The object without external forces should conduct the uniform motion only, It's
true for all objects in general. However it's not true in details. Real body's
are not single point objects.
To use Newtonian mechanics need to simplify each body of model to single point
object. Then it's possible apply these equations to model. Otherwise these
equations are useless.
This is the main big mistake of our modern classical mechanic.
For example an experiment with 2 pencils.
If pencils have repulsive action aligned from their center of mass then it
possible to use Newtonian mechanic for this model. Each of this pencils
congruent to single point object.All pieces of these pencils conduct same
motion.
If these pencils have repulsive action aligned away from their center mass then
these pencils not congruent to single object points. Because each piece of these
pencils conduct it's own cycloidal motion with own trajectory. In this case
Newtonian mechanic doesn't work.
--- In gravitationalpropulsionstevenson@yahoogroups.com, Alex Belov
<abelov0927@...> wrote:
>
> Sir Newton made a Procrustean bed for classical mechanic.
>
> How?
> First law.
> http://en.wikipedia.org/wiki/Newton's_laws_of_motion
> First Law: "An object in motion will stay in motion and an object at rest will
stay at rest unless acted upon by an external force" or "A body persists in a
state of uniform motion or of rest unless acted upon by an external force."
>
> This law describes object(better says - single point object) behavior, which
"persists in a state of uniform motion". However This laws does not cover
situation when object persist in state of cycloidal motion.
> Lagrangian mechanics and Hamiltonian mechanics are a re-formulation of
classical mechanics that combines conservation of momentum with conservation of
energy. Those mechanics don't cover cycloidal motion also.
> http://en.wikipedia.org/wiki/Lagrangian_mechanics
>
> Base on this law the modern classical mechanics cannot explain some of natural
phenomenons.
> One of this:
> http://knol.google.com/k/paradox-of-classical-mechanics-2#
>
> I hope someone can repeat an experiment from this essay in a future.
>
> Thank you.
>
> Alex
>
> --- On Thu, 2/4/10, cambridgephysics <hubrotherfour@...> wrote:
>
> From: cambridgephysics <hubrotherfour@...>
> Subject: Gravitational Propulsion, Re: Please take a look on my theory
> To: gravitationalpropulsionstevenson@yahoogroups.com
> Date: Thursday, February 4, 2010, 9:37 PM
>
>
>
>
>
>
>
> Â
>
>
>
>
>
>
>
>
>
>
>
> Hi,
> Â Â I tried to respond earlier to this experiment but must have made an error
in posting.
> My question has to do with the use of perhaps a round copper rod instead of a
pencil in performing this experiment? This regards physical properties that may
influence results.
> Â Have you any results using anything other than pencils?
> Thankyou,
> George
>
> --- In gravitationalpropul sionstevenson@ yahoogroups. com, "abelov0927"
<abelov0927@ ..> wrote:
> Â Hi.
> Please take a look on my site.
> http://knol. google.com/ k/paradox- of-classical- mechanics- 2
> Â This site contains theory, simulation and real experiment.
> Â The real experiment proves this theory.
> Â However, to prove this theory this experiment should be repeated many times
by other research groups.
> Â
> Please read it and feel free to comment it.
> Â
> Thank You.
>
Sir Newton made a Procrustean bed for classical mechanic.
How? First law. http://en.wikipedia.org/wiki/Newton's_laws_of_motion First Law: "An object in motion will stay in motion and an object at rest will stay at rest unless acted upon by an external force" or "A body persists in a state of uniform motion or of rest unless acted upon by an external force."
This law describes object(better says - single point object) behavior, which "persists in a state of uniform motion". However This laws does not cover situation when object persist in state of cycloidal motion. Lagrangian mechanics and Hamiltonian mechanics are a re-formulation of classical mechanics that combines conservation of momentum with conservation of energy. Those mechanics don't cover cycloidal motion also. http://en.wikipedia.org/wiki/Lagrangian_mechanics
Base
on this law the modern classical mechanics cannot explain some of natural phenomenons. One of this: http://knol.google.com/k/paradox-of-classical-mechanics-2#
I hope someone can repeat an experiment from this essay in a future.
Thank you.
Alex
--- On Thu, 2/4/10, cambridgephysics <hubrotherfour@...> wrote:
From: cambridgephysics <hubrotherfour@...> Subject: Gravitational Propulsion, Re: Please take a look on my theory To: gravitationalpropulsionstevenson@yahoogroups.com Date: Thursday, February 4, 2010, 9:37 PM
Hi,
I tried to respond earlier to this experiment but must have made an error in posting.
My question has to do with the use of perhaps a round copper rod instead of a pencil in performing this experiment? This regards physical properties that may influence results.
Have you any results using anything other than pencils?
Thankyou,
George
--- In gravitationalpropul sionstevenson@ yahoogroups. com, "abelov0927" <abelov0927@. ..> wrote: Hi. Please take a look on my site.
http://knol. google.com/ k/paradox- of-classical- mechanics- 2 This site contains theory, simulation and real experiment. The real experiment proves this theory. However, to prove this theory this experiment should be repeated many times by other research groups.
Here's my thinks about this experiment.
The problem is, the rotating pencil starts 2 motions in one time. Rather than
non-rotating pencil starts just one motion.
For rotating pencil:
For each of motions (translation and rotation) need a force to accelerate the
rotating pencil. The modern classic mechanic uses net force for sum of sub
forces which spending to initiate one of the type of motions.
However, these pencils take just one momentum from each other on ideal
model(from rubber band on real experiment).
Base on third Newtons Law, these pencils must cause forces with same value and
opposite directions.
For Non-Rotating pencil.
Base on third Newton's law, this all net force is spending to accelerate pencil
on just one type - the translation motion.
Base on law of momentum conservation, these pencils must have identical by value
translation velocities with opposite directions in case if both objects(pencils)
conduct translation motion only. However, one of these objects(rotating pencil)
conduct two motions together.
I made my own theory about this case. I think, the nature has one more
standalone motion which I call "Rotation with Translation Movement". Base on
this theory the pencil with this "rotation with translation motion" will take a
lower translation velocity than pencil with translation motion only. This
experiment proves it.(for me)
This paradox may be a clue for many unsolved things in physics world. It's
simple. However, it may completely change a world.
The classical mechanics describe motions base on single point objects.
Here's complex object which conduct complex motion.
Could someone repeat this experiment with high precision equipment?
--- On Thu, 2/4/10, cambridgephysics <hubrotherfour@...> wrote:
From: cambridgephysics <hubrotherfour@...>
Subject: Gravitational Propulsion, Re: Please take a look on my theory
To: gravitationalpropulsionstevenson@yahoogroups.com
Date: Thursday, February 4, 2010, 9:37 PM
Â
Hi,
  I tried to respond earlier to this experiment but must have made an error
in posting.
My question has to do with the use of perhaps a round copper rod instead of a
pencil in performing this experiment? This regards physical properties that may
influence results.
 Have you any results using anything other than pencils?
Thankyou,
George
--- In gravitationalpropul sionstevenson@ yahoogroups. com, "abelov0927"
<abelov0927@. ..> wrote:
 Hi.
Please take a look on my site.
http://knol. google.com/ k/paradox- of-classical- mechanics- 2
 This site contains theory, simulation and real experiment.
 The real experiment proves this theory.
 However, to prove this theory this experiment should be repeated many times by
other research groups.
Â
Please read it and feel free to comment it.
Â
Thank You.
Hi.
How it could be change result?
Anyway.
Base on law of momentum conservation P=cosnt and momentum equation P=mv, the
translation velocity of objects with identical mass must have identical value
and opposite direction. This law of momentum conservation for objects which
conduct translation motions only.
However one of these objects conduct rotation and translation motion together.
This is a standalone phenomenon.
However classical with own law of momentum conservation. However, the modern
classical mechanic has rotation and translation motions only. With their own law
of momentum conservation.
Main Idea of this experiment was show the rotation with translation motion as a
standalone natural phenomenon with own law of momentum conservation. This motion
should be included into classical mechanics book.
Thank you
Alex
--- On Thu, 2/4/10, cambridgephysics <hubrotherfour@...> wrote:
From: cambridgephysics <hubrotherfour@...>
Subject: Gravitational Propulsion, Re: Please take a look on my theory
To: gravitationalpropulsionstevenson@yahoogroups.com
Date: Thursday, February 4, 2010, 9:37 PM
Â
Hi,
  I tried to respond earlier to this experiment but must have made an error
in posting.
My question has to do with the use of perhaps a round copper rod instead of a
pencil in performing this experiment? This regards physical properties that may
influence results.
 Have you any results using anything other than pencils?
Thankyou,
George
--- In gravitationalpropul sionstevenson@ yahoogroups. com, "abelov0927"
<abelov0927@. ..> wrote:
 Hi.
Please take a look on my site.
http://knol. google.com/ k/paradox- of-classical- mechanics- 2
 This site contains theory, simulation and real experiment.
 The real experiment proves this theory.
 However, to prove this theory this experiment should be repeated many times by
other research groups.
Â
Please read it and feel free to comment it.
Â
Thank You.
I tried to respond earlier to this experiment but must have made an error in posting.
My question has to do with the use of perhaps a round copper rod instead of a pencil in performing this experiment? This regards physical properties that may influence results.
Have you any results using anything other than pencils?
Thankyou,
George
--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote: Hi. Please take a look on my site.
http://knol.google.com/k/paradox-of-classical-mechanics-2 This site contains theory, simulation and real experiment. The real experiment proves this theory. However, to prove this theory this experiment should be repeated many times by other research groups.
Hi.
Please take a look on my site.
http://knol.google.com/k/paradox-of-classical-mechanics-2
This site contains theory, simulation and real experiment.
The real experiment proves this theory.
However, to prove this theory this experiment should be repeated many
times by other research groups.
Please read it and feel free to comment it.
Thank You.
The problem is, the rotating pencil starts 2 motions in one time. Rather than
non-rotating pencil starts just one motion.
For rotating pencil:
For each of motions (translation and rotation) need a force to accelerate the
rotating pencil. The modern classic mechanic uses net force for sum of sub
forces which spending to initiate one of the type of motions.
However, these pencils take just one momentum from each other on ideal
model(from rubber band on real experiment).
Base on third Newtons Law, these pencils must cause forces with same value and
opposite directions.
For Non-Rotating pencil.
Base on third Newton's law, this all net force is spending to accelerate pencil
on just one type - the translation motion.
Base on law of momentum conservation, these pencils must have identical by value
translation velocities with opposite directions in case if both objects(pencils)
conduct translation motion only. However, one of these objects(rotating pencil)
conduct two motions together.
I made my own theory about this case. I think, the nature has one more
standalone motion which I call "Rotation with Translation Movement". Base on
this theory the pencil with this "rotation with translation motion" will take a
lower translation velocity than pencil with translation motion only. This
experiment proves it.(for me)
This paradox may be a clue for many unsolved things in physics world. It's
simple. However, it may completely change a world.
The classical mechanics describe motions base on single point objects.
Here's complex object which conduct complex motion.
Could someone repeat this experiment with high precision environment?
--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
<abelov0927@...> wrote:
>
> Hi.Could someone repeat this
> experiment?http://knol.google.com/k/paradox-of-classical-mechanics-2#
> Thank you
>
>
> 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.
> 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
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/pencils1.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/pencils22.jpg>
>
> The following are the snapshots of the experiment dynamics.
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/1m.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/2m.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/3m.jpg>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/4m.jpg> Links to
> experiments movies (avi files)
> Experiment 2_1 <http://mysite.verizon.net/vze27vxm/exp1/exp1.avi>
> Experiment 2_2 <http://mysite.verizon.net/vze27vxm/exp2/exp2.avi>
> Experiment 2_3 <http://mysite.verizon.net/vze27vxm/exp3/exp3.avi>
>
> Links to experiments pictures (zip files)
> Experiment 2_1 <http://mysite.verizon.net/vze27vxm/exp1/exp1_images.zip>
> Experiment 2_2 <http://mysite.verizon.net/vze27vxm/exp2/exp2_images.zip>
> Experiment 2_3 <http://mysite.verizon.net/vze27vxm/exp3/exp3_images.zip>
>
>
> The experiment results The translation velocity of non-rotating pencil
> has value - 2.88 m/s The translation velocity of rotating pencil has
> value - 2.17 m/s The translation velocity of rotating pencil 25% less
> than translation velocity of non-rotating pencil. Base on momentum
> equation (P=m*V) and law of momentum conservation this velocity
> deference should be in range of mass difference. In case if mass of
> pencils has difference about 5-10%, the translation velocity should have
> difference 5-10% also.
> However, the translation velocity has difference 25% on this experiment.
> The translation velocities of these pencils are not equal.
> ConclusionThe 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.
>
3 successful experiments were conducted with 2 pencils. In these experiments pencils with rotation movement have lower velocity than pencils without rotation.
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 following are the snapshots of the experiment dynamics.
The translation velocity of non-rotating pencil has value - 2.88 m/s
The translation velocity of rotating pencil has value - 2.17 m/s
The translation velocity of rotating pencil 25% less than translation velocity of non-rotating pencil.
Base on momentum equation (P=m*V) and law of momentum conservation this velocity deference should be in range of mass difference. In case if mass of pencils has difference about 5-10%, the translation velocity should have difference 5-10% also.
However, the translation velocity has difference 25% on this experiment.
The translation velocities of these pencils are not equal.
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.
Lunar laser ranging: 40 years of high-level science
"Lunar reflectors left by Apollo astronauts have been used continuously since 1969 to study the earth-moon system and test the validity of the laws of physics. Five "corner mirrors" or "retro-reflector arrays" were taken to the moon in 1969 and the early 1970s (Figures 1 and 2). Corner mirrors are important scientific instruments because, when struck precisely by a laser beam, they reflect the beam in a parallel path back to the source of the laser. The arrays left on the moon were used continuously in the lunar laser ranging, or LLR experiment. "
This is kinda burried in a page of data, but may be of interest;
"The FE will also turn on the Russian TEKh-15/DAKON-M IZGIB ("Bend") experiment hardware in the SM to take structural dynamics data during the undocking operation. Afterwards, DAKON will be deactivated again. [IZGIB has the objective to help update mathematical models of the ISS gravitation environment, using accelerometers of the Russian SBI Onboard Measurement System, the GIVUS high-accuracy angular rate vector gyrometer of the SUDN Motion Control & Navigation System and other accelerometers for unattended measurement of micro-accelerations at science hardware accommodation locations - (1) in operation of onboard equipment having rotating parts (gyrodynes, fans), (2) when establishing and keeping various ISS attitude modes, and (3) when performing crew egresses into space and physical exercises.]"
"The flexibility is in response to technological advances. Such flexpath destinations could well include these individual bodies, but also opportunistic sites bereft of deep gravity wells such as NEOs, Lagrange (libration) points in the vicinity of the Earth, and eventually a Martian moon to be used as a teleoperations center for work on the Red Planet. It is recognized that both the flexpath strategy and a strategy directed wholly towards the Moon, as is the baseline plan, were both considered by the Augustine committee to be exploration strategies that were viable."
Russia to develop nuclear-powered spacecraft for Mars mission
"MOSCOW. (Yury Zaitsev, for RIA Novosti) - At its recent meeting, the Presidential Commission for Modernization and Technological Development of Russia's Economy called for developing a transport-energy module with a megawatt-class nuclear propulsion unit.
President Dmitry Medvedev approved the project and promised to find funding for it. Analysts say Russia could restore its status as a leading space power if it scores a breakthrough with nuclear propulsion. Soviet and U.S. nuclear spacecraft programs were marred by a number of accidents."
"Astronomers are announcing today that they have discovered 33 pairs of waltzing black holes in distant galaxies. This result was presented by Dr. Julia Comerford of the University of California, Berkeley in Berkeley, California to the American Astronomical Society meeting in Washington, DC. This result is particularly important because it shows that supermassive black hole pairs are more common than previously known from observations, and because the black hole pairs can be used to estimate how often galaxies merge with each other."
Propelled by light: the promise and perils of solar sailing
"The Society hopes to launch the sail in about a year as part of a three-stage plan to demonstrate the viability of solar sail propulsion, which has never been tested in orbit. The group says it is the only practical technology that might be used for interstellar travel, since the light generates a small but constant pressure that should accelerate a sail to high speeds over time"
Russia plans to save Earth from asteroid hit but NASA unperturbed
"Russia kicked up a celestial storm after the head of its space agency claimed it may send a spacecraft to deflect a large asteroid off its course. Roscosmos chief Anatoly Perminov said this will reduce the possibility of an earth impact.
Though scientists elsewhere are sceptical, there is concern that the Russians might still go ahead because such a 'space cowboy' project can capture people's imagination and attract a lot of sponsors."
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;
"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;
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;
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.
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:
Thus it is inferred that their kinetic energies are identical. Now, lets compare their translation energies and angular momentum.
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:
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.
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:
Where, Pj - linear momentum Lk - angular momentum Ru - unit radius
The law of conservation of momentum for the translation movement with rotation is:
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:
Which implies that:
Assuming this momentum is applied to the angular movement:
So, the angular momentum is:
Translation momentum is then equal to:
Lets sum these parts and check full momentum:
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:
Translation and angular velocities for cylinder 2 equal to:
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:
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:
WhereP - 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.
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.
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.
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).
--- 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 >
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
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?
>
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.jpghttp://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com1.jpghttp://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
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 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.
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.
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:
As it shown on equation their energies are equal. Let's compare their translation and angular momentums.
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:
As it shown on equation their energies are not equal. Let's compare their translation and angular momentums.
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:
note: Pj - linear momentum Lk - angular momentum Ru - unit radius
This is the law of momentum conservation for the translation with rotation movement:
This law has a complex number and it has linear and angular momentums.
Full momentum transfer for this movement has calculation by absolute value:
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:
Full kinetic energy of rotation with translation movement is:
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.
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 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:
As it shown on equation their energies are equal. Let's compare their translation and angular momentums.
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:
As it shown on equation their energies are not equal. Let's compare their translation and angular momentums.
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:
note: Pj - linear momentum Lk - angular momentum Ru - unit radius
This is the law of momentum conservation for the translation with rotation movement:
This law has a complex number and it has linear and angular momentums.
Full momentum transfer for this movement has calculation by absolute value:
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:
Full kinetic energy of rotation with translation movement is:
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.
My assumption this movement have a linear and angular momentums together. This is the law of momentum conservation for the translation with rotation movement:
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:
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:
As it shown on equation their energies are equal. Let's take the derivatives from their parts of energy.
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:
As it shown on equation their energies are not equal. Let's take the derivatives from their parts of energy.
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 observer.
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.
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#
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 momentumvector 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:
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. >
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.
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, à repulsion and à 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. > > > > > > > > > > > > > > >
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, à repulsion and à 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. > > > > > > > > > >
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.
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, à repulsion and à 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. > > > > > >
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