I wrote the interesting paper, and I require feedback. Could you
please check it over. Thanks in advance.

---

20 joules equals 20 joules, right? Well, consider the following:

"Ball A"
work done = 20 joules
force = 10 Newtons
mass = 10 kg
acceleration = 1 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 2 m/s
change in time = 2 s

"Ball B"
work done = 20 joules
force = 10 Newtons
mass = 0.1 kg
acceleration = 100 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 20 m/s
change in time = 2/10 s

Each ball experienced the same force over the same distance. And so,
we can make the following statement.

"Ball A experienced 20 joules of work
and Ball B experienced 20 joules of work"

So, each ball had the same amount of work done on it. Makes sense.
However, if you agree that the above statement is correct, then you
shouldn't be able to deny the validity of the following statement:

"Ball A experienced 10 newtons held for 2 seconds
while Ball B experienced 10 newtons held for 2/10 of a second"

Thus if you agree with the first statement, then "10 newtons held for
2 seconds must equal 10 newtons held for 2/10 of a second"!

Intuitively speaking, that's ridiculous! If you cannot see the
intuitive error present here, then the following analogy may help you.
Consider two classmates, Jack and Jill, both able to hold a one
kilogram brick. Naturally, holding that brick on Earth is
approximately equivalent to maintaining a force of 10 Newtons. Let's
say that Jack held his brick for 20 seconds, and Jill held her brick
for 2 seconds. Now, without pulling out any scientific jargon, who
did the most work? If you try to answer that question in plain
English, then I'm sure you will see the intuitive error.

This leaves the joule system for work in a bit of a muddle, and I
fully agree that I'm not exactly sure how to explain this
short-coming, even though I'm sure I have the start.

We saw from the analogy that, in plain English, Jack did more work
than Jill. Thus we also see that work should be (intuitively
speaking) proportional to force and a duration of time. Using that as
a defintion for work, we find that "W=Ft".

When you learn about physics, and you first encounter the term
"force", you are told that it is a form of energy. Likewise, when you
encounter "work", you are again told that it is a form of energy. It
is true that both are forms of energy, but they are obviously not
equivalent. The difference between the two is never explained. If we
allow "work to equal force multiplied by time" then we have a
wonderful explanation: Force and work are both forms of energy, but
they are apparent in different "time frames". That is, work requires
a duration of real time for an effect to be experienced, meanwhile,
force requires an infinitesimal amount of time to have an effect
experienced.

We can also attack this problem from another angle, and also arrive at
"W=Ft".

Force equals mass times acceleration. Intuitively speaking, it is
blindingly obvious that force should be proportional to mass and to
acceleration. However, why isn't there a "coefficient"? And why not
"mass squared" or "acceleration cubed"? The equation is how it is
because of two things; one, intuitively, it makes sense not to add
extra "factors", and two, it simply gives the "right answers".

Now, let's examine the equation for work, that is "W=½mv²".
Intuitively speaking, it is blindingly obvious that work is
proportional to mass and to velocity. However, we added "factors" to
the equation. Without using scientific or mathematical jargon, I say
that we should be able to describe the equation for work in plain
English, like we did for force. This equation is how it is because of
only one thing; it "works". Meanwhile, if we remove all the extra
"factors", and say that "work equals mass multiplied by velocity"
("W=mv"), then we have again arrived at the equivalent equation\0
"W=Ft".

I said that the equation "W=½mv²" is how it is because of one thing,
it "works". But does it really? Consider dropping a brick from the
height of one meter above the ground. Drop it, and the brick falls.
Now, it is said that when you lift the brick up to one meter, then you
have given the brick a "potential energy". But let's consider two
scenarios, Jack and Jill, each lifting the brick from the ground to
one meter above the earth. Jack lifts it in 20 seconds while Jill
lifts it in 2 seconds. True, the outcome is the same for either
participant. However, in plain english, Jack did more work; he did
the same amount of "useful" work, but he did a whole lot of "useless"
work by taking his time.

Now, work defined as it is today, is wrong intuitively, but
nonetheless, it is a very useful "measuring tool". That is, it
calculates "useful" work, but not "useless" work. And intuitively,
work should encompass both "useful" and "useless" work.

I know that what I call work is called momentum. And so I assert that
work and momentum should be equivalent and synonymous. And I propose
that the real unit for work (that is, force multiplied by time) should
be "P", for Prescott, Joule's middle name. Thus, one Prescott equals
one newton second.

The law of conservation of energy is wrong! There are two reasons
for this:

1) The Joule system is wrong (it only encompasses "useful" work)
2) Attributing potential energy to objects is usually wrong

In reality, energy is being created all around us instantaneously (it
cannot be destroyed instantaneously). When energy is created
instantaneously, its immediate affect on the system will be nothing
(i.e. for forces, the vectors "cancel each other out"). After the
immediate effect, and after a minute amount of real time, this
instantaneous energy will be found to have either done "positive work"
on the system or "negative work"; that is, energy will be added to the
system, or destroyed. Should this instanteous energy be sustained for
a longer duration of real time, then the energy might be found to have
not added or removed any energy from the system (that is, it added the
same amount of energy that was removed).

"Potential energy" should only be called that so long as the potential
cannot disappear without being realised. Consider a log of wood.
Hold it in the air. Nothing is happening. Drop it. It falls to the
Earth, and proportionally, the Earth "falls" to the log. Pick it up
again, and in doing so, we say that we are giving it "potential
energy". But, now, without dropping it, burn the log. The log
disappears, and with it, so does the "potential energy". The
"potential energy" disappeared without being realized. So, either we
say that energy was destroyed, or we say that the log never truly had
a "potential".

Now, consider a battery. Between the anode and the cathode there is a
potential difference. However, can we destroy this potential energy?
No. The potential energy is within the chemical bonds, and the
"destruction" of the chemical will always realize the potential.

---------------------------------------------

Now, I am going to apply work using Prescotts on an electrical
circuit.

***************************
Let's find the average drift velocity:
-------------------------
A is the average (weighted with respect to L)
cross-section of the wire (m²)
n is "free" electrons per unit volume (electrons/m³)
e is the magnitude of charge of an electron
(1.602 * 10^19 C/electron)
v is the average drift velocity of the electrons (m/s)
I is the current in the (C/s)
dq is an infinitesmal amount of charge (C)
dt is an infinitesmal amount of time (s)
dN is an infinitesmal number of electrons (electrons)
-------------------------
(1) dq = e*dN

dN = nAv*dt
(2) dt = dN/(nAv)

(1)/(2) dq/dt = e*dN/(dN/nAv)
I = enAv
v = I/(enA)


***************************
Let's find force:
-------------------------
W_j is the Work in Joules (N*m)
f is the force (N)
s is the distance (m)
V is volts (N*m/C)
-------------------------
W_j = F*s
dW_j = F*v*dt
dW_j/dt = F*v
V*I = F*v

V*I
F = -----
v

= VenA

-------------------------
P is pressure (Pa)
-------------------------
So,
F
V = ---
enA

P
= --
en

We can now omit the use of Joules in the description of Volts. We can
say that "Voltage is the electromagnetic-pressure (creatd by an EMF
source) per density of charge."

Notice that the pressure supplied by an EMF has nothing to do with the
length of the circuit. A battery hooked to a 1 meter circuit of 1cm²
wire uses the same force as a similar battery hooked to a 100 meter
circuit of similar wire! Yet, it's obvious that more *work* is being
done in the 100 meter circuit than in the 1 meter circuit. The reason
why the force is the same while the work isn't is not hard at all to
understand. An EMF source creates "electromagnetic pressure" on the
anode and/or cathode. Once a circuit is started, this electromagnetic
pressure is felt throughout the circuit. You can imagine the
electrons as being dominoes. Whether you have 1 meter of "dominoes"
falling or 100 meters of "dominoes" falling, the initial pressure or
force may be the same, and yet, the amount of work done can be very
different. (This obviously means that energy *isn't* conserved.
That's right.)


***************************
-------------------------
W_p is the Work in Prescotts (N*s)
t is a duration of time (s)
-------------------------
W_p = F*t
= VenA*t


***************************
-------------------------
U is Work (in Prescotts) per Coulomb (N*s/C)
Q is an amount of charge (C)
p is the resistivity of the wire (ohms)
L is the length of the wire (m)
-------------------------
U = W_p/Q
= F/I
= (VenA)/(V/R)
= enAR
= enA*(p*L/A)
= enpL

Now, U is a constant for any given circuit. So, given any circuit, it
takes a constant amount of work to move a Coulomb along the circuit.
Makes sense that it doesn't vary..


***************************
-------------------------
µ is Work (in Prescotts) per Coulomb meter (N*s/(C*m))
-------------------------
µ = dU/dL
= enp

Thus, the rate at which work is done per unit distance depends only on
the material. Makes sense..


***************************
-------------------------
t_c is the average change in time between electron collisions (s)
m_e is the mass of an electron (9.109 * 10^(-31) kg/electron)
-------------------------

Each electron gains m_e*2v of energy before it makes a collision and
losses it's energy. The collision will take place in t_c seconds. U
is the amount of work to move a Coulomb L meters. Thus, in L meters,
there will be L/(v*t_c) number of collisions. So,

L m_e*2v
----- * ------ = enpL
v*t_c e

2m_e
t_c = ----
e²np


which is correct.

---------------------------------------------

3 inventions:
1) The Simple Newton Engine
2) The Semi-Circular Newton Engine
3) The Newton Motor

All three inventions work on Newton's law that "every action has an
equal and opposite reaction." The idea is to harness the "action" and
elimenate the "reaction".

---

|-| 1) The Simple Newton Engine
|P|
| |
| |
|-|

The Simple Newton Engine is simply a cylinder with a piston ("P").
The idea is to force the piston down the shaft either by using
electromagnets or the explosion of gas. (The piston may require
wheels to move about the cylinder.) The cylinder itself will move
forward, and the piston will move down the cylinder. The piston must
be stopped before it slams into the back of the cylinder, either by
friction or by a method which converts the "negative" energy into
something usuable. When the piston has reached the end, it must be
moved to the front of the cylinder, perhaps by a motor.

---

|-| |-| 2) The Semi-Circular Newton Engine
|P| | |
| | | |
\ \ / /
\ \ / /
\ \___/ /
\_____/

The Semi-Circular Newton Engine is like the Simple Newton Engine,
except that the piston moves through a semi-circular loop. Thus, the
"negative energy" changes direction by 90 degrees, and in doing so
becomes usuable energy which can propel the cylinder, or chamber,
further. The internal combustion engine has four parts: the intake
stroke, the compression stroke, the combustion stroke, and the exhaust
stroke. As the piston moves through the Semi-circular Newton Engine,
the combustion stroke for one part of the loop can be the compression
stage for the other side of the loop. That leaves the intake and
exhaust strokes which must fit in.

---

mmmmmmmmmmmmmmmmmmmm
mmmmm ____ mmmmm -- "m" are magnets
mmmm /WWWWWW\ mmmm
mmm /W/ \W\ mmm
mm /W/ mm \W\ mm
m W mmmm W m -- "W" is a wire coil
m |W| mmmmmm |W| m
m |W| mmmmmm |W| m
m W mmmm W m
mm \W\ mm /W/ mm
mmm \W\____/W/ mmm
mmmm \WWWWWW/ mmmm
mmmmm mmmmm
mmmmmmmmmmmmmmmmmmmm

If the magnets are arranged such that the field is perpendicular to
the wire coil, and if a current is set-up in the wire coil, then the
wire coil will either move forward or backward. This set-up could be
used in either the Newton Engines; the wire coil would be the
"piston".

---

3) The Newton Motor

Front view:

--------- -- wire cylinder
| |
/-|\ /|-\ -- frame (holds magnets)
| |mmmmmmmmm| |
| --------- |
_|--mmmmmmmmm--|_

/\
||__ magnets


Side view:

--
/ \ -- wire cylinder

| OO |
||
\ || /
||
__||__ -- frame


The Newton Motor is similar to a regular motor except that there is
only a small portion of the wire exposed to magnets. Thus the frame
experiences a forward movement, while the wire cylinder experiences a
circular motion. Of course, this circular motion can be harnessed to
power a generator.

---------------------------------------------

Consider an Earth that is stationary and is not affected by any
external forces. Alone on the Earth is a hummingbird sitting in its
nest in the world's last tree. The rest of the Earth is totally
lifeless and motionless. Suddenly, the hummingbird, which has a mass
of 5 grams, begins to hover 3 kilometers off the ground. The downward
gravitational force on the hummingbird is given by the equation

F = G*m_b*m_e / r^2

where G is the gravitatiional constant
(6.672 * 10^(-11) Nm^2/kg^2)
m_b is the mass of the bird (0.005 kg)
m_e is the mass of the Earth (5.98 10^24 kg)
r is the distance between the Earth and the bird
(6370 km approx.)

Now, this hummingbird is resilient and has enough energy to hover
above the ground for 10^19 years. It is obvious that the hummingbird
is converting chemical energy into kinetic energy. By doing so, two
things happen; one, the hummingbird is pushed upward, and two, air is
pushed downward. Since the hummingbird is a fair enough distance from
the Earth (3km to be exact), the downward force on the air molecules
never actually reach the ground because it gets distributed amongst
the other air particles. And so, as this force is distributed amongst
billions of molecules, none of them will ever gain a sufficient
velocity to reach the ground.

So, we took care of all the forces, right? Wrong! We only considered
the gravitational force of the earth on the bird. But what about the
gravitational force of the bird on the earth? That force creates an
acceleration of

a = G*m_b / r^2
= 8.221426476641*10^(-27) meters/second^2

After 10^19 years, when the hummingbird returns to its nest, the Earth
will be traveling at a velocity of

a = 8.221426476641*10^(-27) meters/second^2

t = 10^19 years
= 3.1536*10^26 seconds

v = a * t
= 2 meters/second

The Earth was stationary and now it's moving! Can you account
for the energy? Where did the energy to move the Earth come from? We
have already accounted for the bird's energy which simply pushed air.
You see, as the bird was hovering, we could say that the bird is
perpetually falling to the Earth. Likewise, the Earth was perpetually
falling toward the hummingbird. And thus, the vectors of the forces
cancel each other out! Now, I hope you can clearly see and appreciate
that gravity (and other forces) create kinetic energy out of nothing.
Some of you may argue at this point that the bird's chemical energy
was converted to the Earth's kinetic energy. That's quite ridiculous
because, as we saw earlier, the chemical energy of the bird was
transferred to kinetic energy of wind particiles; and so, the chemical
energy is already accounted for.

What does all of this mean? It means that perpetual motion and
free energy devices do not contradict reality!

---------------------------------------------

by Raheman Velji

"Raheman" wrote in message
om...
I wrote the interesting paper, and I require feedback. Could you
please check it over. Thanks in advance.

---

20 joules equals 20 joules, right? Well, consider the following:

"Ball A"
work done = 20 joules
force = 10 Newtons
mass = 10 kg
acceleration = 1 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 2 m/s
change in time = 2 s

"Ball B"
work done = 20 joules
force = 10 Newtons
mass = 0.1 kg
acceleration = 100 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 20 m/s
change in time = 2/10 s

Each ball experienced the same force over the same distance. And so,
we can make the following statement.

"Ball A experienced 20 joules of work
and Ball B experienced 20 joules of work"

So, each ball had the same amount of work done on it. Makes sense.
However, if you agree that the above statement is correct, then you
shouldn't be able to deny the validity of the following statement:

"Ball A experienced 10 newtons held for 2 seconds
while Ball B experienced 10 newtons held for 2/10 of a second"

Thus if you agree with the first statement, then "10 newtons held for
2 seconds must equal 10 newtons held for 2/10 of a second"!

Intuitively speaking, that's ridiculous!
If you cannot see the intuitive error present here, then the following
analogy may help you.
Consider two classmates, Jack and Jill, both able to hold a one
kilogram brick. Naturally, holding that brick on Earth is
approximately equivalent to maintaining a force of 10 Newtons. Let's
say that Jack held his brick for 20 seconds, and Jill held her brick
for 2 seconds. Now, without pulling out any scientific jargon, who
did the most work? If you try to answer that question in plain
English, then I'm sure you will see the intuitive error.

Beware your intuition! Neither Jack nor Jill did any work. Holding a
weight at rest will tire you and you will radiate heat energy (due to
processes within your body that produce the required force) But the
mechanical work produced outside your body is zero.

This leaves the joule system for work in a bit of a muddle, and I
fully agree that I'm not exactly sure how to explain this
short-coming, even though I'm sure I have the start.

We saw from the analogy that, in plain English, Jack did more work
than Jill.

No. They both do zero work.

Thus we also see that work should be (intuitively
speaking) proportional to force and a duration of time. Using that as
a defintion for work, we find that "W=Ft".

When you learn about physics, and you first encounter the term
"force", you are told that it is a form of energy.

If you were told that you were cheated of a proper education in physics.

Likewise, when you
encounter "work", you are again told that it is a form of energy. It
is true that both are forms of energy, but they are obviously not
equivalent. The difference between the two is never explained. If we
allow "work to equal force multiplied by time" then we have a
wonderful explanation: Force and work are both forms of energy, but
they are apparent in different "time frames". That is, work requires
a duration of real time for an effect to be experienced, meanwhile,
force requires an infinitesimal amount of time to have an effect
experienced.

We can also attack this problem from another angle, and also arrive at
"W=Ft".

Force equals mass times acceleration. Intuitively speaking, it is
blindingly obvious that force should be proportional to mass and to
acceleration.

"Intuitively speaking, it is blindingly obvious..." WOW. If only it were
so.

However, why isn't there a "coefficient"? And why not
"mass squared" or "acceleration cubed"? The equation is how it is\0
because of two things; one, intuitively, it makes sense not to add
extra "factors", and two, it simply gives the "right answers".



Now, let's examine the equation for work, that is "W=½mv²".
Intuitively speaking, it is blindingly obvious that work is
proportional to mass and to velocity. However, we added "factors" to
the equation. Without using scientific or mathematical jargon, I say
that we should be able to describe the equation for work in plain
English, like we did for force. This equation is how it is because of
only one thing; it "works". Meanwhile, if we remove all the extra
"factors", and say that "work equals mass multiplied by velocity"
("W=mv"), then we have again arrived at the equivalent equation
"W=Ft".


Are you just renaming momentum to be called "work," and therby sow
confusion? Or does your intuition tell you that momentum and energy are the
same entities?

I said that the equation "W=½mv²" is how it is because of one thing,
it "works". But does it really? Consider dropping a brick from the
height of one meter above the ground. Drop it, and the brick falls.
Now, it is said that when you lift the brick up to one meter, then you
have given the brick a "potential energy". But let's consider two
scenarios, Jack and Jill, each lifting the brick from the ground to
one meter above the earth. Jack lifts it in 20 seconds while Jill
lifts it in 2 seconds. True, the outcome is the same for either
participant. However, in plain english, Jack did more work; he did
the same amount of "useful" work, but he did a whole lot of "useless"
work by taking his time.

Now, work defined as it is today, is wrong intuitively, but
nonetheless, it is a very useful "measuring tool". That is, it
calculates "useful" work, but not "useless" work. And intuitively,
work should encompass both "useful" and "useless" work.

I know that what I call work is called momentum. And so I assert that
work and momentum should be equivalent and synonymous. And I propose
that the real unit for work (that is, force multiplied by time) should
be "P", for Prescott, Joule's middle name. Thus, one Prescott equals
one newton second.

The law of conservation of energy is wrong! There are two reasons
for this:

1) The Joule system is wrong (it only encompasses "useful" work)
2) Attributing potential energy to objects is usually wrong

So when you release a brick and it transfers its acquired energy to your toe
you just smile. Raheman, we don't attribute potential energy to objects, per
se. We atribute it to a mass by virtue of its position in a force field
relative to some other location in that field. We call it "potential
energy" because it exhibits itself ONLY if the mass were to be transfered
from its current location to the new one.

When (and if) you understand your error vis a vis "energy" and "work," you
will reject all you've written below this line. So there is no point to my
making any further comment..

Good luck. And put your intuition aside in favor of more study..
Eli Botkin


In reality, energy is being created all around us instantaneously (it
cannot be destroyed instantaneously). When energy is created
instantaneously, its immediate affect on the system will be nothing
(i.e. for forces, the vectors "cancel each other out"). After the
immediate effect, and after a minute amount of real time, this
instantaneous energy will be found to have either done "positive work"
on the system or "negative work"; that is, energy will be added to the
system, or destroyed. Should this instanteous energy be sustained for
a longer duration of real time, then the energy might be found to have
not added or removed any energy from the system (that is, it added the
same amount of energy that was removed).

"Potential energy" should only be called that so long as the potential
cannot disappear without being realised. Consider a log of wood.
Hold it in the air. Nothing is happening. Drop it. It falls to the
Earth, and proportionally, the Earth "falls" to the log. Pick it up
again, and in doing so, we say that we are giving it "potential
energy". But, now, without dropping it, burn the log. The log
disappears, and with it, so does the "potential energy". The
"potential energy" disappeared without being realized. So, either we
say that energy was destroyed, or we say that the log never truly had
a "potential".

Now, consider a battery. Between the anode and the cathode there is a
potential difference. However, can we destroy this potential energy?
No. The potential energy is within the chemical bonds, and the
"destruction" of the chemical will always realize the potential.

---------------------------------------------

Now, I am going to apply work using Prescotts on an electrical
circuit.

***************************
Let's find the average drift velocity:
-------------------------
A is the average (weighted with respect to L)
cross-section of the wire (m²)
n is "free" electrons per unit volume (electrons/m³)
e is the magnitude of charge of an electron
(1.602 * 10^19 C/electron)
v is the average drift velocity of the electrons (m/s)
I is the current in the (C/s)
dq is an infinitesmal amount of charge (C)
dt is an infinitesmal amount of time (s)
dN is an infinitesmal number of electrons (electrons)
-------------------------
(1) dq = e*dN

dN = nAv*dt
(2) dt = dN/(nAv)

(1)/(2) dq/dt = e*dN/(dN/nAv)
I = enAv
v = I/(enA)


***************************
Let's find force:
-------------------------
W_j is the Work in Joules (N*m)
f is the force (N)
s is the distance (m)
V is volts (N*m/C)
------------------------
W_j = F*s
dW_j = F*v*dt
dW_j/dt = F*v
V*I = F*v

V*I
F = -----
v

= VenA

-------------------------
P is pressure (Pa)
-------------------------
So,
F
V = ---
enA

P
= --
en

We can now omit the use of Joules in the description of Volts. We can
say that "Voltage is the electromagnetic-pressure (created by an EMF
source) per density of charge."

Notice that the pressure supplied by an EMF has nothing to do with the
length of the circuit. A battery hooked to a 1 meter circuit of 1cm²
wire uses the same force as a similar battery hooked to a 100 meter
circuit of similar wire! Yet, it's obvious that more *work* is being
done in the 100 meter circuit than in the 1 meter circuit. The reason
why the force is the same while the work isn't is not hard at all to
understand. An EMF source creates "electromagnetic pressure" on the
anode and/or cathode. Once a circuit is started, this electromagnetic
pressure is felt throughout the circuit. You can imagine the
electrons as being dominoes. Whether you have 1 meter of "dominoes"
falling or 100 meters of "dominoes" falling, the initial pressure or
force may be the same, and yet, the amount of work done can be very
different. (This obviously means that energy *isn't* conserved.
That's right.)


***************************
-------------------------
W_p is the Work in Prescotts (N*s)
t is a duration of time (s)
-------------------------
W_p = F*t
= VenA*t


***************************
-------------------------
U is Work (in Prescotts) per Coulomb (N*s/C)
Q is an amount of charge (C)
p is the resistivity of the wire (ohms)
L is the length of the wire (m)
-------------------------
U = W_p/Q
= F/I
= (VenA)/(V/R)
= enAR
= enA*(p*L/A)
= enpL

Now, U is a constant for any given circuit. So, given any circuit, it
takes a constant amount of work to move a Coulomb along the circuit.
Makes sense that it doesn't vary..


***************************
-------------------------
µ is Work (in Prescotts) per Coulomb meter (N*s/(C*m))
-------------------------
µ = dU/dL
= enp

Thus, the rate at which work is done per unit distance depends only on
the material. Makes sense..


***************************
-------------------------
t_c is the average change in time between electron collisions (s)
m_e is the mass of an electron (9.109 * 10^(-31) kg/electron)
-------------------------

Each electron gains m_e*2v of energy before it makes a collision and
losses it's energy. The collision will take place in t_c seconds. U
is the amount of work to move a Coulomb L meters. Thus, in L meters,
there will be L/(v*t_c) number of collisions. So,

L m_e*2v
----- * ------ = enpL
v*t_c e

2m_e
t_c = ----
e²np


which is correct.

---------------------------------------------

3 inventions:
1) The Simple Newton Engine
2) The Semi-Circular Newton Engine
3) The Newton Motor

All three inventions work on Newton's law that "every action has an
equal and opposite reaction." The idea is to harness the "action" and
elimenate the "reaction".

---

|-| 1) The Simple Newton Engine
|P|
| |
| |
|-|

The Simple Newton Engine is simply a cylinder with a piston ("P").
The idea is to force the piston down the shaft either by using
electromagnets or the explosion of gas. (The piston may require
wheels to move about the cylinder.) The cylinder itself will move
forward, and the piston will move down the cylinder. The piston must
be stopped before it slams into the back of the cylinder, either by
friction or by a method which converts the "negative" energy into
something usuable. When the piston has reached the end, it must be
moved to the front of the cylinder, perhaps by a motor.

---

|-| |-| 2) The Semi-Circular Newton Engine
|P| | |
| | | |
\ \ / /
\ \ / /
\ \___/ /
\_____/

The Semi-Circular Newton Engine is like the Simple Newton Engine,
except that the piston moves through a semi-circular loop. Thus, the
"negative energy" changes direction by 90 degrees, and in doing so
becomes usuable energy which can propel the cylinder, or chamber,
further. The internal combustion engine has four parts: the intake
stroke, the compression stroke, the combustion stroke, and the exhaust
stroke. As the piston moves through the Semi-circular Newton Engine,
the combustion stroke for one part of the loop can be the compression
stage for the other side of the loop. That leaves the intake and
exhaust strokes which must fit in.

---

mmmmmmmmmmmmmmmmmmmm
mmmmm ____ mmmmm -- "m" are magnets
mmmm /WWWWWW\ mmmm
mmm /W/ \W\ mmm
mm /W/ mm \W\ mm
m W mmmm W m -- "W" is a wire coil
m |W| mmmmmm |W| m
m |W| mmmmmm |W| m
m W mmmm W m
mm \W\ mm /W/ mm
mmm \W\____/W/ mmm
mmmm \WWWWWW/ mmmm
mmmmm mmmmm
mmmmmmmmmmmmmmmmmmmm

If the magnets are arranged such that the field is perpendicular to
the wire coil, and if a current is set-up in the wire coil, then the
wire coil will either move forward or backward. This set-up could be
used in either the Newton Engines; the wire coil would be the
"piston".

---

3) The Newton Motor

Front view:

--------- -- wire cylinder
| |
/-|\ /|-\ -- frame (holds magnets)
| |mmmmmmmmm| |
| --------- |
_|--mmmmmmmmm--|_

/\
||__ magnets


Side view:

--
/ \ -- wire cylinder

| OO |
||
\ || /
||
__||__ -- frame


The Newton Motor is similar to a regular motor except that there is
only a small portion of the wire exposed to magnets. Thus the frame
experiences a forward movement, while the wire cylinder experiences a
circular motion. Of course, this circular motion can be harnessed to
power a generator.

---------------------------------------------

Consider an Earth that is stationary and is not affected by any
external forces. Alone on the Earth is a hummingbird sitting in its
nest in the world's last tree. The rest of the Earth is totally
lifeless and motionless. Suddenly, the hummingbird, which has a mass
of 5 grams, begins to hover 3 kilometers off the ground. The downward
gravitational force on the hummingbird is given by the equation

F = G*m_b*m_e / r^2

where G is the gravitatiional constant
(6.672 * 10^(-11) Nm^2/kg^2)
m_b is the mass of the bird (0.005 kg)
m_e is the mass of the Earth (5.98 10^24 kg)
r is the distance between the Earth and the bird
(6370 km approx.)

Now, this hummingbird is resilient and has enough energy to hover
above the ground for 10^19 years. It is obvious that the hummingbird
is converting chemical energy into kinetic energy. By doing so, two
things happen; one, the hummingbird is pushed upward, and two, air is
pushed downward. Since the hummingbird is a fair enough distance from
the Earth (3km to be exact), the downward force on the air molecules
never actually reach the ground because it gets distributed amongst
the other air particles. And so, as this force is distributed amongst
billions of molecules, none of them will ever gain a sufficient
velocity to reach the ground.

So, we took care of all the forces, right? Wrong! We only considered
the gravitational force of the earth on the bird. But what about the
gravitational force of the bird on the earth? That force creates an
acceleration of

a = G*m_b / r^2
= 8.221426476641*10^(-27) meters/second^2

After 10^19 years, when the hummingbird returns to its nest, the Earth
will be traveling at a velocity of

a = 8.221426476641*10^(-27) meters/second^2

t = 10^19 years
= 3.1536*10^26 seconds

v = a * t
= 2 meters/second

The Earth was stationary and now it's moving! Can you account
for the energy? Where did the energy to move the Earth come from? We
have already accounted for the bird's energy which simply pushed air.
You see, as the bird was hovering, we could say that the bird is
perpetually falling to the Earth. Likewise, the Earth was perpetually
falling toward the hummingbird. And thus, the vectors of the forces
cancel each other out! Now, I hope you can clearly see and appreciate
that gravity (and other forces) create kinetic energy out of nothing.
Some of you may argue at this point that the bird's chemical energy
was converted to the Earth's kinetic energy. That's quite ridiculous
because, as we saw earlier, the chemical energy of the bird was
transferred to kinetic energy of wind particiles; and so, the chemical
energy is already accounted for.

What does all of this mean? It means that perpetual motion and
free energy devices do not contradict reality!

---------------------------------------------

by Raheman Velji

"Raheman" wrote in message
om...

We saw from the analogy that, in plain English, Jack did more work
than Jill. Thus we also see that work should be (intuitively
speaking) proportional to force and a duration of time. Using that as
a defintion for work, we find that "W=Ft".

Sorry, but this is incorrect.
Impulse is defined as the product of Force and time.

I=FT

Naturally we see that Impulse is measured in Newton*seconds. Impulse is
mathematically equivalent to momentum.

F=ma; I=mat

v=at; I=mv (p is used to denote momentum) p=mv=ft=I

Whe can easily test the veracity of this equivalence by comparing the units
of momentum and impulse. As stated above, Impulse is measured in
Newton*seconds. Force * Time, or equivalently kg*m*s/s^2 which reduces to
kg*m/s.

Now momentum, being the product of mass and velocity is measured in units of
Kg*m/s (mass*velocity). As you can see, they are mathematically equivalent.

Work is defined as a force applied over a distance. W=Fd. Fd certainly does
not equal Ft.

Work is mathematically equivalent to E.

F=ma; W=mad
v^2=2ad; v^2/2=ad;
W/m=ad;W/m=v^2/2;
W=mv^2/2 (E is used to denote Energy, in this case, kinetic)
E=mv^2/2=mad=W

So, momentum (p) is equivalent to Impulse (I) and
Work (W) is equivalent to Energy (E)

The relationship between Impulse and Energy is a bit different then you
would have it.

E=mv^2/2, rearrange to solve for v, v^2=2E/m

I=mv, square both sides, I^2=m^2v ^2; solve for v; v^2=I^2/m^2

Since v^2=2E/m and v^2=I^2/m^2, then

I^2/m^2=2E/m; If we rearrange to solve for I;

I^2=2Em^2/m or I=(2Em)^-2; Solved for E then,

E=I^2/2m

Cheers,
Jim

"Raheman" skrev i melding om...
I wrote the interesting paper, and I require feedback. Could you
please check it over. Thanks in advance.

---

20 joules equals 20 joules, right? Well, consider the following:

"Ball A"
work done = 20 joules
force = 10 Newtons
mass = 10 kg
acceleration = 1 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 2 m/s
change in time = 2 s

Kinetic energy = 0.5mv^2 = 20 joules = work done = Fs = 10N*2m
Momentum = mv = 20 kg*m/s = 20 Ns = Kt = 10N*2s

"Ball B"
work done = 20 joules
force = 10 Newtons
mass = 0.1 kg
acceleration = 100 m/s²
change in distance = 2 m
initial velocity = 0 m/s
final velocity = 20 m/s
change in time = 2/10 s

Kinetic energy = 0.5mv^2 = 20 joules = work done = Fs = 10N*2m
Momentum = mv = 2 kg*m/s = 2 Ns = Kt = 10N*0.2s


Each ball experienced the same force over the same distance. And so,
we can make the following statement.

"Ball A experienced 20 joules of work
and Ball B experienced 20 joules of work"

So, each ball had the same amount of work done on it. Makes sense.

And that's why the kinetic energies are equal.

However, if you agree that the above statement is correct, then you
shouldn't be able to deny the validity of the following statement:

"Ball A experienced 10 newtons held for 2 seconds
while Ball B experienced 10 newtons held for 2/10 of a second"

And that's why the momentums are different.

Thus if you agree with the first statement, then "10 newtons held for
2 seconds must equal 10 newtons held for 2/10 of a second"!

No. They results in different momentums, and are thus not equal.

Paul