Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[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

 

[A.T.]

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.

Their first prototype had fixed pitch and went above 2x windspeed from a standstill. but the acceleration below windspeed was very low. Maybe your pitch was too high, the wind too low or the assumed inefficiencies worse then in reality. The self start is not really the key claim here anyway. The core question is if it can achieve a steady state above windspeed.

How did you calculate the propeller axial force and reaction torque below windspeed? I did some simulations using http://www.mh-aerotools.de/airfoils/javaprop.htm" libraries (propeller simulation based on blade element theory). This software cannot really handle the the situation at windspeed(zero airspeed) very accurately so I started the cart just above windspeed. I used the actual propeller geometry of the Blackbird and with the folowing parameters:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 2
mass[kg]: 295

it went close to 3 x windspeed in 4.5m/s wind.

WS = true windspeed
wheel_drag = retarding force on wheel needed to turn the prop
net_thrust = prop_thrust - wheel_drag
net_friction = aero_drag + rolling_friction

The pitch was variable and adjusted for best acceleration (see blade_angle bottom left)

 

[A.T.]

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 2
mass[kg]: 295

it went close to 3 x windspeed in 4.5m/s wind.

Since the efficiency parameters are estimates I also checked the sensitivity of the result to those parameters:

 

[OmCheeto]

Maybe your pitch was too high, the wind too low or the assumed inefficiencies worse then in reality.

Way too high according to your chart. I had a pitch of 45 degrees, based probably on what I could visually observe from the original mini-carts.

How did you calculate the propeller axial force and reaction torque below windspeed?

Most embarrassing. I searched for over 2 hours yesterday and could not find the simulation. But anyways, it's moot point, now that we know my prop pitch was incorrect.

 

[A.T.]

Way too high according to your chart. I had a pitch of 45 degrees, based probably on what I could visually observe from the original mini-carts.

The plotted angle is the blade angle at 75% prop radius. Since the blade is twisted it is much higher closer to the root. The transmission ratio between propeller_omega and ground_speed plays also a role here.

 

[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...

Same with me. I guess we should have more ego.
On the other hand, I think the original post suggested downwind, I think implying no angle, so it did seem interesting. I am to busy to spend much time thinking about it, and if someone can clearly and simply explain without trying to sound like a genius, I would like to hear about it.

 

[mender]

Essentially, it's an overdrive gearing system that uses two inputs (air speed and ground speed) to generate an output (cart speed).

See A.T.'s last post on page three and start with the "Under the Ruler Faster Than the Ruler" video.

 

[chingel]

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.

Just for the sake of the argument let's assume 100% efficiency.

In the cart's frame:

power_ground_to_cart = ground_speed x wheel_drag
power_cart_to_air = air_speed x thrust

Are these powers equal as I understand from the previous posts? This means that the propeller takes as many joules from the cart's kinetic energy as it gives to the air particles.

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.

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.

The diagrams I don't understand either. They already have 1,5 windspeed at the beginning? What happens at windspeed? The air is still, the propellers get turned by the wheels, which slows down the cart, the propeller speeds up the cart. With 100% efficiency the speed should stay the same.

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.

Yes, but what does it mean? The ship still keeps speeding up and the laser's power stays the same, doesn't it? Let's assume 100% efficiency at using the photons for propulsion. If the ship goes from 1 km/h to 2 km/h, the power of something doubles due to the formula, but the power of the input engine ie the laser doesn't change. The effect of the small speed increase on the amount of photons hitting is negligible How does the formula exactly work?

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).

When you put gears in there to make the propeller move slower, it displaces less air and also drags the wheels less. 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.

 

[rcgldr]

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).

When you put gears in there to make the propeller move slower, it displaces less air and also drags the wheels less. 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.

The input power = (ground speed relative to cart) x (ground force on the wheels)
The output puter = (thrust speed relative to cart) x (thrust force from the propeller)

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

 

[chingel]

And they are equal as I understand from the post 36? The joules per second it gets from the wheels is the joules per second the propeller uses to propel the air? Why doesn't this mean that any energy that the propeller gives to the air particles trying to propel itself, no matter how fast they are, no matter what gears you use, gets taken from the carts kinetic energy and therefore slows the cart down?

 

[mender]

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.

 

[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.