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

[electrodacus]

1. Both the Zeppelin and the Blackbird are "vehicles" with the same propeller design.
2. Both the Zeppelin and the Blackbird are powered by the energy of the tail wind only.
3. The Blackbird can go faster than the speed of the tail wind (based on the experiments we have seen).
4. Based on the above the Zeppelin should fly faster than the speed of the tail wind..
 

1. Yes you can add a useless propeller to Zeppelin.
2. No. The blackbird is powered by stored energy also.
3. Yes. But just for a limited amount of time not indefinitely as claimed.
4. No. Since Zeppelin has no wheels touching the ground it can not power the propeller.

[PlainName]

like that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent

You keep saying that but don't say what the difference (an apparently massive difference) is. So... I don't have a treadmill and I can't believe that your magic only works with a treadmill. What other means could be used to prove, or not, your intuition? I think you will need to be quite explicit so no-one can be accused of not doing it quite right. If things need to be a certain way up you need to say that up front.

[electrodacus]

like that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent

You keep saying that but don't say what the difference (an apparently massive difference) is. So... I don't have a treadmill and I can't believe that your magic only works with a treadmill. What other means could be used to prove, or not, your intuition? I think you will need to be quite explicit so no-one can be accused of not doing it quite right. If things need to be a certain way up you need to say that up front.

I think my diagram shows clearly a treadmill and not a flat dragged surface.
Check post #351 and I explained why it is very different yesterday.
The intermediary case b) should make it clear why there is a difference between a) and c)
Treadmill is not moving relative to ground the pushed treadmill is clearly moving relative to ground.

[IanB]

Treadmill is not moving relative to ground the pushed treadmill is clearly moving relative to ground.

In both cases the top surface of the treadmill is moving relative to ground.

[IanB]

But yes you can make changes to my diagram like that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent.

a treadmill that is not moving relative to the ground
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
a powered off treadmill moved relative to the ground     
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.

If you cannot change any of the incorrect green words then the two things are equivalent and you should give up.

[bdunham7]

Check post #351 and I explained why it is very different yesterday.
The intermediary case b) should make it clear why there is a difference between a) and c)
Treadmill is not moving relative to ground the pushed treadmill is clearly moving relative to ground.

You didn't explain anything.  You just asserted that it was true and didn't even wave your hands much.

In each case, assuming no slippage, the surface of the treadmill will provide force tangential to and at the circumference of the wheel sufficient to rotate the wheel at 4m/s.  The amount of that force depends on the amount of reaction force that is provided by loading the generator, friction, etc.  If it there is no load or friction, the force will be zero--in all 3 cases.  If there is a load, the force will be whatever force is required to rotate the wheel and that force will be transmitted to the axle and thus the body of the car.  Those two forces, combined with the two resultant reaction forces will also result in torque on the body of the car, which could be a trap for any would-be demonstrator!  But again, this is all the same in all 3 cases.  As I said in my earlier car, all of these interactions are mediated by forces.  So if you think there is a difference between the three cases, show us what additional force is involved.

[electrodacus]

In both cases the top surface of the treadmill is moving relative to ground.

Seems like a circular discussion.
Maybe I can find some different way to explain thing excluding this part for now.

Say we go back to propeller based vehicle.
Say vehicle speed is 0m/s and wind speed is 0m/s
And say vehicle is 1kg
And you have 10Ws of energy that you can chose to use powering the wheel 90% efficiency or propeller 70% efficiency.
What will you chose to get the vehicle to highest possible speed using the limited available energy of 10Ws ? 

[Alex Eisenhut]

You are educated stupid. You are born singular. You need cubic mind. Support Wind Cube or be cursed.
 
Wind Cube only offends the educated stupid.

[IanB]

Say we go back to propeller based vehicle.

No, let's not do that. If you keep trying to divert the question and changing the subject, then we will know you are a troll.

[electrodacus]

I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.

Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%

So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s

Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.

[bdunham7]

I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.

Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%

So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s

Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.

OK, that is easy enough.  Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car.  I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W.  Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather 2.422.24m/s, the airspeed that the propeller sees.  So the force generated will be (3.816W * 0.7 efficiency / 2.422.24m/s) = 1.1041.1925N.  The net force (propeller force minus wheel force) will then be 0.1925N and the vehicle will accelerate at 0.1925m/s².  Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed.  In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) = 2.028 2.385W of additional power for a total of 4.6992 5.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W.  And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.

But I think I've made an additional error, so wait a bit.

Edit:  Should be all good now.  If not, someone point out the errors.

[electrodacus]

I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.

Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%

So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s

Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.

OK, that is easy enough.  Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car.  I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W.  Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather 2.422.24m/s, the airspeed that the propeller sees.  So the force generated will be (3.816W * 0.7 efficiency / 2.422.24m/s) = 1.1041.1925N.  The net force (propeller force minus wheel force) will then be 0.11925N and the vehicle will accelerate at 0.11925m/s².  Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed.  In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) = 2.028 2.385W of additional power for a total of 4.6992 5.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W.  And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.

But I think I've made an additional error, so wait a bit.

Edit:  Should be all good now.  If not, someone point out the errors.

Good effort but you forgot something super important. What is the kinetic energy of the vehicle before and after you did the break ?
Say you maintain that 3.816W break for one second now the vehicle kinetic energy is reduce by 3.816Ws
Now you apply 3.816W * 0.7 = 2.671W for one second and you can see you end up with lower kinetic energy than you started with.
And is irrelevant if you do that for 1 full second or a 1ms or even 1us the end result is that vehicle speed will be reduced.
The wind can not provide anything unless you refer to stored energy. As long as wind speed is lower than vehicle speed there is nothing to gain from that.

Forces have no role here since all you care is vehicle kinetic energy change. If you end up with lower kinetic energy then vehicle has slowed down.
Do you see a difference in kinetic energy gain between a vehicle traveling at 2.24m/s with no wind and one traveling at 4.24m/s and 2m/s wind when power is applied to the propeller ? 

[bdunham7]

Good effort but you forgot something super important.

You're right, I forgot that you're a stone headed cretin that clearly can't comprehend any of this.  I give up.  And I don't give up easily.

Forces have no role here

OK, in a universe where forces have no role in basic mechanics, then I have nothing to offer.  My entire education was a complete waste.

[electrodacus]

OK, in a universe where forces have no role in basic mechanics, then I have nothing to offer.  My entire education was a complete waste.

To be honest education of most people seems to have been a waste of time.  Only knowing the formula and not knowing how to apply or what result you should expect is sort of pointless.
If you think you are frustrated try to be in my place where almost nobody understand how basic physics works including people teaching others.

If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power. Breaking power and propulsion power. As long as breaking power is more than propulsion power vehicle will slow down unless there is something else that can offer propulsion.
That something else could be wind but only if wind speed is higher than vehicle speed when direction is exactly the same or as it is in this particular case earlier stored energy.

What if wind speed was 6.24m/s so higher wind speed than vehicle speed ? Based on how you applied the formula this will be a fairly bad thing.

[Kleinstein]

Blackbird or the Zeppelin can not use any energy from the tail wind when vehicle speed is above that tail wind speed.

The red part is repeated over and over again as an argument, but it is plain wrong. Not so obvious, but still wrong.

The red part is not wrong and there is no trick to make that right as if it was to be right it will violate the conservation of energy.
You always mention force when force has nothing to do with conservation of energy.
You only care about power and since propeller is powered by the wheel generator the power you take from the wheel is the most in ideal case you put in the propeller. In real world propeller output may be just 70% of what is available at the wheel so without stored energy the vehicle could never exceed wind speed.
Because of this misconception you also think that the wheel only vehicle in my diagram can move from left to right when that is not possible and you will never be able to prove something impossible.

You missed / cut out the more important point: for the dicussion the red sentence in question is not an agument, but part in question.
This has nothing to do with physics but plain logic. Using the claim you want to prove is not working.

Forces are importand to calculate power and thus to get the balance of power right.
The wind / treadmill are power sources and you have to include there power in the energy balance.
The power available to the (generator-) wheel is not just coming from the powered wheel / prop, but also from the treadmill / wind.

The proof that it can work is allready there in the tread several times:

The result may be confusing, but the math is simple:
The power needed by the prop to generate a fixed amount of thrust is independent of the wind speed.
The power the generator can produce is force times speed. The speed is wind-speed plus a little from the movement in the air.

Plot the two in a graph (X= speed relative to ground, Y = power)  and the result is clear:
Given enough wind speed the generator can produce more power than the prop needs to pull it.

The power from the wind is there as the wind speed contribution to the speed of the vehicle. For a start you can even neglect the power of the prop going back to the wheels.

[iMo]

I've been still trying to decouple the concept of the propeller and the wheels here.
The Zeppelin has no contact with ground.
Let us imagine an Iceboat - the same construction as the Blackbird (the same body and propeller construction), but running on 3 runners made of steel with sharpened edges, the blades having pretty low forward resistance.
Will the Iceboat travel faster than the tail wind speed?

[Kleinstein]

To travel faster than the wind (in the same direction) you still need grip to the ground. To generate power you need to transfer force to the ground, so simple low friction sliding does not work.
The power harmessed from the wind is force transferred to the ground times wind speed minus losses from the prop from generating vortices and similar distortions in the wind field. So with no force to the ground there is no power to start with.

With an ice sail going sideways, the runners can transfer force (sideways) to the ice easy and thus plenty of power.

[Labrat101]

100 years ago we all thought that the Plus & Minus on a battery that electricity flowed . From Plus to Minus 
Every one was happy with this ..  But it was found to be WRONG it goes from Minus to Plus ..    
Most people now except this as this is what you are told in school / Uni or what ever.
weather right or wrong the world maybe Flat . Depends on your perspective thinking . What you see is not always what is .

 Mother Nature does not have to play by the rules of mea Humans . So what when you see something that you Don't understand your Brain
will try to fit some thing in to stop it blowing a fuse ..
A computer if it does not understand will return ERROR .  No Data . because the Human did not enter it .

Somethings are as is . Appling maths formulators to do quick fix may solve your problem but not be Right solution
Murphy's Law is always true .. Humans will Always Get it round the wrong way.

[cbutlera]

I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.

Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%

So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s

Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.

OK, that is easy enough.  Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car.  I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W.  Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather 2.422.24m/s, the airspeed that the propeller sees.  So the force generated will be (3.816W * 0.7 efficiency / 2.422.24m/s) = 1.1041.1925N.  The net force (propeller force minus wheel force) will then be 0.11925N and the vehicle will accelerate at 0.11925m/s².  Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed.  In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) = 2.028 2.385W of additional power for a total of 4.6992 5.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W.  And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.

But I think I've made an additional error, so wait a bit.

Edit:  Should be all good now.  If not, someone point out the errors.

Good effort but you forgot something super important. What is the kinetic energy of the vehicle before and after you did the break ?
Say you maintain that 3.816W break for one second now the vehicle kinetic energy is reduce by 3.816Ws
Now you apply 3.816W * 0.7 = 2.671W for one second and you can see you end up with lower kinetic energy than you started with.
And is irrelevant if you do that for 1 full second or a 1ms or even 1us the end result is that vehicle speed will be reduced.
The wind can not provide anything unless you refer to stored energy. As long as wind speed is lower than vehicle speed there is nothing to gain from that.

Forces have no role here since all you care is vehicle kinetic energy change. If you end up with lower kinetic energy then vehicle has slowed down.
Do you see a difference in kinetic energy gain between a vehicle traveling at 2.24m/s with no wind and one traveling at 4.24m/s and 2m/s wind when power is applied to the propeller ?

No, it is you who have forgotten something super important - Newton's third law.  Something that I thought that you had finally grasped from my thought experiment with me walking in an aeroplane.  Remember, the energy that I expended pushing against the aeroplane floor was 35 joules.  The aeroplane floor was pushing right back against me (Newton's third law), while moving at 200 m/s.  Which is how it contributed the additional 14,000 joules to my total gain in kinetic energy of 14,035 joules.

In my thought experiment, that 14,000 joule contribution was so great, compared to the 35 joules that my muscles had contributed, that you were unable to pretend that it didn't exist.  Clearly my muscles couldn't have contributed the entire 14,035 joules of my gain in kinetic energy.

Now that much lower speeds are being discussed, you have gone right back to pretending that the 2.385W contribution to the total power from the wind in bdunham7's calculation doesn't exist.  As bdunham7 has shown, once you include that additional power, there is a net gain in kinetic energy, not a net loss as you appear to stubbornly believe.

[Alex Eisenhut]

If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power. Breaking power and propulsion power. As long as breaking power is more than propulsion power vehicle will slow down unless there is something else that can offer propulsion.

Teaching you the difference between breaking and braking is also kind of pointless, I guess. 

But since you like special cases, here's a breaker that brakes, or a brake that breaks.

[Alex Eisenhut]

Question: do you agree with the three red arrows I added to your diagram?

[bdunham7]

To be honest education of most people seems to have been a waste of time.  Only knowing the formula and not knowing how to apply or what result you should expect is sort of pointless.

In your case this is apparent.  Education does not cure stupidity.  What is this 'formula'  you think I applied?

If you think you are frustrated try to be in my place where almost nobody understand how basic physics works including people teaching others.

    I'm sure that they don't "understand" it the way you do.  You seem to have adopted 'conservation of energy' as a religion of some sort without comprehending what it means nor understanding where or how to apply it.

If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power.

Ok, so imagine your 1kg vehicle is travelling at 4.24m/s and you drive it into an infinitely strong and rigid concrete wall.  Calculate the power that the wall provides to stop the vehicle.

What if wind speed was 6.24m/s so higher wind speed than vehicle speed ? Based on how you applied the formula this will be a fairly bad thing.

A bad thing?  So plug in the numbers to whatever formula you think I used and post the numbers.  How bad can they be? 

[electrodacus]

100 years ago we all thought that the Plus & Minus on a battery that electricity flowed . From Plus to Minus 
Every one was happy with this ..  But it was found to be WRONG it goes from Minus to Plus ..    
Most people now except this as this is what you are told in school / Uni or what ever.
weather right or wrong the world maybe Flat . Depends on your perspective thinking . What you see is not always what is .

 Mother Nature does not have to play by the rules of mea Humans . So what when you see something that you Don't understand your Brain
will try to fit some thing in to stop it blowing a fuse ..
A computer if it does not understand will return ERROR .  No Data . because the Human did not enter it .

Somethings are as is . Appling maths formulators to do quick fix may solve your problem but not be Right solution
Murphy's Law is always true .. Humans will Always Get it round the wrong way.

We did not know how the electricity flows as we did not understood what it is and we just chose as a convention a direction. We still use the same convention and that will not affect in any way out ability to predict what happens that is why we still use the old convention.

We also had the wrong impression that earth is flat but we changed that when we found out that is not the case as that makes a big difference unlike the convention for electric current flow.

What you are trying to imply here is wrong. Enough people including me know why this vehicle exceeds wind speed (energy storage) so there is no magic for me. You also do not think there is magic as you apply the formula's wrong and think you have a rational answer but that gets you to wrong predictions like vehicle will never slow down as long as there is wind and that is not the case.

[Alex Eisenhut]

Question: Do you agree with the red arrows I added to your diagram?

[electrodacus]

    I'm sure that they don't "understand" it the way you do.  You seem to have adopted 'conservation of energy' as a religion of some sort without comprehending what it means nor understanding where or how to apply it.

I understand perfectly how to apply it. If you have 10W for propulsion and your propulsion is 70% efficient you will get 7W worth of thrust no matter if your vehicle drives at 2m/s or 4m/s or any other speed if you apply this 7W worth of thrust for one second the vehicle kinetic energy will increase by 7Ws (7 Joules if you prefer).
And if you apply a break of 10W for one second then your vehicle will have lost 10Ws worth of kinetic energy no matter if speed was 2m/s or 10m/s
So is clear that braking (that to English police) to generate power and then use that for accelerating the vehicle will result in a reduced kinetic energy and that will mean reduced speed as mass of the vehicle has no reason to change.

Ok, so imagine your 1kg vehicle is travelling at 4.24m/s and you drive it into an infinitely strong and rigid concrete wall.  Calculate the power that the wall provides to stop the vehicle.

Wrong way of asking the question as there is not enough data to be able to answer that. The question that can easily be answered with the amount of data available is calculate the energy the vehicle acted against the wall and that is fairly simple since we know the vehicle speed and mass we can calculate the kinetic energy of the vehicle 9Ws and so since before hitting the wall vehicle had that and after the wall that will be zero (assuming non elastic collision) then it will be 9Ws.

A bad thing?  So plug in the numbers to whatever formula you think I used and post the numbers.  How bad can they be? 

Your formula is not wrong it the way you interpret the result that is wrong.  You say since there is higher force it means vehicle will accelerate in a certain direction when that is wrong as the force you refer to is force acting against the propeller or against the wheel not against the body of the vehicle.
Force at the wheel is not the same with force on the body when the wheels are acting against two isolated mediums.

The result and conclusion of your force and speed formula needs to be exactly the same as my results using Power and energy.
Is clear to see that using power instead of speed and force has lower chances of making mistakes. As my conclusions are correct and yours (all of you) are wrong.