Over at Good Math, Bad Math, Mark has a takedown of a device purported to move directly downwind, faster than the wind. Wind-Powered Perpetual Motion (and Dave Munger thinks he’s wrong.)
Here’s the video
The objection is simple: when you are traveling at the wind speed, there is no more wind in the cart reference frame, so there’s no force. The treadmill analysis is flawed.
If you’re testing a wind powered vehicle, then in a closed, windless room, putting the vehicle on a treadmill moving at 10mph is not the same thing as putting the vehicle on a stationary surface in a 10mph wind.
By putting it on a treadmill, you haven’t recreated the real-world situation — you always have your wind, and the treadmill doesn’t remove that. You never test the condition of having the wind relative to the cart drop to zero. So while it’s not faked, it’s still a sham.
It shouldn’t be hard to engineer a device such that the wheels rotate faster than the propeller, i.e. whatever the propeller’s rotation rate is for a wind of speed X, the wheel edges move faster than X. Since the wind is always present, the cart will move forward on the treadmill moving at X. Even uphill.
My question is this: if this works, at what speed does the cart stop accelerating?
UPDATE: Or with no wind present, as in the test (On the first viewing I thought they had a fan turned on) what you’re doing is converting treadmill kinetic energy into propulsion by turning the propeller. But you don’t need to have much propulsion to move forward, even uphill. Not a valid test.
Update, Mark II. See the comments — I was viewing this from the mistaken notion that the propeller was acting as a turbine while on the ground and at low speed, which isn’t the case.
This has the implication, I think, that the cart must have enough mass to ensure that the propeller acts as a propeller. My question of what the maximum speed is still stands, because I’m sure it involves fluid mechanics and that’s not something I’ll win should I tangle with it.
I’m confused. Since a treadmil set at 10mph in a still air room is the same as traveling down a road at 10mph with a 10mph tailwind, why do you say “not a valid test”? Simple principles of inertial frames of reference say otherwise. If I put you in those two situations on a bicycle and had you close your eyes, how would you know the difference?
JB
The propeller is going the other way — that’s one thing that is bothering me. With a tailwind, the propeller moves, say, clockwise. But this is the reverse. The propeller moves anti-clockwise to generate thrust, which I don’t think is the same as a simple translation of coordinate systems as with a bicycle.
Hi swansont.
I have a couple comments.
First, inertial frames of reference aren’t defined by any particular mechanism placed within the frames (in this case the bicycle), but rather by the relative motion between objects. What Galileo, Newton and Eintein have explained is that *no mechanism* can differentiate between the frames. This means that there is no instrument or test which can tell us if the wind blowing by us is caused by the ground moving or the air moving — it’s simply impossible. This is why we can test and demonstrate the cart on a treadmill — the frames are identical.
On to the cart mechanism: (all clockwise and counterclockwise references will be when looking from the rear of the cart)
The prop on the cart is geared to the wheels. If you put the cart in a tailwind and lift the wheels off the ground, the prop will spin CCW. In this mode it’s just acting like a typical wind turbine that you see generating electricity up on the hill. Put the cart on the ground in the same tailwind, and the wind begins to blow the cart forward which turns the prop CW.
To repeat, the spinny pinwheel thingy on the back of the cart never acts as a turbine as long as the wheels are rolling on the ground. It always acts as a propeller.
JB
PS: I’m the guy in the above video BTW.
I’m the guy holding the camera. I’m curious, aside from our cart, what other experiment do you think you could devise to distinguish the difference between a moving road in still air vs. moving air over a stationary road? Galileo, Newton, and Einstein could think of none (and they were pretty confident no one else could either).
Keep in mind words like “moving” and “still” are all relative. Even the road in front of your house is only stationary relative to the planet it’s on (which happens to be spinning and hurtling through space).
The prop on the cart is geared to the wheels. If you put the cart in a tailwind and lift the wheels off the ground, the prop will spin CCW. In this mode it’s just acting like a typical wind turbine that you see generating electricity up on the hill. Put the cart on the ground in the same tailwind, and the wind begins to blow the cart forward which turns the prop CW.
OK, this is what I was missing, and is absolutely crucial (IMO) to the explanation. I was envisioning the turbine action at low speeds.
“My question is this: if this works, at what speed does the cart stop accelerating?”
At the speed where equilibrium of forces is reached. All the cart is doing is taking the relative motion between air and ground, and “gearing it up”. The prop has a theoretical speed through the air, defined by its pitch and its speed of rotation. The prop is geared to the wheels, so it is forced to turn as the cart advances. If the prop is turning at a theoretical speed through the air of, say 10 mph, and the actual speed of the cart is only 8 mph, the prop generates a force that accelerates the cart forwards. Once the theoretical prop speed and the real prop speed through the air are equal, the cart will just keep up a steady speed.
let’s say the gearing between prop and wheel is such that the theoretical speed of the prop through the air is half that of the cart along the ground. In that case, if there is a 10 mph tail wind, equilibrium will be attained when the cart reaches 20 mph, since at that speed that prop will be moving through the air at 10 mph. (Of course that is neglecting losses due to friction and unwanted drag. The real speed will be less, but by using efficient bearings and propeller, the losses can be kept very low).
> My question is this: if this works, at what speed does the cart stop accelerating?
If the propellor can act on the air without “stirring it” (that is to say, the only kinetic energy it imparts to the air is that due to the change in air’s velocity along the direction of motion), infinitely fast. Of course, in our 3-D world, pushing on air in this perfectly efficient way is impossible. Associated math to follow:
Assume:
Car with wheels and a propellor on top.
Car’s wheels do not slip on the ground, and their bearings turn with no friction.
Propellor is much larger than rest of car, so interaction with the wind dominated by propellor.
Any batteries, motors, or generators mentioned are 100% efficient. I am using motors and generators because they make calculations easy. Gears are hard.
1-D problem
Velocity of wind relative to ground: Vw
Velocity of car relative to ground: Vc
I’m gonna start with two trivial examples to warm up, then get to the real deal.
1) Wheels free-spinning. Propellor free-spinning.
Equilibrium solution: Vc = Vw
2) Wheels free-spinning. Propellor driven by motor to push against wind, with energy supplied by MAGIC!
Equilibrium solution: Vc > Vw
3) Propellor driven by motor to push against wind, with energy supplied by car’s battery. Generator connected to wheels to charge car’s battery.
Equilibrium (constant velocity, net power into batter = 0) solution? Let’s work it out!
For convenience, we’ll work in the intertial frame of the car (with the positive direction defined as being opposite the direction the ground is moving):
The propellor encounters some mass of wind per unit time dMdt, and changes its velocity by some amount -D.
The resulting force on the car (equal and opposite, and all that) is +D*dMdt.
From the change in the wind’s kinetic energy, we calculate the power supplied to the wind is
Power = (1/2) * dMdt * ( ((vw-vc)-D)^2 – (vw-vc)^2 ).
You should draw yourself a picture to convince yourself my sign convention is correct.
(In the above, we have neglected any sort of turbulence, etc by assuming the only energy imparted to the air is that associated with the change in its net momentum along the direction of motion. Those neglected effects will degrade the performance of the system, and this is a very idealized model of air, but hey – we’re physicists, after all.)
Now, what is the available power supplied to the battery from the generator? In equilibrium the force on the car is zero, so we have a force -D * dMdt and a velocity of the road of -Vc, and a resulting
Power = force * velocity = D * dMdt * Vc.
In equilibrium, the power from the generator must equal the power driving the propellor, so
(1/2) * dMdt * ( ((vw-vc)-D)^2 – (vw-vc)^2 ) = D * dMdt * Vc
work through algebra to find
D = 2 * Vw
Holy cow! There’s no condition on Vc! You can go as fast as you damn please directly downwind! (In this overidealized physics version). To get a finite velocity, we will have to introduce a bit of loss into our system, which we will model as a power loss which is linear in the difference in the velocity between the car and the wind. To make the equations work out nicely I’ll take this power to be L * D * dMdt * (vc-vw). This leads to a revised energy equation:
(1/2) * dMdt * ( ((vw-vc)-D)^2 – (vw-vc)^2 ) = D * dMdt * Vc – L * D * dMdt * (vc-vw)
which we solve to find
Vc = Vw * (1+L)/L – D/2L
Clearly, to get large velocities, we need L small (pushing the air without stirring it up, for example) and D small – the propellor causes a small change in the air velocity.
At large L we have the expected limit Vc -> Vw.
As L -> 0, we have Vc -> (Vw – D/2) / L.
Counterintuitive? Certainly. C’est la vie.
Exercises for the reader:
Check my math. Calculate in your favorite reference frame. Calculate the relativistic version. Develop a better model of the losses in the system than the ad-hoc method employed above.