ThinAirDesigns wrote:

In the case of the DDWFTTW vehicle, the propeller
*speed up* the air relative to itself ('cause that's what propellers
do), but is *slows* the air relative to the ground (which is what has
to happen if we want to harvest energy).

Before you and Rick slink off to repeat this argument elsewhere, I'd
like to say thanks for popping in and giving us your explanation.

You could probably save yourselves some hassle by revising the
introductory paragraph on your explanation page, on the grounds that it
didn't seem to help anyone here to "get it", adding a few fancy
animations and the skateboarder analogy would help too.

ThinAirDesigns wrote:

@Ronald Raygun
The exactly-at-windspeed scenario is a bit of a crazy special case ...

small snip

Therefore the presumption that the wheels must generate 1hp
and the prop must therefore provide 10lb of thrust is a bit silly,

You did notice that it was *you* who brought up the "crazy special
case" of exactly at wind speed. :-)

Indeed. Guilty as charged. But I did it merely in order to illustrate
why I felt it wrong of you to apply the "Power = Force * Speed" formula
using the speed of the input air. I still feel that's so, and moreover
that it's wrong to apply that simple formula at all to cases where the
force in question is applied to objects being accelerated.

It is your own special case that has brought into play the seemingly
odd (but true) case of the 100% efficient propeller which can produce
10lbs of force on 0hp.

Well, I find that hard to take, because surely a prop can only give
thrust as a result of imparting impulse to air, which necessarily
involves accelerating (doing work on) that air, which in turn necessarily
consumes power. If not, why not? It seems to me that a prop giving
thrust must always consume power even if the thrust itself is not
going to do useful work, and even if the fluid's initial speed is zero.
Think of a boat with an outboard engine. Its water prop consumes power
not just when it's pushing the boat through the water, but also when it's
pushing it against an immobile dock. An air prop is in principle just
the same as a water prop, isn't it?

Let me get away from any crazy prop effects by returning to the skater.

Suppose he needs thrust of 60N (that's quite a lot, but I guess his wheels
are very poorly lubricated) to maintain his speed of 2m/s. He applies this
constant and continuous force of 60N to a steady stream of 60kg pedestrians
coming at him at 1m/s, and he grabs hold of each pedestrian just long enough
to accelerate him to 2m/s. The instant he lets go of one guy, he grabs the
next, with a seamless changeover so that his thrust is constant. With an
acceleration of 60N/60kg=1m/s/s, he pushes each pedestrian for 1 second.

Now if you were to apply your simple formula "Power = Force * Speed" to
calculate the power requirement of the skater's arms, in the same way as
were doing to the car's propeller, would you use the force of 60N and
the *arrival* speed of 1m/s? If so, you would get a power of 60W.

This, as I pointed out earlier, is incorrect because if you do the energy
budget, 90J of work is done on each pedestrian, because the pedestrian
travels 1.5m during the second he is being accelerated. Therefore the
mean power needed is not 60W but 90W.

The problem is that "Power = Force * Speed" is no more universally true
than is "Distance = Speed * Time". They only hold where the speeds are
constant. Power is the gradient, with respect to time, of energy,
and energy is the integral, with respect to distance, of force.

Where force is constant, you can say that "Energy = Force * Distance".
When, in addition, distance is linear with time, then energy is also
linear with time, and *then* you can apply the simple formula, but not
generally otherwise.

In the situation of our pedestrians, force is constant but speed is not,
and neither is power. For each pedestrian, as his speed increases from 1
to 2 m/s, the instantaneous power increases from 60W to 120W, and its
mean value is 90W.

We can corroborate the 90W figure by looking at the gain in kinetic
energy of each pedestrian. It starts at 1/2*60kg*(1m/s)^2 when he's
grabbed (that's 30J) and finishes at 1/2*60kg*(2m/s)^2 when he's let
go (that's 120J), so the gain is 90J per pedestrian, corresponding to
90J per second or 90W.

How much of this do you disagree with and why?

Ronald Raygun wrote:
ThinAirDesigns wrote:

@Ronald Raygun
The exactly-at-windspeed scenario is a bit of a crazy special case ...
small snip

Therefore the presumption that the wheels must generate 1hp
and the prop must therefore provide 10lb of thrust is a bit silly,
You did notice that it was *you* who brought up the "crazy special
case" of exactly at wind speed. :-)

Indeed. Guilty as charged. But I did it merely in order to illustrate
why I felt it wrong of you to apply the "Power = Force * Speed" formula
using the speed of the input air. I still feel that's so, and moreover
that it's wrong to apply that simple formula at all to cases where the
force in question is applied to objects being accelerated.

It is your own special case that has brought into play the seemingly
odd (but true) case of the 100% efficient propeller which can produce
10lbs of force on 0hp.

Well, I find that hard to take, because surely a prop can only give
thrust as a result of imparting impulse to air, which necessarily
involves accelerating (doing work on) that air, which in turn necessarily
consumes power.

No. Static thrust does no work any more than a labrador sitting on your
foot and imparting thrust, is doing work.

Thrust is very misleading in fluid dynamics. There are so many ways in
which thrust is easy to measure, but the actual air mass accelerated
both in size and velocity change is not.

In the limit an infinite mass suffering infinitesimal acceleration
generates any thrust you want. For zero power

It is far easier to look overall at the drag profile if the machine, its
speed, and the wind speed and see whether you can using the prop extract
enough work from the air mass to overcome the power to move the vehicle
AND spin the prop to blow air backwards to extract the wind.

It seems to me this concept is about creating a 'dynamic sail'
consisting of a slipstream of air being blown backwards

In a boat, you could have a set of sails on a caterpillar track that
were driven backwards by a screw underwater, and then furled and led
forward to the bow of the ship. Thought experiment only.

Or another way to look at it is that the whole thing represents no more
than wind overdrive. A gearing system.

If I was to pose the question in this form:
"A rope runs through a tunnel at 1m/s. It is driven by a very powerful
motor and serious loads can be applied and it will remain at 1 m/s.
Devise a mechanism that will utilise this rope to drive a cart faster
than 1m/s in the direction the rope is moving'

I think a very simple geared system with one gear being driven by the
rope and the other by the ground wheels, meshed together would end up
with a very high speed of the axles' centres..more than 1m/s.

ISTM this is the same principle.

By having the wheels or the waterscrew you can establish a zero velocity
frame with respect to which the windspeed can be offset to extract
power. The fact that the vehicle is not tied to this frame doesn't
matter, as long as conceptually your 'sail' is.

To my mind it cuts two ways. Theory says it should work, my gut says it
cant,. If the tests are verified tho, it sees to say it can and does work.

It would be a fairly trivial exercise to build a RC model though and try
it out.








If not, why not? It seems to me that a prop giving
thrust must always consume power even if the thrust itself is not
going to do useful work, and even if the fluid's initial speed is zero.
Think of a boat with an outboard engine. Its water prop consumes power
not just when it's pushing the boat through the water, but also when it's
pushing it against an immobile dock. An air prop is in principle just
the same as a water prop, isn't it?


No, props need not consume any power to make thrust In the theoretical
limit.

That they do is down to beating and vortex and sound generation. And
practical size limitations, a very large very thin prop turning very
slowly can generate enormous thrust. Until you start to move, then
thrust times speed dominates and that's the power. A variable pitched
prop in very fine pitch is a lot less draggy and power hungry at slow
speeds an accelerates a plane smartly..as speed builds up, pitch
coarsens to keep up with incoming airflow, and the prop load increases
on the engine. Thrust times speed is here.


Yes it takes power to hover a helicopter, but less than you think. They
have oddly, very long very thin blades, spinning very slowly..if they
were longer, thinner and slower they would take even less power, but
then the aircraft would be infeasibly large.

On Oct 2, 1:08*pm, The Natural Philosopher wrote:

No. Static thrust does no work any more than a labrador sitting on your
foot and imparting thrust, is doing work.

I agree with everything in your post, with one slight clarification
that I expect you'll agree with. Static thrust does no work on the
craft. It does do work on the air however (except of course in the
limit as you approach a 100% efficient prop).

@Ronald Raygun:

What The Natural Philosopher tells you above is the simple truth. Do
the math.

Momentum = MV
Energy = 1/2 MV^2

You have tried to to show that my force/energy analysis is wrong
because those calcs show that in a perfect world 0hp is required at
exactly windspeed. Well, the analysis and equations are correct and
would be wrong if they *didn't* show this.

On Oct 2, 6:24*pm, ThinAirDesigns wrote:

What The Natural Philosopher tells you above is the simple truth. *Do
the math.

Momentum = MV
Energy = 1/2 MV^2

Just to expand on that a bit...

force * time = impulse or momentum = mass * velocity

So: force (or thrust) = velocity * (mass/time)

This means we can get exactly the same thrust if we double the mass
flow rate and accelerate the mass half as much.

Looking at the other equation:

energy = mass * velocity^2 / 2

Power = energy / time.

So power = (mass/time) * velocity^2 / 2

Clearly if you double the (mass/time) term and halve the velocity
term, power is reduced. You can repeat this as many times as you like
- and thus make the required power as small as you'd like without
reducing thrust.

Helicopters do in fact use this principle to hover with far less power
than would be required if they used an airplane propeller. A Cessna
and an R22 have similar powerplants, but a Cessna couldn't dream of
hanging from its prop.

Rick Cavallaro wrote:
On Oct 2, 1:08 pm, The Natural Philosopher wrote:

No. Static thrust does no work any more than a labrador sitting on your
foot and imparting thrust, is doing work.

I agree with everything in your post, with one slight clarification
that I expect you'll agree with. Static thrust does no work on the
craft. It does do work on the air however (except of course in the
limit as you approach a 100% efficient prop).

Indeed, and the 100% efficient prop is of course the infinitely large
one turning infinitely slowly, for generating STATIC thrust.

On Oct 3, 1:29*am, The Natural Philosopher wrote:

Indeed, and the 100% efficient prop is of course the infinitely large
one turning infinitely slowly, for generating STATIC thrust.

Yup - we were going to go with one of those on the Blackbird, but JB
brought up all these silly problems about the height of the pylons,
cost of fiberglass - blah, blah, blah.... : )

So what have you guys done with dennis? Did he just slink away
quietly because he figured out he's probably wrong - but still doesn't
quite understand it.

On 3 Oct, 18:02, Rick Cavallaro wrote:
So what have you guys done with dennis? Did he just slink away
quietly because he figured out he's probably wrong - but still doesn't
quite understand it.

you a troll too?

"Jim K" wrote in message
...
On 3 Oct, 18:02, Rick Cavallaro wrote:
So what have you guys done with dennis? Did he just slink away
quietly because he figured out he's probably wrong - but still doesn't
quite understand it.

you a troll too?

I decided he was a troll.

ThinAirDesigns wrote:

@Ronald Raygun:

What The Natural Philosopher tells you above is the simple truth. Do
the math.

Momentum = MV
Energy = 1/2 MV^2

I figured this out a few hours before reading this.

You have tried to to show that my force/energy analysis is wrong
because those calcs show that in a perfect world 0hp is required at
exactly windspeed. Well, the analysis and equations are correct and
would be wrong if they *didn't* show this.

Ronald Raygun wrote:

ThinAirDesigns wrote:

@Ronald Raygun:

What The Natural Philosopher tells you above is the simple truth. Do
the math.

Momentum = MV
Energy = 1/2 MV^2

I figured this out a few hours before reading this.

Oops, pressed wrong button and the message went out before I'd written
the rest.

The thrust you get is equal to the rate of change of impulse, and this
corresponds to the air speed difference (output air speed minus input air
speed) multiplied by the rate at which mass is put through the prop.

This means that to get a particular thrust value, you can either put
through one particular quantity of air per second at one particular speed
difference, or alternatively you can put through k times that quantity
at 1/k times the air speed difference, for any real positive k.

This impinges on the power requirement, which is equal to the difference
in the input and output airs' kinetic power.

Power = 1/2 (mass throughput rate) * (v2^2-v1^2)

In the special case where v1=0, we get

Power = 1/2 Thrust * v2 [because Thrust = (mass rate)*(v2-v1)]

I concede that this shows that in principle the power needed to produce
a given amount of thrust can be made as small as we want by making the
output air speed as small as we like. Unfortunately the mass throughput
required per unit of time is now equal to thrust/v2, which means that if
we wanted to reduce the power to zero, we'd need to put through an
infinite amount of mass each second.

This is clearly unachievable, and so I must reject your analogy with the
two-by-four plank leaning against the wall, until such time as you can
produce a 2X4 of infinite length and zero density. :-)

You have tried to to show that my force/energy analysis is wrong
because those calcs show that in a perfect world 0hp is required at
exactly windspeed. Well, the analysis and equations are correct and
would be wrong if they *didn't* show this.

Depends what you mean by perfect. You merely stipulated that the components
(propeller, motor, generator) should initially be 100% efficient. A
propeller is 100% efficient if the mechanical power you put into it from the
motor matches the power which comes out, which is represented by the
difference between the before and after kinetic energies of the air being
put through per unit of time. I did not expect it would necessarily have
to have infinite size. If input and output power are zero, its efficiency
is 0/0 which is undefined. How can you say it's 100%? Hardly a perfect
world!

Anyway, let's leave the silly static special case and return to the one
in which I originally took exception to one aspect of your analysis. The
case with the car travelling at 55ft/s in a 27.5ft/s wind, and 10lb of
thrust.

If you're wanting to get 10lb thrust in those circumstances using only
1/2hp, then (if my calculations are right) this is possible only if you play
the same silly game as you've done with the static case, i.e. you would end
up with prop air output speed the same as air input speed, i.e. zero speed
difference and hence a need for infinite air throughput.

I had thought that the purpose of your discussion which included the
analysis was to try to persuade unbelievers (who no doubt think the whole
DDWFTTW idea is impossible, and that your documented demonstrations of it
nevertheless actually working must therefore be elaborate hoaxes). I would
comment that an analysis which only works with infinitely large propellers
is hardly going to be terribly persuasive.

@Ronald Raygun

if we wanted to reduce the power to zero, we'd need to put
through an infinite amount of mass each second.

That's correct -- that's exactly what and infinitely large propeller I
described would do.

This is clearly unachievable,

That's correct, many illustrative physics thought experiments are
unachievable -- that does not mean however that they are not
theoretically correct.

and so I must reject your analogy with the two-by-four plank
leaning against the wall, until such time as you can
produce a 2X4 of infinite length and zero density. *:-)

You just conceded that a propeller can theoretically produce 10lbs of
force while consuming 0hp. The 2x4 can also do this. Sound
comparison. QED

@Ronald Raygun wrote:

I would comment that an analysis which only works
with infinitely large propellers is hardly going to be terribly
persuasive.

You'll kindly notice that it was YOUR analysis that required an
infinitely large propeller. My analysis works just fine as written
and involved no such thing. It's a fact.

Had I picked exactly wind speed to analyize (as YOU did), I would have
picked a different (but equally as accurate) method. Not all things
are perfect for illustrating all situations.

I might add here that you have yet to produce an *accurate* analysis,
let alone a persuasive one, so you might want to go a bit easy on the
high horse.

On Oct 3, 3:34*pm, Ronald Raygun wrote:

If you're wanting to get 10lb thrust in those circumstances using only
1/2hp, then (if my calculations are right) this is possible only if you play
the same silly game as you've done with the static case, i.e. you would end
up with prop air output speed the same as air input speed, i.e. zero speed
difference and hence a need for infinite air throughput.

What on earth are you talking about. What sort of silly math are you
using to compute 1/2 h.p. for zero acceleration of the air!?

I had thought that the purpose of your discussion which included the
analysis was to try to persuade unbelievers (who no doubt think the whole
DDWFTTW idea is impossible, and that your documented demonstrations of it
nevertheless actually working must therefore be elaborate hoaxes). *I would
comment that an analysis which only works with infinitely large propellers
is hardly going to be terribly persuasive.

It would appear that you're now dedicated to not understanding the
basic analysis. So be it.

ThinAirDesigns wrote:

@Ronald Raygun wrote:

I would comment that an analysis which only works
with infinitely large propellers is hardly going to be terribly
persuasive.

You'll kindly notice that it was YOUR analysis that required an
infinitely large propeller. My analysis works just fine as written
and involved no such thing. It's a fact.

Had I picked exactly wind speed to analyize (as YOU did), I would have
picked a different (but equally as accurate) method. Not all things
are perfect for illustrating all situations.

I might add here that you have yet to produce an *accurate* analysis,
let alone a persuasive one, so you might want to go a bit easy on the
high horse.

OK, sorry if I came across that way. I have no expertise in aerodynamics
or propellerology and am trying to understand your analysis at the level
of basic laws of motion using nothing more advanced than high school level
physics, addled by the passage of nearer 4 than 3 decades. I'd be grateful
if you'd care to point out any mistakes in the following.

Let V1 be the air input speed to the prop (27.5ft/s in the example).
Let V2 be the air output speed from the prop (you left this value open in
your example, and I picked 55ft/s for reasons I stated earlier, but let's
leave it open for now).

For convenience, I'm not going to use V2 directly but will instead
use the air speed difference across the propeller:
Let DV be V2-V1.
For further convenience I shall express DV as a multiple of V1:
Let k = DV/V1, or DV = k*V1, which just means V2 = V1*(k+1).
Let F be the thrust required from the prop (10lb in the example).
Let M be the rate at which (air) mass is being put through the prop.

We know that M = F/DV = F/(k*V1).

[For example, if I were to pick V2 in the way I did before, then we
would have k=1 and M=11.7 lb/s]

Let me define kinetic power in the obvious way: If a mass m of air moving
at speed V has kinetic *energy* (1/2)*m*V^2, then a stream of air passing
through a particular area at rate M [lb/s] has kinetic *power* (1/2)*M*V^2.

The kinetic power of the air stream entering the prop is therefore

P1 = (1/2)*M*V1^2

and that of the exiting stream is

P2 = (1/2)*M*V1^2 * (k+1)^2

The kinetic power *gain* of the air stream, as a result of passing through
the prop, in other words the propeller output power, is

DP = P2-P1 = (1/2)*M*V1^2*k*(k+2)

Since M = F/(k*V1), this simplifies to

DP = F*V1*(k+2)/2

If I pick V2=2*V1, as I did earlier, this corresponds to k=1, which gives

DP = (3/2)*F*V1

You will recall that with F=10lb and V1=27.5ft/s, F*V1 = 1/2hp, and so
in the case k=1, we get DP = 3/4hp, the result I reported earlier.

I'm now interested to see what values of k, and hence V2 and M, are implied
by YOUR result for DP, which was 1/2hp, in the same circumstances, i.e.
where F and V1, and therefore F*V1, are given as 10lb, 27.5ft/s, and 1/2hp
respectively.

We have DP = F*V1*(k+2)/2. On the LHS we substitute 1/2hp for DP, and
on the RHS we substitute 1/2hp for F*V1, and we get

1/2hp = 1/2hp * (k+2)/2

and therefore

(k+2)/2 = 1

This means k=0, DV=0, V2=V1, and M=infinite.

Looks like your analysis also requires an infinite propeller.

Rick Cavallaro wrote:

On Oct 3, 3:34 pm, Ronald Raygun wrote:

If you're wanting to get 10lb thrust in those circumstances using only
1/2hp, then (if my calculations are right) this is possible only if you
play the same silly game as you've done with the static case, i.e. you
would end up with prop air output speed the same as air input speed, i.e.
zero speed difference and hence a need for infinite air throughput.

What on earth are you talking about. What sort of silly math are you
using to compute 1/2 h.p. for zero acceleration of the air!?

I've just shown my "silly" math in a parallel posting.
It doesn't purport to compute 1/2 hp for zero acceleration, rather it
*starts with* the 1/2hp your friend gave, together with the given values for
thrust and air input speed, and tries to solve for air output speed.

I had thought that the purpose of your discussion which included the
analysis was to try to persuade unbelievers (who no doubt think the whole
DDWFTTW idea is impossible, and that your documented demonstrations of it
nevertheless actually working must therefore be elaborate hoaxes). I
would comment that an analysis which only works with infinitely large
propellers is hardly going to be terribly persuasive.

It would appear that you're now dedicated to not understanding the
basic analysis. So be it.

No, I'm not, please don't dismiss me as a "dennis". I'm genuinely trying to
understand why I can get the energy budget to add up using your result.

If there's a mistake in my maths or my arithmetic, please help me find it.

On Oct 3, 6:58*pm, Ronald Raygun wrote:

please don't dismiss me as a "dennis". *I'm genuinely trying to
understand why I can get the energy budget to add up using your result.

My apologies. Perhaps I read too much into your explaining
confidently that JB's analysis was wrong (rather than suggesting that
maybe it was wrong, or asking for clarification). I've been through
this thing a whole bunch of times, and that's almost always a dead
giveaway.

If there's a mistake in my maths or my arithmetic, please help me find it..

Fair enough. The mistake is not in your math, but in your assumptions
(or perhaps a simple miscommunication). In your above post you say
"Looks like your analysis also requires an infinite propeller". Yes -
that's right. JB starts with a lossless prop (as he states). I'd
have to go back and look at his analysis, but I suspect he shows that
there's lots of excess power available with that assumption. This
lets us distinguish the difference between an engineering problem and
a problem of violating physical law. If we have excess power in the
no-losses situation, we know it's just down to an engineering problem.

And yes, as you suggest (and as we also have suggested) a lossless
prop will always be of infinite diameter and will have zero delta-
velocity across the disk. An 85% efficient prop however can be
significantly smaller than that (and will have a non-zero delta-V).

One thing to keep in mind is that prop efficiency is typically defined
relative to "V-infinity". This is the free-stream velocity. In the
case of an airplane, it's the plane's airspeed. For this definition
eff = power_out / power_in

power_out = thrust * V_infinity
power_in = torque * rotational_rate

Relative to the aircraft this is the definition that makes sense. The
pilot really doesn't care how much breeze the prop makes. He just
cares how much force it creates at a given airspeed. If you're
designing house fans you wouldn't use that definition for efficiency.
In this case you DO care how much breeze your prop creates.

Of course choosing different definitions for efficiency cannot change
the outcome of an experiment, but it can lead you to an incorrect
result if you use a different definition than what is assumed by the
person that did the analysis.

Does this make sense?

Ronald Raygun wrote:
Rick Cavallaro wrote:

On Oct 3, 3:34 pm, Ronald Raygun wrote:
If you're wanting to get 10lb thrust in those circumstances using only
1/2hp, then (if my calculations are right) this is possible only if you
play the same silly game as you've done with the static case, i.e. you
would end up with prop air output speed the same as air input speed, i.e.
zero speed difference and hence a need for infinite air throughput.
What on earth are you talking about. What sort of silly math are you
using to compute 1/2 h.p. for zero acceleration of the air!?

I've just shown my "silly" math in a parallel posting.
It doesn't purport to compute 1/2 hp for zero acceleration, rather it
*starts with* the 1/2hp your friend gave, together with the given values for
thrust and air input speed, and tries to solve for air output speed.

I had thought that the purpose of your discussion which included the
analysis was to try to persuade unbelievers (who no doubt think the whole
DDWFTTW idea is impossible, and that your documented demonstrations of it
nevertheless actually working must therefore be elaborate hoaxes). I
would comment that an analysis which only works with infinitely large
propellers is hardly going to be terribly persuasive.
It would appear that you're now dedicated to not understanding the
basic analysis. So be it.

No, I'm not, please don't dismiss me as a "dennis". I'm genuinely trying to
understand why I can get the energy budget to add up using your result.


Don't use momentum when you want to calculate energy.


What is happening here, is that the air is slowed with respect to the
ground, and the energy is used to accelerate the car with respect to the
ground.


If it was a wind turbine driving an electric train, you wouldn't have
the conceptual problem, would you?.



If there's a mistake in my maths or my arithmetic, please help me find it.

"The Natural Philosopher" wrote in message
...

What is happening here, is that the air is slowed with respect to the
ground, and the energy is used to accelerate the car with respect to the
ground.

Shame you don't read what I write, I have explained it before but it was
ignored (not really a surprise, thin rick doesn't actually explain anything
that is connected to the problem, he just invents something to deflect the
issue and claims "its too hard for you to understand"..

but its all about energy *and* thrust, if one doesn't work the other can be
ignored.

Say you have a cart travelling at 20 m/s with a wind behind it blowing at 10
m/s

The relative wind at the cart is 10 m/s to the rear.

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

Now put E=mv2 into the picture and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

My simple maths tells me its 4 times the energy you get from stopping the
wind .

Now if anyone can tell me how it stopping enough wind to generate four times
the energy it releases I would be interested.
I would like TNP to explain but he won't.


If it was a wind turbine driving an electric train, you wouldn't have the
conceptual problem, would you?.

I would if the turbine was on the train and using the passing wind to
generate the power or if the train was driven by a prop powered from the
wheels. Other than that they have no relevance at all and its just another
diverting tactic.


And while we are at it can anyone explain what the coupling is between the
energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots faster
flying downwind in a 10 knot wind so none of the energy released by stopping
the wind is absorbed by the plane.

On Oct 4, 3:18*am, "dennis@home" wrote:

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

There's no need to stop the wind behind you. Just slowing it relative
to the ground is sufficient. That means we only need to accelerate it
(by any amount) relative to the cart.

Now put E=mv2 into the picture

I hope that's just a double typo.

"Rick Cavallaro" wrote in message
...
On Oct 4, 3:18 am, "dennis@home" wrote:

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

There's no need to stop the wind behind you. Just slowing it relative
to the ground is sufficient. That means we only need to accelerate it
(by any amount) relative to the cart.

OK so do it for 1 m/s then, you still lose energy from the cart, there is
nowhere else for it to come from.


Now put E=mv2 into the picture

I hope that's just a double typo.


No it is not a typo, the energy of a moving object is mass x velocity
squared.
We can ignore the mass of the air unless you are creating a vacuum
somewhere.

You are confusing it with E=mc2, not really surprising from someone who
doesn't understand physics.
I will give you a clue.. why does it take four times as long to stop a car
from 20 mph as it does from 10 mph rather than twice as long. Please note
that I said as long not as far, brakes approximately dissipate energy (into
heat, they get hot as they are designed to do) at a constant rate.

On 04/10/2010 11:18, dennis@home wrote:

but its all about energy *and* thrust, if one doesn't work the other can
be ignored.

Ah, you're back. Would you care to reply to the post I made a couple of
days ago?

Say you have a cart travelling at 20 m/s with a wind behind it blowing
at 10 m/s

The relative wind at the cart is 10 m/s to the rear.

Now to stop the wind behind you have to accelerate the air going past
the cart to 20 m/s.

Do you? Why is that?

Now put E=mv2 into the picture and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

My simple maths tells me its 4 times the energy you get from stopping
the wind .

I can also stop the wind with a wall, and I don't believe you'd claim
that takes energy. So why do you need to invent that energy input?

And while we are at it can anyone explain what the coupling is between
the energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots
faster flying downwind in a 10 knot wind so none of the energy released
by stopping the wind is absorbed by the plane.

Does a plane have a connection to the ground? I'm thinking generally
speaking no, and when it does it's not via an arrangement such as is
seen on those cars, merely a wheel possibly braked.

The connection to the ground is what makes this work, which is why
thinking of planes is causing you to get it all wrong.

"Clive George" wrote in message
...
On 04/10/2010 11:18, dennis@home wrote:

but its all about energy *and* thrust, if one doesn't work the other can
be ignored.

Ah, you're back. Would you care to reply to the post I made a couple of
days ago?

Say you have a cart travelling at 20 m/s with a wind behind it blowing
at 10 m/s

The relative wind at the cart is 10 m/s to the rear.

Now to stop the wind behind you have to accelerate the air going past
the cart to 20 m/s.

Do you? Why is that?

Now put E=mv2 into the picture and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

My simple maths tells me its 4 times the energy you get from stopping
the wind .

I can also stop the wind with a wall, and I don't believe you'd claim that
takes energy.

It does actually, it creates turbulence and ultimately warms the air from
the energy taken from the wind..

So why do you need to invent that energy input?

The wall isn't moving faster than the wind is it?
Why do you need to try and deflect the issue by bringing in a stationary
wall.
Why don't you try putting the wall on the back of a truck travelling faster
than the wind and explain how that wall extracts energy from the wind? You
can assume an infinite wall if you want.


And while we are at it can anyone explain what the coupling is between
the energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots
faster flying downwind in a 10 knot wind so none of the energy released
by stopping the wind is absorbed by the plane.

Does a plane have a connection to the ground? I'm thinking generally
speaking no, and when it does it's not via an arrangement such as is seen
on those cars, merely a wheel possibly braked.

The connection to the ground is what makes this work, which is why
thinking of planes is causing you to get it all wrong.

Try again, you are the one wrong and the above energy statement proves it.
Go and make a YouTube video with a moving wall that extracts energy from the
tail wind, it should be easy for you as you appear to think a brick wall is
the same as the propeller.


I am more surprised you didn't just say the formula was wrong as I omitted
the constant because it didn't matter.

On 04/10/2010 13:07, dennis@home wrote:


"Clive George" wrote in message
...
On 04/10/2010 11:18, dennis@home wrote:

but its all about energy *and* thrust, if one doesn't work the other can
be ignored.

Ah, you're back. Would you care to reply to the post I made a couple
of days ago?

No answer.

Say you have a cart travelling at 20 m/s with a wind behind it blowing
at 10 m/s

The relative wind at the cart is 10 m/s to the rear.

Now to stop the wind behind you have to accelerate the air going past
the cart to 20 m/s.

Do you? Why is that?

No answer.

Now put E=mv2 into the picture and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

My simple maths tells me its 4 times the energy you get from stopping
the wind .

I can also stop the wind with a wall, and I don't believe you'd claim
that takes energy.

It does actually, it creates turbulence and ultimately warms the air
from the energy taken from the wind..

Ok, no energy input from the wall.

So why do you need to invent that energy input?

The wall isn't moving faster than the wind is it?
Why do you need to try and deflect the issue by bringing in a stationary
wall.
Why don't you try putting the wall on the back of a truck travelling
faster than the wind and explain how that wall extracts energy from the
wind? You can assume an infinite wall if you want.

Speaking of deflection, how about we stop this little sidetrack of yours
and go back to where you still haven't answered my questions from a few
days ago. Those questions aren't there to make you look stupid, they're
intended to lead you to understand what's going on.

And while we are at it can anyone explain what the coupling is between
the energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots
faster flying downwind in a 10 knot wind so none of the energy released
by stopping the wind is absorbed by the plane.

Does a plane have a connection to the ground? I'm thinking generally
speaking no, and when it does it's not via an arrangement such as is
seen on those cars, merely a wheel possibly braked.

The connection to the ground is what makes this work, which is why
thinking of planes is causing you to get it all wrong.

Try again, you are the one wrong and the above energy statement proves it.

Your energy statement is ********.

Go and make a YouTube video with a moving wall that extracts energy from
the tail wind, it should be easy for you as you appear to think a brick
wall is the same as the propeller.

No, the wall is pointing out why your energy argument is wrong.

I am more surprised you didn't just say the formula was wrong as I
omitted the constant because it didn't matter.

Dennis, the thing works, we can see it works, we can do the physics to
demonstrate why it works.
You're starting from a position of "It can't possibly work", and
distorting your knowledge of physics to try and fit that position.
Instead of doing that, how about dropping that assumption, and do the
physics properly?
I've started to lead you through an explanation, but you refused to
address it, complaining about "deflection". Go back to it and actually
look at it, then reply here.

dennis@home wrote:

Say you have a cart travelling at 20 m/s with a wind behind it blowing at
10 m/s

The relative wind at the cart is 10 m/s to the rear.

Yes

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

Yes

Now put E=mv2 into the picture

Near enough. Kinetic energy isn't emm vee squared, it's a half emm vee
squared.

and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

OK

My simple maths tells me its 4 times the energy you get from stopping the
wind .

Yes, near enough. There's 50 J of energy in 1kg of air moving at 10 m/s,
and 200 J (four times as much), in 1kg of air moving at 20 m/s.
The difference, the energy you have to add to the 10 m/s air to accelerate
it to 20 m/s, is 150 J, which is *three* times the 50 J you can get by
stopping 1kg of 10 m/s air.

Now if anyone can tell me how it stopping enough wind to generate four
times the energy it releases I would be interested.
I would like TNP to explain but he won't.

So let me try.

And while we are at it can anyone explain what the coupling is between the
energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

Correct. There is no obvious coupling. That doesn't mean there is no
coupling, it just means the coupling which is there is non-obvious.
I don't know where it is either, but it's in there somewhere.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots faster
flying downwind in a 10 knot wind so none of the energy released by
stopping the wind is absorbed by the plane.

Correct, this is because the airplane, unlike the car, is not on wheels
which are in contact with the ground and which it could couple to a
generator to provide power for its propellers. The airplane only sees
the air and just moves at 100kn relative to it. Obviously if someone
were to accelerate the air all around the plane (e.g. if the weather
system causes a 10kn breeze to start up out of nothing), then the
ground speed of the plane goes up to 110kn, and the plane's kinetic
energy relative to the ground has to go up by 21%. The way this is
achieved is that the plane's headwind temporarily drops from 100kn to
90kn, and there's less air resistance at that speed, so if the engines
are running at the same power, the spare surplus power will accelerate
the plane until the headwind reaches 100kn again.

The key to understanding how the car works lies in the fact that the
car has access both to the ground and to the air. But first, imagine that
the car has no propeller, and that its front axle is coupled to an electric
motor, and its back axle is coupled to a generator, and the generator
powers the electric motor. If there are no losses in the system, we can
tune it so that the forces match (i.e. generator drag matches motor thrust)
and then the powers will also match (the generator would supply exactly the
same amount of energy per second as the motor consumes). The car would then
behave in exactly the same way as if the generator and motor were just
sitting decoupled from the axles, doing nothing. You could tow the car up
to any speed you want using an auxiliary vehicle, or you could use a battery
to run the motor initially, and as soon as you cut the tow rope or switched
the battery out, the car would just gradually slow down due to friction, air
resistance, etc.

In other words, the system is energy-neutral. While the car is running,
the energy coming into the car from the generator goes out again through
the motor, and it just loops round and round. In isolation, the motor is
providing real thrust and doing real work, and the generator is extracting
real energy from the moving road, and applying real drag. But there is no
free lunch. The car is essentially running on incestuous power, the motor's
thrust is balancing the generator's drag so the system if force-neutral,
and the energy coming in via the generator is "free", so long as we don't do
anything with it except essentially give it back to where it came from.

Now we add the propeller, and uncouple the motor from the axle and couple
it to the prop instead. We can use an auxiliary vehicle to tow the car to
a certain speed, or a battery to power the prop, and then switch on the
generator-to-motor coupling and switch off the battery or cut the tow rope.

Same as with the motor driven axle, you should be able to get an energy
neutral system, in which free energy from the moving road is used to
power the propeller, to give the thrust to keep the road moving. Now you
simply need to do the energy budgeting to see at what wind speeds this will
be capable of being self-sustaining.

In other words some of the energy we need to accelerate the air from
10 to 20 m/s is "free" because it comes back in via the wheels. So
long as the free proportion is big enough that the rest is not more
than we get from slowing down of the wind, we should be in business.

As with a sailing boat which pulls energy out of the relative motion
between two different media (air and water), the car also has the benefit
of being on the interface between two media (air and ground), and can
exploit their relative motion to extract energy. An airplane cannot
do that because it's entirely surrounded by just one medium and has
nothing else to grab on to.

"Clive George" wrote in message
o.uk...
On 04/10/2010 13:07, dennis@home wrote:


"Clive George" wrote in message
...
On 04/10/2010 11:18, dennis@home wrote:

but its all about energy *and* thrust, if one doesn't work the other
can
be ignored.

Ah, you're back. Would you care to reply to the post I made a couple
of days ago?

No answer.

Cr@p posting.


Say you have a cart travelling at 20 m/s with a wind behind it blowing
at 10 m/s

The relative wind at the cart is 10 m/s to the rear.

Now to stop the wind behind you have to accelerate the air going past
the cart to 20 m/s.

Do you? Why is that?

No answer.

Obvious to anyone with half a brain.


Now put E=mv2 into the picture and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the
wind
from behind.

My simple maths tells me its 4 times the energy you get from stopping
the wind .

I can also stop the wind with a wall, and I don't believe you'd claim
that takes energy.

It does actually, it creates turbulence and ultimately warms the air
from the energy taken from the wind..

Ok, no energy input from the wall.

So why do you need to invent that energy input?

The wall isn't moving faster than the wind is it?
Why do you need to try and deflect the issue by bringing in a stationary
wall.
Why don't you try putting the wall on the back of a truck travelling
faster than the wind and explain how that wall extracts energy from the
wind? You can assume an infinite wall if you want.

No answer, what a surprise?
You brought in walls and now i have asked a more sensible question about
moving walls you decide to try and deflect again.

there is a simple answer to all this, all you have to do is show where the
energy is coming from.
I have shown that you don't get it from the wind.
the propeller is just a brake on the cart once you exceed the wind speed and
probably before you do as its not 100% efficient.


Speaking of deflection, how about we stop this little sidetrack of yours
and go back to where you still haven't answered my questions from a few
days ago. Those questions aren't there to make you look stupid, they're
intended to lead you to understand what's going on.

You think my use of the conservation of energy is stupid, that makes you
look more stupid.


And while we are at it can anyone explain what the coupling is between
the energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots
faster flying downwind in a 10 knot wind so none of the energy released
by stopping the wind is absorbed by the plane.

Does a plane have a connection to the ground? I'm thinking generally
speaking no, and when it does it's not via an arrangement such as is
seen on those cars, merely a wheel possibly braked.

The connection to the ground is what makes this work, which is why
thinking of planes is causing you to get it all wrong.

Try again, you are the one wrong and the above energy statement proves
it.

Your energy statement is ********.

Go on then explain where I am wrong.
If you can't I think we can take it that the whole thing is just bull.


Go and make a YouTube video with a moving wall that extracts energy from
the tail wind, it should be easy for you as you appear to think a brick
wall is the same as the propeller.

No, the wall is pointing out why your energy argument is wrong.

No the wall is irrelevant to my argument and is a attempt to deflect the
argument.


I am more surprised you didn't just say the formula was wrong as I
omitted the constant because it didn't matter.

Dennis, the thing works, we can see it works, we can do the physics to
demonstrate why it works.

You can't do the energy sums though can you?
It easy enough..

relative speed of wind to cart
speed of real wind
accelerate the relative wind to be faster than the cart so it can actually
slow the real wind.
subtract the energies and prove you can get more energy from slowing the
wind than from speeding up the wind.


You're starting from a position of "It can't possibly work", and
distorting your knowledge of physics to try and fit that position.

Where have I distorted any physics?
I don't need infinite props and stationary walls to distort things and trick
people.

Instead of doing that, how about dropping that assumption, and do the
physics properly?

I have, you go on about thrust, but thrust is useless without energy. Show
where you actually extract energy from the wind and get it onto the cart.

You can't.


I've started to lead you through an explanation, but you refused to
address it, complaining about "deflection". Go back to it and actually
look at it, then reply here.

You are wrong, there I have said what I think of your explanation.
Prove the energy equation and I might believe you, until then you have no
chance.

I also recommend you stop bringing in irrelevant arguments about static
walls, carts on treadmills, etc. none of which answer the energy question,
it makes you look like you are trying to avoid the truth.

"Ronald Raygun" wrote in message
...
dennis@home wrote:

Say you have a cart travelling at 20 m/s with a wind behind it blowing at
10 m/s

The relative wind at the cart is 10 m/s to the rear.

Yes

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

Yes

Now put E=mv2 into the picture

Near enough. Kinetic energy isn't emm vee squared, it's a half emm vee
squared.

and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

OK

My simple maths tells me its 4 times the energy you get from stopping the
wind .

Yes, near enough. There's 50 J of energy in 1kg of air moving at 10 m/s,
and 200 J (four times as much), in 1kg of air moving at 20 m/s.
The difference, the energy you have to add to the 10 m/s air to accelerate
it to 20 m/s, is 150 J, which is *three* times the 50 J you can get by
stopping 1kg of 10 m/s air.

Now if anyone can tell me how it stopping enough wind to generate four
times the energy it releases I would be interested.
I would like TNP to explain but he won't.

So let me try.

And while we are at it can anyone explain what the coupling is between
the
energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

Correct. There is no obvious coupling. That doesn't mean there is no
coupling, it just means the coupling which is there is non-obvious.
I don't know where it is either, but it's in there somewhere.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots faster
flying downwind in a 10 knot wind so none of the energy released by
stopping the wind is absorbed by the plane.

Correct, this is because the airplane, unlike the car, is not on wheels
which are in contact with the ground and which it could couple to a
generator to provide power for its propellers. The airplane only sees
the air and just moves at 100kn relative to it. Obviously if someone
were to accelerate the air all around the plane (e.g. if the weather
system causes a 10kn breeze to start up out of nothing), then the
ground speed of the plane goes up to 110kn, and the plane's kinetic
energy relative to the ground has to go up by 21%. The way this is
achieved is that the plane's headwind temporarily drops from 100kn to
90kn, and there's less air resistance at that speed, so if the engines
are running at the same power, the spare surplus power will accelerate
the plane until the headwind reaches 100kn again.

The key to understanding how the car works lies in the fact that the
car has access both to the ground and to the air. But first, imagine that
the car has no propeller, and that its front axle is coupled to an
electric
motor, and its back axle is coupled to a generator, and the generator
powers the electric motor. If there are no losses in the system, we can
tune it so that the forces match (i.e. generator drag matches motor
thrust)
and then the powers will also match (the generator would supply exactly
the
same amount of energy per second as the motor consumes). The car would
then
behave in exactly the same way as if the generator and motor were just
sitting decoupled from the axles, doing nothing. You could tow the car up
to any speed you want using an auxiliary vehicle, or you could use a
battery
to run the motor initially, and as soon as you cut the tow rope or
switched
the battery out, the car would just gradually slow down due to friction,
air
resistance, etc.

In other words, the system is energy-neutral. While the car is running,
the energy coming into the car from the generator goes out again through
the motor, and it just loops round and round. In isolation, the motor is
providing real thrust and doing real work, and the generator is extracting
real energy from the moving road, and applying real drag. But there is no
free lunch. The car is essentially running on incestuous power, the
motor's
thrust is balancing the generator's drag so the system if force-neutral,
and the energy coming in via the generator is "free", so long as we don't
do
anything with it except essentially give it back to where it came from.

Now we add the propeller, and uncouple the motor from the axle and couple
it to the prop instead. We can use an auxiliary vehicle to tow the car to
a certain speed, or a battery to power the prop, and then switch on the
generator-to-motor coupling and switch off the battery or cut the tow
rope.

Same as with the motor driven axle, you should be able to get an energy
neutral system, in which free energy from the moving road is used to
power the propeller, to give the thrust to keep the road moving. Now you
simply need to do the energy budgeting to see at what wind speeds this
will
be capable of being self-sustaining.

In other words some of the energy we need to accelerate the air from
10 to 20 m/s is "free" because it comes back in via the wheels. So
long as the free proportion is big enough that the rest is not more
than we get from slowing down of the wind, we should be in business.

No it isn't.
To slow the wind you have to accelerate the air passing the prop.
At the same speed as the wind, say 10 m/s there is zero airflow and ignoring
losses it will go forever.

Now go to 11 m/s for the car you have to accelerate the air (using the prop)
by 21% to even match the wind speed.

If you don't accelerate the air to the same speed as the cart or faster it
will just pile up in front of the car's prop and we know that can't happen
if its going to keep going.

You have to accelerate it by more if you want to slow the wind and extract
energy.

When you do the maths you find that you always have to extract more energy
from the air to speed up the air going through the prop than you get from
slowing the wind.

It doesn't matter if you find an unknown coupling mechanism it still doesn't
work as the energy equation is busted.

You can't claim the wheels provide extra energy as there is nowhere for the
energy to come from other than the wind and given the above there isn't any,
i.e. it can only be extracted by slowing the wheels down. That by definition
means slowing the car.


As with a sailing boat which pulls energy out of the relative motion
between two different media (air and water), the car also has the benefit
of being on the interface between two media (air and ground), and can
exploit their relative motion to extract energy.

That only works when sailing across wind, where the speed of the boat
downwind is less than the wind speed, that is not the case with this claim.

An airplane cannot
do that because it's entirely surrounded by just one medium and has
nothing else to grab on to.

On Oct 4, 6:04*am, "dennis@home" wrote:

Obvious to anyone with half a brain.

Well there's your problem.

On Oct 4, 4:33*am, "dennis@home" wrote:

Now put E=mv2 into the picture

I hope that's just a double typo.

No it is not a typo, the energy of a moving object is mass x velocity
squared.

So it's not a double typo. It's a simple case of a double idiot. You
got the formula wrong by a factor of 100% and you seem to think that
simply sticking a "2" in there somehow automatically means "squared".

You are confusing it with E=mc2, not really surprising from someone who
doesn't understand physics.

Sorry, my irony meter exploded on your previous post. I won't be able
to process that sentence until the new one arrives.


why does it take four times as long to stop a car
from 20 mph as it does from 10 mph rather than twice as long. Please note
that I said as long not as far

It doesn't dumbass.

On 04/10/2010 14:04, dennis@home wrote:

I've started to lead you through an explanation, but you refused to
address it, complaining about "deflection". Go back to it and actually
look at it, then reply here.

You are wrong, there I have said what I think of your explanation.

Which bit of what I said is wrong? I hadn't even got as far as dealing
with the wind.

I had my little car with a treadmill. Tell me what was wrong with the
mechanics/physics of that.

dennis@home wrote:

Now put E=mv2 into the picture

... the energy of a moving object is
mass x velocity squared.

Two problems with that Dennis -- "wrong" and "wrong".

The energy of a moving object is NOT "mass x velocity squared", and
that isn't even what you expressed in your equation. Your equation
says that energy is mass x velocity x 2.

Before you go claiming all sorts of wrongness on others parts at least
you should learn the actual equation for energy and then also learn
how equations are written.

http://en.wikipedia.org/wiki/Kinetic_energy














Say you have a cart travelling at 20 m/s with a wind behind it blowing at
10 m/s

The relative wind at the cart is 10 m/s to the rear.

Yes

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

Yes

Now put E=mv2 into the picture

Near enough. *Kinetic energy isn't emm vee squared, it's a half emm vee
squared.

and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

OK

My simple maths tells me its 4 times the energy you get from stopping the
wind .

Yes, near enough. *There's 50 J of energy in 1kg of air moving at 10 m/s,
and 200 J (four times as much), in 1kg of air moving at 20 m/s.
The difference, the energy you have to add to the 10 m/s air to accelerate
it to 20 m/s, is 150 J, which is *three* times the 50 J you can get by
stopping 1kg of 10 m/s air.

Now if anyone can tell me how it stopping enough wind to generate four
times the energy it releases I would be interested.
I would like TNP to explain but he won't.

So let me try.

And while we are at it can anyone explain what the coupling is between
the
energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

Correct. *There is no obvious coupling. *That doesn't mean there is no
coupling, it just means the coupling which is there is non-obvious.
I don't know where it is either, but it's in there somewhere.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots faster
flying downwind in a 10 knot wind so none of the energy released by
stopping the wind is absorbed by the plane.

Correct, this is because the airplane, unlike the car, is not on wheels
which are in contact with the ground and which it could couple to a
generator to provide power for its propellers. *The airplane only sees
the air and just moves at 100kn relative to it. *Obviously if someone
were to accelerate the air all around the plane (e.g. if the weather
system causes a 10kn breeze to start up out of nothing), then the
ground speed of the plane goes up to 110kn, and the plane's kinetic
energy relative to the ground has to go up by 21%. *The way this is
achieved is that the plane's headwind temporarily drops from 100kn to
90kn, and there's less air resistance at that speed, so if the engines
are running at the same power, the spare surplus power will accelerate
the plane until the headwind reaches 100kn again.

The key to understanding how the car works lies in the fact that the
car has access both to the ground and to the air. *But first, imagine that
the car has no propeller, and that its front axle is coupled to an
electric
motor, and its back axle is coupled to a generator, and the generator
powers the electric motor. *If there are no losses in the system, we can
tune it so that the forces match (i.e. generator drag matches motor
thrust)
and then the powers will also match (the generator would supply exactly
the
same amount of energy per second as the motor consumes). *The car would
then
behave in exactly the same way as if the generator and motor were just
sitting decoupled from the axles, doing nothing. *You could tow the car up
to any speed you want using an auxiliary vehicle, or you could use a
battery
to run the motor initially, and as soon as you cut the tow rope or
switched
the battery out, the car would just gradually slow down due to friction,
air
resistance, etc.

In other words, the system is energy-neutral. *While the car is running,
the energy coming into the car from the generator goes out again through
the motor, and it just loops round and round. *In isolation, the motor is
providing real thrust and doing real work, and the generator is extracting
real energy from the moving road, and applying real drag. *But there is no
free lunch. *The car is essentially running on incestuous power, the
motor's
thrust is balancing the generator's drag so the system if force-neutral,
and the energy coming in via the generator is "free", so long as we don't
do
anything with it except essentially give it back to where it came from.

Now we add the propeller, and uncouple the motor from the axle and couple
it to the prop instead. *We can use an auxiliary vehicle to tow the car to
a certain speed, or a battery to power the prop, and then switch on the
generator-to-motor coupling and switch off the battery or cut the tow
rope.

Same as with the motor driven axle, you should be able to get an energy
neutral system, in which free energy from the moving road is used to
power the propeller, to give the thrust to keep the road moving. *Now you
simply need to do the energy budgeting to see at what wind speeds this
will
be capable of being self-sustaining.

In other words some of the energy we need to accelerate the air from
10 to 20 m/s is "free" because it comes back in via the wheels. *So
long as the free proportion is big enough that the rest is not more
than we get from slowing down of the wind, we should be in business.

No it isn't.
To slow the wind you have to accelerate the air passing the prop.
At the same speed as the wind, say 10 m/s there is zero airflow and ignoring
losses it will go forever.

Now go to 11 m/s for the car you have to accelerate the air (using the prop)
by 21% to even match the wind speed.

If you don't accelerate the air to the same speed as the cart or faster it
will just pile up in front of the car's prop and we know that can't happen
if its going to keep going.

You have to accelerate it by more if you want to slow the wind and extract
energy.

When you do the maths you find that you always have to extract more energy
from the air to speed up the air going through the prop than you get from
slowing the wind.

It doesn't matter if you find an unknown coupling mechanism it still doesn't
work as the energy equation is busted.

You can't claim the wheels provide extra energy as there is nowhere for the
energy to come from other than the wind and given the above there isn't any,
i.e. it can only be extracted by slowing the wheels down. That by definition
means slowing the car.

As with a sailing boat which pulls energy out of the relative motion
between two different media (air and water), the car also has the benefit
of being on the interface between two media (air and ground), and can
exploit their relative motion to extract energy.

That only works when sailing across wind, where the speed of the boat
downwind is less than the wind speed, that is not the case with this claim.



An airplane cannot
do that because it's entirely surrounded by just one medium and has
nothing else to grab on to.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

On 04/10/2010 12:47, Doctor Drivel wrote:

"The Natural Philosopher" wrote in message
...

Perpetual motion - what a great concept! AND it's no use to terrorists!

They machine to invent as an anti-gravity machine. That will give you
perpetual motion. Have a weighted wheel. To one side of the wheel has no
gravity the others side does. Weights take the wheel down. No gravity on
the other side means the weight goes up and around and down again. The
wheel turns forever.

You have been reading H G Wells.

A balloon is an anti-gravity machine, so is an aircraft. They defy gravity.

No they are not. Balloons float and aircraft need some sort of motion to
stay up. You might just as well call a boat an anti-gravity machine.

I saw a "perpetual motion" machine once. Weights pushed a wheel
downwards. A balloon (anti-gravity machine) lifted the wheel up on the
other side. A ratchet mechanism took the balloon from the top to the
bottom of the wheel to lift again. It was largish and went on and on
once started - pretty slowly. A large wheel I suppose with gearings
could turn a generator.

Your machine as described would not work. That is not to say that a
series of devices that change size depending on their inclination
couldn't rotate a wheel but it wouldn't go on for ever. They would be
doing work and hence extracting energy from their surroundings. Put such
a machine under water (the better to exploit the difference in buoyancy
and the water would ultimately freeze if there wasn't an outside source
to warm the water. I suppose that in modern parlance you could say
entropy sucks.

On 4 Oct, 16:31, ThinAirDesigns wrote:
dennis@home wrote:
Now put E=mv2 into the picture
... the energy of a moving object is
mass x velocity *squared.

Two problems with that Dennis -- "wrong" and "wrong".

The energy of a moving object is NOT "mass x velocity squared", and
that isn't even what you expressed in your equation. *Your equation
says that energy is mass x velocity x 2.

Before you go claiming all sorts of wrongness on others parts at least
you should learn the actual equation for energy and then also learn
how equations are written.

http://en.wikipedia.org/wiki/Kinetic_energy





Say you have a cart travelling at 20 m/s with a wind behind it blowing at
10 m/s

The relative wind at the cart is 10 m/s to the rear.

Yes

Now to stop the wind behind you have to accelerate the air going past the
cart to 20 m/s.

Yes

Now put E=mv2 into the picture

Near enough. *Kinetic energy isn't emm vee squared, it's a half emm vee
squared.

and work out how much energy it takes to
accelerate the air going past from 10 m/s to 20 m/s so it stops the wind
from behind.

OK

My simple maths tells me its 4 times the energy you get from stopping the
wind .

Yes, near enough. *There's 50 J of energy in 1kg of air moving at 10 m/s,
and 200 J (four times as much), in 1kg of air moving at 20 m/s.
The difference, the energy you have to add to the 10 m/s air to accelerate
it to 20 m/s, is 150 J, which is *three* times the 50 J you can get by
stopping 1kg of 10 m/s air.

Now if anyone can tell me how it stopping enough wind to generate four
times the energy it releases I would be interested.
I would like TNP to explain but he won't.

So let me try.

And while we are at it can anyone explain what the coupling is between
the
energy in the air that is stopped and the cart.
Its all very well saying the energy is lost by the wind but there is no
obvious coupling to the cart which is travelling faster than the wind.

Correct. *There is no obvious coupling. *That doesn't mean there is no
coupling, it just means the coupling which is there is non-obvious.
I don't know where it is either, but it's in there somewhere.

I expect that it goes into turbulence and is just lost. Lets face it an
airplane capable of doing 100 knots doesn't suddenly go 10 knots faster
flying downwind in a 10 knot wind so none of the energy released by
stopping the wind is absorbed by the plane.

Correct, this is because the airplane, unlike the car, is not on wheels
which are in contact with the ground and which it could couple to a
generator to provide power for its propellers. *The airplane only sees
the air and just moves at 100kn relative to it. *Obviously if someone
were to accelerate the air all around the plane (e.g. if the weather
system causes a 10kn breeze to start up out of nothing), then the
ground speed of the plane goes up to 110kn, and the plane's kinetic
energy relative to the ground has to go up by 21%. *The way this is
achieved is that the plane's headwind temporarily drops from 100kn to
90kn, and there's less air resistance at that speed, so if the engines
are running at the same power, the spare surplus power will accelerate
the plane until the headwind reaches 100kn again.

The key to understanding how the car works lies in the fact that the
car has access both to the ground and to the air. *But first, imagine that
the car has no propeller, and that its front axle is coupled to an
electric
motor, and its back axle is coupled to a generator, and the generator
powers the electric motor. *If there are no losses in the system, we can
tune it so that the forces match (i.e. generator drag matches motor
thrust)
and then the powers will also match (the generator would supply exactly
the
same amount of energy per second as the motor consumes). *The car would
then
behave in exactly the same way as if the generator and motor were just
sitting decoupled from the axles, doing nothing. *You could tow the car up
to any speed you want using an auxiliary vehicle, or you could use a
battery
to run the motor initially, and as soon as you cut the tow rope or
switched
the battery out, the car would just gradually slow down due to friction,
air
resistance, etc.

In other words, the system is energy-neutral. *While the car is running,
the energy coming into the car from the generator goes out again through
the motor, and it just loops round and round. *In isolation, the motor is
providing real thrust and doing real work, and the generator is extracting
real energy from the moving road, and applying real drag. *But there is no
free lunch. *The car is essentially running on incestuous power, the
motor's
thrust is balancing the generator's drag so the system if force-neutral,
and the energy coming in via the generator is "free", so long as we don't
do
anything with it except essentially give it back to where it came from.

Now we add the propeller, and uncouple the motor from the axle and couple
it to the prop instead. *We can use an auxiliary vehicle to tow the car to
a certain speed, or a battery to power the prop, and then switch on the
generator-to-motor coupling and switch off the battery or cut the tow
rope.

Same as with the motor driven axle, you should be able to get an energy
neutral system, in which free energy from the moving road is used to
power the propeller, to give the thrust to keep the road moving. *Now you
simply need to do the energy budgeting to see at what wind speeds this
will
be capable of being self-sustaining.

In other words some of the energy we need to accelerate the air from
10 to 20 m/s is "free" because it comes back in via the wheels. *So
long as the free proportion is big enough that the rest is not more
than we get from slowing down of the wind, we should be in business.

No it isn't.
To slow the wind you have to accelerate the air passing the prop.
At the same speed as the wind, say 10 m/s there is zero airflow and ignoring
losses it will go forever.

Now go to 11 m/s for the car you have to accelerate the air (using the prop)
by 21% to even match the wind speed.

If you don't accelerate the air to the same speed as the cart or faster it
will just pile up in front of the car's prop and we know that can't happen
if its going to keep going.

You have to accelerate it by more if you want to slow the wind and extract
energy.

When you do the maths you find that you always have to extract more energy
from the air to speed up the air going through the prop than you get from
slowing the wind.

It doesn't matter if you find an unknown coupling mechanism it still doesn't
work as the energy equation is busted.

You can't claim the wheels provide extra energy as there is nowhere for the
energy to come from other than the wind and given the above there isn't any,
i.e. it can only be extracted by slowing the wheels down. That by definition
means slowing the car.

As with a sailing boat which pulls energy out of the relative motion
between two different media (air and water), the car also has the benefit
of being on the interface between two media (air and ground), and can
exploit their relative motion to extract energy.

That only works when sailing across wind, where the speed of the boat
downwind is less than the wind speed, that is not the case with this claim.

An airplane cannot
do that because it's entirely surrounded by just one medium and has
nothing else to grab on to.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

(Kinetic) energy= mass x velocity squared
http://en.wikipedia.org/wiki/Kinetic..._and_etymology
Been known for a long time.

harry
(Kinetic) energy= mass x velocity squared

A: wrong.

B: your own link says your wrong.

"Clive George" wrote in message
o.uk...
On 04/10/2010 14:04, dennis@home wrote:

I've started to lead you through an explanation, but you refused to
address it, complaining about "deflection". Go back to it and actually
look at it, then reply here.

You are wrong, there I have said what I think of your explanation.

Which bit of what I said is wrong? I hadn't even got as far as dealing
with the wind.

I had my little car with a treadmill. Tell me what was wrong with the
mechanics/physics of that.

Sorry but you and rick are just too stupid to bother with.

Roger Chapman wrote in
:

A balloon is an anti-gravity machine, so is an aircraft. They defy
gravity.


PMFJI - Most people don't know that there are actually some gravity-powered
devices in the world today. Many don't realise that this technology has
actually been around for a while.

In 1782 the Montgolfier brothers built and manned such a device. It was a
hot air balloon!


Al

On 04/10/2010 19:16, dennis@home wrote:


"Clive George" wrote in message
o.uk...
On 04/10/2010 14:04, dennis@home wrote:

I've started to lead you through an explanation, but you refused to
address it, complaining about "deflection". Go back to it and actually
look at it, then reply here.

You are wrong, there I have said what I think of your explanation.

Which bit of what I said is wrong? I hadn't even got as far as dealing
with the wind.

I had my little car with a treadmill. Tell me what was wrong with the
mechanics/physics of that.

Sorry but you and rick are just too stupid to bother with.

We've offered to help you understand, but you won't even try. Are you
scared you might learn something?

"ThinAirDesigns" wrote in message
...
dennis@home wrote:

Now put E=mv2 into the picture

... the energy of a moving object is
mass x velocity squared.

Two problems with that Dennis -- "wrong" and "wrong".

Another one that's too thick to bother with.

I hope everyone notes who it is that doesn't answer the simple but telling
questions.

You can't answer the energy problem because if you did you would have to
admit to fraud.

I think you should publish a list of who agrees with you so we know who the
idiots are.