# Wind powered vehicle (Derivation)

1. Aug 1, 2011

### FireStorm000

This thread is for MATH regarding wind powered vehicles. I don't want to see your opinion. I want facts and math. PERIOD
This is my (attempted) derivation of what a wind powered car is capable of. I may have accidentally derived an infinite energy machine, so if someone would check my work and see what's wrong...
I started with the laws for drag and lift:
T = wing thickness
d = distance
$\rho$ = mass density of air
V = velocity of air relative to turbine (flow velocity)
CD and CL are the coefficients of drag and lift, respectively.
FD=$\rho$* CD * h * r * V2
FL=$\rho$* CL * h * r * V2
and the definition of work:
W = F * d

The power of the turbine is then:
Vtangential = r * $\omega$
P = FL * r * $\omega$, where $\omega$ is rotational velocity in rad/s
Substituting:
P = $\omega$ * $\rho$ * CL * h * r2 * V2

Power to overcome drag is:
P = FD * V
P = $\rho$* CD * h * r * V3

Thus maximum achievable velocity(assuming %100 efficiency):
$\omega$ * $\rho$ * CL * h * r2 * V2 = $\rho$* CD * h * r * V3
V = $\omega$ * r * CL / CD
V = $\omega$ * r * (Lift/Drag Ratio)
for a reasonable turbine lets say: $\omega$ = 1rad/s or 1/s, r = 1m/rad or 1m, L/D = 30 (values range from less than 1 through 60. 30 is considered a reasonable ratio)
V = 1/s * 1m * 30 = 30m/s

At anything less than that extra power is being generated. Changing context a bit, let's consider if we put this turbine on a carousel. Rotating said carousel gives us an initial velocity, thus power. According to my derivation, the power required to overcome drag is less than the power generated by the turbine. The power excess is given by:
Pdiff = $\rho$ * h * r * V2 ($\omega$ * CL * r - CD * V)

This is a problem. It appears I just created an infinite energy machine. To my (albeit slightly out of practice) eye, my derivation appears correct, however this violates half the laws of physics I know of. So what did I do wrong, or, in the unlikely event this is correct, how is this even possible?

Last edited: Aug 1, 2011
2. Aug 1, 2011

### Phrak

I don't follow this too well. I assume 'r' is the radius of some sort of turbine blade.

W = F dot d would be an equation for work where d is the projected distance in the direction of the vector F.

I think you may need two different variables. One for a turbine blade radius and another for distance of motion traveled in the direction of the resultant force, if I understand the meaning of 'r'.

3. Aug 1, 2011

### FireStorm000

yes, this if for a single turbine blade dimensions r and h with coefficient of drag CD, etc.

You are correct, however, the dot product simplifies to F*d because the lift is by definition perpendicular to air flow, and the turbine is assumed to be mounted such that the axis of rotation points into the wind.
Fixed. I was hoping people could follow that the blade follows a circular path of length d based on the radius and angle traversed, and the time derivative would be Vtan = r * $\omega$

4. Aug 1, 2011

### rcgldr

What is h in these formulas?

What seems to be missing is the "pitch" of the turbine. This is the distance wind would flow through the turbine per revolution in a no load situation.

For these carts to work, there needs to be some effective gear ratio between the induced wind speed and the ground speed. An upwind cart using a turbine needs a gear ratio greater than 1, while a downwind cart using a propeller needs a gear ratio less than 1. If the pitch is variable, and if there is 100% efficiency (zero drag) there's no mathematical limit to the speed of the cart (this is idealized and ignores issues like supersonic speeds), as the effective gear ratio approaches 1.

Getting back to the turbine case (which I assume is an upwind cart), as long as there is a true headwind (wrt ground) then wind speed > ground speed (wrt cart), and this allows the efffective gearing to divide wind speed and multiply wind related force when applied to the ground.

This was already done with a downwind vehicle (propeller, not turbine). Link to youtube video showing the "vehicle" accelerating from rest and eventually advancing against the turntable at a fairly fast pace:

Last edited by a moderator: Sep 25, 2014
5. Aug 1, 2011

### xxChrisxx

Problems are as follows:

Assuming %100 efficiency - nat a valid assumption.
Not taking into account the drag of the thing you are trying to drive, you've only looked at the drag from the wings.
I'm assuming h is chord.

What you are infact showing with the OP is that a wing/aerofoil produces more lift than drag, and that the system can generate power. Which shouldn't be surprising.

In reality you'd not only have to take into account the drag from the aerofoils, but the drag caused by the spinning blades presenting an effective 'face' to the wind. Worst case would be a solid disc 90degrees to the wind flow.

This is acutally what happenes when turbines are made too stiff/large for the application. They deflect air around them rather than across the blades.

6. Aug 1, 2011

### FireStorm000

h is the hight/thickness of the blade. h * r gives the cross sectional area A of the blade, as seen in the conventional version of this formula.

I'd just assume the cart is all electric, and 100% efficient, thus gearing becomes a non issue. I did take into account drag on the turbine blades, thus why I have a maximum speed.

I'm speaking entirely from an energy perspective here, in which gearing doesn't matter. In theory, the energy extracted from the airflow is independent of ground velocity, and depends only on the flow velocity across the turbine blades. I have thus far seen no reason that power extracted from the air depends on either the vehicle's or the air's relation in velocity to the ground.

You can't disprove an infinite energy machine with inefficiencies.
let's assume we're driving something with minimal drag compared to the turbines.
There will still be net energy production, which still means infinite energy.

again, h is blade thickness, r*h=A

This is true, but still doesn't address the fact that my equations don't show any relation to wind speed: If the vehicle moves in zero wind, it still produces more power than it consumes, which is the primary issue.

7. Aug 1, 2011

### rcgldr

Part of the dilemma is the 100% effciency. You've described a vehicle that can generate some net amount of power at any speed, but one that consumes zero power in order to be propelled at that speed.

Still there seems to be an issue with the math. Take the case of a spinning propeller in a gravity free environment (but one with air). The drag of the propeller is going to slow it down, regardless of the lift to drag ratio, as long as the drag isn't zero. I don't see how a turbine is going to behave any differently.

Last edited: Aug 1, 2011
8. Aug 1, 2011

### xxChrisxx

He's right that you can't disprove a theoretical PPM by mentioning efficiencies.
However, the scenario doesn't reflect the reality, for the reasons quoted.

You simply can't just ignore a vital power drain by assuming the thing you are driving has no drag.

Last edited: Aug 1, 2011
9. Aug 1, 2011

### Phrak

It's useful to view this in the frame of reference where the air stream is in motion and the turbine is at rest. To be clear, the plane containing the turbine blades is at rest in this frame.

Where d is the distance traveled by the blade per your defintion of d, d is perpendicular to the force due to lift, FL. FLd=0.

Last edited: Aug 1, 2011
10. Aug 1, 2011

### FireStorm000

To make more clear my infinite energy problem, scrap the cart, and picture a turntable with four turbines mounted on it, each hooked up to a generator. Lets say each turbine has four blades. Everything is electric(KE can be converted with 100% efficiency to Electrical energy, and with 100% efficiency back)

Now we invest energy Estart to get the turntable spinning. The max efficiency for the blades is at 10m/s, so a 4m radius turntable gives us a speed of 2.5rad/s. Now, the power extracted from lift on the blades is 16blades * 17W/blade = 270W. The power lost to drag on the blades from the oncoming air is 180W. Power lost to drag from the blades spinning is .18W. The drag from a spinning disk is negligible. Let us assume also that both the table and turbine blades are mounted such that the force of friction is very small. That still leaves us with 90W that apparently came from nowhere.

You misunderstand me. Airflow is parallel to the axis of rotation. The lift is applied tangent to the radius of the blade, and perpendicular to the axis of rotation, applying a torque to the turbine. Because the blade sweeps out an area also tangent to the radius and perpendicular to the axis of rotation, F dot d is nonzero, as both vectors are parallel.

11. Aug 1, 2011

### Phrak

Lift on a body is by definition perpendicular to velocity of a body through fluid.

Somehow I don't think we are communicating well. I'll leave this little problem to someone else. No worries.

12. Aug 1, 2011

### FireStorm000

Oh, I see what you're saying. I'm calling the drag from the turbine spinning drag, since it provides no useful energy. I'm assuming that the blades aren't spinning very fast in relation to the air moving parallel to the axis of rotation. Thus we can simplify the velocity of air moving across the blades to be equal to that of the air velocity relative to the turbine.

13. Aug 1, 2011

### xxChrisxx

14. Aug 1, 2011

### A.T.

With this definition of airflow direction, you cannot assume L/D=30. L/D = 30 is reasonable for the airfoil with respect to the airflow at the airfoil. To achieve that L/D in respect to the prop axis you would need a negative D at the airfoil, for some combinations of omega and V.

And that's why your math breaks down for low axial airflow velocities, where the tangential component is not negligible, compared to the axial component.

Last edited: Aug 1, 2011
15. Aug 1, 2011

### rcgldr

One issue is that a turbine extracts more power from the air than it generates. The speed of the air in the vicinty of the spinning turntable is going to get increased, reducing the relative flow to the turbines, and the overall drag is going to slow down the turntable (the rate woudl depend on it's inertia).

As mentioned above, Betz law states that at best, a turbines power output = 16/27 of power input, about 59.3% efficiency. The downwind powered vehicles use propellers, where efficiency can be around 80%. Using the 2.8x wind speed of the BB cart as an example, from the cart's frame of reference:

W = wind speed
F = force at the ground used to drive propeller

power input from ground = 2.8 x W x F
power output to air at 72% overall efficiency = 1.8 x W x 1.12 F

This results in a net forwards force of .12F, which is consumed by the carts rolling resistance at 2.8 W and aerodynamic drag at 1.8 W.

For an upwind cart to go upwind at -1 x wind speed

W = wind speed
F = equivalent of force at the turbine used to drive wheels

power input from headwind = 2.0 x W x F
power output to ground at 52% overall efficiency = 1.0 x W x 1.04 x F

A net forward force of .04 F, consumed by the cart's rolling resistance at 1.0 W and aerodynamic drag at 2.0 W. The turbine would need to be relatively large to deal with the losses of the cart to acheive -1x wind speed.

Last edited: Aug 1, 2011
16. Aug 1, 2011

### willem2

I don't think Betz' law applies here. Betz' law tells us it's impossible to get all the energy out of the wind that passes through the turbine, but this is not because of drag, but because the air still needs to have velocity left to go through the turbine.

You might have to use a turbine with at least 1/0.59 times the area because of Betz' law, and such a turbine will have more drag as well, but there's no minimum amount of drag.