Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147449 times)

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Offline electrodacus

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Not sure how you all managed to change the subject.

The question is what is the available wind power to a vehicle driving directly downwind.
This question is to prove that no wind power vehicle can exceed wind speed without either external energy source or an energy storage device.

Correct answer is this below but if you disagree please feel free to offer the correct equation.

0.5 * air density * area * (wind speed - vehicle speed)^3


Online bdunham7

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Correct answer is this below but if you disagree please feel free to offer the correct equation.

0.5 * air density * area * (wind speed - vehicle speed)^3

0.5 * air density * area * (wind speed - vehicle speed)^3

The equation for total power available does not have vehicle speed in it.  :)
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Offline electrodacus

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Correct answer is this below but if you disagree please feel free to offer the correct equation.

0.5 * air density * area * (wind speed - vehicle speed)^3

0.5 * air density * area * (wind speed - vehicle speed)^3

The equation for total power available does not have vehicle speed in it.  :)

The equation is correct as I stated for a direct down wind vehicle.
Wind speed relative to a directly downwind vehicle is wind speed - vehicle speed.

Online bdunham7

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The equation is correct as I stated for a direct down wind vehicle.

Nope.

Quote
Wind speed relative to a directly downwind vehicle is wind speed - vehicle speed.

True but irrelevant.  The power theoretically available to the vehicle, if you want to try and use that as the basis for some calculations, depends on the wind speed relative to the ground.
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Offline electrodacus

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True but irrelevant.  The power theoretically available to the vehicle, if you want to try and use that as the basis for some calculations, depends on the wind speed relative to the ground.


Think about this way.

You have a 100% efficient wind turbine so output will be 0.5 * air density * swept area * w^3
Now you put that wind turbine on top of a vehicle and drive directly downwind at say at a quarter wind speed
Then your wind turbine will output this 0.5 * air density * swept area * (w -(w/4))^3
Anyone that has ever calculate the wind power available to a sail vehicle going directly downwind will know the equation
0.5 * air density * area * (wind speed - vehicle speed)^3
That is the reason a sail vehicle can never exceed wind speed directly downwind.

Online bdunham7

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You have a 100% efficient wind turbine so output will be 0.5 * air density * swept area * w^3
Now you put that wind turbine on top of a vehicle and drive directly downwind at say at a quarter wind speed
Then your wind turbine will output this 0.5 * air density * swept area * (w -(w/4))^3
Anyone that has ever calculate the wind power available to a sail vehicle going directly downwind will know the equation
0.5 * air density * area * (wind speed - vehicle speed)^3
That is the reason a sail vehicle can never exceed wind speed directly downwind.

The Blackbird is neither a sail vehicle nor a vehicle with a wind turbine on it.
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Offline fourfathom

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I was wrong -- a fast sailboat *can* beat a balloon directly downwind. 

Here are the polars for one of the 2013 America's Cup races (you may recall that these were extremely fast catamarans):


Look at the polar diagram for 20 kts true windspeed at a true wind angle of 135 degrees (TWS and TWA are ground-referenced values shown here, not the boat-referenced AWS and AWA).  You will see that with a 20 kt wind the boat can sail 45 kts at 135 degrees (180 degrees is directly downwind).  Jibing +/- 45 degrees, to beat the balloon the boat would only have to exceed (20  / 0.707) or 28.28 kts.  I haven't tried to figure out the optimum downwind jibing angle, but at 45 degrees off DDW there is plenty of margin.  By the way, these conditions would give you (on each 180 +/- 45 degree jibe) an AWS of 33.9 kts, AWA 34 degrees.

So this is a jibing boat, similar in many ways to a spinning propeller.
« Last Edit: December 13, 2021, 03:37:37 am by fourfathom »
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Online IanB

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A propeller can create a pressure differential in air and that is where energy is stored.
See below graph   https://en.wikipedia.org/wiki/Axial_fan_design


You misunderstand that graph. It does not show what you think it shows. In fluid flow theory, there is static pressure and total pressure. Total pressure is the sum of the static pressure and the velocity component of the fluid. The total pressure may be higher downstream of the fan, but it is pointing in a direction away from the fan, so there is no way it can act as a store of energy. The static pressure points in all directions, and could potentially be a store of energy behind the fan. Except the static pressure is lower than the surroundings, and therefore it is more like a vacuum dragging the fan backwards. To store energy in a compressible fluid you need to  contain it within walls, as in a storage tank. There is no tank here, there are no walls, there is no containment, therefore no store of pressure energy. Once the air leaves the fan, its energy quickly dissipates in all directions away from the fan.

I showed you a video where someone measured the pressure in the air stream leaving a fan to verify that it is lower than the surroundings. Unless you can show how that video was somehow faked or manipulated, you have to accept experimental evidence as fact.
 

Offline electrodacus

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A propeller can create a pressure differential in air and that is where energy is stored.
See below graph   https://en.wikipedia.org/wiki/Axial_fan_design


You misunderstand that graph. It does not show what you think it shows. In fluid flow theory, there is static pressure and total pressure. Total pressure is the sum of the static pressure and the velocity component of the fluid. The total pressure may be higher downstream of the fan, but it is pointing in a direction away from the fan, so there is no way it can act as a store of energy. The static pressure points in all directions, and could potentially be a store of energy behind the fan. Except the static pressure is lower than the surroundings, and therefore it is more like a vacuum dragging the fan backwards. To store energy in a compressible fluid you need to  contain it within walls, as in a storage tank. There is no tank here, there are no walls, there is no containment, therefore no store of pressure energy. Once the air leaves the fan, its energy quickly dissipates in all directions away from the fan.

I showed you a video where someone measured the pressure in the air stream leaving a fan to verify that it is lower than the surroundings. Unless you can show how that video was somehow faked or manipulated, you have to accept experimental evidence as fact.

Are you able to read a graph? or are you saying that the graph is incorrect ?

Is P2 higher than ambient pressure PA and much higher than P1 ? Thus a pressure differential potential energy.

Online IanB

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Are you able to read a graph? or are you saying that the graph is incorrect ?

Is P2 higher than ambient pressure PA and much higher than P1 ? Thus a pressure differential potential energy.

Yes, the graph is incorrect. For a typical fan in the open air, P2 < PA, as you can verify yourself by experiment.

When you have two pieces of evidence, being (A) a pretty picture someone posted on the internet, and (B) the results of a physical experiment you can perform yourself, you have to take (B) every time. Reality is always more reliable than a picture.

Please do the experiment, so you can satisfy yourself that the picture is wrong.
 

Offline PlainName

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Quote
If it has stored energy say that 0.07mWh I calculated

Stored where?
 

Offline Kleinstein

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You have a 100% efficient wind turbine so output will be 0.5 * air density * swept area * w^3
Now you put that wind turbine on top of a vehicle and drive directly downwind at say at a quarter wind speed
Then your wind turbine will output this 0.5 * air density * swept area * (w -(w/4))^3
Anyone that has ever calculate the wind power available to a sail vehicle going directly downwind will know the equation
0.5 * air density * area * (wind speed - vehicle speed)^3
That is the reason a sail vehicle can never exceed wind speed directly downwind.
This calculation neglegts the force on the turbine. This gives an additional energy as drag force times vehichle speed.
Even for just a passive sail the energy is vehichle speed times force and thus
0.5 * air density * area * (wind speed - vehicle speed)^2 * vehicle speed

The passive sail is however not the best that can be done. So the theoretical available energy can be larger than that. The blackbird vehicle showed one way to do that. A sail boat going at a suitable angle is another way - though it complicates the area part. Anyway it still gets power when the speed component parallel to the wind is faster than the wind speed so there is no inherent zero at that speed. Yes it is sideways, but for a though expoeriment one can consider the sideways area as still part of the vehicle size, so the movement would be only inside (like the fan going in circles).
One could consider the fan blades as sails going zig-zag.  So this kind of circumvents the going straight in the direction of the wind.
 
 

Offline electrodacus

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Quote
If it has stored energy say that 0.07mWh I calculated

Stored where?

Pressure differential created by the propeller.

Online bdunham7

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I was wrong -- a fast sailboat *can* beat a balloon directly downwind. 

Here are the polars for one of the 2013 America's Cup races (you may recall that these were extremely fast catamarans):


Look at the polar diagram for 20 kts true windspeed at a true wind angle of 135 degrees (TWS and TWA are ground-referenced values shown here, not the boat-referenced AWS and AWA).  You will see that with a 20 kt wind the boat can sail 45 kts at 135 degrees (180 degrees is directly downwind).  Jibing +/- 45 degrees, to beat the balloon the boat would only have to exceed (20  / 0.707) or 28.28 kts.  I haven't tried to figure out the optimum downwind jibing angle, but at 45 degrees off DDW there is plenty of margin.  By the way, these conditions would give you (on each 180 +/- 45 degree jibe) an AWS of 33.9 kts, AWA 34 degrees.

So this is a jibing boat, similar in many ways to a spinning propeller.

I knew iceboats can do that, and I suppose the wheeled equivalents like a the Blackbird chassis, but to get near ideal numbers like that those catamarans must have hulls coated with buttered teflon.
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Offline electrodacus

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This calculation neglegts the force on the turbine. This gives an additional energy as drag force times vehichle speed.
Even for just a passive sail the energy is vehichle speed times force and thus
0.5 * air density * area * (wind speed - vehicle speed)^2 * vehicle speed

The passive sail is however not the best that can be done. So the theoretical available energy can be larger than that. The blackbird vehicle showed one way to do that. A sail boat going at a suitable angle is another way - though it complicates the area part. Anyway it still gets power when the speed component parallel to the wind is faster than the wind speed so there is no inherent zero at that speed. Yes it is sideways, but for a though expoeriment one can consider the sideways area as still part of the vehicle size, so the movement would be only inside (like the fan going in circles).
One could consider the fan blades as sails going zig-zag.  So this kind of circumvents the going straight in the direction of the wind.

The turbine is supplied by wind power so you can not add that again.
Everything on a wind only powered vehicle is supplied by the wind power.
You can search research papers or similar and see that any direct down wind sail vehicle powered only by the wind will use this wind power formula
0.5 * air density * area * (wind speed - vehicle speed)^3

For a wind turbine the same formula is used except you do not need that vehicle speed subtracted as the wind turbine is fixed to the ground and this is the ideal wind power so to this you will add turbine and generator efficiency.
I designed my own turbine so I worked with this sort of formulas before.

The passive sail is actually the most efficient and an ideal one will be 100% efficient.
That is why no vehicle exceeding wind speed directly downwind (special case) powered only by the wind can not exceed wind speed.
The way blackbird is doing that is by storing energy in pressure differential while it is below wind speed and then use that stored energy to be able to accelerate well past wind speed.

Sail boat going at an angle is a geometric problem as the sail will still have access to wind speed since it will not be driving faster than wind relative to wind direction. Best case is driving perpendicular to wind direction as then you have always access to same wind power as vehicle is not moving at all in the wind direction.
And while driving at an angle the vehicle kinetic energy is what will store energy and then in an ideal case no friction you can just change direction to direct down wind and maintain that higher wind speed indefinitely (no losses) in real vehicle they will slow down same as blackbird will also slow down.
Since kinetic energy is directly related to vehicle speed as soon as a real sail vehicle changes direction to direct down wind it will start to slow down but on the blackbird energy is stored in a separate energy storage device the pressure differential.
If you have some electrical knowledge the equivalent of kinetic energy is a capacitor and the analog to pressure differential is an inductor.
 

Offline Kleinstein

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The turbine is supplied by wind power so you can not add that again.
Everything on a wind only powered vehicle is supplied by the wind power.
You can search research papers or similar and see that any direct down wind sail vehicle powered only by the wind will use this wind power formula
0.5 * air density * area * (wind speed - vehicle speed)^3

The formular would be OK for the power a turbine can generate on the moving vehincle, but this is not relevant here.

The energy added to a vehicle with sail (usually in the form of kinetic energy or energy available to generate from a genrator at it's wheels) it is the formula I gave. The power is simply force times speed.  Wind force is proportional to the square of relative velocity and the speed is the speed of the vehicle. You need the vehicle speed to get zero power from the sail when not moving at all, which is the trivial case: no speed means no power from a sail. You don't want the house to gain kinetic energy from the wind !

The passive sail is actually the most efficient and an ideal one will be 100% efficient.
No : the sail is not most efficient way to harness the wind, at least not in all cases. It is obviously not at low speed, aspecially 0 speed, as it is 0 efficiency there. It is also not the most efficient way when the vehicle speed approaches the wind speed, as there it is zero efficiency too. The vehicle in question here is shown to be more efficient, as it can still get power from the wind, even at the speed of the wind and a little faster. You don't have to believe it up front, but you also can't exclude it up front from intuition. Doing this is trying a circular argument so it is not a valid argument.

Sail boat going at an angle is a geometric problem as the sail will still have access to wind speed since it will not be driving faster than wind relative to wind direction.
That claim is not true: - by going zig zag with the wind a sail boat and especially an ice sail can go faster than a ballon, just not a straight line. Just look at the graph in the answer before yours. They get some 1.6 times the speed of the wind / ballon when doing a zig zag with a 45 degree angle and even more with a smaller angle like 20 degree.
 

Offline electrodacus

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The formular would be OK for the power a turbine can generate on the moving vehincle, but this is not relevant here.

The energy added to a vehicle with sail (usually in the form of kinetic energy or energy available to generate from a genrator at it's wheels) it is the formula I gave. The power is simply force times speed.  Wind force is proportional to the square of relative velocity and the speed is the speed of the vehicle. You need the vehicle speed to get zero power from the sail when not moving at all, which is the trivial case: no speed means no power from a sail. You don't want the house to gain kinetic energy from the wind !


OK let me try again this time with an example that should give you a better intuition of what happens.

1m^2 swept area wind turbine with 40% efficiency in a 10m/s wind speed   vs   1m^2 sail

If thery speed relative to the ground is zero then of course sail will do nothing and there will just be a static force as on any wall where the wind turbine output power will be 0.5 * 1.2 * 1 * 10^3 * 0.4 = 240W

So stationary wind turbine wins 240W vs 0W

Now install the wind turbine and the sail on a vehicle driving directly downwind at 0.1m/s, 5m/s and 9m/s

0.1m/s
wind turbine  0.5 * 1.2 * 1 * (10-0.1)^3 * 0.4 = 232.87W
sail (can install an electric generator at the wheel say that is 90% efficient)   0.5 * 1.2 * 1 * (10-0.1)^3 * 0.9 = 523.96W

5m/s
wind turbine 0.5 * 1.2 * 1 * (10-5)^3 * 0.4 = 30W
sail               0.5 * 1.2 * 1 * (10-5)^3 * 0.9 = 67.5W

9m/s
Wind turbine 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.24W
sail                0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.54W

 

Online IanB

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1m^2 swept area wind turbine with 40% efficiency in a 10m/s wind speed   vs   1m^2 sail

If thery speed relative to the ground is zero then of course sail will do nothing and there will just be a static force as on any wall where the wind turbine output power will be 0.5 * 1.2 * 1 * 10^3 * 0.4 = 240W

So stationary wind turbine wins 240W vs 0W

Now install the wind turbine and the sail on a vehicle driving directly downwind at 0.1m/s, 5m/s and 9m/s

0.1m/s
wind turbine  0.5 * 1.2 * 1 * (10-0.1)^3 * 0.4 = 232.87W
sail (can install an electric generator at the wheel say that is 90% efficient)   0.5 * 1.2 * 1 * (10-0.1)^3 * 0.9 = 523.96W

5m/s
wind turbine 0.5 * 1.2 * 1 * (10-5)^3 * 0.4 = 30W
sail               0.5 * 1.2 * 1 * (10-5)^3 * 0.9 = 67.5W

9m/s
Wind turbine 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.24W
sail                0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.54W

So if, by your own admission, the sail power is zero when the vehicle is stationary at 0 m/s, how can the sail power suddenly be 524 W when the vehicle is traveling at 0.1 m/s? (Which is nearly stationary.)

Would the sail power be higher or lower when the vehicle is traveling at 0.01 m/s? How about 0.001 m/s?

Can you make a graph of sail power vs vehicle speed? What does it look like?
 

Offline electrodacus

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1m^2 swept area wind turbine with 40% efficiency in a 10m/s wind speed   vs   1m^2 sail

If thery speed relative to the ground is zero then of course sail will do nothing and there will just be a static force as on any wall where the wind turbine output power will be 0.5 * 1.2 * 1 * 10^3 * 0.4 = 240W

So stationary wind turbine wins 240W vs 0W

Now install the wind turbine and the sail on a vehicle driving directly downwind at 0.1m/s, 5m/s and 9m/s

0.1m/s
wind turbine  0.5 * 1.2 * 1 * (10-0.1)^3 * 0.4 = 232.87W
sail (can install an electric generator at the wheel say that is 90% efficient)   0.5 * 1.2 * 1 * (10-0.1)^3 * 0.9 = 523.96W

5m/s
wind turbine 0.5 * 1.2 * 1 * (10-5)^3 * 0.4 = 30W
sail               0.5 * 1.2 * 1 * (10-5)^3 * 0.9 = 67.5W

9m/s
Wind turbine 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.24W
sail                0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.54W

So if, by your own admission, the sail power is zero when the vehicle is stationary at 0 m/s, how can the sail power suddenly be 524 W when the vehicle is traveling at 0.1 m/s? (Which is nearly stationary.)

Would the sail power be higher or lower when the vehicle is traveling at 0.01 m/s? How about 0.001 m/s?

Can you make a graph of sail power vs vehicle speed? What does it look like?

While there is zero power there is a force acting on the sail and so you can stay stationary only if you get good enough brakes and enough tire friction.

You can calculate so for 0.001m/s is 0.5 * 1.2 * 1 * (10-0.001)^3 * 0.9 = 539.8W

Yes you can see the graph in my video at about minute 2 https://youtu.be/4Hol57vTIkE?t=119  there the example is also for a 1m^2 sail and 10kg vehicle ideal case no friction and you can see vehicle speed (blue) vs time and power (green).

Offline Kleinstein

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You can calculate so for 0.001m/s is 0.5 * 1.2 * 1 * (10-0.001)^3 * 0.9 = 539.8W
With some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
This value is obviously wrong (way to high) and shows that the formular used above must be wrong, not just at zero speed.


 

Offline electrodacus

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With some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
This value is obviously wrong (way to high) and shows that the formular used above must be wrong, not just at zero speed.

The formula used above is correct you used the wrong speed to calculate the force on the sail. You need to use 10-0.001 = 9.999m/s
An ideal 1m^2 sail in 10m/s speed will see a 60N force.  I considered the generator 90% efficient in that formula that is why is less than 600W

Online bdunham7

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With some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
This value is obviously wrong (way to high) and shows that the formular used above must be wrong, not just at zero speed.

The formula used above is correct you used the wrong speed to calculate the force on the sail. You need to use 10-0.001 = 9.999m/s
An ideal 1m^2 sail in 10m/s speed will see a 60N force.  I considered the generator 90% efficient in that formula that is why is less than 600W

OK, so 60N of force x 0.001m/s is how much power?
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Offline electrodacus

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OK, so 60N of force x 0.001m/s is how much power?

That is not how you will make the calculation. It will be 60N * (10m/s-0.001m/s).
This sort of things is what also make people think Blackbird can exceed wind speed powered only the the wind with no energy storage.


That 0.001m/s is unreasonable and as a theoretical exercise.
Just made some quick calculation and for a 10kg vehicle with a 1m^2 sail it takes just around 2ms (0.002 seconds) to get from zero to 0.5m/s
« Last Edit: December 14, 2021, 10:24:47 pm by electrodacus »
 

Online bdunham7

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That is not how you will make the calculation. It will be 60N * (10m/s-0.001m/s).
This sort of things is what also make people think Blackbird can exceed wind speed powered only the the wind with no energy storage.

OK, if that is the case, then what is the power at 0 m/s?
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Offline electrodacus

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That is not how you will make the calculation. It will be 60N * (10m/s-0.001m/s).
This sort of things is what also make people think Blackbird can exceed wind speed powered only the the wind with no energy storage.

OK, if that is the case, then what is the power at 0 m/s?

No motion of the vehicle means zero power can be generated at the wheel.
but see my edited comment above where I mentioned it takes just around 0.002 seconds 2ms for a 10kg vehicle with a 1m^2 sail to get from 0m/s to 0.5m/s and so maybe around 10ns that is nanoseconds to get to 0.001m/s (that is the step increment I used in my graph calculator the one in the video).


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