Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[mender]

But you can't just have the ground turn the propeller without slowing the car down by an equal amount. You can't use the moving ground when driving a car to turn a generator without slowing the car down.

Yes you can:
http://www.youtube.com/watch?v=LMgDvC5lqsY&feature=related

 

[chingel]

Because the energy to power the cart is from the wind. After the cart passes, the kinetic energy of the wind is lower; the prop slows down the speed of the air mass (wind) relative to the ground.

If there wasn't a wind, you'd be right. The wind is the external energy source; no wind = no energy source to power the cart.

Yes the propeller slows down the wind, but it takes energy to change the particles speed and the propeller has to do work.

I have seen the videos that it is possible, but I fail to understand it. I was saying you can't do that to the explanation that it somehow just extracts the energy from the moving ground, you can't just extract energy like that, it would slow you down unless something else is compensating for it.

 

[rcgldr]

Since ground speed > thrust speed, then ground force can be < thrust force, even after losses due to power conversion. I already gave an example with some made up number in post #36:

post 36

And they are equal as I understand from the post 36?

But they wouldn't have to be. For an example including 10% loss of power and force:

80 lbs x 25 mph > 90 lbs x 20 mph
80 lbs x 25 mph = 2000 lb mph = 5.333 hp
90 lbs x 20 mph = 1800 lb mph = 4.800 hp

power output is 0.533 hp less than power input, but net force is +10 lbs (more thrust than opposing force from wheels).

 

[A.T.]

In the cart's frame:

power_ground_to_cart = ground_speed x wheel_drag
power_cart_to_air = air_speed x thrust

This means that the propeller takes as many joules from the cart's kinetic energy

Nope. Read again:

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.

You can't use the moving ground when driving a car to turn a generator without slowing the car down.

Yes I can. Read again:

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.

This one is not slowing down either, despite the fact that the reels are braking, in order to turn the blue gear:

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

Now replace the blue gear with propeller and the ruler with the airmass.

The diagrams I don't understand either. They already have 1,5 windspeed at the beginning? What happens at windspeed?

Draw it at windspeed yourself, for exercise. The apparent wind will be horizontal. It is even easier to produce a forward unconstrained force then, with the blade at less pitch.

I can't see how it can bypass taking energy from the cart to propel the air particles. They are standing still to the propeller and to move them it needs energy from the cart, no matter what gearing you use to make it slower or faster.

Nope. Read again:

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.

 

[A.T.]

Yes the propeller slows down the wind, but it takes energy to change the particles speed and the propeller has to do work.

No, in the ground frame, where the propeller slows down the air, the air is doing work on the propeller. Check that blade_force dot blade_velocity > 0 in the diagram. It doesn't take energy to slow down something.

 

[chingel]

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.

OK in the cart's frame the propeller isn't taking energy from the cart, but rather it is trying to accelerate the cart to match the speed of the ground due to the drag. In the end I think it applies in any case that any drag on the wheels will make the cart and ground try to match their speeds.

Of course the net force determines if the cart slows down or speeds up. I am trying to understand where does it come from. I am saying that you cannot just use gears on a car to multiply a force and then apply it back to the road. This wouldn't have an effect, it doesn't matter because the energy stays the same. So that the net thrusting force of the cart cannot simply come from the fact that the force from the wheels is multiplied and is greater at the propeller.

Draw it at windspeed yourself, for exercise. The apparent wind will be horizontal. It is even easier to produce a forward unconstrained force then, with the blade at less pitch.

I am having trouble. The wind created will be horizontal, but since at windspeed air is still to the cart and once the propeller is spinning it is moving faster than the air particles and therefore accelerating air particles instead of the other way around.

No, in the ground frame, where the propeller slows down the air, the air is doing work on the propeller. Check that blade_force dot blade_velocity > 0 in the diagram. It doesn't take energy to slow down something.

How is the air doing work on the propeller when it is at windspeed? From the ground's frame, they are moving at the same speed and then the propeller uses energy to change the speed of the air particles. The propeller doesn't know what it is doing, the air particles are sitting still in its way and it has to push them away, which takes energy.

 

[cjl]

OK in the cart's frame the propeller isn't taking energy from the cart, but rather it is trying to accelerate the cart to match the speed of the ground due to the drag. In the end I think it applies in any case that any drag on the wheels will make the cart and ground try to match their speeds.

Of course the net force determines if the cart slows down or speeds up. I am trying to understand where does it come from. I am saying that you cannot just use gears on a car to multiply a force and then apply it back to the road. This wouldn't have an effect, it doesn't matter because the energy stays the same. So that the net thrusting force of the cart cannot simply come from the fact that the force from the wheels is multiplied and is greater at the propeller.

The key part here (and what differentiates what the cart is doing from "multiplying the force and applying it back to the road") is that the air and the ground are going different speeds relative to the cart. As a result, you can indeed use a gear ratio to get a net force out.

I am having trouble. The wind created will be horizontal, but since at windspeed air is still to the cart and once the propeller is spinning it is moving faster than the air particles and therefore accelerating air particles instead of the other way around.

From the cart's frame of reference, yes. To analyze it further in this frame, the ground is moving at some speed relative to the cart. This turns the wheels, which are connected to the propeller. Clearly, this requires some force at the wheels, so this will cause a power drain on the cart equal to the force times the groundspeed. Depending on the efficiency of the propeller, some of the power is then put into the air. However, the air around the cart is still, and the air through the propeller is moving slowly compared to the groundspeed. Since power is force times velocity, the force from the propeller can be very large (since the velocity of air relative to the prop is small). Even if the power is smaller, the force is larger, and thus, the cart accelerates.

How is the air doing work on the propeller when it is at windspeed? From the ground's frame, they are moving at the same speed and then the propeller uses energy to change the speed of the air particles. The propeller doesn't know what it is doing, the air particles are sitting still in its way and it has to push them away, which takes energy.

From the ground frame, the air is traveling at some velocity, and thus has kinetic energy. After the cart passes, the air is moving slower relative to the ground, thus, the energy that was in the wind was transferred to the cart. The mechanism is somewhat irrelevant - all that really matters is that the wind was slowed by the cart's passing, thus by simple conservation of energy, the wind did work on the cart.

 

[A.T.]

OK in the cart's frame the propeller isn't taking energy from the cart, but rather it is trying to accelerate the cart to match the speed of the ground...

The propeller accelerates the cart to match the airspeed of the cart with the propeller pitch.

I am saying that you cannot just use gears on a car to multiply a force and then apply it back to the road.

The DDWFTTW cart doesn't apply it back to the road. It applies the greater force to the air, which moves slower than the ground in the carts frame.

So that the net thrusting force of the cart cannot simply come from the fact that the force from the wheels is multiplied and is greater at the propeller.

It comes exactly from that fact. Your "So" doesn't make sense, because you described a different mechanism before that sentence.

...once the propeller is spinning it is moving faster than the air particles and therefore accelerating air particles...

No, in the ground frame it is always slowing down the air particles. Your "therefore" doesn't make sense, because your conclusion doesn't follow from the premise. The propeller blade can have a higher speed than the air, and still slow down the air. See red tracer in the ground frame clip:

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

How is the air doing work on the propeller when it is at windspeed?

Draw the vector diagram for the blade when the cart_speed = true_wind. Check the dot product of blade_force and blade_velocity.

From the ground's frame, they are moving at the same speed and then the propeller uses energy to change the speed of the air particles. The propeller doesn't know what it is doing, the air particles are sitting still in its way and it has to push them away, which takes energy.

If you keep confusing yourself by jumping between reference frames, you will never get it.

 

[chingel]

From the cart's frame of reference, yes. To analyze it further in this frame, the ground is moving at some speed relative to the cart. This turns the wheels, which are connected to the propeller. Clearly, this requires some force at the wheels, so this will cause a power drain on the cart equal to the force times the groundspeed. Depending on the efficiency of the propeller, some of the power is then put into the air. However, the air around the cart is still, and the air through the propeller is moving slowly compared to the groundspeed. Since power is force times velocity, the force from the propeller can be very large (since the velocity of air relative to the prop is small). Even if the power is smaller, the force is larger, and thus, the cart accelerates.

I don't understand how does the force matter? Some of the power is given to the air that is motionless around the cart, why in the world does the force matter? I keep on reading it is important, but it doesn't make sense to me. If you use gears to multiply the force, the propeller moves slower and displaces less air, but any joules that the propeller will give to the air particles must come directly from the wheels. Several times equations have come up showing the power at the propeller is the same as the power at the wheels, meaning the J/sec the propeller gives to the air is exactly the same as the drag at the wheels. Now how can the force matter if it uses exactly as much energy for propulsion as it gets from the drag?

The DDWFTTW cart doesn't apply it back to the road. It applies the greater force to the air, which moves slower than the ground in the carts frame.

In the case of the car, at 100% efficiency, you can run the energy you get by drag through transmissions, slow it down, multiply the force several times, convert it to electricity or something else and use it for propulsion. The best you could get out of it is the same you took out of the car's kinetic energy by drag, no matter how many times you multiply the force in the process. But in the case of the wind cart for some reason the force matters, even though several times equations have been posted showing that the power is the same at both ends.

No, in the ground frame it is always slowing down the air particles. Your "therefore" doesn't make sense, because your conclusion doesn't follow from the premise. The propeller blade can have a higher speed than the air, and still slow down the air. See red tracer in the ground frame clip:

Ok but how it is slowing down the air? When the cart is at windspeed, the particles are just sitting there motionless around the cart. If the propeller whirls around, it has to push them away and uses energy for that. Yes the wind will move backwards but due to the work the propeller does. To me it seems that the energy to do the work can only come from the wheels, causing drag, and as the power equations show, the energy taken and the energy used up per second iare equal.

In the diagram, is the tangential blade velocity the speed at which the propeller is spinning? How did you get the blade force angle and length of the vector?

 

[A.T.]

I don't understand how does the force matter?

This guy here seems to have the answer:

Of course the net force determines if the cart slows down or speeds up.

Some of the power ...

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

...meaning the J/sec the propeller gives to the air is exactly the same as the drag at the wheels...

Comparing power (J/s) with force (drag) is meaningless.

In the case of the car,

Which car?

But in the case of the wind cart for some reason the force matters,

The net force always determines the acceleration.

... energy ...work ...energy ...work ... power...energy ... energy

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

In the diagram, is the tangential blade velocity the speed at which the propeller is spinning?

tangential_blade_velocity = radial_position * propeller_angular_velocity

It varies with the radial position and depends on the transmission ratio between wheels and propeller. I chose a simple example where it equals the cart speed, so in the ground frame the airfoil moves at 45° relative to driving direction.

How did you get the blade force angle and length of the vector?

The direction of the blade_force depends on the lift/drag ratio of the airfoil. In the diagram it is about 5, which is very pessimistic. Efficient airfoils achieve more than 20.

angle_apparent_wind_to_blade_force = atan(lift/drag)

The magnitude shown in the diagram is irrelevant, as it doesn't affect the fact that the air is doing work on the blade in the ground frame, and pushing it forward. If the magnitude is too low to overcome friction in the real world than you simply make the blades bigger.

 

[chingel]

What I meant is that since the cart is moving faster than the wind, which can be seen from the countless videos, there must be a net thrusting force coming from somewhere. What I am trying to say is that just multiplying the force from the wheels shouldn't matter and I am trying to get an explanation why it does.

Imagine you are freewheeling on a regular car, except it has no air resistance and can convert motion to electricity or do pretty much anything else at 100% efficiency. There is a machine on board that produces photons and it has a flywheel connected to the wheels. The device can shoot out photons with more energy at a lower rate, increasing the thrusting force per photon, or shoot them out at any other combination. Do you agree that in such a scenario the best that is achievable at 100% efficiency is that the cart will maintain it's speed?

In the case of the wind cart, the fact that the force is multiplied somehow causes a net thrusting force on the cart, despite the fact that the power at both ends is the same. Let's make the car shoot out photons with more force but at a lower rate, increasing the force. Does the hypothetical car start accelerating? Then why does the wind cart accelerate when you multiply force, after all, the energy lost by drag and the energy used for propulsion are equal in both scenarios?

Does the reference frame matter regarding the power at the input and output?

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

They should be still equal as far as I understood. If you just think about it, any energy used by the propeller to propel the air particles has to come from the wheels, because that is what moves the propeller.

Another question I have regarding the diagram. Shouldn't the path of the blade be at a right angle to the wind and vehicle speed when it is moving directly downwind? Or is the propeller supposed to not point directly upwind?

When the cart is at windspeed, you are right the wind will be horizontal and the wind will cause the blade to move forward, but, since in the cart's frame, the air is motionless around the cart and the propeller is moving around, the propeller is accelerating the air particles, not the other way around. Every action has an equal and opposite reaction, the propeller is at an angle and throws the particles in one direction and therefore the cart in the other. It takes energy to do that and the energy comes from the wheels. Any energy given to the air particles trying to propel itself is exactly the same energy that is taken from the cart, as the equations show, the joules per second at both ends are equivalent.

 

[cjl]

Any energy given to the air particles trying to propel itself is exactly the same energy that is taken from the cart, as the equations show, the joules per second at both ends are equivalent.

Yes, but the force isn't. Since the propeller is acting on the slow moving air, while the wheels are rolling on the fast moving ground (and power is force multiplied by velocity), the propeller is applying more force to the air than the ground is applying to the wheels (even though the power is the same).

I do think you're confusing yourself with reference frames here though. Pick one and stick with it - you can't jump around between them at will, especially since the energy and power balance is frame-dependent.

 

[A.T.]

there must be a net thrusting force coming from somewhere

net_thrust = prop_thurst - wheel_force

multiplying the force from the wheels shouldn't matter

It does matter becuse it makes:

prop_thurst > wheel_force

And that makes:

net_thrust > 0

Imagine you are freewheeling on a regular car, except it has no air resistance and can convert motion to electricity or do pretty much anything else at 100% efficiency. There is a machine on board that produces photons and it has a flywheel connected to the wheels. The device can shoot out photons with more energy at a lower rate, increasing the thrusting force per photon, or shoot them out at any other combination.

Why not add a flux compensator and some magnets, to make it even more complicated? It is simple classical mechanics that you don't get. So how does it help bringing photon thrusters into it? The air molecules are not massless and they are not produced on board.

You should read this very nice power analysis by uart based on molecules:
https://www.physicsforums.com/showthread.php?p=2063418

Does the reference frame matter regarding the power at the input and output?

Of course

They should be still equal as far as I understood. If you just think about it, any energy used by the propeller to propel the air particles has to come from the wheels, because that is what moves the propeller.

No, the wheels just turn the propeller. But the total motion of the propeller in the ground frame is not a pure rotation.

Another question I have regarding the diagram. Shouldn't the path of the blade be at a right angle to the wind and vehicle speed when it is moving directly downwind?

No, not in the ground frame.

Or is the propeller supposed to not point directly upwind?

The propeller axis is parallel to the true wind. The blades are not moving parallel to the true wind in the ground frame.

When the cart is at windspeed, you are right the wind will be horizontal and the wind will cause the blade to move forward, but, since in the cart's frame, the air is motionless around the cart and the propeller is moving around, the propeller is accelerating the air particles, not the other way around. Every action has an equal and opposite reaction, the propeller is at an angle and throws the particles in one direction and therefore the cart in the other. It takes energy to do that and the energy comes from the wheels. Any energy given to the air particles trying to propel itself is exactly the same energy that is taken from the cart, as the equations show, the joules per second at both ends are equivalent.

No. Read it again and repeat aloud:


In the cart's frame the cart doesn't have any kinetic energy

In the cart's frame the cart doesn't have any kinetic energy

In the cart's frame the cart doesn't have any kinetic energy

...

 

[rcgldr]

What I meant is that since the cart is moving faster than the wind, which can be seen from the countless videos, there must be a net thrusting force coming from somewhere.

The net thrusting force is due to a combination of the tail wind and the propeller. If the wind speed is +10 mph, and the cart is moving at +15 mph, then the relative speed of the wind is -5 mph with respect to the cart, while the relative speed of the ground is -15 mph with respect to the cart. The ground speed is 3 times that of the air speed. The cart is designed to take advantage of this difference in speed, since the wheels interact with the faster moving ground and the propeller interacts with the slower moving air. If the effective gearing of the propeller is 1/2 of the ground speed, it would be accelerating the wind from -5 mph to -7.5 mph (1/2 of the -15 mph of the ground speed). Since the speed is cut in half at the propeller, the force could be doubled and power output would be the same as power input (if there were no losses). Since there are losses, either the force and/or speed of the thrust from the propeller will be a bit less, but if the cart is efficient, the speed can be enough to generate more thrust than the opposing force from the wheels, and in the case of BB, where the propeller's pitch (no load thrust speed) is set to 80% of the ground speed, equality between losses and the opposing forces occur at about 3.5x wind speed.

Does the reference frame matter regarding the power at the input and output?

Since power = force x speed, and since the reference frame affects the speed, the reference frame affects the magnitude of the power, but it doesn't affect the calculated forces (or the net force).

 

[OmCheeto]

hmmm...

This is odd. Why, if the cart is traveling directly downwind, would the wheels experience lateral load?

4:30 in the following video.

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

Seems to me that the wheels would only experience lateral load if the cart was tacking into the wind.

 

[A.T.]

This is odd. Why, if the cart is traveling directly downwind, would the wheels experience lateral load?

It doesn't experience lateral load when it is traveling straight directly downwind. It experiences lateral load when it turns. He even explains that in the video: A bicycle tilts when it turns, so it doesn't have high lateral loads.

 

[chingel]

The Atom Man example in entirely different, he is actually saying that the generator uses less power to hurl those air molecules than the power that the generator delivers to thrust itself forward. Now if such a thing were true and I could somehow understand it, the increased power would indeed make the cart go forward.

He uses the thrust from the propeller to power the generator. Why does he multiply the propellers thrust with the vehicles speed to get the generators power? Shouldn't he multiply it by the propellers speed, i.e. the speed at which he propels the molecules?

With the ruler example, I can understand what happens. If the ruler is slippery and it takes a few seconds for the cart to catch up with the ruler, at ruler speed, when the ruler is stationary above the cart, the wheels that are touching the ground are going at speed x, the ruler also goes at speed x. Since the wheels are thinner in the middle, let's say their diameter is half of the parts' that touch the ground, they move at speed x/2. Since the big wheel is against the thinner parts of the wheels, it's surface also moves at speed x/2. If the big wheel and the ruler touch, the ruler is going faster than the big wheel, therefore friction will try to speed up the cart to match the ruler speed with the speed of the big wheel. I don't know how to apply this analogy to the wind cart, since at windspeed the wind is sitting still around the cart (in the cart's frame) and the propeller is moving faster, giving energy to the air, not as in the case of the ruler powered vehicle.

 

[A.T.]

The Atom Man example in entirely different,

It is exactly the same.

he is actually saying that the generator uses less power to hurl those air molecules than the power that the generator delivers to thrust itself forward.

The generator is not using power, it is generating power. Atom man is using that power to hurl molecules and create thrust that is twice the drag of the generator.

Why does he multiply the propellers thrust with the vehicles speed to get the generators power?

Because the generator works at the ground, so the ground speed is relevant for the generator power. He doesn't multiply the entire propeller thrust, just half of it, because he wants the generator drag to be only half of the prop thrust, so that he can still accelerate.

With the ruler example, I can understand what happens. If the ruler is slippery and it takes a few seconds for the cart to catch up with the ruler, at ruler speed, when the ruler is stationary above the cart, the wheels that are touching the ground are going at speed x, the ruler also goes at speed x. Since the wheels are thinner in the middle, let's say their diameter is half of the parts' that touch the ground, they move at speed x/2. Since the big wheel is against the thinner parts of the wheels, it's surface also moves at speed x/2. If the big wheel and the ruler touch, the ruler is going faster than the big wheel, therefore friction will try to speed up the cart to match the ruler speed with the speed of the big wheel.

Good so far.

I don't know how to apply this analogy to the wind cart,

As you correctly noted the top point of the blue gear will be moving slower than the ruler, if the cart moves at ruler speed. So the ruler will still push the cart forward. To translate this to the propeller you have to consider the point where the streamline intersects with the propeller blade. This point is also moving slower than the wind, if the cart moves at wind speed:

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

since at windspeed the wind is sitting still around the cart (in the cart's frame) and the propeller is moving faster, giving energy to the air, not as in the case of the ruler powered vehicle.

What do you mean by "not as in the case of the ruler powered vehicle"? You didn't even mention energy or power in our analysis of the ruler cart, so I don't know where you see the difference in that regard. You should stay consistent in the methods that you use to analyze the two systems, if you want to compare them.

Why don't you apply your energy arguments (that you use on the wind cart) to the ruler cart. You will find that the ruler cart is impossible, so your energy arguments are flawed.

You can also apply your velocity-matching arguments (that you use on the ruler cart) to the wind cart. You will find that is it trivially possible.

 

[chingel]

To translate this to the propeller you have to consider the point where the streamline intersects with the propeller blade. This point is also moving slower than the wind, if the cart moves at wind speed:

Could you explain it more a little bit? Where is the point, what streamline, where do they intersect?

 

[A.T.]

Could you explain it more a little bit? Where is the point, what streamline, where do they intersect?

In the http://www.youtube.com/watch?v=FqJOVHHf6mQ" the stream line is the horizontal column of air (blue with dots). The location where it meets the blade is moving slower than the air (in the ground frame), so the air is pushing forward on the blade. This corresponds to the top of the blue gear that moves slower than the ruler.

 

[chingel]

Why is the blade moving up? How long can it keep going like that? Is there a similar point when the blade is rotating instead of moving up?

 

[A.T.]

Why is the blade moving up?

For the airfoil the vertical axis in the diagram is the circumferential direction of the helix that the airfoil actually moves along. This is shown right at the begin.

 

[chingel]

When I look at a helix from the side shouldn't I see it going up, then down, then up again etc? But then again it shouldn't matter in which direction the blade is going, it is always at a right angle to the wind in the cart's frame of reference?

In that sense the cart could go faster than the wind, but it is still very confusing, because there has to be an explanation for the energy questions I had. The generator uses power to hurl air molecules, the cart gains kinetic energy, but the generator is getting power from the cart's kinetic energy in the first place (in the cart's frame the energy will cause the ground and cart to match their speed, it doesn't matter, the amount of energy is the same and the effect is basically the same, the energy makes the cart slow down relative to the ground in either frame).

I have a question, let's say we are in zero gravity and we have zero air resistance. We have a machine that always outputs 500W at 100% efficiency. It is trying to accelerate a 1 kg weight. If we use different gear ratios, will the weight have different speeds at 1 second after the machine started running? Since no energy is lost in the transmission, all the 500W will go to the weight's kinetic energy, is that right? At 1 second the weight's kinetic energy will be 500J no matter what gears you use, or not? If we use 2:1 gear ratio, the weight will move half the distance, but since the machine also feels less resistance due to the mechanical advantage, it will speed up until it does work at a rate 500 J/s. Is that right?

 

[A.T.]

When I look at a helix from the side...

The diagram in not from the side. For the airfoil the vertical axis in the diagram is the circumferential direction.

But then again it shouldn't matter in which direction the blade is going, it is always at a right angle to the wind in the cart's frame of reference?

Yes, but in the ground frame is moving at some other angle, depending on the radial position and transmission ratio. In the diagram it is 45°.

but it is still very confusing, because there has to be an explanation for the energy questions I had.

Your questions make no sense. You confuse energy/power from different reference frames. You confuse energy/power with forces. etc.

I have a question, let's say we are in zero gravity and we have zero air resistance. We have a machine that always outputs 500W at 100% efficiency. It is trying to accelerate a 1 kg weight. If we use different gear ratios, will the weight have different speeds at 1 second after the machine started running? Since no energy is lost in the transmission, all the 500W will go to the weight's kinetic energy, is that right? At 1 second the weight's kinetic energy will be 500J

No, you cannot have a constant 500W kinetic energy increase if you accelerate from rest. That would mean infinite force at the begin:

F = P / v

...no matter what gears you use, or not?

No, you said that you want 500W kinetic energy increase. That determines you gear ratio to provide the needed force (see formula above)

 

[rcgldr]

But then again it shouldn't matter in which direction the blade is going, it is always at a right angle to the wind in the cart's frame of reference?

Only at when the cart is at wind speed, and that ignores induced flow. In reality, at some speed below wind speed, induced flow will be zero relative to the ground and that is the only case when blade is moving at "right angle" to apparent wind. At faster than wind speed, the blade moves forward as well as at a right angle relative to the air.

In that sense the cart could go faster than the wind, but it is still very confusing, because there has to be an explanation for the energy questions I had. The generator uses power to hurl air molecules, the cart gains kinetic energy, but the generator is getting power from the cart's kinetic energy in the first place.

Energy is never obtained from the cart's kinetic energy. The power input and output are from the moving ground (earth) and/or the moving air depending on frame of reference. From the ground frame of reference, the air is being slowed down, so the air's kinetic energy is the power source. From the air's frame of reference, the Earth is being slowed down (a very tiny amount), so the Earth's kinetic energy is the power source. From the cart's frame of reference, it's a combination, and dependent on the cart's speed; once the cart is moving faster than the wind, then it takes kinetic energy from the earth, and adds kinetic energy to the air (relative to the cart).

Again using the cart's current velocity as a frame of reference:

power_input = ground_speed x ground_force_applied_to_wheels
power_output = thrust_speed x thrust_force_applied_to_air

For the BB, the propeller pitch is set to 80% of the ground speed. Because of losses in the power conversion through the drive train and at the propeller, the power output to the air is less than the power input from the ground. Assume the power losses are 10%, that means that 90% of the power_input is output to the air. The BB is very efficient otherwise, so the thrust force only needs to be a bit more than the opposing ground force for the BB reach it's max speed of 2.8x wind speed, where rolling resistance + aero drag + opposing ground force are equal and opposing thrust force. The 0.8 effective gearing allows the ground force to be multiplied by 1.25. For example, the losses may reduce the force multiplier to 1.12, but 12% more thrust force than opposing ground force allows the BB to go downwind faster than the wind. I don't know the actual numbers involved with forces and losses, only that under good conditions, the BB has gone 2.8x wind speed.

update - the last "official record speed" I see for BB mentions 2.8x wind speed. I don't know what the issue was for the reported 3.48x wind speed, other than it wasn't considered an "official record".

 

[Fun Value]

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...

Well, I am just a ninth grade Physical Science and Math Teacher, so I guess I will probably never have a half-decent brain. Is it ok for me to comment or ask questions on this site? I was thinking that if I could understand this, then maybe I could teach it to my students.
How can one expect a half-decent brain to understand it if a wholly-decent brain is not sure it works without actually doing an experiment to see if it will work?
I think it is reasonable to question whether it would work. I have taken several Physics courses and never recall coming across the problem. I think it natural for many to think that if a load is applied to wheels that are in some way matched to wind speed, the wheels would slow down. On the other hand all of my ninth grade students seem to understand how a sail boat can theoretically go much faster than the wind. When they asked me if it was possible to go faster downwind, I thought probably some sort of arrangement of gearing attached to a wind - mil should work. Yet as I think about even the videos (which I finally watched) and even as I think about the explanations, seems like sometimes I get it and sometimes I don't.

 

[rcgldr]

I think it natural for many to think that if a load is applied to wheels that are in some way matched to wind speed, the wheels would slow down.

The load on the wheels is not related to wind speed, instead the load is related to the torque it takes to drive the propeller. As mentioned before this load translates into an opposing force from the ground, and this force times the the ground speed (relative to the cart) is the input power used to drive the propeller. In the meantime, the propeller interacts with the slower moving air (and not the ground), whenever the cart operates with a tailwind. This allows an effective gearing where speed is divided and torque is multiplied to result in more force from the propeller but at a lower air speed. The torque it takes to drive the propeller is related to the mass and the rate of acceleration of the air affected by the propeller (and the efficiency of the propeller).

From a ground frame of reference, the thrust produced from the propeller slows down the wind, and that is the source of energy that is consumed by the losses in the cart which are related to rolling resistance, aerodynamic drag, and losses in the drivetrain.

An example of another device that operated on a similar principle is the Brennan torpedo, which used wires that were retracted to drive propellers that drove the torpedo forwards. It was geared so that the rate of wire retraction was greater than the torpedo's forward speed (relative to a water based frame of reference), which allowed the propellers to produce more thrust than the opposing force of the retracting wires:

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

Continuing with this comparason. If there was a stream of water and the torpedo was moving down stream (and steering wasn't an issue), then the wires could be attached to a fixed base, and the torpedo would move faster than the stream. For example, a Brennan torpedo moving at 15 mph in a 10 mph stream. From the torpedoe's frame of reference, wire speed is -15 mph, water speed is -5 mph, which allows the effective wire to propeller gearing to almost 3:1, multiplying the force, dividing the speed, ...

 

[chingel]

The multiplying the force explanation makes no sense to me. So if I have the 500W engine, will using a 2:1 gear ratio make the weight accelerate faster than using a 1:1 gear ratio? After all, it also multiplies the force. I would think it does not make it accelerate faster, since nothing is lost in the transmission, all the 500W go to the weight's kinetic energy. After 1 s the weight will have 500J of kinetic energy, it's speed will be E=1/2*m*v^2; v=(2/m)^0,5=2^0,5

F=ma, but if you think about it, the gear at the weight moves slower, therefore it feels less resistance from the weight, because due to the slower speed it is trying to accelerate it slower. Then the mechanical advantage makes the engine feel even less force. So the engine speeds up until it is doing 500J of work. Is that wrong or not?

No, you cannot have a constant 500W kinetic energy increase if you accelerate from rest. That would mean infinite force at the begin:

F = P / v

But that would mean that the force would have to be infinite for any power, since the v is zero.

 

[rcgldr]

The multiplying the force explanation makes no sense to me. So if I have the 500W engine, will using a 2:1 gear ratio make the weight accelerate faster than using a 1:1 gear ratio?

Do you drive a real car or motorcycle? Does the car accelerate faster in first gear than it does in top gear, and is the maximum speed you can go much lower in first gear than it is in top gear?

You're focusing too much on the power issue. The only issue with power is that total power output must be less than total power input since there are always losses with most real devices.

What allows the cart to work is that the wheels interact with the ground while the propeller interacts with the air. If the air is moving with respect to the ground, then the cart can take advantage of this when going downwind, because the speed of the air will be less than the speed of the ground, no matter how fast the cart is moving. If the cart was a perfect machine, there would be no limit to how fast it could go because the difference in speed between the air and the ground wouldn't be affected by the cart. A real cart does slow down the air, but how much it needs to slow down the air depends on how efficient the cart is. The cart uses gearing (advance ratio) to take advantage of the difference in speed between air and ground so that the combined force between the wind and propeller is greater than the opposing force of the wheels that drive the propeller, and that extra bit of force allows the cart to accelerate to a faster speed unitl the other opposing forces related to rolling resistance, aerodynamic drag, and drive train losses, cancel out that extra bit of force.

 

[Fun Value]

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.

Also, for some of us, probably most of us, it takes a lot of repetition until certain facts become concrete in our minds. Still, for me, when I am not thinking correctly, it is amazing that the wind powered car can go against (or should I say “against?”) the very wind that is propelling it - kind of a concrete foundation in my mind.
While I am at it, I think a number of students might have many other ideas. For example they might wonder if a vortex is created by the propeller thereby bringing air into the center at an angle from outside the system. Maybe someone will eventually be able to demonstrate whether this and many other things happen using a wind tunnel.
I think it is good for students to try to analyze such a problem, try to figure it out for themselves before having it explained so they can see some of those false foundations upon which they have been building.
Finally I hope that I will be able to remember why I had difficulty so I can teach others.
I can admit that some things in science have become so simple that I have forgotten what I used to think before I understood it – forgotten some of the reasons students don’t get it.