DDWFTTW Turntable Test: 5 Min Video - Is It Conclusive?

[schroder]

Schroder, thank-you for making this thread so lively, and I hope the name-calling will be retracted or otherwise dealt with.

May I ask, how did you originally know that DDWFTTW would be impossible? (I must admit, even though I am studying for my third physics degree, it wasn't obvious to me that any particular physical laws are contradicted by such devices.) I was also wondering, do you agree with Einstein (http://www.physclips.unsw.edu.au/jw/sailing.html" ) that sail boats can tack diagonally across the wind in such a way that their velocity made good (that is, the component going directly downwind) exceeds the wind speed (even though the boat itself obviously cannot go DDWFTTW)?

Thank you for contributing to this subject minus any personal attacks. I wish the entire thread could (have) been conducted in this manner.
I never knew, for certain that DDWFTTW was impossible, but certainly it struck a chord with me that something is wrong with all the explanations and there is no outdoor evidence to support it. On the Turntable, it can be shown that the combination of translational and rotational motion results in a velocity of roughly 1.41 x the tread velocity. This is undeniable. I have been seeking an explanation of where the necessary energy comes from and I am convinced I have found it in the mechanical heterodyne.
I certainly do agree with Einstein concerning relativity and I also agree with the Sydney yacht club about 18 footers! I have sailed these same craft, but out of Perth. The very website you directed me to contains this gem of information:

Sailing downwind (parallel to the wind, like the boat at left) is easy to understand: the wind blows into the sails and pushes against them. The wind is faster than the boat so the air is decelerated by the sails. The sails push backwards against the wind, so the wind pushes forward on the sails. But for a boat with normal sails, the catch is that, downwind, you can only ever sail more slowly than the wind. Which is usually boring, even with a spinnaker.

Directly down wind is always slower than the wind! Yes, by tacking at an angle, you can achieve velocities greater than wind speed. But you are moving at an angle. Also, the VMG claim needs some interpretation. VMG is usually delineated with respect to a line, not a point. This is very important to understand. If a draw a line perpendicular to the direction of the wind, and another line down wind, also perpendicular to wind direction, I can always be moving at the same angle of attack in traveling from one line to the next. If I can get up to a velocity greater than wind speed while traveling at that angle, it is not inconceivable that I can cross that line before the wind does. But, if the contest were between two points, I would need to change tack somewhere in mid-course which would cost a great deal of time and it really is inconceivable that I could reach a point that is directly down wind before the wind does. So I exercise caution when I see the term “VMG” as it does need to be clarified if we are talking about a line or a point.
Back to the carts, I have never seen an explanation of where the required energy comes from, have you? The best the proponents can offer is: “There is a lot of energy in that moving air mass”. Certainly, that is true, but just how the cart extracts that energy has never been explained. Uart’s explanation in Post #856 is typical. He starts out by saying the propeller requires a certain amount of power in order to provide a certain amount of thrust. He then goes on to say that this power is provided by the wheels. He concludes by saying that if the propeller is sufficiently efficient, it can both provide the thrust AND power the wheels! I think even the first-year physics students will have a hard time buying into that!

My explanation about mechanical resonance might actually help the cause of the believers in DDWFTTW if it could be shown that the cart is capable of this in the outdoors. Unfortunately, I think that it is impossible for the cart to achieve resonance with the earth!

 

[vanesch]

Directly down wind is always slower than the wind!

Agreed... with a sailing boat.

 

[vanesch]

On the Turntable, it can be shown that the combination of translational and rotational motion results in a velocity of roughly 1.41 x the tread velocity. This is undeniable.

Wait a minute ! Are you finally saying here, that the relative velocity of the cart wrt the turning table is, after all, faster than the relative velocity of the turning table wrt the ground ?

You've changed your mind on this one then ?

Because a while ago you maintained that the relative velocity of the cart wrt the table was SLOWER ? That was a wrong claim then, or not ?

 

[Subductionzon]

Schroder you are getting a bit closer now that you understand VMG. A boat in the water may have difficulty getting a VMG to be higher than wind speed, but an iceboat has no such problem. An iceboat can achieve a VMG of about three times wind speed. They do not dawdle when they come about. The spinning prop on the cart acts as a sail that is on a close reach (I hope that is the right term). Its advantage is that it makes the cart act like it is on a continuous tack with the wind. If you followed that sailing link you would also see claims for highly efficient sailboats that have a VMG higher than wind speed.

 

[uart]

Uart’s explanation in Post #856 is typical. He starts out by saying the propeller requires a certain amount of power in order to provide a certain amount of thrust. He then goes on to say that this power is provided by the wheels. He concludes by saying that if the propeller is sufficiently efficient, it can both provide the thrust AND power the wheels! I think even the first-year physics students will have a hard time buying into that!

My explanation about mechanical resonance might actually help the cause of the believers in DDWFTTW if it could be shown that the cart is capable of this in the outdoors. Unfortunately, I think that it is impossible for the cart to achieve resonance with the earth!

However resonance is NOT a source of energy. Resonance allows stored energy to build up cycle by cycle in a vibrational system, but it is not a source of energy and therefore your claim is ungrounded. Actually resonance is an often used claim of "free energy" nuts to explain why their nonsense systems really-truly can achieve over unity performance etc. The "water fuelled car" crowd (and I mean the fruitloop ones who actually believe they can use a battery to power the water to H2, O2 conversion and then run the car and charge that battery by combusting the H2, O2 mix) often talk about using "resonance" of the H2O molecules in their electrolysis units to explain why they can achieve this amazing feat. They talk as if resonance is some type of magical source of free energy. So you're in really good company with this claim now Schroder.

Further, the idea that the 1.41 proves that this has something to do with a Bessel null is unfounded. By changing the propeller pitch that value could be decreased (without the added damping of adding a friction block) so there's another desperate grab at a "theory" out the Window. Also I'm well aware of the Bessel DE and it's occurrence is vibrational systems with cylindrical symmetry, but please show us the Bessel DE here and how it relates to the variables under consideration (turntable speed and cart speed).

 

[cesiumfrog]

If I can get up to a velocity greater than wind speed while traveling at that angle, it is not inconceivable that I can cross that line before the wind does. But, if the contest were between two points, I would need to change tack somewhere in mid-course which would cost a great deal of time and it really is inconceivable that I could reach a point that is directly down wind before the wind does. So I exercise caution when I see the term “VMG” as it does need to be clarified if we are talking about a line or a point.

Schroder, you're in Perth? G'day from Canberra!

I agree that the "propeller driven by the wheels" reasoning is nonsense, and we also agree that a real sailboat can cross a line before the wind does. You said that the reason why the boat couldn't beat the wind to a point is because it loses too much time and momentum when it has to change tack.

Then I was http://www.celestiamotherlode.net/catalog/images/screenshots/various/fic_orionsarm_Orion_s_Arm_Ringworld_1__sbowers.jpg" " with a uniform wind blowing through its axis. A sailboat on the inside would take a helical path right around the halo, and wouldn't ever need to change tack, so do you agree with me that (in principle) it could thereby beat the wind to a point?

 

[uart]

This is the story of Atom Man, a miniature sized superhero from the planet Zogg. Planet Zogg is much like Earth, the laws of physics are the same except that their "air" molecules are heavier and slower than ours, so when the wind blows hard the motion of these molecules is predominately just downwind with not very much random thermal motion.

Now Atom Man and his arch rival Mini-Schroder both have downwind carts that they can ride in, and they like to race each other on windy days. After much racing and tweaking both Atom Man and Mini-Schroder have tuned their carts so that they travel downwind with very little friction. They both attain almost the same speed as the wind, so their finishes are generally pretty closely contested.

Atom Man is very small and nimble, he can even see individual molecules of gas in the air. When he gets into his downwind cart to race he can see the molecules of air colliding with the back of his cart and bouncing off, providing the change in momentum to accelerate the cart forward. As his cart increases to closer to the wind speed he sees these molecules colliding with ever less relative speed and imparting ever less momentum, so he never can quite reach the speed of the wind, let alone exceed it.

One day while Atom Man is racing Mini-Schroder downwind he gets a great idea. The race is neck a neck, with both carts doing almost exactly the wind speed, when Atom Man notices that the air molecules are now almost stationary relative to him and he can easily reach out and grab them. So he starts grabbing these molecules as fast as he can and hurling them backwards to create a forward thrust (in much the same way as a spacewalking astronaut could throw a spanner to create thrust if he had no other propulsion available). This idea works and Atom Man actually gets his cart to faster than wind speed, resoundingly beating Mini-Schroder in the process.

The victory is short lived however. Mini-Schroder immediately protests to the DWC Academy about Atom Mans cheating tactics. The complaint is upheld, and rightly so. After all Atom Man was using up much of his own stored energy (from his internal batteries, you see Atom Man is battery powered) in the process of hurling those atoms. The Academy's rules strictly state that the only allowed source of power in these races is the wind.

Atom Man is not so keen to give up on the idea of DWFTTW just yet, so he asks the Academy to rule on whether he can continue to use his air molecule throwing trick if he obtains the power to throw those molecules from a generator running off the wheels of the cart instead of from his internal batteries. After some deliberation the Academy decides that this is ok and violates no rules. Meanwhile Mini-Schroder is not the least bit concerned about the ruling. He knows full well that Atoms Mans generator will create drag at the wheel and he "knows" (without any calculation or anything, he just knows ok) that this drag will exceed any thrust that Atom Man is able to produce by hurling air molecules.

Atom Man is not so sure about that, he knows a little about basic Physics so he does the analysis. He figures out that if he throws "n" molecules per second, with each molecule having a mass "m" and leaving at velocity "v" then the average forward thrust will be equal to the average rate of change of momentum. That is, atom mans produced thrust is :

$$F = n m v$$

Now Atom Man knows that if his strategy is to be successful he must be sure that the drag from his generator is less than this amount, so he decides to use only half of this extra thrust in powering the generator. The theoretical maximum power that his generator can deliver is therefore :

$$P = \frac{1}{2} F w = \frac{1}{2} n m v w$$

Where "w" is the speed of the cart (and approximately the speed of the wind when atom man starts his hurling).

Finally Atom man calculates how much power he requires to hurl those molecules. The KE given to each molecule is $\frac{1}{2} m v^2$, and this is done "n" times per second, so the average power required is :

$$P = \frac{1}{2} n m v^2$$

Since $\frac{1}{2} n m v w > \frac{1}{2} n m v^2$ whenever $w > v$ Atom Man sees that he can easily exceed the wind speed whenever the wind speed is greater than his throwing speed. The result is successful and Atom Man now can always beat Mini Schroder in these races.

Mini Schroder again complains to the Academy, this time he even accuses Atom Man of witchcraft, but nobody will take him seriously. Eventually he storms off and falls into a Bessel null, never to be heard of again. Everyone else lives happily ever after.

 

[schroder]

Schroder, you're in Perth? G'day from Canberra!

I agree that the "propeller driven by the wheels" reasoning is nonsense, and we also agree that a real sailboat can cross a line before the wind does. You said that the reason why the boat couldn't beat the wind to a point is because it loses too much time and momentum when it has to change tack.

Then I was http://www.celestiamotherlode.net/catalog/images/screenshots/various/fic_orionsarm_Orion_s_Arm_Ringworld_1__sbowers.jpg" " with a uniform wind blowing through its axis. A sailboat on the inside would take a helical path right around the halo, and wouldn't ever need to change tack, so do you agree with me that (in principle) it could thereby beat the wind to a point?

G’day mate! I have spent some time in the Sydney area as well for Lockheed Martin Space Systems when they have a new satellite launch or a transfer to do. Have you anything to do with the big dish at Parkes, by any chance?
I don’t understand all there is to the “ring world” but I can agree that if the boat could stay on the same tack on a helical path it could reach a point as easily as it could reach a line.

 

[vanesch]

I agree that the "propeller driven by the wheels" reasoning is nonsense

Why ?

Have you considered the calculation in posts around 472-480 ?

Let us consider the essential parts. We place ourselves in the reference frame of the cart.

The model is given by the following elements:
*) the cart is rolling with a velocity v_cart on the floor (that is to say, in this reference frame, the floor is passing by at a velocity of -v_cart) with a wheel (or a set of wheels).

*) The cart propeller sees an incoming air speed of v_in and will give it an outgoing velocity of v_out.

*) The outgoing velocity of the air is essentially determined by the rotation velocity of the propeller (and its pitch) which is proportional to the rotation of the wheels, and hence to v_cart. We have hence as a model: v_out = K v_cart.

*) the air is moving relative to the floor with a velocity v_wind

In the ground frame, we have the velocity vectors in black ; in the cart frame, which is moving with a velocity v_cart to the right, we have the velocity vectors in red.

We assume that the cart is moving DWFTTW initially, and calculate the force balance on the entire thing. If the overall force is to the left, the cart will slow down ; if it is to the right, then the cart will speed up (or remain steady if we include extra drag).

We take the symbols v_cart, v_in, v_out and v_wind to represent positive numbers (the magnitudes of the displayed vectors).

The transformation from the ground (OXYZ) to the cart (oxyz) frame for velocities comes down to subtracting the vector v_cart from the black velocity vectors, right ?

We see then that the cart velocity is 0 (v_cart - v_cart), which is logical ;
that the ground velocity is (0 - v_cart) is v_cart to the left,

and that the (incoming) air velocity is v_wind - v_cart. Given that we oriented v_in already to the left, this becomes: v_in = v_cart - v_wind (magnitudes, we suppose that v_cart > v_wind).

As the rotation velocity of the wheel is proportional to v_cart, we assume that v_out = K v_cart, and we suppose that K is such that v_out > v_in.

So we have a propeller that sees an incoming air mass with velocity v_in, and it accelerates it to v_out. If the mass of air per unit of time that is thus accelerated is given by m, then we have, per unit of time, a mass m that goes from v_in to v_out.

Now, that means that the MOMENTUM gained by that air per unit of time is:
m (v_out - v_in), right. Now, to do so, Newton tells us that we have to exert a force on the air which is equal to m (v_out - v_in). That's the force that the propeller must exert on the air. Newton, again, tells us that this means that the air must exert a force F_air on the propeller (and on the cart) which is the negative of this. However, as the velocities v_out and v_in were oriented to the left, this comes down to a force TO THE RIGHT equal to:

F_air = m (v_out - v_in).

That's the force that the air exerts on the propeller. It PULLS the cart in the forward direction as long as v_out > v_in.

Now, as the air is accelerated, it didn't only gain momentum, it also gained kinetic energy. Per unit of time, the KE gained by the air is the kinetic energy after - kinetic energy before:

P = 1/2 m (v_out^2 - v_in^2)

That is the POWER that the propeller must deliver to the air. So an ideal propeller will at least need to *receive* this power P. On the axle of the propeller, the mechanical power P needs to be delivered.

Where does that power come from ? Well, the wheel is rolling on the floor and is mechanically connected to the propeller. If the propeller TAKES mechanical power P, that means that on the axle of the wheel, rotating at an angular rotation velocity w, a power is extracted. It means that a *torque* is exercised on the wheel against the rotation velocity w. The wheel is turning CCW, so the propeller, through the mechanical coupling, exercises a T torque in the CW sense (to extract power P). P = w x T. The propeller exerts a force, through this torque, on the ground, such that T = r x F with r the radius of the wheel. This force is oriented forward to have the torque act CW. So the force that the wheel exerts on the ground is forward with a magnitude of F = T / r = P / (w x r)
Now, we have that v_cart = w x r, so we can write this as:

F = P / v_cart

The *reaction force* of the ground on the wheel will be of the same magnitude, but opposite orientation F_wheel:

F_wheel = P / v_cart to the left.

The two forces that act upon our cart in the horizontal direction are:

F_air = m (v_out - v_in) to the right and:

F_wheel = P / v_cart = 1/2 m (v_out^2 - v_in^2) / v_cart to the left.

So the total forward force is given by F_air - F_wheel. If this is a positive number, then the net force will accelerate the cart in the forward direction, otherwise it will make it slow down, right ?

F_tot = m (v_out - v_in) - 1/2 m (v_out^2 - v_in^2) / v_cart

Now, the rest is algebra: we fill in:

v_out = K x v_cart:

F_tot = m x { K x v_cart - v_in - 1/2 (K^2 x v_cart^2 - v_in^2) / v_cart }

v_in = v_cart - v_wind:

F_tot = m x { K x v_cart - (v_cart - v_wind)
- 1/2 (K^2 x v_cart^2 - (v_cart - v_wind)^2) / v_cart }

F_tot = m x { (K-1) x v_cart + v_wind -
1/2 (K^2 x v_cart - v_cart + 2 v_wind - v_wind^2/v_cart) }

F_tot = m x { (K - 1 - K^2/2 + 1/2) v_cart + v_wind^2/(2 v_cart) }

F_tot = m x { - 1/2 (1-K)^2 v_cart + v_wind^2/(2 v_cart) }

F_tot = m / 2 {v_wind^2 - (1-K)^2 v_cart^2 } / v_cart

We see that we have a forward force if:

|1 - K| v_cart < v_wind, or v_cart < v_wind / |1-K|

So if we take K = 0.333, say, then we have a forward force as long as
v_cart < 1.5 x v_wind.


What's wrong with this calculation ?

 

[vanesch]

This is the story of Atom Man, a miniature sized superhero from the planet Zogg. Planet Zogg is much like Earth, the laws of physics are the same except that their "air" molecules are heavier and slower than ours, so when the wind blows hard the motion of these molecules is predominately just downwind with not very much random thermal motion.

[ ... ]
Mini Schroder again complains to the Academy, this time he even accuses Atom Man of witchcraft, but nobody will take him seriously. Eventually he storms off and falls into a Bessel null, never to be heard of again. Everyone else lives happily ever after.

Brilliant !

 

[spork]

there is no outdoor evidence to support it.

Not only IS there outdoor evidence to support DDWFTTW, but Jack Goodman took video of his demonstration and I suspect you've seen it on the web.

On the Turntable, it can be shown that the combination of translational and rotational motion results in a velocity of roughly 1.41 x the tread velocity. This is undeniable.

Undeniable!? Hardly. It has also been shown that they can result in other speeds.

I have been seeking an explanation of where the necessary energy comes from...

So you decided to ignore all the correct and lucid descriptions presented and lock your teeth into this ludicrous notion of a mechanical heterodyne!?

Directly down wind is always slower than the wind!

In a traditional sailboat.

Back to the carts, I have never seen an explanation of where the required energy comes from, have you?

Have I? YES. In fact I've described it many times, and I refuse to believe you haven't seen it. Ignored it - perhaps.

just how the cart extracts that energy has never been explained

COMPLETELY UNTRUE.

My explanation about mechanical resonance might actually help...

Sorry, but misinformation does not help science.

I agree that the "propeller driven by the wheels" reasoning is nonsense

Sorry, but the propeller IS driven by the wheels. This can be very easily shown.

 

[swerdna]

I haven’t had much time lately to work on and test “real” wind/water devices. Have built a couple of water test models but have had no success with them so far. Built a small paddle boat and a floating paddle wheel using the “Brennan method” but neither worked. They were only very quickly and roughly put together so expect better results when I have more time to make them properly.

 

[ThinAirDesign]

This is the story of Atom Man,

<snip>

Mini Schroder again complains to the Academy, this time he even accuses Atom Man of witchcraft, but nobody will take him seriously. Eventually he storms off and falls into a Bessel null, never to be heard of again. Everyone else lives happily ever after.

So freakin' perfect.

When I was trying to imagine how a prop could extract energy from the air by doing work on the air, I thought of a similar illustration to your above -- thinking of an arm reaching out and pushing on floating bowling balls. It really helped me understand.

I hope you don't mind if I copy your post and use it as an illustration on other forums when needed.

Very nice.

JB

 

[cesiumfrog]

I can agree that if the boat could stay on the same tack on a helical path it could reach a point as easily as it could reach a line.

Schroder, are you a satellite engineer? (My area is currently theoretical/computational GR, nothing to do with Parkes or Tidbinbilla.)

Imagine if we built a long double-helix shaped roller-coaster track outside, aligned so that the wind blows directly along its axis. Do you agree that, if I release some "sail cars" at one end of the track (neglecting gravity and friction for a moment), they could get to the other end faster than the wind?

I was looking at the design for the proposed DDWFTTW cart, and I noticed something interesting about the way that the wheels are connected to the fan: no matter how fast the cart is going (nor what speed the wind blows at) the propeller-sails always trace out a fixed helical path so as to always maintain the same tack. Do you see where I'm going with this?

 

[schroder]

Atom Man is not so sure about that, he knows a little about basic Physics so he does the analysis. He figures out that if he throws "n" molecules per second, with each molecule having a mass "m" and leaving at velocity "v" then the average forward thrust will be equal to the average rate of change of momentum. That is, atom mans produced thrust is :

$$F = n m v$$

Now Atom Man knows that if his strategy is to be successful he must be sure that the drag from his generator is less than this amount, so he decides to use only half of this extra thrust in powering the generator. The theoretical maximum power that his generator can deliver is therefore :

$$P = \frac{1}{2} F w = \frac{1}{2} n m v w$$

Where "w" is the speed of the cart (and approximately the speed of the wind when atom man starts his hurling).

Finally Atom man calculates how much power he requires to hurl those molecules. The KE given to each molecule is $\frac{1}{2} m v^2$, and this is done "n" times per second, so the average power required is :

$$P = \frac{1}{2} n m v^2$$

Since $\frac{1}{2} n m v w > \frac{1}{2} n m v^2$ whenever $w > v$ Atom Man sees that he can easily exceed the wind speed whenever the wind speed is greater than his throwing speed. The result is successful and Atom Man now can always beat Mini Schroder in these races.

You obviously missed your true calling; you should be writing children’s nursery tales. Do you really think that the sequence of steps you have presented will hold up under scrutiny? This is a physics forum, after all!
Let’s look at your “calculations” in a bit more detail: You start out by saying the propeller, in order to grab those air molecules, needs to exert a certain amount of Force equal to mvn/s. Let’s put some numbers in there to bring this fantasy to life, shall we? Let’s use those heavy air molecules you mentioned that weigh 1 kg each and let’s start out at wind velocity of 10 m/s and we need to throw 10 of those molecules back every second, so n = 10. Fair enough? That will be (1kg) (10m/s) (10 molecules/s) = 100 Newtons of Force. This is the amount of Force the propeller needs to provide to throw those heavy air molecules around.
Now, you speak of “deciding” to use just half of the force that the propeller needs to provide?
Why don’t we just calculate the energy and power we need to provide to the propeller? KE = 1/2 mv^2 so at 10m/s the KE is 50 Joules. The power = Energy/time or Energy x n = 50 J x 10 = 500 Watts. So our generator needs to be able to provide 500 Watts to the propeller.
Now, would you care to continue this story? Where do you propose to find this 500 Watts? Do you propose to use just half of the Force needed at the propeller to run the generator? Do you believe that a motor which requires 500 Watts of power input to produce 100 Newtons of Force can be tapped into to use 50 Newtons of that Force to run a genset which then in turn produces the needed 500 Watts? Do you honestly believe in what you have written or are you being dishonestly deceptive? In either case, I think you should forget about physics altogether and concentrate on your true calling!

 

[schroder]

Schroder, are you a satellite engineer? (My area is currently theoretical/computational GR, nothing to do with Parkes or Tidbinbilla.)

Imagine if we built a long double-helix shaped roller-coaster track outside, aligned so that the wind blows directly along its axis. Do you agree that, if I release some "sail cars" at one end of the track (neglecting gravity and friction for a moment), they could get to the other end faster than the wind?

I was looking at the design for the proposed DDWFTTW cart, and I noticed something interesting about the way that the wheels are connected to the fan: no matter how fast the cart is going (nor what speed the wind blows at) the propeller-sails always trace out a fixed helical path so as to always maintain the same tack. Do you see where I'm going with this?

Yes. Satellites and deep space probes.

I see where you are going but I am not sure that the resistance encountered by the sails traveling on the helical tracks would be equivalent to the resistance of the propeller-sails. I’m not sure there is any equivalence between the two.

 

[mheslep]

Sailing downwind (parallel to the wind, like the boat at left) is easy to understand: the wind blows into the sails and pushes against them. The wind is faster than the boat so the air is decelerated by the sails. The sails push backwards against the wind, so the wind pushes forward on the sails. But for a boat with normal sails, the catch is that, downwind, you can only ever sail more slowly than the wind. Which is usually boring, even with a spinnaker.

Directly down wind is always slower than the wind!

The propeller blades in these demonstrations are moving across the wind vector, not moving directly before it.

 

[spork]

In either case, I think you should forget about physics altogether and concentrate on your true calling!

I'll just file that under IRONY

Yes. Satellites and deep space probes.

NONSENSE.

 

[vanesch]

Let’s look at your “calculations” in a bit more detail: You start out by saying the propeller, in order to grab those air molecules, needs to exert a certain amount of Force equal to mvn/s. Let’s put some numbers in there to bring this fantasy to life, shall we? Let’s use those heavy air molecules you mentioned that weigh 1 kg each and let’s start out at wind velocity of 10 m/s and we need to throw 10 of those molecules back every second, so n = 10. Fair enough? That will be (1kg) (10m/s) (10 molecules/s) = 100 Newtons of Force.

Ok.

This is the amount of Force the propeller needs to provide to throw those heavy air molecules around.

No, the propeller doesn't "need an amount of force". Its interaction with the air *results* in an interaction force, which is this 100 Newtons. The propeller exerts 100 N on the air, and hence the air exerts, by reaction, 100 N on the propeller.

Now, you speak of “deciding” to use just half of the force that the propeller needs to provide?

No, Atom Man wonders whether he can have an ENERGY balance if he doesn't have larger drag generated than half of this force by the generator. So he DECIDES to build a generator that will have half of this force as a drag, and is then going to find out how much energy he can get from that generator.

Why don’t we just calculate the energy and power we need to provide to the propeller?

But that is exactly what he's doing! He's just trying to find out what is the CONDITION for the drag needed not to exceed half of the propulsion force of the propeller.

KE = 1/2 mv^2 so at 10m/s the KE is 50 Joules. The power = Energy/time or Energy x n = 50 J x 10 = 500 Watts. So our generator needs to be able to provide 500 Watts to the propeller.

Exactly.

Now, would you care to continue this story? Where do you propose to find this 500 Watts? Do you propose to use just half of the Force needed at the propeller to run the generator?

Well, that's simple. The car is running at 25 m/s (say that is wind speed, and the car is runing at windspeed). The wheel is hence turning at a rotation rate which is such that at its rim, the velocity is 25 m/s. (let's not go deep down the trench here, please) Now, if you put a generator on the axle of the wheel which
delivers 500 W, then that will generate a DRAG equal to 500 W / 25m/s = 20 N.

So powering a generator with the wheel gives you a drag force of 20 N, far below the limit of 50 N we set ourselves.

So we cause a drag force of 20 N by the wheels, while the propeller pulls forward with 100 N. Net force on the car: 80 N forward.

EDIT: I just realize that you gave a windspeed of 10 m/s while I took arbitrarily 25 m/s. No problem, the 10 m/s is handled next...

Do you believe that a motor which requires 500 Watts of power input to produce 100 Newtons of Force can be tapped into to use 50 Newtons of that Force to run a genset which then in turn produces the needed 500 Watts?

Yes, of course. What do you think is the drag caused by a generator delivering 500 W on an axle of a wheel that runs at 25 m/s ?

Imagine that this is not a FDDW car, but rather a normal sailing car (to remove temporarily any mental block), and that the user has installed a generator on the axle of the wheels to power the headlights and make the stereo play. What do you think is the drag force that such a generator will generate if it delivers 500 W and if the car is going at 25 m/s ?

EDIT: Or still another way of seeing this: imagine that you are an engineer and that you have to design a BRAKE for a car that will run at 25 m/s and that will need a braking force of 20 N. How much power are you going to dissipate in the brake and how much cooling do you have to foresee ? Answer:
to have a braking force of 20 N (drag) on a wheel that runs at 25 m/s, no matter what I do, I will have to dissipate 20 N x 25 m/s = 500 W of heat.
Say that you are an engineer working in the aeronautical industry, and that you have to design the brakes for an airplane that will touch down at 25 m/s and that you need a braking force of 100 N. What cooling do you need ? If you think that only 500 W will be dissipated, and you design the cooling of the brakes that way, *your brakes will melt* and you will have committed a professional error. 100 N at 25 m/s will dissipate 2500 W ! So this is important to understand for an engineer.

Now, to come back to Atom Mans' case, as we are throwing the balls at 10 m/s, and the condition he found was that we should throw them SLOWER than the wind speed (at ground level) for his condition of half drag force, then this means that we should just arrive at 50 N drag when the cart (and the wind) is actually doing 10 m/s. Let us check.

We still need 500 W, but this time the wheels are only running at 10 m/s. The drag that a 500 W generator will cause this time is 500 W / 10m/s = 50 N. Indeed.
When we throw the balls at exactly wind speed (on the ground) when the cart is running at wind speed, the drag force is half of the propeller force.But, but but, you say, there MUST be something wrong, I don't know what, but the result CAN'T be that I have a thing that *generates* me 100 N, and that only *drags* 50 N, because such a thing would speed up all by itself, right ? It is a PMM, right ? I don't know where, but you MUST be playing a trick on me somewhere ?

Let us have Silly Man, who has heard over the whole thing, and thinks he's now going to become rich with a PMM based upon Atom Mans invention. He puts a similar cart on a large track on a windless day, and tries to speed up "for nothing". Just to get going, he asks his assistant to speed him up to 20 m/s. He will throw the molecule-balls also at 10 m/s. "It's even easier" he thinks, now that this is a windless day, the molecules come in at 20 m/s!
So he STOPS the molecules, and throws them back at 10 m/s. He needs 500 W for the throwing, and he will have a drag on the wheels of 500W / 20m/s = 25 N, but he will have a forward force from his throwing of 100 N, so he will "speed up" for nothing, right ?
Not right. This time he STOPS the molecules which CAME IN at 20 m/s. So he has to stop 10 molecules of 1 kg per second, and he will have to exercise hence a force of 200 N on them to stop them. By reaction, this will cause a backward force of 200 N on him, and hence he has: 100 N forward (throwing), 200 N backward (stopping), 25 N backward (drag) = 125 N backward! His car quickly stops.

Mm, he says, that's normal. The balls come in faster than I can throw them, so in fact I slow them down. That's not a good idea. So this time he asks his assistent to have the car only sped up at 5 m/s. Then the balls come in at 5 m/s and he will accelerate them to 10 m/s. Let's see what this does. At 5 m/s, to generate 500 W, you need a drag of 500 W / 5 m/s = 100 N.
The balls come in at 5 m/s and he STOPS them. So this will result in a backward force of 50 N. He has hence: 100 N forward, 50 N backward, 100 N drag backward = 50 N backward. Again, his car stops. Because the car is too slow of course, he says.

At 8 m/s, we have: drag: 500 W / 8m/s = 62.5 N ; stopping: 80 N, so we have:
100 N forward (throwing), 80 N backward (stopping), 62.5 N drag = 42.5 N backward, stops again!

And for years, Silly Man tries to find the right combination of throwing speed, car speed, etc... and ends up ruined and locked up in an asylum.

The trick was that because of the wind, the molecule balls came in GENTLER (in our case, were even at rest) in the car than that they could be thrown out, while nevertheless throwing them out at a velocity lower than ground speed. That is due to the fact that there was wind.

If we look at the energy balance in the ground frame: the affected molecules had initially 10 m/s of speed (wind), and by the throwing back, they ended up at rest. So those molecules lost 500 W. That's where, ultimately, the power comes from...
If originally, the molecules don't have any kinetic energy in the ground frame, you won't be able to use it.

 

[uart]

Let’s look at your “calculations” in a bit more detail: You start out by saying the propeller, in order to grab those air molecules, needs to exert a certain amount of Force equal to mvn/s. Let’s put some numbers in there to bring this fantasy to life, shall we? Let’s use those heavy air molecules you mentioned that weigh 1 kg each and let’s start out at wind velocity of 10 m/s and we need to throw 10 of those molecules back every second, so n = 10. Fair enough? That will be (1kg) (10m/s) (10 molecules/s) = 100 Newtons of Force. This is the amount of Force the propeller needs to provide to throw those heavy air molecules around.
Now, you speak of “deciding” to use just half of the force that the propeller needs to provide?
Why don’t we just calculate the energy and power we need to provide to the propeller? KE = 1/2 mv^2 so at 10m/s the KE is 50 Joules. The power = Energy/time or Energy x n = 50 J x 10 = 500 Watts. So our generator needs to be able to provide 500 Watts to the propeller.
Now, would you care to continue this story? Where do you propose to find this 500 Watts? Do you propose to use just half of the Force needed at the propeller to run the generator? Do you believe that a motor which requires 500 Watts of power input to produce 100 Newtons of Force can be tapped into to use 50 Newtons of that Force to run a genset which then in turn produces the needed 500 Watts? Do you honestly believe in what you have written or are you being dishonestly deceptive? In either case, I think you should forget about physics altogether and concentrate on your true calling!

Ok let's do those calculations again, but properly this time.

First up you want 100N thrust, that's a bit more than such a little cart needs but I'll go with that figure anyway. You want to throw at 10m/s (same as wind speed) when the previous calculations clearly showed that it was more efficient to throw more mass but at a lower speed, but I'll go with that figure too.

Thrust = (10 kg/sec) * (10 m/sec) = 100 N
Power required = 1/2 * (10 kg/sec) * (10 m/sec)^2 = 500 Watts
Power Available = F v = (100 N) * (10 m/sec) = 1000 Watts

Where's the problem with that Schroder?

Lets do it even more efficiently. Let's develop our thrust by moving more mass per second but at a lower speed. Say the wind speed is 10 m/s as before, but this time we throw 20 kg/sec but at a reduced speed of 5 m/s. Now we have :

Thrust = (20 kg/sec) * (5 m/sec) = 100 N (same as before).
Power Required = 1/2 * (20 kg/sec) * (5 m/sec)^2 = 250 Watts.
Power Available = F v = (100 N) * (10 m/s) = 1000 Watts.

So it's even better this time wouldn't you agree?

 

[rcgldr]

Now Atom Man knows that if his strategy is to be successful he must be sure that the drag from his generator is less than this amount, so he decides to use only half of this extra thrust in powering the generator.

But this isn't the case. What Atom man needs to know is the power output versus power input for a propeller, and the minimal amount of thrust required to overcome the aerodynamic drag from a small apparent headwind on the cart and the rolling resistance from larger apparent head speed from the ground (rolling resistance) to obtain a specific maximum speed. Chances are that the prop chosen will generate more thrust than is minimally needed, but this will be opposed by the wheels that drive the prop, and the ratio of opposing force / thrust is < 1 if the prop and related gearing are efficient enough. So the ratio of opposing force / thrust remains about the same, but since the drag related factors remain about the same, the ratio (thrust - opposing force) / (drag factors) increases with prop thrust.

Let’s use those heavy air molecules you mentioned that weigh 1 kg each and let’s start out at wind velocity of 10 m/s and we need to throw 10 of those molecules back every second, so n = 10. That will be (1kg) (10m/s) (10 molecules/s) = 100 Newtons of Force.

The speed of the molecules thrown back doesn't need to be 10 m/s. 4 m / sec would probably be sufficient. The cart is only attempting to go DDWFTTW, not twice as fast as the wind. So restated:

(1kg) (4m/s) (10 molecules/s) = 40 Newtons.
KE = 1/2 mv^2 so at 4 m/s the KE is 8 Joules.
power output = Energy/time or Energy x n = 8 J x 10 = 80 watts.

Assume the cart is moving at 12 m/s down wind.
Assume prop and gearing efficiency to total about 69.4%.
So power input needs to be (1/.694) = 1.44 times power output = 144 watts.
At 12 m / s, the required driving force = (144 / 12) = 12 Newtons.

So we have 40 Newtons of thrust opposed by 12 Newtons of force related to driving the prop, leaving 28 Newtons of force to compenstate for the opposing aerodynamic drag of the apparent 2 m/s headwind, and the rolling resistance from the 12 m / sec forward speed of the cart.

 

[schroder]

Thrust = (10 kg/sec) * (10 m/sec) = 100 N
Power required = 1/2 * (10 kg/sec) * (10 m/sec)^2 = 500 Watts
Power Available = F v = (100 N) * (10 m/sec) = 1000 Watts

Where's the problem with that Schroder?


So it's even better this time wouldn't you agree?

You and Vanesch should get together and write some joint effort fairy tales! Between his “propeller that does not require any Force” and your “power available” you two can rewrite the book of physics, as long as it is in a Fairy Tale! Please remember to start you stories with “Once upon a time” so the reader will know what to expect.
You write this, and ask me what is wrong with it?

Power Available = F v = (100 N) * (10 m/sec) = 1000 Watts

Where's the problem with that Schroder?

Do you mean that you do not know what is wrong with it, or are you just testing me?
The power required to turn the propeller is clearly 500 Watts from your own calculation:
Thrust = (10 kg/sec) * (10 m/sec) = 100 N
Power required = 1/2 * (10 kg/sec) * (10 m/sec)^2 = 500 Watts

Now where is the power available coming from? You mean the Total power expended in the isolated system of the propeller and the air as the propeller works on the air and vice versa? And to what part of the system is this expended power available, pray tell? I can answer that for you: The power is expended; it is no longer “available”, GONE, FINISHED, and KAPUT!

But that is not all! You still have not shown where the 500 Watts is coming from to throw those heavy air molecules around. You and Vanesch are playing a game here, thinking that maybe no one is out there who can see through it. I find this extremely boorish and just plain stupid. If DDWFTTFW means that much to you, that you will come on here and post blatant lies and falsehoods, and if this Physics Forum is so afraid of coming out and challenging these lies, then this Forum is losing credibility steadily. I know there are people on here who can see that what you are posting is garbage. You cannot show a source of energy for the cart at velocity greater than wind velocity, so you resort to this type of circular argument, which any first year physics student can see through. It is getting OLD and TIRESOME.

My plan is to continue investigating the heterodyne. I generally think it is best to take notice when numbers crop up on their own and then try to see if those numbers fit any meaningful known relationship. When the ratio of 2.4 came up, I automatically thought of the first carrier drop out of a Bessel null, and when 1.41 came up I naturally think out peak and RMS values. Also, from experience with rotating machinery I am very familiar with beats, harmonics and vibrations. All the indications are there for some sort of heterodyne. Looking at Swerdna’s video, the cart first advances in one direction until it stands still momentarily on the tread with the linear velocity of the wheel exactly the same as the linear velocity of the tread. That is the very definition of a beat frequency and it cannot stay there for long! One of them, the cart or the tread, must drop out of that beat and of course it is the cart. The drop out is very similar to a carrier drop out at a Bessel null. If you have ever done a carrier drop out you would immediately see the resemblance here in the mechanical system. Now the cart starts running in the opposite direction at a ratio of 1 : 2.4 which is the modulus of a first Bessel null! I am not making any of this up, these are the actual numbers obtained by analyzing the video. I cannot ignore these numbers. Do I have a complete theory or a complete explanation of what is happening here? No, obviously not. The combination of the cart and the turntable is very interesting and I intend to look into this until I understand it completely. In the meanwhile, I will no longer respond to your and Vanesch’s blatant falsehoods as they are not worth one more moment of my time.
When I have my heterodyne model complete, I may post a link to it here, or I may not. Meanwhile feel free to lie and hoodwink your way to fame on myth busters, but I seriously doubt they are still interested in DDWFTTFW, especially after I fill them in.

My turn to roll on the floor laughing!

 

[vanesch]

You and Vanesch should get together and write some joint effort fairy tales! Between his “propeller that does not require any Force” and your “power available” you two can rewrite the book of physics, as long as it is in a Fairy Tale! Please remember to start you stories with “Once upon a time” so the reader will know what to expect.
You write this, and ask me what is wrong with it?

Power Available = F v = (100 N) * (10 m/sec) = 1000 Watts

Where's the problem with that Schroder?

Do you mean that you do not know what is wrong with it, or are you just testing me?

There is of course nothing wrong with it. What do you think is wrong with it ?

The power required to turn the propeller is clearly 500 Watts from your own calculation:
Thrust = (10 kg/sec) * (10 m/sec) = 100 N
Power required = 1/2 * (10 kg/sec) * (10 m/sec)^2 = 500 Watts

Yes, that is correct. We need to have 500 W to feed the motor of the propeller/throwing machine/...

Tell me, a generator that DELIVERS 500 W and is powered by the axle of a wheel that rolls at 10 m/s, HOW MUCH DRAG FORCE DOES THAT CAUSE ?

(answer: 50 N).

Other question: if you want a drag force of 50 N by BRAKING, how much POWER do you dissipate in the brake if the car is going at 10 m/s ? Could you, pretty please, answer that question ?

In the meanwhile, I will no longer respond to your and Vanesch’s blatant falsehoods as they are not worth one more moment of my time.

I fully understand that. The mechanics is so evident, and the proof so easy, that it becomes embarrassing to try to talk yourself out of it, that's it ? You don't want to enter again in something like your 10 + 2 = 8 debacle, I understand. Who was it saying that you are the kind of person that would even maintain that 1 + 1 = 3, rather than admit to be wrong ? In fact, you warned us from the start: "I cannot accept to be wrong", you said a while back. You know, sometimes, even though it seems insurmountable, admitting that you've been wrong all the time can be a liberating experience. You should try it. We all know it now here. There's no point trying to hide it. It will give you probably a great sense of relief.

In the mean time, in as much as you are the engineer you claim to be, you should ponder about the question of the brakes... How much heat must I be able to evacuate from a brake that gives me 50 N of drag on a car that runs at 10 m/s... If you have that one wrong, then you might make a lot of professional errors. Now, maybe you wouldn't make an error there if you could liberate yourself from your mental block.

 

[vanesch]

My plan is to continue investigating the heterodyne. I generally think it is best to take notice when numbers crop up on their own and then try to see if those numbers fit any meaningful known relationship.

You mean, that if a phone number contains 31415, that you think that you're calling a circle ?

BTW, tell us when your turntable is ready, and when your 3 tachymeters are installed, to compare it to my prediction T3 = R/r (T1 - T2). I'm "curious"

 

[spork]

You and Vanesch should get together and write some joint effort fairy tales!

I honestly don't know if you can see the irony, or you're really that misguided.

You and Vanesch are playing a game here, thinking that maybe no one is out there who can see through it. I find this extremely boorish and just plain stupid. If DDWFTTFW means that much to you, that you will come on here and post blatant lies and falsehoods, and if this Physics Forum is so afraid of coming out and challenging these lies, then this Forum is losing credibility steadily.

If you look at this simply as a numbers game, it should be perfectly clear that everyone but you sees Vanesch's posts for what they are - accurate and truthful. It amazes me that you continue to dig a deeper and deeper hole for yourself.

My plan is to continue investigating the heterodyne.

By all means. Please also investigate HHO, pyramid power, crystals, and coriolis as an explanation for the cart's performance.

I generally think it is best to take notice when numbers crop up on their own and then try to see if those numbers fit any meaningful known relationship.

Better physics through numerology.

The combination of the cart and the turntable is very interesting and I intend to look into this until I understand it completely.

In that case I hope you're a young man - and have lots of schooling ahead of you.

 

[ThinAirDesign]

Meanwhile feel free to lie and hoodwink your way to fame on myth busters, but I seriously doubt they are still interested in DDWFTTFW, especially after I fill them in.

Well, I can only *hope* that you will do as you say and "fill them in".

Since I fear that you're quackery will get lost amongst all the other quackery that the MythBusters get inundated with (see their own forum for examples), I have an offer for you: Write up your paper or whatever it is that you are going to "fill them in" with, post (or PM) it and I will make sure it gets to the appropriate Executive Producer. We have an actual dialog with these folks and we can make sure your particular form of quackery get's it's due with them.

Truly schroder -- every good episode needs a foil and if you're credentialed and as experienced as you say, if your position has the support of the "Academy" as you say, and the thought of making us look like the boobs you believe we are holds any appeal, we need each other. We need to produce a skeptic who comes off as smart, educated and convincing, who is also willing to take an *on air* stand on this -- so far the people willing to criticize haven't met those filters.

JB

PS: Remember, when it comes to your referenced "fame on MythBusters", neither spork no I even expect to be mentioned on any such show -- the right skeptic however would likely actually get the on-air "fame", not us. After all, everyone remembers the pilot who said the plane wouldn't take off on the POAT episode.

 

[The Dagda]

I cannot ignore these numbers. Do I have a complete theory or a complete explanation of what is happening here? No, obviously not. The combination of the cart and the turntable is very interesting and I intend to look into this until I understand it completely. In the meanwhile, I will no longer respond to your and Vanesch’s blatant falsehoods as they are not worth one more moment of my time.
When I have my heterodyne model complete, I may post a link to it here, or I may not. Meanwhile feel free to lie and hoodwink your way to fame on myth busters, but I seriously doubt they are still interested in DDWFTTFW, especially after I fill them in.

My turn to roll on the floor laughing!

That comes off as very arrogant you are aware of the level of education of some of these people, they're not fobbing you off with pseudoscience, this is standard physics for a very good reason.

Still can't say I'm surprised, when in a hole...

 

[spork]

Remember, when it comes to your referenced "fame on MythBusters", neither spork no I even expect to be mentioned on any such show -- the right skeptic however would likely actually get the on-air "fame", not us.

Agreed, but that's fame I could do without.

 

[Subductionzon]

Schroder, I seriously suggest that you look at the recent posts of vanesh again. It still irks me when I watch spork's video for his Mythbusster's challenge when I hear one of my quotes trying to debunk Jack Goodman's video. You have left so many crazy statements that they could haunt you for years.

 

[mender]

My plan is to continue investigating the heterodyne. I generally think it is best to take notice when numbers crop up on their own and then try to see if those numbers fit any meaningful known relationship. When the ratio of 2.4 came up, I automatically thought of the first carrier drop out of a Bessel null, and when 1.41 came up I naturally think out peak and RMS values.

(snip)

I am not making any of this up, these are the actual numbers obtained by analyzing the video. I cannot ignore these numbers. Do I have a complete theory or a complete explanation of what is happening here? No, obviously not. The combination of the cart and the turntable is very interesting and I intend to look into this until I understand it completely. In the meanwhile, I will no longer respond to your and Vanesch’s blatant falsehoods as they are not worth one more moment of my time.

Schoder, you might save yourself a lot of trouble if you can get Ynot to repeat his test at a couple of different turntable speeds and then see if the data still holds the pattern that you're looking for.

I'm curious, you really don't know that the energy to run this thing is coming from the wind? I think I can help you work through that issue if you are willing to go step by step.

 

[spork]

It still irks me when I watch spork's video for his Mythbusster's challenge when I hear one of my quotes trying to debunk Jack Goodman's video.

Sorry about that. We just pulled out a few of the classic quotes at random. I don't think I could tell you who said anyone of them.

 

[Subductionzon]

Sorry about that. We just pulled out a few of the classic quotes at random. I don't think I could tell you who said anyone of them.

No problem, it keeps me humble, not humber

 

[ThinAirDesign]

LOL Sub ... I'm the guy who picked though and pulled out those quotes and until now I had no idea one of those was yours.

JB

 

[spork]

LOL Sub ... I'm the guy who picked though and pulled out those quotes and until now I had no idea one of those was yours.

Well, it's a good testimony to considering new real-world evidence and keeping an open mind even when you start out with strong intuitive feelings on the subject. Imagine if everyone was willing to do that.

 

[Subductionzon]

Well, it's a good testimony to considering new real-world evidence and keeping an open mind even when you start out with strong intuitive feelings on the subject. Imagine if everyone was willing to do that.

Then we would have hardly any fun at all. Where would threads like this be without schroder, humber and now 3bodyproblem? Of course these people have possibly served a good purpose keeping this topic alive for others to find and worry over.