Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[spork]

That's an awesome way to visualize it

I think that's my favorite explanation yet. It makes it really obvious how it works.

Thanks. But interestingly enough, not a word about it from chingel. I thought I was answering his question. Doesn't it at least deserve a "you're full of it"?

 

[mender]

There was this reply:

The cart would indeed move along the jelly faster than the jelly. I am having difficulty imagining that when you get the cart up to jelly speed and then engage the propeller, how does the force apply, how does it not use it's own energy to propel itself?

It almost sounds like a clarification might be in order; I'm wondering what chingel means by "it's own energy". To my way of thinking, the only "energy" the cart would have would be from the motion imparted by the jello.

 

[spork]

There was this reply...

It almost sounds like a clarification might be in order; I'm wondering what chingel means by "it's own energy". To my way of thinking, the only "energy" the cart would have would be from the motion imparted by the jello.

Turns out I'm the fool. My apologies to Chingel. I missed that part because he started out by responding to another post.

Let's take it point by point:

The cart would indeed move along the jelly faster than the jelly.

Good. Now we replace the jello with air (i.e. replace the moving jello with wind). Done. We're going directly downwind faster than the wind.

I am having difficulty imagining that when you get the cart up to jelly speed and then engage the propeller...

No, no, no. We don't get it up to speed and then engage anything. The wheels are rolling against the ground at all times. The prop is geared to the wheels at all times. The prop is effectively "geared" into the jello at all times. This is now a simple kinematic puzzle like the yo-yo. We don't have to think about where the energy comes from this way.

When a 200 lb man pulls on the string of a 3 oz yo-yo, you don't worry where the energy comes from. By the same token, we don't care how big a monster it takes to push that jello along. He just does it - no matter what it takes. All we have to do is ask what motions will the wheels, prop, and cart follow when encased in that jello that moves along over the road.

how does the force apply

In the case of the jello, the jello just pushes on the back of the prop. When the jello pushes on the back of the prop, it will tend to want to turn the prop counter-clockwise (as seen from behind). But it will also want to push the prop along in the direction the jello is going. If it does push it along in the direction the jello is going, the prop HAS to turn clockwise as seen from behind - because it's geared to the wheels that way.

So there's a torque from the jello trying to turn the prop CCW and a torque on the prop-shaft (from the wheels) trying to turn it CW. Because we geared it extremely low (100 wheel turns for a singe prop turn), we know which one will win. The CCW jello torque on such a low pitched prop will be minimal. But the geared down torque from the wheels in the CW direction will be immense. So the prop moves down-jello with the jello, but it also screws its way through the jello very gradually.

how does it not use it's own energy to propel itself?

The energy comes from the giant monster playing with his food.

In the case of the wind, the energy come from uneven heating of the Earth's surface (in other words - the sun - which also has a whole lot more energy than our little cart needs).

 

[A.T.]

I'm wondering what chingel means by "it's own energy".

People who have problems understanding the energy balance in DDWFTTW usually have a flawed understanding of the energy concept itself. The whole point of such brainteasers it to challenge naively intuitive folksy notions about physics, and lead to a more precise understanding. The later however requires accepting that you have a flawed understanding of certain general concepts in physics (like energy), and not just a problem with understanding this particular example.

 

[rcgldr]

I'm wondering what chingel means by "it's own energy".

In his earlier posts, it seemed he had the idea that the somehow the cart was extracting it's own kinetic energy as part of the input power. I thought I explained the actual source of energy in an earlier post, but in case he missed it I'll restate it here.

In all frames of reference the cart's own kinetic energy is never an input, it's always the end result of extracting energy from either the wind or the earth, depending on the frame of reference, and is designed to take advantage of the fact that there is a difference in speed between the wind and the ground, even when the cart's own speed is greater (up to a limit, which for the BB, is about 2.8x wind speed using a ground frame of reference).

From a ground frame of reference, even when the cart is moving at or faster than wind speed, the thrust from propeller (relative to ground) is moving slower than wind speed (relative to ground), and that thrust is slowing down (a portion of) the tail wind (relative to ground), and that slowing of the tail wind is the source of energy.

From the air's frame of reference, the forward force (from the wheels) on the surface of the Earth is slowing down the Earth (relative to the air) a very tiny bit, and that is the source of energy that allows the propeller to accelerate the air, and accelerate the cart until it reaches it's maximum speed.

The DDWFTTW cart is like a hybrid car, except that the braking energy is being used to drive the propeller as opposed to charging a battery. This only works if there's a tail wind that allows the cart to utilize effective gearing between wheels and propeller so that propeller thrust is greater than braking force, while at the same time propeller power output is less than braking power input, due to the greater difference in speed between cart and ground versus cart and air (when the cart is moving downwind).

The DUWFTTW (upwind) cart is like a wind mill, where the turbine (propeller) energy is being used to drive the wheels as opposed to driving a generator. This only works if there's a head wind that allows the cart to utilize effective gearing between turbine and wheels so that wheel driving force is greater than turbine drag, while at the same time wheel driving power output is less than turbine power input, due to the greater difference in speed between cart and air versus cart and ground (when the cart is moving upwind). Aerodynamic drag is greater in this case, so the maximum speed will be less than that of a DDWFTTW cart going downwind.

 

[spork]

People who have problems understanding the energy balance in DDWFTTW usually have a flawed understanding of the energy concept itself. The whole point of such brainteasers it to challenge naively intuitive folksy notions about physics, and lead to a more precise understanding. The later however requires accepting that you have a flawed understanding of certain general concepts in physics (like energy), and not just a problem with understanding this particular example.

That's a very good point, and really helps to explain the almost violent reactions we sometimes get. It's not just that this goes against intuition. It challenges people on a deeper level.

 

[rcgldr]

I think that part of the issue is statements like a DDWFTTW cart outruns the wind that powers it, when in fact it's the difference in speed between wind and ground that powers the cart, and that the thrust from the wheel driven propeller never outruns the wind, but instead always slows down the wind, even when the cart itself is moving faster than the wind.

After this, it's just a case of explaining the design of a DDWFTTW cart, mostly how effective gearing between wheels and propeller allow the cart to achieve faster than wind speed as long as there's a difference in speed between a tailwind and the ground.

 

[chingel]

How does the fulcrum example apply to the under the ruler example, where the center of the wheel is between the pivot point and the point where the force is applied? How does the center of the wheel move faster there?

The other examples are easier to understand and they should work. But I still have a problem imagining what happens when I am on a cart at windspeed and then I engage the propeller? Why doesn't all the energy used for propulsion of the air particles cause exactly as much drag at the wheels?

 

[spork]

I think that part of the issue is statements like a DDWFTTW cart outruns the wind that powers it...

Agreed. That's why I try and correct people, explaining that our cart outpaces the wind. It doesn't outrun the wind any more than a swimmer can outrun the water. We're always immersed in it.

I still have a problem imagining what happens when I am on a cart at windspeed and then I engage the propeller? Why doesn't all the energy used for propulsion of the air particles cause exactly as much drag at the wheels?

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.

 

[A.T.]

Why doesn't all the energy used for propulsion of the air particles cause exactly as much drag at the wheels?

Because there is no reason it should.

 

[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 betwen 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.