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

Currently, the only "energy problem" here is the one where you
incorrectly state that the the formula for energy is:

E=mv2
and
... the energy of a moving object is
mass x velocity squared.

Just for your own education, here is the REAL equation for energy:

1/2*M*V^2 or ONE HALF mass x velocity squared.

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

On 04/10/2010 19:30, dennis@home wrote:

"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.

I'm not sure whether to be amused at the heroic efforts you're making to
avoid discussing why this thing works, insulting everybody who
understands it and repeatedly demonstrating just how pigheaded you can
be, or saddened that somebody who presumably ought to have the education
required to understand the problem cannot swallow his pride, approach it
rationally, and learn something rather than just shouting.

The "you're an idiot" usenet style you're using only works if you're
right, and the people you're arguing have very effectively demonstrated
that you're not. You don't even need to demonstrate it, you can just
work it out from first principles, but you refuse to.

"Clive George" wrote in message
...
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?

You have repeatedly failed to answer the energy question and used
diversionary tactics, are you scared others will learn something?

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.

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

Who says you have to STOP the wind? You just need to SLOW it to some
extent, to do that you just have to accelerate SOME air past the cart.

"Clive George" wrote in message
...
On 04/10/2010 19:30, dennis@home wrote:

"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.

I'm not sure whether to be amused at the heroic efforts you're making to
avoid discussing why this thing works, insulting everybody who understands
it and repeatedly demonstrating just how pigheaded you can be, or saddened
that somebody who presumably ought to have the education required to
understand the problem cannot swallow his pride, approach it rationally,
and learn something rather than just shouting.

The "you're an idiot" usenet style you're using only works if you're
right, and the people you're arguing have very effectively demonstrated
that you're not. You don't even need to demonstrate it, you can just work
it out from first principles, but you refuse to.

Look back through the posts and you will find that it was not I that started
calling people idiots and stupid because they didn't understand you
hypocrite.

Just answer the easy energy problem and everything will be fine.
Unless you do I will still be calling you a fraud.
You could just go and post your cr@p in an American news group where I am
sure you will find many more supporters.

"ThinAirDesigns" wrote in message
...
dennis@home
You can't answer the energy problem because if
you did you would have to admit to fraud.

Currently, the only "energy problem" here is the one where you
incorrectly state that the the formula for energy is:

E=mv2
and
... the energy of a moving object is
mass x velocity squared.

Just for your own education, here is the REAL equation for energy:

1/2*M*V^2 or ONE HALF mass x velocity squared.

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


Do you think the constant makes any difference.
I have news for you it doesn't.
I also stated that I had ignored it because it didn't matter.

have you worked out why it takes four times as long to stop a car that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

"Andy Burns" 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.

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

Who says you have to STOP the wind? You just need to SLOW it to some
extent, to do that you just have to accelerate SOME air past the cart.

They like using special cases and it makes it easy to understand.
Exactly the same applies to some and to slow.
This is because they appear on *both sides* of the equation and you lose
energy whatever combination you use.
Try putting in some figures and it will be obvious.

"AL_n" wrote in
:

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


PS - Sorry; I should have read the original post and watched the video. I
was assuming this thread was about "anti-gravity" pseudo-science.

I thought I was being clever. :-7

Al

On 04/10/2010 20:12, dennis@home wrote:


"Clive George" wrote in message
...
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?

You have repeatedly failed to answer the energy question and used
diversionary tactics, are you scared others will learn something?

Um, I was there first.

Short answer : No, you don't need to provide the energy input you claim.

Longer answer : Thinking about the energy in the way you're doing will
confuse things. I will admit I'm not happy about dealing with the fluid
flows and energy going into them, and can't easily explain why your
analysis is wrong, which is why I prefer to deal with forces and then
add distance later. When you do that you discover your analysis must be
wrong.

So, are you prepared to go with the forces?

On Oct 4, 12:28*pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

On Oct 4, 1:08*pm, Clive George wrote:

I will admit I'm not happy about dealing with the fluid
flows and energy going into them, and can't easily explain why your
analysis is wrong...

If Dennis will post his analysis, or give us a link to it, I'll be
happy to show where he goes wrong.

"Clive George" wrote in message
o.uk...
On 04/10/2010 20:12, dennis@home wrote:


"Clive George" wrote in message
...
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?

You have repeatedly failed to answer the energy question and used
diversionary tactics, are you scared others will learn something?

Um, I was there first.

Short answer : No, you don't need to provide the energy input you claim.

Longer answer : Thinking about the energy in the way you're doing will
confuse things. I will admit I'm not happy about dealing with the fluid
flows and energy going into them, and can't easily explain why your
analysis is wrong, which is why I prefer to deal with forces and then add
distance later. When you do that you discover your analysis must be wrong.

So, are you prepared to go with the forces?

Like you are prepared to go with the energy.
if you can't understand something as simple as the energy transfers why
should I think you get the forces correct?
Anyway as I previously stated you can't do forces in isolation to energy,
either both work or nothing does.


Here's a good forces one for you as someone introduced a wall earlier..

take a perfect wall, that is infinite, rigid and infinitesimally thin..

now take on air molecule from one side and put it on the other..

There you are infinite force.

I wonder what energy was used (well I don't but others will).

"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes four
times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.
It really is that simple.. rick doesn't have a clue, which is why he can't
answer the energy questions.

The Natural Philosopher wrote:

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

Why not? Aren't they inextricably linked?

I'm only using momentum indirectly in order to get an expression for the
propeller exit air speed V2 in terms of the parameters which JB's
example gave (namely power in from wheels, thrust, car speed, wind speed,
and hence V1 which is the relative wind speed seen by the car). The thrust
figure implies (via the momentum equations) a particular inverse
proportional relationship between "M dot" and "V2-V1", which I solve for
V2 so that I can use the expression in the kinetic energy equations for the
air before and after it has been through the prop.

What is happening here, is that the air is slowed with respect to the
ground,

I'm fully aware of that and don't have a problem with it. In the car
frame, the air is being accelerated (backwards) from V1 to V2, and in the
ground frame this corresponds to it being slowed by the same difference
V2-V1. Perhaps it would be less misleading to say that in both frames
the air is being decelerated with reference to the direction in which the
car is moving, but while the deceleration in the ground frame is from a
positive value to a less positive value (i.e. towards zero), in the car
frame it is from a negative value to a more negative value (i.e. away from
zero).

Thus the kinetic energy of the wind being decelerated actually increases
in the car frame, while in the ground frame it decreases.

and the energy is used to accelerate the car with respect to the
ground.

Yes, it could in principle be used to accelerate the car, but (and
I may have misunderstood) I took JB's analysis to refer to a steady
state car velocity, where all the acceleration of the car has already
happened and the car is now at terminal velocity, with the energy loss
from air resistance being what balances the energy harvested from the
wind.

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

What conceptual problem? :-)

On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.
It really is that simple.. rick doesn't have a clue, which is why he
can't answer the energy questions.

This brings to mind a conversation with my real ale pub landlord about
the chevrons that mark some motorways that declare 'keep 2 chevrons
apart'. To my mind, they are too close together, but he maintains that
any modern car will stop within that distance as all the cars are
slowing at the same rate.

I was bringing my wife back from a Scout training camp in the Lakes,
when the small van in front of me sent smoke out from his tyres, Skidded
out of the outside lane, into the middle lane, turned 90 degrees to the
right and arrested his speed by driving slap bang into the crash
barrier. Needles to say, I had to drive into him.

Dave

dennis@home
the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes four
times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.

Hmmm -- I wonder what would happen if car manufacturers decided to
size their brakes such that they could easily dissipate the heat from
say a 100mph to 50mph panic stop. Those same brakes could then also
easily dissipate the required amount of heat from 50mph to 0mph and
thus the vehicle would only take twice the time to stop from the
higher speed rather than four times. In this "new, enlightened
manufacturer" scenario, the limit for typical panic stop would then
not be heat dissipation at the brakes, but rather tire/pavement
coefficient of friction.

Someone should build a car with brakes like this. Oh, wait --
EVERYONE DOES!!!!!!!!!!!!

It really is that simple.. rick doesn't have a clue,

You on the other hand must not have a car. Due to aerodynamic drag,
on the braking limit your car will take less time to brake from 100mph
to 50mph than it will from 50mph to 0mph. Try it with a stopwatch
sometime.

"ThinAirDesigns" wrote in message
...
dennis@home
the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes
four
times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.

Hmmm -- I wonder what would happen if car manufacturers decided to
size their brakes such that they could easily dissipate the heat from
say a 100mph to 50mph panic stop. Those same brakes could then also
easily dissipate the required amount of heat from 50mph to 0mph and
thus the vehicle would only take twice the time to stop from the
higher speed rather than four times. In this "new, enlightened
manufacturer" scenario, the limit for typical panic stop would then
not be heat dissipation at the brakes, but rather tire/pavement
coefficient of friction.

Someone should build a car with brakes like this. Oh, wait --
EVERYONE DOES!!!!!!!!!!!!

Balls!

I can't wait for the first manufacturer to get sued because he deliberately
makes his brakes less efficient at low speeds just to satisfy your crazy
idea about how brakes work.

In the real world the bit from 100 to fifty takes longer than the bit from
50-0 as anyone who has driven will be able to tell you.


It really is that simple.. rick doesn't have a clue,

Now we also have proof that thin air doesn't have a clue.

To think that anyone believes these people might have invented something
that breaks the laws of physics amazes me, they can't even apply the
simplest physics to the simplest cases like car brakes and get the correct
answer. I blame the education system for not educating them.

You on the other hand must not have a car. Due to aerodynamic drag,
on the braking limit your car will take less time to brake from 100mph
to 50mph than it will from 50mph to 0mph. Try it with a stopwatch
sometime.


My car isn't a brick and I specifically choose slow speeds to avoid
aerodynamic effects you dumbass.

Doctor Drivel:
They defy gravity they do not go down, they go up.

This one is *a f***ing idiot

A lighter than air balloon is bowing to gravity rather than defying
it. Gravity is actually forcing it UP.

JB

On 04/10/2010 21:55, dennis@home wrote:


"Clive George" wrote in message
o.uk...
On 04/10/2010 20:12, dennis@home wrote:


"Clive George" wrote in message
...
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?

You have repeatedly failed to answer the energy question and used
diversionary tactics, are you scared others will learn something?

Um, I was there first.

Short answer : No, you don't need to provide the energy input you claim.

Longer answer : Thinking about the energy in the way you're doing will
confuse things. I will admit I'm not happy about dealing with the
fluid flows and energy going into them, and can't easily explain why
your analysis is wrong, which is why I prefer to deal with forces and
then add distance later. When you do that you discover your analysis
must be wrong.

So, are you prepared to go with the forces?

Like you are prepared to go with the energy.
if you can't understand something as simple as the energy transfers why
should I think you get the forces correct?

Well, you haven't actually pointed out where my forces bit is wrong yet,
whereas I reckon your energy thing is wrong because it doesn't need to
accelerate to 20ms-1.

Anyway as I previously stated you can't do forces in isolation to
energy, either both work or nothing does.

Indeed. So, tell me where I've gone wrong with my forces.

On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..

True, but that's not reached in any modern brake. That's one of the
reasons cars have ABS.

so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.

Therefore false.

So it takes four times as long to stop.

False.

A modern car brakes at a fairly constant force, limited by tyre grip.
v = u + at, a is the same, one of u and v is zero, therefore twice the
speed = twice the time. Twice as long.

s = ut + 1/2 a*t*t
t is doubled, s is quadrubled, therefore four times the distance.

That's o-level physics isn't it?

dennis@home
In the real world the bit from 100 to fifty takes longer than the bit from
50-0 as anyone who has driven will be able to tell you.

LOL -- the stopwatch doesn't lie. You should try one sometime.

My car isn't a brick and I specifically choose slow speeds to avoid
aerodynamic effects you dumbass

That's a pretty neat trick how you avoid "aerodynamic effects". Care
to share how slow you have to go to avoid those?

JB

On 04/10/2010 23:07, Dave wrote:
On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.
It really is that simple.. rick doesn't have a clue, which is why he
can't answer the energy questions.

This brings to mind a conversation with my real ale pub landlord about
the chevrons that mark some motorways that declare 'keep 2 chevrons
apart'. To my mind, they are too close together, but he maintains that
any modern car will stop within that distance as all the cars are
slowing at the same rate.

He's right, they're not the full stopping distance apart, and they won't
work if something suddenly stops faster than it normally can.

"Clive George" wrote in message
...
On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..

True, but that's not reached in any modern brake. That's one of the
reasons cars have ABS.

Balls.
Cars have anti lock brakes because most drivers can't cope with difficult
conditions and you are trying to deflect your inability to understand simple
physics yet again.


so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.

Therefore false.

So it takes four times as long to stop.

False.

A modern car brakes at a fairly constant force, limited by tyre grip.

For a car with cr@p tyre in the wet that might be true, but not any decent
car, Oh I forget you don't do decent cars with proper suspension and tyres
over there.
Its funny that something like a Jag can pull 1.3G braking while you crappy
cars only do about 0.8G, They both use similat tyres so I guess you don't
push the pedal hard enough or you have cr@p brakes.
Don't ever get close to a European car as you will go up his arse if he
brakes.

v = u + at, a is the same, one of u and v is zero, therefore twice the
speed = twice the time. Twice as long.

s = ut + 1/2 a*t*t
t is doubled, s is quadrubled, therefore four times the distance.

That's o-level physics isn't it?

Its primary school but the kids get it right unlike you.

On 05/10/2010 00:08, dennis@home wrote:


"Clive George" wrote in message
...
On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...

On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..

True, but that's not reached in any modern brake. That's one of the
reasons cars have ABS.

Balls.
Cars have anti lock brakes because most drivers can't cope with
difficult conditions and you are trying to deflect your inability to
understand simple physics yet again.

"one of the reasons".

so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.

Therefore false.

So it takes four times as long to stop.

False.

A modern car brakes at a fairly constant force, limited by tyre grip.

For a car with cr@p tyre in the wet that might be true, but not any
decent car

Not true. The limiting factor for a single stop for any modern car will
be grip, not brakes. Show me evidence to the contrary if you have any.
Once you're racing or descending mountains, life gets hotter, but that's
not what we're talking about.

Oh I forget you don't do decent cars with proper suspension
and tyres over there.

Where is "over there"? What sort of car do you think I drive? You might
like to take a look at the address I use when posting to usenet and the
groups I read for a clue to some of that answer.

Its funny that something like a Jag can pull 1.3G braking

Cite.

while you
crappy cars only do about 0.8G,

Why are my "crappy cars" any different to yours?

They both use similat tyres

Cite.

so I guess
you don't push the pedal hard enough or you have cr@p brakes.
Don't ever get close to a European car as you will go up his arse if he
brakes.

I do try to avoid getting too close to people, but that's nothing to do
with then being European. Of course if I'm driving I'll tend to be as
close to a European car as you can get...

v = u + at, a is the same, one of u and v is zero, therefore twice the
speed = twice the time. Twice as long.

s = ut + 1/2 a*t*t
t is doubled, s is quadrubled, therefore four times the distance.

That's o-level physics isn't it?

Its primary school but the kids get it right unlike you.

Nope.

dennis@home wrote:

"Ronald Raygun" wrote in message
...

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.

Yes.

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.

Two questions here. (1) What do you mean by "by 21%"? 21% of what?
Of the 10 m/s ground wind speed? Of the 1 m/s headwind? Of the 11 m/s
car speed? (2) Why?

You could accelerate the 1 m/s wind as seen by the car to 6 m/s if
you wanted, or you could accelerate to 11 m/s, or to any value you like.
The question is what the effect would be. In these two examples you'd
be accelerating the apparent wind by 5 or 10 m/s, and therefore you'd be
slowing the ground wind by 5 or 10 to 5 or 0 m/s. In the first case you
would harvest 75%, and in the second 100% of the ground wind's kinetic
energy, though there may be (I don't know, I'm speculating) thermodynamic
reasons why going for anywhere near 100% extraction is not going to be as
helpful as we may hope. Remember, all you need is your prop to generate
enough thrust to get your generator to provide enough power. The wheels
are being driven at 11 m/s, remember.

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.

Eh? Are you trying to accelerate the air in the wrong direction?

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

No. The ground air is moving in the same direction as the car. The
relative wind seen by the car is moving in the opposite direction.
The propeller is accelerating the air in a backwards direction (just
like an airplane would do), so the air is going into the prop from the
front of the car at 1 m/s towards the prop, and is coming out the back
of the prop at more than 1 m/s away from the prop. If you accelerate it
by 2 m/s to 3 m/s, this means that the speed of air which has been through
the prop, viewed from the ground, will now be travelling at 8 m/s because
it has been slowed down by 2 m/s.

Remember: The wind is being slowed (in the sense of losing kinetic energy)
from the point of view of a ground observer. For an observer on the car,
the wind is being speeded up (in the sense of gaining kinetic 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.

Don't be so sure. Try listing all the items whose energy status is
changing, and remember that for any objects which have kinetic
energy, the energy quantities depend on where you are observing from.

You can't claim the wheels provide extra energy as there is nowhere for
the energy to come from other than the wind

Not true. This energy comes from the thrust of the propeller pushing
the car forwards. Same as it did when the motor was couple to the other
axle.

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.

Well, actually a sailing boat *can* sail downwind (but not directly
downwind) faster than the wind, but that's beside the point here. The
point is that it is the presence of two media in relative motion which
make it all possible. That applies to the car too. Without relative
motion it would be literally powerless.

By the way, suppose you are in a boat in the middle of a wide river.
There is no wind. Can you sail ashore?

dennis@home wrote:

For a car with cr@p tyre in the wet that might be true, but not any decent
car, Oh I forget you don't do decent cars with proper suspension and tyres
over there.
Its funny that something like a Jag can pull 1.3G braking while you crappy
cars only do about 0.8G, They both use similat tyres so I guess you don't
push the pedal hard enough or you have cr@p brakes.
Don't ever get close to a European car as you will go up his arse if he
brakes.

What the hell do you mean by 1.3G and 0.9G?

Could it be that a car can decelerate at 0.9 or 1.3 times that "g" which
is normally given as 9.81 m/s^2? Say, that's a constant, isn't it?
Doesn't constant deceleration mean that the speed reduces linearly with
time? That for twice the speed we therefore have twice the stopping time?
That stopping *distance* therefore varies with the square of speed?

On Oct 4, 2:11*pm, "dennis@home" wrote:

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..

And there we have the proof that you don't understand cars OR
physics. Any car on the road can brake at any rate from zero to wheel
lockup from 20 mph to zero. The most effective braking is just shy of
wheel lockup. Thus you could easily achieve that constant braking
force from 20 to 0. Braking from 20 to 0 will take twice as long as
braking from 10 to 0.


so as a car travelling twice as fast has four times the energy it takes four
times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.

Wrong.


It really is that simple.. rick doesn't have a clue, which is why he can't
answer the energy questions.

I think it's fair to say that when I designed and built the Blackbird
I effectively answered the energy questions. If you don't think I can
answer some specific energy question then by all means ask it. I
think we all know why you won't do that.

You do realize that everyone here consistently claims that you're
consistently wrong - right?

On Oct 4, 3:37*pm, "dennis@home" wrote:

To think that anyone believes these people might have invented something
that breaks the laws of physics amazes me

Wrong again monkey-boy. Everyone here but you understands that what
we build demonstrates some fairly basic and well understood principles
of physics in a pretty counter-intuitive way. You're the only one
that seems to think our vehicle would have to break the laws of
physics.

One thing that's become crystal clear is that Dennis is to be used
strictly for entertainment purposes. To be fair - I was warned.

Raygun,

I tried to address your questions somewhat carefully. No response?

"Ronald Raygun" wrote in message
...

8 a load of cr@p

Don't be so sure. Try listing all the items whose energy status is
changing, and remember that for any objects which have kinetic
energy, the energy quantities depend on where you are observing from.


Don't forget that the *only* place it makes any sense from is the cart.
It is using the energy, it is generating the energy from the conditions
around it.
Now do the physics properly and admit you are wrong.


You can't claim the wheels provide extra energy as there is nowhere for
the energy to come from other than the wind

Not true. This energy comes from the thrust of the propeller pushing
the car forwards. Same as it did when the motor was couple to the other
axle.

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.

Well, actually a sailing boat *can* sail downwind (but not directly
downwind) faster than the wind, but that's beside the point here. The
point is that it is the presence of two media in relative motion which
make it all possible. That applies to the car too. Without relative
motion it would be literally powerless.

By the way, suppose you are in a boat in the middle of a wide river.
There is no wind. Can you sail ashore?


Yet another diversionary tactic with some total unrelated effect of the
current generating wind.

Anyway I am bored with your constant repeating of what is obviously wrong
and the diversionary tactics employed by you, and thin rick. Consider
yourself plonked along with the rest of the idiots.

in 992059 20101004 184231 ThinAirDesigns wrote:
harry
(Kinetic) energy= mass x velocity squared

A: wrong.

B: your own link says your wrong.

Your "your" is wrong too ;-)

On 4 Oct, 21:55, "dennis@home" wrote:
"Clive George" wrote in message

o.uk...





On 04/10/2010 20:12, dennis@home wrote:

"Clive George" wrote in message
...
On 04/10/2010 19:16, dennis@home wrote:

"Clive George" wrote in message
news:jpudnXfsMYW9fzTRnZ2dnUVZ7rmdnZ2d@brightvi ew.co.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?

You have repeatedly failed to answer the energy question and used
diversionary tactics, are you scared others will learn something?

Um, I was there first.

Short answer : No, you don't need to provide the energy input you claim.

Longer answer : Thinking about the energy in the way you're doing will
confuse things. I will admit I'm not happy about dealing with the fluid
flows and energy going into them, and can't easily explain why your
analysis is wrong, which is why I prefer to deal with forces and then add
distance later. When you do that you discover your analysis must be wrong.

So, are you prepared to go with the forces?

Like you are prepared to go with the energy.
if you can't understand something as simple as the energy transfers why
should I think you get the forces correct?
Anyway as I previously stated you can't do forces in isolation to energy,
either both work or nothing does.

Here's a good forces one for you as someone introduced a wall earlier..

take a perfect wall, that is infinite, rigid and infinitesimally thin..

now take on air molecule from one side and put it on the other..

There you are infinite force.

I wonder what energy was used (well I don't but others will).- Hide quoted text -

- Show quoted text -

Force x distance = energy.
If you have a suspended weight it exerts a force.
However it's potential energy is only released when it starts to
move. Think of the weight in a weight driven clock.

Rick Cavallaro wrote:

Raygun,

I tried to address your questions somewhat carefully.

I'm grateful.

No response?

Not quite yet.

Dave wrote:
On 04/10/2010 22:11, dennis@home wrote:


"Rick Cavallaro" wrote in message
...
On Oct 4, 12:28 pm, "dennis@home" wrote:

have you worked out why it takes four times as long to stop a car
that is
going twice as fast yet.
You can use either formula to find out as the constant doesn't matter.

Asked an answered. You claimed it takes 4 times as long to stop a car
that's going 20 mph vs. one that's going 10 mph. My answer: no it
doesn't. A normal car will take twice as long.

And there we have the proof you don't understand energy.

For anyone that cares..

the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes
four times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.
It really is that simple.. rick doesn't have a clue, which is why he
can't answer the energy questions.

This brings to mind a conversation with my real ale pub landlord about
the chevrons that mark some motorways that declare 'keep 2 chevrons
apart'. To my mind, they are too close together, but he maintains that
any modern car will stop within that distance as all the cars are
slowing at the same rate.

I was bringing my wife back from a Scout training camp in the Lakes,
when the small van in front of me sent smoke out from his tyres, Skidded
out of the outside lane, into the middle lane, turned 90 degrees to the
right and arrested his speed by driving slap bang into the crash
barrier. Needles to say, I had to drive into him.

Dave
Two chevrons is the reaction time, in the hope his brakes are no better
than yours.

A car length per 30 mph is no bad guide. I try to leave a lot more.

When I test drove a jaguar, I did a full emergency stop from 130mph.
Took nearly a mile.

ThinAirDesigns wrote:
dennis@home
the brakes use friction to turn kinetic energy to heat
they have a maximum rate at which they can do this..
so as a car travelling twice as fast has four times the energy it takes four
times as long to dissipate the heat in the brakes.
So it takes four times as long to stop.

Hmmm -- I wonder what would happen if car manufacturers decided to
size their brakes such that they could easily dissipate the heat from
say a 100mph to 50mph panic stop. Those same brakes could then also
easily dissipate the required amount of heat from 50mph to 0mph and
thus the vehicle would only take twice the time to stop from the
higher speed rather than four times. In this "new, enlightened
manufacturer" scenario, the limit for typical panic stop would then
not be heat dissipation at the brakes, but rather tire/pavement
coefficient of friction.

Someone should build a car with brakes like this. Oh, wait --
EVERYONE DOES!!!!!!!!!!!!


Er. no.

Most production saloons will be starting to fade on just a single
application of the brakes from top speed.

Your jaguars, porsches, and to an extent beemers and mercs will do a bit
better, as will some of the better hot hatches.

But brakes that will do continuous full power almost indefinitely are
only found on the very best sports cars. Or on the race track.

It really is that simple.. rick doesn't have a clue,

You on the other hand must not have a car. Due to aerodynamic drag,
on the braking limit your car will take less time to brake from 100mph
to 50mph than it will from 50mph to 0mph. Try it with a stopwatch
sometime.


But a lot more distance..:-)

The Natural Philosopher gurgled happily, sounding
much like they were saying:

Hmmm -- I wonder what would happen if car manufacturers decided to size
their brakes such that they could easily dissipate the heat from say a
100mph to 50mph panic stop. Those same brakes could then also easily
dissipate the required amount of heat from 50mph to 0mph and thus the
vehicle would only take twice the time to stop from the higher speed
rather than four times. In this "new, enlightened manufacturer"
scenario, the limit for typical panic stop would then not be heat
dissipation at the brakes, but rather tire/pavement coefficient of
friction.

Someone should build a car with brakes like this. Oh, wait -- EVERYONE
DOES!!!!!!!!!!!!

Er. no.

Most production saloons will be starting to fade on just a single
application of the brakes from top speed.

Your jaguars, porsches, and to an extent beemers and mercs will do a bit
better, as will some of the better hot hatches.

But brakes that will do continuous full power almost indefinitely are
only found on the very best sports cars. Or on the race track.

If you've got brakes that are fading from a single 100mph stop, you
desperately need to get them fixed properly - because they most certainly
should not be doing so, no matter what you're driving.

Unless, of course, you've bought the cheapest ****test pattern disks and/
or pads, in which case it's probably to be expected but don't blame the
car's manufacturer.

Adrian wrote:
The Natural Philosopher gurgled happily, sounding
much like they were saying:

Hmmm -- I wonder what would happen if car manufacturers decided to size
their brakes such that they could easily dissipate the heat from say a
100mph to 50mph panic stop. Those same brakes could then also easily
dissipate the required amount of heat from 50mph to 0mph and thus the
vehicle would only take twice the time to stop from the higher speed
rather than four times. In this "new, enlightened manufacturer"
scenario, the limit for typical panic stop would then not be heat
dissipation at the brakes, but rather tire/pavement coefficient of
friction.

Someone should build a car with brakes like this. Oh, wait -- EVERYONE
DOES!!!!!!!!!!!!

Er. no.

Most production saloons will be starting to fade on just a single
application of the brakes from top speed.

Your jaguars, porsches, and to an extent beemers and mercs will do a bit
better, as will some of the better hot hatches.

But brakes that will do continuous full power almost indefinitely are
only found on the very best sports cars. Or on the race track.

If you've got brakes that are fading from a single 100mph stop, you
desperately need to get them fixed properly - because they most certainly
should not be doing so, no matter what you're driving.


I am sorry, but that is he reality of a brand new car of the sort of
shopping trolley variety with a decent load on board.


You have a touching faith in manufacturers.



Unless, of course, you've bought the cheapest ****test pattern disks and/
or pads, in which case it's probably to be expected but don't blame the
car's manufacturer.


Nope. A pair of unvented single pot sliding caliper disks and a set of
drums on the back, which is what most average small cars have, is only
capable or about one emergency high speed stop, especially when 4 or 5
up.. and then the brakes are shot. The steel disk and drums are not
capable of dissipating the energy, they must store it as heat increase,
and until they cool, there is bugger all left.

That's why you have vented cross drilled disks on sports cars, and so on.

If there was no advantage to them, on the same weight of car (or less)
why do they get fitted? Just for show, I suppose you think.

On Oct 5, 3:37*am, The Natural Philosopher
wrote:

When I test drove a jaguar, I did a full emergency stop from 130mph.
Took nearly a mile.

Wow - that's truly atrocious! Were you sleeping at the time? I
guarantee I could do that in 1/4th that distance without waking my
passenger in my old Mazda. Based on this data point it's pretty hard
to take this following claim of yours seriously:

Your jaguars, porsches, and to an extent beemers and mercs will do a bit
better, as will some of the better hot hatches.

Rick Cavallaro gurgled happily, sounding much like
they were saying:

When I test drove a jaguar, I did a full emergency stop from 130mph.
Took nearly a mile.

Wow - that's truly atrocious! Were you sleeping at the time? I
guarantee I could do that in 1/4th that distance without waking my
passenger in my old Mazda.

Quite.

A quick google for high speed stopping distances found this -
http://www.volvoclub.org.uk/pdf/Spee...gDistances.pdf

Braking distance from 140mph? 230m.

The formula used by the HC is, allegedly, x^2 ÷ 20 + x = Total stopping
distance in feet where x = speed in mph.

30mph = 75ft
40mph = 120ft
50mph = 150ft
130mph = 975ft

The formula used for thinking distance is simpler - 10ft per 10mph.
So subtract that, and you're at 845ft or ~280m.

Either way, TNP's mile or 1600m seems a rather huge under-estimate.

On Oct 5, 3:37*am, The Natural Philosopher wrote:

When I test drove a jaguar, I did a full emergency stop from 130mph.
Took nearly a mile.

I'm still stunned by this statement. Your full emergency stop seems
to have been about 0.1G

Aero drag would probably put you in that ballpark from 130 down to 100
mph without touching the brakes.