Forget planes on treadmills, here's a sail-cart out-sailing the wind!

Miscellaneous and Personal Stuff I Must Share
21

Hey, one at at time we’ll get them all. :slight_smile:

Yes and yes. A simple change in gearing will do it and the propeller does indeed have to drive the wheels when going upwind.

Now, a simple gear change will do just as I say, but it won’t create a “practical” (if that term has any meaning in this case LOL) upwind vehicle. In the case of the Blackbird, changing the gearing (or simply placing smaller drive wheels on it) will cause it to back up directly into the wind.

Here’s what going on in this downwind/upwind gearing issue: when the vehicle has a relative tailwind (startup phase) the angled propeller blade are actually trying to rotate the rotor in the opposite direction than the wheels are forcing it to turn. Check out the following video: Pay particular attention to the direction of propeller rotation and blade pitch at ~0:15 seconds into the video. Considering that the vehicle is in a pronounced tailwind at this point (see 0:20 for wind direction) it’s not hard to figure out that this wind is trying to turn the angled blades CCW (from the rear). The bluff drag of the entire vehicle acting through the wheels is forcing the rotor to turn CW.

The winner in this ‘force on the blades vs force on the wheels’ battle is determined by the gearing between the prop and the wheels. Adjust the ratio in one direction and the blades will win, the rotor will turn CCW and the vehicle will go (back up) directly upwind. Adjust it the other way and the wheels win the battle and force the rotor to turn CW and downwind we go.

To make a truly efficient upwind vehicle, the propeller blades need to be replaced with turbine blades (different camber location on the blades). We will be building such a rotor over this winter and intend to set a new directly upwind record in the spring. The current record is about 60% of windspeed and we believe we can rather easily bump this up to greater than 100%

That is about as simple and perfectly accurate description as I have ever heard. Kudos – hope you don’t mind if I steal that. :slight_smile:

JB

22

I expect royalties in the amount of 15% of all revenues generated by your use of my phrasing. :stuck_out_tongue:

I see what you meant by “simple gearing change” now, thanks.

23

I apologize that my link in my most recent post was screwed up. Here is the working link to the video I was referring to.

24

You know what, I’m starting to think this idea is actually crazy enough to work. Consider:

  1. You have a wind turbine on the ground. Plainly it can extract energy from the difference between its airspeed and its groundspeed (the latter being zero).

  2. You replace your wind turbine with a sail-cart that has a dynamo attached to its wheels that will dump its power into a battery. Plainly it will run downwind and transfer energy to the battery. There will be a steady-state groundspeed depending on the wind speed and the dynamo. We will conduct our experiments in a mathematically ideal world where energy conversions are 100% efficient. :smiley:

  3. You replace your sail with a propellor that is gimmicked not to be spun by the wind (a ratchet will do). Plainly the cart will still run downwind. A propellor may make an indifferent sail but it is still superior to no sail at all. Again, there will be a steady-state groundspeed and a net transfer of energy to the battery.

  4. You remove the battery and connect the dynamo to a motor that drives the prop. The energy that formerly went to the battery is now transferred to the cart as kinetic energy. Plainly the cart runs downwind faster than in 3.

  5. You dispense with the whole electrical generation system and couple the wheels mechanically to the prop. There is no conceptual difference from 4.

The obvious question arises as to what the steady-state groundspeed is. We know that in the case of a sail-cart it cannot be greater than the windspeed (that is, the directly-downwind vector component cannot be greater than the windspeed). But we have a reason for this: a sail experiencing no airspeed generates no thrust. To assume that the limit speed of the propellor cart is also the windspeed is to beg the question. So let’s consider further…

a) A given prop-driven aeroplane at a given power setting has a steady-state airspeed.

b) The same aeroplane’s groundspeed when heading directly downwind is simply the sum of its airspeed and the wind speed.

c) The aeroplane is experiencing a subjective headwind though - aeroplanes that do not are indifferent fliers. :smiley:

d) If the aeroplane be granted some means of extracting energy from its groundspeed, it is plain that both its groundspeed and its airspeed will drop. It is not a given, though, that the airspeed will fall to zero.

e) If the extracted energy be fed back into the prop, the groundspeed and airspeed will increase again (not necessarily back to its original speed, but that needn’t matter).

f) There is plainly a steady state where all of the energy spent on maintaining airspeed is being extracted from groundspeed. This is not a free lunch - we have groundspeed to spare only because we are travelling with the wind. But the aeroplane is still experiencing a subjective headwind.

Conceptually we could model this aeroplane as a kite-like device, except that instead of being held in place by the string, it is carrying an infinite supply of infinitely thin massless inextensible string (available from any good mathematical concepts store :smiley: ) on a spool coupled to the propellor. Thus as the plane is blown downwind the unwinding string spins the spool which turns the propellor to maintain airspeed, allowing the plane to face downwind and make positive airspeed.

But owing to the high price of massless inextensible string, it’s simpler to let the plane fly just low enough for its undercarriage to run along the ground and couple the wheels to the prop mechanically.

But as we’re taxying, the wings are now a waste of space…

Counter-instinctual is right, but the more I think about it the more plausible it sounds. :cool:

25

The friction force between the wheels and the ground (theoretically) occurs where V=0. The equation for Power is Force*Velocity. Since V=0, it is impossible to derive any power from the wheels. I’ve read the thread, but haven’t had the time or inclination to investigate myself, so I can’t say how it works. However, anyone claiming that the wheels are driving the prop resulting in acceleration is either (1) crappy at explaining things or (2) full of shit.

26

I truly don’t follow you here.

If what you are saying is that given a high enough coefficient of friction, the velocity of the portion of a tire in contact with the pavement will be zero relative to the pavement, I agree – but don’t follow the relevence.

Except the velocity of the vehicle isn’t zero in the scenario I posted. It’s velocity is 55ft per sec.

Surely you wouldn’t argue that a traditional automobile is doing no work when it’s traveling 55ft per sec just because in the frame of the road the velocity of the point at the bottom of it’s tires is zero.

Let’s work on (1) together, 'cause (2) won’t work out well for you if pick that one.

JB

27

No, the velocity of the tire relative to the ground at the point of contact is 0. The relevance is that if velocity at the point of contact is 0, is that there is no possibility of net work being done by the friction force. This is high school physics.

Of course not. The engine is applying a torque to a rotating axle. Torque*rotational velocity=work. I whole heartedly deny that, using the normal assumptions of friction, the ground is doing work on the car.

28

You may want to review your high school physics. I’m not sure why you’re so caught up in the importance of v = 0. The relative velocity between the surface and a single point on the wheel is zero instantaneously during contact, but non-zero before and after.

What’s especially peculiar, is that you understand the relationship between work, torque, and rotational velocity. Well… the ground applies a torque to the vehicle’s axle.

Thought experiment. You have a wheel powered generator and the SDMB’s favorite physics contraption: a treadmill. If you touch the generator’s wheel to the surface of the treadmill (you are holding the generator stationary in your frame of reference), would you not agree that the treadmill would spin the wheel and your generator would produce power? The treadmill is doing work on your generator. How is that different from the mindboggling “DDWFTTW vehicle.”

29

I think I’ve found an analogue that will make it a bit more intuitive.

Consider a vehicle that has a sail attached to a mast. The mast is mounted on a dolly that means it can slide from one end of the vehicle to the other (whilst remaining upright).

The vehicle sets off downwind, starting with the sail in the forward position, but as it moves, a rope wrapped around the rear axle winds the mast and sail along the rail toward the rear of the vehicle. Now, not only is the wind pushing the vehicle downwind, the vehicle is also pushing the sail upwind. The result has to be an increase in forward speed.

At any one moment, the vehicle with the propeller is equivalent to one with a conventional sail being moved backward against the wind.

30

Right, but the only velocity that matters is the velocity at the application of the force. Since it is 0, that means no net work. If you treat the car as a system, no energy is lost from friction and no energy is gained. That’s the point. It is impossible to derive energy from the ground. The energy has to come from the wind.

Yes, but it isn’t doing net work, it is just converting one type of energy to another (linear to rotational).

Because the earth isn’t a treadmill? If your frame of reference were the wheel in this case, it would be no different from a rocket powered car. There would appear to be no net energy transfer from the ground to the wheel. All the energy would come from the force of your hand * the linear velocity.

I don’t deny that you can use a treadmill or some such thing to transfer energy. I deny that you can use static friction to transfer energy from the ground to a system in the reference frame of the earth.

31

Wow – just wow.

JB

32

I don’t find that to be a good analogue, although I’m beginning to believe that the overall concept is possible.

In your analogue, you aren’t providing any motive power from the wheels to the sail, but the “air cart” designers are. A propellor, although it has airfoil elements, is not just a passive surface like a sail. The wind isn’t just pushing against the propellor like you would a sail surface.

33

What formulation of work are you using that is dependent on the velocity? We can consider work as a change of energy or evaluate work done by a mechanical force (force x distance). In either case, terms may exist which are completely independent of instantaneous velocity. In the problem at hand, the best work analysis is a torque applied through a revolution.

In the end the energy does come from the wind, but the ground is used to convert it.

It is moving energy (work) from a larger system, the entire craft’s kinetic energy, to a smaller system, the massless axle and generator. From the generator’s perspective, the ground is doing work (by applying a torque).

The system isn’t in the reference frame of the earth - it’s in the reference frame of a mass of air moving over the earth. In the example, the moving surface of the treadmill is earth’s frame of reference. My hand and the wheel are the wind and craft’s frame of reference.

Again, as you mentioned earlier, the energy is coming from the wind (my hand), it is just that part of that energy is harnessed via a rolling wheel and static friction. The energy can then be used to push off against my hand, or against the wind via a propeller.

34

Intellectually, I’m convinced this is possible. But it won’t be real until the Mythbusters have done it.

35

Yes I am. In the device I described, the rear axle winches the mast backwards toward the rear of the vehicle. Moving the sail backwards pushes it against the wind. OK, only a bit, but we’re talking about principles, not practicalities.

Granted, that can only be done for a short time, until the mast reaches the rear of the vehicle, which is why a propellor works better - it’s equivalent to a continuously moving sail

36

There’s not much point as it’s already been done.

37

OK, I just made this out of Lego.
Here is an example of an object that moves forward faster across a surface than the object that is pushing it.

You push the vehicle with the rack, moving the vehicle forward, but the wheels drive the rack backward - net result, the vehicle moves forward faster relative to the ground than the rack pushing it is moving relative to the ground. Can I go to bed now?

38

treis – let’s get practical for a moment and leave the theoretical behind.

If we place a rubber tire linked directly to a generator on the end of an arm and lower the tire to the pavement from a steadily propelled platform, can generate electricity?

Thanks

JB

39

The problem is that people will think of that as slowing the vehicle down by exactly the amount you would gain from the generator.

People are thinking of this as a vehicle trying to propel itself using power extracted purely and solely from its own motion. Obviously that would be impossible, but it isn’t one of those cases.
There is a net energy input from the wind, which could be exploited if this was a stationary generator and the wind was passing by it, but can still be exploited by vehicle being propelled by it, relative to the ground.

40

That’s fine, and under most circumstances it’s true.

My question to treis it just meant as a starting point since we are so far apart at the moment.

JB