## RE: [free_energy] my exercise and yours

Expand Messages
• ... This message came from kloskie@hotmail.com. I am guessing from the style and the tone that this message must also be from Paul Baughn, although I was
Message 1 of 23 , Mar 1, 2002
> If you drop a object from a 100 meter height,
> given there is no resistance.
>
> how long will it take to hit
> the ground?
>
> How fast will it be going when it hits?
>

This message came from kloskie@.... I am guessing from the style
and the tone that this message must also be from Paul Baughn, although I was
confused for a bit. Paul, am I right? Are you using two or three different
e-mail accounts?

I haven't been asking these questions in order to embarrass you.

I have been trying to demonstrate where your theoretical knowledge is shaky.
I have repeatedly stated that your fluid flow calculations are in error; I
have been trying to go back to first principles to demonstrate to you *why*
they are in error.

On this level, the conversation is not about me or you. You called me a
"dimwit". That's an opinion, and you are welcome to it. Your opinion of my
intelligence doesn't make one bit of difference to me, nor does it make a
single shard of difference to that dropped object.

You calculations about that 100 meter drop are, in point of incontrovertable
fact, wrong.

When an object at rest is suddenly subjected to a constant acceleration a,
its velocity after time T will be

V = a T

The distance it will travel is

d = 1/2 a T^2

This is elementary stuff; with constant acceleration the velocity is the
time integral of acceleration, and the distance is the time integral of
velocity.

Paul -- for the third time: Look in the first couple of chapters of your
physics text. It will be there. Guaranteed. Failing that, ten seconds of
searching the web gave me this: http://physics.webplasma.com/physics04.html

The same equations I've given are there; if you look at them, keep in mind
that for simplicity I've intentionally been ignoring Vo and Xo (the initial
velocity and position) by setting them to zero.

Applying them here, where a falling object is subjected to a constant
acceleration of 9.8 m/s^2 gives a falling time of 4.52 seconds, and a final
velocity of 44.27 meters per second.

It appears that you attempted to apply those equations, and made the
following error:

d = 1/2 a T^2 transforming to T = sqrt(2d/a)

You somehow have decided that T = d/a.

This is just plain wrong. d is in meters; a is m/s^2. Dividing d by a
gives you dimensions of sec^2. You can't give a time in seconds squared; it
has to come out to seconds. Any calculation which does so *has* to be
wrong.

I have been at pains to explain what the principles and equations are when I
calculate a numerical result. You have not returned the courtesy. I ask
again: Please justify, with theoretical analysis and/or references that I
can check, why you think an object dropped from 100 meters in a 9.8 m/s^2
gravitational field, will be moving at 100 m/s^2 [sic] after 10.2 seconds.

And you simply *must* be open to the possibility that you are wrong.

In fact, we need to thrash out once and for all, why do you think you can
call a velocity "100 m/sec^2" Velocity is *defined* as a change in distance
divided by a change in time;
• ... My answer would go as follows: the terminal velocity (v) is given by: v = sqaureroot( 2 * g * h ) = squareroot ( 2 * 9.8 m/s^2 * 100 m ) = squareroot (
Message 2 of 23 , Mar 1, 2002
--- In free_energy@y..., "kloskie" <kloskie@h...> wrote:
> If you drop a object from a 100 meter height,
> given there is no resistance.
>
> how long will it take to hit
> the ground?
>
> How fast will it be going when it hits?

My answer would go as follows:

the terminal velocity (v) is given by:

v = sqaureroot( 2 * g * h )
= squareroot ( 2 * 9.8 m/s^2 * 100 m )
= squareroot ( 1960 m^2/s^2 )
= 44.3 m/s

(note that the units of velocity are m/s, not m/s^2)

the time required to hit the ground is calculated by dividing the
final velocity by the rate of acceleration caused by Erath's gravity:

t = v/g
= (44.3 m/s) / (9.8 m/s^2)
= 4.52 s

Someone please correct me if I did this incorrectly.

>
> *******now heres yours************
>
>
> if you drop a object.
> and it takes 22.54 seconds to hit the ground.
>
> How fast will it be going when it hits the ground?
> how high was it dropped from.
>
> I would have included my work but Im not sure that
> you are not trying to get people to do your homework.
>

Oooh, Oooh, let me take this one too. This is an easy problem. If
we assume that the object is falling due to the Earth's gravitational
field, then the object is accelerating at 9.8 m/s^2. Therefore
(assuming that the object had zero velocity before it began
accelerating, and assuming negligible air resisitance), the object's
terminal velocity would equal the amount of time it accelerated
multiplied by the rate of acceleration:

v = t * g
= 22.54 s * 9.8 m/s^2
= 2201 m/s

Is this right?

Leo
• you ever go fishing bob. sometimes when you use rotten bait your catch will be greater than when you use fresh bait. it depends on the fish. I intentionally
Message 3 of 23 , Mar 1, 2002
you ever go fishing bob.
sometimes when you use rotten bait your catch will be greater
than when you use fresh bait.
it depends on the fish.

I intentionally gave you wrong answers to see if I could get you to
admit to the correctness or validity of my proposed system.

this mission was successful.

this clip is in reference to bobs post # 2621
here he gives a problem and then he gives his solution to
the problem.
the problem is a mere mass vs mass problem.
the correct answer will provide the distance that
mass 1 will pull mass 2.

it is a simple problem.
note: he is using "frictionless wheels"
**********************************
note also:
water flowing through a smooth wall plastic pipe
is met with a resistance of only .002 which can be
expressed as a percentage of the gravity that causes
resistance in the form of friction.

ie..a
1000 kg mass of fluid
vs a mass equivalent to
.002 percent of the 1000 kg mass.
***************************************
to continue with bobs exercise.

and the mass has a velocity of 1 m/s

Here's the question: At the exact moment that the rising weight comes
to a halt, how high will it be over the ground? How long will it take
to get there?

The height of the 1kg mass at the moment the masses stop moving will
be 51 meters.
The time it will take for the 1kg mass to reach that height will be
102 seconds.
*** this is correct ***
this proves that bob knows physics.

******************
so bob is saying here that a 1000kg mass that has a velocity of 1
meter per second can lift a 1kg mass to a height of 51meters.
even though the 1000kg mass is only traveling at a velocity
of 1 meter per second when it starts to pull the 1 kg mass.
and it will take 102 seconds to lift it 51meters.

******************

although this exercise can in no way show a comparison to the system
in which we are discussing.
mainly because you cannot tie ropes to water.
and in the system we are discussing the fluid is in constant
circulation.
and circulating through different sized pipes.
and it has two different constant velocities.

here he wants to use an example that describes a
stationary mass and a mass with velocity.
and the two masses can only travel at the same velocity.
once they are both in motion.
because of the rope that binds there movement to one another.

in my re-structured example that would have provided a better
description of the fluids in motion I had both masses moving at a
velocity of 1m/sec.
were anyone to look at the illustration
( that I provide at the bottom of this post )
it clearly shows that neither of the masses in question are
stationary.
and it clearly shows that the two masses in question will
have two separate velocities.
and it clearly shows that the two masses will be traveling through
two different size pipes.

in bobs calculations he first uses Kinetic energy.
ie..

The Kinetic Energy of the 1000kg mass moving at 1 m/s is
0.5 M V^2 = 500 Joules.
500 * 1 = 500 Joules.

I would like to add here that.
if observed while the system is in operation.
the 1 kg mass moving at a velocity of 10m/sec is as follows.

The Kinetic Energy of the 1kg mass moving at 10 m/s is
0.5 M V^2 = 500 Joules.
.5 * 100 = 50 Joules.

here it is clear that the 1000kg mass has (10) times the energy
of the 1kg mass
ie..

500J vs 50J

we could have stopped at this point.
and assured ourselves that the 1000kg mass had more energy.
than the 1kg mass.

however in true skeptic form he then moves on to convert the Kinetic
energy to a distance..
ie..

The 500 Joules of Kinetic Energy in the moving 1000kg mass will be
converted
entirely into the Potential Energy of the lifted 1 kg mass.

500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.

Do the division, and height = 500/9.8 meters = 51.02 meters.
...this is correct....

so he has used KE to provide distance.

he then moves on to the second mass
and here he uses the F=m*a equation.
his answer here states clearly that the 1kg mass
will cause a deceleration of 0.0098 m/s^2
the velocity of the 1000kg mass is 1m/sec
in a situation where there is no friction.
as in "frictionless wheels"
the first second the 1000kg mass will decelerate to 0.9902m/sec^2
the second second the 1000kg mass will decelerate to 0.9804/sec^2
so on and so on.
....this is correct...

The pulley system will exert a constant 9.8 Newtons
(1kg * 9.8 m/s^2) of horizontal force on the 1000kg mass.

F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2

.... this is correct.....

****************
Here's the exercise: Attach the 1kg mass to a length of massless rope.
(Every physics lab has an infinite supply of massless rope.) Run the
rope
though a frictionless pulley way up high. (The frictionless pulleys
are in
the cabinet next to the frictionless rope locker.) Run the rope
through
another pulley at ground level, so that you pull horizontally on the
rope to
raise the weight off the floor.

The door opens, and in comes a 1000kg shopping cart, rolling on the
perfectly level floor on frictionless wheels at 1 m/s. Lasso the cart
with
the free end of the rope.

As the rope tightens, the moving 1000kg mass starts to lift the 1kg
mass.
The work that it is doing -- or, if you insist, the resistance that
it is
feeling, although I cringe to say it that way -- causes the shopping
cart to
slow down. Eventually the shopping cart will come to a halt.

At that moment, of course, the 1kg weight will start to return
earthward,
dragging the shopping cart backwards. But let's ignore that for a
moment.

Here's the question: At the exact moment that the rising weight comes
to a
halt, how high will it be over the ground? How long will it take to
get
there?

I sincerely hope that you don't regard this calculation as
irrelevant. It
is absolutely central to understanding your pipe system.

The answers to my two questions are:

The height of the 1kg mass at the moment the masses stop moving will
be 51
meters.

The time it will take for the 1kg mass to reach that height will be
102
seconds.

I am either right or wrong. Which is it? And if I am wrong, what are
the

==============

My calculations:
VERY IMPORTANT.
NOTE ( BOBS ) CALCULATIONS.
NOT MINE.

The Kinetic Energy of the 1000kg mass moving at 1 m/s is 0.5 M V^2 =
500
Joules.
..... CORRECT .....

The 500 Joules of Kinetic Energy in the moving 1000kg mass will be
converted
entirely into the Potential Energy of the lifted 1 kg mass.

500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.

Do the division, and height = 500/9.8 meters = 51.02 meters.

.... ALSO CORRECT ....

The pulley system will exert a constant 9.8 Newtons (1kg * 9.8 m/s^2)
of
horizontal force on the 1000kg mass.

F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2

..... ALSO CORRECT ....

We want to know the time necessary for that constant acceleration to
change the velocity by 1 m/s, so we use the equation

V = a T which means that T = V / A = 1 / 0.0098 = 102.04 seconds.

..... ALSO CORRECT ......
at this time I would like for all to note that bob knows his math.
and he is capable of performing math in many ways.
he has demonstrated that he can think on many levels when he is
trying to prove a point.

to continue..

fluid traveling through a pipe is subject only to friction
and viscosity if the pipe is level in orientation to the earths
center.
I had hoped that you would have made a issue out of this.

in the below situation gravity of 9.8m/sec^2 is normally used.
in order to use the below equation in a pipe you must use a
"degree" of gravity.
this degree of gravity is applied to the mass.
this degree is .002 % of 9.8m/sec^2
in a smooth wall plastic pipe.
find it yourself.

or 0.0196m/sec^2
thus the weight of the mass when used in the equation
will be the same and the amount of gravity used will be
a percentage of the gravity constant.
in a work equation Force is equal to mass * gravity.

F = m*g
F = 1000kg * 0.0196
F = 19.6 kg
thus
E = 19.6 J
to use your brain bob is to realize that if the amount
of friction is reduced then the force required to move a
object is also reduced.
thus a 1000kg mass will have a resistance to motion
in the form of friction of 19.6 N
in other words the force that pushes back will be 19.6 N
opposite the direction of the force to move the object.

thus a force of 19.6 N can put the mass in motion.
thus to move the 1000kg mass 1 meter in 1 second will
require a amount of energy of 19.6 N * 1m = 19.6 J

heres a energy comparison.

energy required to "slide" the 1000kg mass 1 meter in 1 second

E = F = mg = 1000kg * .0196m/sec^2 = 19.6 N * 1m = 19.6 J

energy required to "lift" the 1kg to a height of 10 meters in 1 second

E = F = mg = 1kg * 9.8m/sec^2 = 9.8 N * 10m = 98 J

there is the perfect energy comparison.
try it to compare mass movements that have friction.
which my system has.
you will see that the answer is correct.

to use the exact same equation to observe the two situations.
and to use the exact amounts of elements that will effect
the movement of the two fluids.
will render a correct comparison.
and it will render a comparison that can be clearly understood.
unless you dont want others to be capable of understanding the
comparison which I believe is your motive.
as others have mentioned.

according to the above comparison the energy required to
slide the 1000kg mass is less than the energy required to
lift the 1kg mass
its there in plain physics equations.
and it cannot be denied by anyone who can use physics to determine
the amounts energy of mass in motion.

given that you know how to use physics.
and that you know how to use physics in many ways.

I stand on proof that supports the following....

rather than do things a simple way you choose to do things in
some strange way in a effort to try to confuse me.

you do not want to reveal a true energy comparison.

it is clear that you are not trying to deliver a honest evaluation.

to listen to your applied math and comparisons will deliver a
comparison that has no visible result.

ie....

I have been trying to demonstrate where your theoretical knowledge is
shaky.
I have been trying to go back to first principles to demonstrate to
you *why* they are in error.

**** even at the expense of using confusion techniques ****

And you simply *must* be open to the possibility that you are wrong.
**** I am always open to that ****
are you.

also a final point...
rather than agree to use the equation you and those of your type
will state that it is incorrect.
or you will simply QUIT because you know that if you continue
my system will be proven to be valid.
• im sure you meant v=220.892m/sec instead of v=2201 m/sec but you didnt include how high I will give you that. ok that would be simple to find now that you know
Message 4 of 23 , Mar 1, 2002
im sure you meant v=220.892m/sec

but you didnt include how high
I will give you that. ok
that would be simple to find now that you know the
final velocity.

--- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:
> --- In free_energy@y..., "kloskie" <kloskie@h...> wrote:
> > If you drop a object from a 100 meter height,
> > given there is no resistance.
> >
> > how long will it take to hit
> > the ground?
> >
> > How fast will it be going when it hits?
> > answer....v = 100.0000000005 m/sec^2
>
>
>
> My answer would go as follows:
>
> the terminal velocity (v) is given by:
>
> v = sqaureroot( 2 * g * h )
> = squareroot ( 2 * 9.8 m/s^2 * 100 m )
> = squareroot ( 1960 m^2/s^2 )
> = 44.3 m/s
>
> (note that the units of velocity are m/s, not m/s^2)
>
> the time required to hit the ground is calculated by dividing the
> final velocity by the rate of acceleration caused by Erath's
gravity:
>
> t = v/g
> = (44.3 m/s) / (9.8 m/s^2)
> = 4.52 s
>
> Someone please correct me if I did this incorrectly.
>
>
>
> >
> > *******now heres yours************
> >
> >
> > if you drop a object.
> > and it takes 22.54 seconds to hit the ground.
> >
> > How fast will it be going when it hits the ground?
> > how high was it dropped from.
> >
> > I would have included my work but Im not sure that
> > you are not trying to get people to do your homework.
> >
>
> Oooh, Oooh, let me take this one too. This is an easy problem. If
> we assume that the object is falling due to the Earth's
gravitational
> field, then the object is accelerating at 9.8 m/s^2. Therefore
> (assuming that the object had zero velocity before it began
> accelerating, and assuming negligible air resisitance), the
object's
> terminal velocity would equal the amount of time it accelerated
> multiplied by the rate of acceleration:
>
> v = t * g
> = 22.54 s * 9.8 m/s^2
> = 2201 m/s
>
> Is this right?
>
> Leo
• Paul, thanks, you are correct about my typo. I showed the correct setup, but then entered the numbers incorrectly into my calculator (multi-tasking at work
Message 5 of 23 , Mar 1, 2002
Paul, thanks, you are correct about my typo. I showed the correct
setup, but then entered the numbers incorrectly into my calculator
(multi-tasking at work takes it toll). So we agree that:

v = t * g
= 22.54 s * 9.8 m/s^2
= 220.9 m/s (4 significant figures)

The original height of the object (h) can then be calculated by
converting (on paper) the kinetic energy back into gravitational
potential energy:

initial E_pot = final E_kin

--- In free_energy@y..., "gearjammer123us" <paul@w...> wrote:
> im sure you meant v=220.892m/sec
>
> but you didnt include how high
> I will give you that. ok
> that would be simple to find now that you know the
> final velocity.
>
>
>
>
> --- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:
> > --- In free_energy@y..., "kloskie" <kloskie@h...> wrote:
> > > If you drop a object from a 100 meter height,
> > > given there is no resistance.
> > >
> > > how long will it take to hit
> > > the ground?
> > > answer....t= 10.2040816327 seconds
> > >
> > > How fast will it be going when it hits?
> > > answer....v = 100.0000000005 m/sec^2
> >
> >
> >
> > My answer would go as follows:
> >
> > the terminal velocity (v) is given by:
> >
> > v = sqaureroot( 2 * g * h )
> > = squareroot ( 2 * 9.8 m/s^2 * 100 m )
> > = squareroot ( 1960 m^2/s^2 )
> > = 44.3 m/s
> >
> > (note that the units of velocity are m/s, not m/s^2)
> >
> > the time required to hit the ground is calculated by dividing
the
> > final velocity by the rate of acceleration caused by Erath's
> gravity:
> >
> > t = v/g
> > = (44.3 m/s) / (9.8 m/s^2)
> > = 4.52 s
> >
> > Someone please correct me if I did this incorrectly.
> >
> >
> >
> > >
> > > *******now heres yours************
> > >
> > >
> > > if you drop a object.
> > > and it takes 22.54 seconds to hit the ground.
> > >
> > > How fast will it be going when it hits the ground?
> > > how high was it dropped from.
> > >
> > > I would have included my work but Im not sure that
> > > you are not trying to get people to do your homework.
> > >
> >
> > Oooh, Oooh, let me take this one too. This is an easy problem.
If
> > we assume that the object is falling due to the Earth's
> gravitational
> > field, then the object is accelerating at 9.8 m/s^2. Therefore
> > (assuming that the object had zero velocity before it began
> > accelerating, and assuming negligible air resisitance), the
> object's
> > terminal velocity would equal the amount of time it accelerated
> > multiplied by the rate of acceleration:
> >
> > v = t * g
> > = 22.54 s * 9.8 m/s^2
> > = 2201 m/s
> >
> > Is this right?
> >
> > Leo
• ... Paul -- I think I ve had enough. I was trying to go through a step-by-step exercise to approach the concepts needed to analyze your system. In response
Message 6 of 23 , Mar 1, 2002
>
> you ever go fishing bob.
> sometimes when you use rotten bait your catch will be greater
> than when you use fresh bait.
> it depends on the fish.
>
> I intentionally gave you wrong answers to see if I could get you to
> admit to the correctness or validity of my proposed system.
>
> this mission was successful.

Paul -- I think I've had enough. I was trying to go through a step-by-step
exercise to approach the concepts needed to analyze your system.

In response you first came up with a bunch of answers that were wrong.

When I pointed out they were wrong, you went off on a vitriolic diatribe

Now you are claiming that *one* wrong answer -- the 10.2 seconds for an
object to drop 100 meters, which you did by making an invalid calculation of
t = d/a which exactly matches numerous invalid calculations that you've done
before -- was done incorrectly intentionally, in order to bait me.

On top of that, now you turn around and post my message again, copying down
each step with your evaluation "CORRECT" after each step, praising me like I
am a good little boy for my valiant efforts when it is fact glaringly --
even grotesquely -- obvious that you haven't the faintest chance of doing
them yourself.

A number of people have complimented me on my patience and cordiality as I
have pursued this with you.

I now very patiently and cordially, and without any rancor or anger, tell
you that you are so full of ordure that your eyes must be brown.
• Hello All, Sorry for posting this twice. I accidentally posted the message below before completing the calculation of the initial height of the object. See
Message 7 of 23 , Mar 1, 2002
Hello All,

Sorry for posting this twice. I accidentally posted the message
below before completing the calculation of the initial height of the
object. See below for final steps.

--- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:
> Paul, thanks, you are correct about my typo. I showed the correct
> setup, but then entered the numbers incorrectly into my calculator
> (multi-tasking at work takes it toll). So we agree that:
>
> v = t * g
> = 22.54 s * 9.8 m/s^2
> = 220.9 m/s (4 significant figures)
>
> The original height of the object (h) can then be calculated by
> converting (on paper) the kinetic energy back into gravitational
> potential energy:

initial E_pot = final E_kin

m * g * h = 0.5 * m * v^2

g * h = 0.5 * v^2

h = 0.5 * v^2 / g

= 0.5 * (220.9 m/s)^2 / (9.8 m/s^2)

= 2490 m

I hope that I caught all of the typos this time.

P.S. I saw your free-energy fluid flow design -- very nicely
done. But I still stand by my conclusion that you can not extract
more than the equivalent of E=mgh from the kinetic energy of the
fluid flow.

>
>
>
> --- In free_energy@y..., "gearjammer123us" <paul@w...> wrote:
> > im sure you meant v=220.892m/sec
> > instead of v=2201 m/sec
> >
> > but you didnt include how high
> > I will give you that. ok
> > that would be simple to find now that you know the
> > final velocity.
> >
> >
> >
> >
> > --- In free_energy@y..., "chudslayer" <BordersChess@h...> wrote:
> > > --- In free_energy@y..., "kloskie" <kloskie@h...> wrote:
> > > > If you drop a object from a 100 meter height,
> > > > given there is no resistance.
> > > >
> > > > how long will it take to hit
> > > > the ground?
> > > > answer....t= 10.2040816327 seconds
> > > >
> > > > How fast will it be going when it hits?
> > > > answer....v = 100.0000000005 m/sec^2
> > >
> > >
> > >
> > > My answer would go as follows:
> > >
> > > the terminal velocity (v) is given by:
> > >
> > > v = sqaureroot( 2 * g * h )
> > > = squareroot ( 2 * 9.8 m/s^2 * 100 m )
> > > = squareroot ( 1960 m^2/s^2 )
> > > = 44.3 m/s
> > >
> > > (note that the units of velocity are m/s, not m/s^2)
> > >
> > > the time required to hit the ground is calculated by dividing
> the
> > > final velocity by the rate of acceleration caused by Erath's
> > gravity:
> > >
> > > t = v/g
> > > = (44.3 m/s) / (9.8 m/s^2)
> > > = 4.52 s
> > >
> > > Someone please correct me if I did this incorrectly.
> > >
> > >
> > >
> > > >
> > > > *******now heres yours************
> > > >
> > > >
> > > > if you drop a object.
> > > > and it takes 22.54 seconds to hit the ground.
> > > >
> > > > How fast will it be going when it hits the ground?
> > > > how high was it dropped from.
> > > >
> > > > I would have included my work but Im not sure that
> > > > you are not trying to get people to do your homework.
> > > >
> > >
> > > Oooh, Oooh, let me take this one too. This is an easy
problem.
> If
> > > we assume that the object is falling due to the Earth's
> > gravitational
> > > field, then the object is accelerating at 9.8 m/s^2.
Therefore
> > > (assuming that the object had zero velocity before it began
> > > accelerating, and assuming negligible air resisitance), the
> > object's
> > > terminal velocity would equal the amount of time it
accelerated
> > > multiplied by the rate of acceleration:
> > >
> > > v = t * g
> > > = 22.54 s * 9.8 m/s^2
> > > = 2201 m/s
> > >
> > > Is this right?
> > >
> > > Leo
• bob im having a little problem with your calculation below (((( The 500 Joules of Kinetic Energy in the moving 1000kg mass will be converted entirely into the
Message 8 of 23 , Mar 3, 2002
bob
im having a little problem with your calculation below
((((
The 500 Joules of Kinetic Energy in the moving 1000kg mass will be
converted
entirely into the Potential Energy of the lifted 1 kg mass.

500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.
))))

I get the following.

500 kg-m^2/s^2 = 1kg * 9.8 m/sec^2 * h
500/1 = 1*9.8*10
500/1 = 9.8*10
500 = 98

500 and 98 are not equal.

--- In free_energy@y..., "gearjammer123us" <paul@w...> wrote:
> you ever go fishing bob.
> sometimes when you use rotten bait your catch will be greater
> than when you use fresh bait.
> it depends on the fish.
>
> I intentionally gave you wrong answers to see if I could get you to
> admit to the correctness or validity of my proposed system.
>
> this mission was successful.
>
> this clip is in reference to bobs post # 2621
> here he gives a problem and then he gives his solution to
> the problem.
> the problem is a mere mass vs mass problem.
> the correct answer will provide the distance that
> mass 1 will pull mass 2.
>
> it is a simple problem.
> note: he is using "frictionless wheels"
> **********************************
> note also:
> water flowing through a smooth wall plastic pipe
> is met with a resistance of only .002 which can be
> expressed as a percentage of the gravity that causes
> resistance in the form of friction.
>
> ie..a
> 1000 kg mass of fluid
> vs a mass equivalent to
> .002 percent of the 1000 kg mass.
> ***************************************
> to continue with bobs exercise.
>
> and the mass has a velocity of 1 m/s
>
> Here's the question: At the exact moment that the rising weight
comes
> to a halt, how high will it be over the ground? How long will it
take
> to get there?
>
>
> The height of the 1kg mass at the moment the masses stop moving
will
> be 51 meters.
> The time it will take for the 1kg mass to reach that height will be
> 102 seconds.
> *** this is correct ***
> this proves that bob knows physics.
>
> ******************
> so bob is saying here that a 1000kg mass that has a velocity of 1
> meter per second can lift a 1kg mass to a height of 51meters.
> even though the 1000kg mass is only traveling at a velocity
> of 1 meter per second when it starts to pull the 1 kg mass.
> and it will take 102 seconds to lift it 51meters.
>
> ******************
>
> although this exercise can in no way show a comparison to the system
> in which we are discussing.
> mainly because you cannot tie ropes to water.
> and in the system we are discussing the fluid is in constant
> circulation.
> and circulating through different sized pipes.
> and it has two different constant velocities.
>
> here he wants to use an example that describes a
> stationary mass and a mass with velocity.
> and the two masses can only travel at the same velocity.
> once they are both in motion.
> because of the rope that binds there movement to one another.
>
> in my re-structured example that would have provided a better
> description of the fluids in motion I had both masses moving at a
> velocity of 1m/sec.
> were anyone to look at the illustration
> ( that I provide at the bottom of this post )
> it clearly shows that neither of the masses in question are
> stationary.
> and it clearly shows that the two masses in question will
> have two separate velocities.
> and it clearly shows that the two masses will be traveling through
> two different size pipes.
>
> in bobs calculations he first uses Kinetic energy.
> ie..
>
> The Kinetic Energy of the 1000kg mass moving at 1 m/s is
> 0.5 M V^2 = 500 Joules.
> 500 * 1 = 500 Joules.
>
> I would like to add here that.
> if observed while the system is in operation.
> the 1 kg mass moving at a velocity of 10m/sec is as follows.
>
> The Kinetic Energy of the 1kg mass moving at 10 m/s is
> 0.5 M V^2 = 500 Joules.
> .5 * 100 = 50 Joules.
>
> here it is clear that the 1000kg mass has (10) times the energy
> of the 1kg mass
> ie..
>
> 500J vs 50J
>
> we could have stopped at this point.
> and assured ourselves that the 1000kg mass had more energy.
> than the 1kg mass.
>
> however in true skeptic form he then moves on to convert the
Kinetic
> energy to a distance..
> ie..
>
> The 500 Joules of Kinetic Energy in the moving 1000kg mass will be
> converted
> entirely into the Potential Energy of the lifted 1 kg mass.
>
> 500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.
>
> Do the division, and height = 500/9.8 meters = 51.02 meters.
> ...this is correct....
>
> so he has used KE to provide distance.
>
> he then moves on to the second mass
> and here he uses the F=m*a equation.
> his answer here states clearly that the 1kg mass
> will cause a deceleration of 0.0098 m/s^2
> the velocity of the 1000kg mass is 1m/sec
> in a situation where there is no friction.
> as in "frictionless wheels"
> the first second the 1000kg mass will decelerate to 0.9902m/sec^2
> the second second the 1000kg mass will decelerate to 0.9804/sec^2
> so on and so on.
> ....this is correct...
>
> The pulley system will exert a constant 9.8 Newtons
> (1kg * 9.8 m/s^2) of horizontal force on the 1000kg mass.
>
> F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2
>
> .... this is correct.....
>
>
>
> ****************
> Here's the exercise: Attach the 1kg mass to a length of massless
rope.
> (Every physics lab has an infinite supply of massless rope.) Run
the
> rope
> though a frictionless pulley way up high. (The frictionless pulleys
> are in
> the cabinet next to the frictionless rope locker.) Run the rope
> through
> another pulley at ground level, so that you pull horizontally on
the
> rope to
> raise the weight off the floor.
>
> The door opens, and in comes a 1000kg shopping cart, rolling on the
> perfectly level floor on frictionless wheels at 1 m/s. Lasso the
cart
> with
> the free end of the rope.
>
> As the rope tightens, the moving 1000kg mass starts to lift the 1kg
> mass.
> The work that it is doing -- or, if you insist, the resistance that
> it is
> feeling, although I cringe to say it that way -- causes the
shopping
> cart to
> slow down. Eventually the shopping cart will come to a halt.
>
> At that moment, of course, the 1kg weight will start to return
> earthward,
> dragging the shopping cart backwards. But let's ignore that for a
> moment.
>
> Here's the question: At the exact moment that the rising weight
comes
> to a
> halt, how high will it be over the ground? How long will it take to
> get
> there?
>
> I sincerely hope that you don't regard this calculation as
> irrelevant. It
> is absolutely central to understanding your pipe system.
>
> The answers to my two questions are:
>
> The height of the 1kg mass at the moment the masses stop moving
will
> be 51
> meters.
>
> The time it will take for the 1kg mass to reach that height will be
> 102
> seconds.
>
> I am either right or wrong. Which is it? And if I am wrong, what
are
> the
>
> ==============
>
> My calculations:
> VERY IMPORTANT.
> NOTE ( BOBS ) CALCULATIONS.
> NOT MINE.
>
> The Kinetic Energy of the 1000kg mass moving at 1 m/s is 0.5 M V^2
=
> 500
> Joules.
> ..... CORRECT .....
>
> The 500 Joules of Kinetic Energy in the moving 1000kg mass will be
> converted
> entirely into the Potential Energy of the lifted 1 kg mass.
>
> 500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.
>
> Do the division, and height = 500/9.8 meters = 51.02 meters.
>
> .... ALSO CORRECT ....
>
> The pulley system will exert a constant 9.8 Newtons (1kg * 9.8
m/s^2)
> of
> horizontal force on the 1000kg mass.
>
> F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2
>
> ..... ALSO CORRECT ....
>
>
> We want to know the time necessary for that constant acceleration
to
> change the velocity by 1 m/s, so we use the equation
>
> V = a T which means that T = V / A = 1 / 0.0098 = 102.04 seconds.
>
> ..... ALSO CORRECT ......
> at this time I would like for all to note that bob knows his math.
> and he is capable of performing math in many ways.
> he has demonstrated that he can think on many levels when he is
> trying to prove a point.
>
> to continue..
>
> fluid traveling through a pipe is subject only to friction
> and viscosity if the pipe is level in orientation to the earths
> center.
> I had hoped that you would have made a issue out of this.
>
> in the below situation gravity of 9.8m/sec^2 is normally used.
> in order to use the below equation in a pipe you must use a
> "degree" of gravity.
> this degree of gravity is applied to the mass.
> this degree is .002 % of 9.8m/sec^2
> in a smooth wall plastic pipe.
> find it yourself.
>
> or 0.0196m/sec^2
> thus the weight of the mass when used in the equation
> will be the same and the amount of gravity used will be
> a percentage of the gravity constant.
> in a work equation Force is equal to mass * gravity.
>
> F = m*g
> F = 1000kg * 0.0196
> F = 19.6 kg
> thus
> E = 19.6 J
> to use your brain bob is to realize that if the amount
> of friction is reduced then the force required to move a
> object is also reduced.
> thus a 1000kg mass will have a resistance to motion
> in the form of friction of 19.6 N
> in other words the force that pushes back will be 19.6 N
> opposite the direction of the force to move the object.
>
> thus a force of 19.6 N can put the mass in motion.
> thus to move the 1000kg mass 1 meter in 1 second will
> require a amount of energy of 19.6 N * 1m = 19.6 J
>
> heres a energy comparison.
>
> energy required to "slide" the 1000kg mass 1 meter in 1 second
>
> E = F = mg = 1000kg * .0196m/sec^2 = 19.6 N * 1m = 19.6 J
>
> energy required to "lift" the 1kg to a height of 10 meters in 1
second
>
> E = F = mg = 1kg * 9.8m/sec^2 = 9.8 N * 10m = 98 J
>
> there is the perfect energy comparison.
> try it to compare mass movements that have friction.
> which my system has.
> you will see that the answer is correct.
>
> to use the exact same equation to observe the two situations.
> and to use the exact amounts of elements that will effect
> the movement of the two fluids.
> will render a correct comparison.
> and it will render a comparison that can be clearly understood.
> unless you dont want others to be capable of understanding the
> comparison which I believe is your motive.
> as others have mentioned.
>
>
> according to the above comparison the energy required to
> slide the 1000kg mass is less than the energy required to
> lift the 1kg mass
> its there in plain physics equations.
> and it cannot be denied by anyone who can use physics to determine
> the amounts energy of mass in motion.
>
> given that you know how to use physics.
> and that you know how to use physics in many ways.
>
> I stand on proof that supports the following....
>
> rather than do things a simple way you choose to do things in
> some strange way in a effort to try to confuse me.
> the above is your tools.
>
> you do not want to reveal a true energy comparison.
>
>
> it is clear that you are not trying to deliver a honest evaluation.
>
> to listen to your applied math and comparisons will deliver a
> comparison that has no visible result.
>
>
> ie....
>
> I have been trying to demonstrate where your theoretical knowledge
is
> shaky.
> I have been trying to go back to first principles to demonstrate to
> you *why* they are in error.
>
> **** even at the expense of using confusion techniques ****
>
>
> And you simply *must* be open to the possibility that you are wrong.
> **** I am always open to that ****
> are you.
>
> also a final point...
> rather than agree to use the equation you and those of your type
> will state that it is incorrect.
> or you will simply QUIT because you know that if you continue
> my system will be proven to be valid.
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