Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[spork]

Because there is no reason it should.

That is of course a completely correct and accurate answer. But I know you realize people get stuck thinking it should because they've stored an answer to a similar problem and then given it no more thought. The similar problem would have you using the wheels to turn a prop that pushes you forward, but not in the wind.

 

[chingel]

Again, keep in mind that we don't engage the propeller at wind speed. It's geared to the wheels from the very start. But to answer your second question - "Why doesn't all the energy used for propulsion of the air particles cause exactly as much drag at the wheels?" - that's because the cart is moving over the ground faster than it's going through the air. 10 lbs of thrust x 10 mph (through the air) is how much energy we need. 10 lbs of thrust x 20 mph (over the ground) is how much energy we get. The wheels constitute the long end of the lever. They move further over the ground than the prop moves through the air. Like any lever, we can get more force over a short distance at the shorter end of the lever.

But if we did engage it at windspeed, it should also work?

How did you get the numbers that 10 x 20 is how much energy we get?

Another thought I had, where the propellers are parallel to the wind for example. The wind pushes the cart up to windspeed, and if it is efficient enough, it will get up to almost exactly windspeed with the propeller doing a lot of work. So basically the wind pushes the cart, but it also pushes the propeller, because they are connected to the wheels. If we then direct the propeller backwards, the force will push the cart forward, but then the wind isn't pushing on the cart anymore, it is offering a little air resistance because the cart moves faster than the wind, the propeller does work against the air, getting it's energy from the wheels, which should make the cart slow down, or not?

I am still wondering how does the fulcrum example work with the ruler cart, where the middle of the wheel is on a shorter section of the fulcrum? The fulcrum doesn't apply there for some reason, but why? Or does it still apply?

 

[rcgldr]

But if we did engage it at windspeed, it should also work?

Yes, but the prop makes a lousy sail, so it gets to windspeed quicker if you let the prop do it's job while getting up to wind speed, although the initial start is always slow.

How did you get the numbers that 10 x 20 is how much energy we get?

That's the input power value, 10 lbs 20 mph = 200 lb mph = 5.333 horsepower = 3977 watts. The energy per second would be 2933 lb ft or 3977 Newton meters (joules). The mentioned output power is 1/2 of that (10 x 10).

If we then direct the propeller backwards, the force will push the cart forward, but then the wind isn't pushing on the cart anymore, it is offering a little air resistance because the cart moves faster than the wind, the propeller does work against the air, getting it's energy from the wheels, which should make the cart slow down.

Although the force at the wheels is an opposing (braking) force, the effective gearing multiplies the force at the wheels and divides the speed. In the example above with a 10 mph tailwind, the input force and input speed from the ground to the wheel was 10 lbs, 20 mph. If there were no losses, the output force from the prop to the air could be 20 lbs, at 10 mph, if the tailwind was greater than 10 mph, but with a 10 mph tailwind, as the cart approaches 20 mph, the output force to the prop would decrease to zero, which would also decrease the opposing force on the ground to zero for a zero loss cart (since the wheels are now driving the propeller with zero force).

For the BB, effective gearing from wheels to prop is .8 = 1/1.25, so the force is multiplied by 1.25 and speed divided by 1.25 before losses. A zero loss cart could achieve 5x wind speed with these numbers, but we also know it's actual speed in one of it's runs was "only" 2.8x wind speed. With a 10 mph tailwind, that means 28 mph ground speed. Input power could be = 28 mph x 100 lbs, output power = 22.4 mph x 125 lbs in a no loss case. However there are losses. So output power might be 20 mph x 119 lbs (15% loss). That's 19 lbs of net forward thrust (used to overcome rolling resistance and 18 mph worth of aerodynamic drag), with the tailwind being slowed down by 2 mph, the true energy source from a ground based frame of reference. That 19 lbs of net downwind force is the result of the 2 mph upwind change in airspeed, regardless of the frame of reference. From a ground frame of refernce, the input power from the wind being slowed down is 19 lbs x 10 mph = .507 hp = 378 watts.

I am still wondering how does the fulcrum example work with the ruler cart, where the middle of the wheel is on a shorter section of the fulcrum?

For that cart, the fulcrum effect is due to the ratio of the inner and outer diameters of the spools on the bottom of the cart. In the videos, the inner diameters are 1/2 the outer diametsrs, so it's a cart with an effective gearing of .5. It goes down ruler at 2x ruler speed or upground at -1x ground speed. If the ratio was 2/3, it would go down ruler at 3x ruler speed or upground at -2x ground speed, faster than the difference in speed between ruler and ground regardless of which one moves.

 

[A.T.]

I am still wondering how does the fulcrum example work with the ruler cart, where the middle of the wheel is on a shorter section of the fulcrum? The fulcrum doesn't apply there for some reason, but why? Or does it still apply?

What is "fulcrum example"? Do you mean the lever analogy shown here:

[PLAIN]http://img375.imageshack.us/img375/4229/dwfttwwheel09dashedleve.gif

You can represent any gearbox as levers. Here for the ruler cart:

[URL]http://forums.randi.org/imagehosting/26573493a5038e7a04.gif[/URL]

This is by Brian-M from Randi forum:
http://forums.randi.org/showthread.php?p=4252145

 

[chingel]

That's the input power value, 10 lbs 20 mph = 200 lb mph = 5.333 horsepower = 3977 watts. The energy per second would be 2933 lb ft or 3977 Newton meters (joules). The mentioned output power is 1/2 of that (10 x 10).

Yes but how did he get it? Did he multiply the propeller force by the wheel speed?

Although the force at the wheels is an opposing (braking) force, the effective gearing multiplies the force at the wheels and divides the speed. In the example above with a 10 mph tailwind, the input force and input speed from the ground to the wheel was 10 lbs, 20 mph. If there were no losses, the output force from the prop to the air could be 20 lbs, at 10 mph, if the tailwind was greater than 10 mph, but with a 10 mph tailwind, as the cart approaches 20 mph, the output force to the prop would decrease to zero, which would also decrease the opposing force on the ground to zero for a zero loss cart (since the wheels are now driving the propeller with zero force).

Yes, but you can apply brakes to a wheel in several different places. If you apply them further away from the center, you need less force, if you apply them closer to the center you need more force, but when force x speed is the same, the wheel will slow down by the same amount in the same amount of time.


With the fulcrum example I meant viewing the contact point with the ground as the pivot point of the fulcrum. How to make it work viewing it like that? When viewing the hexagonal yo-yo as a lever where the pivot point is on the ground, which was important for understanding the wind cart, how does the ruler cart go faster than the ruler?

 

[A.T.]

When viewing the hexagonal yo-yo as a lever where the pivot point is on the ground,

Only in the reference of the ground. In the frame of the airmass(string), the air(string) is the fulcrum. The position of a lever's pivot is frame dependent, just like the flow of kinetic energy. If you base explanations on such frame dependent concepts, you will not understand the general mechanism, but rather just what one specific observer sees.

which was important for understanding the wind cart, how does the ruler cart go faster than the ruler?

Just like the yo-yo. You can replace the ruler with a string, winded around the reel, and diverted around the blue gear.

 

[A.T.]

"Wind powered" cart is a bit of a mis-nomer, and is part of the reason for the confusion.

"Wind" is ambiguous. Some people confuse "wind" and "air", or "true wind" and "relative wind". But in the context of "Wind power", the term "wind" means "air movement relative to the ground".

The source of the power for DDWFTTW carts is the difference in speed between wind and ground,

It's better to say: The source of the power for DDWFTTW carts is the difference in speed between air and ground.

 

[spork]

"Wind powered" cart is a bit of a mis-nomer, and is part of the reason for the confusion. The source of the power for DDWFTTW carts is the difference in speed between wind and ground, and all of this has been explained in previous posts in this thread.

In the frame of the ground, this is a wind powered cart. Even a sailboat, on anything but a direct downwind run, is powered by the difference in speed between the water and air.

 

[chingel]

I think that the extracting energy from the difference explanation isn't very good. While technically true, it is like saying a car extracts energy from the different speeds of the flywheel and the ground, assuming 1:1 gear ratio. It doesn't really explain anything. The real source of energy is what makes the flywheel turn or what makes the wind blow. The explanation that you can simply extract energy from a difference doesn't really mean anything to me.

Isn't it true that the wind is supplying the energy in all reference frames? In the cart's frame, the wind pushes on the blade that is constricted to rotate and move slower than the wind if the cart moves forward. Maybe in the wind's reference frame it is different, but still the blade is sort of moving slower than the wind and the wind pumps into it and by action-reaction moves it forward. Of course it needs the ground to be there to be able to push on it, but isn't still the wind providing the power?

schroder, if you pull a yo-yo by the string, when the string is under the middle part of the yo-yo, it will climb up the string, moving faster than the string. Please try it out and see for yourself. If instead of pulling it, you put a parachute at the end of the string which pulls it, the yo-yo will move faster than the parachute. Do you agree?

 

[spork]

Here is a link I found useful:

http://www.greglondon.com/tumbleweed/tumbleweed.pdf

I haven't looked at his latest version of that document, but the first version was so horribly riddled with basic errors and really totally complete wrongness, that I have a hard time believing this version is worth reading.

 

[rcgldr]

Isn't it true that the wind is supplying the energy in all reference frames?

No, as mentioned before, from the cart's frame of reference or the air's frame of reference, slowing down the Earth's surface (which is moving backwards relative to the frame of reference) a tiny amount (due to the forward force applied by the wheels) is the source of energy.

What you need to understand in that example is even if the yo-yo moves faster than the parachute, the parachute is moving slower than the wind.

and likewise, the thrust from the propeller on a DDWFTTW cart is moving slower than the wind, even when the cart is moving faster. Previous posts in this thread explained this in more detail.

 

[A.T.]

I think that the extracting energy from the difference explanation isn't very good.

Extracting energy from the velocity difference of two masses, by reducing that velocity difference

is a very good general explanation, because it holds true in every reference frame.

Isn't it true that the wind is supplying the energy in all reference frames?

wind = velocity difference of air & ground

 

[chingel]

Let's not go down the drain too quickly and make too radical statements. This topic can be confusing and I think we should discuss IMO the most intuitive example slowly with him first, if he is finding the propeller cart confusing as well.

No, as mentioned before, from the cart's frame of reference or the air's frame of reference, slowing down the Earth's surface (which is moving backwards relative to the frame of reference) a tiny amount (due to the forward force applied by the wheels) is the source of energy.

and likewise, the thrust from the propeller on a DDWFTTW cart is moving slower than the wind, even when the cart is moving faster. Previous posts in this thread explained this in more detail.

Isn't the cart moving forward by pushing the ground with the wheels? The propeller is constrained to move slower than the wind, the wind pushes on the propeller, which pushes the wheels, which pushes the ground, not the other way around? Does it slow the ground down in the cart's frame or accelerate in the other way? Doesn't applying force through wheels to move forward make the Earth rotate the other way a little?

Isn't the point discussed earlier on the propeller moving slower than the wind still moving slower than the wind in the cart's frame, or doesn't negative velocity mean going slower?schroder, do you agree that the yo-yo, and only the yo-yo, would move faster than the wind if it is pulled by a parachute? Let's just assume we can make the yo-yo light enough, the parachute big and efficient enough etc? What about the cart that uses many parachutes rotating around the chassis opening and closing when needed? Do you think that is different or would that also move faster than the wind?

 

[rcgldr]

Isn't the cart moving forward by pushing the ground with the wheels?

Not a DDWFTTW cart. Instead the propeller's backwards thrust into the tailwind results in the tailwind and thrust pushing the cart forwards, while the wheels driving the propeller generate a forward force onto the ground (with the ground generating an equal and opposing force backwards onto the wheels). As mentioned before, the effective gearing of the BB cart is .8 and would multiply the force at the wheels to the propeller by 1.25 and divide the speed at the wheels to the propeller by 1.25 if there were no losses.

Does it slow the ground down in the cart's frame?

Yes, the Earth's surface is moving backwards relative to the cart and the forward force of the wheels "slows" the Earth down by a tiny amount (relative to the cart).

 

[A.T.]

the wind pushes on the propeller, which pushes the wheels, which pushes the ground, not the other way around?

If by "pushing" you mean "exerting a force on" then it is like you describe and also the other way around. See Newton's 3rd law.

Does it slow the ground down in the cart's frame

Yes, in the cart's frame the ground moves opposite to the force from the cart. But the Earth is so heavy that this is negligible.

 

[spork]

Explaining physics to schroder is equally futile. We already have a thread where dozens of people, including PF mentors, tried to explain the treadmill equivalence to him for 60+ pages:
https://www.physicsforums.com/showthread.php?p=2067372

I'm pretty sure the PF mentors also explained to me that the treadmill was not equivalent when this topic first appeared here.

 

[A.T.]

When the belt is moving under the plane, it doesn't have rotational kinetic energy, because it isn't rotating, it is moving in a straight line. It has rotational kinetic energy when it is rotating or turning around the ends of the treadmill. I cannot imagine how the plane would use the rotational energy of the belt in the part that is rotating.

What plane? You are confusing brainteasers here.

But otherwise you're right. His argument is beyond silly. As if it made a difference for the cart on the belt, if the belt loops or is just very long and moving in a straight line. It is quite simple: The top belt part moves at a constant velocity relative to ground, therefore the rest frame of top belt part is just as inertial as the rest frame of the ground.

 

[chingel]

These are just details. Let's assume the yo-yo has enough mass so that it wouldn't fly off in the wind, let's assume that the inner diameter is half of the outer diameter, so that the yo-yo moves twice the speed of the string pulling it. Now let's assume the wind is 10 m/s, and the parachute pulls by a force of 2 N when it goes at 9 m/s, ie 1 m/s slower than the wind. Now let's assume that the yo-yo is big enough that it requires a constant 2 N force to keep it's speed when moving at 18 m/s while combating rolling friction and air resistance. This yo-yo would move at a speed of 18 m/s. Do you agree?

 

[spork]

link goes to a deleted file, although there are 4 video links there.

Inside the box to the left of the video links, click where it says "Play flash full screen"

 

[ZapperZ]

This thread has been trimmed, pruned, cleaned up, and reopened. I don't want to have to do this again.

If you notice a member that continually spews nonsense, and continually refuses to learn, it is NOT your responsibility to handle such things and escalate the situation. Please REPORT a post and bring it to the attention of the Mentors. Let US do the dirty work. Otherwise, the thread is in danger of either being closed, or completely removed.

Zz.

 

[rcgldr]

yo-yo ... parachute

This could work (depends on how much drag there is between parachute and ground), but the yo-yo eventually reaches the parachute and the process stops.

Isn't the cart moving forward by pushing the ground with the wheels?

In case it wasn't made clear, when the cart is moving forwards, the wheels on the cart are turning the propeller against the wind, so that the propeller is generating thrust which pushes the air backwards (or slows it down). The external forces are the air pushing forwards against the propeller, and the ground pushing backwards against the wheels (the wheels are using a braking force to drive the propeller). The effective gearing combined with a tailwind allow the force with the air to be greater, but at a lower speed, than the opposing force with the ground, while still allowing for some losses in the system.

 

[OmCheeto]

This thread has been trimmed, pruned, cleaned up, and reopened. I don't want to have to do this again.

If you notice a member that continually spews nonsense, and continually refuses to learn, it is NOT your responsibility to handle such things and escalate the situation. Please REPORT a post and bring it to the attention of the Mentors. Let US do the dirty work. Otherwise, the thread is in danger of either being closed, or completely removed.

Zz.

No!

Should we start the process of solving this problem mathematically, and with real physics?

Or should we continue to wait for video's of new and improved DDWFTTW vehicles? They've built a full size version, so I think it might be time.

But I would very much like to see an over-sized tumbleweed, and a thread-spool vehicle. Images of https://www.physicsforums.com/showpost.php?p=1809929&postcount=8" run through my head when I think of such things.

I've learned much over the last few years, and might be able to solve this over the next 10 weekends.

 

[spork]

Should we start the process of solving this problem mathematically, and with real physics?

Done - many times, and in many ways.

 

[OmCheeto]

Done - many times, and in many ways.

I'll take that as a yes.



There go my weekends, again...

Postulate #1: A vehicle with near zero rolling resistance, and simply a sail, should be able achieve near wind speed velocity.

Advancement of the problem: Convert the sail into a windmill/propeller, connected to the wheels via gearing, to propel said device, faster than the wind.

...

I'll be back in ~3.34664351851852 days.

 

[spork]

I'll take that as a yes.

Personally, I think you should take that as a no. Why set out from scratch to give this a proper physics and math treatment when it's already been done many times.

The most likely outcome of course is getting the proper math and physics completely wrong - as so many have done.

I'll be back in ~3.34664351851852 days.

Alrighty then. I'll be here to let you know where you went wrong.

 

[A.T.]

Should we start the process of solving this problem mathematically, and with real physics?

Here some material on this:

# ^ Drela, Mark. "Dead-Downwind Faster Than The Wind (DFTTW) Analysis". http://www.boatdesign.net/forums/attachments/propulsion/28167d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddw2.pdf.

# ^ Gaunaa, Mac; Øye, Stig; Mikkelsen, Robert (2009). "Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion". Marseille, France: Proceedings EWEC 2009. http://orbit.dtu.dk/getResource?recordId=241183&objectId=2&versionId=1.

# ^ Bauer, Andrew (1969). "Faster Than The Wind". Marina del Rey, California: First AIAA. Symposium on Sailing. http://projects.m-qp-m.us/donkeypus...aster-Than-The-Wind-The-Ancient-Interface.pdf.

 

[AVReidy]

Certain high-performance sailboats can travel faster than the wind when sailing downwind due to the apparent wind created by their speed. These boats don't travel directly downwind, however. They need to be angled to take advantage of the apparent wind they create, and traveling directly downwind would probably result in the boat reaching a sort of equilibrium (maximum downwind speed) where the force of the apparent wind on the sail is equal to the force of the wind.

 

[spork]

Certain high-performance sailboats can travel faster than the wind when sailing downwind due to the apparent wind created by their speed.

There's no question that high performance sailboats can tack their way downwind to beat a free floating balloon to a downwind mark. It's easy to show exactly how they do this. Whether it's due to the "apparent wind created by their speed" becomes sort of a philosophical question. In the end, they are doing exactly what the blades of the prop on the downwind cart does. They are exploiting the energy available at the interface of the true wind and the surface (water or ground). The idea that they "make their own wind" is a sort of hand wavy thing that tends to be thrown around by sailors. Yes, the apparent wind experienced by the sail is different than the true wind seen over the water. But there's no free lunch - just some free geometry.

These boats don't travel directly downwind, however. They need to be angled to take advantage of the apparent wind they create, and traveling directly downwind would probably result in the boat reaching a sort of equilibrium (maximum downwind speed) where the force of the apparent wind on the sail is equal to the force of the wind.

Not quite sure what you mean here when you say the force of the apparent wind on the sail is equal to the force of the wind.

 

[AVReidy]

I don't fully believe they get any more speed from the apparent wind, but I did read this once. If they could gain speed from their speed, then they would continue to speed up indefinitely, so it's kind of violating the laws more than perpetual motion does.

 

[rcgldr]

apparent wind

A sailcraft doesn't get any direct benefit from an apparent (head) wind that increases with the sailcraft's speed, other than the sail may be optimized (orientation and shape (curvature)) for a specific range of apparent wind.

Here's a summary of the math. Assuming a sailcraft holds a constant heading θ relative to the true wind (θ = zero means in the direction of the true wind), then the apparent wind can be split into two components:

apparent_crosswind = wind_speed x sin(θ)
apparent_headwind = sailcraft_speed - (wind_speed x cos(θ))

Note the crosswind component is constant, regardless of the speed of the sailcraft. Only the apparent headwind is related to the sailcraft speed.

The sailcraft's speed can be split into two components:

sailcraft_crosswind_speed = sailcraft_speed x sin(θ)
sailcraft_downwind_speed = sailcraft_speed x cos(θ) ... (negative value means upwind)

All of the thrust from the sail on a sailcraft is due to diversion of the apparent crosswind. The aerodynamic drag is related to the apparent headwind. An ideal sail diverts the apparent wind changing it's direction to directly aft of the sailcraft without decreasing the magnitude of the wind (zero loss). An actual sail will divert the apparent wind generating enough thrust to overcome total drag and also slow down the true wind enough to extract the power it takes to maintain the sailcraft's speed.

Link to a pdf with info about iceboats and telemetry from runs.

iceboat_performance.pdf

The telemetry for one run used a heading of 30 degrees left of downwind, wind speed of 18 mph, iceboat speed of 70 mph, apparent wind of 55.15 mph made up of 9 mph apparent crosswind, 54.41 mph apparent headwind, downwind component of speed 60.6 mph (3.37 x wind speed).