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Wind Power Vehicle Traveling Down Wind Faster Than The Wind

by yn3
Tags: faster, power, traveling, vehicle, wind
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chingel
#37
Jun9-11, 06:17 AM
P: 261
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.
#38
Jun9-11, 10:06 AM
P: 4,021
Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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:




Quote Quote by chingel View Post
energy input has to equal energy output
But force output can greater that force input.

Quote Quote by chingel View Post
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.
#39
Jun9-11, 10:23 AM
P: 4,021
Quote Quote by rcgldr View Post
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
#40
Jun9-11, 06:11 PM
P: 11
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 im 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, wouldnt 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? im sure if that make sense...
rcgldr
#41
Jun9-11, 10:51 PM
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Quote Quote by poont2 View Post
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.
#42
Jun10-11, 04:14 AM
P: 4,021
Quote Quote by poont2 View Post
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.

Quote Quote by poont2 View Post
But one thing im 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, wouldnt 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)

Quote Quote by poont2 View Post
does this meen that the force from the wheel would greatly overpower the this effect? im 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
#43
Jun10-11, 06:06 AM
P: 261
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.
#44
Jun10-11, 06:54 AM
P: 4,021
Quote Quote by chingel View Post
Regarding power = force x speed. Is the speed in the equation the speed of the object that force is applied to?
Yes
Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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.



Quote Quote by chingel View Post
You can't just use gears to do more work,
Correct. But you can use gears to get more speed:



Quote Quote by chingel View Post
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?

Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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
#45
Jun10-11, 03:13 PM
P: 261
Quote Quote by A.T. View Post
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.
#46
Jun10-11, 05:07 PM
P: 4,021
Quote Quote by chingel View Post
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.
Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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

Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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:



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



Here a nice animation that transforms one into the other:




Quote Quote by chingel View Post
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.




Quote Quote by chingel View Post
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.

Quote Quote by chingel View Post
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:

rcgldr
#47
Jun10-11, 05:25 PM
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Quote Quote by chingel View Post
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 space ship increases, the number of photons per second that hit the space ship "sail" decreases.

Quote Quote by chingel View Post
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
#48
Jun11-11, 12:33 PM
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Quote Quote by OmCheeto View Post
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
#49
Jun11-11, 10:12 PM
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Quote Quote by Redbelly98 View Post
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
#50
Jun11-11, 11:25 PM
P: 206
Quote Quote by RCP View Post
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.
#51
Jun12-11, 03:08 AM
P: 4,021
Quote Quote by OmCheeto View Post
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 JavaProp 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)
Attached Thumbnails
183_plots.png  
A.T.
#52
Jun12-11, 06:14 AM
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Quote Quote by A.T. View Post
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:





Attached Thumbnails
170_Cd.png   170_Crr.png   170_trans_eff.png  
OmCheeto
#53
Jun12-11, 01:48 PM
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Quote Quote by A.T. View Post
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.
#54
Jun12-11, 02:39 PM
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Quote Quote by OmCheeto View Post
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.


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