"Adrian" wrote in message
...
Rick Cavallaro gurgled happily, sounding much like


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

Rick will dismiss that, it states the stopping distances as..

Initial speed (mph) 140 100 70
Braking Distance (m) 230 118 58

Which just happens to be agree with what I said and he called me stupid for
saying it.
Nothing that proves he is wrong will be listened too.
He will lie about it and claim it is wrong or just plain ignore it and start
a new diversionary tactic.

Of course anyone that knows physics will see the energy equations sitting
there telling you what happened.


Anyway here is a good one to test ones brain regarding energies (nothing to
do with brakes)..

If a car travelling at 100 mph hits an immovable object (say a bridge
buttress) is that worse than two identical cars hitting each other head on
at 50 mph each?

How about the same car hitting the immovable object at 50 mph?

On 05/10/2010 13:11, dennis@home wrote:


"Adrian" wrote in message
...
Rick Cavallaro gurgled happily, sounding much like


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

Rick will dismiss that, it states the stopping distances as..

Initial speed (mph) 140 100 70
Braking Distance (m) 230 118 58

Which just happens to be agree with what I said and he called me stupid
for saying it.
Nothing that proves he is wrong will be listened too.
He will lie about it and claim it is wrong or just plain ignore it and
start a new diversionary tactic.

You're being monumentally stupid again. I could carry on laughing at
you, but I'll take pity and explain where you went wrong. "how long to
stop" is being taken as time, not distance. Twice the time for the car
to stop from twice the speed.

Clive George gurgled happily, sounding much
like they were saying:

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

Rick will dismiss that, it states the stopping distances as..

Initial speed (mph) 140 100 70
Braking Distance (m) 230 118 58

Which just happens to be agree with what I said and he called me stupid
for saying it.
Nothing that proves he is wrong will be listened too. He will lie about
it and claim it is wrong or just plain ignore it and start a new
diversionary tactic.

You're being monumentally stupid again. I could carry on laughing at
you, but I'll take pity and explain where you went wrong. "how long to
stop" is being taken as time, not distance. Twice the time for the car
to stop from twice the speed.

Apart from the time-to-stop being pretty much irrelevant, TNP's claim
which started this subthread was for a "mile" to stop, with Rick's
rejoinder being "1/4th that distance".

Dennis's undoubted stupidity is a side issue here.

On Oct 5, 5:11*am, "dennis@home" wrote:

Rick will dismiss that, it states the stopping distances as..

*Initial speed (mph) * * * *140 100 70
*Braking Distance (m) * * 230 118 58

Which just happens to be agree with what I said and he called me stupid for
saying it.

If you really want to sound stupid you should keep claiming you know
what I'll do.

I don't dismiss the data above - and it has nothing to do with what
you said. You may recall that you made the claim for stops from 20
mph and 10 mph.

He will lie about it and claim it is wrong or just plain ignore it and start
a new diversionary tactic.

Yes, so far my "diversionary tactics" have been to answer your
ridiculous assertions head-on. Remember a couple of pages ago when I
agreed to look at your analysis if you posted it (or a link to it)?
Interesting that you completely ignored that after begging for us to
respond to it.

Your powers of trolling are truly impressive.


If a car travelling at 100 mph hits an immovable object (say a bridge
buttress) is that worse than two identical cars hitting each other head on
at 50 mph each?

How about the same car hitting the immovable object at 50 mph?

At least learn to pose the question properly. You're embarrassing
yourself more than necessary.

On 05/10/2010 13:26, Adrian wrote:
Clive gurgled happily, sounding much
like they were saying:

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

Rick will dismiss that, it states the stopping distances as..

Initial speed (mph) 140 100 70
Braking Distance (m) 230 118 58

Which just happens to be agree with what I said and he called me stupid
for saying it.
Nothing that proves he is wrong will be listened too. He will lie about
it and claim it is wrong or just plain ignore it and start a new
diversionary tactic.

You're being monumentally stupid again. I could carry on laughing at
you, but I'll take pity and explain where you went wrong. "how long to
stop" is being taken as time, not distance. Twice the time for the car
to stop from twice the speed.

Apart from the time-to-stop being pretty much irrelevant, TNP's claim
which started this subthread was for a "mile" to stop, with Rick's
rejoinder being "1/4th that distance".

I'm talking about earlier posts than that, not TNP's lack of knowledge
of where the brake pedal is or Ricks mention of 1/4 the distance without
waking his passenger, ie not full on braking.

The Natural Philosopher
Er. no.

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

Yes, and that doesn't change a whit of what I said -- ALL (developed
countries) production car brakes are sized such that they are traction
limited during a typical (non repeating) top speed panic stop. They
now install ABS precisely to prevent the driver from sliding the tires
in a panic stop, even on dry surface. Do many of these brakes begin
to fade during such a stop -- yes .. and I never said different, but
it's trivial to take this fade into account during the sizing
process. Brake fade is not a binary 'brakes - no brakes' situation.
Initially, the onset of fade can be overcome simply with more pressure
to produce the same braking force -- but that curve goes to hell very
quickly. Fortunately, a panic stop is over very quickly.

As an experienced autocross and production class track racer,I would
place any size wager that from any speed on any of the above vehicles,
with ABS off one can slide the tires during any phase of said panic
stop. Traction limited. QED

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

I never once used or implied "indefinitely" or "almost indefinitely".
I described a panic stop.

The Natural Philosopher
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.

You use a lot of vague and relative terms (getting more vague and
relative as people press you).

Do brakes get hot -- yes. Do brakes fade - yes. Will a production
car loaded to the gills perform a panic stop from top speed and still
be traction limited. YES.

That last point is all that is relevent to Dennis the Dumbass'
comment.

Oh, and there may be "bugger left" after that above panic stop, but in
our country that car will still have plenty of brakes to safely
perform more normal duties even right after. Our US NHTSA tests and
requires this. Perhaps your country will allow a car to be brakeless
right after a panic stop (but I doubt it), but ours will not. You
can find many links to NHTSA brake testing.

"ThinAirDesigns" wrote in message
...
The Natural Philosopher
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.

You use a lot of vague and relative terms (getting more vague and
relative as people press you).

Hypocrite.


Do brakes get hot -- yes. Do brakes fade - yes. Will a production
car loaded to the gills perform a panic stop from top speed and still
be traction limited. YES.

That last point is all that is relevent to Dennis the Dumbass'
comment.

Do the maths and you will see what I said was correct, I know you are pretty
dumb and but even you should be able to do it.


Oh, and there may be "bugger left" after that above panic stop, but in
our country that car will still have plenty of brakes to safely
perform more normal duties even right after.

TNP is not very bright, that's why he believes what you say about how to
sail faster downwind.
I notice that all the people that do believe you are a bit thick.

I notice you ignore the table in the pdf showing braking performance that
happens to agree with constant energy dissipation like I said happens and
not constant force like you said happens.
Constant energy dissipation means that the g force on the car is not
constant and they don't take twice as long to stop from double the speed as
I also said they don't and you called me stupid for saying because you
incorrectly think they do.
The more you and rick say the less convinced I am that you and rick know
anything other than bull****ting.

Dennis the Dumbass:
I notice you ignore the table in the pdf showing braking performance that
happens to agree with constant energy dissipation like I said happens and
not constant force like you said happens.

I never saw such a .pdf Please repost the link

On 05/10/2010 15:12, dennis@home wrote:


"ThinAirDesigns" wrote in message
...
The Natural Philosopher
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.

You use a lot of vague and relative terms (getting more vague and
relative as people press you).

Hypocrite.


Do brakes get hot -- yes. Do brakes fade - yes. Will a production
car loaded to the gills perform a panic stop from top speed and still
be traction limited. YES.

That last point is all that is relevent to Dennis the Dumbass'
comment.

Do the maths and you will see what I said was correct, I know you are
pretty dumb and but even you should be able to do it.


Oh, and there may be "bugger left" after that above panic stop, but in
our country that car will still have plenty of brakes to safely
perform more normal duties even right after.

TNP is not very bright, that's why he believes what you say about how to
sail faster downwind.
I notice that all the people that do believe you are a bit thick.

I notice you ignore the table in the pdf showing braking performance
that happens to agree with constant energy dissipation like I said
happens and not constant force like you said happens.
Constant energy dissipation means that the g force on the car is not
constant and they don't take twice as long to stop from double the speed

Wow, you really believe that?

Rick Cavallaro wrote:

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

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).

I agree, and I think I know where to look to resolve this.

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).

He did indeed state that, but of course I had no way of knowing that
this necessarily meant the propeller would have to be infinite in size.

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.

Indeed that is what he does. The problem was that my initial analysis,
such as it was, came to the same conclusion as his (that there is plenty
of spare power available), but his gave twice as much spare power as mine.

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.

I completely agree.

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.

That's a pity, isn't it, because the object of the analysis was to explore
the feasibilty of building a *real* (and therefore finite size) vehicle
that can outrun the wind. It seems to me that therefore it would have been
far better to say "let's ignore the losses and see how much spare power
there's going to be, and then think about whether it's enough" than to say,
as he did, "let's assume there will be no losses, that all the components
are 100% efficient, ..." especially if that was said with full knowledge
that this would require an infinite propeller. Moreover if it was said
in a forum where most readers would not have that specialist knowledge,
it's not a very effective way to try to explain things to non-experts.

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?

Sort of, er, um, to be honest, no, not really, by which I mean it doesn't
make sense to me because I don't understand propellers well enough to be
comfortable with the implications associated with various definitions of
their efficiency.

Frankly, I don't think efficiency is really the main problem giving
rise to my misunderstanding. I think it's more of a relativity problem,
i.e. looking at things in different inertial frames and then trying to
tie them together.

The other day you had a quibble with something TNP said about static
propellers doing no work, and you rightly pointed out that while a
hovering helicopter's rotor thrust is doing no work on the helicopter
itself, it is nevertheless doing work on the air. The same thing I
think is happening to me when I'm trying to look at what work the
car's propeller is doing on the air, and how this relates to the
work the prop's thrust is doing on the car.

I'm hopeful that with some more concentrated thought I can sort out
the misunderstanding, and I've a feeling the prop efficiency issue
is a red herring and that its complications can be side-stepped by
going back to the (very useful) skateboard analogy.

Rick Cavallaro wrote:
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:


Ok a bit of an exagerration, but its 6 seconds of full braking, and you
travel a LOT further than two chevrons in that time.

Its about 1/10th of a mile I guess.


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 wrote:
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


yeah. I hold my hand up to that. About 1/10th of a mile.. 176 yards.
Give or take.


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

ThinAirDesigns wrote:
The Natural Philosopher
Er. no.

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

Yes, and that doesn't change a whit of what I said -- ALL (developed
countries) production car brakes are sized such that they are traction
limited during a typical (non repeating) top speed panic stop. They
now install ABS precisely to prevent the driver from sliding the tires
in a panic stop, even on dry surface. Do many of these brakes begin
to fade during such a stop -- yes .. and I never said different, but
it's trivial to take this fade into account during the sizing
process. Brake fade is not a binary 'brakes - no brakes' situation.
Initially, the onset of fade can be overcome simply with more pressure
to produce the same braking force -- but that curve goes to hell very
quickly. Fortunately, a panic stop is over very quickly.

As an experienced autocross and production class track racer,I would
place any size wager that from any speed on any of the above vehicles,
with ABS off one can slide the tires during any phase of said panic
stop. Traction limited. QED

Nope. not all cheap cars have ABS, and many can not lock the wheels at
much beyond 30mph - there sis no requirement that they can.




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

I never once used or implied "indefinitely" or "almost indefinitely".
I described a panic stop.

And I am saying that brakes will lose efficiency perceptibly in ONE stop
from full speed. Its when your foot is mashed to the floor and there is
no wheel locking and you are still doing 40mph down hill with 4 people
up, and not much is happening..that you start to understand..

ThinAirDesigns wrote:
The Natural Philosopher
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.

You use a lot of vague and relative terms (getting more vague and
relative as people press you).

Do brakes get hot -- yes. Do brakes fade - yes. Will a production
car loaded to the gills perform a panic stop from top speed and still
be traction limited. YES.

No.

That last point is all that is relevent to Dennis the Dumbass'
comment.


And its false.

Oh, and there may be "bugger left" after that above panic stop, but in
our country that car will still have plenty of brakes to safely
perform more normal duties even right after. Our US NHTSA tests and
requires this. Perhaps your country will allow a car to be brakeless
right after a panic stop (but I doubt it), but ours will not. You
can find many links to NHTSA brake testing.

Now add in load and incline. Does they test specify that?

On 05/10/2010 17:21, The Natural Philosopher wrote:

Now add in load and incline. Does they test specify that?

Ooh, you've added in descending to this. Is this based on an incident
where you came downhill using the brakes rather too much, then had to
actually stop?

Preheating the brakes on a descent isn't what we're talking about.

(nearest I've had to that was descending a pass in Austria in an
overladen C15 van with trailer. Still stopped at the red light, just had
a large amount of smoke emerging from behind the wheels. I've learned
since then...)

The Natural Philosopher
many can not lock the wheels at much beyond 30mph -

I make no claim of knowledge of cars from your country (because I
don't know what country that is) but I still believe you're full of
it. For sure in the US that is a statement that is simply not true --
we can lock up the tires of our cars at any speed -- experience and
physics both tell me so.

Considering that the only difference between sliding the tires at
30mph and sliding them at 60mph is stopping the added rotational
energy of the hub/wheel/tire (which is relatively small), on what
grounds should we believe your statement -- I present that it flies in
the face of simply physics.

Again, please explain why a car that can lock up the wheels at 30mph
will have a significantly more difficult time locking the wheels at
60mph or even 100mph. Does the rubber somehow grip the road more at
100mph?

JB

ThinAirDesigns wrote:
The Natural Philosopher
many can not lock the wheels at much beyond 30mph -

I make no claim of knowledge of cars from your country (because I
don't know what country that is) but I still believe you're full of
it. For sure in the US that is a statement that is simply not true --
we can lock up the tires of our cars at any speed -- experience and
physics both tell me so.

Considering that the only difference between sliding the tires at
30mph and sliding them at 60mph is stopping the added rotational
energy of the hub/wheel/tire (which is relatively small), on what
grounds should we believe your statement -- I present that it flies in
the face of simply physics.

Again, please explain why a car that can lock up the wheels at 30mph
will have a significantly more difficult time locking the wheels at
60mph or even 100mph. Does the rubber somehow grip the road more at
100mph?


Nope. The heat build up in the disks and tyres soon reduces the
frictional efficiency of the pad/disk contact and then the fluid starts
to boil off any moistire inside the hydraulics.. ..

ABS makes it worse, because instead of the tyres getting hot and ripping
as they lock,. the pads and disks have to absorb the whole heat load.

Lets say our one ton car is stopping from 130 mph in 6 seconds..


Kinetic energy is 180,000 kilogram meters per second squared. so an
average work rate of 30,000 kilogram meters per second..although MOST of
the energy is dissipated early when speed is higher.

So about 30KW of heat..that's ten electric fires flat out, going into
the braking system...

with the worst being the first second..power is retardation force times
velocity.

If you have a 400bhp car that stops faster than it accelerates, its
dumping more than 400bhp into its brakes while it does it.





JB

The Natural Philosopher

People don't slam the brakes on at 130mph unless they are racing.

Stop trying to change your stated scenario:

YOUR QUOTE:
...many can not lock the wheels at much beyond 30mph

My question is and has been this:
Other than stopping the additional rotation mass of the hub/whee/tire
combo, give a reason why brakes that are capable of generating enough
force to slide the tires at 30mph would be incapable of sliding the
tires at any speed "much beyond 30mph"?

You've tried to dance around and go off in directions like "incline"
etc. that are not included in your original claim. Now you're
attempting to slither into questions of driver behavior and how they
might or might not react in a panic stop.

Discuss the physics please. You've claimed that many cars ***CAN
NOT*** lock their wheels at even say 50mph. I'm asking for a rational
physics reason for your claim that they are *incapable* of this.

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

People don't slam the brakes on at 130mph unless they are racing.

Depends how far ahead the camera (or stripy volvo) that they just spotted
is, doesn't it?

The Natural Philosopher wrote:

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

****ing hell, an Intercity 125 or 225 can stop in 1.25 miles, that's at
least 350 tons.

On Oct 5, 9:04*am, Ronald Raygun
wrote:

That's a pity, isn't it, because the object of the analysis was to explore
the feasibilty of building a *real* (and therefore finite size) vehicle
that can outrun the wind.

I don't see that as a pity at all. I think his analysis does that
quite nicely. And it breaks it into two very convenient parts. The
first part proves that it's only an engineering problem (no laws of
physics need changing), and the second part shows that it can be done
with readily available gear.

*It seems to me that therefore it would have been
far better to say "let's ignore the losses and see how much spare power
there's going to be, and then think about whether it's enough" than to say,
as he did, "let's assume there will be no losses, that all the components
are 100% efficient, ..."

Go back and read his analysis again. He did exactly as you suggest.

especially if that was said with full knowledge
that this would require an infinite propeller. *Moreover if it was said
in a forum where most readers would not have that specialist knowledge,
it's not a very effective way to try to explain things to non-experts.

There's no need to imagine or be concerned with infinite propellers.
If the first part of the analysis showed that you needed a 100%
efficient propeller you'd probably just dismiss the whole idea right
there as uninteresting at best. If you wanted to consider what a 100%
efficient propeller was - you'd come to find that it had to be
infinitely long. But an 85% propeller, which does just fine, is quite
realistic.


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?

Sort of, er, um, to be honest, no, not really, by which I mean it doesn't
make sense to me because I don't understand propellers well enough to be
comfortable with the implications associated with various definitions of
their efficiency.

Sorry, the definition used relates to "actuator disk theory" - which
is a method of analyzing propellers. Basically it's kind of like
saying what if we could push air with some specified efficiency
without regard to how we do it specifically (i.e. kind of ignore the
propeller). So this would seem to go in the direction you're looking
for - which is to try and understand the thing on first principles -
rather than having to understand propeller design. In any event, it's
unfortunate if the prop terms and efficiency terms were confusing. I
assure you that's the opposite of our intention.

Frankly, I don't think efficiency is really the main problem giving
rise to my misunderstanding. *I think it's more of a relativity problem,
i.e. looking at things in different inertial frames and then trying to
tie them together.

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.


The other day you had a quibble with something TNP said about static
propellers doing no work, and you rightly pointed out that while a
hovering helicopter's rotor thrust is doing no work on the helicopter
itself, it is nevertheless doing work on the air.

I don't think there was a quibble. I was merely clarifying a point
that he also readily agreed with.

I'm hopeful that with some more concentrated thought I can sort out
the misunderstanding, and I've a feeling the prop efficiency issue
is a red herring and that its complications can be side-stepped by
going back to the (very useful) skateboard analogy.

I assure you the prop efficiency thing was only intended as a way to
do the analysis. But if you think the skateboard analogy will be more
intuitive let's have at it. I'll ask you to lead.

Rick Cavallaro wrote:

On Oct 5, 9:04 am, Ronald Raygun
wrote:

That's a pity, isn't it, because the object of the analysis was to
explore the feasibilty of building a *real* (and therefore finite size)
vehicle that can outrun the wind.

I don't see that as a pity at all. I think his analysis does that
quite nicely.

I'm sure that's what it was meant to do.

But at first it doesn't, because his analysis made the explicit assumption
of a loosless propeller, and hence the implicit assumption of an infinite
propeller. Yes, he slipped in a rider at the end allowing for 85%
efficiency, but in a subsequent exchange of views he was adamant
that, in the initial analysis, he was not simply ignoring losses in the
calculations despite acknowledging that they would of course exist, but
instead he was assuming there were none, that all components were 100%
efficient. This philosophical distinction seemed important to him, even
though (it seems to me) the implication of this gets in the way of
promoting understanding, given that most readers (here) would be unaware
that 100% efficiency implies infinite size.

When we got to discussing the static case and when I said:

## Suppose car speed and wind speed are *both* 55ft/s. The
## wheels are still delivering 1hp. The headwind is zero, so you
## would calculate the power needed by the prop to produce 10lb
## of thrust as 0hp. Can't be right.

His reply was:

# But it is right, because at that point in the calculations we are
# still describing the scenario using theoretical, lossless components.
# We've defined the transmission method as lossless and the propeller
# also as 100% efficient.

And it breaks it into two very convenient parts. The
first part proves that it's only an engineering problem (no laws of
physics need changing), and the second part shows that it can be done
with readily available gear.

That's a good approach, but the prop efficiency stuff ought to have gone
into the engineering part, whereas he seems to have it initially in the
physics part. It seems to me that the physics analysis would be more
illuminating if it allowed us to work with finite values of air throughput.

It seems to me that therefore it would have been
far better to say "let's ignore the losses and see how much spare power
there's going to be, and then think about whether it's enough" than to
say, as he did, "let's assume there will be no losses, that all the
components are 100% efficient, ..."

Go back and read his analysis again. He did exactly as you suggest.

I have done, and he didn't. See above.

especially if that was said with full knowledge
that this would require an infinite propeller. Moreover if it was said
in a forum where most readers would not have that specialist knowledge,
it's not a very effective way to try to explain things to non-experts.

There's no need to imagine or be concerned with infinite propellers.
If the first part of the analysis showed that you needed a 100%
efficient propeller you'd probably just dismiss the whole idea right
there as uninteresting at best. If you wanted to consider what a 100%
efficient propeller was - you'd come to find that it had to be
infinitely long. But an 85% propeller, which does just fine, is quite
realistic.

OK

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?

Sort of, er, um, to be honest, no, not really, by which I mean it doesn't
make sense to me because I don't understand propellers well enough to be
comfortable with the implications associated with various definitions of
their efficiency.

Sorry, the definition used relates to "actuator disk theory" - which
is a method of analyzing propellers. Basically it's kind of like
saying what if we could push air with some specified efficiency
without regard to how we do it specifically (i.e. kind of ignore the
propeller). So this would seem to go in the direction you're looking
for - which is to try and understand the thing on first principles -
rather than having to understand propeller design. In any event, it's
unfortunate if the prop terms and efficiency terms were confusing. I
assure you that's the opposite of our intention.

What actually got me going was that he stated the prop's power output
would equal thrust times the incoming airspeed. I thought, by considering
the work it had to do on the air, that it should be more than that.
something like thrust times a speed value somewhere (probably halfway)
between the incoming and outgoing air speeds.

If that is wrong, I'd like to know why it's wrong. I'm sure you could
just tell me, but please don't just yet, as it'll be better for my
soul if I can work it out myself.

Frankly, I don't think efficiency is really the main problem giving
rise to my misunderstanding. I think it's more of a relativity problem,
i.e. looking at things in different inertial frames and then trying to
tie them together.

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.

Indeed.

The other day you had a quibble with something TNP said about static
propellers doing no work, and you rightly pointed out that while a
hovering helicopter's rotor thrust is doing no work on the helicopter
itself, it is nevertheless doing work on the air.

I don't think there was a quibble. I was merely clarifying a point
that he also readily agreed with.

It was a minor correction; I call that a quibble. If you understand a
quibble to be more of a dispute, that's not what I meant.

I'm hopeful that with some more concentrated thought I can sort out
the misunderstanding, and I've a feeling the prop efficiency issue
is a red herring and that its complications can be side-stepped by
going back to the (very useful) skateboard analogy.

I assure you the prop efficiency thing was only intended as a way to
do the analysis. But if you think the skateboard analogy will be more
intuitive let's have at it. I'll ask you to lead.

OK. Watch this space.

"Rick Cavallaro" wrote in message
...

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.


For once I can agree with rick (shock + horror), use the same frame of
reference and do the sums.


try this example..

assume a wind of -10 m/s at the cart
assume the ground is doing -20 m/s


What does this mean you ask..

well its a cart going at twice the wind speed down wind.
Its the same as the wind being 10 m/s and the cart travelling down wind at
20 m/s which I will use for the example as it makes the maths easier. You
can use any other figures you like, the conclusions are the same.

This gives a kinetic energy to the relative wind passing the cart of 100
units (-10 x -10) but do notice it is going backwards and is being generated
by ricks prop.

Ah look at the energy needed, we are expending energy to make the wind go
backwards.

However we are not slowing the real wind at all (it is still -10 m/s) so it
is not losing any energy to anywhere

This is impossible.
There is no energy input.

Now rick will say the energy comes from the wheels and drives the prop..

so in this steady state the wheels must be providing the 100 units of energy
but its not slowing the real wind so there can't be any energy coming from
the wind. In actual fact the energy is going into the prop and none is
coming back.

Opps that's impossible too you can't just create energy from nowhere ( not
even using wheels and props).



so lets increase the speed of the prop to throw more air back and slow the
wind to extract energy as rick says he does..

So we increase the prop wind to -11 m/s , the energy used has gone up from
100 to 121 = 21 extra units (using the energy = constant x mass x velocity
squared formula and remembering two negatives multiplied are a positive).


So what has happened to the real wind? Yes it has gone from -10 m/s to -11
m/s and its energy has changed by 121-100=21 extra units.

Remember we are using a single frame of reference, the cart.

Do I see a problem here? is 21 greater than 21? no so we have no extra
energy extracted from the wind by the cart.

What does it say..

well even with perfect conditions there are no circumstances using any
device (even a 100% efficient prop) where you can extract more energy by
slowing the wind when the device is moving faster than the wind in the
direction of the wind than the energy it takes to do it.
(This isn't a surprise to a physicist as its just conservation of energy.)


Well is this what really happens? probably not, after all this is the
perfect case, lets look at some of the problems rick appears to avoid in the
real world..

To stop the physical impossibility of the air piling up behind the the cart
would have to increase speed to maintain the steady state of the wind speed
from the prop being the same as the wind, if it doesn't turbulence will
occur even with a 100% efficient prop. (The excess air has to go somewhere
and the prop has no control over it) there is no energy being input so this
can't happen.

To actually overcome losses the prop would have to accelerate the air to a
higher speed than in the steady state this results in the pressure build up
behind the prop and that causes turbulence as the air trys to even its self
out, this loses energy so more energy is needed by the prop, its impossible
to get this energy back as it was shown above.
there are friction losses,
etc.

Are there any values of speed where it works and you can actually extract
energy?
well yes, if the cart is going slower than the wind you can get energy from
the wind, this is not really surprising.

dennis@home wrote:

"Rick Cavallaro" wrote in message
...

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.

For once I can agree with rick (shock + horror), use the same frame of
reference and do the sums.


try this example..

assume a wind of -10 m/s at the cart
assume the ground is doing -20 m/s


What does this mean you ask..

Do we? It seems obvious enough.

well its a cart going at twice the wind speed down wind.
Its the same as the wind being 10 m/s and the cart travelling down wind at
20 m/s which I will use for the example as it makes the maths easier. You
can use any other figures you like, the conclusions are the same.

OK

This gives a kinetic energy to the relative wind passing the cart of 100
units (-10 x -10)

To what wind? There is -10 m/s wind all around the cart, but we're only
interested in that part of it which is going through the prop. We need to
know how much that is. Suppose you put 6kg of air through it every second.
Then the kinetic energy of those 6kg of air before they go through the prop
will be (1/2)x(6kg)x(-10m/s)^2 which is 300 kg m^2/s^2, or 300 J.

but do notice it is going backwards and is being
generated by ricks prop.

No, it's there already, it's the headwind resulting from the cart's
ground speed being 10 m/s more than the wind's ground speed. What the
prop does is make the 10 m/s backwards relative wind go even faster
backwards.

Ah look at the energy needed, we are expending energy to make the wind go
backwards.

Yes, of course we are expending energy to make the relative wind go faster
backwards, but how much energy? To what speed are you wanting the prop
to accelerate the wind? Suppose you want to accelerate that 6kg of air
from -10 m/s to -20 m/s.

This would increase its kinetic energy to 1200 J, which means adding 900 J
to the 300 J it already had.

However we are not slowing the real wind at all (it is still -10 m/s) so
it is not losing any energy to anywhere

We are slowing both the relative wind ("slowing" in the sense of making
its speed more negative, i.e. from -10 m/s to -20 m/s), and also slowing
the real wind (from +10 m/s to 0 m/s). But beware, you are switching
frames when changing from relative to real wind. Both winds experience
a reduction in speed by 10 m/s, but in the two frames, the kinetic energy
changes which accompany the speed changes are not the same. In the moving
frame it increases by 900 J (from 300 J to 1200 J), in the stopped ground
frame it decreases by 300 J (from 300 J to 0 J).

Momentum change rules tell us that if you accelerate 6kg by 10 m/s each
second, this will give 60 N of thrust. If you didn't have the propeller
and instead had a mule pulling the cart using a string with 60 N tension,
without accelerating the cart, and if the cart is moving at 20 m/s, then
the mule would be doing 1200 J of work each second (i.e. delivering
1200 W of power). Obviously if you're applying a force to the cart and it
isn't accelerating, then the force must be doing something else, and in
this case it is operating a generator driven by the wheels. The generator's
1200 W power output could be used to boil an on-board kettle. Or we could
divert 900 W of this power to run a motor to drive the prop to provide the
thrust to make the mule redundant. The spare 300 W is available for losses
and air drag.

Is it a coincidence that the spare 300 J is the same energy as the real
wind is giving up?

Doctor Drivel
It does not go down it goes UP.

Of course the lighter than air balloon is going up; it has no choice
-- gravity is forcing it UP.

Lighter than air craft ... slaves to gravity.

On Oct 6, 3:02*am, Ronald Raygun wrote:

I'm sure that's what it was meant to do... But at first it doesn't

That's exactly what it does. *At first* it considers a 100% efficient
device - as you suggest. He then looks at the excess energy available
and answers the engineering question - "is this enough energy to make
it work in the real world?"

You're getting stuck on the notion that a 100% efficient propeller
would have to be infinite in size. Why? 100% efficient ball bearings
and zero rolling resistance tires can no more easily exist in the real
world than a prop of infinite diameter. Exactly how large a prop
would you allow in the first part of the proof when he is simply
trying to see if a DDWFTTW cart would violate the laws of physics?

because his analysis made the explicit assumption
of a loosless propeller, and hence the implicit assumption of an infinite
propeller. *Yes, he slipped in a rider at the end allowing for 85%
efficiency

In other words, he did precisely what you suggest. He starts with
lossless components, determines that no laws of physics would be
violated, and then addresses the real-world questions. In the very
beginning of the analysis he explains that he's starting with lossless
components, but that losses will be dealt with later. There's nothing
hidden or tricky about this.

the implication of this gets in the way of
promoting understanding, given that most readers (here) would be unaware
that 100% efficiency implies infinite size.

100% efficient components don't exist in the real world. I think he
assumes that everyone but Dennis would recognize that.



When we got to discussing the static case and when I said: ...
His reply was: ...

Yes, and his reply is correct.

That's a good approach, but the prop efficiency stuff ought to have gone
into the engineering part, whereas he seems to have it initially in the
physics part.

The imperfect prop is introduced in the engineering part. The perfect
prop is assumed in the physics part. There's no other way to follow
the very approach you suggest.

*It seems to me that the physics analysis would be more
illuminating if it allowed us to work with finite values of air throughput.

Doctor Drivel wrote:

"ThinAirDesigns" wrote in message
...
Doctor Drivel
It does not go down it goes UP.

Of course the lighter than air balloon is going up; it has no choice
-- gravity is forcing it UP.

So if defies gravity.

Something doesn't defy gravity just by going up.

Suppose you have a jar half full of syrup. The syrup occupies
the bottom half. There is air in the top half.

Now turn the jar upside down. It's OK, the lid is closed.
What heppens next is that gravity forces the syrup down and the air up.
So they both *obey* gravity. The air does not defy it.

A helium filled balloon goes up because gravity is pulling on the
air around and above it more strongly than on the balloon itself.

Put the balloon in a car, and a football on the back seat. Drive
round a corner quickly, and the football will shoot off to one side,
and the balloon to the other. Both items are obeying, not gravity
this time, but a similar phenomenon, inertial acceleration.

dennis@home wrote:



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

"Rick Cavallaro" wrote in message

...

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.

For once I can agree with rick (shock + horror), use the same frame of
reference and do the sums.


try this example..

assume a wind of -10 m/s at the cart
assume the ground is doing -20 m/s


What does this mean you ask..

Do we? It seems obvious enough.

well its a cart going at twice the wind speed down wind.
Its the same as the wind being 10 m/s and the cart travelling down wind
at
20 m/s which I will use for the example as it makes the maths easier.
You can use any other figures you like, the conclusions are the same.

OK

This gives a kinetic energy to the relative wind passing the cart of 100
units (-10 x -10)

To what wind? There is -10 m/s wind all around the cart, but we're only
interested in that part of it which is going through the prop. We need
to know how much that is.

No we don't.

Don't we? I thought you said we were *giving* kinetic energy to the
wind, expending energy. That means we'd have to make it go faster.
We can only make air go faster if we push it, and if we make 2kg of
air go faster, this will take twice as much energy as making only
1 kg of air go similarly faster. So we do need to know how much air
we're making faster.

I assumed, by the way, that you're referring to air which the propeller
is making go faster. If, on the other hand you mean the air which the
structure of the car is pushing out of its way (kind of like the bow wave
of a boat), then that's different. That's just ordinary air drag which
we are for the moment teating as neglible.

Suppose you put 6kg of air through it every second.
Then the kinetic energy of those 6kg of air before they go through the
prop
will be (1/2)x(6kg)x(-10m/s)^2 which is 300 kg m^2/s^2, or 300 J.

Suppose you put ten times that mass it doesn't matter.
Its a perfect prop the mass on either side is the same, it doesn't change
the mass in the flow so you can ignore it.

No you can't. Each kg of air has kinetic energy of 50 J if it's moving
at 10 m/s. If you speed it up to 12 m/s, it will have 72 J of energy.
So if you're putting 1 kg through per second, you'll be imparting 22 W of
power, and if you're putting through 2 kg, it's 44 W, so you need to
know both input speed and output speed and also quantity per second.

but do notice it is going backwards and is being
generated by ricks prop.

No, it's there already, it's the headwind resulting from the cart's
ground speed being 10 m/s more than the wind's ground speed. What the
prop does is make the 10 m/s backwards relative wind go even faster
backwards.

So where is that energy from then if not the prop.

The prop is not *making* energy, it is *converting* (mechanical) energy
from the electric motor which drives it, into kinetic energy which
it is adding to the wind's. The electric motor in turn derives its electric
energy from the generator, which converts mechanical energy from the wheels
into electricity. The mechanical energy which drives the wheels essentially
comes from the thrust generated by the propeller, and it is this fact which
initially makes people think that something is being got for nothing, that
the *energy* for creating the thrust comes from the thrust itself.

That's why it's probably easier to work with the skateboarder.

dennis@home wrote:

"Ronald Raygun" wrote in message
news

I can also do it in a moving frame, for example in that moving at the
hero's initial speed, and get the same answer. I can even do it in a
frame which accelerates with the hero, and still get the same answer,
but it isn't really very interesting, and dennis wouldn't understand the
inertial forces adjustment.

After seeing your reply to my post where you changed frames of reference
and said I was changing them when I clearly did not I don't think you
understand it at all.

Before I move to the next step, are you happy with the analysis so far?

You shouldn't be, not until you understand what the same frame of
reference means.
It might help to understand what physical effects a perfect propeller
means too.

I didn't change frames. I did the analysis in just the one frame,
with all measurements relative to the ground. I showed the workings,
which confirmed that the energies at the beginning add up to the same
as at the end.

Then I also did the whole analysis of the same situation in a different
frame, and then in a third one too, and got the energies to balance
there too, but I merely reported this without showing the workings.

Did you see my subsequent post in which the hero is able to power
a 120 W lightbulb by using only 90 W of muscle power, plus the
spare 30W harvested from slowing down the plebs? What did you
think of it?

I hope I'm not speaking too soon, as that post has not yet passed
Rick's inspection.

Ronald Raygun wrote:

OK, then. Let's add deliberate energy-wasting drag.

As before, the pleb is accelerated backwards for 1 second and, relative
to the ground, is brought to a stop in a distance of 0.5m.

As before, the pleb loses 30 J of kinetic energy, but the hero's
kinetic energy does not change.

This time, the hero's arms are exerting 60 N of force over a distance
of only 1.5 m, so they are doing 90 J of work.

The energy balances: 90 J stored energy plus 30 J kinetic energy from
the pleb give us the 120 J for the hero's lightbulbs.

Assuming Rick approves this, the next step is trivial. We replace the
hero's arm with an electric robot arm programmed to push the pleb.
We plug this robot arm into the generator instead of one of the two
60W lightbulbs, and replace the other with a 30W bulb. That way the
arm gets the same 90W as the hero's arm had.


Now, how does this all work if we want to look at it, as dennis
seems to, entirely from a frame of reference moving with the hero.

In the hero's frame, obviously Vh and Sh and Eh are zero.

Ap = -A
Vp = -A t + Vp0
Sp = -1/2 A t^2 + Vp0*t
Ep = 1/2 M Vp^2

Initial condition:
Vp0 = -1 m/s

Reminder:
A = 1 m/s^2
M = 60kg

Calculate:
Vp1 = -2 m/s
Sp1 = -1.5 m
Ep0 = 30 J
Ep1 = 120 J
Arm work = 90 J

Now we have a problem getting the energies to balance. The hero has done
90 J work, but now instead of gaining 30 J from the pleb, we're losing
90 J to him. Dear oh dear oh dear, our energy budget has been completely
depleted to zero and there's nothing left for the lightbulb.

There's something wrong!

No there isn't.

It's easy to forget what we took for granted when our reference frame was
the ground. Notice how once we moved the frame to the hero, his own
acceleration, speed, distance, and kinetic energy are all zero, as were
those of the ground before we moved. Well, now that the frame is moving
with the hero, the ground's motions are of interest.

The wheel drag is a force which the ground is exerting on the hero's
wheels, but of course the same force is pushing back against the ground,
moving it forward, i.e. tending to reduce the relative speed between the
two. The ground is hugely massive, of course, but we do slow it down
ever so slightly, so we need to take into account the way its kinetic
energy changes.

The Earth is initially moving at speed Ve0 (equal to -2 m/s which we'll
denote by unsuffixed V) in the hero's frame. We put a nominal figure on
the earth's mass (which for convenience we model as flat and thin so we
don't need to worry about rotational effects). Let's say its mass is H
(for huge). Its kinetic energy is therefore

Ee = 1/2 H Ve^2

We are slowing down the Earth using a puny 60 N of force, giving an
acceleration of -60N/H. Another way of expressing this is (M/H)*A.

Ve = M/H A t + V

We calculate:
Ve1 = (M/H m/s + V) m/s
Ee0 = 1/2 H V^2
Ee1 = 1/2 H (M/H m/s + V)^2

The energy difference Ee1-Ee0
= 1/2 H ((M/H m/s)^2 + 2 M/H m/s V + V^2 - V^2)
= 1/2 M (M/H m/s + 2 V m/s)

Because H is huge compared to M, we can ignore the M/H term since it'll be
vanishingly small compared to 2V. The earth is losing M V m/s = 120 J of
kinetic energy.

Now it balances again: The hero puts in 90 J of work, the Earth contributes
120 J, the pleb takes away 90 J, and so we have 120 J left over for the
lightbulb.

On 7/10/2010 10:51 a.m., Ronald Raygun wrote:
Doctor Drivel wrote:

wrote in message
...
Doctor Drivel
It does not go down it goes UP.

Of course the lighter than air balloon is going up; it has no choice
-- gravity is forcing it UP.

So if defies gravity.

Something doesn't defy gravity just by going up.

Suppose you have a jar half full of syrup. The syrup occupies
the bottom half. There is air in the top half.

Now turn the jar upside down. It's OK, the lid is closed.
What heppens next is that gravity forces the syrup down and the air up.
So they both *obey* gravity. The air does not defy it.

A helium filled balloon goes up because gravity is pulling on the
air around and above it more strongly than on the balloon itself.

Put the balloon in a car, and a football on the back seat. Drive
round a corner quickly, and the football will shoot off to one side,
and the balloon to the other. Both items are obeying, not gravity
this time, but a similar phenomenon, inertial acceleration.


I can't believe Drivel is as stupid as he appears. Therefore I think he's
pulling everyone's legs.

"Ronald Raygun" wrote in message
...

Read what I said in the last post you replied to.
I just showed that you never get any energy back from slowing the wind when
you are travelling down wind even with perfect conditions and a perfect
prop.
It doesn't matter at all what size the prop is (that is air mass) which you
appear to think it does.

In the absence of getting any energy back it doesn't matter in the slightest
about the mechanics including the mechanics extracting the energy from the
wheels and giving it to the prop.
All that happens is that you add in the various energy loses like
turbulence, friction, the braking done by taking energy form the wheels and
the whole thing just slows down.

I am not going to go through all your points one by one as you fail to grasp
even the simplest perfect case of just a perfect prop.
If that can't extract any energy from the wind the rest doesn't matter, it
is the *only* *source* of energy that you have.

The energy from the wheels is just stored energy you have already extracted
from somewhere else, it is not a *source* of energy.

You can bull**** like rick if you want, I will just put you on the bull****
list.

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

"Ronald Raygun" wrote in message
news

I can also do it in a moving frame, for example in that moving at the
hero's initial speed, and get the same answer. I can even do it in a
frame which accelerates with the hero, and still get the same answer,
but it isn't really very interesting, and dennis wouldn't understand the
inertial forces adjustment.

After seeing your reply to my post where you changed frames of reference
and said I was changing them when I clearly did not I don't think you
understand it at all.

Before I move to the next step, are you happy with the analysis so far?

You shouldn't be, not until you understand what the same frame of
reference means.
It might help to understand what physical effects a perfect propeller
means too.

I didn't change frames. I did the analysis in just the one frame,
with all measurements relative to the ground. I showed the workings,
which confirmed that the energies at the beginning add up to the same
as at the end.


You changed them from my example and got them wrong, there was no slowing of
the wind in my example and you just slowed it by using a different frame of
reference. Check it if you want, I know that there was no slowing of the
wind. Which is why there was no energy being extracted and that was the
absolute best case for you to get energy from the wind.
Then I also did the whole analysis of the same situation in a different
frame, and then in a third one too, and got the energies to balance
there too, but I merely reported this without showing the workings.

Did you see my subsequent post in which the hero is able to power
a 120 W lightbulb by using only 90 W of muscle power, plus the
spare 30W harvested from slowing down the plebs? What did you
think of it?

I hope I'm not speaking too soon, as that post has not yet passed
Rick's inspection.

"Gib Bogle" wrote in message
...
On 7/10/2010 10:51 a.m., Ronald Raygun wrote:
Doctor Drivel wrote:

wrote in message
...
Doctor Drivel
It does not go down it goes UP.

Of course the lighter than air balloon is going up; it has no choice
-- gravity is forcing it UP.

So if defies gravity.

Something doesn't defy gravity just by going up.

Suppose you have a jar half full of syrup. The syrup occupies
the bottom half. There is air in the top half.

Now turn the jar upside down. It's OK, the lid is closed.
What heppens next is that gravity forces the syrup down and the air up.
So they both *obey* gravity. The air does not defy it.

A helium filled balloon goes up because gravity is pulling on the
air around and above it more strongly than on the balloon itself.

Put the balloon in a car, and a football on the back seat. Drive
round a corner quickly, and the football will shoot off to one side,
and the balloon to the other. Both items are obeying, not gravity
this time, but a similar phenomenon, inertial acceleration.


I can't believe Drivel is as stupid as he appears. Therefore I think he's
pulling everyone's legs.

Does that make him go faster than pulling their arms?
We do need to maximise this free energy in this thread.

dennis@home wrote:

"Ronald Raygun" wrote in message
...

Read what I said in the last post you replied to.
I just showed that you never get any energy back from slowing the wind
when you are travelling down wind even with perfect conditions and a
perfect prop.

You strung a lot of words together, but in a way which didn't make
clear what you were trying to say. I'm unable to follow your argument
because your words don't convey a clear statement of what facts you're
establishing and what conclusions you're drawing from them.

The main thing is that you need to get the energy budget to balance.
It doesn't matter which frame you do it in, but for each balancing
exercise you have to stay in the same frame.

It would help if you used terminology which is obvious to understand
and appropriate to the frame you're using. For example "real wind" or
"true wind" is something which really only makes sense in the ground
frame. The corresponding term for the cart frame would be "apparent
wind" or "relative wind".

If you have a cart going at 30mph in a 20mph wind, i.e. the cart is
10mph faster than the wind, then the real wind is 20mph and the
relative wind is -10mph. OK?

When you were saying the wind was not being slowed, did you mean the
prop was just freewheeling? So that it was not changing the speed
of the air at all? Then it would be consuming no power. That's not
interesting.

What we want the prop to do is to provide thrust. To do this it has
to slow down the wind, for example from its incoming -10mph to an
outgoing -15 or even -20mph. You see that -20 is less than -10,
which is why I call it "slowing down". You may prefer to think of
it as speeding up from 10 to 20 (both backwards). That's fine, because
after all for kinetic energy purposes we square the velocity so any
minus sign will disappear.

It is clear that in this situation, and in the cart frame of reference,
you are not reducing the relative wind's kinetic energy, but increasing
it. So you can't harvest any energy from it. To make it work, the balance
of kinetic energy has to come from somewhere else, and as I explained
elsewhere, it comes from slowing down the ground.

That's why it's easier to work in the ground frame, where the ground
does not move, but the real wind is slowed down (for example from 20mph
to 15mph, or even to 10mph, or even to zero).

dennis@home wrote:

"Gib Bogle" wrote in message
...

I can't believe Drivel is as stupid as he appears. Therefore I think
he's pulling everyone's legs.

Does that make him go faster than pulling their arms?
We do need to maximise this free energy in this thread.

I think he's pulling the legs up to make them defy gravity.

Ronald Raygun wrote:

Ronald Raygun wrote:

OK, then. Let's add deliberate energy-wasting drag.

As before, the pleb is accelerated backwards for 1 second and, relative
to the ground, is brought to a stop in a distance of 0.5m.

As before, the pleb loses 30 J of kinetic energy, but the hero's
kinetic energy does not change.

This time, the hero's arms are exerting 60 N of force over a distance
of only 1.5 m, so they are doing 90 J of work.

The energy balances: 90 J stored energy plus 30 J kinetic energy from
the pleb give us the 120 J for the hero's lightbulbs.

Assuming Rick approves this, the next step is trivial. We replace the
hero's arm with an electric robot arm programmed to push the pleb.
We plug this robot arm into the generator instead of one of the two
60W lightbulbs, and replace the other with a 30W bulb. That way the
arm gets the same 90W as the hero's arm had.

But now let's go back a step.

I've shown that our hero can keep his 120W lamps lit by putting in
only 75% of the power himself, and harvesting the other 25% by stealing
30J of kinetic energy from one pleb each second. But these ratios are
specific to the assumption that he's reducing each pleb's speed to zero.

I've been a victim of my intuition by thinking that if I take as
much kinetic energy as I can from the pleb (i.e. all of it), then
that's the best I can do in terms of getting the pleb to contribute
to the energy budget.

It's easy to see that if the hero were to push each pleb for 2s instead of
1s, that this would reduce the pleb's speed from 1 m/s to -1 m/s, and so the
pleb ends up with the same kinetic energy he started with. That means
there's no contribution available from the pleb, and the hero would need
to do 100% of the work himself. He might as well leave the plebs alone
and crank his generator by hand.

It gets worse. If he pushes for 3s, the pleb's KE ends up *more* than
it was, and the hero has to do 125% of the work needed to keep his bulbs
lit.

So let's go the other way. Push for only half a second. The pleb's speed
is reduced by half, but his kinetic energy therefore by 3/4 and the hero
needs to do only 62.5% of the work. And if we push for only 1/4 s, the hero
needs to do 56.25% of the work.

So, if instead of stealing 100% of a pleb's speed each second, we instead
stole 50% of the speed from two plebs each second, we'd get more energy out.
If we take 1/N of the speed away from N plebs each second, we get the same
thrust, but get more power from them, and in the limit, when we let N go to
infinity, the hero ends up needing to do only 50% of the work.

But of course this would mean letting delta V go towards zero and there
may not be enough plebs in the hero's immediate vicinity, and he'd need
to spread his arms ever wider. This corresponds to making the propeller
bigger and bigger.

With the benefit of hindthought (is there such a word?) this is obvious,
since given that KE varies with the square of speed, the optimum *rate* of
energy extraction is if the speed difference spans the steepest slope of
the energy curve. That's near your starting speed.

In conclusion, gentlemen, I believe I've now shown that you were right and
I was wrong, and so I apologise again for the tone I took.

It's true what they say. The best (well perhaps not the best, but a good)
way to learn about something is to write about it.

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

"Ronald Raygun" wrote in message
...

Read what I said in the last post you replied to.
I just showed that you never get any energy back from slowing the wind
when you are travelling down wind even with perfect conditions and a
perfect prop.

You strung a lot of words together, but in a way which didn't make
clear what you were trying to say. I'm unable to follow your argument
because your words don't convey a clear statement of what facts you're
establishing and what conclusions you're drawing from them.


That's because i am trying to convey simple physics to people that obviously
have no understanding of physics.
If they did understand physics I wouldn't need to do it.


The main thing is that you need to get the energy budget to balance.
It doesn't matter which frame you do it in, but for each balancing
exercise you have to stay in the same frame.

Rubbish.
Stick to one frame, you obviously have trouble when you change frames just
like rick does.


It would help if you used terminology which is obvious to understand
and appropriate to the frame you're using. For example "real wind" or
"true wind" is something which really only makes sense in the ground
frame. The corresponding term for the cart frame would be "apparent
wind" or "relative wind".

If you have a cart going at 30mph in a 20mph wind, i.e. the cart is
10mph faster than the wind, then the real wind is 20mph and the
relative wind is -10mph. OK?

Just like I said in mine except I chose 20 and 10.
That is -10 and -10 referenced to the cart and hence to the prop.
See the problem now? they are both the same, you are travelling at twice the
wind speed and are not slowing the wind.
You are not extracting any energy.
This is possible with a perfect prop, there are no losses.


Now if you speed up the prop and the cart (I don't care how it doesn't
matter) and you add 1 to the speed what has happened?


When you were saying the wind was not being slowed, did you mean the
prop was just freewheeling? So that it was not changing the speed
of the air at all? Then it would be consuming no power. That's not
interesting.

Of course its interesting, you are travelling at twice the wind speed and
not consuming energy to do it, this is what a perfect prop does no losses.
Its the equivalent of a machine that just accelerates air through a tube
creating no turbulence, etc. (They don't exist but lets not stop reality
getting in the way of physics).


What we want the prop to do is to provide thrust.

No we want the prop to extract energy from the wind, we don't care how it
does this.

To do this it has
to slow down the wind, for example from its incoming -10mph to an
outgoing -15 or even -20mph. You see that -20 is less than -10,
which is why I call it "slowing down". You may prefer to think of
it as speeding up from 10 to 20 (both backwards). That's fine, because
after all for kinetic energy purposes we square the velocity so any
minus sign will disappear.

It is clear that in this situation, and in the cart frame of reference,
you are not reducing the relative wind's kinetic energy, but increasing
it. So you can't harvest any energy from it. To make it work, the
balance
of kinetic energy has to come from somewhere else, and as I explained
elsewhere, it comes from slowing down the ground.

So you feel its OK to just switch frames of reference to get the answer you
want then?
Do you go by the name rick or thin air when things are going bad then?

That's why it's easier to work in the ground frame, where the ground
does not move, but the real wind is slowed down (for example from 20mph
to 15mph, or even to 10mph, or even to zero).


Well if it doesn't work in the frame relative to the cart we can always
create a new one where it does.
Shame that that is breaking the laws of physics.

Anyway now we have established that to do it you have to change frames of
reference I think we can just forget it and find some amusement elsewhere.

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

"Ronald Raygun" wrote in message
...

Read what I said in the last post you replied to.
I just showed that you never get any energy back from slowing the wind
when you are travelling down wind even with perfect conditions and a
perfect prop.

You strung a lot of words together, but in a way which didn't make
clear what you were trying to say. I'm unable to follow your argument
because your words don't convey a clear statement of what facts you're
establishing and what conclusions you're drawing from them.


That's because i am trying to convey simple physics to people that obviously
have no understanding of physics.
If they did understand physics I wouldn't need to do it.


The main thing is that you need to get the energy budget to balance.
It doesn't matter which frame you do it in, but for each balancing
exercise you have to stay in the same frame.

Rubbish.
Stick to one frame, you obviously have trouble when you change frames just
like rick does.


It would help if you used terminology which is obvious to understand
and appropriate to the frame you're using. For example "real wind" or
"true wind" is something which really only makes sense in the ground
frame. The corresponding term for the cart frame would be "apparent
wind" or "relative wind".

If you have a cart going at 30mph in a 20mph wind, i.e. the cart is
10mph faster than the wind, then the real wind is 20mph and the
relative wind is -10mph. OK?

Just like I said in mine except I chose 20 and 10.
That is -10 and -10 referenced to the cart and hence to the prop.
See the problem now? they are both the same, you are travelling at twice the
wind speed and are not slowing the wind.
You are not extracting any energy.
This is possible with a perfect prop, there are no losses.


Now if you speed up the prop and the cart (I don't care how it doesn't
matter) and you add 1 to the speed what has happened?


When you were saying the wind was not being slowed, did you mean the
prop was just freewheeling? So that it was not changing the speed
of the air at all? Then it would be consuming no power. That's not
interesting.

Of course its interesting, you are travelling at twice the wind speed and
not consuming energy to do it, this is what a perfect prop does no losses.
Its the equivalent of a machine that just accelerates air through a tube
creating no turbulence, etc. (They don't exist but lets not stop reality
getting in the way of physics).


What we want the prop to do is to provide thrust.

No we want the prop to extract energy from the wind, we don't care how it
does this.

To do this it has
to slow down the wind, for example from its incoming -10mph to an
outgoing -15 or even -20mph. You see that -20 is less than -10,
which is why I call it "slowing down". You may prefer to think of
it as speeding up from 10 to 20 (both backwards). That's fine, because
after all for kinetic energy purposes we square the velocity so any
minus sign will disappear.

It is clear that in this situation, and in the cart frame of reference,
you are not reducing the relative wind's kinetic energy, but increasing
it. So you can't harvest any energy from it. To make it work, the
balance
of kinetic energy has to come from somewhere else, and as I explained
elsewhere, it comes from slowing down the ground.

So you feel its OK to just switch frames of reference to get the answer you
want then?
Do you go by the name rick or thin air when things are going bad then?

That's why it's easier to work in the ground frame, where the ground
does not move, but the real wind is slowed down (for example from 20mph
to 15mph, or even to 10mph, or even to zero).


Well if it doesn't work in the frame relative to the cart we can always
create a new one where it does.
Shame that that is breaking the laws of physics.

Anyway now we have established that to do it you have to change frames of
reference I think we can just forget it and find some amusement elsewhere.