Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147399 times)

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Offline electrodacus

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Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fration ( vehicle speed / wind speed) is.

When you are start moving slowly in to wind direction you need to apply brakes so some small part of the power will be used to increase the vehicle kinetic energy while acceleration and the difference will be burned by the friction brakes (assuming that is what you use to slow down the vehicle).
So if you use no brakes then vehicle will use all that wind power to accelerate the vehicle (increase kinetic energy).
Thus as soon as the vehicle moves the power is either all used to increase vehicle kinetic energy thus super fast get to a large fraction of the wind speed or if you want to maintain that low speed constantly then you need to do something with all that wind power available either convert to heat using friction brakes or generate electricity and store that in some sort of energy storage.

I think this is important to understand so I will try to insist on this point.
Say wind speed is 10m/s and vehicle is 100kg and has a 1m^2 sail driving directly down wind.
When you are stationary no work is done and we all agree with that.
Now say you want the vehicle to get to 1m/s and maintain that speed after it got there.
Kinetic energy for that vehicle at 1m/s will be 0.5 * 100kg * 1^2 = 50Ws
So with 600W available at the start and 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W available when vehicle gets to 1m/s the vehicle need about 100ms to get from 0m/s to 1m/s
And if you want to stop accelerating at that point you will need to setup a generator at the wheel that can take 437.4W and do whatever it wants with them you can heat water or air or store that in a battery but as long as you take out those 437.4W the vehicle can maintain 1m/s and not accelerate or decelerate.
That 437.4W of wind power will be available to vehicle as long as wind matains 10m/s
If you understand this example and explanation you should be able to understand my entire explanation about how blackbird works.

I wanted to answer the other questions but I feel we need to concentrate on this and once this part is understood the rest should be simple.

Online bdunham7

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That 437.4W of wind power will be available to vehicle as long as wind matains 10m/s
If you understand this example and explanation you should be able to understand my entire explanation about how blackbird works.

I wanted to answer the other questions but I feel we need to concentrate on this and once this part is understood the rest should be simple.
 

Try doing your math with the units!
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Offline electrodacus

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Try doing your math with the units!

Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.

Online bdunham7

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Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.

Take the expression " 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W" and put the actual units in everywhere the number refers to units. 

So the 0.5 refers to area of the sail, so 0.5m2, I don't know offhand what the 1.2 or 1 refer to, the (10-1)3 refers to speed, so (9m/s)3 becomes 729m3s-3 and so on, then your result will be some monstrosity of units until you simplify it.  If it doesn't simplify down to a usable unit, then your formula or math are wrong.
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Offline electrodacus

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Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.

Take the expression " 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W" and put the actual units in everywhere the number refers to units. 

So the 0.5 refers to area of the sail, so 0.5m2, I don't know offhand what the 1.2 or 1 refer to, the (10-1)3 refers to speed, so (9m/s)3 becomes 729m3s-3 and so on, then your result will be some monstrosity of units until you simplify it.  If it doesn't simplify down to a usable unit, then your formula or math are wrong.

Sorry OK I see what you mean. Here is the equation with units.

0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W

Online bdunham7

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Sorry OK I see what you mean. Here is the equation with units.

0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W

OK, now if you could just explain what each of the terms is, actually just the 1.2kg/m3 term--I see the 1m2 is area and the others are the speed of the wind and the sail.  Then instead of magically arriving at 'Watts' at the end, do all the operations with the units and see if what you get simplifies to watts or not.
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Offline electrodacus

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Sorry OK I see what you mean. Here is the equation with units.

0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W

OK, now if you could just explain what each of the terms is, actually just the 1.2kg/m3 term--I see the 1m2 is area and the others are the speed of the wind and the sail.  Then instead of magically arriving at 'Watts' at the end, do all the operations with the units and see if what you get simplifies to watts or not.

The 0.5 is a constant has no unit
1.2kg/m^3 is the air density
1m^2 is the equivalent are of the sail
then there is the wind speed and vehicle speed in m/s
I use all international system of unit so I do not need to be as careful as if using all sort of strange unrelated units.
And yes that result is in Watt's

Online bdunham7

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The 0.5 is a constant has no unit
1.2kg/m^3 is the air density
1m^2 is the equivalent are of the sail
then there is the wind speed and vehicle speed in m/s
I use all international system of unit so I do not need to be as careful as if using all sort of strange unrelated units.
And yes that result is in Watt's

Where are you getting that the result is in watts?  And are you saying that the force on a sail is just the mass of the air x its speed x the area of the sail?  Where do you get that?
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Offline Kleinstein

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Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fraction ( vehicle speed / wind speed) is.

When you are start moving slowly in to wind direction you need to apply brakes so some small part of the power will be used to increase the vehicle kinetic energy while acceleration and the difference will be burned by the friction brakes (assuming that is what you use to slow down the vehicle).
So if you use no brakes then vehicle will use all that wind power to accelerate the vehicle (increase kinetic energy).
Thus as soon as the vehicle moves the power is either all used to increase vehicle kinetic energy thus super fast get to a large fraction of the wind speed or if you want to maintain that low speed constantly then you need to do something with all that wind power available either convert to heat using friction brakes or generate electricity and store that in some sort of energy storage.

I think this is important to understand so I will try to insist on this point.
If all the power or only a small farction is actually used makes a big difference. The problem with the assumption all the theoretically possible power would be added to the kinetic energy is that it is rather hard (essentially impossible) to do this.

With a 100 kg vehicle to add the first 0.5 Ws to the kinetic energy it would need a speed of 0.1 m/s.  If the sail would provide a full 500 W to add to the kineatic energy, this would be only 1 ms to reach 0.1 m/s. It is only a low speed, but 0.1 m/s   / 1 ms is still 100 m/s² and thus a bit more than 10 times gravity for the acceleration needed (on averge). Things get even more carzy when looking at lower speed / shorter time.   So there must be an error in the calculation.

The error in this example is in the idea that a "sail" would be 100 % energy efficient in converting to kinetic energy. Energy is still conserved, but most is converted to heat and not to the kinetic energy of the vehicle.

I somehow have the feeling the concept of force is not really understood. Using a power for the drag is an indication.  Drag is a force and not a power.
Except from this error I have not seen a problem with the units. The desire to do the calculations with examples instead of the letters is often found with beginners who try to use intuation instead of math. However even beginners are usually way better in realizing that they may be wrong.
 

Online bdunham7

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Except from this error I have not seen a problem with the units.

Well, sure, it's not the only error, but you have to start somewhere.  This example appears to come out as kg*m2/s3, does that come out to watts somehow? 

A newton is 1 kg*m/s2, so then we have N*m/s, so I'll be damned, it is watts.  Oh well.  Maybe he's right.   :)
« Last Edit: December 16, 2021, 11:50:55 pm by bdunham7 »
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Offline electrodacus

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If all the power or only a small farction is actually used makes a big difference. The problem with the assumption all the theoretically possible power would be added to the kinetic energy is that it is rather hard (essentially impossible) to do this.

With a 100 kg vehicle to add the first 0.5 Ws to the kinetic energy it would need a speed of 0.1 m/s.  If the sail would provide a full 500 W to add to the kineatic energy, this would be only 1 ms to reach 0.1 m/s. It is only a low speed, but 0.1 m/s   / 1 ms is still 100 m/s² and thus a bit more than 10 times gravity for the acceleration needed (on averge). Things get even more carzy when looking at lower speed / shorter time.   So there must be an error in the calculation.

The error in this example is in the idea that a "sail" would be 100 % energy efficient in converting to kinetic energy. Energy is still conserved, but most is converted to heat and not to the kinetic energy of the vehicle.

I somehow have the feeling the concept of force is not really understood. Using a power for the drag is an indication.  Drag is a force and not a power.
Except from this error I have not seen a problem with the units. The desire to do the calculations with examples instead of the letters is often found with beginners who try to use intuation instead of math. However even beginners are usually way better in realizing that they may be wrong.

We are discussing an ideal case so no friction losses.
And yes it will take less than one ms to accelerate to 0.1m/s 0.85ms as there are 600W when you start with 10m/s wind speed and 1m^2 sail ideal case.
But 600W is not infinite or incredible and less than 1ms means there is not much energy at all
You are maybe concerned in real life about the sort of forces that will require but you probably forget that in real life all materials deform hopefully just elastic deformation if vehicle was designed correctly.
Think about just a wheel with an air filled tire and perfect brakes applied so that wheel can not rotate. Due to tire elasticity you can still push the vehicle without needing infinite force
In ideal case this elasticity and thus any energy storage in them is ignored but in real life you can not get rid of them the same way you can not get rid of friction.
So that 0.5Ws (half a joule) is just nothing and will likely be absorbed by some part deformation like mentioned rubber on a wheel.
So from theory you think there are huge forces but in practice those are fairly small since they are dampened by all the elastic deformations in a system. Even steel will have some amount of deformation and yes that require higher forces but still nothing close to infinite or what you will get when you try to do a theoretical ideal case.
That is why power and energy are much more intuitive to use as they will not seem ridiculous like around 600W for less than 1ms or 0.5Ws seems super small.

A sail is the most efficient device you can use if you want wind power converted into kinetic energy and an ideal sail will be 100% efficient that is why this calculation demonstrates that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or a external energy source.

Even a real sail will be fairly close to 100% efficient same way a wheel is super efficient and very close to 100%. A propeller by contrast is super inefficient but the advantage in this particular case is that propeller can increase the pressure differential and thus a way of storing energy.
I feel you are the closest here in understanding this. Hopefully my clarifications on why those large theoretical forces are of no concern in a real system are good enough if not let me know and I will try to add more details.
I'm fairly bad at explaining things.



Online IanB

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Just a quick summary on the calculations and units:

The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)

We have been neglecting the drag coefficient in the discussions so far.

The transmitted power where force and speed are in the same direction is then given by: (power) = (force) x (speed)

The wind pressure is given by: (wind pressure) = (one half) x (air density) x (apparent wind velocity)^2

The drag coefficient varies a lot, but for a flat, perpendicular surface it could be about 1.1 or so.

The apparent wind velocity is the velocity experienced by the sail.

The density of air is about 1.2 kg/m3, so for a 10 m/s wind hitting a 1 m2 flat surface moving downwind at 1 m/2, the wind pressure would be 0.5 x 1.2 x (10-1)^2 x 1.1 ~= 53 N

The transmitted power would then be 53 N x 1 m/s = 53 W.
 
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Offline electrodacus

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Just a quick summary on the calculations and units:

The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)

We have been neglecting the drag coefficient in the discussions so far.

The transmitted power where force and speed are in the same direction is then given by: (power) = (force) x (speed)

The wind pressure is given by: (wind pressure) = (one half) x (air density) x (apparent wind velocity)^2

The drag coefficient varies a lot, but for a flat, perpendicular surface it could be about 1.1 or so.

The apparent wind velocity is the velocity experienced by the sail.

The density of air is about 1.2 kg/m3, so for a 10 m/s wind hitting a 1 m2 flat surface moving downwind at 1 m/2, the wind pressure would be 0.5 x 1.2 x (10-1)^2 x 1.1 ~= 53 N

The transmitted power would then be 53 N x 1 m/s = 53 W.

Yes all seems correct other than for ideal case I will use 1 as drag coefficient
Then you have 0.5 * 1.2 * (10-1)^2 * 1 = 48.6N
Then 48.6N * 9m/s = 437.4W so same exact result I got.
« Last Edit: December 17, 2021, 01:50:25 am by electrodacus »
 

Online IanB

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Then 46.8N * 9m/s = 437.4W so same exact result I got.

But the 46.8 N is not moving at 9 m/s. If you make the sail move at 9 m/s, then the apparent wind velocity is going to be 1 m/s.

In that case: Force on sail = 0.5 * 1.2 * (10-9)^2 * 1 = 0.6 N

Power = 0.6 N * 9 m/s = 5.4 W

You cannot have a high wind pressure on the sail, and also make the sail move at the speed of the wind. That is double accounting, and it does not work.
 

Online bdunham7

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Then 46.8N * 9m/s = 437.4W so same exact result I got.

That would be the energy that is dissipated, the wind energy that is not transmitted to the vehicle.

The remainder, 46.8N * 1m/s, goes to the vehicle.
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Online bdunham7

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Just a quick summary on the calculations and units:

The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)

That omits the laminar drag portion, but I'm guessing perhaps that is negligible in this case?  (I don't actually know)

And that does appear to be the equation, more or less, that he is using.  I didn't recognize it because of the way he was expressing it, but it works out--as I showed above!
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Offline electrodacus

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Then 46.8N * 9m/s = 437.4W so same exact result I got.

But the 46.8 N is not moving at 9 m/s. If you make the sail move at 9 m/s, then the apparent wind velocity is going to be 1 m/s.

In that case: Force on sail = 0.5 * 1.2 * (10-9)^2 * 1 = 0.6 N

Power = 0.6 N * 9 m/s = 5.4 W

You cannot have a high wind pressure on the sail, and also make the sail move at the speed of the wind. That is double accounting, and it does not work.

That will be if you completely ignored the wind. So if there was no wind and vehicle was driving at 1m/s then this will be the power it will need to supply to his motor to counteract drag  0.5 * 1.2 * 1 * 1^3 = 0.6W
0.6W / 1N = 0.6N

Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.
In the other direction if there is no wind and this vehicle with 1m^2 frontal area needs to drive at 9m/s then it needs 437.4W to maintain that speed just to counter the drag.  So that is how much an electric bike will use at this speed if frontal area will be 1m^2 (usually is less than half that but this is just an example with round numbers).
 

Online IanB

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Just a quick summary on the calculations and units:

The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)

That omits the laminar drag portion, but I'm guessing perhaps that is negligible in this case?  (I don't actually know)

And that does appear to be the equation, more or less, that he is using.  I didn't recognize it because of the way he was expressing it, but it works out--as I showed above!

The equation works in general for any shape of object. The area is calculated in some standard way (usually the cross section facing the wind), and the drag coefficient accounts for the shape of the object. Basically, all the complex geometry, skin friction, turbulent effects and everything else is wrapped up in the Cd value. It can vary a lot, from as little as 0.01 to as much as 2.
 

Online IanB

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Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.

If you want to get 437 W from a vehicle moving at 1 m/s then you can use the formula: power = force x velocity

You have:

437 W = force x 1 m/s

So the force on the sail would have to be 437 N.

From the sail equation:

Force = 0.5 x (air density) x (apparent wind velocity)^2 x (area of sail)

437 N = 0.5 x 1.2 kg/m3 x va^2 x 1

va^2 = 437 / (0.5 x 1.2) = 728

va = 27 m/s

This tells that the wind speed would have to be 27 + 1 = 28 m/s.
 

Online bdunham7

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The equation works in general for any shape of object. The area is calculated in some standard way (usually the cross section facing the wind), and the drag coefficient accounts for the shape of the object. Basically, all the complex geometry, skin friction, turbulent effects and everything else is wrapped up in the Cd value. It can vary a lot, from as little as 0.01 to as much as 2.

What I meant is that there is another term in the full equation, the laminar drag.  This is the low-speed drag component dependent on speed, the viscosity of the medium and constant based on the shape and size of the object. the viscosity of the medium. 
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Offline electrodacus

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Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.

If you want to get 437 W from a vehicle moving at 1 m/s then you can use the formula: power = force x velocity

You have:

437 W = force x 1 m/s

So the force on the sail would have to be 437 N.

From the sail equation:

Force = 0.5 x (air density) x (apparent wind velocity)^2 x (area of sail)

437 N = 0.5 x 1.2 kg/m3 x va^2 x 1

va^2 = 437 / (0.5 x 1.2) = 728

va = 27 m/s

This tells that the wind speed would have to be 27 + 1 = 28 m/s.

I feel like you see the air something less relevant than the road. Maybe just because it is invisible and low density.

If you apply no brakes at all then those 437W will all be available to accelerate the vehicle. Sort of seems you agree with that then not sure why if you agree with that you do not agree with the fact that you need to apply 437W of braking in order to maintain speed.
Ideal vehicle needs no power to maintain speed so you need to take all that available wind power and convert it into heat through friction brakes or use a generator to take advantage of that in some other ways.

On the other side you can also look at a vehicle driving at 9m/s with no wind so there is a 9m/s apparent wind and that 9m/s apparent wind requires you to provide 437W in order to be able to maintain vehicle speed at 9m/s
If you had a 9m/s vehicle and a 9m/s wind from the back then 0m/s apparent wind you need 0W so no power to maintain vehicle speed.
I know something seems maybe too good to be true but it is the reality and can be tested.

Online bdunham7

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If you apply no brakes at all then those 437W will all be available to accelerate the vehicle.

No, it won't.  You don't accelerate a vehicle with power, you do it with force.  At 1m/s with a 10m/s wind, you have your 46.8 Newtons available to accelerate the vehicle, which at 1m/s represents 46.8 watts of power.  The remainder of the wind's power will be dissipated.  Of course that number changes as the speed change.  And b/t/w, from an earlier post, 46.8 * 9 = 421.2, not 437.4.  Not a big issue, but it was making something not look right.
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Online IanB

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What I meant is that there is another term in the full equation, the laminar drag.  This is the low-speed drag component dependent on speed, the viscosity of the medium and constant based on the shape and size of the object. the viscosity of the medium.

As the "drag coefficient", I think the Cd value includes that. The Cd value may not exactly be a constant, and may vary with wind speed, but within the region it applies, it is a kind of "catch all" for all the indirect friction effects.

The viscosity of the medium would get included in the Reynolds number, and the Cd may then be correlated against the Reynolds number.
 

Offline electrodacus

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No, it won't.  You don't accelerate a vehicle with power, you do it with force.  At 1m/s with a 10m/s wind, you have your 46.8 Newtons available to accelerate the vehicle, which at 1m/s represents 46.8 watts of power.  The remainder of the wind's power will be dissipated.  Of course that number changes as the speed change.  And b/t/w, from an earlier post, 46.8 * 9 = 421.2, not 437.4.  Not a big issue, but it was making something not look right.

If you look back at earlier example I stated that it takes around 100ms to get to 1m/s and the vehicle kinetic energy at 1m/s is 50Ws
So according to your theory it will take more like one full second for this vehicle to get to 1m/s instead of just 100ms.

Sorry is 48.6N it was just a typo that I will correct. Thanks for noticing.

Offline Brumby

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I think I'm gonna regret this.....

.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or a external energy source.
Taking the above literally - which is what you have been doing all along - and you will never be able to see the full mechanism.  The above is the "obvious" bit - but it's not the full story.

If I were to change your statement to something more correct, it would look like this:
Quote
.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or an additional energy source.

Here is the key - there IS an additional energy source, but it is not as obvious as wind on a sail (which is why "intuition" is a really bad influence - really bad.).  It is, however, fully contained within the system consisting of only the Blackbird, the wind and the ground.  It is not external to that system and it is not an energy storage mechanism.

If you cannot OPENLY consider such a possibility and work through the relevant calculations, you will never understand.

Such closed-mindedness would have kept the study of physics from progressing any further than Newton.
 


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