Merged DDWFTTW in a “Real” Wind

[jduffy77]

You should listen to what everyone is trying to tell you, because in this one instance at least, everyone is right! The thrust force generated by the propeller pushes the air one way, at 2.2 m/s, relative to the cart, and pushes the cart the opposite way, at 9 m/s relative to the ground. The 350 Watts in my calculation is adequate to push the air at 2.2 m/s ( 160 N x 2.2 m/s ~ 350 Watts) . But it is not adequate to push the 295 kg car over the ground at 9 m/s (160 N x 9 m/s = 1,440 Watts)

Maybe now you can see why I base my analysis around energy and power as well as forces, as just looking at forces alone does not tell you the whole story. In this case, if there was no other source of energy, it is clear the cart could not work. For example, if the cart was trying to run in still air, it should be very obvious that it isn’t going anywhere for long because it does not have the energy needed to do so.

In the case we are considering, with a wind blowing at 7 m/s and the cart moving along due to the Good Lord towing it up there and releasing it, there is some energy available from slowing down the wind.

The calculation for propeller thrust showed you the mass of air that gets slowed down is about 72 kg and is slows by 2.2 m/s with respect to the ground. From that you can see that the air mass loses about 935 J/s. Earlier, I rounded that up to a nice 1000 Watts of additional power available to the cart, if that power could be 100% transferred to the cart. The 100% transference is absurd, but even if it were allowed to happen, that makes the total power available to the cart 1,350 Watts and it needs 1,440 Watts to move nine meters in one second. If it can’t move that nine meters in one second, it cannot produce the 350 Watts at the propeller, and that is the end of that.

Notice, in this analysis I didn’t even bother to mention the fact that in order to get that 350 Watts on the propeller, the cart needed to come up with somewhere between 750 and 850 Watts to overcome efficiency losses. Remember, we are in the ground frame, so no ground power allowed. The only source of energy is whatever kinetic energy the cart has been given to it by the Holy Tow Vehicle, which has towed and departed. So the cart has to dig into its own kinetic energy to get the propeller working on the air. Some 500 Watts (500 J/s) is being lost in that process, slowing the cart down. Now to get back up to 9 m/s it needs that 1,440 Watts, plus the 500 Watts it burned off in this rather asinine assault on conservation of energy.

Well, as all assaults on conservation of energy are doomed to fail, so it is with this one. There simply is not enough energy available from the wind to power the cart to move the required 9 m/s over the ground to spin the propeller and do the work to slow down the air and get the energy from the wind to power the cart to move the required 9 m/s over the ground to spin the propeller to do the . . . . . .By now maybe you see what the problem is? No matter how many times our carteers deny it, they are in fact proposing just another in a long long line of perpetual motion machines.
The math says they are wrong. The NALSA data says they are wrong, and the treadmill has not one blessed thing to do with wind power.

Just pointing out for anyone who doesn't want to slog through it again, Hyevpranggloss has made the same glaringly obvious error he always does.

ETA: The real reason he talks about power and forces at the same time is so he can mix things up inappropriately. It is the same reason he writes in prose instead of equations. So he can hide his silliness. All you have to do is read through it carefully and his problems become obvious.

 

[3point14]

The math says they are wrong.

No it doesn't.

And even if it does, then a reasonable man, having seen the concept practically proven, would accept that there is a flaw in the maths.

Reality trumps model. Every. Single. Time.

 

[Modified]

So by exactly the same token, when I carry a briefcase across a room I am doing no work on it.

Right.

That is not what 69dodge said at all. It would be more like "when I carry a briefcase across a room by the handle, the handle does no work on the rest of the briefcase", which is a more interesting question.

 

[69dodge]

But in the cart's case, it is a trivial matter that in the FOR where the ground is stationary, there is a force of 156 N (or 160) acting on the cart over a distance of 9 meters in one second. So that is 1404 (or 1440) Joules of work is done on the cart.

There is work being done on the cart at that rate.

The effective power is derived from the "Engine Power" formula on that page previously linked. It is the amount of power effectively converted to thrust AFTER aerodynamic losses (turbulence, rotational losses etc.) are taken out.

It's the effective work done at the prop, and it gets added to the KE loss of the air to account for the total work done on the cart above.

Would all of the above still hold if the propeller were being spun at the same RPM by an engine rather than by the wheels rolling over the ground?

When you say that the "total work" done on the cart per second is 1440 Joules, I gather that you are referring not to the net work done, but rather to the total positive work done, via the propeller. And to get the net work done on the cart, we'd need to add the negative work done on the cart by the ground, via the wheels. Is that correct?

 

[69dodge]

You should listen to what everyone is trying to tell you, because in this one instance at least, everyone is right! The thrust force generated by the propeller pushes the air one way, at 2.2 m/s, relative to the cart, and pushes the cart the opposite way, at 9 m/s relative to the ground. The 350 Watts in my calculation is adequate to push the air at 2.2 m/s ( 160 N x 2.2 m/s ~ 350 Watts) . But it is not adequate to push the 295 kg car over the ground at 9 m/s (160 N x 9 m/s = 1,440 Watts)

I don't know why you keep mentioning the mass of the cart. It doesn't matter except insofar as it determines the amount of rolling resistance and the magnitude of any acceleration. It doesn't determine the sign of the acceleration.

In the case we are considering, with a wind blowing at 7 m/s and the cart moving along due to the Good Lord towing it up there and releasing it, there is some energy available from slowing down the wind.

The calculation for propeller thrust showed you the mass of air that gets slowed down is about 72 kg and is slows by 2.2 m/s with respect to the ground. From that you can see that the air mass loses about 935 J/s. Earlier, I rounded that up to a nice 1000 Watts of additional power available to the cart, if that power could be 100% transferred to the cart. The 100% transference is absurd, but even if it were allowed to happen, that makes the total power available to the cart 1,350 Watts and it needs 1,440 Watts to move nine meters in one second.

Why does it need 1440 Watts to move nine meters in one second? If I were trying to calculate how much power it needs, I'd figure out how much air resistance and rolling resistance there was, and see how much power is needed to overcome those two. What has your 1440 Watts to do with them?

Remember, we are in the ground frame, so no ground power allowed.

I think you are being inconsistent. We are in the ground frame, so the ground isn't moving. But it exerts a torque on the wheels of the cart, and the wheels are moving (turning). If the 160-N force of the air on the propeller can somehow be the cause of (160 N)(9 m/s) = 1440 W that needs to be accounted for, even though the air isn't moving at 9 m/s, simply because the cart is moving at 9 m/s, then the torque of the ground on the wheels (multiplied by their angular speed) should be allowed to account for some of that 1440 W, because the wheels are turning, even though the ground is stationary.

 

[Dr. Keith]

I don't need to worry at all about what you say because all the evidence I
need is in the NALSA data;

Cool. Let's see it then.

and I have the TM cart.

No you don't. Never have, never will.

Silly rabbit, carts are for adults.

 

[Dr. Keith]

I still have all that data Spork,

so do something with it already.

Or is it over your head? Let me guess, you don't really understand the data do you? Aww, that's kinda sad. Do you know anyone who can help, maybe a middle school kid with a graphing calculator.

 

[Llyricist]

Would all of the above still hold if the propeller were being spun at the same RPM by an engine rather than by the wheels rolling over the ground?

yes

When you say that the "total work" done on the cart per second is 1440 Joules, I gather that you are referring not to the net work done, but rather to the total positive work done, via the propeller. And to get the net work done on the cart, we'd need to add the negative work done on the cart by the ground, via the wheels. Is that correct?

No, the negative work done by the wheels (I don't think that's a strictly accurate description, but we know what we mean) only effects the amount of positive work necessary to do the job. If you put a 1/2 HP motor and fuel on the cart and replaced the pilot with one that was the ~10-20 lbs lighter to make up the difference in weight, you could get the same performance with only 72 N of thrust over 9 meters, or 650 J. The air would lose 462 J and the engine would add 188 J /s of effective power at the shaft.

If desired, I can spell out the necessary math.

 

[Dr. Keith]

It's worth more than that, Tobermoryphil.

A stupid car that goes up a treadmill is worth more than $1000? Really, that is pretty damn delusional. Even I'd it existed and even if it worked. Which it doesn't.

 

[Dr. Keith]

Trenchant denial of my cart and the posited descriptions of what it may be ( though you also say it does not exist at all) only makes the wide-eyed wonder seen in your cart videos all the more amusing.

It doesn't exist. You are ting and know that it will never exist. We are laughing as you squirm around like a petulant child caught in a poorly thought out lie.

Nothing is going to stop that video from being posted where and when I choose. A simple compilation of the NALSA data will show all but the ardent that you have no such ddw record, NALSA support, or not.

simple, but not yet done. Pissing in the toilet is simple, I've done it twice this morning. But you can't get the data to make your point after how many weeks?

And the only thing preventing the video from existing is the lack of a decent cgi budget. The cart doesn't exist. Never has never will.

 

[Dr. Keith]

[qimg]http://www.internationalskeptics.com/forums/attachment.php?attachmentid=22119&stc=1&d=1305114178[/qimg]
.

Oh my god they did it in a tornado!

Or the raw data has some collection artifacts. Hmm. I'm going with tornado, it will make the movie better.

 

[jduffy77]

Why does it need 1440 Watts to move nine meters in one second? If I were trying to calculate how much power it needs, I'd figure out how much air resistance and rolling resistance there was, and see how much power is needed to overcome those two. What has your 1440 Watts to do with them?

It is a little hard to figure out because his error is so basic that you can't believe what he is doing. He needs to count the work done at air/propeller twice. He does this by counting it in two separate reference frames. His 1440 Watts is coming essentially form multiplying values which should not have any relation to each other. Like Price Gallon Milk = Price of Quart * Air Temp.

 

[Llyricist]

It is a little hard to figure out because his error is so basic that you can't believe what he is doing. He needs to count the work done at air/propeller twice. He does this by counting it in two separate reference frames. His 1440 Watts is coming essentially form multiplying values which should not have any relation to each other. Like Price Gallon Milk = Price of Quart * Air Temp.

No, actually it is correct (sort of, I got 1404J but he rounded the numbers up), it refers to my example way back about 9 pages ago. The resistance was calculated, and so was the necessary negative work to gain the necessary thrust.

 

[Llyricist]

Oh my god they did it in a tornado!

Or the raw data has some collection artifacts. Hmm. I'm going with tornado, it will make the movie better.

Yeah I love how he threw an example out there without any context, or time units, or anything.

 

[Llyricist]

And this says your ddw record is a farce.

As with so many other even more obvious examples that nobody was to see, the cart's performance nose-dives as it approaches 0 degrees with respect to the wind, and improves again when that difference increases.
[qimg]http://www.internationalskeptics.com/forums/attachment.php?attachmentid=22117&stc=1&d=1305107294[/qimg]

We have no way of knowing from that chart what the cart's heading is with respect to the wind.

 

[profpangloss]

I don't know why you keep mentioning the mass of the cart. It doesn't matter except insofar as it determines the amount of rolling resistance and the magnitude of any acceleration. It doesn't determine the sign of the acceleration..

Why should it bother you that I mentioned the mass of the cart (295 kg)? I didn't use it in the calculations. Does it bother you that a cart of 295 kg mass moving at 9 m/s is supposedly powered by a 72 kg mass of air moving at 7 m/s? If it does, that is a good sign! Maybe you are just beginning to see how ridiculous this ddwfttw claim is?

Why does it need 1440 Watts to move nine meters in one second? If I were trying to calculate how much power it needs, I'd figure out how much air resistance and rolling resistance there was, and see how much power is needed to overcome those two. What has your 1440 Watts to do with them?.

Why? Because work = Force x Distance, and Power = Work / Time. Since Velocity = Distance / Time, Power = Force x Velocity. All very well established in the fundamentals of physics. The cart must move at 9 m/s under a force of 160 N. The Power is 1,440 Watts. There is no disputing that, and if you argue against this again, you just embarrass yourself.

I think you are being inconsistent. We are in the ground frame, so the ground isn't moving. But it exerts a torque on the wheels of the cart, and the wheels are moving (turning). If the 160-N force of the air on the propeller can somehow be the cause of (160 N)(9 m/s) = 1440 W that needs to be accounted for, even though the air isn't moving at 9 m/s, simply because the cart is moving at 9 m/s, then the torque of the ground on the wheels (multiplied by their angular speed) should be allowed to account for some of that 1440 W, because the wheels are turning, even though the ground is stationary.

NO! We are in the ground frame, so the ground is not moving, and by definition it has no energy. Please do not embarrass yourself any further in attempting to dispute that fundamental fact. If the wheels are rolling it is because some other energy has been used to get them to roll against the resistance of the ground. Any torque at the wheels is entirely due to whatever Force and energy is driving the cart against the resistance offered by the ground. The ground resistance does help to develop that torque, in that it facilitates the transformation of the cart's kinetic energy ( energy of motion) into rotational energy at the wheels. The ground resistance acts as a transform only, it cannot add any energy to the cart. On the contrary, the cart loses kinetic energy in the form of heat as it works against the resistance of the ground. NONE of the 1,440 Watts can be accounted for by an "ground energy". As I have shown, the cart's kinetic energy, in this particular case we are looking at, can only provide the 350 Watts at the propeller ( but the cart loses an additional 500 Watts to heat). If the wind cannot make up the shortage to get to 1,440 Watts, (and it can't) the cart does not work, it slows down. This should come as no surprise to anyone, since the carteers are claiming voodoo physics.
All my numbers are out there, easy to understand while all you are hearing from the carteers is nonsense. You decide what you want to believe.

 

[Llyricist]

No, the negative work done by the wheels (I don't think that's a strictly accurate description, but we know what we mean) only effects the amount of positive work necessary to do the job. If you put a 1/2 HP motor and fuel on the cart and replaced the pilot with one that was the ~10-20 lbs lighter to make up the difference in weight, you could get the same performance with only 72 N of thrust over 9 meters, or 650 J. The air would lose 462 J and the engine would add 188 J /s of effective power at the shaft.

If desired, I can spell out the necessary math.

And for the energy balance in that case, we have Inputs of energy amounting to 462 J from the air, and 280 J from the engine. For a total of 742 J.

Losses in the 80% efficient drive train are 56 J
Aerodynamic losses at the prop are 36 J
Aerodynamic drag at 31 J
Rolling resistance at 521 J
For a total of 644 J

Leaving 98 J for the added KE of the cart after acceleration.

 

[69dodge]

Why? Because work = Force x Distance, and Power = Work / Time. Since Velocity = Distance / Time, Power = Force x Velocity. All very well established in the fundamentals of physics. The cart must move at 9 m/s under a force of 160 N. The Power is 1,440 Watts.

NO! We are in the ground frame, so the ground is not moving, and by definition it has no energy. [...] NONE of the 1,440 Watts can be accounted for by an "ground energy".

Yes, power is force times velocity. But which velocity? The force is exerted between the air and the cart. And you are using the velocity of the cart, which is receiving the force, not that of the air, which is providing the force. Fine. All well and good. But power is also torque times angular speed. The torque is exerted between the ground and the wheels. So you should use the angular speed of the wheels, which are receiving the torque, not that of the ground, which is providing the torque. (The ground doesn't even have an angular speed. It's not turning; the wheels are.)

The ground is not moving, you say, so it has no energy to provide to the cart. True enough. But, by the same token, the ground remains not moving, so neither can it remove any energy from the cart. A large part of your 1440 Watts of power is used to overcome the backward force of the ground on the wheels. So, according to you, the ground can remove energy from the cart, but it can't add any?

I still say you're being inconsistent.

 

[Llyricist]

Yes, power is force times velocity. But which velocity? The force is exerted between the air and the cart. And you are using the velocity of the cart, which is receiving the force, not that of the air, which is providing the force. Fine. All well and good. But power is also torque times angular speed. The torque is exerted between the ground and the wheels. So you should use the angular speed of the wheels, which are receiving the torque, not that of the ground, which is providing the torque. (The ground doesn't even have an angular speed. It's not turning; the wheels are.)

The ground is not moving, you say, so it has no energy to provide to the cart. True enough. But, by the same token, the ground remains not moving, so neither can it remove any energy from the cart. A large part of your 1440 Watts of power is used to overcome the backward force of the ground on the wheels. So, according to you, the ground can remove energy from the cart, but it can't add any?

I still say you're being inconsistent.

Strictly speaking, all that backward force is actually occurring in the wheels, bearings, sprockets, chain, and propeller themselves. In the form of resistance to the rotation of the axles.

Much of that energy is lost in the form of heat radiation, some is gained in the form of heat in the objects, but unused for work, but a good portion is gained in the form of rotational energy in the prop.

 

[69dodge]

How'd you get 1000 Joules per second? I get 1250.

(1/2)(72 kg)(9 m/s)² - (1/2)(72 kg)(6.8 m/s)² = 1251.36 Joules.

I figured out my mistake here.

I used 9 m/s as the speed of the air, instead of 7 m/s.