Leo,

I unfortunatly don't have time to adress everything because the
computer I am using at the moment is at the local library and it only
gives out two half hour time slots. I would be more than happy to get
into this with you, I only ask you be direct and give me some pointed
questions.

First off let me state I will not lighten up. Nothing against you
personally, I rather enjoy your posts. I am quite passionate about
physics however, I feel it is the most vital of things to learn, and
also feel it would benifit everyone no matter what field they are in
to learn the principles of it as long as what is learned are the
foundation principles. I really (as I said) detest it when self
purported educated people perpetuate pat sayings and conventional
thinking that is completely untrue, because it's an easy thing to do.
It really shows nothing but an acceptence of conventional ( and
someone elses I might add) thought, with there being very little
criticle thinking. Your defence of that was really nothing but an
excuse. That being said...

There are three undisputable observations regarding gravity. They are;
1. Gravity attracts mass
2. The greater the mass the greater the attraction
3. The closer an object is to an object with measurable gravity, the
greater the attraction. (note this one in paticular)
That being said,


                 ****

                 The background microwave radiation shows that the
universe is very
                 flat on the COSMIC scale. But curved space has been
detected on
                 smaller scales, as evidenced by the ability of massive
bodies
                 (superclusters) to bend the light from more distant
bodies in the
                 line of sight. Haven't you seen the astronomical
photographs of the
                 lensing effect? Curving of space has even been
detected around our
                 sun, as shown by the behavoir of Mercury's orbital
precesion in
                 accord with general relativity.

Light being attracted to a large body  does not prove space is curved.
1. light has mass
2. Mass has gravity
2. Gravity attracts mass

Mercuries path does not prove it either.

Something must be pointed out here for anyone who may get confused.
When we are talking about space we are refering to space having an
unseen structure ie. the sub-stratuum. Not a complete vacuum. Gravity
cannot curve nothingness. Any observations of a large mass curving
space around it are indistinguashable from point three in the
beginning of this message. The closer an object-the greater the
attraction. Again, this is a real fact, and does not in any way prove
WHAT gravity is,

Darn I have to go, will post more later

--- In free_energy@y..., "michaelcamer" <michaelcamer@y...> wrote:
> Leo,
>
> I unfortunatly don't have time to adress everything because the
> computer I am using at the moment is at the local library and it
only
> gives out two half hour time slots. I would be more than happy to
get
> into this with you, I only ask you be direct and give me some
pointed
> questions.
>
> First off let me state I will not lighten up. Nothing against you
> personally, I rather enjoy your posts.


****
Michael, thank you, I enjoy these scientific discussion too.  When I
said "lighten up" I did so because of your use of the word "stupid",
which merely indicates an emotional reaction without conveying any
information.  If you think that the rubber-sheet analogy for the
curving of space by mass is deficient, just tell us why, and maybe
propose a more useful model.  But remember that the analogy is just
an aid for those of us who have trouble visualizing 4-D hyperspace.
We don't all have the time to become well-versed enough in modern
physics to discuss it at the level of rigour that you apparantly
want.  Those who wish to seriously apply relativity of course use the
actual equations.
****


  I am quite passionate about
> physics however, I feel it is the most vital of things to learn,
and
> also feel it would benifit everyone no matter what field they are
in
> to learn the principles of it as long as what is learned are the
> foundation principles. I really (as I said) detest it when self
> purported educated people perpetuate pat sayings and conventional
> thinking that is completely untrue,


****
I agree that there is much over-simplification in the presentation of
science to the public.  But what concerns me even more is the anti-
science emotionalism and low level of scientific literacy in the
general public (I refer to the US).  And I think that the excessive
over-simplification in science teaching is not the cause of the
illiteracy, but rather the other way around.

As far as any "pat sayings" on my part, were you implying that
general relativity is "completely untrue"?  Or was it only the rubber-
sheet analogy that you object to?
****



  because it's an easy thing to do.
> It really shows nothing but an acceptence of conventional ( and
> someone elses I might add)


****
So a critical thinker should not mention anyone else's theories, even
when these theories are well supported by experiments?  I should make
up my own theory before I discuss physics with anyone?  Gee, if I had
known that, I would not have wasted my time taking physics courses in
college.
****


  thought, with there being very little
> criticle thinking. Your defence of that was really nothing but an
> excuse.


****
An excuse for what?  I was simply mentioning a theory, that along
with quantum mechanics, forms the basis of the modern scientific
worldview.  I don't object to people being creative in their
speculations, but please don't tell me that I am wrong for mentioning
the work of those who have come before us.
****


  That being said...
>
> There are three undisputable observations regarding gravity. They
are;
> 1. Gravity attracts mass
> 2. The greater the mass the greater the attraction
> 3. The closer an object is to an object with measurable gravity,
the
> greater the attraction. (note this one in paticular)
> That being said,
>
>
>                 ****
>
>                 The background microwave radiation shows that the
> universe is very
>                 flat on the COSMIC scale. But curved space has been
> detected on
>                 smaller scales, as evidenced by the ability of
massive
> bodies
>                 (superclusters) to bend the light from more distant
> bodies in the
>                 line of sight. Haven't you seen the astronomical
> photographs of the
>                 lensing effect? Curving of space has even been
> detected around our
>                 sun, as shown by the behavoir of Mercury's orbital
> precesion in
>                 accord with general relativity.
>
> Light being attracted to a large body  does not prove space is
curved.
> 1. light has mass

****
Photons have mass?  I always heard that nothing with mass could
attain the speed of light.  According to special relativity, doesn't
the mass of an object continually increase as its speed approaches
the speed of light?  If so, wouldn't the mass become infinite in
order for the object to actually reach the speed of light? (I don't
claim to thoroughly understand special relativity, so I could be
wrong on this last point).
****



> 2. Mass has gravity
> 2. Gravity attracts mass
>
> Mercuries path does not prove it either.


****
Not Mercury's path, but the precession of the planet's spin.
Measurements show that its precession does not obey classical
mechanics, but does obey the theory of general relativity.  From what
I understand, the deviation from classical behavoir is attributed to
curving of space-time by the sun's mass (although the effect is
negligable for the other planets, which are too far away).

Leo C.
****


>
> Something must be pointed out here for anyone who may get confused.
> When we are talking about space we are refering to space having an
> unseen structure ie. the sub-stratuum. Not a complete vacuum.
Gravity
> cannot curve nothingness. Any observations of a large mass curving
> space around it are indistinguashable from point three in the
> beginning of this message. The closer an object-the greater the
> attraction. Again, this is a real fact, and does not in any way
prove
> WHAT gravity is,
>
> Darn I have to go, will post more later

Leo,

Sorry, I have to be really short with this. Just not enough time.
Regarding,

> Photons have mass? I always heard that nothing with mass could
               attain the speed of light.

Yes photons have mass. Anything with energy has mass. Photons have no
rest mass, but photons are not at rest.

Michael

materials needed.

(1) 20 ft lenght of 2" pvc pipe.
(2) 2" 90 degree elbows
(1) 1 gallon bucket.
(1) 1" section of 2" pvc pipe
attach the two 90 degree elbows to the 20 ft pipe so that the hole is
facing up.
attach the 1" section to one of the elbows.

take your garden hose and fill the pipe with water.

fill your 1 gallon bucket with water.

calculate the mass of the water inside the pipe.

now pour the water in your bucket into the 1" section
pour it verry slowly.

determine the amount of energy required to lift the water into the
pipe.

notice that same amount of energy comes out at the other end in the
form of potential energy.


you have expended 1 unit of energy.
and have had a return of 1 unit of energy.

meanwhile you have moved the mass of water between the elbows.

calculate the velocity of the moving mass of water.
find the ke of that moving mass.

you have a overunity device.

if you put 1 unit of energy in.
and get 1 unit of energy out.

that is a equalibrium.

if there is (any) movement of water in between the elbows
that is overunity.

Dear Sir

I should hope that a larger pump would be required to reverse the direction of flow.

the pump would be acting on a large moving mass and would first have to stop the flow

in order to reverse the flow.

then the pump would need to lift the water to a height at the end of the pipe.

I truly do not believe that a same size and flow rated pump could reverse the direction.

as the energy needed would be greater.

the pump that lifts the water is doing just that lifting it.

a pump that would reverse the flow direction would have to push the long mass of water and then lift the water.

there can be no denial that the reversing pump would require more energy.

its simple physics.

trying to get some to understand simple physics is a great task.

it seems that you can think using physics properties vs physics laws.

I commend you.

you have your own brain.

  mohammed alkhamis <moh_alkhamis@...> wrote:

Hello,


Think of reversing the water from the other end, and see if it does need
more energy what I mean is if you put a pump in the outlet end should this
pump be bigger than the lifting pump to reverse the flow of water, if so the
energy must be different ,,

Mohammed Alkhamis
http://us.geocities.yahoo.com/gb/view?member=energy_is_free
http://www.geocities.com/energy_is_free/






>From: "damian_thornus"
>To: free_energy@yahoogroups.com
>Subject: [free_energy] Re: Gravity
>Date: Fri, 07 Jun 2002 16:41:11 -0000
>
>Dear Sir
>thank you for your reply.
>
> > like the excessive amounts of energy in the pipe flow situation
> > where the energy that is used to maintain a flow is outweighed
> > numerous times by the energy of the fluid that is contained in the
> > flow.
> > ke=1/2mv^2
> >
> > we know its there but were affraid to do any thing with it.
>
>
>****
>We know excess energy is there? It seems like several knowledgable
>people recently analysed Paul Gearjammer's flow energy over-unity in
>detail (on theoretical grounds) and found it unconvincing.
>****
>
>(on theoretical grounds)
>
>
>1) instead of using a liquid consider a solid.
>at any moment of time.
>note: not continous only a fraction of time to observe the potential
>of the kinetic energy in the moving mass.
>
>surely this moving mass can not break the second law of
>thermodynamics by not containing a amount of kinetic energy.
>
>and render your model as a slug of ice vs water.
>instead of using propellers use a rack atop the slug of ice.
>
>mount a pinion to the top of the pipe.
>
>freeze the water as it enters the pipe.
>and melt the ice as it leaves the pipe.
>
>the energy required to lift the water up into the pipe would be
>energy 1.
>
>the energy of the moving slug of ice would be energy 2
>
>the shaft torque from the pinion would be that of the slug of ice
>moving.
>and can easily be calculated using ke=1/2mv^2
>
>show the energies durring a 1 second time frame.
>reduce your findings to input and output
>horse power or kilo watts durring that 1 second.
>compare the two.
>
>by taking only a small portion of this kinetic energy through a load
>of some type would not stop the flow and would return the original
>energy put in to keep the flow moving.
>
>by no means is there more energy put in than can be taken out.
>
>of course the pipe thingy does not freeze the water or melt it so
>please do not include this into your calculations.
>
>also set the ice flow resistance to the same resistance as that of
>water flowing through a pipe.
>
>it would be against physics to deny that there is more energy that
>can be taken vs the energy put in.
>
>this is what I refer to when I speak of our environment and our
>leaders and as you say our culture dumbing us or numbing us to the
>reality of matters concerning things such as overunity.
>
>I have read the responces to the pipe thing and have not found a
>sencible reason as to why it would not work.
>
>there have been some really screwy reasons presented like the time
>when the energy required to lift the water into the pipe and the
>energy of the water comming out of the pipe were compared to show
>that it could not work.
>this comparison removed the advantage of any energy from the massive
>amount of water moving through the pipe.
>when this water that moves through the pipe is the main reason that
>there is excessive energy.
>
>once again " the dumbing "
>
>I personaly would love to see a real reason why it would not work.
>
>I am assumming that each propeller would feel the push and pull of
>the entire mass of water.
>as this is how fluids present themselves to objects in a environment
>that will not allow the fluid to move around the object as in a river.
>the direction of flow by using a pipe would not allow the water to
>take a less resistive path.
>as water flows over and around a rock in a riverbed.
>
>the propellers are trapping amounts of kinetic energy taking energy
>from the flow .
>
>given that a pipe that is completely filled with agregate will still
>present a flow capacity and this agregate has all but stopped the
>flow this only reveals that the agregate has absorbed the energy of
>the flow.
>
>it appears when reading these reasons that there is a lack of
>understanding concerning fluid behavior.
>
>and that there is a continous pointing to the laws used to give
>reason for its not being a workable or reasonable idea.
>
>what I would like to see would be a presentation of a reasonably
>inteligent explanation of the two energies involved.
>
>input and output.
>not just quotations from some physics book or personal theories.
>
>I have noticed the great detail given in other cases where is the
>great detail in this case?
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>--- In free_energy@y..., "chudslayer" wrote:
> > --- In free_energy@y..., "damian_thornus"
>wrote:
> > > relativity would require that all things are relative.
> >
> > ****Leo's reply****
> > General relativity requires that MOTION is relative, not that "all
> > things are relative".
> > ****
> >
> >
> > > thus a explosion in space would have to expand in all directions.
> > > thus space would have to be in all directions.
> > > if our universe is expanding along a flat plane then there is no
> > > space outside of this flat plane.
> >
> >
> > ****
> > You apparantly misunderstood what was meant by the statement
> > that "the curvature of space-time is flat" (as indicated by
>analysis
> > of the microwave background radiation). This does not mean that
> > space is flat in the sense of being 2-D. It means that there is
>very
> > little curvature of 3-D space into higher dimension(s). A pretty
> > abstract idea for sure, so most people have trouble with this.
> > ****
> >
> >
> >
> > > and there would have to be either a force that holds all galaxies
> > in
> > > that plane or the galaxies that leave that plane become non-
> > existant.
> > > we are speaking of a box of sorts.
> > > I find this hard to believe.
> > > but who are we to say what shape the universe is.
> > > when we have not been able to leave our own solar system.
> > > and will not as long as energy is a product.
> > > like the excessive amounts of energy in the pipe flow situation
> > > where the energy that is used to maintain a flow is outweighed
> > > numerous times by the energy of the fluid that is contained in
>the
> > > flow.
> > > ke=1/2mv^2
> > >
> > > we know its there but were affraid to do any thing with it.
> >
> >
> > ****
> > We know excess energy is there? It seems like several knowledgable
> > people recently analysed Paul Gearjammer's flow energy over-unity
>in
> > detail (on theoretical grounds) and found it unconvincing. And no
> > one has reported any experimental data on any prototypes. We're
> > not "afraid" of over-unity technologies, we are just waiting for
> > reproducable evidence. With all the claims out there of O/U
>devices
> > (MEG, Lutec, Bearden's motor, Adam motor, SmartPack, etc), we
>should
> > see some on the market soon, right?
> > ****
> >
> >
> > > we like mr cammer stated are being dumbed or numbed by our
> > > environment and those who we have relied upon to set rules to
>live
> > by.
> > > these rules are killing us off every day we just cant see it.
> > > but we will soon begin to feel the results of our ignorance.
> >
> >
> > ****
> > I would agree that cultural and environmental pollution
>are "dumbing
> > us down". But I think science and reason are our greatest hope for
> > saving ourselves from the human tendency to descend into
> > superstition, intolerance, and selfishness. Don't blame technology
> > for our ills -- it is just a tool.
> >
> > Leo C.
> > ****
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > --- In free_energy@y..., Autymn D. C. wrote:
> > > > Received: 06/6/2002 11.58
> > > >
> > > > >1. Space and time ARE the same thing. They consist of
>difference
> > > which
> > > > how?
> > > >
> > > > >2. Space is not curved. Expansion theory requires a flat
> > universe.
> > > > >Already observed. Believe it or not mass curving space does
>not
> > > > Expansion in hyperspace requires a curved universe. It's just
> > that
> > > the
> > > > NASA high-altitude balloon experiments meant to measure the
> > > universe's
> > > > curvature via inhomogenities in the microwave background found
> > just
> > > about
> > > > no curvature because the expansion had been so great, and the
> > view
> > > of the
> > > > background past the universe's observable horizon at ~14 Gly
>away
> > > is
> > > > outside our lightcone. Space can still be locally curved by
> > > gravity and
> > > > straightened by magnetism.
> > > > >explain gravity, since BOTH mass and space are assumed to
> > consist
> > > of
> > > > The space needs to be curved in a fourth dimension, so the
>rubber
> > > > sheet/ball bearing model doesn't quite work as its argued
> > > circularly.
> > > >
> > > > -Aut
>




_________________________________________________________________
Join the world’s largest e-mail service with MSN Hotmail.
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>
> materials needed.
>
> (1) 20 ft lenght of 2" pvc pipe.
> (2) 2" 90 degree elbows
> (1) 1 gallon bucket.
> (1) 1" section of 2" pvc pipe
> attach the two 90 degree elbows to the 20 ft pipe so that the hole is
> facing up.
> attach the 1" section to one of the elbows.
>

Your analysis of your supposed over-unity demonstrator is, of course,
inaccurate.

When all relevant factors are taken into account, a more complete analysis
shows that there is a net loss of useful energy, with some being "lost", as
usual, to entropy.

You see, you made a mistake when you say "notice that same amount of energy
comes out at the other end in the form of potential energy."  'Tain't so.

On the feed side, you have that 1" standpipe, which you put in to act as a
funnel -- it keeps the water you are trying to feed in from the bucket from
just sloshing over the sides of the already full elbow.

But on the exit side, you don't have a matching 1" standpipe.

Thus, you have to lift the water on the feed side one inch higher (to get it
into the funnel) than the system will lift it on the exit side twenty feet
away.  So, you are putting in more potential energy than you are getting
out.

<shrug> Sorry.

>So, you are putting in more potential energy than you are getting
> out.

Im glad you pointed that out.
I would be lifting 1 gallon of water 1" higher.

the remaining energies are at a balance concerning the amounts of
input energy lifted and the amounts of potential energy output.

1 gallon of water = 231 cu inches
231 cu inches of water weighs = 8.33 lbs
8.33 lbs = 3.78 kg
the force required to lift the 1 gallon.
F=mg=3.78kg X 9.8m/sec^2= 37.044 N
if I lift 3.78 kg to a height of 1"
1" = 0.0254 m
W=37.044 N X 0.0254 m = 0.9409176 J
if this is carried out in a time frame of 60 seconds.
P = 0.9409176/60= 0.01568196 J/minute

and .000261366 J/sec

the distance that the mass of water in the pipe would move would be.

231 cu in / 3.14 area =  73.56 inches/minute

the weight of water in the pipe would be 27.20 lb

the kinetic energy of the 27.20 lbs moving a distance of 73.56 inches
in one minute.
27.20 lb = 12.33 kg
1/2 of 12.33 kg = 6.165 kg
73.56 inches = 6.13 ft
6.13 ft = 1.86 m
velocity = 1.86 m/minute
v= 0.0311404 m/sec
v^2 = 0.0311404 m/sec^2 = .0009697245122m/sec

ke = 1/2mv^2
ke = 6.165 kg X .0009697245122 m/sec  = .005978351618 J

at any second durring the one minute the kinetic energy of the
moving mass of water will be .005978351618 J


the additional energy required each second to lift the water the
additional 1"
.000261366 J
the constant energy of the moving mass of water
.005978351618 J

the excess energy = .005716985618 J


> When all relevant factors are taken into account, a more complete
>analysis
> shows that there is a net loss of useful energy, with some
>being "lost", as
> usual, to entropy.

perhaps you could further explain.
or go into a little more detail with a few numbers.
I could say that the moon is made of cheese.
or that I can pick myself up with my bootstraps.
but that bears no meaning in physics or science.
data gathering and acurately applied math are always much better than
lip service.





--- In free_energy@y..., "Bob Dubner" <rdubner@c...> wrote:
> >
> > materials needed.
> >
> > (1) 20 ft lenght of 2" pvc pipe.
> > (2) 2" 90 degree elbows
> > (1) 1 gallon bucket.
> > (1) 1" section of 2" pvc pipe
> > attach the two 90 degree elbows to the 20 ft pipe so that the
hole is
> > facing up.
> > attach the 1" section to one of the elbows.
> >
>
> Your analysis of your supposed over-unity demonstrator is, of
course,
> inaccurate.
>
> When all relevant factors are taken into account, a more complete
analysis
> shows that there is a net loss of useful energy, with some
being "lost", as
> usual, to entropy.
>
> You see, you made a mistake when you say "notice that same amount
of energy
> comes out at the other end in the form of potential
energy."  'Tain't so.
>
> On the feed side, you have that 1" standpipe, which you put in to
act as a
> funnel -- it keeps the water you are trying to feed in from the
bucket from
> just sloshing over the sides of the already full elbow.
>
> But on the exit side, you don't have a matching 1" standpipe.
>
> Thus, you have to lift the water on the feed side one inch higher
(to get it
> into the funnel) than the system will lift it on the exit side
twenty feet
> away.  So, you are putting in more potential energy than you are
getting
> out.
>
> <shrug> Sorry.

From:        damian_thornus, damian_thornus@...

>v^2 = 0.0311404 m/sec^2 = .0009697245122m/sec
m^2/s^2

>ke = 6.165 kg X .0009697245122 m/sec  = .005978351618 J
m^2/s^2
You didn't divide by two.  It's .002989... J

>the additional energy required each second to lift the water the
>additional 1"
>.000261366 J
>the constant energy of the moving mass of water
>.005978351618 J
>the excess energy = .005716985618 J
That is from pouring the water at a greater height.  The .002989... J is
being converted into the .0002614... J for the entirety of the pouring,
then into the additional energy of the falling water, then into the drag.

>data gathering and acurately applied math are always much better than
>lip service.
So why don't you have them?

-Aut

From:        damian_thornus, damian_thornus@...

>determine the amount of energy required to lift the water into the
>pipe.
>notice that same amount of energy comes out at the other end in the
>form of potential energy.

>calculate the velocity of the moving mass of water.
>find the ke of that moving mass.

>if there is (any) movement of water in between the elbows
>that is overunity.
You are now and forever an illiterate retard whose head should be smashed
against a wall.  The KE is included in (1) to cause (2) subtracted by
drag.

before: mgh + .5mv^2 + m_2gh_2 =
         mgh +      0 + m_2gh_2 = E_0
  after: mgh + (.5Mv^2 - drag) =
         mgh + (.5m_2v_2^2 + .5mv^2 - drag) = E

h_2 > h
v_2 > v
E_0 = E
m_2gh_2 > .5m_2v_2^2 + .5mv^2

-Aut

Dear Sir
thank you for your reply.

like the excessive amounts of energy in the pipe flow situation
where the energy that is used to maintain a flow is outweighed 
numerous times by the energy of the fluid that is contained in the 
flow.
ke=1/2mv^2

we know its there but were affraid to do any thing with it.

****
We know excess energy is there? It seems like several knowledgable 
people recently analysed Paul Gearjammer's flow energy over-unity in 
detail (on theoretical grounds) and found it unconvincing. 
****

(on theoretical grounds) 


1) instead of using a liquid consider a solid.
at any moment of time.
note: not continous only a fraction of time to observe the potential
of the kinetic energy in the moving mass.

surely this moving mass can not break the second law of 
thermodynamics by not containing a amount of kinetic energy.

and render your model as a slug of ice vs water.
instead of using propellers use a rack atop the slug of ice.

mount a pinion to the top of the pipe.

freeze the water as it enters the pipe.
and melt the ice as it leaves the pipe.

the energy required to lift the water up into the pipe would be 
energy 1.

the energy of the moving sl
ug of ice would be energy 2

the shaft torque from the pinion would be that of the slug of ice
moving.
and can easily be calculated using ke=1/2mv^2

show the energies durring a 1 second time frame.
reduce your findings to input and output 
horse power or kilo watts durring that 1 second.
compare the two.

by taking only a small portion of this kinetic energy through a load 
of some type would not stop the flow and would return the original 
energy put in to keep the flow moving.

by no means is there more energy put in than can be taken out.

of course the pipe thingy does not freeze the water or melt it so 
please do not include this into your calculations.

also set the ice flow resistance to the same resistance as that of 
water flowing through a pipe.

it would be against physics to deny that there is more energy that 
can be taken vs the energy put in.

this is what I refer to when I speak of our environment and our 
leaders and as you say our culture dumbing us or numbing us to the 
reality of matters concerning things such as overunity.

I have read the responces to the pipe thing and have not found a 
sencible reason as to why it would not work.

there have been some really screwy reasons presented like the time 
when the energy required to lift the water into the pipe and the 
energy of the water comming out of the pipe were compared to show 
that it could not work.
this comparison removed the advantage of any energy from the massive 
amount of water moving through the pipe.
when this water that moves through the pipe is the main reason that 
there is excessive energy.

once again " the dumbing " 

I personaly would love to see a real reason why it would not work.

I am assumming that each propeller would feel the push and pull of 
the entire mass of water.
as this is how fluids present themselv
es to objects in a environment 
that will not allow the fluid to move around the object as in a river.
the direction of flow by using a pipe would not allow the water to 
take a less resistive path.
as water flows over and around a rock in a riverbed.

the propellers are trapping amounts of kinetic energy taking energy 
from the flow .

given that a pipe that is completely filled with agregate will still 
present a flow capacity and this agregate has all but stopped the 
flow this only reveals that the agregate has absorbed the energy of 
the flow.

it appears when reading these reasons that there is a lack of 
understanding concerning fluid behavior.

and that there is a continous pointing to the laws used to give 
reason for its not being a workable or reasonable idea.

what I would like to see would be a presentation of a reasonably 
inteligent explanation of the two energies involved.

input and output.
br>not just quotations from some physics book or personal theories. 

I have noticed the great detail given in other cases where is the 
great detail in this case?








 









 





























--- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:

--- In free_energy@y..., "damian_thornus" <damian_thornus@y...> 

wrote:

relativity would require that all things are relative.

****Leo's reply****
General relativity requires that MOTION is relative, not that "all 
things are relative".
****

thus a explosion in space would have to expand in all directions.
thus space would have to be in all directions.
if our universe is expanding along a flat plane then there is no 
space outside of this flat plane.

****
You apparantly misunderstood what was meant by the statement 
that "the curvature of space-time is flat" (as indicated by 

analysis 

of the microwave background radiation). This does not mean that 
space is flat in the sense of being 2-D. It means that there is 

very 

little curvature of 3-D space into higher dimension(s). A pretty 
abstract idea for sure, so most people have trouble with this.
****

and there would have to be either a force that holds all galaxies 

in 

that plane or the galaxies that leave that plane become non-

existant.

we are speaking of a box of sorts.
I find this hard to believe.
but who are we to say what shape the universe is.
when we have not been able to leave our own solar system.
and will not as long as energy is a product.
like the excessive amounts of energy in the pipe flow situation
where the energy that is used to maintain a flow is outweighed 
numerous times by the energy of the fluid that is contained in 

the 

flow.
ke=1/2mv^2

we know its there but were affraid to do any thing with it.

****
We know excess energy is there? It seems like several knowledgable 
people recently analysed Paul Gearjammer's flow energy over-unity 

in 

detail (on theoretical grounds) and found it unconvincing. And no 
one has reported any experimental data on any prototypes. We're 
not "afraid" of over-unity technologies, we are just waiting for 
reproducable evidence. With all the claims out there of O/U 

devices 

(MEG, Lutec, Bearden's motor, Adam motor, SmartPack, etc), we 

should 

see some on the market soon, right?
****

we like mr cammer stated are being dumbed or numbed by our 
environment and those who we have relied upon to set rules to 

live 

by.

these rules are killing us off every day we just cant see it.
but we will soon begin to feel the results of our ignorance.

****
I would agree that cultural and environmental pollution 

are "dumbing 

us down". But I think science and reason are our greatest hope for 
saving ourselves from the human tendency to descend into 
superstition, intolerance, and selfishness. Don't blame technology 
for our ills -- it is just a tool.

Leo C.
****

--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:

Received: 06/6/2002 11.58

1. Space and time ARE the same thing. They consist of 

difference 

which 

how?

2. Space is not curved. Expansion theory requires a flat 

universe. 

Already observed. Believe it or not mass curving space does 

not 

Expansion in hyperspace requires a curved universe. It's just 

that 

the 

NASA high-altitude balloon experiments meant to measure the 

universe's 

curvature via inhomogenities in the microwave background found 

just 

about 

no curvature because the expansion had been so great, and the 

view 

of the 

background past the universe's observable horizon at ~14 Gly 

away 

is 

outside our lightcone. Space can still be locally curved by 

gravity and 

straightened by magnetism.

explain gravity, since BOTH mass and space are assumed to 

consist 

of 

The space needs to be curved in a fourth dimension, so the 

rubber 

sheet/ball bearing model doesn't quite work as its argued 

circularly.

-Aut

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I thought so.
all you really know how to do is run the lip.
correct spelling and be a complete pain.

> You didn't divide by two.  It's .002989... J

the kinetic energy of the 27.20 lbs moving a distance of 73.56 inches
in one minute.
27.20 lb = 12.33 kg
1/2 of 12.33 kg = 6.165 kg

as for your un-labled energy balance.
I will need a while to figure out what you were doing.




--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> From:        damian_thornus, damian_thornus@y...
>
> >v^2 = 0.0311404 m/sec^2 = .0009697245122m/sec
> m^2/s^2
>
> >ke = 6.165 kg X .0009697245122 m/sec  = .005978351618 J
> m^2/s^2
> You didn't divide by two.  It's .002989... J
>
> >the additional energy required each second to lift the water the
> >additional 1"
> >.000261366 J
> >the constant energy of the moving mass of water
> >.005978351618 J
> >the excess energy = .005716985618 J
> That is from pouring the water at a greater height.  The .002989...
J is
> being converted into the .0002614... J for the entirety of the
pouring,
> then into the additional energy of the falling water, then into the
drag.
>
> >data gathering and acurately applied math are always much better
than
> >lip service.
> So why don't you have them?
>
> -Aut

my opinion is, not here

your opinion is here but where is you math backing up your opinion?






--- In free_energy@y..., David Sligar <audax22@e...> wrote:
> Sorry, Damian -- Overunity may exist somewhere, my opinion is, not
here.
>  Not even close.  Now, instead of spending an hour doing a
quantitative
> analysis with drawings, etc.,  I'll just challenge you:  build your
> device, accurately measure the energy in vs. energy out, and draw
your
> own conclusions.  First, you're going to have to learn about
metrology.
>  That may be more than you're willing to do, but if you do it
right,
> you'll benefit immeasureably from the education.  Remember, you've
made
> this argument.  The burden of proof belongs to you, and only to
you.
>   Respectfully,   David Sligar
>
> damian_thornus wrote:
>
> >Dear Sir
> >thank you for your reply.
> >
> >>like the excessive amounts of energy in the pipe flow situation
> >>where the energy that is used to maintain a flow is outweighed
> >>numerous times by the energy of the fluid that is contained in
the
> >>flow.
> >>ke=1/2mv^2
> >>
> >>we know its there but were affraid to do any thing with it.
> >>
> >
> >
> >****
> >We know excess energy is there? It seems like several knowledgable
> >people recently analysed Paul Gearjammer's flow energy over-unity
in
> >detail (on theoretical grounds) and found it unconvincing.
> >****
> >
> >(on theoretical grounds)
> >
> >
> >1) instead of using a liquid consider a solid.
> >at any moment of time.
> >note: not continous only a fraction of time to observe the
potential
> >of the kinetic energy in the moving mass.
> >
> >surely this moving mass can not break the second law of
> >thermodynamics by not containing a amount of kinetic energy.
> >
> >and render your model as a slug of ice vs water.
> >instead of using propellers use a rack atop the slug of ice.
> >
> >mount a pinion to the top of the pipe.
> >
> >freeze the water as it enters the pipe.
> >and melt the ice as it leaves the pipe.
> >
> >the energy required to lift the water up into the pipe would be
> >energy 1.
> >
> >the energy of the moving slug of ice would be energy 2
> >
> >the shaft torque from the pinion would be that of the slug of ice
> >moving.
> >and can easily be calculated using ke=1/2mv^2
> >
> >show the energies durring a 1 second time frame.
> >reduce your findings to input and output
> >horse power or kilo watts durring that 1 second.
> >compare the two.
> >
> >by taking only a small portion of this kinetic energy through a
load
> >of some type would not stop the flow and would return the original
> >energy put in to keep the flow moving.
> >
> >by no means is there more energy put in than can be taken out.
> >
> >of course the pipe thingy does not freeze the water or melt it so
> >please do not include this into your calculations.
> >
> >also set the ice flow resistance to the same resistance as that of
> >water flowing through a pipe.
> >
> >it would be against physics to deny that there is more energy that
> >can be taken vs the energy put in.
> >
> >this is what I refer to when I speak of our environment and our
> >leaders and as you say our culture dumbing us or numbing us to the
> >reality of matters concerning things such as overunity.
> >
> >I have read the responces to the pipe thing and have not found a
> >sencible reason as to why it would not work.
> >
> >there have been some really screwy reasons presented like the time
> >when the energy required to lift the water into the pipe and the
> >energy of the water comming out of the pipe were compared to show
> >that it could not work.
> >this comparison removed the advantage of any energy from the
massive
> >amount of water moving through the pipe.
> >when this water that moves through the pipe is the main reason
that
> >there is excessive energy.
> >
> >once again " the dumbing "
> >
> >I personaly would love to see a real reason why it would not work.
> >
> >I am assumming that each propeller would feel the push and pull of
> >the entire mass of water.
> >as this is how fluids present themselves to objects in a
environment
> >that will not allow the fluid to move around the object as in a
river.
> >the direction of flow by using a pipe would not allow the water to
> >take a less resistive path.
> >as water flows over and around a rock in a riverbed.
> >
> >the propellers are trapping amounts of kinetic energy taking
energy
> >from the flow .
> >
> >given that a pipe that is completely filled with agregate will
still
> >present a flow capacity and this agregate has all but stopped the
> >flow this only reveals that the agregate has absorbed the energy
of
> >the flow.
> >
> >it appears when reading these reasons that there is a lack of
> >understanding concerning fluid behavior.
> >
> >and that there is a continous pointing to the laws used to give
> >reason for its not being a workable or reasonable idea.
> >
> >what I would like to see would be a presentation of a reasonably
> >inteligent explanation of the two energies involved.
> >
> >input and output.
> >not just quotations from some physics book or personal theories.
> >
> >I have noticed the great detail given in other cases where is the
> >great detail in this case?
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >--- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:
> >
> >>--- In free_energy@y..., "damian_thornus" <damian_thornus@y...>
> >>
> >wrote:
> >
> >>>relativity would require that all things are relative.
> >>>
> >>****Leo's reply****
> >>General relativity requires that MOTION is relative, not
that "all
> >>things are relative".
> >>****
> >>
> >>
> >>>thus a explosion in space would have to expand in all directions.
> >>>thus space would have to be in all directions.
> >>>if our universe is expanding along a flat plane then there is no
> >>>space outside of this flat plane.
> >>>
> >>
> >>****
> >>You apparantly misunderstood what was meant by the statement
> >>that "the curvature of space-time is flat" (as indicated by
> >>
> >analysis
> >
> >>of the microwave background radiation).  This does not mean that
> >>space is flat in the sense of being 2-D.  It means that there is
> >>
> >very
> >
> >>little curvature of 3-D space into higher dimension(s).  A pretty
> >>abstract idea for sure, so most people have trouble with this.
> >>****
> >>
> >>
> >>
> >>>and there would have to be either a force that holds all
galaxies
> >>>
> >>in
> >>
> >>>that plane or the galaxies that leave that plane become non-
> >>>
> >>existant.
> >>
> >>>we are speaking of a box of sorts.
> >>>I find this hard to believe.
> >>>but who are we to say what shape the universe is.
> >>>when we have not been able to leave our own solar system.
> >>>and will not as long as energy is a product.
> >>>like the excessive amounts of energy in the pipe flow situation
> >>>where the energy that is used to maintain a flow is outweighed
> >>>numerous times by the energy of the fluid that is contained in
> >>>
> >the
> >
> >>>flow.
> >>>ke=1/2mv^2
> >>>
> >>>we know its there but were affraid to do any thing with it.
> >>>
> >>
> >>****
> >>We know excess energy is there?  It seems like several
knowledgable
> >>people recently analysed Paul Gearjammer's flow energy over-unity
> >>
> >in
> >
> >>detail (on theoretical grounds) and found it unconvincing.  And
no
> >>one has reported any experimental data on any prototypes.  We're
> >>not "afraid" of over-unity technologies, we are just waiting for
> >>reproducable evidence.  With all the claims out there of O/U
> >>
> >devices
> >
> >>(MEG, Lutec, Bearden's motor, Adam motor, SmartPack, etc), we
> >>
> >should
> >
> >>see some on the market soon, right?
> >>****
> >>
> >>
> >>>we like mr cammer stated are being dumbed or numbed by our
> >>>environment and those who we have relied upon to set rules to
> >>>
> >live
> >
> >>by.
> >>
> >>>these rules are killing us off every day we just cant see it.
> >>>but we will soon begin to feel the results of our ignorance.
> >>>
> >>
> >>****
> >>I would agree that cultural and environmental pollution
> >>
> >are "dumbing
> >
> >>us down".  But I think science and reason are our greatest hope
for
> >>saving ourselves from the human tendency to descend into
> >>superstition, intolerance, and selfishness.  Don't blame
technology
> >>for our ills -- it is just a tool.
> >>
> >>Leo C.
> >>****
> >>
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> >>>
> >>>>Received:    06/6/2002 11.58
> >>>>
> >>>>>1. Space and time ARE the same thing. They consist of
> >>>>>
> >difference
> >
> >>>which
> >>>
> >>>>how?
> >>>>
> >>>>>2. Space is not curved. Expansion theory requires a flat
> >>>>>
> >>universe.
> >>
> >>>>>Already observed. Believe it or not mass curving space does
> >>>>>
> >not
> >
> >>>>Expansion in hyperspace requires a curved universe.  It's just
> >>>>
> >>that
> >>
> >>>the
> >>>
> >>>>NASA high-altitude balloon experiments meant to measure the
> >>>>
> >>>universe's
> >>>
> >>>>curvature via inhomogenities in the microwave background found
> >>>>
> >>just
> >>
> >>>about
> >>>
> >>>>no curvature because the expansion had been so great, and the
> >>>>
> >>view
> >>
> >>>of the
> >>>
> >>>>background past the universe's observable horizon at ~14 Gly
> >>>>
> >away
> >
> >>>is
> >>>
> >>>>outside our lightcone.  Space can still be locally curved by
> >>>>
> >>>gravity and
> >>>
> >>>>straightened by magnetism.
> >>>>
> >>>>>explain gravity, since BOTH mass and space are assumed to
> >>>>>
> >>consist
> >>
> >>>of
> >>>
> >>>>The space needs to be curved in a fourth dimension, so the
> >>>>
> >rubber
> >
> >>>>sheet/ball bearing model doesn't quite work as its argued
> >>>>
> >>>circularly.
> >>>
> >>>>-Aut
> >>>>
> >
> >
> >
> >To drop of the list, send email to:
> >free_energy-unsubscribe@e...
> >
> >Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >

From:        damian_thornus, damian_thornus@...

>all you really know how to do is run the lip.
>correct spelling and be a complete pain.
No, I maintain accuracy.

>the kinetic energy of the 27.20 lbs moving a distance of 73.56 inches
>in one minute.
Uh, no, all of the water didn't move 73.56" or even that close.  All of
the water moved a total of only one to two inches depending on the
sections you're counting.

>I will need a while to figure out what you were doing.
Potential energy was converted into kinetic when you thought they existed
separately.

-Aut

From:  "damian_thornus" <damian_thornus@y...>
Date:  Sat Jun 8, 2002  7:43 pm
Subject:  Re: a overunity model

>data gathering and acurately applied math are always much better
>than lip service.

Paul, this "damian_thornus" identity doesn't fool anybody.

You've had your pipe dream (pun intended) explained to you countless
times by countless people, and they've all told you it doesn't
produce free energy.  Just because you are incapable of understanding
the thermodynamics and the math that doesn't mean everybody else is.

I agree whole heartedly with your statement: "data gathering and
acurately applied math are always much better than lip service".
I've said it before and I say it again: Build the freakin' thing!
Take measurements!  Produce this free energy!  Prove us and centuries
of thermodynamics wrong!

Come on.  What's stopping you?

>
> >So, you are putting in more potential energy than you are getting
> > out.
>
> Im glad you pointed that out.
> I would be lifting 1 gallon of water 1" higher.
>
> the remaining energies are at a balance concerning the amounts of
> input energy lifted and the amounts of potential energy output.


Hi, Paul.  It hadn't occurred to me that you were you.  But when you dropped
the capital letters, and started doing calculations out to 10 significant
digits, well, it had to be you.

You know, I can put you on to a program called uBasic, written by a math
professor named Yuji Kida, from Japan.  It works just like regular Basic,
except that you can set it to print out results to a couple of thousand
digits.

Anyway, look: No calculations needed.  Physics 101.  Add the second 1-inch
pipe section to the other end of your 20-foot pipe, so that the machine is
symmetrical.  Make sure it is mostly level.  Fill it with some impossible
theoretical fluid that is utterly frictionless, so that the liquid comes to
within a half-inch or so of the top of each 1-inch section.  Mark the levels
in each 1-inch segment, so that you know where the starting points are.

Okay?

Now, fill a thimble with the same fluid. Carefully lower that thimble into
the north 1-inch section of pipe, so that it is just barely above the
surface.  Now, dump it in.

At the moment the thimblefull of liquid starts to move, it is some small
distance above the rest level of the liquid.  As it starts pushing down on
the fluid in the pipe, its weight starts pushing at the mass of fluid in the
pipe.  Under that impetus, the fluid starts to move southward.

At the south end, fluid will start moving up past the rest level, which
reduces the net southward force on the fluid.

When the incoming thimblefull is halfway down past the rest level, the net
force on the fluid becomes zero.  All of the potential energy from the
thimblefull of fluid has been converted to the kinetic energy of the moving
fluid. Inertia keeps it moving, although a moment later, it will start
slowing down as the kinetic energy starts being converted to potential
energy at the south end.

It will continue to slow down until it comes to a halt.  At the moment it
comes to a halt, the level of fluid at the north end will be exactly at the
starting point, and at the south end there will be exactly one thimblefull
of water above the starting point.

The process will now reverse.

Congratulations.  We have invented the pendulum. Because the system is a
lossless thought experiment, it will oscillate forever.  There is a
technical term for this with fluids; it is called, "sloshing."

In the real world, hardly anything will happen.  Turbulence, friction, and
surface tension will all work against us.  Add a thimble of water into 20
feet of 2-inch PVC pipe, and basically nothing at all will happen.  Start
the water sloshing in some other fashion and it will come to a halt very
rapidly.

Look in any first-year physics text for the equations governing pendulums.
When the weight is as high as it goes, the potential energy is maximized and
the kinetic energy is zero.  When the weight is in the bottom middle of its
swing, the potential energy is at a minimum, and the kinetic energy is
maximized.  The difference in potential energies and the maximum kinetic
energy are identical.  This isn't just theoretical; it has been thoroughly
and repeatedly demonstrated in high school and college labs for centuries.

This latest thought of yours works just that.

so you vector off of your 1st mistake with another mistake.
you are proving that you only cut and paste articles.

the entire mass of the water in the 20 ft pipe will be moving for a
period of 60 seconds.
the 73.56 inches is the 1 gallon of water that would move
into the 20 foot pipe.

1 gallon = 231 cu inches
the cross sectional area of the 2" pipe is 3.14 sq inches.
231 / 3.14  = 73.56 cubic inches

> all you really know how to do is run the lip.
> correct spelling and be a complete pain.
> No, I maintain accuracy.

no you maintain an illusion.








--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> From:        damian_thornus, damian_thornus@y...
>
> >all you really know how to do is run the lip.
> >correct spelling and be a complete pain.
> No, I maintain accuracy.
>
> >the kinetic energy of the 27.20 lbs moving a distance of 73.56
inches
> >in one minute.
> Uh, no, all of the water didn't move 73.56" or even that close.
All of
> the water moved a total of only one to two inches depending on the
> sections you're counting.
>
> >I will need a while to figure out what you were doing.
> Potential energy was converted into kinetic when you thought they
existed
> separately.
>
> -Aut

yes you are the person who insits that the energy comparison of the
energy required to

1) lift the water or fluid into the pipe
and
2) the 4.5 meter slug of water that exits the end of the pipe

is a intelligent reason for pauls pipe system not to work.

how crude.

in the above you neglected the entire moving mass of fluid.
do you think that the mass of moving fluid can possibly not
contain an amount of energy given it has motion?

now you have sudgested we use a thimble full of a mysterious fluid.
and then you sudgest we stop the fluid from moving which naturaly
would remove all of its kinetic energy.

why didnt you break it down to using atomic particles.
you could have reduced the amount of output energy this way
and it might fit into your illusion much better.

your use of a pendulum is offtrack here.
1) this pipe flow system does not reverse the fluids direction.

the fluid flows due to a differential pressure.
that of the height of the water inlet and the height of the water
outlet.

the possible differential pressure used in my overunity example
would give only a collumb pressure of 1" or 0.0361 psi

however this extremely small pressure will move the entire mass of
fluid inside the pipe.

that is just simple hydraulics.

mr mohamed has sudgested that in pauls pipe system the energy
required to pump water from the end of the pipe and then up to the
height where water is placed in the upstand or verticle pipe
might be more than the energy required to just lift water into the
upstand or verticle pipe.

you being a extremist should understand this.

by the way do you have bootstraps.








--- In free_energy@y..., "Bob Dubner" <rdubner@c...> wrote:
> >
> > >So, you are putting in more potential energy than you are getting
> > > out.
> >
> > Im glad you pointed that out.
> > I would be lifting 1 gallon of water 1" higher.
> >
> > the remaining energies are at a balance concerning the amounts of
> > input energy lifted and the amounts of potential energy output.
>
>
> Hi, Paul.  It hadn't occurred to me that you were you.  But when
you dropped
> the capital letters, and started doing calculations out to 10
significant
> digits, well, it had to be you.
>
> You know, I can put you on to a program called uBasic, written by a
math
> professor named Yuji Kida, from Japan.  It works just like regular
Basic,
> except that you can set it to print out results to a couple of
thousand
> digits.
>
> Anyway, look: No calculations needed.  Physics 101.  Add the second
1-inch
> pipe section to the other end of your 20-foot pipe, so that the
machine is
> symmetrical.  Make sure it is mostly level.  Fill it with some
impossible
> theoretical fluid that is utterly frictionless, so that the liquid
comes to
> within a half-inch or so of the top of each 1-inch section.  Mark
the levels
> in each 1-inch segment, so that you know where the starting points
are.
>
> Okay?
>
> Now, fill a thimble with the same fluid. Carefully lower that
thimble into
> the north 1-inch section of pipe, so that it is just barely above
the
> surface.  Now, dump it in.
>
> At the moment the thimblefull of liquid starts to move, it is some
small
> distance above the rest level of the liquid.  As it starts pushing
down on
> the fluid in the pipe, its weight starts pushing at the mass of
fluid in the
> pipe.  Under that impetus, the fluid starts to move southward.
>
> At the south end, fluid will start moving up past the rest level,
which
> reduces the net southward force on the fluid.
>
> When the incoming thimblefull is halfway down past the rest level,
the net
> force on the fluid becomes zero.  All of the potential energy from
the
> thimblefull of fluid has been converted to the kinetic energy of
the moving
> fluid. Inertia keeps it moving, although a moment later, it will
start
> slowing down as the kinetic energy starts being converted to
potential
> energy at the south end.
>
> It will continue to slow down until it comes to a halt.  At the
moment it
> comes to a halt, the level of fluid at the north end will be
exactly at the
> starting point, and at the south end there will be exactly one
thimblefull
> of water above the starting point.
>
> The process will now reverse.
>
> Congratulations.  We have invented the pendulum. Because the system
is a
> lossless thought experiment, it will oscillate forever.  There is a
> technical term for this with fluids; it is called, "sloshing."
>
> In the real world, hardly anything will happen.  Turbulence,
friction, and
> surface tension will all work against us.  Add a thimble of water
into 20
> feet of 2-inch PVC pipe, and basically nothing at all will happen.
Start
> the water sloshing in some other fashion and it will come to a halt
very
> rapidly.
>
> Look in any first-year physics text for the equations governing
pendulums.
> When the weight is as high as it goes, the potential energy is
maximized and
> the kinetic energy is zero.  When the weight is in the bottom
middle of its
> swing, the potential energy is at a minimum, and the kinetic energy
is
> maximized.  The difference in potential energies and the maximum
kinetic
> energy are identical.  This isn't just theoretical; it has been
thoroughly
> and repeatedly demonstrated in high school and college labs for
centuries.
>
> This latest thought of yours works just that.

From:        damian_thornus, damian_thornus@...

>so you vector off of your 1st mistake with another mistake.
-of
I've made no mistakes.  Don't try to make yourself appear smart by using
mathematical terms, especially those you were given, in expressions.
>you are proving that you only cut and paste articles.
You wrote no article.

>the entire mass of the water in the 20 ft pipe will be moving for a
>period of 60 seconds.
>the 73.56 inches is the 1 gallon of water that would move
>into the 20 foot pipe.
The water still only needs to move one inch to flow out of the fitting.
Why do you have it anyway?

>no you maintain an illusion.
Reality is your illusion.

-Aut

> I've made no mistakes.
you stated in your first mistake that I neglected to divide the mass
by 2
when you divided the already divided mass and then gave your responce.
possibly in order to reduce the amount of output energy.


> The water still only needs to move one inch to flow out of the
fitting.

no it only needs to move a fraction of an inch to flow out of the
pipe.
flowing water is not bound by any type of linear measurement.
it is not solid.
you would not need to move a entire inch of the water before any
would come out of the other end.

> Why do you have it anyway?
why do I have what? the water?

> Reality is your illusion.
no reality is my reality.

my illusion would be to try to get a reasonably acurate
reply from someone in this group.

I have supplied a fantastically simple overunity example or model
and there has not been one reply that bears the slightest concern
nor the slightest difficulty in the return of a rebuttal.

why is this?






--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> From:        damian_thornus, damian_thornus@y...
>
> >so you vector off of your 1st mistake with another mistake.
> -of
> I've made no mistakes.  Don't try to make yourself appear smart by
using
> mathematical terms, especially those you were given, in expressions.
> >you are proving that you only cut and paste articles.
> You wrote no article.
>
> >the entire mass of the water in the 20 ft pipe will be moving for
a
> >period of 60 seconds.
> >the 73.56 inches is the 1 gallon of water that would move
> >into the 20 foot pipe.
> The water still only needs to move one inch to flow out of the
fitting.
> Why do you have it anyway?
>
> >no you maintain an illusion.
> Reality is your illusion.
>
> -Aut

>
> yes you are the person who insits that the energy comparison of the
> energy required to
>
> 1) lift the water or fluid into the pipe
> and
> 2) the 4.5 meter slug of water that exits the end of the pipe
>
> is a intelligent reason for pauls pipe system not to work.
>
> how crude.

Paul-Damian -- you are really quite remarkable.

You set up a pipe, with two elbows, open ends up, with one additional inch
of pipe attached to one of them.  At least, that was my impression.

To make a point, I changed the pipe slightly, by adding a second one-inch
segment to the other elbow.  My point was to show that there is no excess
energy.  Potential energy gets converted to kinetic energy, and then back to
potential energy.  The principles involved are identical to a pendulum
swinging back and forth.

Sure, what I described doesn't match what you described.  But that doesn't
automatically make it wrong.  You are saying I am wrong because I didn't say
what you said, but I am afraid that's less than totally convincing.

It's true that I didn't say what you said.  That doesn't make me wrong, you
know; if I am wrong, I must've made a mistake.  Point to it.

As for your description, I used the slightly altered design to try to lead
you to your error.  Even without the extra one-inch segment on the 90-degree
elbow at the south end, your "overunity" demonstration is neglecting the
fact that in order for water to leave by the elbow it has to be lifted.
Your description assumes that no work is done lifting water out of that
hole -- and that's your error.

From:        damian_thornus, damian_thornus@...

>you stated in your first mistake that I neglected to divide the mass by 2
>when you divided the already divided mass and then gave your responce.
It wasn't already divided because I checked it.  Stop misspelling
response.  You multiplied and squared, but you didn't divide.

>no it only needs to move a fraction of an inch to flow out of the pipe.
The inch is counted from the inch fitting.  The poured water moves down
an inch to meet the fitting's capacity, which is balanced by the
outflowing water having the energy to move about an inch in the same pipe.

>why do I have what? the water?
the fitting

>my illusion would be to try to get a reasonably acurate
accurate
>reply from someone in this group.
Most replies from people other than you are very accurate.

>I have supplied a fantastically simple overunity example or model
No, you haven't.
>and there has not been one reply that bears the slightest concern
There is hardly any concern with underunity models.
>nor the slightest difficulty in the return of a rebuttal.
Any rebuttal would be very easy, obviously.

-Aut

*******
You set up a pipe, with two elbows, open ends up, with one additional
inch
of pipe attached to one of them. At least, that was my impression.

To make a point, I changed the pipe slightly, by adding a second one-
inch
segment to the other elbow.
*******
your changing the pipe was your error.
*******
My point was to show that there is no excess
energy.
*******
there could be no excess energy in your example.
*******
Potential energy gets converted to kinetic energy, and then back to
potential energy.
*******
at least you are in agreement that there is kinetic energy in the
water as it flows to the opposite end.

here is how it works.
1) you supply work to lift the water.
2) this water then has potential energy.
3) you pour the water in the pipe.
4) the water flows/moves to the opposite end.
""""""""""""""""""""
5) as this water moves it ( has ) kinetic energy.
""""""""""""""""""""
6) because the water levels in the two ( elbows ) are equal
    the additional water pressure due to the additional
    height of the water in the 1" section pushes all the
    other water down the first elbow and through the pipe
    and up the second elbow.

7) as the water comes out of the second elbow it has
    potential energy in the exact same amount as the energy
    required to lift that volume of water to the exact same height
    of the first elbow
8) although there is a additional amount of energy required to
    lift the water the additional 1" in height the kinetic energy
    of the entire mass of fluid is greater than that additional
    energy.

it may not be in this example because the pipe is so small and so
short however this still proves that there is a amount of energy
contained in a flow of fluid.


***********
The principles involved are identical to a pendulum
swinging back and forth.
***********
they are in your example not in mine.

here is where ( you ) state that only the water comming out of the
pipe is where any energy is availiable.

message 2896


Okay, then, Paul -- educate me some more.  I accept your description,
which I have shown involves 519 kilowatts of gravitational potential
work due to water falling through 10 meters of 48-inch pipe, pushing
water horizontally through a 1000-meter stretch of 48-inch pipe,
making 54 kilowatts of moving water available at the exit end.  This
system moves energy from the upstand to the exit end with only 90
percent losses.

your resilient nature when inteligence is in need is quite convincing
to some probably but not me.

above you have agreed that there is water that is flowing through a
1000 meter pipe.
this is 2.5 million pounds would you like to confirm your belief
that there can not be energy in a moving mass this size.
you claim 90% losses.

> the flow velocity can be found by dividing the
> discharge volume by the area of the pipes cross section.
> 5.2885868026 / 1.1674540318 = 4.5300171643m/sec

ke = 1/2mv^2
this means 1/2 of the mass times the velocity of the mass squared
2,500,000 lbs = 1,133,975 kg
1/2 of 1,133,975 kg = 566987.5 kg
the mass has a velocity of 4.5300171643 m/sec

the velocity of the mass squared = 20.5210555089

ke = 11635181.96 J

I suppose that your obvious neglect of this energy was a mistake
as you appear to be a inteligent person.

in physics there are two types of energy
potential and kinetic.
you only seem to have knowledge of potential energy.

otherwise you would have noticed the massive amounts of
energy in the moving mass.
******************
here is proof and you wrote it
you believe it when it is in your favor
******************
  Gravitational potential energy is mass times g time height; so the
change in gravitational energy each second is 11,700 kg * 9.8 m/sec^2
* 4.53 meters.  That comes out to be 519,000 joules per second: 519
kilowatts.


please tell me how 519,000 Joules can produce 11,635,181.96 Joules.


<shrug a little harder>




















--- In free_energy@y..., "Bob Dubner" <rdubner@c...> wrote:
> >
> > yes you are the person who insits that the energy comparison of
the
> > energy required to
> >
> > 1) lift the water or fluid into the pipe
> > and
> > 2) the 4.5 meter slug of water that exits the end of the pipe
> >
> > is a intelligent reason for pauls pipe system not to work.
> >
> > how crude.
>
> Paul-Damian -- you are really quite remarkable.
>
> You set up a pipe, with two elbows, open ends up, with one
additional inch
> of pipe attached to one of them.  At least, that was my impression.
>
> To make a point, I changed the pipe slightly, by adding a second
one-inch
> segment to the other elbow.  My point was to show that there is no
excess
> energy.  Potential energy gets converted to kinetic energy, and
then back to
> potential energy.  The principles involved are identical to a
pendulum
> swinging back and forth.
>
> Sure, what I described doesn't match what you described.  But that
doesn't
> automatically make it wrong.  You are saying I am wrong because I
didn't say
> what you said, but I am afraid that's less than totally convincing.
>
> It's true that I didn't say what you said.  That doesn't make me
wrong, you
> know; if I am wrong, I must've made a mistake.  Point to it.
>
> As for your description, I used the slightly altered design to try
to lead
> you to your error.  Even without the extra one-inch segment on the
90-degree
> elbow at the south end, your "overunity" demonstration is
neglecting the
> fact that in order for water to leave by the elbow it has to be
lifted.
> Your description assumes that no work is done lifting water out of
that
> hole -- and that's your error.

YOUR WRONG YOUNG LADY
From: damian_thornus, damian_thornus@y...

>v^2 = 0.0311404 m/sec^2 = .0009697245122m/sec
m^2/s^2

>ke = 6.165 kg X .0009697245122 m/sec = .005978351618 J
m^2/s^2
******You didn't divide by two. It's .002989... J*******

the mass is 27.20 lb = 12.33 kg
1/2 of 12.33 kg = 6.165 kg

ke = 6.165 kg X .0009697245122 m/sec = .005978351618 J

if you re-divide it like you did you get

ke = 3.0825 kg X .0009697245122 m/sec = .002989...J
*********You didn't divide by two. It's .002989... J*******








--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> From:        damian_thornus, damian_thornus@y...
>
> >you stated in your first mistake that I neglected to divide the
mass by 2
> >when you divided the already divided mass and then gave your
responce.
> It wasn't already divided because I checked it.  Stop misspelling
> response.  You multiplied and squared, but you didn't divide.
>
> >no it only needs to move a fraction of an inch to flow out of the
pipe.
> The inch is counted from the inch fitting.  The poured water moves
down
> an inch to meet the fitting's capacity, which is balanced by the
> outflowing water having the energy to move about an inch in the
same pipe.
>
> >why do I have what? the water?
> the fitting
>
> >my illusion would be to try to get a reasonably acurate
> accurate
> >reply from someone in this group.
> Most replies from people other than you are very accurate.
>
> >I have supplied a fantastically simple overunity example or model
> No, you haven't.
> >and there has not been one reply that bears the slightest concern
> There is hardly any concern with underunity models.
> >nor the slightest difficulty in the return of a rebuttal.
> Any rebuttal would be very easy, obviously.
>
> -Aut

From:        damian_thornus, damian_thornus@...

>YOUR WRONG YOUNG LADY
The word is "you're", illiterate.

>the mass is 27.20 lb = 12.33 kg
>1/2 of 12.33 kg = 6.165 kg

>ke = 6.165 kg X .0009697245122 m/sec = .005978351618 J
Your calculations were so extraneous and out-of-order that you didn't
leave the 1/2 in the KE expression.  You divided by two before the
expression when the "6.165 kg" figure had no application.  You should
have written this:

KE = .5(12.33 kg)(.009697 m^2/s^2)

starting over:

>if I lift 3.78 kg to a height of 1"
>1" = 0.0254 m
>W=37.044 N X 0.0254 m = 0.9409176 J

>if this is carried out in a time frame of 60 seconds.
>and .000261366 J/sec
This figure doesn't matter.

>the distance that the mass of water in the pipe would move would be.
>231 cu in / 3.14 area = 73.56 inches/minute
Your minute was how long it took to lift the gallon, which didn't matter.
  You didn't say it would also take a minute to pour the water in.

>at any second durring the one minute the kinetic energy of the
>moving mass of water will be .005978351618 J
Multiply that by 60 to get the total energy, .3587 J.  But you know what
stupid thing you did?  You confused the minute it took to lift the gallon
of water (what's wrong with your arm?) with the minute it took to
displace the water in the pipe.  You DO NOT choose the time that the
water displaces that water in the pipe; the time, which comes from
velocity from gravity's acceleration from the pipe and fitting's height,
is fixed.  Because you chose a minute, and because all of the water
couldn't be poured at once, you included more mass than was available in
the KE.

You also forgot that the velocity from the water falling 1" would also be
fixed:

v = at
s = s_0 + vt + .5at^2
0 = .0254 m + .5gt^2
rt(-.0508 m/g) = t
v = g rt(-.0508 m/g)
v^2 = (-.0508 m)g = .4978 m^2/s^2

So the KE of the entire gallon falling is .9409 J, the same as your
figure for lifting, not .005978 J.  Because the elbows are also raised,
the energy is more than that.  The same .9409 J will be transferred to
the water in the pipe, that will displace the same amount of water as the
gallon.

>the kinetic energy of the 27.20 lbs moving a distance of 73.56 inches
>in one minute.
That's 1.868 m, and that divided by v (.7055 m/s) is 2.648 s, not one
minute.  However, the time assumes that there is no additional mass to
accelerate.  The kinetic energy goes into all of the water now:

KE = .9409 J = .5(12.33 + 3.78 kg)v^2 => v = .3418 m/s => t = 5.465 s

You don't choose the velocity or time, you fool.  They're average, btw.
Eric says that I'm not allowed to say that your head should be smashed
against a wall because it could start a flamewar.  That's not fair, is
it?  I mean, only one person would be flamed...

-Aut

Bob,
I like the way you explain issues that arise on this
web site.  For that reason, I am directing my problem
to you(although it is open to all) as follows:-

"Please give me an expression of the time (t) it takes
for water flowing through a pipeline to reach steady
state flow condition, when the initial velocity inside
the pipeline is 0 m/s - do it in metric."

Assume the following:-
1)Length of pipeline = L meters
2)Inside Diameter of pipeline = D meters
3)Head difference = H meters ( to remain constant)
4)Gravity acceleration = g m/s/s
5)Roughness coefficient(Hazen William) of pipeline = c
6)Give any other assumptions

Also, Please give me the expression for the amount of
water that is discharged before the steady state is
reached.

I am looking at a situation where water from a tank is
discharged through a pipeline of Length L meters and
diameter D meters, to a point that is H meters lower
than the level of water in the tank.  I am assuming
that at time t=0, the pipeline is full of water and
the palm of my hand that is blocking the pipe outlet
is experiencing pressure equivalent to H meters of
water.  Then, I instantaneously remove the palm of my
hand from the pipe outlet and the following happen:-

1) The mass of water inside the entire pipeline is
accelerated from rest to a terminal velocity = V m/s

2) Water then continues to flow steadily at velocity =
V m/s as long as the level of water inside the tank
remains the unchanged.

What I request you is to give the expression for the
time taken for the water to complete the acceleration
phase, before steady state conditions are reached, and
the amount of water that will have been discharged in
the process.

The reason I am requesting you this is because I am
working on a an idea originally floated by Paul.
However, Paul decided to introduce turbines in the
pipeline and he was ruthlessly dealt with.  Therefore,
before I spill my beans for scrutiny, I would like you
to assist as requested above.

The overunity demonstration below that you have very
ably answered is very close to my Idear.

Thanking you in advance.

Joseph






--- Bob Dubner <rdubner@...> wrote:
<HR>
<html><body>


<tt>
><BR>
> materials needed.<BR>
><BR>
> (1) 20 ft lenght of 2" pvc pipe.<BR>
> (2) 2" 90 degree elbows<BR>
> (1) 1 gallon bucket.<BR>
> (1) 1" section of 2" pvc pipe<BR>
> attach the two 90 degree elbows to the 20 ft pipe
so that the hole is<BR>
> facing up.<BR>
> attach the 1" section to one of the
elbows.<BR>
><BR>
<BR>
Your analysis of your supposed over-unity demonstrator
is, of course,<BR>
inaccurate.<BR>
<BR>
When all relevant factors are taken into account, a
more complete analysis<BR>
shows that there is a net loss of useful energy, with
some being "lost", as<BR>
usual, to entropy.<BR>
<BR>
You see, you made a mistake when you say "notice
that same amount of energy<BR>
comes out at the other end in the form of potential
energy."  'Tain't so.<BR>
<BR>
On the feed side, you have that 1" standpipe,
which you put in to act as a<BR>
funnel -- it keeps the water you are trying to feed in
from the bucket from<BR>
just sloshing over the sides of the already full
elbow.<BR>
<BR>
But on the exit side, you don't have a matching
1" standpipe.<BR>
<BR>
Thus, you have to lift the water on the feed side one
inch higher (to get it<BR>
into the funnel) than the system will lift it on the
exit side twenty feet<BR>
away.  So, you are putting in more potential
energy than you are getting<BR>
out.<BR>
<BR>
<shrug> Sorry.<BR>
<BR>
</tt>


<br>
<tt>
To drop of the list, send email to:<BR>
free_energy-unsubscribe@egroups.com </tt>
<br>

<br>
<tt>Your use of Yahoo! Groups is subject to the <a
href="http://docs.yahoo.com/info/terms/">Yahoo! Terms
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</body></html>


__________________________________________________
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>
> "Please give me an expression of the time (t) it takes
> for water flowing through a pipeline to reach steady
> state flow condition, when the initial velocity inside
> the pipeline is 0 m/s - do it in metric."
>
> Assume the following:-
> 1)Length of pipeline = L meters
> 2)Inside Diameter of pipeline = D meters
> 3)Head difference = H meters ( to remain constant)
> 4)Gravity acceleration = g m/s/s
> 5)Roughness coefficient(Hazen William) of pipeline = c
> 6)Give any other assumptions

Joseph -- thanks for the compliment.

I have never learned how to do fluid mechanics calculations, so I can't
answer your question without doing some research.  I did poke around on the
Web a little, and learned enough to know that I don't think you've
constrained the problem enough, although I may be wrong.  I think you need
to put in the pipe's slope as well.

The Web site at http://www.lmnoeng.com seems to have a tremendous amount of
useful information; in particular, they have an on-line calculator that
gives steady-state flow and velocity.
http://www.lmnoeng.com/hazenwilliams.htm

If I am reading this correctly, in the Hazen-Williams model the friction
coefficient regarded as constant, so a first cut at calculating the
acceleration time would involve setting up a scenario, finding the
steady-state velocity, assuming that there is a total resistive force equal
to the head pressure, and then work that back into an acceleration curve,
assuming that the resistive force is proportional to the velocity at any
moment.

A better solution, it appears, would be to use the Darcy-Weisbach friction
loss equation and actually do an integration on a simulated pipe system.
The equation looks amenable to a formal mathemetical analysis, or else it
wouldn't be hard to write a numerical integrator.

<grin> But having dangled all that in front of you, I am afraid I can't
afford right now to educate myself into a ersatz "expert" to help you out.
I could do a bunch of calculations, sure, but this kind of thing requires
that first you design stuff and then build it and test it to make sure that
both your understanding and calculations seem reasonable.  And I am not
equipped to go out to the back yard and hook pipes together and make
mudpies.

But I will do this, if you don't know the math to handle it yourself. Create
the pipe of your choice.  Use http://www.lmnoeng.com/hazenwilliams.htm to
figure out the final velocity Vf and the head loss hf.  Give us those
numbers (along with the other parameters for pipe diameter and so on) and
I'll then show you how to come up with the acceleration curve, which will
give you the other numbers you need.

you could obtain a copy of cranes technical paper 410
http://www.vervante.com/crane/
>>>>>>>>>>>>>
Crane Technical Paper No. 410 (TP-410) is the quintessential guide to
understanding the flow of fluid through valves, pipe and fittings,
enabling you to select the correct equipment for your piping system.
Originally developed in 1942, the latest edition of Crane TP-410
serves as an indispensable technical resource for specifying
engineers, designers and engineering students. TP-410 is authored by
Crane Valve Group (CVG), one of the world's leading suppliers of
valve products and services. Also available in Metric Edition
>>>>>>>>>>>>>

Price: $33.00
Pages: 132
Date Published: 06/1998
ISBN: 1400527120

around page 8 to 21 I believe.
I can upload a page or two for reference material.

from your post I take it that you are implementing the water hammer
effect of the kinetic energy of the moving mass.

there is an interesting shock wave that will transverse your pipe
in this situation.
which will lead to materials failures.
the shock wave is the result of the kinetic energy of the mass
being absorbed instantly by the pipe.
which places a large amount of stress on the pipe.


and I believe that there is a slight bit of compression that occurs
upon impact.
thus the shock wave.


perhaps bob could use a copy of this paper also I understand now why
he has such a hard time visualizing and comprehending fluids.
he has never studied them.











--- In free_energy@y..., Joseph Kanyugi <jkanyugi2000@y...> wrote:
> Bob,
> I like the way you explain issues that arise on this
> web site.  For that reason, I am directing my problem
> to you(although it is open to all) as follows:-
>
> "Please give me an expression of the time (t) it takes
> for water flowing through a pipeline to reach steady
> state flow condition, when the initial velocity inside
> the pipeline is 0 m/s - do it in metric."
>
> Assume the following:-
> 1)Length of pipeline = L meters
> 2)Inside Diameter of pipeline = D meters
> 3)Head difference = H meters ( to remain constant)
> 4)Gravity acceleration = g m/s/s
> 5)Roughness coefficient(Hazen William) of pipeline = c
> 6)Give any other assumptions
>
> Also, Please give me the expression for the amount of
> water that is discharged before the steady state is
> reached.
>
> I am looking at a situation where water from a tank is
> discharged through a pipeline of Length L meters and
> diameter D meters, to a point that is H meters lower
> than the level of water in the tank.  I am assuming
> that at time t=0, the pipeline is full of water and
> the palm of my hand that is blocking the pipe outlet
> is experiencing pressure equivalent to H meters of
> water.  Then, I instantaneously remove the palm of my
> hand from the pipe outlet and the following happen:-
>
> 1) The mass of water inside the entire pipeline is
> accelerated from rest to a terminal velocity = V m/s
>
> 2) Water then continues to flow steadily at velocity =
> V m/s as long as the level of water inside the tank
> remains the unchanged.
>
> What I request you is to give the expression for the
> time taken for the water to complete the acceleration
> phase, before steady state conditions are reached, and
> the amount of water that will have been discharged in
> the process.
>
> The reason I am requesting you this is because I am
> working on a an idea originally floated by Paul.
> However, Paul decided to introduce turbines in the
> pipeline and he was ruthlessly dealt with.  Therefore,
> before I spill my beans for scrutiny, I would like you
> to assist as requested above.
>
> The overunity demonstration below that you have very
> ably answered is very close to my Idear.
>
> Thanking you in advance.
>
> Joseph
>
>
>
>
>
>
> --- Bob Dubner <rdubner@c...> wrote:
> <HR>
> <html><body>
>
>
> <tt>
> ><BR>
> > materials needed.<BR>
> ><BR>
> > (1) 20 ft lenght of 2" pvc pipe.<BR>
> > (2) 2" 90 degree elbows<BR>
> > (1) 1 gallon bucket.<BR>
> > (1) 1" section of 2" pvc pipe<BR>
> > attach the two 90 degree elbows to the 20 ft pipe
> so that the hole is<BR>
> > facing up.<BR>
> > attach the 1" section to one of the
> elbows.<BR>
> ><BR>
> <BR>
> Your analysis of your supposed over-unity demonstrator
> is, of course,<BR>
> inaccurate.<BR>
> <BR>
> When all relevant factors are taken into account, a
> more complete analysis<BR>
> shows that there is a net loss of useful energy, with
> some being "lost", as<BR>
> usual, to entropy.<BR>
> <BR>
> You see, you made a mistake when you say "notice
> that same amount of energy<BR>
> comes out at the other end in the form of potential
> energy."  'Tain't so.<BR>
> <BR>
> On the feed side, you have that 1" standpipe,
> which you put in to act as a<BR>
> funnel -- it keeps the water you are trying to feed in
> from the bucket from<BR>
> just sloshing over the sides of the already full
> elbow.<BR>
> <BR>
> But on the exit side, you don't have a matching
> 1" standpipe.<BR>
> <BR>
> Thus, you have to lift the water on the feed side one
> inch higher (to get it<BR>
> into the funnel) than the system will lift it on the
> exit side twenty feet<BR>
> away.  So, you are putting in more potential
> energy than you are getting<BR>
> out.<BR>
> <BR>
> <shrug> Sorry.<BR>
> <BR>
> </tt>
>
>
> <br>
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>
> <br>
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I also forgot to include the position of our solar system within our
galaxy.
please do not forget this as it is an important factor in your
world.


Your minute was how long it took to lift the gallon, which didn't
matter.
You didn't say it would also take a minute to pour the water in.

tick tock, tick tock, tick tock.
rinnnnnnnnnnnng.

time an interesting " word "





--- In free_energy@y..., Autymn D. C. <lysdexia@j...> wrote:
> From:        damian_thornus, damian_thornus@y...
>
> >YOUR WRONG YOUNG LADY
> The word is "you're", illiterate.
>
> >the mass is 27.20 lb = 12.33 kg
> >1/2 of 12.33 kg = 6.165 kg
>
> >ke = 6.165 kg X .0009697245122 m/sec = .005978351618 J
> Your calculations were so extraneous and out-of-order that you
didn't
> leave the 1/2 in the KE expression.  You divided by two before the
> expression when the "6.165 kg" figure had no application.  You
should
> have written this:
>
> KE = .5(12.33 kg)(.009697 m^2/s^2)
>
> starting over:
>
> >if I lift 3.78 kg to a height of 1"
> >1" = 0.0254 m
> >W=37.044 N X 0.0254 m = 0.9409176 J
>
> >if this is carried out in a time frame of 60 seconds.
> >and .000261366 J/sec
> This figure doesn't matter.
>
> >the distance that the mass of water in the pipe would move would
be.
> >231 cu in / 3.14 area = 73.56 inches/minute
> Your minute was how long it took to lift the gallon, which didn't
matter.
>  You didn't say it would also take a minute to pour the water in.
>
> >at any second durring the one minute the kinetic energy of the
> >moving mass of water will be .005978351618 J
> Multiply that by 60 to get the total energy, .3587 J.  But you know
what
> stupid thing you did?  You confused the minute it took to lift the
gallon
> of water (what's wrong with your arm?) with the minute it took to
> displace the water in the pipe.  You DO NOT choose the time that
the
> water displaces that water in the pipe; the time, which comes from
> velocity from gravity's acceleration from the pipe and fitting's
height,
> is fixed.  Because you chose a minute, and because all of the water
> couldn't be poured at once, you included more mass than was
available in
> the KE.
>
> You also forgot that the velocity from the water falling 1" would
also be
> fixed:
>
> v = at
> s = s_0 + vt + .5at^2
> 0 = .0254 m + .5gt^2
> rt(-.0508 m/g) = t
> v = g rt(-.0508 m/g)
> v^2 = (-.0508 m)g = .4978 m^2/s^2
>
> So the KE of the entire gallon falling is .9409 J, the same as your
> figure for lifting, not .005978 J.  Because the elbows are also
raised,
> the energy is more than that.  The same .9409 J will be transferred
to
> the water in the pipe, that will displace the same amount of water
as the
> gallon.
>
> >the kinetic energy of the 27.20 lbs moving a distance of 73.56
inches
> >in one minute.
> That's 1.868 m, and that divided by v (.7055 m/s) is 2.648 s, not
one
> minute.  However, the time assumes that there is no additional mass
to
> accelerate.  The kinetic energy goes into all of the water now:
>
> KE = .9409 J = .5(12.33 + 3.78 kg)v^2 => v = .3418 m/s => t = 5.465
s
>
> You don't choose the velocity or time, you fool.  They're average,
btw.
> Eric says that I'm not allowed to say that your head should be
smashed
> against a wall because it could start a flamewar.  That's not fair,
is
> it?  I mean, only one person would be flamed...
>
> -Aut

>
> perhaps bob could use a copy of this paper also I understand now why
> he has such a hard time visualizing and comprehending fluids.
> he has never studied them.
>

Paul -- it's time to give it a rest.

Your statement is, in fact, not true, which means you are batting a
thousand.  I have studied fluids, but on a very elementary level.  An
accurate theoretical answer to Joseph's question would require a non-linear
analysis, and very possibly a numerical simulation.  This is something that
I could do, but doing a good job of it would take several hours of study,
and I can't afford to do that today.

The level on which you've been attempting to function on doesn't require
anywhere near that level of sophistication.  And, so far, you've managed to
generate so many incorrect 10-digit numbers that the mind boggles.  You're
so far off base, you're not even wrong.

Fluids are tricky.  Handled accurately, there are a lot of non-linear, and
as you've amply demonstrated, non-intuitive aspects.  You just don't have a
good enough grasp of the background to apply the principles you've been
digging out.

And I think I can prove it. I've offered you a couple of challenges before,
in a dual attempt to show you that you don't yet have the tools necessary,
and to show you the path you need to follow to get a handle on those tools.

Here's another of those challenges:  A simple frictionless ramp.  The
vertical height of the ramp is one meter.  The base of the ramp is ten
meters.  At the top of the ramp is a one-kilogram mass.  Assume the ideal
case: no friction, no wind resistance.  Use the value 9.80 meters per
second-squared as the gravitational constant.

Release the mass, letting it slide freely down the frictionless ramp.

Calculate, to three significant digits:

1) The change in gravitational potential energy as the mass slides from the
top of the ramp down to the bottom.

2) The time in seconds for the mass to slide to the bottom of the ramp.

3) The final velocity of the mass as it exits the ramp.

4) The kinetic energy of the mass as it exits the ramp.

Hit me with some numbers.

I am not trying to trick you, or embarrass you.  But I believe that you
don't have the training or tools to handle even this very simple analysis.
I think that if you attempt to do those calculations, you will come up with
the wrong answers.  If that's the case, then you simply have no business
whatsoever trying to analyze fluid systems, which are incredibly more
complex than a weight sliding down an ideal frictionless ramp.

Damian,

The device you described sounds easy to build.  Have you tried
building it?  If so, did you see any evidence of the over-unity
effect?

Leo C.


--- In free_energy@y..., "damian_thornus" <damian_thornus@y...> wrote:
> materials needed.
>
> (1) 20 ft lenght of 2" pvc pipe.
> (2) 2" 90 degree elbows
> (1) 1 gallon bucket.
> (1) 1" section of 2" pvc pipe
> attach the two 90 degree elbows to the 20 ft pipe so that the hole
is
> facing up.
> attach the 1" section to one of the elbows.
>
> take your garden hose and fill the pipe with water.
>
> fill your 1 gallon bucket with water.
>
> calculate the mass of the water inside the pipe.
>
> now pour the water in your bucket into the 1" section
> pour it verry slowly.
>
> determine the amount of energy required to lift the water into the
> pipe.
>
> notice that same amount of energy comes out at the other end in the
> form of potential energy.
>
>
> you have expended 1 unit of energy.
> and have had a return of 1 unit of energy.

&&&&&&&&&&
Damian, actually, you don't get all of the potential energy back.
You initially had to lift the entire gallon of water to a height at
least equal to the top of the input elbow.  But after the pipe is
full, the vast majority of the water ends up at a level below the
input elbow (in the horizontal 20' pipe, near the ground).
&&&&&&&&&&

>
> meanwhile you have moved the mass of water between the elbows.
>
> calculate the velocity of the moving mass of water.
> find the ke of that moving mass.
>
> you have a overunity device.

&&&&&&&&&&
It would be over-unity only if the kinetic energy released exceeded
the gravitational potential energy lost by the drop in height from
the top of the elbow to the horizontal 20' pipe.  Classical physics
says that the kinetic energy released could not exceed the lost
potential energy.  If you don't accept classical physics, I guess the
next step is to build the device and try it out.

You have everything to gain by testing your idea, and nothing to
lose.  Don't worry about the Men-In-Black.  After all, what would be
the point of them going after you, since the secret is already out,
and anyone could build the device.  If it works, you would be a hero
and famous.  And even if you can't patent it, you could still scale
it up and generate yopur own electricity.

Leo C.
&&&&&&&&&&


>
> if you put 1 unit of energy in.
> and get 1 unit of energy out.
>
> that is a equalibrium.
>
> if there is (any) movement of water in between the elbows
> that is overunity.