correct.

Can you cite any actual person that 'designs any kind of vehicle?  Is not the designer of the Blackbird such a person?  Do you think he agrees with you?

Quote
therms of drag

 |O :-DD

Why not 'pints of drag'?  Or perhaps 'acres of drag'.

You need to get your units correct.  Drag is furlongs squared per fortnight • firkins.  I believe the term for a drag unit is the RuPaul.   ^-^
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Offline electrodacus

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There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.


The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part.  It would still need a good explaination if the 2 parts together are actually the maximum avialable power.

So both formulas can be correct, but are made for different problems.  For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.

We are now talking about direct upwind not direct down wind.
For direct down wind you have w-v as wind power available to vehicle
For direct upwind you have w+v as power needed by the vehicle to overcome drag.

Both a wind turbine and a sail will have available the same wind power but the wind turbine is limited by Betz limit around 59% while a sail has no such limitation as wind speed behind a sail can be zero thus a sail is much more efficient in theory 100%
So while stationary a vehicle with a wind turbine can at most have 59% of that ideal Wind power equation while a sail can have 100% in ideal case.
Of course if vehicle is not moving (brakes applied) then none of that potential power is used by the vehicle.
Thus if you wanted to move that sail vehicle against wind (same as a cyclist is a sail) it requires that potential wind power + some extra to move.
My earlier example is perfectly correct and for a bicycle to move at 1km/h against a 230km/h head wind it will require a minimum of 66kW and gears can not help with anything as a cyclist may be able to provide 300W for minutes and maybe peak around 1kW it will never be able to provide 66kW and so never be able to move against a 230km/h head wind let a lone the possibility that 300W is enough to do that as the wrong formula and that calculator will imply.

Offline IanB

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There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.


The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part.  It would still need a good explaination if the 2 parts together are actually the maximum avialable power.

So both formulas can be correct, but are made for different problems.  For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.

Both a wind turbine and a sail will have available the same wind power but the wind turbine is limited by Betz limit around 59% while a sail has no such limitation as wind speed behind a sail can be zero thus a sail is much more efficient in theory 100%
So while stationary a vehicle with a wind turbine can at most have 59% of that ideal Wind power equation while a sail can have 100% in ideal case.

The direct downwind sail analysis is here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830

Such a sail can be at most 15% efficient.

If a sail is traveling directly downwind at the speed of the wind, then (w-v) is zero, (w-v)^2 is zero, (w-v)^3 is zero.  Any way you look at it, if the sail is moving at the same speed as the wind, then the power will be zero.
 

Offline gnuarm

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Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics)


They add vehicle speed and wind speed for force part of the equation but not for the power part ?

This will be very easy to explain if you apply the correct formula correctly. 

The drag equation is Fd = 1/2 · Cd · A · p · v^2

The v is relative to the medium so Fd = 1/2 · Cd · A · p · (vw + vo)^2

That's the force created by the vehicle moving through the air.

The power to move the vehicle is measured at the point of the wheels in contact with the ground.  There the force is equal to the drag force, so the power is P = Fd · vo

You don't include the wind velocity because the wheels don't see the wind velocity only the drag force. 

Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero.  In that case vo has approximated zero and so the power required to move the car has approximated zero. 


Quote
How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.

This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.

Check this https://www.omnicalculator.com/sports/cycling-wattage  use 50kg cyclist weight 20kg bike weight 1km/h (0.277m/s) bike speed and 230km/h (63.88m/s) headwind.
The Cd*A for tops is 0.408
So 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2 * 0.277 = 284.9  less than they get 300W with no elevation but they add the other less significant parts as riling resistance
So yes you think 300W is enough to deal with 230km/h head wind ? If so you never experienced strong winds (I'm sure nobody here experienced 230km/h).
The most I have ever experienced is 110km/h gusts (not even continues wind speed) and this is 8x less damaging than 230km/h

Correct answer for that is 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2*(63.88 + 0.277) = 66kW
Unless you are a super hero with super human strength you will not going to pedal against a 230km/h wind at any speed not even 1km/h

How much power does that formula say is required for a zero bike speed?   The correct answer is zero.  I think anyone who understand work and power understand there is no work if there is no motion, so no power also.  Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into.  We could make windmills without any movement!  That would be a big plus!!!
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Offline electrodacus

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The direct downwind sail analysis is here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830

Such a sail can be at most 15% efficient.

If a sail is traveling directly downwind at the speed of the wind, then (w-v) is zero, (w-v)^2 is zero, (w-v)^3 is zero.  Any way you look at it, if the sail is moving at the same speed as the wind, then the power will be zero.

No vehicle can move directly upwind powered only by wind without energy storage (Fact).
Yes that is correct a sail can not exceed wind speed directly down wind and the equation is correctly predicting that.


Offline electrodacus

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Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero.  In that case vo has approximated zero and so the power required to move the car has approximated zero. 


How much power does that formula say is required for a zero bike speed?   The correct answer is zero.  I think anyone who understand work and power understand there is no work if there is no motion, so no power also.  Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into.  We could make windmills without any movement!  That would be a big plus!!!

That is not the correct answer. You assume there is brake. Remove the brake and think again what will be the power needed to maintain around zero speed while you have a headwind.
You will need to remove the brakes in order to move so at very low speed you have no brakes to anchor you to the ground.

Online bdunham7

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Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero.  In that case vo has approximated zero and so the power required to move the car has approximated zero. 

I'm afraid reductio ad absurdum isn't going to work here.  You can't convincingly reduce an argument to an absurdity when it already is an absurdity.
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 gnuarm

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Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero.  In that case vo has approximated zero and so the power required to move the car has approximated zero. 


How much power does that formula say is required for a zero bike speed?   The correct answer is zero.  I think anyone who understand work and power understand there is no work if there is no motion, so no power also.  Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into.  We could make windmills without any movement!  That would be a big plus!!!

That is not the correct answer. You assume there is brake. Remove the brake and think again what will be the power needed to maintain around zero speed while you have a headwind.
You will need to remove the brakes in order to move so at very low speed you have no brakes to anchor you to the ground.

I didn't say anything about a brake.  I said zero motion.  Work is force over a distance.  You can push as hard as you want against an object and if it does not move you have done no work on it.  You may have sweated up a storm, but that's internal inefficiencies.  Actually, I can lean into a wall without any real effort on my part.  I'm just using my weight and gravity to provide the force.

I would construct some examples to show you there is no work being done, but it really gets tiresome that you just can't understand the basic concepts. 

So you think that holding an object in a fixed position against a force requires performing work on the object?  Do you understand what is meant by "performing work" in the physics sense?  If the wind is blowing, and something is pushing against the object, but not moving it against the wind, is the something doing work? 
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Offline electrodacus

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    • electrodacus
I didn't say anything about a brake.  I said zero motion.  Work is force over a distance.  You can push as hard as you want against an object and if it does not move you have done no work on it.  You may have sweated up a storm, but that's internal inefficiencies.  Actually, I can lean into a wall without any real effort on my part.  I'm just using my weight and gravity to provide the force.

I would construct some examples to show you there is no work being done, but it really gets tiresome that you just can't understand the basic concepts. 

So you think that holding an object in a fixed position against a force requires performing work on the object?  Do you understand what is meant by "performing work" in the physics sense?  If the wind is blowing, and something is pushing against the object, but not moving it against the wind, is the something doing work?

I remove the brakes from a bicycle you can sit on that and then in a 230km/h head wind and see if you can have a low speed close to zero relative to ground. In fact you can keep the brakes and you will still have no change against 230km/h wind.
A wall is anchored to ground.

Offline Kleinstein

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If a sail would be more efficient than the wind turbine, there would be more sails and less turbines. :-DD

To show than an equation is correct, one can not use that same equation as an argument. That's not how logic works - sorry.

My earlier example is perfectly correct and for a bicycle to move at 1km/h against a 230km/h head wind it will require a minimum of 66kW and gears can not help with anything as a cyclist may be able to provide 300W for minutes and maybe peak around 1kW it will never be able to provide 66kW and so never be able to move against a 230km/h head wind let a lone the possibility that 300W is enough to do that as the wrong formula and that calculator will imply.

I don't hink the high power makes sense: We all know mechanical power is force times speed.

With a constant power at a low speed this would need an awful lot of force - more force the slower you go. With this logic it would be impossible the slowly walk against an even weak head-wind, as it would need near inifite force at a very slow speed.
 

Offline electrodacus

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    • electrodacus

I don't hink the high power makes sense: We all know mechanical power is force times speed.

With a constant power at a low speed this would need an awful lot of force - more force the slower you go. With this logic it would be impossible the slowly walk against an even weak head-wind, as it would need near inifite force at a very slow speed.

What if the bicycle traveled at 231km/h ? Will you then agree that it needed 66kW to do so ?
If you agree with the above why do you think there is a difference between vehicle moving or air moving ?
Yes the force needed is ridiculous that is why in real world it will not happen the bike will just slip and be pushed in the wind direction.

You can slow walk with low head wind speed but not against a 230km/h wind that will not be possible.
Against a weak wind you can.  For a bicycle you can slowly bike against 35 to 40km/h depending on fit you are and how long you need to do that.
Every time the head wind doubles in speed you need 8x more power so 80km/h (8x more power than at 40km/h) is already impossible even for best most fit cyclist.


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