Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[yn3]

There was and still is a hot debate on the matter of a wind power car that travels down the wind faster than the wind. As a part of larger interest in the subject I put the theory for this case, it can be found in my website
https://sites.google.com/site/yedidianeumeier/"
I welcome your comments

 

[mheslep]

There are several past threads on the subject containing thousands of posts:
https://www.physicsforums.com/showthread.php?t=284134&highlight=wind
https://www.physicsforums.com/showthread.php?t=272437&highlight=DDWFTTW
https://www.physicsforums.com/showthread.php?t=270041&highlight=DDWFTTW

 

[yn3]

"There are several past threads on the subject containing thousands of posts" Indeed so, and all the previous threads are closed, at list one of them because of improper language. My experience with published theory on this subject suggest that I can contribute to the matter. I hope that this thread will remain civil and to the point. I am welcoming your feedback.

 

[rcgldr]

The most recent thread includes links to videos of a working full scale (human pilot) model that accomplished about 28 mph speed with a 10 mph tail wind:

https://www.physicsforums.com/showthread.php?t=390801

Videos of this model before the outer skin was added can be seen here:

http://www.youtube.com/user/TraderTurok

Some sailcraft, such as a "Skeeter" ice boat, when tacking downwind (at an angle, not directly downwind), can also achieve a net downwind speed greater than the wind.

 

[yn3]

Thanks, you are right I am very familiar with Rick blackbird. NALSA actually ratified a 2.8X speed record on his behalf, you can see it in http://www.nalsa.org/index.htm
Note that unlike sail car, prop-turbine wind car does not tack the wind and thus a race of these cars could take place on a regular closed loop race track. I intend to develop the theory for all direction car and hopefully build one myself.

 

[Curl]

Anyone with a half-decent brain can ponder this question for a few minutes and realize that YES, it is possible to make a vehicle go faster than the wind, directly down the wind, solely on wind power.

It is also possible to go directly against the wind, but this one is even easier to figure out.

 

[yn3]

Well, the "half-decent brain can ponder this question for a few minutes and realize that YES" seems at that point like http://en.wikipedia.org/wiki/Egg_of_Columbus" . I personally communicated with two distinguished professors, one of which share the Nobel Prize with Al Gore and is currently advising the president on energy matters, the other one run a blog about physics where he explained why it cannot be done. Both admitted to me that now they are confused, they don't know how it can be explained. Now, leave this aside and proceed to ask the following. To move with (down) the wind the car uses a propeller driven by the wheels, to move upwind one needs a wind turbine the shaft of which drives the wheels. How fast can this car go? Now, what happens if you go in an angle to the wind? Is there an angle where the propeller is inefficient and the turbine likewise so that it is so to say a blind spot to the car, meaning an angle it cannot drive or is it a point of overlapping manning both propeller and turbine are effective, if so what is the maximum velocity at this angle? All this can be readily answered, however it will take me a bit more than few minutes to put the equations run the simulation type it neatly and post it.

 

[mheslep]

... two distinguished professors, one of which share the Nobel Prize with Al Gore

That of course is the the highly political Peace prize, and not for physics or the other sciences.

 

[Academic]

Well, I guess I don't have half-decent brain power because I thought about for a few minutes and couldn't come up with it. Damn my less than half decent brain...

 

[yn3]

"That of course is the the highly political Peace prize, and not for physics or the other sciences. " Indeed so but the guy, according to his bio is
Professor of Nuclear Engineering and Ph.D (1998) & MA (1986) Harvard University (Physics),
AB (1984) Cornell University (Physics)
The AB is in his bio it may be a typo for BA but you see my point.

"Damn my less than half decent brain" Can I offer you half of mine? My wife says my brain is full of useless physics and otherwise useless historical knowledge.

To the point, humor and gossip is always great but what about discussing the all direction car?

 

[rcgldr]

> Angle to wind ...

Once at a significant angle, a standard sail setup would be better. In a DDWFTTW vehicle, the ground force opposes the forward motion of the vehicle, because that force is used by the wheels to drive the propeller. In a sailcraft, except for drag or rolling resistance, the ground force is perpendicular to the forward motion of the sailcraft, and consumes no energy, and this would be a better setup when moving at an angle to the wind.

 

[yn3]

"Once at an angle, a standard sail setup would be better" This is a qualitative statement. What angle do you have in mind 1deg? 10 degrees? However, let's wait and see what other thinks.

 

[A.T.]

To move with (down) the wind the car uses a propeller driven by the wheels, to move upwind one needs a wind turbine the shaft of which drives the wheels. How fast can this car go?

In terms of wind-speed-multiple there is no particular limit in either direction. It just depends on efficiency.

in terms speed relative to the air, there are practical limits related to the gas properties. I doubt the sound barrier could be broken.

in terms speed relative to the ground there the speed of light limit

Now, what happens if you go in an angle to the wind? Is there an angle where the propeller is inefficient and the turbine likewise so that it is so to say a blind spot to the car, meaning an angle it cannot drive or is it a point of overlapping manning both propeller and turbine are effective,

The blind spot for faster than wind would be at 90°. But with cyclic pitch control propeller/turbine ranges would overlap.

 

[rcgldr]

The top class ice boats, called Skeeters, would out perform the DDWFTTW Blackbird, given sufficient room to tach at an angle to the wind (30 to 40 degrees). Based on information about the Skeeter in this pdf file:

http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

I calculated that a Skeeter's downwind component of speed was 3.36x wind speed in an 18 mph wind (over 60.6 mph downwind component while at taching at 30 degrees offset from wind). The ratio would be higher in a 10 mph wind due to less aerodynamic drag from the apparent headwind. See post # 28 in this thread:

https://www.physicsforums.com/showthread.php?t=283813

 

[mender]

There were runs as high as 3.48x WS but were not used for the record because they were borderline. The team wanted an absolutely indisputable run for the initial record.

The purpose of establishing the record was more about proof of DDW travel by a recognized independent authority than the highest number achievable - I'm sure that others will go for the big numbers in years to come.

 

[yn3]

Surprise surprise: That is if my calculations are true, and rest assure that I am and will double checking them for a while.
The propeller wind car can do extremely well in side winds. Assuming that the propeller shaft can be rotated sideways, (which is not possible in the current ThinAirDesigns car), a car that can do 3XW down the wind can do more than 4XW with 90 degrees wind, the propeller will be swayed about 70 degrees sideways to 20 degrees off the wind. Moreover, the calculations show that even with the fix front thrust propeller, the aforementioned car capable of 3XW down the wind will do still better than 2XW at 45 degrees to the wind. This part of my calculation should be able to be confirmed or rejected by the experience of ThiAirDesigns, I am sure that they took some runs with angle to the wind. mender what say you?

 

[ThinAirDesign]

Moreover, the calculations show that even with the fix front thrust propeller, the aforementioned car capable of 3XW down the wind will do still better than 2XW at 45 degrees to the wind. This part of my calculation should be able to be confirmed or rejected by the experience of ThiAirDesigns...

We never ran the vehicle in anger in full 45 cross, but we have data showing ~2.5x in sustained 30-40 cross.

JB

 

[yn3]

"We never ran the vehicle in anger in full 45 cross, but we have data showing ~2.5x in sustained 30-40 cross."
That's seems to support my claim, if I know your transmission efficiency and the maximum down the wind speed ratio, I know it is around 3.5 but would like to know your simulation prediction, I then could run my calculations to provide a so called polara, the maximum ratio as function of angle.

 

[RCP]

Sorry to see this topic dropped. I was looking to seeing Yn3's maximum angle. Rumor has it that with it's fixed prop Blackbird functions best directly downwind. Any thoughts on that?

 

[ThinAirDesign]

RCP, it's not the 'fixed pitch' nature of the Blackbird (which has gone DDWFTTW in both fixed pitch and variable pitch configurations) that makes it function best DDW, it's the circular nature of it's sail path.

Fixed or standard 'variable pitch' makes no difference in it's off axis performance. Even if it were equipped with a 'cyclic pitch' hub, it would still perform best DDW -- but the off axis performance would degrade slower off DDW.

But the you knew that already -- I'm just responding to the thread to answer your question on the record.

JB

 

[vanesch]

Here we go again :-)

As others said, you do not deserve your marks on mechanics 101 if you can't figure out that this is trivially possible to go faster than the wind. The elementary theory of it (of the order of an exercise of medium difficulty in said mechanics 101 course) has been posted several times and it is quite disturbing if people with a degree which should have mastered such a course cannot understand it almost immediately, worse, dispute it. You could just as well dispute that things heavier than air can fly.

The long, locked threads cited earlier must contain about everything that can possibly be discussed about it.

 

[chingel]

I have read a dozen of pages on this topic but I cannot understand it. Could someone please explain it or give a link to an explanation? Isn't it true that once you are up to wind speed and use the wheels start to power the propeller, the energy will come from the kinetic energy of the cart and since the propeller cannot propel air at 100% efficiency some energy will be lost and shouldn't the cart slow down? I simply don't understand how it works.

 

[OmCheeto]

I have read a dozen of pages on this topic but I cannot understand it.

I read them all.

Could someone please explain it or give a link to an explanation? Isn't it true that once you are up to wind speed and use the wheels start to power the propeller, the energy will come from the kinetic energy of the cart and since the propeller cannot propel air at 100% efficiency some energy will be lost and shouldn't the cart slow down? I simply don't understand how it works.

It's meant to be an exercise in aerodynamics. Break the problem down into a free body diagram and figure out if it is possible or not.

 

[mender]

I have read a dozen of pages on this topic but I cannot understand it. Could someone please explain it or give a link to an explanation? Isn't it true that once you are up to wind speed and use the wheels start to power the propeller, the energy will come from the kinetic energy of the cart and since the propeller cannot propel air at 100% efficiency some energy will be lost and shouldn't the cart slow down? I simply don't understand how it works.

If there was no wind, the cart would have to be 100% efficient just to maintain speed, which of course in the real world isn't possible. Since the air is moving wrt the ground, the thrust at the propeller only has to be a portion of what it would be if there was no wind. Given reasonable losses to friction and such, a properly designed cart is quite capable of doing so.

Did that help?

 

[chingel]

If there was no wind, the cart would have to be 100% efficient just to maintain speed, which of course in the real world isn't possible. Since the air is moving wrt the ground, the thrust at the propeller only has to be a portion of what it would be if there was no wind. Given reasonable losses to friction and such, a properly designed cart is quite capable of doing so.

Did that help?

I can understand that the thrust has to be a portion of what it would have to be when there is no wind, but nevertheless, no matter how little thrust, it has to come from the kinetic energy of the moving cart, I would think. Then the propeller uses some of that energy to accelerate air particles in one direction which accelerates the cart in the other direction. Isn't it true that the propeller cannot accelerate air particles with more force than it is getting from the wheels? What am I missing?

 

[Uglybb]

A simple wikipedia search would have given you all the answers

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

The formula and maths etc ... everything its all there!

 

[mender]

I can understand that the thrust has to be a portion of what it would have to be when there is no wind, but nevertheless, no matter how little thrust, it has to come from the kinetic energy of the moving cart, I would think. Then the propeller uses some of that energy to accelerate air particles in one direction which accelerates the cart in the other direction. Isn't it true that the propeller cannot accelerate air particles with more force than it is getting from the wheels? What am I missing?

You're forgetting that the thrust at the prop can be higher than the drag at the wheels but only because the ground is moving past the cart faster than the air. If there is no wind, the air is moving past the cart at the same speed.

The cart essentially gears down the force at the wheels and turns the prop slower than the ground speed but with more force against the air.

The drag at the wheels is x, the speed across the ground is v; since the prop is acting on the air, with no losses, the thrust from the prop can be 2x when it is turning at a rate to move the air back at half the ground speed (v/2). Same principle as a lever; the force at one end can be much greater than at the other but the distance is less.

Better?

 

[mender]

A simple wikipedia search would have given you all the answers

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

The formula and maths etc ... everything its all there!

But even with all the answers, sometimes it has to be explained in the right order and the right way for the light bulb to go on.

 

[chingel]

You're forgetting that the thrust at the prop can be higher than the drag at the wheels but only because the ground is moving past the cart faster than the air. If there is no wind, the air is moving past the cart at the same speed.

The cart essentially gears down the force at the wheels and turns the prop slower than the ground speed but with more force against the air.

The drag at the wheels is x, the speed across the ground is v; since the prop is acting on the air, with no losses, the thrust from the prop can be 2x when it is turning at a rate to move the air back at half the ground speed (v/2). Same principle as a lever; the force at one end can be much greater than at the other but the distance is less.

Better?

I'm sorry I do not understand how. How can the thrust be higher than the drag at the wheel? Aren't they are mechanically connected? Move air back at v/2 relative to what, the cart or the ground? If you use gears to make the propeller move faster than the wheels, it will displace more air but also drag the wheels more, because it does more work and needs more energy, if you make it spin slower it provides less acceleration but also causes less drag, at least the way I understand it.

 

[mender]

How can the thrust be higher than the drag at the wheel? Aren't they are mechanically connected?

Thrust is a force, and drag is a force. You do know that by using a lever or gears, you can put x amount of force into one end and get 2x, 3x, or 10x force out of the other end right?

Gears are mechanically connected yet change the amount of force put in at one end to a different force at the other; no magic, just leverage or mechanical advantage.

Move air back at v/2 relative to what, the cart or the ground?

Relative to the cart.

If you use gears to make the propeller move faster than the wheels, it will displace more air but also drag the wheels more, because it does more work and needs more energy, if you make it spin slower it provides less acceleration but also causes less drag, at least the way I understand it.

But we use gears to slow down the propeller, displacing less air but supplying more force for the same amount of work taken from the wheels.

For the sake of clarity, let's take losses out of the picture for a moment.

The wind is blowing at 10 mph. The cart, without the propeller engaged, gets pushed up to 10 mph. The pitch of the prop and the gearing of the cart is set up to push air back relative to the cart at half the speed of the wheels. The prop is engaged; the air around the cart gets pushed back at 5 mph, half the speed of the wheels. Since the force at the wheels is geared down, the force at the prop is twice as much as the drag at the wheels and the cart accelerates. F=MA so if the net force acting on the cart is positive, as it is so far, the cart will accelerate.

The cart gets to 15 mph; the prop is now turning fast enough to push air back at 7.5 mph and the force at the prop is still higher than at the wheels, so the cart will still keep accelerating. About now you're thinking that if that is true,the cart will keep accelerating forever; it won't, because the gear ratio determines the top speed in accordance to the speed of the wind.When the cart gets to 20 mph, the prop is pushing air back at 10 mph and the wind is still blowing at 10 mph, meaning that the prop is no longer exerting any force on the air, it's just freewheeling. With this gear ratio, the absolute best the cart can travel across the ground is twice the speed of the wind.

Now the wind drops to 7.5 mph; what happens to the cart? At 20 mph, it is still pushing air back at 10 mph but with the wind at only 7.5 mph, that means that instead pushing air back, it is actually trying to drag air forward at 2.5 mph. The net force acting on the cart is now negative, so the only thing that can happen is that the cart will slow down. What speed does it stabilize at? Again, twice the speed of the wind, or 15 mph.

Gearing the wheel speed down by half (0.5) to power the propeller gives a theoretical top speed of twice the wind speed. A different ratio will give a different theoretical multiple of the wind speed. Gearing the cart down less means the prop is spinning faster but with less force; it has the potential to faster for the same wind speed but it needs to be more efficient because the prop is supplying less force than before to overcome the real world losses.

 

[RCP]

@mender: I always like your basic explanations for how the cart works best. Find them reassuring in terms of how I perceive it. (And I wouldn't confess this elsewhere for fear of casting doubts on your understanding. )

I know it's in the blog somewhere, but bet you know the gear ratio of BB. What was it again? And using your explanation above, what would be the max BB could maintain steady state above ws? I may have seen this laid out many times before, but after 4k pages of such forgive an old guy if it overloads my RAM. The info is there, but often inaccessible.

After much thought I can finally envision BB leaving the packing popcorn in the dust. With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

I'm speaking hypothetically above, as I know the NALSA rules take gusts and direction into consideration. But it was a first of its kind test for a sport's club, so might it not be in the realm of possibility they got some things wrong?

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

 

[chingel]

Thank you for your explanations.

Here is where I still have a problem. Doesn't energy matter also? Yes we can use gears to multiply the force, but doesn't the energy stay the same? Eg isn't the energy you put into at one end the same what you get from the other end, no matter what gears you use between? Otherwise you could just connect some gears to a wheel and then connect it to another wheel and multiply the force. You could do that, but doesn't the energy stay the same?

To me it seems that it takes a certain amount of energy to make the propeller move the air and that energy can only come from the kinetic energy of the cart, no matter what gears you use. If you use gears to make the propeller slower, the force at the wheels is lower, but since they move more in the same time, don't they expend the same amount of energy as the propeller gets?

 

[mheslep]

The air stream 'moves' the propeller more than the other way around. Or, in terms of energy, the propeller takes kinetic energy from the air stream and converts it to the kinetic energy of the platform.

 

[mender]

I know it's in the blog somewhere, but bet you know the gear ratio of BB. What was it again? And using your explanation above, what would be the max BB could maintain steady state above ws? I may have seen this laid out many times before, but after 4k pages of such forgive an old guy if it overloads my RAM. The info is there, but often inaccessible.

IIRC, the advance ratio is 0.8, and to get the theoretical top speed, you use this formula: 1/(1 - advance ratio). So for BB it would be 1/(1 - 0.8) or 1/0.2, which is five times the speed of the wind. Speaks pretty highly of the team when they managed to get runs in the 70% of TTS (theoretical top speed) range with their first vehicle on their second outing.

After much thought I can finally envision BB leaving the packing popcorn in the dust. With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

So if a heavy tin can could beat the popcorn just by being at ws when the wind dies slightly, why do you have a hard time visualizing BB beating the (maybe) 10 mph popcorn when it is going almost 20 mph faster even without the wind dying? The only chance the popcorn (or a leaf) would have of beating a DDWFTTW vehicle is if a sudden gust of wind propelled the popcorn (or leaf) to speed before the cart has a chance to accelerate, making the BB's momentum a disadvantage.

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

As was the one in the original video by Goodman. Tested and proven as a TM cart, and would have beaten the popcorn handily even if they started at the same time.

It's nice (and still fun!) to run through this without the waters being muddied.

 

[mender]

Thank you for your explanations.

Here is where I still have a problem. Doesn't energy matter also? Yes we can use gears to multiply the force, but doesn't the energy stay the same? Eg isn't the energy you put into at one end the same what you get from the other end, no matter what gears you use between? Otherwise you could just connect some gears to a wheel and then connect it to another wheel and multiply the force. You could do that, but doesn't the energy stay the same?

To me it seems that it takes a certain amount of energy to make the propeller move the air and that energy can only come from the kinetic energy of the cart, no matter what gears you use. If you use gears to make the propeller slower, the force at the wheels is lower, but since they move more in the same time, don't they expend the same amount of energy as the propeller gets?

Yes, absolutely! That's what we've been talking about the whole time, the fact that the energy input from the wheels is used by the prop, and as such the energy output can never be more than the input. The force output certainly can, just like a lever and that's what we've also been talking about levers.

As mheslep pointed out, when the prop pushes air back relative to the cart it is reducing the speed of the air over the ground (reducing the kinetic energy of the wind) and using that energy to propel the cart (increasing the cart's kinetic energy). Since a real world cart is never 100% efficient, the cart will never reach the theoretical top speed but instead will travel at some percentage of it.

The Blackbird reached 3.5x the wind speed, which is 70% of its theoretical top speed of 5x wind speed; pretty impressive!

 

[rcgldr]

Doesn't energy matter also?

This was covered in the previous threads. Note that

power = force x speed

For the wheels, the point of application of force is the ground, which is moving backwards relative to the cart. For the propeller, the point of application of force is the air, which is moving at (ground speed relative to cart) - (wind speed relative to ground) = wind speed relative to cart.

As an example, say the wind speed relative to ground is +10 mph, and that the cart is moving at +25 mph (downwind). Ground speed relative to the cart is -25 mph. Wind speed relative to the cart is -15 mph. If the effective advance ratio is .8, then the propeller would produce thrust at -20 mph relative to the cart if there was no load. This ratio means that with zero losses, the force at the propeller can be 1.25 (25/20) times the opposing force at the wheels. Assume the force at the wheels is 80 lbs, then the propeller could produce up to 100 lbs of thrust with no losses:

power = 80 lbs x 25 mph (wheels) = 100 lbs x 20 mph (propeller)

This results in an ideal net forward force of 20 lbs. The real thrust force and speed will be less, but as long as the net force is greater than rolling resistance and drag the cart accelerates, and in the case of the BB cart, it reaches a max around 3.5x wind speed.

The propeller makes things a bit more complicated than just gearing. The propeller's pitch is part of the effective gearing (and advance ratio). The propeller's size (width and length) affects how much thrust force is produced for a given thrust speed (relative to the air) and pitch. A long (large diameter) propeller will generally be more efficient.

 

[chingel]

I'm just a regular student and to be honest I never quite understood how power = force x speed. I have always thought that power means how many joules in a second you can do. The speed of what it is in the formula? When it takes me certain energy to produce a force of 1 N and I apply the force on a 0,5 kg object it moves faster than when I apply the same force on a 1 kg object. It's not like I double my power output, isn't it? Because I am using the same energy per second, the object just offers less resistance and moves faster, in no way it shouldn't increase my power output in my understanding.

I guess I am stuck on the energy concept. Is there a way to explain it in terms of energy? The propeller and the wheels are connected, energy input has to equal energy output, the energy that it takes from the kinetic energy of the cart it gives to the air particles trying to propel itself, but since the energies are equal they should cancel out, energy shouldn't change no matter what gears you use. Some extra energy has to come from somewhere.

 

[A.T.]

I'm just a regular student and to be honest I never quite understood how power = force x speed.

It's just the time derivate of:

work = force x distance

since:

power = work / time
velocity = distance / time

you have:

power = force x velocity

This form allows you to analyze energy balance instantaneously, which is handy for analyzing accelerating things.

I guess I am stuck on the energy concept. Is there a way to explain it in terms of energy?

First you have to pick an inertial reference frame to do your energy analysis using work = force x distance. For the energy balance in the reference frame of the cart check this video at 1:05:

energy input has to equal energy output

But force output can greater that force input.

The propeller and the wheels are connected, energy input has to equal energy output, the energy that it takes from the kinetic energy of the cart it gives to the air particles trying to propel itself,

About which reference frame are you talking here?

 

[A.T.]

The propeller makes things a bit more complicated than just gearing. The propeller's pitch is part of the effective gearing (and advance ratio).

Here is a diagram that shows how propeller pitch and advance ratio are connected:

 

[poont2]

OMG this diagram just blew up my brain, lol. what is that gear and what is that grey vertical teeth? Let me spend some time to understand it, I guess I have less than half of a brain of normal human...

But one thing I am wondering, when the car with propeller is stationary with no wind, and wind start to blow (same direction and the car's direction). If the propeller is not connected to the wheel, wouldn't the propeller spin in the opposite angular direction that you want it to spin? does this meen that the force from the wheel would greatly overpower the this effect? I am sure if that make sense...

 

[rcgldr]

What is that gear and what is that grey vertical teeth?

The gear represents the wheels. The grey vertical bar represents the drivetrain connecting the gear (wheels) and the propeller.

When the car with propeller is stationary with no wind, and wind start to blow ...

For a downwind cart in that situation the propeller act's as a "bluff body" (a sail), blocking the wind. This results in a forward force on the cart, which results in an backwards force (torque) on the wheels, and the turning wheels torque is passed through the drivetrain to cause the propeller to turn.

One exception case is if there isn't enough grip at the wheels to handle the wind or a sudden gust, where it's possible for the cart to move forwards, but with the propeller causing the wheels to spin (slide) backwards. As the cart picks up speed, this reverse torque from the propeller will diminish, and eventually the wheels will regain grip unless the grip factor is very low.

 

[A.T.]

OMG this diagram just blew up my brain, lol. what is that gear and what is that grey vertical teeth?

As rcgldr said, the gear is the wheel, the gray rack is the transmission, that couples wheel and propeller rotation. It is important to point out, that the vertical direction in the diagram, corresponds to the circumferential direction for the airfoil, which is spinning not just going up.

But one thing I am wondering, when the car with propeller is stationary with no wind, and wind start to blow (same direction and the car's direction). If the propeller is not connected to the wheel, wouldn't the propeller spin in the opposite angular direction that you want it to spin?

Correct. See the wheel rotation for CASE C - HELD IN AIR: the wheels have no ground contact so the rotor is freewheeling as a turbine and turns the wheels "backwards" (as if the cart would go upwind)

does this meen that the force from the wheel would greatly overpower the this effect? I am sure if that make sense...

Yes, the torque transmitted from the wheels is greater than the aerodynamic torque, so the rotor turns as a propeller, against the aerodynamic torque.

You can also put it the other way around, like it is shown in the diagram (CASE C - START UP): The axial aerodynamic force on the propeller(blue arrow) that pushes the cart downwind is greater than the ground reaction force (red arrow) caused by the circumferential aerodynamic force(or torque) on the propeller.

In reality the axial aerodynamic force on the propeller(blue arrow) is of course also supported by the aerodynamic force (bluff body drag) on the chassis, supporting the self start.

 

[chingel]

Regarding power = force x speed. Is the speed in the equation the speed of the object that force is applied to? Doesn't the speed of the object depend on how much resistance it offers to the force, not on the power? If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going, to him it doesn't even matter if it hits anything, it is just outputting at a constant power, the ship speeds up, but the power doesn't change. Basically I just don't understand the equation.

The force at the propeller is bigger, but since it is moving slower, it uses up as much energy as the wheels give. You can't just use gears to do more work, where does the cart get the extra energy? This thing is frying my brain.

When you are going downwind and reach wind speed the tangential airflow component and the downwind force component is zero, so they shouldn't increase the speed. Instead the wheels start driving the propeller, slowing the cart down. How does it work?

 

[A.T.]

Regarding power = force x speed. Is the speed in the equation the speed of the object that force is applied to?

Yes

Doesn't the speed of the object depend on how much resistance it offers to the force,

No. The instantaneous speed doesn't. The instantaneous acceleration does.

If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going, to him it doesn't even matter if it hits anything, it is just outputting at a constant power, the ship speeds up, but the power doesn't change.

The power in power = force x speed here is the change of the ships kinetic energy per time, not the laser power. Since KE grows with velocity squared, the input of KE increases even if the acceleration is constant.

The force at the propeller is bigger, but since it is moving slower, it uses up as much energy as the wheels give.

In the cart frame, the air moves slower than the ground. So the cart can do negative work on the fast ground (harvest energy) with a low force, and use that energy to do positive work on the slow air with a large force. The force difference is the net thrust.

In the ground frame the air is doing positive work on the cart, while the cart uses the ground as a fulcrum to multiply the airs speed.

https://www.youtube.com/watch?v=g8bxXRQtcMY

You can't just use gears to do more work,

Correct. But you can use gears to get more speed:

https://www.youtube.com/watch?v=E7vcQcIaWSQ

where does the cart get the extra energy?

Which extra energy? You have not even stated in which reference frame you are doing your energy analysis?

When you are going downwind and reach wind speed the tangential airflow component and the downwind force component is zero,

Wrong. At wind speed the propeller is spinning in still air, so the airflow at the blade is mostly tangential. The airfoil generates lift perpendicular to the airflow so the downwind force component is not zero.

Instead the wheels start driving the propeller, slowing the cart down.

The wheels are turning the propeller right from the start, not from windspeed on. But they are not slowing the cart down, because the prop's axial force is greater than their retarding force.

 

[chingel]

The power in power = force x speed here is the change of the ships kinetic energy per time, not the laser power. Since KE grows with velocity squared, the input of KE increases even if the acceleration is constant.

Does this mean that you cannot calculate the power of the laser knowing the force the laser applies and the speed at which the ship moves? Does it also mean that the rate of growth of the ships kinetic energy will be exponential as the speed keeps increasing? Why doesn't the equation using the propeller's thrust and windspeed apply to the change of the air particles kinetic energy per time instead of the propellers power? The equation is very confusing to me.

In the video the cart certainly does move faster than the blue chain if it has a rod stuck in the moving chains, but I cannot understand how it is the same with the DDWFTTW.

If the cart is at windspeed, the wind doesn't push the propeller or the cart, the propeller gets moved only by the wheels. This means that any energy that the propeller uses when turning comes out of the wheels and thus the carts kinetic energy. When you are driving a car you can't just use the moving ground to do negative work and use gears to lever it, any energy you extract will slow you down exactly as much. Electric cars use breaking energy to generate electricity, they can't just do it while cruising without energy loss. There has to be some other effect at play with the wind cart because they somehow can go faster than the wind and I cannot understand what it is.

 

[A.T.]

Does this mean that you cannot calculate the power of the laser knowing the force the laser applies and the speed at which the ship moves?

I think you would at least have to know how much of the light is reflected. But you cannot assume that the increase of the ships KE per time equals the laser power.

Does it also mean that the rate of growth of the ships kinetic energy will be exponential as the speed keeps increasing?

Classicaly KE grows with speed squared, not exponentially.

Why doesn't the equation using the propeller's thrust and windspeed apply to the change of the air particles kinetic energy per time instead of the propellers power?

In the ground frame the equation does apply as you suggest:

power_air_to_cart = true_wind_speed x thrust

In the cart's frame:

power_ground_to_cart = ground_speed x wheel_drag
power_cart_to_air = air_speed x thrust

The equation is very confusing to me.

You have to realize that KE is a frame dependent quantity. Every explanation based on KE will be valid only for a certain frame. A different observer will see a completely different energy balance. So maybe it is not that useful to ask for "explanations in terms of energy". But if you do, you should at least specify the reference frame.

In the video the cart certainly does move faster than the blue chain if it has a rod stuck in the moving chains, but I cannot understand how it is the same with the DDWFTTW.

Put a big sail in the middle of the rod, stick one end into the ground and the other end will go twice the windspeed.

If the cart is at windspeed, the wind doesn't push the propeller

Why not? The blades are spinning. They have a relative airflow perpendicular to true wind direction. It is trivial to produce a downwind force with an airfoil set at the right angle. And with an efficient airfoil that downwind force can be 20 times greater than the tangential drag of the blade, that brakes the wheels.

the propeller gets moved only by the wheels.

That's is true in the cart's frame, where the propeller is only rotating. Here the ground moves, and turns the propeller via the wheels.

In the ground frame you can see the wheels & transmission as mere kinematic constraint for the blades, that enforce a certain helical path for the airfoils. The blades are pushed by the air along that helical path:

[PLAIN]http://img811.imageshack.us/img811/4922/propellervectors.png

Here for comparison a sailcraft on broad reach, constrained laterally by the keel/skates/wheels:

[PLAIN]http://img253.imageshack.us/img253/6694/downwindvectorsen3.png

Here a nice animation that transforms one into the other:

https://www.youtube.com/watch?v=UGRFb8yNtBo

This means that any energy that the propeller uses when turning comes out of the wheels and thus the carts kinetic energy.

Your statements about energy are meaningless without specifying which reference frame you mean.

In the cart's frame the cart doesn't have any KE. But the ground has plenty of it and gives it to the air, via wheels and propeller.

In the ground frame the propeller is taking KE from the air, so it is an energy input to the cart. To confirm this check the vectors of blade_velocity and blade_force in the diagram above. They are at less than 90° (their dot product is positive), so the air is doing positive work on the blade. Check also the red tracer in the very first clip of the below animation. It shows how the air gets slowed down in the ground frame.

https://www.youtube.com/watch?v=FqJOVHHf6mQ

When you are driving a car you can't just use the moving ground to do negative work and use gears to lever it, any energy you extract will slow you down exactly as much.

You will not slow down, if you use the extracted energy to push against something that moves slower relative to you than the ground. The net force determines if you slow down, not just the force from the ground.

There has to be some other effect at play with the wind cart because they somehow can go faster than the wind and I cannot understand what it is.

There is no "other effect". It is just gearing. You should try to understand rigid examples of such gearing first:

https://www.youtube.com/watch?v=k-trDF8Yldc

 

[rcgldr]

If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going ...

From the Earth's frame of reference, as the speed of the spaceship increases, the number of photons per second that hit the spaceship "sail" decreases.

If the cart is at windspeed, the wind doesn't push the propeller or the cart, the propeller gets moved only by the wheels.

The wind pushes against the air being propelled by the propeller. That air just aft of the propeller experiences some amount of compression, generating equal and opposing forces, upwind againts the true wind, and downwind against the propeller. That downwind force on the propeller corresponds to the thrust generated by the propeller, which due to the effective gearing (advance ratio) produces this thrust at a lower speed (relative to the cart), than the force at the wheels from the ground. This allows the thrust to be greater than the opposing force from the ground while at the same time using less power because of the reduced speed of that thrust. The reduced speed is possible because the wind speed is always less than the ground speed when the cart is moving downwind, regardless of the speed of the cart (relative to the ground).

 

[Redbelly98]

It's meant to be an exercise in aerodynamics. Break the problem down into a free body diagram and figure out if it is possible or not.

Better yet, let the car be moving at the same speed as the wind, so that the air drag is zero. Then figure out if the acceleration can be positive, in which case faster than the wind is possible.

 

[OmCheeto]

Better yet, let the car be moving at the same speed as the wind, so that the air drag is zero. Then figure out if the acceleration can be positive, in which case faster than the wind is possible.

Oh! I never thought of that. Quite a while back, I ran a computer simulation on the device from a standstill, and it only reached 70% of wind speed. Though it was a fixed prop pitch simulation. It appears that the full scale model has variable pitch blades. Back to the drawing board.

 

[ThinAirDesign]

With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

RCP, if you'll try to remember (or even read the NALSA rules again), you will see that during the 10 second timing period the BB cart had to be *accelerating* -- that is it had to exit the timing period *faster* than it entered. The tin can that is beating the wind during a quick lull is doing *just the opposite* -- it is slowing down.

The requirement for acceleration through the timing period was added *precisely* to make it impossible for the BB to 'coast' through a lull to get the record.

so might it not be in the realm of possibility they got some things wrong?

Like what?

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

OMG RCP -- "clever enough"? You think the reason we didn't put RC on our small model was because we weren't "clever" enough?. That's freakin' funny. ROFLAO. Goodman put RC on and was accused of using the battery that was onboard to power the cart. THAT'S why we never made one RC. We wanted to be able to say correctly that there were no batteries on board at all.

But then you've known this over and over -- you just can't keep it all together.

JB