A rolling&nbsp;disk&nbsp;on a straight line surface.

This upcoming mission of the Shuttle to the ISS will at last address many of the issues that have been ongoing and of major concern to me. For many years I have had fears that the poor design of the CMGs (Gyroscopes) on board the ISS could end up in disaster. Well this launch is not only going to bring up a spare CMG, it's taking up two spares! I am very happy that NASA has decided to do this.They have also adopted some of my suggestions about making some smaller gyros eventually that could be lifted by smaller craft.

Will be a shame to lose the Shuttles when they are retired, because they were unique in their ability to lift so much mass into orbit. Nothing else is big enough to lift one of those massive gyros, let alone 2 of them. So we will lose an important space exploration tool when the Shuttles are gone. Here is a link to one story about this mission, and a few quotes from the link for your review;

Atlantis mission to fortify future of the space station
http://www.spaceflightnow.com/shuttle/sts129/091114preview/

"With just six missions left on NASA's shuttle manifest between now and the end of fiscal 2010, Atlantis' mission is one of two devoted primarily to delivering critical spare parts and equipment - orbital replacement units, or ORUs - that are too large to be delivered by European, Russian or Japanese cargo ships.

Mounted on pallets in Atlantis' payload bay are two spare control moment gyroscopes, used to control the station's orientation in space; a high pressure oxygen tank for the station's airlock; and a spare pump module, nitrogen tank and an ammonia reservoir for the lab's cooling system.

The pallets also carry a replacement robot arm latching end effector, or mechanical hand; a spare power cable spool used by the arm's mobile transporter; a solar array battery charge-discharge unit; and a device used to prevent potentially dangerous electrical arcs between the station and the electrically charged extreme upper atmosphere."

Mystery

_______________
--- In gravitationalpropulsionstevenson@yahoogroups.com, "mysterygravity" <mysterystevenson1@...> wrote:

Again the CMG's are becoming an issue on the ISS. Again I feel it
is possible to supply supplimental gyroscopic systems before it is too late, if that is within our political will. The issues are being blamed
on computer glitches and on and on, but stresses on the entire station
are being "amassed". Here is a current link as to breaking news,but
think there will be more;

http://www.spaceflightnow.com/shuttle/sts117/070612cmgs/
http://www.spaceflightnow.com/shuttle/sts117/070612cmgs/

There is more about recent testing of the one down CMG and yet sources
are continuously mentioning the 4 Gyro system as though all
function...See near the end of this BBC story;

http://bbsnews.net/article.php/2007051521254433
http://bbsnews.net/article.php/2007051521254433

Here are some older links to our posts on similar subjects ;

http://tech.groups.yahoo.com/group/gravitationalpropulsionstevenson/message/45
http://tech.groups.yahoo.com/group/gravitationalpropulsionstevenson/message/45

http://tech.groups.yahoo.com/group/gravitationalpropulsionstevenson/message/86
http://tech.groups.yahoo.com/group/gravitationalpropulsionstevenson/message/86

Please note that there is a copy of one of my communications to NASA
concerning this issue from last year in our private files area in the
"Time Capsule" in the group. Also in the public links area of the group is a link to a free video link with it's own player that has as one of
it's channels a NASA link for live viewing of what is happening.There
are continuing issues almost hourly and so am holding back on
indications that are now unfolding...

Note ; unrelated that a sister group to this group has been
started and soon will release invitations to join in the expansion.

Mystery [B-)]
____________________________

#207

From: "abelov0927" <abelov0927@...>
Date: Wed Oct 28, 2009 12:21 am
Subject: Prove existing classical mechanics laws with additional natural phenomenon.

abelov0927

Offline

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9 #

A Rotation with Translation Movement is standalone natural phenomenon

For this experiment, two identically thin cylinders which are initially static to the observer are taken. These cylinders are attached with internal mechanical springs that induce a repulsive action between them.

Two experiments are to be conducted.

The first experiment.

An action that repels the two cylinders is initiated. This action is induced from their center of mass. Now the cylinders are observed to move in directions opposite to each other. Based on the law of conservation of momentum, the linear velocities of the cylinders are equal. The kinetic energies of the cylinders are deduced using the below derivation:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+0(\text{no rotation})$

$v_1=v_2\Rightarrow (m\frac{v_1^2}{2}+0) = (m\frac{v_2^2}{2}+0) \text{ , }T_1=T_2$

Thus it is inferred that their kinetic energies are identical. Now, lets compare their translation energies and angular momentum.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2 = mv_2 \text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }\omega_1=\omega_2=0$

$mv_1=mv_2\Rightarrow P_1=P_2\text{ , }0=0\Rightarrow L_1=L_2$

It is inferred that their momentums are identical as well.

This experiment thereby proves that the two cylinders exhibit a symmetric action due to the initial repulsive inducement of their translation movement.

The second experiment.

Initiate the same repulsing action as done in experiment 1, but induce the action in different positions. Induce one cylinder from its center mass and the other from its edge. On observing the movement of the cylinders, the following are noticed:

The two cylinders move in opposite directions

The cylinder that is induced from its center mass alone rotates

Assuming their linear velocities are identical according to the law of conservation of momentum, their kinetic energies are determined:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}(\text{has rotation})$

$v_1=v_2\text{ , }0\ne\omega_2\text{ }\Rightarrow \text{ } (m\frac{v_1^2}{2}+0) \ne (m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}) \text{ , }T_1\ne T_2$

From the above, equation it is inferred that their kinetic energies are not identical. Now, as in the previous experiment, their translation energies and angular momentum are compared.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2=mv_2\text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }0\ne\omega_2$

$mv_1=mv_2\Rightarrow P_1=P_2 \text{ , }0\ne I\omega_2\Rightarrow L_1 \ne L_2$

Hence it is deduced that their linear momentums are identical. The values of the angular momentums of the two cylinders are found to deviate and the cylinder initiated from its center mass alone has an angular momentum.

Thus it is assumed that the action is not symmetric, this assumption is made by relating the action to the initial repulsing event. But this presumption fails to follow the law of conservation of angular momentum.

These deviations are however based on faulty assumptions. The linear velocities of these cylinders are not identical.

A single event can not induce two different movements in a single object. One event can trigger only one kind of movement in each object, in this case the cylinders.

The first experiment conforms to this rule, as there is only one translational movement that is induced.

In the second experiment, one cylinder exhibits only one type of movement, whereas, two types of movements are observed for the other cylinder. Though both the cylinders are stimulated using only one event which implies that each cylinder must exhibit only one type of movement, one cylinder exhibits translation movement and rotational movements. These two movements are thus considered as new type of movements. These movements are hence standalone natural phenomenon. So it follows, the movement should have its momentum and follow its own conservation of momentum.

Assuming the movement has a linear and angular momentum, the total momentum of rotation with translation movement is:

$P_f= \sum P_j +\frac{1}{R_u}[\sum L_k]$

Where, Pj - linear momentum Lk - angular momentum Ru - unit radius

The law of conservation of momentum for the translation movement with rotation is:

$\sum P_j +\frac{1}{R_u}[\sum L_k] = Const$

This movement has two components: rotation and translation. How are these two correlated to the momentum? To answer this lets consider the following diagram of a rotating body.

The mass of the rotating body is concentrated on its radius (Ru). An initial force Pf strikes the body which is at a distance h from the center of mass of the body. It is known that the moment of inertia is:

$I=m(R_u)^2$

Which implies that:

$R_u=\sqrt{\frac{I}{m}}$

Assuming this momentum is applied to the angular movement:

$P_f\frac{h}{R_u+h}$

So, the angular momentum is:

$L = P_f\frac{hR_u}{R_u+h}$

Translation momentum is then equal to:

$P = P_f-P_f\frac{h}{R_u+h} = P_f\frac{R_u}{R_u+h}$

Lets sum these parts and check full momentum:

$P_f=P_f\frac{R_u}{R_u+h}+\frac{1}{R_u}\times P_f\frac{hR_u}{R_u+h}=P_f(\frac{R_u}{R_u+h}+\frac{h}{R_u+h})=P_f$

Adding both these equations, the total momentum is determined.

Following the new law of momentum conservation for translation with rotation movement, the translation and angular velocities of the cylinders have a different value on experiment 2.

Translation and angular velocities for cylinder 1 equal to:

$v_1=P_f\frac{1}{m}$

$\omega_1=0$

Translation and angular velocities for cylinder 2 equal to:

$v_2=P_f\frac{R_u}{m(R_u+h)}$

$\omega_2 = P_f\frac{hR_u}{I(R_u+h)}$

$R_u=\sqrt{\frac{I}{m}}$

Hence, based on the previous statement translation with rotation movement corresponds to the consolidated translation and rotation movements. It is thus inferred that one of the movement is the result of the other primary complex movement.

The above experiments are simulated.

The experiment 1.

The simulation elicits results that relate to the theory.

Initial results: Before start

Values during experiment:

The experiment 2.

Initial results:

Before start

Values during experiment:

Let's simulate same result from experiment using additional torque.

Initial results:
Before start

Values during experiment:

It is inferred that the experiment 1 and 2 with additional torque produced the same results.

The red rod in experiment 2 follows these equations:

$P = m_r\times v_r=F_i\times t_i$

$L=I\times \omega=F_i\times t_i \times R$

Where, P - rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti - initial pulse force time L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, R - rod's unit radius

The red rod in experiment follows these equations:

$P = m_r\times v_r=F_i\times t_i_1$

$L=I\times \omega=\tau \times t_i_2$

Where P - rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti1 - initial pulse force, ti2 - initial torque time L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, tau - torque.

These experiments use different parameters for the equations of angular momentum. For this equation the simulator for experiment 2 uses same initial pulse force. However, for experiment with additional torque simulator uses another parameter (torque). The torque should have a pair on real world. However, the simulator can trigger unpaired torque for one body.

experiment 1 + additional torque = experiment 2

The two rods have same mass and moment of inertia.
However, in experiment 2, the rotating rod exhibits extra torque, which is simulated easily.

Observe experiment 3

The simulator applies extra translation force for rotation rod, which is not the case in real world. A fraction of applied force is spent for rotation movement (torque) then this fraction of force should be detected from the force applied for tranalation. The sum of these rotation and translation fractions of applied force should be equal to applied force for non-rotating rod. (3-rd Newton's law)

The translation velocities of these rods should vary. But they are observes, to be equal on simulator. Because simulator is unaware about rotation with translation movement and hence, the independent rotation and translation movements make it difficult to make inferences from experiment 2. The classical mechanics should include new standalone translation with rotation movement to describe natural phenomenon correctly.

The following are the animations recorded for experiment 2:

This animation follows classical mechnics laws:

This animation follows theory of standalone rotation with translation movement.

The Natural Experiment 2.

3 successful experiments were conducted with 2 pencils.
In these experiments pencils with rotation movement have lower velocity than pencils without rotation.

The theory is CORRECT.

The simulator is WRONG.

Equipment: 2 pencils thread and thin rubber band 3''

The rubber band is repulsing 2 objects (2 pencils).

The mass of the rubber band is much less that the mass of the pencil

The followng are the snapshots of the experiment dynamics.

Links to experiments movies (avi files)
Experiment 2_1
Experiment 2_2
Experiment 2_3

Links to experiments pictures (zip files)
Experiment 2_1
Experiment 2_2
Experiment 2_3

Conclusion

The experiments using the pencils prove the theory that rotation with translation movement is standalone natural phenomenon. This new movement is shown to have its own law of momentum conservation. This theory conforms to the existing classical mechanics laws. This new theory that is framed through this experiment proves existing classical mechanics laws and gives additional natural phenomenon explanations.

email: abelov0927@...

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#206

From: "Jim S" <mysterystevenson1@...>
Date: Tue Oct 6, 2009 7:34 pm
Subject: GRAVITY NEWS 11

mysterysteve...

Offline

GRAVITY NEWS 11

Gravity news resumes as before...

Miniature gravity detector could peer inside planets.
http://www.newscientist.com/article/mg20327206.400-miniature-gravity-detector-could-peer-inside-planets.html
PEERING beneath the surface of Mars and other planets to reveal buried geological features could get easier, thanks to a nifty new silicon gadget.
The device, called a gravity gradiometer, has been designed to measure how much the force of gravity changes from place to place, enabling it to map a planet's gravitational field

NRAO Discovery Explains Gravitational Phenomenon
http://www.stpns.net/view_article.html?articleId=106532610632776013037
"Measuring the curvature of space caused by gravity is one of the most sensitive ways to learn how Einstein's theory of General Relativity relates to quantum physics. Uniting gravity theory with quantum theory is a major goal of 21st-Century physics, and these astronomical measurements are a key to understanding the relationship between the two," said Sergei Kopeikin of the University of Missouri.

Celestial 'surfing' offers hope of cheap and efficient space travel
http://www.timesonline.co.uk/tol/news/science/space/article6829727.ece
He added: "I like to think of them as similar to ocean currents, except they are gravitational currents." Just as ships could drift around the world on ocean currents, so spacecraft could coast the solar system. "You could do it essentially for free, just using a little bit of fuel for course corrections," Professor Ross said.

Some interesting concepts for review in interstellar propulsion;
http://www.k2climb.net/news.php?id=18513

Qinetiq to supply propulsion system
http://www.theengineer.co.uk/Articles/312879/Qinetiq+to+supply+propulsion+system.htm
Qinetiq announced today it will supply a solar-electric propulsion system for the European Space Agency's (ESA's) BepiColombo spacecraft mission to Mercury.
The defence company has been awarded a �23m contract by EADS Astrium to provide the technology.

Extensive Spiral Corrugations (Saturn's rings, gravitational effect)
http://www.nasa.gov/mission_pages/cassini/multimedia/pia11664.html
These images and others like them are only possible around the time of Saturn's equinox which occurs every half-Saturn-year (equivalent to about 15 Earth years). The illumination geometry that accompanies equinox lowers the sun's angle to the ring plane, significantly darkens the rings, and causes out-of-plane structures to cast long shadows across the rings. Cassini's cameras have spotted not only the predictable shadows of some of Saturn's moons (see PIA11657), but also the shadows of newly revealed vertical structures in the rings themselves (see PIA11665).

GOCE Delivering Data For Best Gravity Map Ever
http://www.spacedaily.com/reports/GOCE_Delivering_Data_For_Best_Gravity_Map_Ever_999.html
Following the launch and in-orbit testing of the most sophisticated gravity mission ever built, ESA's GOCE satellite is now in 'measurement mode', mapping tiny variations in Earth's gravity in unprecedented detail.

Enjoy all,

Mystery

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#205

From: "abelov0927" <abelov0927@...>
Date: Fri Sep 11, 2009 5:08 am
Subject: Re: Program glitch or discovery?

abelov0927

Offline

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

I made 3 successful experiments with 2 pencils.
All these experiment shows different velocities between pencils.
The theory is CORRECT.

The simulator is WRONG.
Here is some snapshots from experiment, which shows experiment dynamic.

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> Example from real world.
> (Thank you to D H)
>
> Thrusters can be used in pairs to help eliminate this cross-coupling between translational and rotational acceleration. This is not always possible. For example, consider the Mars Climate Orbiter. From http://www.jamesoberg.com/mars/loss.html
>
> "The use of jet thrusters for attitude control raises further operational issues. In a world of perfect symmetry and unlimited payload size and budget, a spacecraft could rotate cleanly about its center of mass (often carelessly called its center of gravity) if opposing ends were equipped with jets and if those jets pointed in opposite directions and were set at right angles to the axis to be turned. In the real world, rotational jets may not be arranged in such a theoretically perfect alignment. ... On the Mars Climate Orbiter, four separate clusters of jets were located around the vehicle's waist. However, because of the large solar array extending from one side, the craft's center of mass did not coincide with the center point of the waist. Thus there was a significant imbalance each time these small thrusters fired."
>
> The Mars Climate Orbiter needed to use its attitude jets once or twice a day to dump angular momentum from its momentum wheels. The use of these attitude jets also made the vehicle accelerate translationally. Failing to account for this properly was the ultimate cause of the demise of the Mars Climate Orbiter. The mission controllers did try to account for this undesired acceleration, but there was a units mixup. The vehicle reported the undesired acceleration as a number in English units. The controllers interpreted that number to be in metric units.
>
> "According to a JPL spokesman, every maneuver intended to dump momentum added a velocity error of about 0.001 meter per second, on a probe that was traveling at a rate of tens of kilometers per second. These deflections themselves were not the problem, but their incorrect modeling was, when the computer was told the spacecraft had received a force of four or five times as great as it really had."
>
> The end result: The Mars Climate Orbiter entered Mars atmosphere, something it was never intended to do. The vehicle most likely blew up 70 kilometers or so above Mars.
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:
>
> Hi
> I updated my site.
> I add steps how to build these experiments on working model program (demo version)
> Please take a look.
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics/1xmqm1l0s4ys/9#
> Thanks
> Alex
>

#204

From: "abelov0927" <abelov0927@...>
Date: Mon Aug 31, 2009 2:49 am
Subject: Re: Program glitch or discovery?

abelov0927

Offline

Example from real world.
(Thank you to D H)

Thrusters can be used in pairs to help eliminate this cross-coupling between
translational and rotational acceleration. This is not always possible. For
example, consider the Mars Climate Orbiter. From
http://www.jamesoberg.com/mars/loss.html

"The use of jet thrusters for attitude control raises further operational
issues. In a world of perfect symmetry and unlimited payload size and budget, a
spacecraft could rotate cleanly about its center of mass (often carelessly
called its center of gravity) if opposing ends were equipped with jets and if
those jets pointed in opposite directions and were set at right angles to the
axis to be turned. In the real world, rotational jets may not be arranged in
such a theoretically perfect alignment. ... On the Mars Climate Orbiter, four
separate clusters of jets were located around the vehicle's waist. However,
because of the large solar array extending from one side, the craft's center of
mass did not coincide with the center point of the waist. Thus there was a
significant imbalance each time these small thrusters fired."

The Mars Climate Orbiter needed to use its attitude jets once or twice a day to
dump angular momentum from its momentum wheels. The use of these attitude jets
also made the vehicle accelerate translationally. Failing to account for this
properly was the ultimate cause of the demise of the Mars Climate Orbiter. The
mission controllers did try to account for this undesired acceleration, but
there was a units mixup. The vehicle reported the undesired acceleration as a
number in English units. The controllers interpreted that number to be in metric
units.

"According to a JPL spokesman, every maneuver intended to dump momentum added a
velocity error of about 0.001 meter per second, on a probe that was traveling at
a rate of tens of kilometers per second. These deflections themselves were not
the problem, but their incorrect modeling was, when the computer was told the
spacecraft had received a force of four or five times as great as it really
had."

The end result: The Mars Climate Orbiter entered Mars atmosphere, something it
was never intended to do. The vehicle most likely blew up 70 kilometers or so
above Mars.

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:

  Hi
  I updated my site.
  I add steps how to build these experiments on working model program (demo
version)
Please take a look.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics/1xmqm1l0s4ys/\
9#
  Thanks
  Alex

#203

From: "abelov0927" <abelov0927@...>
Date: Fri Aug 28, 2009 9:28 pm
Subject: Re: Program glitch or discovery?

abelov0927

Offline

Hi
I updated my site.
I add steps how to build these experiments on working model program (demo
version)
Please take a look.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#
Thanks
Alex



--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
> I did my own simulations.
>
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#
> The experiment 2 has extra rotation for one rod. This additional torque come
into one rod only. This torque doesn't have pair with negative value.
> I made another simulation where substitute initial momentum displacement to
additional torque. And I got same result.
>
> experiment 1 + additional torque = experiment 2
>
> This experiment 2 show how generate additional torque for Isolated System. The
initial pulse force displacement between experiment 1 and 2 should not produce
extra torque or angular momentum.
>
> Adding one torque without its own pair is breaking law of angular momentum
conservation.
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@> wrote:
> >
> > Base on this theory
> >
[url]http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1\
l0s4ys/9#[/url]
> > Simulator got this glitch.
> > Is it discovery?
> >
>

#202

From: "abelov0927" <abelov0927@...>
Date: Thu Aug 27, 2009 2:43 am
Subject: Re: Program glitch or discovery?

abelov0927

Offline

I did my own simulations.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#
The experiment 2 has extra rotation for one rod. This additional torque come
into one rod only. This torque doesn't have pair with negative value.
I made another simulation where substitute initial momentum displacement to
additional torque. And I got same result.

experiment 1 + additional torque = experiment 2

This experiment 2 show how generate additional torque for Isolated System. The
initial pulse force displacement between experiment 1 and 2 should not produce
extra torque or angular momentum.

Adding one torque without its own pair is breaking law of angular momentum
conservation.

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
> Base on this theory
>
[url]http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1\
l0s4ys/9#[/url]
> Simulator got this glitch.
> Is it discovery?
>

#201

From: "abelov0927" <abelov0927@...>
Date: Tue Aug 18, 2009 5:27 am
Subject: Program glitch or discovery?

abelov0927

Offline

Base on this theory
[url]http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1\
l0s4ys/9#[/url]
Simulator got this glitch.
Is it discovery?

#200

From: "abelov0927" <abelov0927@...>
Date: Tue Aug 18, 2009 3:59 am
Subject: Re: Gravitational Propulsion, A Rotation with Translation Movement is standalone natural phenomenon(version 2)

abelov0927

Offline

Hi
I reply message from other group.
It would be helpfull for you.
Have a fun.
Alex
...
Hi Alex,

I'm not sure to understand all what you mean and may be I'm wrong on some
points.
So here is my simulation to clarify the problem (a hard work :-):
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.wm2d

If you have not Workingmodel software (it is free), here is the video (6Mb,
probably long to download):
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.avi

Here are 3 jpg snapshots. N�0 is the initial position. N�1 is the state just
after the pulse. N�2 is some seconds later.
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com0.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com1.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com2.jpg

The speed vector of each rod is displayed in the animation. The impulse force it
not displayed. It provides 100 N during 0.01s between the rods, at points of
same coordinates (= the position of the center of mass of the green rod).

In the red and green small windows, you can monitor the speeds, kinetic energies
and momenta of each rod. The default equations of the program are used.

In the small blue windows, I have put the equations to obtain the angular speed,
the moment of inertia and the angular momentum of each rod, calculated from the
common center of mass of the system.
This center of mass is displayed in the simulation (it is calculated by the
program, it's not a fixed point. It is at rest because there is no external
force acting onto the system. It is the mid-point of the line joining the
centers of mass of the two rods).
To simplify the calculi, the positions of the rods are chosen in order the
common center of mass to be at position x,y = 0,0.

With this simulation, I have discovered that from the common center of mass C,
each rod possesses a "hidden" angular momentum: each rod flies horizontally away
one another but one is above and the other under the horizontal x axis
containing their common center of mass. Thus the angle delimited by the
horizontal x axis and the line joining the centers of mass of the two rods
(crossing at C), decreases when the distance of the rods from the origin
increases. This variation of the angle is to be considered as an angular
velocity. Then from it, we can calculate the moment of inertia of each rod and
its angular momentum. The result is displayed in the small blue windows.

IMPORTANT : We see that the sum of the angular momenta of the two rods,
calculated in the referential frame of their common center of mass, is equal but
with opposite sign, to the angular momentum of the only rotating rod, calculated
in its proper frame.
Thus by adding these two momenta, we find zero. I guess the key of the problem
lies around this, but it is not yet completely clear for me.

Fran�ois

--- In gravitationalpropulsionstevenson@yahoogroups.com, Katriel Porth
<batl4etrnity@...> wrote:

I have a few simple questions...
  Very concisely, please briefly and simplisticly explain the following...

  1. What is gravity? How does it work and what does it affect?

  2. What is antigravity? How does it work and what does it affect?

Shalom and thank you.
  -Katriel

#199

From: "mysterystevenson1" <mysterystevenson1@...>
Date: Sat Aug 15, 2009 5:38 am
Subject: Re: Gravitational Propulsion, A Rotation with Translation Movement is standalone natural phenomenon(version 2)

mysterysteve...

Offline

http://www.geocities.com/mysterystevenson1/Definitions.html

Hello Katriel,

I only noticed your post just now, but would like to add a partial answer to your question # 2 as to antigravity.

We have this in our files and at a few places on the web, but thought to quickly repeat it for those that may have not seen it.

ANTI-GRAVITY ;
Any effect produced by a multitude of devices to repel, nullify, or interact with gravity, in order to propel said devices.For more see;
http://tech.groups.yahoo.com/group/antigravity

This definition is also located on one site that is about to be deleted along with all sites on Geocities and so would be a good time to copy this group of definitions if you have not done so at;

That link will very soon be gone!

That is only a very broadbased definition of antigravity and does need much modification to become encyclopedic in nature. There is also an antigravitational natural effect that is not described by that definition, however it may be very closely inclusive within that definition as to effect without being actually caused of course by a manmade device. I will attempt to offer more on this in the future as well as some descriptive data as to gravity, Q#1. This is just a quick reply, for now, until I have time to go into more depth. Perhaps some of the others would like to offer their comments in the meantime especially abelov0927 , forgive me if you intended this question only for abelov0927 .

Mystery

--- In gravitationalpropulsionstevenson@yahoogroups.com, Katriel Porth <batl4etrnity@...> wrote:

I have a few simple questions...

Very concisely, please briefly and simplisticly explain the following...

1. What is gravity? How does it work and what does it affect?

2. What is antigravity? How does it work and what does it affect?
Shalom and thank you.

-Katriel

#198

From: Katriel Porth <batl4etrnity@...>
Date: Fri Aug 14, 2009 9:27 pm
Subject: Re: Gravitational Propulsion, A Rotation with Translation Movement is standalone natural phenomenon(version 2)

batl4etrnity

Offline

I have a few simple questions...

Very concisely, please briefly and simplisticly explain the following...

1. What is gravity? How does it work and what does it affect?

2. What is antigravity? How does it work and what does it affect?

Shalom and thank you.

-Katriel

--- On Mon, 8/10/09, abelov0927 <abelov0927@...> wrote:

From: abelov0927 <abelov0927@...>
Subject: Gravitational Propulsion, A Rotation with Translation Movement is standalone natural phenomenon(version 2)
To: gravitationalpropulsionstevenson@yahoogroups.com
Date: Monday, August 10, 2009, 11:14 AM

A Rotation with Translation Movement is standalone natural phenomenon.

Statement of proof.

Let's take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.
Let's make two experiments.

http://knol. google.com/ k/alex-belov/ paradox-of- classical- mechanics- 2/1xmqm1l0s4ys/ 9#

The first experiment.

The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+0(\text{no rotation})$

$v_1=v_2\Rightarrow (m\frac{v_1^2}{2}+0) = (m\frac{v_2^2}{2}+0) \text{ , }T_1=T_2$

As it shown on equation their energies are equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2 = mv_2 \text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }\omega_1=\omega_2=0$

$mv_1=mv_2\Rightarrow P_1=P_2\text{ , }0=0\Rightarrow L_1=L_2$

As it shown on equation their momentums are equal.
For this experiment these cylinders have symmetric action as translation movement relatively to initial event (repulsing) and observer.

The second experiment.

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let's assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}(\text{has rotation})$

$v_1=v_2\text{ , }0\ne\omega_2\text{ }\Rightarrow \text{ } (m\frac{v_1^2}{2}+0) \ne (m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}) \text{ , }T_1\ne T_2$

As it shown on equation their energies are not equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2=mv_2\text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }0\ne\omega_2$

$mv_1=mv_2\Rightarrow P_1=P_2 \text{ , }0\ne I\omega_2\Rightarrow L_1 \ne L_2$

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.
This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can't start own rotating without symmetrical action.

This mean the assumption about identical linear velocity was wrong.

One even can't reproduce two type of movements together for one object. One event can reproduce only one type of movement per object.
On experiment 1 both objects have only one translation movement.
The experiment 2 shows only one type of movements for one object and two types of movements for other object. However these objects taked only one event(repulsing) . It means it should be one type of movements per object and translation with rotation movements for this experiment should be described as new type of moments.
The experiment 2 shows situation where the translation with rotation movement as a standalone natural phenomenon. This movement should have own momentum and law of momentum conservation.

My assumption this movement have a linear and angular momentums together.

The full momentum of rotation with translation movement is:

$P_f= \sum P_j +\frac{1}{R_u}[\sum L_k]i$

note: Pj - linear momentum Lk - angular momentum Ru - unit radius

This is the law of momentum conservation for the translation with rotation movement:

$\sum P_j +\frac{1}{R_u}[\sum L_k]i = Const$

This law has a complex number and it has linear and angular momentums.

Full momentum transfer for this movement has calculation by absolute value:

$Z^2=[\sum P_j]^2 +[\frac{1}{R_u}\sum L_k]^2$

For example: if reverse time back for experiment 2 then one cylinders angular L2 and translation P2 momentums compensated by another cylinders only translation momentum P1.
For this case equation between momentums is:

$(P_1)^2=(P_2)^2 +(\frac{1}{R_u}L_2)^2$

Full kinetic energy of rotation with translation movement is:

$E=m\frac{v^2}{2}+I\frac{w^2}{2}$

Follow law of momentum conservation and using complex number for translation with rotation momentum, the cylinders translation velocities have a different value on experiment 2.

======

Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

email: abelov0927@gmail. com

#197

From: "abelov0927" <abelov0927@...>
Date: Mon Aug 10, 2009 4:14 pm
Subject: A Rotation with Translation Movement is standalone natural phenomenon(version 2)

abelov0927

Offline

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

A Rotation with Translation Movement is standalone natural phenomenon.

Statement of proof.

Let's take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.
Let's make two experiments.

The first experiment.

The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+0(\text{no rotation})$

$v_1=v_2\Rightarrow (m\frac{v_1^2}{2}+0) = (m\frac{v_2^2}{2}+0) \text{ , }T_1=T_2$

As it shown on equation their energies are equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2 = mv_2 \text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }\omega_1=\omega_2=0$

$mv_1=mv_2\Rightarrow P_1=P_2\text{ , }0=0\Rightarrow L_1=L_2$

As it shown on equation their momentums are equal.
For this experiment these cylinders have symmetric action as translation movement relatively to initial event (repulsing) and observer.

The second experiment.

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let's assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}(\text{has rotation})$

$v_1=v_2\text{ , }0\ne\omega_2\text{ }\Rightarrow \text{ } (m\frac{v_1^2}{2}+0) \ne (m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}) \text{ , }T_1\ne T_2$

As it shown on equation their energies are not equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2=mv_2\text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }0\ne\omega_2$

$mv_1=mv_2\Rightarrow P_1=P_2 \text{ , }0\ne I\omega_2\Rightarrow L_1 \ne L_2$

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.
This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can't start own rotating without symmetrical action.

This mean the assumption about identical linear velocity was wrong.

One even can't reproduce two type of movements together for one object. One event can reproduce only one type of movement per object.
On experiment 1 both objects have only one translation movement.
The experiment 2 shows only one type of movements for one object and two types of movements for other object. However these objects taked only one event(repulsing). It means it should be one type of movements per object and translation with rotation movements for this experiment should be described as new type of moments.
The experiment 2 shows situation where the translation with rotation movement as a standalone natural phenomenon. This movement should have own momentum and law of momentum conservation.

My assumption this movement have a linear and angular momentums together.

The full momentum of rotation with translation movement is:

$P_f= \sum P_j +\frac{1}{R_u}[\sum L_k]i$

note: Pj - linear momentum Lk - angular momentum Ru - unit radius

This is the law of momentum conservation for the translation with rotation movement:

$\sum P_j +\frac{1}{R_u}[\sum L_k]i = Const$

This law has a complex number and it has linear and angular momentums.

Full momentum transfer for this movement has calculation by absolute value:

$Z^2=[\sum P_j]^2 +[\frac{1}{R_u}\sum L_k]^2$

For example: if reverse time back for experiment 2 then one cylinders angular L2 and translation P2 momentums compensated by another cylinders only translation momentum P1.
For this case equation between momentums is:

$(P_1)^2=(P_2)^2 +(\frac{1}{R_u}L_2)^2$

Full kinetic energy of rotation with translation movement is:

$E=m\frac{v^2}{2}+I\frac{w^2}{2}$

Follow law of momentum conservation and using complex number for translation with rotation momentum, the cylinders translation velocities have a different value on experiment 2.

======

Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

email: abelov0927@...

#196

From: "abelov0927" <abelov0927@...>
Date: Mon Aug 3, 2009 3:11 am
Subject: Re: Multidimensional physics

abelov0927

Offline

My assumption this movement have a linear and angular momentums together.
This is the law of momentum conservation for the translation with rotation movement:

$\sum P_j +[\sum L_k]i = Const$

This law has a complex number and it has linear and angular momentums.
One of these parts linear or angular should be imaginary number. It depends on observer which frame of reference he uses.

Full momentum transfer for this movement has calculation by absolute value:

$Z^2=[\sum P_j]^2 +[\sum L_k]^2$

---

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#195

From: "abelov0927" <abelov0927@...>
Date: Thu Jul 30, 2009 3:20 am
Subject: Re: Multidimensional physics

abelov0927

Offline

The rotated cylinder center mass has slowest linear velocity than cylinder without rotation on repulsion action.

The statement of proof.

Let's take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.
Let's make two experiments.

The first experiment.
The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(no_._._.rotation)$

$T_2=m\frac{v_2^2}{2}+0(no_._._.rotation)$

$v_1=v_2\Rightarrow T_1=T_2$
As it shown on equation their energies are equal.
Let's take the derivatives from their parts of energy.

$\dot E_t_1 = mv_1$

$\dot E_t_2 = mv_2$

$v_1=v_2\Rightarrow mv_1=mv_2$
As it shown on equation their momentums are equal.
This experiment action is symmetric relatively to observer.

The second experiment.
The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let's assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let's measure their kinetic energies. The full kinetic energy is equal to:
$T_1=m\frac{v_1^2}{2}+0(no_._._.rotation)$

$T_2=m\frac{v_2^2}{2}+I\frac{\omega^2}{2}(has_._._.rotation)$

$v_1=v_2_._._._.\omega_1\ne\omega_2\Rightarrow T_1\ne T_2$

As it shown on equation their energies are not equal.
Let's take the derivatives from their parts of energy.

$\dot E_t_1 = mv_1_._._._._._. \dot E_r_1=0$

$\dot E_t_2=mv_2_._._._._._. \dot E_r_2=I\omega_2$

$v_1=v_2_._._._.\omega_1\ne\omega_2\Rightarrow mv_1=mv_2_._._._. 0\ne I\omega_2$

This assumption broke law of angular momentum conservation. The body can't start own rotating without symmetrical action.
This mean the assumption about identical linear velocity was wrong.

These experiments is explaining the translation with rotation movement is standalone natural phenomenon. And law of momentum conservation should cover this movement.

My assumption this movement have a linear and angular momentum together.
This is the law of momentum conservation for translation with rotation movement.

$[\sum P_1_i]^2 +[\sum L_1_i] ^2 = [\sum P_2_i]^2 +[\sum L_2_i] ^2$

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#194

From: "abelov0927" <abelov0927@...>
Date: Sat Jul 25, 2009 2:30 am
Subject: Multidimensional physics

abelov0927

Offline

The classical mechanic laws were written for one dimension interactions.
I found some cases where these laws should be corrected.
This means the classical mechanics laws should be reviewed for multidimensional
interactions.
One of these simple visual examples is - the cylinder which takes a momentum
asymmetrically (away from his center mass).
The classical physics disallow convert translation momentum to angular momentum.
Because the classical physic allow to convert one type of movement to ONLY one
other. But it can't be start two movements together.
However Mother Nature gives us many examples where one type of movement
interacts to two types of movements together. The problem with cylinder - one of
them. This is not covering by classical mechanics laws. It just substitutes
solutions for any of these types of movements. We should choose this type of
movements first. This is the problem of classical mechanic laws. The momentum
conserve, however the generic law of momentum conservation for multidimensional
movement is covering linear and angular momentums
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#

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#193

From: "abelov0927" <abelov0927@...>
Date: Thu Jul 23, 2009 11:29 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

Let's solve simple problem.

Given a long thin cylinder with length L, radius R and mass M.

Case 1.
This cylinder takes momentum P into his center mass.

Case 2.
This cylinder takes momentum P on a some distance from his center mass.
The cylinder start rotation with translation movement.

How many times the translation velocity for case 2 is lower than translation velocity for case 1?

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

If calculate this problem using classical mechanics laws, the linear momentum should be conserve. However, these cases take different energies. The cylinder with translation and rotation movements will take two parts (translation and rotation) of energies. The cylinder with translation takes only one( translation) part. These translation kinetic energy parts should be equal.
Where this rotation kinetic energy come from? From nowhere? What about angular momentum? How it starts rotating if all momentum transfers to translation part?

If assume for case 2 the incoming momentum vector dividing for two parts initially then it will explain everything. The first part � translation momentum gives ability the cylinder moves. The second part � angular momentum gives ability the cylinder rotates. The sum of these parts kinetic energies is equal to full translation kinetic energy from case1. The case 2 cylinder translation velocity lower than case 1 translation cylinder velocity.
The equation translation velocity is:

$V_c_a_s_e_2=V_c_a_s_e_1\sqrt{\frac{E_t}{E_t+E_r}}$
Et - kinetic energy translation part

Er - kinetic energy rotation part

Let's make a simple experiment with pencil.

Case 1: Let's hit the pencil on his middle � The pencil fly away on a long distance.
Case 2: Let's hit the pencil on his edge � The pencil starts rotation and fly away not so far.

The classical mechanics solution says � For both cases the pencil should fly away on a long distance with same high translation velocity. No matter how fast it is rotating.

Is it really true? I don't see this result.

---
--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> Another good example:
> Let's repulse a long cylinder inside Isolated System. The repulse should hit the cylinder away from object center mass. This hit energy splits for two object movements. This cylinder starts rotation with translation movement. The Isolated System starts move to opposite cylinder translation movement direction. During this object movement let transform this cylinder to a sphere. This sphere is rotating faster than cylinder because this object moment of inertia less than cylinder moment of inertia. However this rotation does not make any sense for translation momentum transfer. The sphere translation momentum is equal to cylinder translation momentum. However from cylinder repulse action the system takes higher translation momentum than sphere has. The sphere translation momentum won't be enough to stop the Isolated System.
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
>
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> This is a good example is shown transfer between angular and linear momentums. Unfortunately the simulator does not allow giving different angular velocities for squares.
> Anyway, on this model easy to understand these squares with different angular velocities will take a different linear velocities on the end of action.
> Base on my theory about frame of reference conversion for complicated movement, the law of momentum conservation works, however for correct calculation need use imaginary part of velocity.
> The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.
>

#192

From: "abelov0927" <abelov0927@...>
Date: Wed Jul 22, 2009 12:57 am
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

Another good example:
Let's repulse a long cylinder inside Isolated System. The repulse should hit the
cylinder away from object center mass. This hit energy splits for two object
movements. This cylinder starts rotation with translation movement. The Isolated
System starts move to opposite cylinder translation movement direction. During
this object movement let transform this cylinder to a sphere. This sphere is
rotating faster than cylinder because this object moment of inertia less than
cylinder moment of inertia. However this rotation does not make any sense for
translation momentum transfer. The sphere translation momentum is equal to
cylinder translation momentum. However from cylinder repulse action the system
takes higher translation momentum than sphere has. The sphere translation
momentum won't be enough to stop the Isolated System.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#


--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
   This is a good example is shown transfer between angular and linear momentums.
Unfortunately the simulator does not allow giving different angular velocities
for squares.
  Anyway, on this model easy to understand these squares with different angular
velocities will take a different linear velocities on the end of action.
  Base on my theory about frame of reference conversion for complicated movement,
the law of momentum conservation works, however for correct calculation need use
imaginary part of velocity.
     The momentum is conserve. The frame of reference is changing. This frame of
reference exchange gives the system ability to move.

#191

From: "abelov0927" <abelov0927@...>
Date: Sun Jul 19, 2009 10:30 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

This is a good example is shown transfer between angular and linear momentums.

Unfortunately the simulator does not allow giving different angular velocities for squares.
Anyway, on this model easy to understand these squares with different angular velocities will take a different linear velocities on the end of action.
Base on my theory about frame of reference conversion for complicated movement, the law of momentum conservation works, however for correct calculation need use imaginary part of velocity.

The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> A few philosophy thinks. Collisions may be classified in two groups.
> Explicit and implicit. 1. Explicit collisions � happens between
> objects which conducts a simple movements relativity to each other.
> For example. A collision between rolling body and wall on the surface.
> Law of momentum conservation is working. 2. Implicit collisions �
> happens between objects which conducts a complicate movements relativity
> to each other.
> For example. A collision between a rolling body and a surface (It's
> kind of weird thing)
> This it may happen in 2 cases.
> a. Between these objects distortions. Rolling friction.
> b. These objects may have shared points to each other during
> movement action. Coupling through construction elements. For ideal model
> the rolling body is conducting rotation with translation movement
> without any collisions on straight line surface. The rolling body with
> complicated movement has a parallel tangential line to the surface.
> If use same frame of reference for implicit and explicit collisions the
> momentum will have a different value. To avoid this law of momentum
> conservation problem, the model can implement two possibilities.
> a. The model should transform own frame of reference.
> If a platform (the surface) is a center of frame of reference then the
> calculated average velocity of rolling ring is higher than rolling
> ring's center mass velocity.
> If reverse this frame of reference and put rolling ring into center then
> the platform velocity should be reduced for calculations. From the other
> word this frame of reference should base on law of momentum
> conservation. Base on this frame of reference momentum measurement other
> parameters (like velocity) should be readjusted. b. The model should
> include a dark matter. This dark matter with own mass gives ability to
> law of momentum conservation works without frame of reference
> replacement.
> For rolling ring model can happening two things. The dark matter may
> reduce the platform mass or be between objects to compensate extra
> momentum.
> =====================================================================
> Are cosmology theories having a same problem?
> Is new frame of reference with law of momentum conservation core can
> eliminate this problem?
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> abelov0927@ wrote:
> >
> > Quote from forum.
> > "Follow by law of momentum conservation
> > V0x(mx(n-1))=V1x(all mass)
> > If discount one element from the ring then and average speed for (n-1)
> > elements is changing.
> > The rest or ring elements have average velocity is
> > V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
> > conserve. The average velocity is changing.
> > P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this
> model
> > has. One ring's element stays on the ground with velocity zero. To
> > solve this problem let reverse frame of reference for this model. The
> > platform moves horizontally with mass M and velocity V. The ring with
> > mass n*m stays and does rotation movement.
> > For vertical plane (a wall) the platform has a translation momentum
> > P=MV. However, the translation momentum for horizontal plane (the
> > platform surface) counts ring's element, which is joining to the
> > surface with velocity V. In this case the translation momentum is
> > P=(M+m)V. The average velocity is equal to V.
> > How the momentum can be different for these planes? Answer is
> horizontal
> > and vertical planes have a different frame of reference. To reach same
> > momentum P=MV. The frame reference center should move with velocity
> V/n.
> > WAIT A SECOND!
> > Are these vertical and horizontal frames of references move relatively
> > to each other?
> > THIS IS THE ONE.
> > If action starts from zero velocity on one frame of reference and this
> > action finish with zero velocity for another frame of reference then
> > relativity to the first frame of reference the system will continue to
> > move on the end of action! The momentum is conserve. The frame of
> > reference is changing. This frame of reference exchange gives the
> system
> > ability to move.
> >
> > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> > abelov0927@ wrote:
> > >
> > > For this model, � repulsion and � collision comes on different
> > planes.
> > >
> > > The vertical plane doesn't have any contacts with the rolling body.
> > > Otherwise the horizontal plane always has a physical contact with
> > rolling body.
> > >
> > > From vertical plane point of view, all elements of rolling body have
> a
> > movement.
> > > If calculate a rolling body translation momentum with respect to
> > vertical plane then all elements of rolling body must be included.
> > >
> > > Otherwise if look on horizontal plane, one element of rolling body
> > stays on the surface all the time.If calculate a rolling body
> > translation momentum with respect to horizontal plane then one element
> > of rolling body must be included to the mass of horizontal plane.
> > >
> > >
> > >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > qm1l0s4ys/9#
> > >
> > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
> "abelov0927"
> > abelov0927@ wrote:
> > > >
> > > > Hi.
> > > > I think this is a clue for understanding.
> > > >
> > > > The element with zero values of linear velocity on the surface is
> > too small. Base on math the geometrical size of this element strives
> to
> > zero limit. However, this is the physical element, and it has its own
> > geometrical size.
> > > >
> > > >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > qm1l0s4ys/9#
> > > >
> > > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
> > "abelov0927" <abelov0927@> wrote:
> > > > >
> > > > >
> > > > >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > \
> > > > > qm1l0s4ys/9
> > > > >
> >
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> \
> > \
> > > > > mqm1l0s4ys/9> #
> > > > >
> > > > > No objects with mass in translation movement can transfer
> momentum
> > > > > without collision action. However, rotation with translation
> > movement
> > > > > gives unique situation when object touches the surface without
> > momentum
> > > > > transfer on a rolling body linear movement direction.
> > > > > The idea is very simple.
> > > > >
> > > > >
> > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> > > > >
> > > > >
> > > > >
> > > > > If spit a rolling ring to small parts set of n elements
> > (1,2,3,...,n)
> > > > > with mass m then each of them conducts linear and circular
> > movement on
> > > > > surface. Good example is �Caterpillar tracks (Continuous
> > track).
> > > > > <http://en.wikipedia.org/wiki/Continuous_track>
> > > > >
> > > > > Each piece has constant angular velocity. This is the perfect
> > picture
> > > > > from UPENN site
> > > > >
> >
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> \
> > \
> > > > > bsection4_1_4_3.html> .
> > > > >
> > > > >
> > > > >
> >
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> \
> > \
> > > > > n.gif>
> > > > >
> > > > > Each piece of ring has variable linear velocity at a surface
> > point. Once
> > > > > per circle each element of ring stop on surface. This (fig. 3)
> > shows the
> > > > > red bit trajectory. At the surface point, this element has
> linear
> > > > > velocity value equal to zero.
> > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > > > > This (fig.2 ) shows all action between ring and surface.
> > > > >
> > > > > This action has 3 phases.
> > > > >
> > > > >
> > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > > > >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > > > > 1. The ring and the surface are getting momentums.
> > > > > The ring has mass n*m
> > > > > The surface has mass M
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > 2. Cut and hold the red bit on the surface (The red bit has
> > velocity
> > > > > zero V=0 on surface. The red bit doesn't transfer momentum to
> the
> > ground
> > > > > on rings linear movement direction)
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > 3. The pseudo chain transfer momentum to the ground. The chain
> has
> > mass
> > > > > (n-1)*m
> > > > > The surface has mass M+m
> > > > >
> > > > > As it has shown on (fig. 2), the ring with mass n*m and the
> > surface with
> > > > > mass M takes momentums on phase 1. On phase 2 the ring must be
> > broken in
> > > > > one location and one element with zero values of linear velocity
> > holding
> > > > > by the surface. This is not meaning to stop the whole ring at
> one
> > time.
> > > > > This mean stops the red piece and cut the ring at the same time.
> > The
> > > > > ring transforms to chain and lose the mass to new value (n-1)*m.
> > On
> > > > > phase 3 the surface with mass M+m holds just one element of ring
> > and
> > > > > other elements of pseudo chain with mass (n-1)*m is continuing
> > movement
> > > > > by his own trajectories until they rich the ground. The very
> > important
> > > > > thing is: phase 1 objects are different from phase 3 objects.
> > > > >
> > > > >
> > > > > What will happen on an end of this action?
> > > > >
> > > > > Will this surface return to be initial velocity(V0=0)?
> > > > >
> > > > > In problem complexity
> > > > >
> > > > > Each element of chain won't stop at the same time. Each element
> > has
> > > > > different momentum but same mass m. If join each stopped element
> > to the
> > > > > surface, then the surface mass increase faster than chain return
> > > > > momentum. This means the surface mass growth will help ground to
> > keep
> > > > > his own momentum.
> > > > >
> > > > > My suggestion:
> > > > > As it was described at beginning of this text, rotation with
> > translation
> > > > > movement gives unique situation when object touches the surface
> > without
> > > > > momentum transfer on a rolling body linear movement direction.If
> > ring
> > > > > has set of n elements then for the surface point only ring's set
> > of n-1
> > > > > elements are always moving. However, one element with zero
> values
> > of
> > > > > velocity stands down. For this particular case, this set of n-1
> > > > > elements (broken thin ring or pseudo chain) not equivalent to
> set
> > of n
> > > > > elements (initial ring). Base on law of momentum conservation
> net
> > > > > momentum for set n-1 elements would have same initial momentum,
> > but
> > > > > momentum density will change for each element of this set n-1.
> The
> > > > > chain will transfer his own momentum to the surface on the end
> of
> > action
> > > > > on phase 3. Base on law of momentum conservation:
> > > > >
> > > > > P = m*V = const
> > > > >
> > > > > m1*V1 = m2*V2
> > > > >
> > > > > The surface with new mass will take a velocity V1 from set of
> > elements
> > > > > n-1. This velocity is different from initial surface velocity
> V0,
> > > > > because the surface mass has been changed.
> > > > >
> > > > > What is the platform end up velocity equation?
> > > > >
> > > > > (Thank you to video_ranger, who provided equation for end up
> > velocity)
> > > > >
> > > > > "If the platform is completely free to move (say floating in
> outer
> > > > > space) momentum conservation requires that it will end up with a
> > > > > positive forward velocity:
> > > > >
> > > > > V1=(n*m/(M+n*m) )*V
> > > > >
> > > > > Kinetic energy is not conserved because as each link slaps down
> on
> > the
> > > > > surface some energy is converted to heat.
> > > > >
> > > > > For a full ring rolling at constant velocity there's no
> horizontal
> > force
> > > > > between the bottom of the ring and the surface but that requires
> > the
> > > > > ring to be balanced (rotationally symmetric). As links become
> > missing
> > > > > from the circle that's no longer true so the succeeding links
> that
> > hit
> > > > > the surface do have a forward pull on them accelerating the
> > platform
> > > > > forward."
> > > > >
> > > > >
> > > > >
> > > > > Follow the law of momentum conservation, an isolated system with
> > > > > transformed a rolling body may have linear velocity more than
> > zero.
> > > > > The surface will return to initial velocity V=0 if rest of the
> > chain
> > > > > could increase momentum on cut action. However, it's nonsense.
> > Base on
> > > > > law of momentum conservation the chain should keep same initial
> > ring's
> > > > > momentum.
> > > > >
> > > >
> > >
> >
>

#190

From: "abelov0927" <abelov0927@...>
Date: Thu Jul 16, 2009 6:26 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

A few philosophy thinks.

Collisions may be classified in two groups. Explicit and implicit.

1. Explicit collisions � happens between objects which conducts a simple movements relativity to each other.
For example. A collision between rolling body and wall on the surface. Law of momentum conservation is working.

2. Implicit collisions � happens between objects which conducts a complicate movements relativity to each other.
For example. A collision between a rolling body and a surface (It's kind of weird thing)
This it may happen in 2 cases.
a. Between these objects distortions. Rolling friction.
b. These objects may have shared points to each other during movement action. Coupling through construction elements.

For ideal model the rolling body is conducting rotation with translation movement without any collisions on straight line surface. The rolling body with complicated movement has a parallel tangential line to the surface.
If use same frame of reference for implicit and explicit collisions the momentum will have a different value.

To avoid this law of momentum conservation problem, the model can implement two possibilities.
a. The model should transform own frame of reference.
If a platform (the surface) is a center of frame of reference then the calculated average velocity of rolling ring is higher than rolling ring's center mass velocity.
If reverse this frame of reference and put rolling ring into center then the platform velocity should be reduced for calculations. From the other word this frame of reference should base on law of momentum conservation. Base on this frame of reference momentum measurement other parameters (like velocity) should be readjusted.

b. The model should include a dark matter.

This dark matter with own mass gives ability to law of momentum conservation works without frame of reference replacement.
For rolling ring model can happening two things. The dark matter may reduce the platform mass or be between objects to compensate extra momentum.

=====================================================================
Are cosmology theories having a same problem?
Is new frame of reference with law of momentum conservation core can eliminate this problem?

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> Quote from forum.
> "Follow by law of momentum conservation
> V0x(mx(n-1))=V1x(all mass)
> If discount one element from the ring then and average speed for (n-1)
> elements is changing.
> The rest or ring elements have average velocity is
> V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
> conserve. The average velocity is changing.
> P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model
> has. One ring's element stays on the ground with velocity zero. To
> solve this problem let reverse frame of reference for this model. The
> platform moves horizontally with mass M and velocity V. The ring with
> mass n*m stays and does rotation movement.
> For vertical plane (a wall) the platform has a translation momentum
> P=MV. However, the translation momentum for horizontal plane (the
> platform surface) counts ring's element, which is joining to the
> surface with velocity V. In this case the translation momentum is
> P=(M+m)V. The average velocity is equal to V.
> How the momentum can be different for these planes? Answer is horizontal
> and vertical planes have a different frame of reference. To reach same
> momentum P=MV. The frame reference center should move with velocity V/n.
> WAIT A SECOND!
> Are these vertical and horizontal frames of references move relatively
> to each other?
> THIS IS THE ONE.
> If action starts from zero velocity on one frame of reference and this
> action finish with zero velocity for another frame of reference then
> relativity to the first frame of reference the system will continue to
> move on the end of action! The momentum is conserve. The frame of
> reference is changing. This frame of reference exchange gives the system
> ability to move.
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> abelov0927@ wrote:
> >
> > For this model, � repulsion and � collision comes on different
> planes.
> >
> > The vertical plane doesn't have any contacts with the rolling body.
> > Otherwise the horizontal plane always has a physical contact with
> rolling body.
> >
> > From vertical plane point of view, all elements of rolling body have a
> movement.
> > If calculate a rolling body translation momentum with respect to
> vertical plane then all elements of rolling body must be included.
> >
> > Otherwise if look on horizontal plane, one element of rolling body
> stays on the surface all the time.If calculate a rolling body
> translation momentum with respect to horizontal plane then one element
> of rolling body must be included to the mass of horizontal plane.
> >
> >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> qm1l0s4ys/9#
> >
> > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> abelov0927@ wrote:
> > >
> > > Hi.
> > > I think this is a clue for understanding.
> > >
> > > The element with zero values of linear velocity on the surface is
> too small. Base on math the geometrical size of this element strives to
> zero limit. However, this is the physical element, and it has its own
> geometrical size.
> > >
> > >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> qm1l0s4ys/9#
> > >
> > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
> "abelov0927" <abelov0927@> wrote:
> > > >
> > > >
> > > >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > > > qm1l0s4ys/9
> > > >
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> \
> > > > mqm1l0s4ys/9> #
> > > >
> > > > No objects with mass in translation movement can transfer momentum
> > > > without collision action. However, rotation with translation
> movement
> > > > gives unique situation when object touches the surface without
> momentum
> > > > transfer on a rolling body linear movement direction.
> > > > The idea is very simple.
> > > >
> > > >
> > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> > > >
> > > >
> > > >
> > > > If spit a rolling ring to small parts set of n elements
> (1,2,3,...,n)
> > > > with mass m then each of them conducts linear and circular
> movement on
> > > > surface. Good example is �Caterpillar tracks (Continuous
> track).
> > > > <http://en.wikipedia.org/wiki/Continuous_track>
> > > >
> > > > Each piece has constant angular velocity. This is the perfect
> picture
> > > > from UPENN site
> > > >
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> \
> > > > bsection4_1_4_3.html> .
> > > >
> > > >
> > > >
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> \
> > > > n.gif>
> > > >
> > > > Each piece of ring has variable linear velocity at a surface
> point. Once
> > > > per circle each element of ring stop on surface. This (fig. 3)
> shows the
> > > > red bit trajectory. At the surface point, this element has linear
> > > > velocity value equal to zero.
> > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > > > This (fig.2 ) shows all action between ring and surface.
> > > >
> > > > This action has 3 phases.
> > > >
> > > >
> > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > > >
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > > > 1. The ring and the surface are getting momentums.
> > > > The ring has mass n*m
> > > > The surface has mass M
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > 2. Cut and hold the red bit on the surface (The red bit has
> velocity
> > > > zero V=0 on surface. The red bit doesn't transfer momentum to the
> ground
> > > > on rings linear movement direction)
> > > >
> > > >
> > > >
> > > >
> > > > 3. The pseudo chain transfer momentum to the ground. The chain has
> mass
> > > > (n-1)*m
> > > > The surface has mass M+m
> > > >
> > > > As it has shown on (fig. 2), the ring with mass n*m and the
> surface with
> > > > mass M takes momentums on phase 1. On phase 2 the ring must be
> broken in
> > > > one location and one element with zero values of linear velocity
> holding
> > > > by the surface. This is not meaning to stop the whole ring at one
> time.
> > > > This mean stops the red piece and cut the ring at the same time.
> The
> > > > ring transforms to chain and lose the mass to new value (n-1)*m.
> On
> > > > phase 3 the surface with mass M+m holds just one element of ring
> and
> > > > other elements of pseudo chain with mass (n-1)*m is continuing
> movement
> > > > by his own trajectories until they rich the ground. The very
> important
> > > > thing is: phase 1 objects are different from phase 3 objects.
> > > >
> > > >
> > > > What will happen on an end of this action?
> > > >
> > > > Will this surface return to be initial velocity(V0=0)?
> > > >
> > > > In problem complexity
> > > >
> > > > Each element of chain won't stop at the same time. Each element
> has
> > > > different momentum but same mass m. If join each stopped element
> to the
> > > > surface, then the surface mass increase faster than chain return
> > > > momentum. This means the surface mass growth will help ground to
> keep
> > > > his own momentum.
> > > >
> > > > My suggestion:
> > > > As it was described at beginning of this text, rotation with
> translation
> > > > movement gives unique situation when object touches the surface
> without
> > > > momentum transfer on a rolling body linear movement direction.If
> ring
> > > > has set of n elements then for the surface point only ring's set
> of n-1
> > > > elements are always moving. However, one element with zero values
> of
> > > > velocity stands down. For this particular case, this set of n-1
> > > > elements (broken thin ring or pseudo chain) not equivalent to set
> of n
> > > > elements (initial ring). Base on law of momentum conservation net
> > > > momentum for set n-1 elements would have same initial momentum,
> but
> > > > momentum density will change for each element of this set n-1. The
> > > > chain will transfer his own momentum to the surface on the end of
> action
> > > > on phase 3. Base on law of momentum conservation:
> > > >
> > > > P = m*V = const
> > > >
> > > > m1*V1 = m2*V2
> > > >
> > > > The surface with new mass will take a velocity V1 from set of
> elements
> > > > n-1. This velocity is different from initial surface velocity V0,
> > > > because the surface mass has been changed.
> > > >
> > > > What is the platform end up velocity equation?
> > > >
> > > > (Thank you to video_ranger, who provided equation for end up
> velocity)
> > > >
> > > > "If the platform is completely free to move (say floating in outer
> > > > space) momentum conservation requires that it will end up with a
> > > > positive forward velocity:
> > > >
> > > > V1=(n*m/(M+n*m) )*V
> > > >
> > > > Kinetic energy is not conserved because as each link slaps down on
> the
> > > > surface some energy is converted to heat.
> > > >
> > > > For a full ring rolling at constant velocity there's no horizontal
> force
> > > > between the bottom of the ring and the surface but that requires
> the
> > > > ring to be balanced (rotationally symmetric). As links become
> missing
> > > > from the circle that's no longer true so the succeeding links that
> hit
> > > > the surface do have a forward pull on them accelerating the
> platform
> > > > forward."
> > > >
> > > >
> > > >
> > > > Follow the law of momentum conservation, an isolated system with
> > > > transformed a rolling body may have linear velocity more than
> zero.
> > > > The surface will return to initial velocity V=0 if rest of the
> chain
> > > > could increase momentum on cut action. However, it's nonsense.
> Base on
> > > > law of momentum conservation the chain should keep same initial
> ring's
> > > > momentum.
> > > >
> > >
> >
>

#189

From: "abelov0927" <abelov0927@...>
Date: Fri Jul 17, 2009 5:05 am
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

Let imagine observer and two objects (A and B). One object A with simple
movement can move only into one plane. The observer in own frame of reference
sees only this plane also.
Another object B with complicated movement can move in two planes and observer
can see only one of these planes.
The object A has position without movement for observer. The object B has a
movement. The object B has component velocity for each plane. However, observer
can see only one component velocity of objects B . The object A takes momentum
from object B after collision and objects A velocity may be higher than
inspected.
a. The observer can assume the object A has a huge mass. Or if observer knows
the object B mass then he can assume include dark matter for collision process.
b. If frame of reference base on law of momentum conservation then using all
know parameters easy to calculate another plane velocity component of object B
which is invisible for observer. This velocity component for observer has
imaginary character. From the other words, the frame of reference use law of
momentum conservation to readjust another object parameters.


A few words about movement without external forces. The object can rotate on own
center mass without external forces. However, if alternate simple and
complicated movement on object collisions then for observer frame of references
it may possible exchange rotation to translation movement.
Everything Is Relative.

#188

From: "abelov0927" <abelov0927@...>
Date: Tue Jul 14, 2009 10:02 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

Quote from forum.
"Follow by law of momentum conservation
V0x(mx(n-1))=V1x(all mass)
If discount one element from the ring then and average speed for (n-1) elements is changing.
The rest or ring elements have average velocity is V'0=(n/(n-1))xV0!"

This is absolutely correct. The momentum is conserve. The average velocity is changing.
P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model has. One ring's element stays on the ground with velocity zero.

To solve this problem let reverse frame of reference for this model. The platform moves horizontally with mass M and velocity V. The ring with mass n*m stays and does rotation movement.
For vertical plane (a wall) the platform has a translation momentum P=MV. However, the translation momentum for horizontal plane (the platform surface) counts ring's element, which is joining to the surface with velocity V. In this case the translation momentum is P=(M+m)V. The average velocity is equal to V.
How the momentum can be different for these planes? Answer is horizontal and vertical planes have a different frame of reference. To reach same momentum P=MV. The frame reference center should move with velocity V/n.
WAIT A SECOND!
Are these vertical and horizontal frames of references move relatively to each other?

THIS IS THE ONE.
If action starts from zero velocity on one frame of reference and this action finish with zero velocity for another frame of reference then relativity to the first frame of reference the system will continue to move on the end of action!

The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> For this model, � repulsion and � collision comes on different planes.
>
> The vertical plane doesn't have any contacts with the rolling body.
> Otherwise the horizontal plane always has a physical contact with rolling body.
>
> From vertical plane point of view, all elements of rolling body have a movement.
> If calculate a rolling body translation momentum with respect to vertical plane then all elements of rolling body must be included.
>
> Otherwise if look on horizontal plane, one element of rolling body stays on the surface all the time.If calculate a rolling body translation momentum with respect to horizontal plane then one element of rolling body must be included to the mass of horizontal plane.
>
>
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> >
> > Hi.
> > I think this is a clue for understanding.
> >
> > The element with zero values of linear velocity on the surface is too small. Base on math the geometrical size of this element strives to zero limit. However, this is the physical element, and it has its own geometrical size.
> >
> > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
> >
> > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@> wrote:
> > >
> > >
> > > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> > > qm1l0s4ys/9
> > > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> > > mqm1l0s4ys/9> #
> > >
> > > No objects with mass in translation movement can transfer momentum
> > > without collision action. However, rotation with translation movement
> > > gives unique situation when object touches the surface without momentum
> > > transfer on a rolling body linear movement direction.
> > > The idea is very simple.
> > >
> > >
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> > >
> > >
> > >
> > > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> > > with mass m then each of them conducts linear and circular movement on
> > > surface. Good example is �Caterpillar tracks (Continuous track).
> > > <http://en.wikipedia.org/wiki/Continuous_track>
> > >
> > > Each piece has constant angular velocity. This is the perfect picture
> > > from UPENN site
> > > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> > > bsection4_1_4_3.html> .
> > >
> > >
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> > > n.gif>
> > >
> > > Each piece of ring has variable linear velocity at a surface point. Once
> > > per circle each element of ring stop on surface. This (fig. 3) shows the
> > > red bit trajectory. At the surface point, this element has linear
> > > velocity value equal to zero.
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > > This (fig.2 ) shows all action between ring and surface.
> > >
> > > This action has 3 phases.
> > >
> > >
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > > 1. The ring and the surface are getting momentums.
> > > The ring has mass n*m
> > > The surface has mass M
> > >
> > >
> > >
> > >
> > >
> > > 2. Cut and hold the red bit on the surface (The red bit has velocity
> > > zero V=0 on surface. The red bit doesn't transfer momentum to the ground
> > > on rings linear movement direction)
> > >
> > >
> > >
> > >
> > > 3. The pseudo chain transfer momentum to the ground. The chain has mass
> > > (n-1)*m
> > > The surface has mass M+m
> > >
> > > As it has shown on (fig. 2), the ring with mass n*m and the surface with
> > > mass M takes momentums on phase 1. On phase 2 the ring must be broken in
> > > one location and one element with zero values of linear velocity holding
> > > by the surface. This is not meaning to stop the whole ring at one time.
> > > This mean stops the red piece and cut the ring at the same time. The
> > > ring transforms to chain and lose the mass to new value (n-1)*m. On
> > > phase 3 the surface with mass M+m holds just one element of ring and
> > > other elements of pseudo chain with mass (n-1)*m is continuing movement
> > > by his own trajectories until they rich the ground. The very important
> > > thing is: phase 1 objects are different from phase 3 objects.
> > >
> > >
> > > What will happen on an end of this action?
> > >
> > > Will this surface return to be initial velocity(V0=0)?
> > >
> > > In problem complexity
> > >
> > > Each element of chain won't stop at the same time. Each element has
> > > different momentum but same mass m. If join each stopped element to the
> > > surface, then the surface mass increase faster than chain return
> > > momentum. This means the surface mass growth will help ground to keep
> > > his own momentum.
> > >
> > > My suggestion:
> > > As it was described at beginning of this text, rotation with translation
> > > movement gives unique situation when object touches the surface without
> > > momentum transfer on a rolling body linear movement direction.If ring
> > > has set of n elements then for the surface point only ring's set of n-1
> > > elements are always moving. However, one element with zero values of
> > > velocity stands down. For this particular case, this set of n-1
> > > elements (broken thin ring or pseudo chain) not equivalent to set of n
> > > elements (initial ring). Base on law of momentum conservation net
> > > momentum for set n-1 elements would have same initial momentum, but
> > > momentum density will change for each element of this set n-1. The
> > > chain will transfer his own momentum to the surface on the end of action
> > > on phase 3. Base on law of momentum conservation:
> > >
> > > P = m*V = const
> > >
> > > m1*V1 = m2*V2
> > >
> > > The surface with new mass will take a velocity V1 from set of elements
> > > n-1. This velocity is different from initial surface velocity V0,
> > > because the surface mass has been changed.
> > >
> > > What is the platform end up velocity equation?
> > >
> > > (Thank you to video_ranger, who provided equation for end up velocity)
> > >
> > > "If the platform is completely free to move (say floating in outer
> > > space) momentum conservation requires that it will end up with a
> > > positive forward velocity:
> > >
> > > V1=(n*m/(M+n*m) )*V
> > >
> > > Kinetic energy is not conserved because as each link slaps down on the
> > > surface some energy is converted to heat.
> > >
> > > For a full ring rolling at constant velocity there's no horizontal force
> > > between the bottom of the ring and the surface but that requires the
> > > ring to be balanced (rotationally symmetric). As links become missing
> > > from the circle that's no longer true so the succeeding links that hit
> > > the surface do have a forward pull on them accelerating the platform
> > > forward."
> > >
> > >
> > >
> > > Follow the law of momentum conservation, an isolated system with
> > > transformed a rolling body may have linear velocity more than zero.
> > > The surface will return to initial velocity V=0 if rest of the chain
> > > could increase momentum on cut action. However, it's nonsense. Base on
> > > law of momentum conservation the chain should keep same initial ring's
> > > momentum.
> > >
> >
>

#187

From: "abelov0927" <abelov0927@...>
Date: Sat Jul 11, 2009 6:02 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

For this model, � repulsion and � collision comes on different planes.

The vertical plane doesn't have any contacts with the rolling body.
Otherwise the horizontal plane always has a physical contact with rolling body.

From vertical plane point of view, all elements of rolling body have a movement.
If calculate a rolling body translation momentum with respect to vertical plane
then all elements of rolling body must be included.

Otherwise if look on horizontal plane, one element of rolling body stays on the
surface all the time.If calculate a rolling body translation momentum with
respect to horizontal plane then one element of rolling body must be included to
the mass of horizontal plane.


http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
> Hi.
> I think this is a clue for understanding.
>
> The element with zero values of linear velocity on the surface is too small.
Base on math the geometrical size of this element strives to zero limit.
However, this is the physical element, and it has its own geometrical size.
>
>
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#
>
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@> wrote:
> >
> >
> > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> > qm1l0s4ys/9
> > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> > mqm1l0s4ys/9> #
> >
> > No objects with mass in translation movement can transfer momentum
> > without collision action. However, rotation with translation movement
> > gives unique situation when object touches the surface without momentum
> > transfer on a rolling body linear movement direction.
> > The idea is very simple.
> >
> >
> >   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> >
> >
> >
> > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> > with mass m then each of them conducts linear and circular movement on
> > surface. Good example is �Caterpillar tracks (Continuous track).
> > <http://en.wikipedia.org/wiki/Continuous_track>
> >
> > Each piece has constant angular velocity. This is the perfect picture
> > from UPENN site
> > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> > bsection4_1_4_3.html> .
> >
> >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> > n.gif>
> >
> > Each piece of ring has variable linear velocity at a surface point. Once
> > per circle each element of ring stop on surface. This (fig. 3) shows the
> > red bit trajectory. At the surface point, this element has linear
> > velocity value equal to zero.
> >   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > This (fig.2 ) shows all action between ring and surface.
> >
> > This action has 3 phases.
> >
> >
> >   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > 1. The ring and the surface are getting momentums.
> > The ring has mass n*m
> > The surface has mass M
> >
> >
> >
> >
> >
> > 2. Cut and hold the red bit on the surface (The red bit has velocity
> > zero V=0 on surface. The red bit doesn't transfer momentum to the ground
> > on rings linear movement direction)
> >
> >
> >
> >
> > 3. The pseudo chain transfer momentum to the ground. The chain has mass
> > (n-1)*m
> > The surface has mass M+m
> >
> > As it has shown on (fig. 2), the ring with mass n*m and the surface with
> > mass M takes momentums on phase 1. On phase 2 the ring must be broken in
> > one location and one element with zero values of linear velocity holding
> > by the surface. This is not meaning to stop the whole ring at one time.
> > This mean stops the red piece and cut the ring at the same time. The
> > ring transforms to chain and lose the mass to new value (n-1)*m. On
> > phase 3 the surface with mass M+m holds just one element of ring and
> > other elements of pseudo chain with mass (n-1)*m is continuing movement
> > by his own trajectories until they rich the ground. The very important
> > thing is: phase 1 objects are different from phase 3 objects.
> >
> >
> > What will happen on an end of this action?
> >
> > Will this surface return to be initial velocity(V0=0)?
> >
> > In problem complexity
> >
> > Each element of chain won't stop at the same time. Each element has
> > different momentum but same mass m. If join each stopped element to the
> > surface, then the surface mass increase faster than chain return
> > momentum. This means the surface mass growth will help ground to keep
> > his own momentum.
> >
> > My suggestion:
> > As it was described at beginning of this text, rotation with translation
> > movement gives unique situation when object touches the surface without
> > momentum transfer on a rolling body linear movement direction.If ring
> > has set of n elements then for the surface point only ring's set of n-1
> > elements are always moving. However, one element with zero values of
> > velocity stands down.    For this particular case, this set of n-1
> > elements (broken thin ring or pseudo chain) not equivalent to set of n
> > elements (initial ring). Base on law of momentum conservation net
> > momentum for set n-1 elements would have same initial momentum, but
> > momentum density will change for each element of this set n-1. The
> > chain will transfer his own momentum to the surface on the end of action
> > on phase 3. Base on law of momentum conservation:
> >
> > P = m*V = const
> >
> > m1*V1 = m2*V2
> >
> > The surface with new mass will take a velocity V1 from set of elements
> > n-1. This velocity is different from initial surface velocity V0,
> > because the surface mass has been changed.
> >
> > What is the platform end up velocity equation?
> >
> > (Thank you to video_ranger, who provided equation for end up velocity)
> >
> > "If the platform is completely free to move (say floating in outer
> > space) momentum conservation requires that it will end up with a
> > positive forward velocity:
> >
> > V1=(n*m/(M+n*m) )*V
> >
> > Kinetic energy is not conserved because as each link slaps down on the
> > surface some energy is converted to heat.
> >
> > For a full ring rolling at constant velocity there's no horizontal force
> > between the bottom of the ring and the surface but that requires the
> > ring to be balanced (rotationally symmetric). As links become missing
> > from the circle that's no longer true so the succeeding links that hit
> > the surface do have a forward pull on them accelerating the platform
> > forward."
> >
> >
> >
> > Follow the law of momentum conservation, an isolated system with
> > transformed a rolling body may have linear velocity more than zero.
> > The surface will return to initial velocity V=0 if rest of the chain
> > could increase momentum on cut action. However, it's nonsense. Base on
> > law of momentum conservation the chain should keep same initial ring's
> > momentum.
> >
>

#186

From: "abelov0927" <abelov0927@...>
Date: Fri Jul 10, 2009 2:31 am
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

A rolling disk on a straight line surface.

The diagram has shown a rolling disk on straight line surface. This movement has rolling and static frictions. This disk contains n sectors with mass m. The (centre mass) CM disk has an initial translation velocity V. The red sector has linear velocity zero. Base on previous explanations, only n-1 sector on the rolling disk has velocity more then zero. These moving sectors transfer linear momentum dP to the surface with mass M+m per time frame dt.

If use classical model: The disk with linear velocity V transfer momentum to the surface. The total velocity on the end of action is:

(equation 1)

If discount the red sector with zero linear velocity: The total velocity on the end of action is:

(equation 2)

The velocity difference between these two models is:

(equation 3)

If sectors geometrical size strives to zero, then sectors mass strive to zero also. For these velocity difference equation, it gives a zero result.

(equation 4)

However, the physical elements have its own geometrical size and end up velocity may be count by (equation 2)

Velocity difference between classical and this modern models is:

(equastion 3)

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> These thoughts can be used for a rolling body.
>
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/rollingfriction.jpg>
> The diagram has shown a rolling disk which contains n sectors with mass
> m. This (centre mass) CM disk has a translation velocity V. The red
> sector has linear velocity zero. Base on previous explanations, only n-1
> sector on the rolling disk has velocity more then zero. These moving
> sectors transfer linear momentum dP to the surface with mass M+m per
> time frame dt. If use classical model: The disk with linear velocity V
> transfer momentum to the surface. The total velocity on the end of
> action is:
> V1=((n*m)/(n*m+M))*V If discount the red sector with zero linear
> velocity: The total velocity on the end of action is:
> V1'=(((n-1)*m)/((n-1)*m+M+m))*V. The velocity difference between
> these two models is:
> V1-V1'=m/(n*m+M)*V If sectors geometrical size strives to zero,
> then sectors mass strive to zero also. For velocity difference equation,
> it gives a zero result.
> V1-V1'=0. At a math point of view with a very small sector with
> zero linear velocity the surface takes equation
> V1=((n*m)/(n*m+M))*V. However, the physical elements have its own
> geometrical size and end up velocity for this action is:
> V1'=(((n-1)*m)/((n-1)*m+M+m))*V Velocity difference is:
> V1-V1'=m/(n*m+M)*V
> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> abelov0927@ wrote:
> >
> >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > qm1l0s4ys/9
> >
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> \
> > mqm1l0s4ys/9> #
> >
> > No objects with mass in translation movement can transfer momentum
> > without collision action. However, rotation with translation movement
> > gives unique situation when object touches the surface without
> momentum
> > transfer on a rolling body linear movement direction.
> > The idea is very simple.
> >
> >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> >
> >
> >
> > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> > with mass m then each of them conducts linear and circular movement on
> > surface. Good example is �Caterpillar tracks (Continuous track).
> > <http://en.wikipedia.org/wiki/Continuous_track>
> >
> > Each piece has constant angular velocity. This is the perfect picture
> > from UPENN site
> >
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> \
> > bsection4_1_4_3.html> .
> >
> >
> >
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> \
> > n.gif>
> >
> > Each piece of ring has variable linear velocity at a surface point.
> Once
> > per circle each element of ring stop on surface. This (fig. 3) shows
> the
> > red bit trajectory. At the surface point, this element has linear
> > velocity value equal to zero.
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > This (fig.2 ) shows all action between ring and surface.
> >
> > This action has 3 phases.
> >
> >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > 1. The ring and the surface are getting momentums.
> > The ring has mass n*m
> > The surface has mass M
> >
> >
> >
> >
> >
> > 2. Cut and hold the red bit on the surface (The red bit has velocity
> > zero V=0 on surface. The red bit doesn't transfer momentum to the
> ground
> > on rings linear movement direction)
> >
> >
> >
> >
> > 3. The pseudo chain transfer momentum to the ground. The chain has
> mass
> > (n-1)*m
> > The surface has mass M+m
> >
> > As it has shown on (fig. 2), the ring with mass n*m and the surface
> with
> > mass M takes momentums on phase 1. On phase 2 the ring must be broken
> in
> > one location and one element with zero values of linear velocity
> holding
> > by the surface. This is not meaning to stop the whole ring at one
> time.
> > This mean stops the red piece and cut the ring at the same time. The
> > ring transforms to chain and lose the mass to new value (n-1)*m. On
> > phase 3 the surface with mass M+m holds just one element of ring and
> > other elements of pseudo chain with mass (n-1)*m is continuing
> movement
> > by his own trajectories until they rich the ground. The very important
> > thing is: phase 1 objects are different from phase 3 objects.
> >
> >
> > What will happen on an end of this action?
> >
> > Will this surface return to be initial velocity(V0=0)?
> >
> > In problem complexity
> >
> > Each element of chain won't stop at the same time. Each element has
> > different momentum but same mass m. If join each stopped element to
> the
> > surface, then the surface mass increase faster than chain return
> > momentum. This means the surface mass growth will help ground to keep
> > his own momentum.
> >
> > My suggestion:
> > As it was described at beginning of this text, rotation with
> translation
> > movement gives unique situation when object touches the surface
> without
> > momentum transfer on a rolling body linear movement direction.If ring
> > has set of n elements then for the surface point only ring's set of
> n-1
> > elements are always moving. However, one element with zero values of
> > velocity stands down. For this particular case, this set of n-1
> > elements (broken thin ring or pseudo chain) not equivalent to set of n
> > elements (initial ring). Base on law of momentum conservation net
> > momentum for set n-1 elements would have same initial momentum, but
> > momentum density will change for each element of this set n-1. The
> > chain will transfer his own momentum to the surface on the end of
> action
> > on phase 3. Base on law of momentum conservation:
> >
> > P = m*V = const
> >
> > m1*V1 = m2*V2
> >
> > The surface with new mass will take a velocity V1 from set of elements
> > n-1. This velocity is different from initial surface velocity V0,
> > because the surface mass has been changed.
> >
> > What is the platform end up velocity equation?
> >
> > (Thank you to video_ranger, who provided equation for end up velocity)
> >
> > "If the platform is completely free to move (say floating in outer
> > space) momentum conservation requires that it will end up with a
> > positive forward velocity:
> >
> > V1=(n*m/(M+n*m) )*V
> >
> > Kinetic energy is not conserved because as each link slaps down on the
> > surface some energy is converted to heat.
> >
> > For a full ring rolling at constant velocity there's no horizontal
> force
> > between the bottom of the ring and the surface but that requires the
> > ring to be balanced (rotationally symmetric). As links become missing
> > from the circle that's no longer true so the succeeding links that hit
> > the surface do have a forward pull on them accelerating the platform
> > forward."
> >
> >
> >
> > Follow the law of momentum conservation, an isolated system with
> > transformed a rolling body may have linear velocity more than zero.
> > The surface will return to initial velocity V=0 if rest of the chain
> > could increase momentum on cut action. However, it's nonsense. Base on
> > law of momentum conservation the chain should keep same initial ring's
> > momentum.
> >
>

#185

From: "abelov0927" <abelov0927@...>
Date: Tue Jul 7, 2009 11:49 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

These thoughts can be used for a rolling body.

The diagram has shown a rolling disk which contains n sectors with mass m. This (centre mass) CM disk has a translation velocity V. The red sector has linear velocity zero. Base on previous explanations, only n-1 sector on the rolling disk has velocity more then zero. These moving sectors transfer linear momentum dP to the surface with mass M+m per time frame dt.

If use classical model: The disk with linear velocity V transfer momentum to the surface. The total velocity on the end of action is:

V1=((n*m)/(n*m+M))*V

If discount the red sector with zero linear velocity: The total velocity on the end of action is:

V1'=(((n-1)*m)/((n-1)*m+M+m))*V.

The velocity difference between these two models is:

V1-V1'=m/(n*m+M)*V

If sectors geometrical size strives to zero, then sectors mass strive to zero also. For velocity difference equation, it gives a zero result.

V1-V1'=0.

At a math point of view with a very small sector with zero linear velocity the surface takes equation

V1=((n*m)/(n*m+M))*V.

However, the physical elements have its own geometrical size and end up velocity for this action is:

V1'=(((n-1)*m)/((n-1)*m+M+m))*V

Velocity difference is:

V1-V1'=m/(n*m+M)*V

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
>
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> qm1l0s4ys/9
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> mqm1l0s4ys/9> #
>
> No objects with mass in translation movement can transfer momentum
> without collision action. However, rotation with translation movement
> gives unique situation when object touches the surface without momentum
> transfer on a rolling body linear movement direction.
> The idea is very simple.
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
>
>
>
> If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> with mass m then each of them conducts linear and circular movement on
> surface. Good example is �Caterpillar tracks (Continuous track).
> <http://en.wikipedia.org/wiki/Continuous_track>
>
> Each piece has constant angular velocity. This is the perfect picture
> from UPENN site
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> bsection4_1_4_3.html> .
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> n.gif>
>
> Each piece of ring has variable linear velocity at a surface point. Once
> per circle each element of ring stop on surface. This (fig. 3) shows the
> red bit trajectory. At the surface point, this element has linear
> velocity value equal to zero.
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> This (fig.2 ) shows all action between ring and surface.
>
> This action has 3 phases.
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> 1. The ring and the surface are getting momentums.
> The ring has mass n*m
> The surface has mass M
>
>
>
>
>
> 2. Cut and hold the red bit on the surface (The red bit has velocity
> zero V=0 on surface. The red bit doesn't transfer momentum to the ground
> on rings linear movement direction)
>
>
>
>
> 3. The pseudo chain transfer momentum to the ground. The chain has mass
> (n-1)*m
> The surface has mass M+m
>
> As it has shown on (fig. 2), the ring with mass n*m and the surface with
> mass M takes momentums on phase 1. On phase 2 the ring must be broken in
> one location and one element with zero values of linear velocity holding
> by the surface. This is not meaning to stop the whole ring at one time.
> This mean stops the red piece and cut the ring at the same time. The
> ring transforms to chain and lose the mass to new value (n-1)*m. On
> phase 3 the surface with mass M+m holds just one element of ring and
> other elements of pseudo chain with mass (n-1)*m is continuing movement
> by his own trajectories until they rich the ground. The very important
> thing is: phase 1 objects are different from phase 3 objects.
>
>
> What will happen on an end of this action?
>
> Will this surface return to be initial velocity(V0=0)?
>
> In problem complexity
>
> Each element of chain won't stop at the same time. Each element has
> different momentum but same mass m. If join each stopped element to the
> surface, then the surface mass increase faster than chain return
> momentum. This means the surface mass growth will help ground to keep
> his own momentum.
>
> My suggestion:
> As it was described at beginning of this text, rotation with translation
> movement gives unique situation when object touches the surface without
> momentum transfer on a rolling body linear movement direction.If ring
> has set of n elements then for the surface point only ring's set of n-1
> elements are always moving. However, one element with zero values of
> velocity stands down. For this particular case, this set of n-1
> elements (broken thin ring or pseudo chain) not equivalent to set of n
> elements (initial ring). Base on law of momentum conservation net
> momentum for set n-1 elements would have same initial momentum, but
> momentum density will change for each element of this set n-1. The
> chain will transfer his own momentum to the surface on the end of action
> on phase 3. Base on law of momentum conservation:
>
> P = m*V = const
>
> m1*V1 = m2*V2
>
> The surface with new mass will take a velocity V1 from set of elements
> n-1. This velocity is different from initial surface velocity V0,
> because the surface mass has been changed.
>
> What is the platform end up velocity equation?
>
> (Thank you to video_ranger, who provided equation for end up velocity)
>
> "If the platform is completely free to move (say floating in outer
> space) momentum conservation requires that it will end up with a
> positive forward velocity:
>
> V1=(n*m/(M+n*m) )*V
>
> Kinetic energy is not conserved because as each link slaps down on the
> surface some energy is converted to heat.
>
> For a full ring rolling at constant velocity there's no horizontal force
> between the bottom of the ring and the surface but that requires the
> ring to be balanced (rotationally symmetric). As links become missing
> from the circle that's no longer true so the succeeding links that hit
> the surface do have a forward pull on them accelerating the platform
> forward."
>
>
>
> Follow the law of momentum conservation, an isolated system with
> transformed a rolling body may have linear velocity more than zero.
> The surface will return to initial velocity V=0 if rest of the chain
> could increase momentum on cut action. However, it's nonsense. Base on
> law of momentum conservation the chain should keep same initial ring's
> momentum.
>

#184

From: "abelov0927" <abelov0927@...>
Date: Sun Jul 5, 2009 10:36 pm
Subject: Re: Paradox of Classical Mechanics

abelov0927

Offline

Hi.
I think this is a clue for understanding.

The element with zero values of linear velocity on the surface is too small.
Base on math the geometrical size of this element strives to zero limit.
However, this is the physical element, and it has its own geometrical size.

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
>
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> qm1l0s4ys/9
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> mqm1l0s4ys/9> #
>
> No objects with mass in translation movement can transfer momentum
> without collision action. However, rotation with translation movement
> gives unique situation when object touches the surface without momentum
> transfer on a rolling body linear movement direction.
> The idea is very simple.
>
>
>   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
>
>
>
> If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> with mass m then each of them conducts linear and circular movement on
> surface. Good example is �Caterpillar tracks (Continuous track).
> <http://en.wikipedia.org/wiki/Continuous_track>
>
> Each piece has constant angular velocity. This is the perfect picture
> from UPENN site
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> bsection4_1_4_3.html> .
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> n.gif>
>
> Each piece of ring has variable linear velocity at a surface point. Once
> per circle each element of ring stop on surface. This (fig. 3) shows the
> red bit trajectory. At the surface point, this element has linear
> velocity value equal to zero.
>   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> This (fig.2 ) shows all action between ring and surface.
>
> This action has 3 phases.
>
>
>   <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> 1. The ring and the surface are getting momentums.
> The ring has mass n*m
> The surface has mass M
>
>
>
>
>
> 2. Cut and hold the red bit on the surface (The red bit has velocity
> zero V=0 on surface. The red bit doesn't transfer momentum to the ground
> on rings linear movement direction)
>
>
>
>
> 3. The pseudo chain transfer momentum to the ground. The chain has mass
> (n-1)*m
> The surface has mass M+m
>
> As it has shown on (fig. 2), the ring with mass n*m and the surface with
> mass M takes momentums on phase 1. On phase 2 the ring must be broken in
> one location and one element with zero values of linear velocity holding
> by the surface. This is not meaning to stop the whole ring at one time.
> This mean stops the red piece and cut the ring at the same time. The
> ring transforms to chain and lose the mass to new value (n-1)*m. On
> phase 3 the surface with mass M+m holds just one element of ring and
> other elements of pseudo chain with mass (n-1)*m is continuing movement
> by his own trajectories until they rich the ground. The very important
> thing is: phase 1 objects are different from phase 3 objects.
>
>
> What will happen on an end of this action?
>
> Will this surface return to be initial velocity(V0=0)?
>
> In problem complexity
>
> Each element of chain won't stop at the same time. Each element has
> different momentum but same mass m. If join each stopped element to the
> surface, then the surface mass increase faster than chain return
> momentum. This means the surface mass growth will help ground to keep
> his own momentum.
>
> My suggestion:
> As it was described at beginning of this text, rotation with translation
> movement gives unique situation when object touches the surface without
> momentum transfer on a rolling body linear movement direction.If ring
> has set of n elements then for the surface point only ring's set of n-1
> elements are always moving. However, one element with zero values of
> velocity stands down.    For this particular case, this set of n-1
> elements (broken thin ring or pseudo chain) not equivalent to set of n
> elements (initial ring). Base on law of momentum conservation net
> momentum for set n-1 elements would have same initial momentum, but
> momentum density will change for each element of this set n-1. The
> chain will transfer his own momentum to the surface on the end of action
> on phase 3. Base on law of momentum conservation:
>
> P = m*V = const
>
> m1*V1 = m2*V2
>
> The surface with new mass will take a velocity V1 from set of elements
> n-1. This velocity is different from initial surface velocity V0,
> because the surface mass has been changed.
>
> What is the platform end up velocity equation?
>
> (Thank you to video_ranger, who provided equation for end up velocity)
>
> "If the platform is completely free to move (say floating in outer
> space) momentum conservation requires that it will end up with a
> positive forward velocity:
>
> V1=(n*m/(M+n*m) )*V
>
> Kinetic energy is not conserved because as each link slaps down on the
> surface some energy is converted to heat.
>
> For a full ring rolling at constant velocity there's no horizontal force
> between the bottom of the ring and the surface but that requires the
> ring to be balanced (rotationally symmetric). As links become missing
> from the circle that's no longer true so the succeeding links that hit
> the surface do have a forward pull on them accelerating the platform
> forward."
>
>
>
> Follow the law of momentum conservation, an isolated system with
> transformed a rolling body may have linear velocity more than zero.
> The surface will return to initial velocity V=0 if rest of the chain
> could increase momentum on cut action. However, it's nonsense. Base on
> law of momentum conservation the chain should keep same initial ring's
> momentum.
>

#183

From: "abelov0927" <abelov0927@...>
Date: Fri Jul 3, 2009 4:02 am
Subject: Paradox of Classical Mechanics

abelov0927

Offline

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

No objects with mass in translation movement can transfer momentum without collision action. However, rotation with translation movement gives unique situation when object touches the surface without momentum transfer on a rolling body linear movement direction.

The idea is very simple.

If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with mass m then each of them conducts linear and circular movement on surface. Good example is �Caterpillar tracks (Continuous track).

Each piece has constant angular velocity. This is the perfect picture from UPENN site.

Each piece of ring has variable linear velocity at a surface point. Once per circle each element of ring stop on surface. This (fig. 3) shows the red bit trajectory. At the surface point, this element has linear velocity value equal to zero.

This (fig.2 ) shows all action between ring and surface.

This action has 3 phases.

1. The ring and the surface are getting momentums.
The ring has mass n*m
The surface has mass M

2. Cut and hold the red bit on the surface (The red bit has velocity zero V=0 on surface. The red bit doesn't transfer momentum to the ground on rings linear movement direction)

3. The pseudo chain transfer momentum to the ground.

The chain has mass (n-1)*m
The surface has mass M+m

As it has shown on (fig. 2), the ring with mass n*m and the surface with mass M takes momentums on phase 1. On phase 2 the ring must be broken in one location and one element with zero values of linear velocity holding by the surface. This is not meaning to stop the whole ring at one time. This mean stops the red piece and cut the ring at the same time. The ring transforms to chain and lose the mass to new value (n-1)*m. On phase 3 the surface with mass M+m holds just one element of ring and other elements of pseudo chain with mass (n-1)*m is continuing movement by his own trajectories until they rich the ground. The very important thing is: phase 1 objects are different from phase 3 objects.

What will happen on an end of this action?

Will this surface return to be initial velocity(V0=0)?

In problem complexity

Each element of chain won't stop at the same time. Each element has different momentum but same mass m. If join each stopped element to the surface, then the surface mass increase faster than chain return momentum. This means the surface mass growth will help ground to keep his own momentum.

My suggestion:

As it was described at beginning of this text, rotation with translation movement gives unique situation when object touches the surface without momentum transfer on a rolling body linear movement direction.If ring has set of n elements then for the surface point only ring's set of n-1 elements are always moving. However, one element with zero values of velocity stands down.

For this particular case, this set of n-1 elements (broken thin ring or pseudo chain) not equivalent to set of n elements (initial ring). Base on law of momentum conservation net momentum for set n-1 elements would have same initial momentum, but momentum density will change for each element of this set n-1. The chain will transfer his own momentum to the surface on the end of action on phase 3.

Base on law of momentum conservation:

P = m*V = const

m1*V1 = m2*V2

The surface with new mass will take a velocity V1 from set of elements n-1. This velocity is different from initial surface velocity V0, because the surface mass has been changed.

What is the platform end up velocity equation?

(Thank you to video_ranger, who provided equation for end up velocity)

"If the platform is completely free to move (say floating in outer space) momentum conservation requires that it will end up with a positive forward velocity:

V1=(n*m/(M+n*m) )*V

Kinetic energy is not conserved because as each link slaps down on the surface some energy is converted to heat.

For a full ring rolling at constant velocity there's no horizontal force between the bottom of the ring and the surface but that requires the ring to be balanced (rotationally symmetric). As links become missing from the circle that's no longer true so the succeeding links that hit the surface do have a forward pull on them accelerating the platform forward."

Follow the law of momentum conservation, an isolated system with transformed a rolling body may have linear velocity more than zero.

The surface will return to initial velocity V=0 if rest of the chain could increase momentum on cut action. However, it's nonsense. Base on law of momentum conservation the chain should keep same initial ring's momentum.

#182

From: "mysterystevenson1" <mysterystevenson1@...>
Date: Wed Jul 1, 2009 8:54 pm
Subject: Re: Would rolling body transformation can help find a clue?

mysterysteve...

Offline

Curious, some of the properties of this design echo some of my concerns with
a "Space Elevator". While the size factor is many magnitudes different, some
factors are similar. When I calculated the length of a space elevator Ribbon or
Cable due to orbital mechanics and the rotation of the Earth, I came up with a
length close to 100,000 miles just to reach an altitude of about 200 miles, in a
counter-rotational direction encircling the Earth many times before reaching
that altitude, and as far as the theoretical "space anchor" well that would be
even further by far. This would be needed just to counter shear factors, however
even supposing great advancements in nano tech, there just is nothing on the
drawing boards that could possibly withstand the stress of the weight/mass of
the cable alone, let alone an elevator.
      I know that this is a bit off topic from the original concept you mention,
however, one of the factors that I contemplated using to alter the force factors
was inclusion of CMGs (Giant Gyros) along the cable to alter attitude and hence
certain shearing factors. Never really came up with a way to make the numbers
doable, however factors did begin to improve. Have you ever considered the
addition of Gyros in your design?CMGs can be altered as wished to influence
attitude as needed.

Mystery

#181

From: "abelov0927" <abelov0927@...>
Date: Wed Jul 1, 2009 6:30 pm
Subject: Re: Would rolling body transformation can help find a clue?

abelov0927

Offline

Just keep in mind. This model has 3 phases. Bodies at phase1 is different from
bodies at phase3. The law of momentum conservation works, but it gives different
result for bodies with different mass.

This is the comment from other forum.

"If the platform is completely free to move (say floating in outer space)
momentum conservation requires that it will end up with a positive forward
velocity V=(nm/(M+nm) )v. Kinetic energy is not conserved because as each link
slaps down on the surface some energy is converted to heat.

For a full ring rolling at constant velocity there's no horizontal force between
the bottom of the ring and the surface but that requires the ring to be balanced
(rotationally symmetric). As links become missing from the circle that's no
longer true so the succeeding links that hit the surface do have a forward pull
on them accelerating the platform forward."
Please read and look on knol site for details.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#


--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
> link to problem site
>
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#
>
> The idea is very simple.
>
> If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with
mass m then each of them conduct linear and circular movement on surface.
>
> Each piece has constant angular velocity. Each piece of ring has variable
linear velocity at surface point. Once per circle each element of ring stop on
surface. At surface point this element has linear velocity value equal to zero.
>
> The ring must be broken in one location and one element of chain which has a
zero value of linear velocity holding by the surface. This is not mean to stop
the whole ring at this time. This mean stop the red piece and cut the ring at
the same time. The surface holds just one element of ring and other elements of
chain is continuing movement by own trajectories.
>
> If calculate net linear momentum of these elements then this net should be
equal to this ring initial linear momentum. But one of these elements is stop
already and net momentum will be for n-1 elements. In this case one element has
been join to the surface and mass is M(suface) + m(element). Set of elements has
a mass equal to (n-1)*m now. It's change initial condition. The surface is still
keeping same momentum and increase own mass. But chain (set of elements n-1)
should hold same ring initial momentum.
>
>
> Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m
is equal to the ring initial linear momentum?
>
> Would set of n-1 elements return whole ring momentum back to surface?
>
> Will this surface return to initial velocity?
>
>
> In problem complexity
>
> Each element of chain won't stop at the same time. Each element has different
momentum but same mass m. If join each stopped element to the surface, then
surface mass increase faster than chain return momentum. This means the surface
mass growth will help ground to keep own momentum.
>
>
>
> My suggestion:
>
> If ring has set of n elements then for surface point only ring's set of n-1
elements is always move. But one element with zero value of velocity stand down
and it will be always part to surface.
>
> For this particular case, this set of n-1 elements (broken thin ring or chain)
not equivalent to set of n elements (initial ring). Base on law of momentum
conservation net momentum for set n-1 elements would have same initial momentum,
but momentum density will change for each element of this set n-1. The surface
will take all elements momentums back when they'll stop. Base on law of momentum
conservation, from same momentum the body with higher mass will take lower
velocity then body with lower mass.
>
> Initially:
> The ring has mass n*m
> The surface has mass M
>
> After ring to chain conversion:
> The chain has mass (n-1)*m
> The surface has mass M+m
>
> P = m*V = const
>
>
> m1*V1 = m2*V2
>
> The surface with new mass will take a velocity V1 from set of elements n-1.
This velocity is different from initial surface velocity V0, because the surface
mass has been changed.
>
> Follow the law of momentum conservation, even all elements of broken ring will
stop on the surface, this surface will have linear velocity more than zero.
>
> The surface will return back to initial velocity V=0 if rest of the chain
could increase momentum on cut action. But it's nonsense. Base on law of
momentum conservation the chain should keep same initial ring's momentum.
>

#179

From: "abelov0927" <abelov0927@...>
Date: Sat Jun 27, 2009 6:48 pm
Subject: Re: Would rolling body transformation can help find a clue?

abelov0927

Offline

mirror of this problem page is:
http://mysite.verizon.net/vze27vxm/index.htm

#178

From: "abelov0927" <abelov0927@...>
Date: Sat Jun 27, 2009 3:14 am
Subject: Would rolling body transformation can help find a clue?

abelov0927

Offline

link to problem site
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4y\
s/9#

The idea is very simple.

If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with mass
m then each of them conduct linear and circular movement on surface.

Each piece has constant angular velocity. Each piece of ring has variable linear
velocity at surface point. Once per circle each element of ring stop on surface.
At surface point this element has linear velocity value equal to zero.

The ring must be broken in one location and one element of chain which has a
zero value of linear velocity holding by the surface. This is not mean to stop
the whole ring at this time. This mean stop the red piece and cut the ring at
the same time. The surface holds just one element of ring and other elements of
chain is continuing movement by own trajectories.

If calculate net linear momentum of these elements then this net should be equal
to this ring initial linear momentum. But one of these elements is stop already
and net momentum will be for n-1 elements. In this case one element has been
join to the surface and mass is M(suface) + m(element). Set of elements has a
mass equal to (n-1)*m now. It's change initial condition. The surface is still
keeping same momentum and increase own mass. But chain (set of elements n-1)
should hold same ring initial momentum.


Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m is
equal to the ring initial linear momentum?

Would set of n-1 elements return whole ring momentum back to surface?

Will this surface return to initial velocity?


In problem complexity

Each element of chain won't stop at the same time. Each element has different
momentum but same mass m. If join each stopped element to the surface, then
surface mass increase faster than chain return momentum. This means the surface
mass growth will help ground to keep own momentum.



My suggestion:

If ring has set of n elements then for surface point only ring's set of n-1
elements is always move. But one element with zero value of velocity stand down
and it will be always part to surface.

For this particular case, this set of n-1 elements (broken thin ring or chain)
not equivalent to set of n elements (initial ring). Base on law of momentum
conservation net momentum for set n-1 elements would have same initial momentum,
but momentum density will change for each element of this set n-1. The surface
will take all elements momentums back when they'll stop. Base on law of momentum
conservation, from same momentum the body with higher mass will take lower
velocity then body with lower mass.

Initially:
The ring has mass n*m
The surface has mass M

After ring to chain conversion:
The chain has mass (n-1)*m
The surface has mass M+m

P = m*V = const


m1*V1 = m2*V2

The surface with new mass will take a velocity V1 from set of elements n-1. This
velocity is different from initial surface velocity V0, because the surface mass
has been changed.

Follow the law of momentum conservation, even all elements of broken ring will
stop on the surface, this surface will have linear velocity more than zero.

The surface will return back to initial velocity V=0 if rest of the chain could
increase momentum on cut action. But it's nonsense. Base on law of momentum
conservation the chain should keep same initial ring's momentum.