neuvophysics.com

E= something squared or
what?

E=md or E=Fd ? Science
says that it's E=Fd. I strongly disagree for a very good reason or two.

Fundamental
consideration

In every case the true
value of the energy deemed available from an object in constant motion (E_{k})
is NOT able to be inferred from the impulse event which must have occurred to
cause the constant motion in the first place. Such impulse events are in other
words accelerative / decelerative events. All such accelerations from the
obviously violent to the gentle pushes and gravitational freefall events are
less noticeable but decidedly non linear hyperbolic events.

In all such cases, initial
energies are being used and lost to the environment. Therefore the curved
portions of these events is not permitted to be used to confirm or deny the
value of any E_{k} which must only be derived from an analysis of the
linear TV or momentum of any object whatsoever. That includes the first two
seconds of gravitational freefall and ditto for the sidewise pushing case.

There is a caveat in that
even though it must be considered that -with force being constant- E_{k}
=m.F.t with 't' being brought to terms of one second and then the F.t is equal
to the distance travelled in one second; the true evaluation of work done to
evince a similar E_{k} is always derivable in the distance travelled
regardless of the time taken but that distance is only being brought to terms
for the empirical requirement of the equations.

REASON 1

F=ma and F=m
g Note:
>< means equal to for the purposes at hand. The heights are at or near
earth surface. ---vacuum supposed.

THEOREM to disprove E=mgh
at W=1kg

1/ Fact: m><W=F_{g}
per kg AT ANY HEIGHT!

2/ Fact:
'g' =
9.81m.s.s FROM ANY HEIGHT!

3/ Fact: F_{g} per
kg =
g ---Therefore the equation
F=ma --- F_{g}=1*
g ---
g=F_{g}/1 derives
9.81m.s.s and according to the WEP that is the acceleration rate in the Earth's
gravitational field for all objects at all proposed heights. This can a only be
due to a fact---

4/ Fact: When mass is 1kg
then F_{g}=
g regardless of the height!

So both height and
acceleration rate are thought to become factors and supposedly E=m
gh via
E=Fd =mad.

However if
m><W=
F_{g}
then E=F_{g}*h actually becomes E=md as proposed above and E=mad and
mgh are precluded.
Note: m is 1kg and 'd' actually becomes 9.81m. This is
because in all cases mass/weight is the total force of gravity via C=F per m/m
(C is for constant g. F is in

In the same vein E=mv^2
can't be used when v is unity. That makes the equation very suspicious.

Or they consider that if
g=F/m
and F=9.81 and m=10kg then
'g' would be 0.981m.s.s which is ridiculous,
so they decided the need to bring in height. So then if E=mgh we now get
E=10*9.81*1 we get E=98.1J. So because that also works for
g=E/mh it
might appear to be the only reasonable solution. NO! We can just use the height
and leave the acc rate out.

The savage argument here
is that the acc rate causes an exponential increase in energy (even if it
doesn't match up with E=mv^2 lol!) NO!
The SEP sees no difference in the energies derivable in the gravitational
inertial case or the sidewise case. They are the same. A force by any other
name is just the same. The only difference is at impact because the
gravitational force is still acting as the object is splatting!

REASON 2

It is a well understood
fact that energy is highly relatable to mass and that includes work done by the
motion of any mass. Have you heard of mE equivalence?

When we analyse the
derivation of equations we must look at relatable dimensions and not just pull
the ideas out of any hat that suits our needs, as we saw above. E.g. we notice
from the close relationships in physics that energy and work definitely have
dimensions of mass, velocity and distance. On the other hand we should note
that force is one step removed from any dimensions of energy. Force gives the
acceleration which leads to the velocity and or distance. Therefore force has
no dimensions of energy, so we can state the following.

1/ To have E_{k}, E_{p}
or work you MUST have mass!

2/ When the choice of
equation comes down to either--- E_{k} (W)= F.d with mass implied or---
M.d with force implied then it becomes the equation which defers to the
dimensions which must be the true form. Therefore it is E=m.d which is true.
Note:
A force is implied but the value is irrelevant.

Also with E=Fd when mass
is zero there is no E_{p}. However with the true equation just
declared--- E=md--- then we can see that when Force is zero there is still E_{p}
in any system, and not only in a gravitational field.

It is also a fact that
even E=mv^2 is really E=md^2 brought to terms of one second. Therefore we again
notice the E=md part -because we are disputing the squared functor
specifically- any derivation from the ignorantly derived E=Fd into E=m.a.d and
its gravitational derivative--- E=m.g.h is specious because the dimensional
aspects of the terms and units have not been heeded either.

This E=m.a.d farce is all
contrived to declare E=mv^2 However we find that E=m.g.h only ever supports
that squared equation when d is 10m and mass is 1kg. Say mass is five and d is
12. E= 5*10*12 = 600 and the answer from E=mv^2 is 5*TV =5*14^2 = 980?? However
it is of profound interest that E=m.h and E=m.v GET ALMOST THE SAME RESULT
EVERY TIME! The answer to
almost has to do with the 9.81 instead of 10
problem.

Have a think of this: A car
weighing 2000kg travelling at 30m.s will have an energy of 1.8 mJ by E=mv^2. Even
halving that; it remains ridiculous. E=mv gives an energy of only 60kJ. However
with E=2mv we get 120kJ which would be more believable to the motor engineers? Most
real life cars are in that 30-60kw energy/power range.
Note: The difference
between energy and power is academic. A J.s is a w.s except that the Joule is
always tied to one volt, while the watt is free. So kw is the same as kJ.

You see; when the energy
derived by the electrodynamic joule is passed down the electromechanical chain,
it all works well for motors and linear drivers etc. It is only when it all gets down to

Having said that -and as I
have stated elsewhere- I am (was) tending to ascribe to the Idea that E=2mv and
2md. The reason available is that it can be understood that energy is being
injected into the falling object at a constant force over time*. So we can
consider the following equation to calculate that. E=m.F.t ---So if t=1sec then
d is 4.95m so we either have E=m.F.t/2 is equivalent to E=md and mv ---or we
leave E=m.F.t and end up with E=2md and 2mv**. That seems to sit better with
the electrodynamic Joule where 1amp and 1volt through 1ohm will move a 1/2kg
object through 1m in 1sec. (because the 1amp is a Joule.sec) and the 1amp and
1volt will give 1 watt (instantaneous) and one Joule second will therefore
derive one watt second. I must say here that the Joule is not a watt-second nor
visa versa.
Note: the calcs didn't seem to support moving a whole kg through
that distance! E=mv^2 is way too much energy to contemplate here though; even if
the values become more than unity of course, lol.

*This is
just expanding E=md--- the 'F.t' part just derives the 'd' to be considered in
E=md. If you study the fundamental consideration clause at the top, you might
be of the understanding that E=md or mv ('d' per second for equations) might be
just fine.

**That
means E=2m.d in the sidewise case also or the SEP would be violated. Please
note that we don't just invent equations to make our maths work. -just like

We're all familiar with

To get at the truth you need
to derive the result yourself.

Many consider that
gravitational force is greater than mechanical inertial force because of height
derived potential energy. No!! Here's why not---

If I mimic the gravity
situation of opposing forces at rest, and apply a constant force (being physically
withheld) against an object in the sidewise case, then (when I start the clock
and remove the holding mechanism) the situation will be exactly the same as in
the gravitational freefall case, and even the same as though I had just simultaneously
clicked the stopwatch and started pushing with the same force to begin the
experiment. Therefore:
Whether the force is already existing and being
withheld, or whether it has an immediate beginning is of no consequence. It is
the same as the gravitational freefall case. SEP is upheld.
Note: Refer
to conclusion below.

Of course I know that TV
is computed by v=gt! Now to the gravitational freefall--- We have measured 10
(9.81)

**Please
refer to the dispute re Fd or md.

OR THEN THERE'S THIS---

SEP says that we can turn
that whole deal sidewise and apply a similar mechanical force of 10

The sleight of hand
sidewise math goes like this. A force of 10N causes a mass of 10N to move a
distance of 10m with a terminal velocity of 10m.s WTH

So they erroneously compute.
E=10*10 =100J WTH

To arrive at that
conclusion they have deviously turned 1kg of
vertical weight into 10N of
horizontal mass! They have just translated Force into mass without even
blinking!
Note: Weight is only conditionally equivalent to Force and mass in
the gravitational inertial case
--- not at all in the sidewise mechanical
inertial case.

If they don't use that
sleight of hand (by an inability to comprehend or conceptualize such laws and
terms or whatever) then they can easily stoop to this one.

They correctly propose
this: If any constant force causes a movement of 1kg through 10 meters it will derive
10 J--- and also

A constant (any) force of
10N on a 1kg mass causes a terminal velocity of 10 m.s over ten meters.
Therefore--- in one second. So by E=mv we get 10J.

Both of those are correct
but they are just two DIFFERENT ways to derive the
same work/energy.

But then they go and
stupidly multiply the two energies together to derive the specious 100J WTH.

Now not having concluded
the validity of either of those sleights of hand, they now go further into
stupid math land and discover an energy equation called E=m.g.h. or is it all
to provide very dubious support for that ridiculous equation to begin with.
They would have us believe that we live in a world which can allow E=md and
also E=mad or mfd. Such a conclusion is insane!

In any case this E=mgh equation (along with either
of the squared functor derivatives) should ring alarm bells because they won't
work at unity values of E- d- h- g- F etc. This is all because Newton et al
came up with the idea that there is some sort of extra potential energy of
height residing in any lifted object in a gravitational field. Duh! Ever tried
squaring 1, or multiplying three ones together?
Hahahahahahahahahahahahahahahahahahahahahahah!

CONCLUSION

Take the sidewise case. I
can move an object sidewise all over the shop. I can press it up against the
withholding device of the experiment but once released,
it doesn't matter at
all what the force is. It only matters what the distance the object of mass
moved or the EQUVALENCE of terminal velocity of momentum it arrived at.

So the SEP declares
that to be ditto for the gravity case.

Now gravity has a
different problem in that the force is still being applied when the falling
object hits the ground. That's OK we can sort that out by the fact that in the
inelastic theoretical case the Earth is forced to move infinitesimally. That
would be the same in the sidewise case if I slammed an object against a wall
with the same continuing force I used to accelerate it.

In both cases variable
density and elasticity causes a variety of observable results but as I have
stated in the thesis--- Any negative change in the acceleration rate (such as
landing, slamming into etc.) immediately voids the previous analysis or
experiment. The inelastic (hard object and Earth) calculations including the energy
calc (which is exactly as expected) are carried out in the thesis. It turns out
to be that its upon landing that a doubling of the energy is realised because
of the continuing force and when considering energy of deceleration then the
energy used is calculated as if it were being decelerated at 9.81m/s/s. In
other words a collision into another similar object would accelerate the new
object to the same velocity but over the next distance beyond that the force is
still acting so we get a new acceleration and we now have two energies. 1/ of
momentum 2/ of acceleration over the comparable distance to the new momentum of
double the velocity and therefore double the energy. So E=2mv is fairly correct
for freefall into other objects including the earth.

Argument for vis viva Sqrt reduced to single units of value.

Argument to be upheld
(earth
surface, vacuum, frictionless, no getting in the weeds of any irrelevant energy
work conceptualism and also; constant forces are at work unless specified. Note
that freefall occurs to the conceptual point of momentum/or distance only.)

Active premise 1/ The
weight of an object is exactly similar to the mass. The said weight is a true
measure of the constantly acting gravitational force.

DISAGREE--- Go fishing or
take up a different pursuit.

AGREE--- Go to next.

2/ The gravitational force
'g' is 9.80665

3/ The Weak equivalence
principle (WEP) declares this 'g' force to be directly proportional to weight,
which can be truly and conveniently stated as 9.81Newtons (N) per kg. In that
regard all objects of various weights will fall at the same rate (by WEP). Even
though 'g' is ALMOST EQUIVALENT TO BUT NOT EQUAL TO the weight and mass of 1kg,
we will now evaluate that 'g' force as 10Newtons per kg weight and mass; or for
now, just plain (10N) for clarity.

Under Newtonian
measurements, it was observed that an object free falling under the singular
and continuous force of gravity reached a terminal velocity (v) of 10m.s. after
one second. From this we can derive an acceleration rate of 10m.s.s

In a gravity nul in space
or sidewise on earth The 'g' force NO LONGER EXISTS. Therefore the

*If you continue to argue
then I suggest you just go back to the classroom and teach the standard physics
because you are not a scientist. Scientists get serious brain trauma from
analysing concepts like this in depth. Lol

If you are still in doubt
let me be very clear: The biggest mistake that

4/ There are (9.81)
Newtons/kg in 1kgF/kg. Therefore the 'g' force of 9.81N is exactly equivalent
to 1kgF.

DISAGREE--- Go fishing
m'boy!

AGREE--- keep going.

5/ A gravitational force
of 1kgF acting on a 1kg object will cause an acceleration rate which is exactly
equivalent to 3/ above. The pertinent result is that 1kgF results in a terminal
velocity of 10m.s after a fall of about half that. Therefore the energy in the
fall (exactly before any ground impact) is calculable by either E=mv E=10J---
or W=md W=10J.

DISAGREE--- You've come so
far. Learn some mathematical and comprehension skills.

AGREE--- keep up the good
work. OK I'm being patronizing but
please!! The next problem has been let slide for 500 years already!

6/ E=10J. Therefore if we
turn the whole deal on its side in space then according to the SEP the energy
calculation is the same. A 1kg object being pushed by a constantly acting force
of 1kgF will cause a terminal velocity (after one second over a distance of 10) of 10m.s and energy is
calculable by E=mv and ditto as work done by E=md (FORCE AND TIME IRELAVENT).
E=10J

7/ It is now obvious that
there is NO SUCH THING AS POTENTIAL ENERGY OF HEIGHT OTHER THAN WHAT IS
POTENTIALLY AVAILABLE FROM THE 'g' FORCE
ITSELF. By a simple computation we can see that (with all else being equal) the
energy in either a freefall or a sidewise push of ten meters is 10J and not the
100J that the

8/
THE GIANT PROBLEM:
The reason is; that for 500 years the folks have missed the fact that the 10
Newtons is actually the
unity 'g' force which HAS ALREADY BEEN
MULTIPLIED BY TEN IN THE UNIT CONVERSION SO if you miss that YOU END UP WITH A
FALSE MATHEMATICAL EQUATION OF E=10N times 10m =100J. That's not at all correct
because that equation is actually the erroneous E=Fd WTH. Even the correct
equations of E=mv and E=md end up with the specious quantification of E=10N
times 10m WTH. So now a mass is measured in

DISAGREE--- OK then also
evaluate this!

9/ One method (stated in
the literature) for deriving one Joule is that a force of 1N pushing for one
second sidewise on a mass of 1kg derives 1Joule of energy. Such a specious
equation can only get derived from the SEP if you make the stupid mistake of
changing the 'g' force from 10

DISAGREE--- So you're
prolly one of the peeps who reckon that m.s.s should be written like--- m.s^{2}.
That can only lead to another specious derivative of the Joule. The problem
with all that is; that m.s.s IS NOT A FRIGGING EQUATION--- it's an expression
of the change of rate over time; just like the expression of the change of rate
over distance is expressed in m.s.m. How do you square that huh?

DISAGREE--- looks like
another 500 years LMAO.

AGREE--- Welcome to the
truth of vis viva square rooted.

How the E=mv^{2}
was derived.

Back in the 1600s when
Christiaan Huygens was mucking about with his balls, it became noticed by
Leibniz and Bernoulli that objects being dropped into clay* appeared to contain
a squared amount of energy, and due to that it was an easy delusion to consider
that the rate of speed increase in a gravitational field appeared to be by a
squared function as well. That was pretty much the case but Huygens didn't know
about force yet.

So it became the decided
opinion of Mr H that the energy in the accelerating object was being increased
by a squared rate as well, so from that mistake he thought incorrectly that
energy could be derived as mass by velocity squared.

That was actually from a
conceptual mistake because he didn't understand the constant force of gravity
involved. That really meant that a constant force was building energy at a
constant rate which may have APPEARED TO BE SQUARED to him. However if we
consider a constant force instead we should understand that the energy is not
increasing at a squared rate and neither is velocity because the new force per
second is adding energy to that which had already arrived in the system the
second before and so forth and that is the same amount. It is force which
causes the concept of energy and not just velocity per se -because we have the
mass to consider for energy to be derived- but note should be taken that the
rate of increase of velocity and energy is proportional and because the latter
is also proportional to mass, that leads to E=mv. There appears to be a problem
for the distance for E=md or E=1*0.45. that's not true at all. There is no
problem because whenever the falling object reaches the distance of 1m the work
energy will have been done. The rest of the work gets done at the bottom during
the splat. The object at instantaneously evaluated momentum retains a component
of about half the original potential energy of height as Work potential. If you
fall to the ground your body gets to do some serious work. That equation isn't
the object here.

So we now know that the
gravitational acceleration rate is a straight line on the graph at about
10m.s.s. and energy is increased at a steady rate and is calculable by E=mv.
The height to time graph is the one that's curved. Don't get confused.

Unfortunately for science
and Mr Newton who was communicating with Mr H, the squared functor for energy
stuck.

*I have
analysed this clay problem at length and have debunked the conclusions arrived
at.

Problems with energy conservation and a possible violation

Everyone should understand
that momentum gets conserved but energy is not conserved with E=1/2mv^{2}.
The reasons given are laughable. Maybe energy is lost to the environment and
other places in the system. No that's not the reason. That's just losses which results
in less than expected momentum and energy and we are SUPPOSED TO IGNORE THAT
FOR THE EXERCISE.

For example Someone posed
this example: A frictionless skater with a mass of one and a velocity of 10
skates up behind another skater and gives them a big hug and they both go
skating off with a combined mass of 2 and a velocity of 5. Momentum is
conserved. In The case of E=P=mv then energy is also conserved.

Let's try vis viva--- E=mv^{2}
will do. First with one skater E=1*10^{2} =100J.

Now for the huggers. E=1*5^{2}.
There is actually an energy conservation violation.

Please don't make an
excuse that the energy is being dissipated due to losses. You idiot! We are
dealing with theoretical lossless functions all round! You can't just bring in
entropy whenever you want. There are no clay balls in this theoretical world.
All objects are perfectly elastic and they share their momentums, rebounds and
energies perfectly. There is a mathematical violation of the mechanical energy
conservation law.

It doesn't work you morons
--- not even halved. Get a grip and fix this! Yes ad hominem time. It's been
nearly 500 years!