disaster warning

A plane is going to crash into the Statue of Liberty. I advise
everybody NOT to board planes.

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Inventions:
1a) The "Wheel" Newton Motor
1b) The "Seesaw" Newton Motor
1c) The "Simple" Newton DC Motor
2a) The "Simple" Newton Engine
2b) The "Horseshoe" Newton Engine
2c) Electromagnetic Propulsion for the Newton Engines

These five inventions (1a,1b,1c,2a,2b) 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", or convert the
"reaction" into something useable. All inventions work without
affecting the environment. That is, they don't need a road to push
off of like cars, and they don't have to spew out gases like a plane
or a space shuttle. They propel themselves *internally*. A prime
example of a device that propels itself internally is a UFO. Even
though I don't know if UFOs exist, the way they are said/seen to move
in the space/sky without affecting their environment makes me believe
that if they do exist, they must propel themselves internally. I
invision that these inventions can best be put to use to propel space
crafts (esp. for escaping the Earth's gravity).

(must be read using a "fixed-size font" to view diagrams)

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-1a) The "Wheel" Newton Motor=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Side view:

m
-=-
- | -
|
/ \ | / \ ----- wheel with magnets
\ | / installed on the outside
/ \ | / \
\|/
m|------------*-----------|m
/|\ forward --
\ / | \ /
/ | \
\ / | \ /
|
- | -
-=-
||||||M2 m M1|||||| --- electromagnets (coils)
||||||M2 M1||||||
-------------------------- ---base


The magnets "m" are connected to a wheel (which is connected to the
base [not shown]), whereas the electromagnets "M1" and "M2" are
fastened to the base.

When one of the magnets "m" reach the bottom, an electric current is
sent through both electromagnets, creating magnetic poles "M1" and
"M2". "M1" should repel "m" while "M2" should attract "m". The force
on the magnet "m" will cause the wheel to turn (in this diagram, that
would be in a clock-wise direction). Meanwhile, the forces on the
electromagnets "M1" and "M2" will cause the base to move in the
opposite direction (forward). Once the magnet "m" has moved
sufficiently far away, the electromagnets "M1" and "M2" should turn
off so that the next magnet "m" may come into position. The base will
experience a force in one direction, creating useful propulsion, while
the wheel can be hooked to a generator whose electrical output can be
used to add more power to the electromagnets.

Also, a motor may be needed to be connected to the wheel to start its
rotation, or to maintain it. Since the electromagnetics are
constantly being turned on and off, it may be better to plug them into
a capacitor, rather than just plugging it into a source which turns on
and off. Note that the magnets "m" on the wheel could just as well be
electromagnets.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-1b) The "Seesaw" Newton Motor-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Top view:


M1a---M2a

m1
\
\ /\
\ ||
o --seesaw ||
\ forward
\
\
m2


M1b---M2b


"M1a", "M1b", "M2a", "M2b", "m1", "m2" are all electromagnets. "M1a",
"M1b", "M2a", and "M2b" are fastened to the base, while "m1" and "m2"
are connected to a "seesaw" whose pivot ("o") is connected to the
base.

When "M1a" and "m1" are nearly touching an electric current is sent
through "M1a", "M1b", and "m1". "M1a" should repel "m1" while "M1b"
should attract "m1". Thus, both "M1a" and "M1b" will experience a
force in the forward direction, while the seesaw swings around
bringing "m2" close to "M2a". As, "M2a" and "m2" are close now, an
electric current will pass through "M2a", "M2b", and "m2". "M2a"
should repel "m2" while "M2b" should attract "m2". Again, "M2a" and
"M2b" will experience a force in the forward direction while the
seesaw swings back to its starting position to repeat the cycle.

Notice that if the seesaw swings so hard that "m1" hits "M1a" or "m2"
hits "M2a" then the force of the collision will cause a forward
movement. However, if "m1" hits "M1b" or "m2" hits "M2b", then the
collision will slow the forward motion. One could get by this hurdle
by installing a brake in the pivot to stop the complete swing of the
seesaw, and thus avoid a collision with the back electromagnets.

Again, since the electromagnetics are constantly being turned on and
off, it may be better to plug them into a capacitor, rather than just
plugging it into a source which turns on and off.

I've tried to make this invention but I haven't been able to build an
electromagnet that's stronger than 0.08 Tesla. That's not strong
enough for my rough design. Also, when two electromagnets are
orientated so that they should repel each other, they don't always
repel each other. That's because of the ferromagnetic material. This
happened to me when my electromagnets were touching. So, no
prototype.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-1c) The "Simple" Newton DC Motor=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Front view:

--------- -- wire wheel
| |
/-|\ /|-\ -- frame (holds magnets)
| |mmmmmmmmm| |
| --------- |
_|--mmmmmmmmm--|_ X forward
(into paper)
/\
||__ magnets


Side view:

--
/ \ -- wire cylinder

| OO | forward --
||
\ || /
||
__||__ -- base


The Simple Newton DC Motor is similar to a regular "simple" DC motor
except that there is only a portion of the wire exposed to a magnetic
field. Thus the base experiences a forward movement, while the wire
wheel experiences a circular motion (in the "side view", the wire
wheel would move clock-wise). The forward motion of the base can be
used to propel the entire motor (and its load). And of course, the
circular motion of the wheel can be harnessed to power a generator,
whose electrical output can then be fed back into the motor.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-2a) The "Simple" Newton Engine=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The Simple Newton Engine is simply a cylinder with a piston in it.
The piston may require wheels to move inside the cylinder.

-----------------------------------------------------------------------------
Step 1:
The idea is to force the piston down the shaft either by using
electromagnets or the explosion of gas.

| ___cylinder
| ||
| \/
|/-------------
|| #| forward --
|\-------------
| /\
| ||__ piston ("#")
|
|--start
-----------------------------------------------------------------------------

-----------------------------------------------------------------------------
Step 2:
The cylinder itself will move forward as the piston moves down the
cylinder.

--
| ___ The cylinder moves "forward"...
| ||
| \/
| /-------------
| | # |
| \-------------
| /\
| ||__ ...as the piston moves "back" through the cylinder
| --
|--start
-----------------------------------------------------------------------------

-----------------------------------------------------------------------------
Step 3:
In fractions of a second, the piston will have arrived at the "back"
of the cylinder. The piston must be stopped before it slams into the
back of the cylinder, because if it does, then the energy of the
piston will cancel out the "forward" velocity of the cylinder. So,
the energy of the piston must be removed (by friction, e.g. brakes on
the wheels) or harnessed (a method which converts the "negative"
energy of the piston into something useable).

|
|
|
| /-------------
| | # |
| \-------------
| /\
| ||__The piston must be stopped before it hits the "back".
|
|--start
-----------------------------------------------------------------------------

-----------------------------------------------------------------------------
Step 4:
When the piston has reached the end, it must be moved to the front of
the cylinder, perhaps by hooking it to a chain which is being pulled
by a motor. Perhaps, the piston can be removed from the cylinder when
it is being transferred to the front, and thus leave the cylinder free
so that another piston can shoot through it.

|
|
|
| /-------------
| |# |
| \-------------
|
|
|
|--start
-----------------------------------------------------------------------------

-----------------------------------------------------------------------------
The engine has moved and gained velocity, and is now ready to restart
at Step 1.

|
|
|
| /-------------
| | #|
| \-------------
|
|
|
|--start
-----------------------------------------------------------------------------

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-2b) The "Horseshoe" Newton Engine-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Top view (cross-section):


|-| |-|
piston ("#")-- |#| | |
| | | | /\
| | | | ||
| | | | --chamber ||
| \ / | forward
\ \ / /
\ \_ _/ /
\_ ----- _/
-___-

The Horseshoe Newton Engine is like the Simple Newton Engine, except
that the piston moves through a semi-circular loop. Thus, the
"negative energy" of the piston changes direction by 90 degrees, and
in doing so becomes useable energy which can propel the chamber
further. This is best done by allowing the piston to hit the end of
the chamber, thus transferring its energy into the forward movement of
the chamber. 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 Horseshoe 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 be fit in somehow.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-2c) Magnetic Propulsion for the Newton Engines=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Cross-section:

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 "m" are arranged such that the field is perpendicular
to the wire, and if a current is set-up in the wire coil, then the
wire coil will either move forward or backward. This set-up can be
used in either of the Newton Engines; the wire coil would be the
"piston" and the magnets would be part of the "cylinder" or "chamber".
The wire coil would need wheels on the side so that it could move
about inside the cylinder or chamber.

(It should be noted that both Newton Engines create a small amount of
force for a relatively minute amount of time. In my mind, they'd only
be effective if many are used simultaneously. For example, I imagine
that it wouldn't be too hard for either Newton Engines to have a burst
of 5N for a tenth of a second. Building a unit of tens of thousands
of such Newton Engines would create a combined force of 5000N,
assuming that the Engine can "reload" in 0.9 seconds.)

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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 5 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
(6.37 * 10^6 m approx.)

Now, this hummingbird is resilient and has enough energy to hover
above the ground for 10^18 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 (5km 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+5)^2
= 8.195661332 * 10^(-27) meters/second^2

After 10^18 years, when the hummingbird returns to its nest, the Earth
will be traveling at a velocity of

a = 8.195661332 * 10^(-27) meters/second^2

t = 10^18 years
= 3.1536 * 10^27 seconds

v = a * t
= 2.584583758 meters/second

The Earth was stationary and now it's moving at more than two
meters per second! Can you account for that? 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 was 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 instantaneously 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 air particles; and so, the chemical
energy is already accounted for.

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

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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 give the same result as 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. (Even if you
were to replace Jack and Jill with two tables, and rested the bricks
on the tables, work is still being done. The tables are "maintaining"
a force, and likewise, the books are "maintaining" a force.)

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,
we can say that "W=Ft".

If we allow "work to equal force multiplied by time" then we can say
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 multiplied by 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
"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", and it *WORKS*
with the non-intuitive equation "W=½mv²". That is, it calculates
"useful" work, but not "useless" work. But 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. I relegate the old, traditional meaning for work
to the term "typical useful work" or just "typical work".

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. (I
have never seen it be destroyed instantaneously). When energy is
created instantaneously, its immediate affect on the system is 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 removed. 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 realized. Consider a balloon of
hydrogen. The hydrogen has a mass of M. Now, if we cause all the
hydrogen to undergo fusion, then we'd be left with a balloon full of
helium and a whole lot of energy. The mass of the helium would be
approx. 0.992*M. There's a drop in mass. But gravitational potential
energy is proportional to mass. So, where did that minute, but
measurable, amount of potential energy go?!? It got turned into
various forms of energy, e.g. heat, light, sound. Do these forms of
energy have a gravitational potential energy? I don't think so; sound
definitely doesn't. So where did that gravitational potential energy
go?!? I don't know. There's definitely less. So, either we say that
potential energy was destroyed without being realized, or we say that
the hydrogen balloon never truly had a "potential". (I'd go with the
second one.)

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Now, I am going to apply work using prescotts on an electrical
circuit. (This section could very well be wrong.)

***************************
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.

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p.s. Two masses (e.g. stars) with sufficient velocities can pass by
each other without colliding and both gain speed. (As I said above,
gravity can create energy.) I believe that that might be the cause
for the seeming acceleration of the expansion of the universe, not
"dark energy". Just a guess..