In the operation of a medieval battle flail,which is shaft with a chain attached to one end and ball attached to the other end of the chain, I have noticed that,
energy is conserved, angular momentum is conserved, but linear momentum is
not.
The example is as follows-
You start with the end of 1 m rod connected to a 1 m essentially massless chain, with a 2kg steel ball connected to the other end of the chain.
A man with superhuman strength holds the end of the rod without the chain at
the origin of the x and y axis and positions the rod at 45 degrees with respect
to the x axis, the ball rests on the x axis 1.414 m from the origin, the chain is held
straight and perpendicular to the rod.
The man now applies an angular accelerating force to the rod while keeping one end of rod at the origin of the x and y axis and keeping the ball while accelerating in a perfectly circular path with a radius of 1.414 m.
In this case the chain Will be maintained at 45 degrees to the trajectory of the ball.
In order to maintain the circular trajectory of the ball the accelerating force in line
with the trajectory of the ball must equal the centrifugal force Acton on the ball, with the
chain maintained at 45 degrees centrifugal force equals accelerating force.
Heres the important part the rod is angularly accelerated from a position of
45 degrees to 135 degrees and the ball is accelerated from 1.414 m on the x
axis in a circular path to 1.414 m on the y axis moving in th -x direction with a velocity of z.
The key lies here in the fact that the force acting on the chain at the end of the rod is 1.414 times the force acting on accelerating the ball.
Working with momentum in the x direction with the rod at 135 degrees or 45 degrees the
momentum in the x direction or F*t is a minimum of .707*1.414 or the same force acting on accelerating the ball. The real important part is when the rod is at 90
degrees the accelerating F*t in the x direction is 1.414 times the F*t acting on
accelerating the ball.
Therefore the accelerating the steel ball of the flail to a velocity of z in the x direction produces more of a reactionary force on the x axis than straight line acceleration of the ball to a velocity of z in the x direction without the flail.
Anyone beg to differ?
QUOTE (Carl Fischbach+Jul 23 2007, 05:40 AM)
I have noticed that, energy is conserved, angular momentum is conserved, but linear momentum is not.
Momentum conservation is a fundamental law of physics. This forum is allegedly set up for advancing new theories not crass misunderstandings of existent theories.
Momentum conservation is a fundamental law of physics. This forum is allegedly set up for advancing new theories not crass misunderstandings of existent theories.
Sorry I offended you with my work. I thought if you accelerate a steel ball with a flail from a rest to a velocity of z in the -x direction then decelerate it in a straight
line to zero without the flail you would be left with a net force in the +x direction.
I also thought that if situation could possibly exist where you have imbalanced forces after acceleration of a mass from zero to z and deceleration to zero, would
be of interest to people in physics.
line to zero without the flail you would be left with a net force in the +x direction.
I also thought that if situation could possibly exist where you have imbalanced forces after acceleration of a mass from zero to z and deceleration to zero, would
be of interest to people in physics.
QUOTE (Dallas+Jul 23 2007, 09:25 AM)
Momentum conservation is a fundamental law of physics.
Even the AWT is based on such law. But what if such law can be even violated? We should realize, the momentum law is valid just on the certain time moment. But at the case of strong spacetime curvature, even the subtle distance interval will result into the substantial time interval. Such insight makes the momentum conservation law pretty local thing by the same way, like the equivalence principle. By AWT the inertia is the result of Aether density gradients. Just because the mass/energy distribution of Aether is violated, the volume of Aether exhibits the density gradients, therefore it's behaves like the real matter. If you'll smooth these gradients again, you'll obtain an apparently empty space again. The energy density required for keeping such space undulating should correspond exactly the mass density obtained by such undulation. The zero mass-energy density of space is therefore just one of many possible states of aether fluctuations (just very, very improbable). The cleaning up such space would require very effectively working Maxwell's daemon, but it doesn't violate the energy conservation law in principle: if you remove the energy from the space, you will remove even the mass. Such process violates anything, just the probability of states. Note, the thermodynamic arrow of time is based on the inertial diffusion of Aether particles. If you remove such inertia, you'll remove all reasons for violation of the second thermodynamical law as well.
For example, just let imagine, what will happen, if some black hole swallows some matter. Will the resulting mass remain exactly additive, after then? Providing that every possible matter loss (via accretion radiation, change of black hole torque, etc will be considered). For example, we can imagine, the gravitational pressure inside of black hole will restrict the motion of Aether particles by such way, the inertia density inside of such black hole will effectively cease to zero and the black hole will virtually swallow the matter without the rest. We should consider, every law in the nature has limited validity scope, so it can be violated in sufficiently extreme conditions.
Of course, the deeper explanation of momentum conservation law violation would open the door for causal explanation of Universe existence from virtually nothing, so it should be handled with care.
Even the AWT is based on such law. But what if such law can be even violated? We should realize, the momentum law is valid just on the certain time moment. But at the case of strong spacetime curvature, even the subtle distance interval will result into the substantial time interval. Such insight makes the momentum conservation law pretty local thing by the same way, like the equivalence principle. By AWT the inertia is the result of Aether density gradients. Just because the mass/energy distribution of Aether is violated, the volume of Aether exhibits the density gradients, therefore it's behaves like the real matter. If you'll smooth these gradients again, you'll obtain an apparently empty space again. The energy density required for keeping such space undulating should correspond exactly the mass density obtained by such undulation. The zero mass-energy density of space is therefore just one of many possible states of aether fluctuations (just very, very improbable). The cleaning up such space would require very effectively working Maxwell's daemon, but it doesn't violate the energy conservation law in principle: if you remove the energy from the space, you will remove even the mass. Such process violates anything, just the probability of states. Note, the thermodynamic arrow of time is based on the inertial diffusion of Aether particles. If you remove such inertia, you'll remove all reasons for violation of the second thermodynamical law as well.
For example, just let imagine, what will happen, if some black hole swallows some matter. Will the resulting mass remain exactly additive, after then? Providing that every possible matter loss (via accretion radiation, change of black hole torque, etc will be considered). For example, we can imagine, the gravitational pressure inside of black hole will restrict the motion of Aether particles by such way, the inertia density inside of such black hole will effectively cease to zero and the black hole will virtually swallow the matter without the rest. We should consider, every law in the nature has limited validity scope, so it can be violated in sufficiently extreme conditions.
Of course, the deeper explanation of momentum conservation law violation would open the door for causal explanation of Universe existence from virtually nothing, so it should be handled with care.
Something from nothing sounds good to me.
QUOTE (Carl Fischbach+Jul 23 2007, 04:17 PM)
Something from nothing sounds good to me.
Unfortunately the only mechanism of such creation known so far is the phase transition of matter, i.e. something like the precipitation of droplets from saturated vapor.
This is not really the formation of something from nothing, this is just a cheating from causual point of view, despite of how deeply is such concept generalized.
Unfortunately the only mechanism of such creation known so far is the phase transition of matter, i.e. something like the precipitation of droplets from saturated vapor.
This is not really the formation of something from nothing, this is just a cheating from causual point of view, despite of how deeply is such concept generalized.
Zephir, you should write a disclaimer before you say anything else that says "AWT is my own crank theory, which not many people believe". Carl is trying to find a real solution to describe a real experiment. AWT cannot help, and will only confuse him if he does not know what it is.
QUOTE (Latrosicarius+Jul 23 2007, 05:14 PM)
Carl is trying to find a real solution to describe a real experiment.
OK, he can post such example into Homework section, after then.
OK, he can post such example into Homework section, after then.
QUOTE
OK, he can post such example into Homework section, after then.
Well, have you considered that some homeworks could be excellent for you?
One small correction to this is experiment to get the ball to follow a circle while
being accelerated, instead of the ball being at rest initially it has to be moving
at a very small velocity in the +y direction when it crosses the x axis,and the rod
is at 45 degrees when acceleration is started, this does not effect the results.
being accelerated, instead of the ball being at rest initially it has to be moving
at a very small velocity in the +y direction when it crosses the x axis,and the rod
is at 45 degrees when acceleration is started, this does not effect the results.
QUOTE (Lalbatros+Jul 23 2007, 07:19 PM)
Well, have you considered that some homeworks could be excellent for you?
Sorry, this is irrelevant question here as well. You can use PM for questions of individual character.
Sorry, this is irrelevant question here as well. You can use PM for questions of individual character.
Zephir,
This is too late now: you answered!
I told you there should be moderation on this site ...
I would stop this kind of stupid post, and you be silenced ...
This is too late now: you answered!
I told you there should be moderation on this site ...
I would stop this kind of stupid post, and you be silenced ...
Carl Fischbach,
Your story is very complicated to read.
Could you try to make a drawing, or could you ask Zephir to make one for you?
Anyway, momentum conservation should make no doubt.
You need to think about it, making the description clearer will help you too.
Your story is very complicated to read.
Could you try to make a drawing, or could you ask Zephir to make one for you?
Anyway, momentum conservation should make no doubt.
You need to think about it, making the description clearer will help you too.
Carl,
it appears to me that linear momentum never occurs in this setup. From what I can see only angular momentum comes into play. Unless the ball traverses the x axis without describing an arc this setup is pure angular momentum.
it appears to me that linear momentum never occurs in this setup. From what I can see only angular momentum comes into play. Unless the ball traverses the x axis without describing an arc this setup is pure angular momentum.
To the best of my knowledge when accelerating a mass from a velocity of zero
regardless of the path it is accelerated and decelerated back to a velocity of zero
the net resultant forces in both the x and y direction must be zero. In the example
I've described with accelerating of the ball with the flail from 0 m/s in the x direction in a 90 degree circular arc to z m/s in the -x direction with 0 m/s
in the y direction then decelerating in a straight line in the x direction to 0 m/s with out the flail there is a net resultant force in the +x direction.
regardless of the path it is accelerated and decelerated back to a velocity of zero
the net resultant forces in both the x and y direction must be zero. In the example
I've described with accelerating of the ball with the flail from 0 m/s in the x direction in a 90 degree circular arc to z m/s in the -x direction with 0 m/s
in the y direction then decelerating in a straight line in the x direction to 0 m/s with out the flail there is a net resultant force in the +x direction.
First off, a perfectly circular path for the ball is impossible. The path is an elongated ellipsoid. The shaft must also move in relation to the changing orbit of the ball in order to maintain change or maintain the orbital speed and direction.
Secondly the angular force needed to create the movement of the ball is offset by the person or machine swinging the haft. No one and nothing can create the movement of a morning star without the movement of the morning star, moving them in turn.
Secondly the angular force needed to create the movement of the ball is offset by the person or machine swinging the haft. No one and nothing can create the movement of a morning star without the movement of the morning star, moving them in turn.
Carl,
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
Of course, critical investigation is another essential element of the Scientific Process.
As recommended above, diagrams would help clarify your presentation.
Since the process described occurs in 3-D space (+time), multiple views and projections at multiple instances will probably be needed.
At least, this could ameliorate potential for ambiguity (in the reader's mind) about the context and view-points in the use of otherwise clear-seeming descriptive phrases.
Be sure to include transfers of linear and angular momentum between the ball&chain and the swinger {person's body} as a whole, and also between the swinger and the swinger's anchor {the comparatively large-massed Earth}.
By the way, I suspect that the overlooked momentum transfers from the ball&chain to the swinger, integrated over trajectories, will account for the apparent failure of linear momentum conservation.
Consider an equivalent, unanchored, gravity-free example:
<> Provide a cylindrically symmetrical Space Capsule with a manipulative utility arm attached to an end.
<> Let the ball's mass be large, perhaps 10% of the Capsule's.
<> Let the manipulative arm (like the ball&chain's stick) be considered of comparatively small mass and moment of inertia.
<> Let the Capsule use its manipulating Arm to slowly keep the Ball swinging in a circle.
<> Carefully calculate the motion of the swinger (i.e. the Space Capsule), as well as the motion of the Ball.
<> Determine whether the Capsule's motions are what conservation of both linear and angular momentum would predict. If so, then the discrepancies that you calculated in your on-Earth example, are likely attributable to neglect of momentum transfers to your swinger and the Earth.
NOTE illusory immobility of a large anchor like the Earth:
<>Although momentum transfers between the Ball and the Earth (via the swinger) are predicted to be equal and opposite, the velocity changes for each object is inversely proportionate to its mass.
<>In other words, the Earth's change in velocity due to this momentum transfer from the Ball, is truly negligible, giving the illusion that there is no momentum transfer to the Earth in your example.
Check it out for yourself.
And keep on making note of apparent discrepancies in the Sciences.
And raise the issues to get answers if you cannot resolve them.
That is how you gain knowledge in a field -- and methodologies to gain more.
And sometimes you may discover something altogether new.
One would think that snobbery would have no place in the Sciences.
But it exists here as in any other discipline.
As a teaching tool, I can only see snobbery aiding other would-be snobs.
But I suppose it has value in toughening your hide.
You know that throughout your life you will face such responses from snobs,
whether in raucous reaction to errors [which do occur],
or in ignorant or careless response to novel concepts.
Good luck,
Robby B
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
Of course, critical investigation is another essential element of the Scientific Process.
As recommended above, diagrams would help clarify your presentation.
Since the process described occurs in 3-D space (+time), multiple views and projections at multiple instances will probably be needed.
At least, this could ameliorate potential for ambiguity (in the reader's mind) about the context and view-points in the use of otherwise clear-seeming descriptive phrases.
Be sure to include transfers of linear and angular momentum between the ball&chain and the swinger {person's body} as a whole, and also between the swinger and the swinger's anchor {the comparatively large-massed Earth}.
By the way, I suspect that the overlooked momentum transfers from the ball&chain to the swinger, integrated over trajectories, will account for the apparent failure of linear momentum conservation.
Consider an equivalent, unanchored, gravity-free example:
<> Provide a cylindrically symmetrical Space Capsule with a manipulative utility arm attached to an end.
<> Let the ball's mass be large, perhaps 10% of the Capsule's.
<> Let the manipulative arm (like the ball&chain's stick) be considered of comparatively small mass and moment of inertia.
<> Let the Capsule use its manipulating Arm to slowly keep the Ball swinging in a circle.
<> Carefully calculate the motion of the swinger (i.e. the Space Capsule), as well as the motion of the Ball.
<> Determine whether the Capsule's motions are what conservation of both linear and angular momentum would predict. If so, then the discrepancies that you calculated in your on-Earth example, are likely attributable to neglect of momentum transfers to your swinger and the Earth.
NOTE illusory immobility of a large anchor like the Earth:
<>Although momentum transfers between the Ball and the Earth (via the swinger) are predicted to be equal and opposite, the velocity changes for each object is inversely proportionate to its mass.
<>In other words, the Earth's change in velocity due to this momentum transfer from the Ball, is truly negligible, giving the illusion that there is no momentum transfer to the Earth in your example.
Check it out for yourself.
And keep on making note of apparent discrepancies in the Sciences.
And raise the issues to get answers if you cannot resolve them.
That is how you gain knowledge in a field -- and methodologies to gain more.
And sometimes you may discover something altogether new.
One would think that snobbery would have no place in the Sciences.
But it exists here as in any other discipline.
As a teaching tool, I can only see snobbery aiding other would-be snobs.
But I suppose it has value in toughening your hide.
You know that throughout your life you will face such responses from snobs,
whether in raucous reaction to errors [which do occur],
or in ignorant or careless response to novel concepts.
Good luck,
Robby B
Lets take a circle with a radius of 1.414 m and place it over an x y axis, with
the center of the circle at the origin of the axis. Starting at where the circle
intersects the +x axis the ball is made to follow the pathway laid out by the circle
as it moves in a counterclockwise direction by forces acting on the ball.
The ball is given .1 m/s velocity in the +y direction where the circle intersects
the +x axis. It is given that if you have an accelerating force acting on the ball
in the same direction as the instantaneous direction of travel of the ball, the only
force that is needed to keep the ball in a circular direction of travel is a centrifugal
force acting perpendicular to the instantaneous direction of travel of the ball, which
is governed by the equation F=mv^2/r. If you maintain the accelerating force acting
in the direction of travel of the ball the same as the centrifugal force, the ball
will have to follow a circular pathway according to known laws.If the chain connected to the ball is kept at 45 degrees with with respect to the instantaneous
direction of travel the ball, and 45 degrees with respect to the perpendicular of the instantaneous direction of travel of the ball and a force of 1.414*s is applied to the end of the chain where s equals the centrifugal force of the ball, the chain will produce a centrifugal force s perpendicular to the direction of travel of the ball and a an accelerating force s in the direction of travel of the ball. If the chain is 1m
long ,the rod is 1m long and radius of the balls circular path is 1.414 m the chain
will be maintained at 90 degree with respect to the rod,where it is attached at
the end of the rod. If the end of the rod is maintained at the origin during
angular acceleration of the rod, the chain must be maintained at 90 degrees
with respect to the rod during acceleration. The inertial forces in the x direction experienced by the person swinging the flail must equal the inertia experienced
by the person linearly decelerating ball in the x direction.
the center of the circle at the origin of the axis. Starting at where the circle
intersects the +x axis the ball is made to follow the pathway laid out by the circle
as it moves in a counterclockwise direction by forces acting on the ball.
The ball is given .1 m/s velocity in the +y direction where the circle intersects
the +x axis. It is given that if you have an accelerating force acting on the ball
in the same direction as the instantaneous direction of travel of the ball, the only
force that is needed to keep the ball in a circular direction of travel is a centrifugal
force acting perpendicular to the instantaneous direction of travel of the ball, which
is governed by the equation F=mv^2/r. If you maintain the accelerating force acting
in the direction of travel of the ball the same as the centrifugal force, the ball
will have to follow a circular pathway according to known laws.If the chain connected to the ball is kept at 45 degrees with with respect to the instantaneous
direction of travel the ball, and 45 degrees with respect to the perpendicular of the instantaneous direction of travel of the ball and a force of 1.414*s is applied to the end of the chain where s equals the centrifugal force of the ball, the chain will produce a centrifugal force s perpendicular to the direction of travel of the ball and a an accelerating force s in the direction of travel of the ball. If the chain is 1m
long ,the rod is 1m long and radius of the balls circular path is 1.414 m the chain
will be maintained at 90 degree with respect to the rod,where it is attached at
the end of the rod. If the end of the rod is maintained at the origin during
angular acceleration of the rod, the chain must be maintained at 90 degrees
with respect to the rod during acceleration. The inertial forces in the x direction experienced by the person swinging the flail must equal the inertia experienced
by the person linearly decelerating ball in the x direction.
We have to assume there is no gravity otherwise this won't work. Or place
the flail in a horizontal position. I assumed a gravity free environment ridiculous oversight on my part.
the flail in a horizontal position. I assumed a gravity free environment ridiculous oversight on my part.
Aside from gravity, the ball cannot be accelerated in a circle using a flexible chain. Using a spinning mount would allow the handle to be planted in a solid position once the ball is moving. At that point friction/gravity would determine the orbit path of the ball.
The main point being, the altering orbit of the ball changes the leverage applied to the swinger. Thus changing where the balance of inertia is going. The transfer of inertia during acceleration is not constant, it peaks and ebbs as the ball reaches differing parts of it's path. If you are going to show that there is excess energy being created or missing energy not accounted for, then the effects of the changing orbit have to be taken into account.
Take a look through Free Energy News. Aside from every crackpot scheme to free electrictyor perpetual motion, they also have a lot of good advice on how to sanity check experiments and ideas. Half the people there spend their time debunking crackpot ideas. The other half keeps thinking up new ones. The one that would probably interest you would be the centerfugal engine (reactionless drive).
The main point being, the altering orbit of the ball changes the leverage applied to the swinger. Thus changing where the balance of inertia is going. The transfer of inertia during acceleration is not constant, it peaks and ebbs as the ball reaches differing parts of it's path. If you are going to show that there is excess energy being created or missing energy not accounted for, then the effects of the changing orbit have to be taken into account.
Take a look through Free Energy News. Aside from every crackpot scheme to free electrictyor perpetual motion, they also have a lot of good advice on how to sanity check experiments and ideas. Half the people there spend their time debunking crackpot ideas. The other half keeps thinking up new ones. The one that would probably interest you would be the centerfugal engine (reactionless drive).
See the differences here:
[Corvidae]
Aside from gravity, the ball cannot be accelerated in a circle using a flexible chain.
The first quote is probably from someone who talks but does never solves something, never learns, and just dreams without feedback.
Often these people are called "creative" "open-minded" people on this forum.
The second quote is from a physicist who simply knows his basics, maybe even an engineer.
Often these people are called "mainstream" "narrow minded" people on this forum.
You could imagine what verbiage Zephir could have produced to answer this question!
Do you need to read Popper before builduig a bridge?
QUOTE
[Robby Bloom]
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
QUOTE (->
| QUOTE |
| [Robby Bloom] I applaud your initiative in presenting your proposal "non-conservation of linear momentum". Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}. |
[Corvidae]
Aside from gravity, the ball cannot be accelerated in a circle using a flexible chain.
The first quote is probably from someone who talks but does never solves something, never learns, and just dreams without feedback.
Often these people are called "creative" "open-minded" people on this forum.
The second quote is from a physicist who simply knows his basics, maybe even an engineer.
Often these people are called "mainstream" "narrow minded" people on this forum.
You could imagine what verbiage Zephir could have produced to answer this question!
Do you need to read Popper before builduig a bridge?
I'm not expert in physics, I believe my analysis is correct, until conclusively
proven otherwise. The chain is essentially massless by the way.
I am an expert in free energy though. I've looked up every article on free energy
news and I wouldn't be in any rush to condemn all those new energy systems.
Energy is a field of interest of mine, and been researching new energy systems
and free energy systems for the past 15 years. I work on designs for improved
efficiency of internal combustion engines and new free energy machines.
I'd like to get something done with improving the efficiency of the internal combustion engine,without serious money you can't get anything done.
As far as free energy is concerned I can build you a free energy machine
today they have been documented to work I've analyzed the designs I
know how they work and why they work, the problem is I would be
setting myself up as the biggest target ever.The problem with the people
in the free energy field they don't think about the consequences of their
work. If you unveiled a free energy machine in the world today the
stock market would crash, there would be mass layoffs Kodak I think
is an example of that, not to mention national security concerns.
On the flip side we are wrecking the planet with fossil fuels the cost
of fossil fuels is going to bankrupt the west in the end, supply and demand.
You want a working free energy machine I will give you one.For those of
you who are not familiar with Thomas Townsendbrown,I'm not positive that spelling is correct, he discovered that a charged high voltage capacitor produces a very small but net thrust towards the positive plate. They have implemented this technology in the wing of a B2 bomber to give it incredible range which is not possible with just fossil fuels.
proven otherwise. The chain is essentially massless by the way.
I am an expert in free energy though. I've looked up every article on free energy
news and I wouldn't be in any rush to condemn all those new energy systems.
Energy is a field of interest of mine, and been researching new energy systems
and free energy systems for the past 15 years. I work on designs for improved
efficiency of internal combustion engines and new free energy machines.
I'd like to get something done with improving the efficiency of the internal combustion engine,without serious money you can't get anything done.
As far as free energy is concerned I can build you a free energy machine
today they have been documented to work I've analyzed the designs I
know how they work and why they work, the problem is I would be
setting myself up as the biggest target ever.The problem with the people
in the free energy field they don't think about the consequences of their
work. If you unveiled a free energy machine in the world today the
stock market would crash, there would be mass layoffs Kodak I think
is an example of that, not to mention national security concerns.
On the flip side we are wrecking the planet with fossil fuels the cost
of fossil fuels is going to bankrupt the west in the end, supply and demand.
You want a working free energy machine I will give you one.For those of
you who are not familiar with Thomas Townsendbrown,I'm not positive that spelling is correct, he discovered that a charged high voltage capacitor produces a very small but net thrust towards the positive plate. They have implemented this technology in the wing of a B2 bomber to give it incredible range which is not possible with just fossil fuels.
I'm not sure I understand properly your system, drawings would help.
As I understand, the chain is not in the horizontal plane to begin with. Any centripetal force exerted by the chain will cause the chain to approach the horizontal plane (assuming no gravity) so your system cannot work as I think you've described it.
Correct me if I've missed something...
As I understand, the chain is not in the horizontal plane to begin with. Any centripetal force exerted by the chain will cause the chain to approach the horizontal plane (assuming no gravity) so your system cannot work as I think you've described it.
Correct me if I've missed something...
I'm not familiar with how to attach files to this system so I'll do a small
diagram here. Assume no gravity. Ignore the minus signs I had to use them to get the editor to work for the drawing.
----------------------------------------y axis
----------------------------------------*
-------------------------------------------------------* B
----------------------------------------* A---------------------------* C x axis
The rod is between point A and B
The chain is between point B and C
The weight is at point C
Point A is at the origin of x y axis.
We will assume that rod and chain are close to massless
The rod is 1 m long
The chain is 1 m long
Distance A C is 1.414 m
The ball is given an initial velocity in the +y direction of .1 m/s 0 m/s in the
x direction as the ball crosses the x axis at point C the ball, chain and rod
are in the position shown above and acceleration of the ball begins here.
For simplification of the problem we will apply the accelerating force at the end
of the rod at point C in line with the chain and perpendicular to the rod. This
accelerating force remains at the end of the rod perpendicular to it in line
with the chain thought out acceleration of the ball.
The rod pivots at point A, acceleration begins when the rod is at 45 degrees
and continues until the the rod is in the 135 degree position now the ball
is crossing the +y axis with velocity z in the -x direction 0 m/s in the y
direction, the ball is instantly detached from the chain at this point, where
it is decelerated in a straight line in the x direction.
For energy to be conserved in this example the force at the end of the rod
has to be 1.414 times the accelerating force on the ball. So at the end
of the rod it feels like you are accelerating a heavier object than you really are.
Thats where the imbalanced momentum comes in.
diagram here. Assume no gravity. Ignore the minus signs I had to use them to get the editor to work for the drawing.
----------------------------------------y axis
----------------------------------------*
-------------------------------------------------------* B
----------------------------------------* A---------------------------* C x axis
The rod is between point A and B
The chain is between point B and C
The weight is at point C
Point A is at the origin of x y axis.
We will assume that rod and chain are close to massless
The rod is 1 m long
The chain is 1 m long
Distance A C is 1.414 m
The ball is given an initial velocity in the +y direction of .1 m/s 0 m/s in the
x direction as the ball crosses the x axis at point C the ball, chain and rod
are in the position shown above and acceleration of the ball begins here.
For simplification of the problem we will apply the accelerating force at the end
of the rod at point C in line with the chain and perpendicular to the rod. This
accelerating force remains at the end of the rod perpendicular to it in line
with the chain thought out acceleration of the ball.
The rod pivots at point A, acceleration begins when the rod is at 45 degrees
and continues until the the rod is in the 135 degree position now the ball
is crossing the +y axis with velocity z in the -x direction 0 m/s in the y
direction, the ball is instantly detached from the chain at this point, where
it is decelerated in a straight line in the x direction.
For energy to be conserved in this example the force at the end of the rod
has to be 1.414 times the accelerating force on the ball. So at the end
of the rod it feels like you are accelerating a heavier object than you really are.
Thats where the imbalanced momentum comes in.
It's becoming clearer. Which energies are you trying to conserve? Have you resolved forces to analyse what the y position of the ball will be at point of release?
The energies I'm tying to conserve are the work at the end of the rod acting on the chain attached there as the rod moves through 90 arc with the increase of energy of the ball with it's increased velocity. Since the ball moves in a circular path, due to the fact the centrifugal force exerted by the ball is always matched by the vector of force of the chains pull on the ball, the ball will be release at 1.414 m on the y axis as it crosses the y axis.
QUOTE (Carl Fischbach+Jul 24 2007, 06:49 PM)
The energies I'm tying to conserve are the work at the end of the rod acting on the chain
The rod does no work, since work done = force x distance and the rod doesn't move in the direction of the force. (Unless the rod is accelerating the chain in magnitude as well as direction.)
I thought the y-axis was the axis of rotation? Do you mean as it crosses the z-axis?
I think you need to somehow form an equation for the vector of the force on the chain. It will be changing direction and magnitude as the rotation speeds up.
Unfortunately, I don't believe the system is nearly as simple as you are trying to model it.
The rod does no work, since work done = force x distance and the rod doesn't move in the direction of the force. (Unless the rod is accelerating the chain in magnitude as well as direction.)
QUOTE
the ball will be release at 1.414 m on the y axis as it crosses the y axis.
I thought the y-axis was the axis of rotation? Do you mean as it crosses the z-axis?
I think you need to somehow form an equation for the vector of the force on the chain. It will be changing direction and magnitude as the rotation speeds up.
Unfortunately, I don't believe the system is nearly as simple as you are trying to model it.
You are correct the work is actually being done to the end of the chain which
happens to be connected to the end of the rod. The x y axis remain
in a constant position the rod the chain the ball move counterclockwise within the
x y axis
The diagram below shown the final position of the ball rod and chain after
acceleration is complete
----------------------------------------------- y axis
----------------------------------------------- * Q
------------------------------ * T
----------------------------------------------- * A ------------------------------- x axis
A is the pivot point and origin where the end of the rod is attached
T is the other end of the rod where the chain is attached
Q is the position of the ball when acceleration of the ball by the chain is finished
and the ball is released
The distance A Q is 1.414 m
It is possible to keep the chain at 45 degrees to the instantaneous direction of
travel of the ball as it accelerates by maintaining the pull on the chain at 1.414 times the centrifugal force created by the accelerating ball. When the inward
force created by the 45 degree vector of the pull of the chain matches the
centrifugal force no outward acceleration of the ball can take place and
the trajectory of the ball must follow a circular path.
happens to be connected to the end of the rod. The x y axis remain
in a constant position the rod the chain the ball move counterclockwise within the
x y axis
The diagram below shown the final position of the ball rod and chain after
acceleration is complete
----------------------------------------------- y axis
----------------------------------------------- * Q
------------------------------ * T
----------------------------------------------- * A ------------------------------- x axis
A is the pivot point and origin where the end of the rod is attached
T is the other end of the rod where the chain is attached
Q is the position of the ball when acceleration of the ball by the chain is finished
and the ball is released
The distance A Q is 1.414 m
It is possible to keep the chain at 45 degrees to the instantaneous direction of
travel of the ball as it accelerates by maintaining the pull on the chain at 1.414 times the centrifugal force created by the accelerating ball. When the inward
force created by the 45 degree vector of the pull of the chain matches the
centrifugal force no outward acceleration of the ball can take place and
the trajectory of the ball must follow a circular path.
QUOTE (Carl Fischbach+Jul 24 2007, 09:15 PM)
It is possible to keep the chain at 45 degrees to the instantaneous direction of
travel of the ball as it accelerates by maintaining the pull on the chain at 1.414 times the centrifugal force created by the accelerating ball. When the inward
force created by the 45 degree vector of the pull of the chain matches the
centrifugal force no outward acceleration of the ball can take place and
the trajectory of the ball must follow a circular path.
Ok, so no work is done by the system. Work will be done to speed the system up, but you are choosing not to analyse that bit. Once the ball is being swung in a circle, if the rod and chain stay in a plane perpendicular to the xz plane then no work is done. This is because all forces are perpendicular to the motion. So I don't know which energies you are equating.
travel of the ball as it accelerates by maintaining the pull on the chain at 1.414 times the centrifugal force created by the accelerating ball. When the inward
force created by the 45 degree vector of the pull of the chain matches the
centrifugal force no outward acceleration of the ball can take place and
the trajectory of the ball must follow a circular path.
Ok, so no work is done by the system. Work will be done to speed the system up, but you are choosing not to analyse that bit. Once the ball is being swung in a circle, if the rod and chain stay in a plane perpendicular to the xz plane then no work is done. This is because all forces are perpendicular to the motion. So I don't know which energies you are equating.
If check the instantaneous work done to the end of the chain connected to the rod
for an angular rotation of .001 degrees of the ball and rod you will find that the
ratio
instantaneous work done to the end of the chain to instantaneous work done to the ball is
((.001*1.414*2*pi)/360)/((.707*1.414 *1.414*.001*2*pi)/360)
which equals 1
The final speed of the ball z is not required to be calculated to prove
an imbalance of momentum in the x direction .
for an angular rotation of .001 degrees of the ball and rod you will find that the
ratio
instantaneous work done to the end of the chain to instantaneous work done to the ball is
((.001*1.414*2*pi)/360)/((.707*1.414 *1.414*.001*2*pi)/360)
which equals 1
The final speed of the ball z is not required to be calculated to prove
an imbalance of momentum in the x direction .
QUOTE (Robby Bloom+Jul 23 2007, 08:54 PM)
Carl,
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
Of course, critical investigation is another essential element of the Scientific Process.
One would think that snobbery would have no place in the Sciences.
But it exists here as in any other discipline.
...SNIP...
As a teaching tool, I can only see snobbery aiding other would-be snobs.
But I suppose it has value in toughening your hide.
You know that throughout your life you will face such responses from snobs,
whether in raucous reaction to errors [which do occur],
or in ignorant or careless response to novel concepts.
Good luck,
Robby B
This forum needs more thoughtful and encouraging critiques like this. Too bad this person doesn't seem to be a member.
I applaud your initiative in presenting your proposal "non-conservation of linear momentum".
Presentation of proposed theory is an essential element of the Scientific Process {ref: Popper and others}.
Of course, critical investigation is another essential element of the Scientific Process.
One would think that snobbery would have no place in the Sciences.
But it exists here as in any other discipline.
...SNIP...
As a teaching tool, I can only see snobbery aiding other would-be snobs.
But I suppose it has value in toughening your hide.
You know that throughout your life you will face such responses from snobs,
whether in raucous reaction to errors [which do occur],
or in ignorant or careless response to novel concepts.
Good luck,
Robby B
This forum needs more thoughtful and encouraging critiques like this. Too bad this person doesn't seem to be a member.
ubavontuba,
Sorry to say that the medieval battle flail will not shake physics.
Engineers are using classical mechanics daily to simulate systems which are much more complicated than the medieval battle flail. Refering to Popper on such a simple question is quite odd. If you have the time, you would better solve the question.
I remember a documentary on TV also claiming strange things about the medieval battle flail and its mechanics. I wonder what the link is with this thread.
Sorry to say that the medieval battle flail will not shake physics.
Engineers are using classical mechanics daily to simulate systems which are much more complicated than the medieval battle flail. Refering to Popper on such a simple question is quite odd. If you have the time, you would better solve the question.
I remember a documentary on TV also claiming strange things about the medieval battle flail and its mechanics. I wonder what the link is with this thread.
You can't be serious.
I came with this idea independently I was looking for ways where momentum
might not be conserved so the idea of a weight on the end of a string connected
to a rod came to mind. I looked for an example that would graphically
illustrate this idea so I thought of the battle flail.
I came with this idea independently I was looking for ways where momentum
might not be conserved so the idea of a weight on the end of a string connected
to a rod came to mind. I looked for an example that would graphically
illustrate this idea so I thought of the battle flail.
Carl Fischbach,
It is much easier to understand your last post than the first one.
Now I understand that your aim was to contradict a basic physical law and that you elaborated a complicated scheme to try this.
The problem is that by elaborating your scheme you just made it a bit more complicated to pinpoint where you introduced a flaw, but in no way you would come with a counterexample to momentum conservation.
You should also understand that by analysing your system with classical mechanics you will never be able to prove anything. The only way to prove something would be to perform a clear experiment. And since momentum conservation has never been contradicted, even in much more complicated or extreme situations, I give you no chance at all for the smallest success.
Momentum conservation and other conservation laws have not only been verified in-depth experimentally, but they have been analysed and understood in extreme-depth theoretically.
Personally, I consider that the Noether theorem is the culminating analysis of conservation laws. This (proven) theorem relates any conservation law with an existing symmetry of physics.
For example, the energy conservation is related to the time-invariance symmetry: doing an experiment now or doing it next year will give the same results, given the same conditions. The theorem says that this implies the conservation of energy.
The monmentum conservation is related to the space-invariance symmetry: performing an experience here or performing the same experiement 1000 km away from here will deliver exactly the same results. The theorem says that this implies the conservation of (linar) momentum.
There are other applications of the Noether theorem.
This theorem is however based on some assumption. But these assumption are also very well verified and very hard to contradict. As a consequence, I think that the weakest point maybe the validity of a symmetry rather than these other assumptions of the Noether theorem. For example, even if the link with Noether is not immediate, the CPT symmetry is known to be imperfect and this implies imperfections in some quantum conservations law, to speak very roughly.
I hope that you might be interrested by this fascinating topics and that you will forget about finding a contradiction of momentum conservation.
It is much easier to understand your last post than the first one.
Now I understand that your aim was to contradict a basic physical law and that you elaborated a complicated scheme to try this.
The problem is that by elaborating your scheme you just made it a bit more complicated to pinpoint where you introduced a flaw, but in no way you would come with a counterexample to momentum conservation.
You should also understand that by analysing your system with classical mechanics you will never be able to prove anything. The only way to prove something would be to perform a clear experiment. And since momentum conservation has never been contradicted, even in much more complicated or extreme situations, I give you no chance at all for the smallest success.
Momentum conservation and other conservation laws have not only been verified in-depth experimentally, but they have been analysed and understood in extreme-depth theoretically.
Personally, I consider that the Noether theorem is the culminating analysis of conservation laws. This (proven) theorem relates any conservation law with an existing symmetry of physics.
For example, the energy conservation is related to the time-invariance symmetry: doing an experiment now or doing it next year will give the same results, given the same conditions. The theorem says that this implies the conservation of energy.
The monmentum conservation is related to the space-invariance symmetry: performing an experience here or performing the same experiement 1000 km away from here will deliver exactly the same results. The theorem says that this implies the conservation of (linar) momentum.
There are other applications of the Noether theorem.
This theorem is however based on some assumption. But these assumption are also very well verified and very hard to contradict. As a consequence, I think that the weakest point maybe the validity of a symmetry rather than these other assumptions of the Noether theorem. For example, even if the link with Noether is not immediate, the CPT symmetry is known to be imperfect and this implies imperfections in some quantum conservations law, to speak very roughly.
I hope that you might be interrested by this fascinating topics and that you will forget about finding a contradiction of momentum conservation.
QUOTE (Carl Fischbach+Jul 25 2007, 08:51 AM)
...I was looking for ways where momentum might not be conserved so the idea of a weight on the end of a string connected to a rod came to mind...
Of course, part of weight momentum will be accumulated into potential energy of string tension. But I don't see the place for violation of momentum law.
The other thing is, every conservation law is local in time, i.e. valid just in situation, where other conditions doesn't change with change of conditions. The violation of these conditions can be tricky. For example, if you're shaking the foam inside of closed vessel, the foam becomes more and more dense. Therefore the momentum of foam "remembers" its history.
Of course, part of weight momentum will be accumulated into potential energy of string tension. But I don't see the place for violation of momentum law.
The other thing is, every conservation law is local in time, i.e. valid just in situation, where other conditions doesn't change with change of conditions. The violation of these conditions can be tricky. For example, if you're shaking the foam inside of closed vessel, the foam becomes more and more dense. Therefore the momentum of foam "remembers" its history.
Zephir:
Can you post anything else than BS?
Can you post anything else than BS?
QUOTE (Lalbatros+Jul 25 2007, 11:46 AM)
Can you post anything else than BS?
Nope, until you explain us, why my post is BS. If you omit such explanation from your post, it's just you, who is spreading the alogical exclamations here.
Nope, until you explain us, why my post is BS. If you omit such explanation from your post, it's just you, who is spreading the alogical exclamations here.
Total BS and word salad:
Not one word makes sense.
QUOTE
[Zephir]
The other thing is, every conservation law is local in time, i.e. valid just in situation, where other conditions doesn't change with change of conditions. The violation of these conditions can be tricky. For example, if you're shaking the foam inside of closed vessel, the foam becomes more and more dense. Therefore the momentum of foam "remembers" its history.
The other thing is, every conservation law is local in time, i.e. valid just in situation, where other conditions doesn't change with change of conditions. The violation of these conditions can be tricky. For example, if you're shaking the foam inside of closed vessel, the foam becomes more and more dense. Therefore the momentum of foam "remembers" its history.
Not one word makes sense.
QUOTE (Carl Fischbach+Jul 24 2007, 11:28 PM)
If check the instantaneous work done to the end of the chain connected to the rod
for an angular rotation of .001 degrees of the ball and rod you will find that the
ratio
instantaneous work done to the end of the chain to instantaneous work done to the ball is
((.001*1.414*2*pi)/360)/((.707*1.414 *1.414*.001*2*pi)/360)
which equals 1
What's the equation you've used here? And where have each of the substituted numbers come from? Show all your steps - if you're asking us to look for a mistake you've made (which is the only way to prove you are right; by nobody being able to disprove it) then you need to show everything you have done.
for an angular rotation of .001 degrees of the ball and rod you will find that the
ratio
instantaneous work done to the end of the chain to instantaneous work done to the ball is
((.001*1.414*2*pi)/360)/((.707*1.414 *1.414*.001*2*pi)/360)
which equals 1
What's the equation you've used here? And where have each of the substituted numbers come from? Show all your steps - if you're asking us to look for a mistake you've made (which is the only way to prove you are right; by nobody being able to disprove it) then you need to show everything you have done.
Zephir,
This is not totally right:
You shoudl have written:
You shoudl have written:
Nope, until you explain me, why my post is BS.
... since you are the only one here not to see why you wrote pure BS.
I will try and make myself clearer here.
You can't equate linear momentum and angular momentum. Angular momentum is tied to radius and linear momentum isn't. So I've converted angular accelerating force into separate x and y component forces which would give the exact identical effect as angular accelerating force. Were only interested in the x
component of pull on the end chain during acceleration of the ball and the x
component of the linear deceleration of the ball. The integral of ( force as a function of time) for the x components must be equal for acceleration and
deceleration of the ball otherwise momentum is not conserved.
I'm not quite sure what you mean by 'angular accelerating force.'
When a mass is given angular acceleration the components break down to a tangential and radial component. Is this what you are talking about?
I'm not quite sure what you mean by 'angular accelerating force.'
When a mass is given angular acceleration the components break down to a tangential and radial component. Is this what you are talking about?
The ball starts accelerating at point C at 0 degrees and is accelerated by the chain
in an arc at a 1.414 m radius to 90 degrees and is released from the chain at Q.
Ok, I must have missed this in your previous posts.
You seem like a smart person, I'm going to ask you at this point to work this out and post the equations. This setup seems fairly simple.
Since the work done to the end of the
chain is the same as done to the ball and time is the same, the integral of (force as a function of time) for the x component of the force acting on the end of the chain during acceleration is greater than the integral of (force as a function of time) deceleration of the ball where distances,forces and time of acceleration acting
on the ball are the same for acceleration and deceleration.
If this is any clearer.
This isn't clear. The way I'm reading it is that the accelerating force is greater than the decelerating force. Well no kidding. You really need to work these equations out and post them. There's a reason why mathematics is the language of physics.
This is not totally right:
QUOTE
Nope, until you explain us, why my post is BS.
You shoudl have written:
QUOTE (->
| QUOTE |
| Nope, until you explain us, why my post is BS. |
You shoudl have written:
Nope, until you explain me, why my post is BS.
... since you are the only one here not to see why you wrote pure BS.
To settle this idea once in for all that laws in physics are safe.
Consider this theory that I come up with independently.
An isolated body that contains energy where it's energy content is determined
by outgoing energy leaving the body and incoming radiation absorbed by the body,
a temperature gradient exists within the body where the work done on moving or vibrating molecules by force of gravity causes a temperature increase in the body as you go from the outside to the center of the body.
This theory explains explains the heating of the atmosphere with decreasing altitude, and why interiors of planets remain hot.
There is an easy experiment to test this theory you take a column of air
and isolate it in an air tight container, then you take the container and isolate it
within a vacuum flask and measure the temperate difference between the
bottom and top of the column of air.
After having this theory for a while on searching the web I found this
web site www.firstgravitymachine.com/dancingmolecules.phtml where this
German guy has taken this theory on step further and built an energy machine
that exploits this principle and generates energy at 100% efficiency
contrary to the second law of thermodynamics. The site doesn't sound like much ,
but the site is very well done and this page is definitely worth looking at.
Consider this theory that I come up with independently.
An isolated body that contains energy where it's energy content is determined
by outgoing energy leaving the body and incoming radiation absorbed by the body,
a temperature gradient exists within the body where the work done on moving or vibrating molecules by force of gravity causes a temperature increase in the body as you go from the outside to the center of the body.
This theory explains explains the heating of the atmosphere with decreasing altitude, and why interiors of planets remain hot.
There is an easy experiment to test this theory you take a column of air
and isolate it in an air tight container, then you take the container and isolate it
within a vacuum flask and measure the temperate difference between the
bottom and top of the column of air.
After having this theory for a while on searching the web I found this
web site www.firstgravitymachine.com/dancingmolecules.phtml where this
German guy has taken this theory on step further and built an energy machine
that exploits this principle and generates energy at 100% efficiency
contrary to the second law of thermodynamics. The site doesn't sound like much ,
but the site is very well done and this page is definitely worth looking at.
The dilemna is:
- refuting all cranks around the web
- studying physics
... up to you.
- refuting all cranks around the web
- studying physics
... up to you.
First of all unless I'm reading your set up incorrectly the only momentum you should be dealing with is angular. Unless the ball is released, at which point it has been given linear momentum. If the ball is being accelerated and is released at point A instead of B where point A will give a greater tangential velocity, then the linear momentum given at these two points will differ. Second, the claims of a unity device or a device which is a 'free energy' device are dubious at best. Build the machine and show me, then I will believe you. By free energy do you mean an over unity device? Also with regards to your post about an Isolated body having energy transfer-- in thermodynamics 'Isolated' means that energy doesn't cross the boundary of the system. Therefore you might want to tell us what you mean by 'isolated.'
I will try and make myself clearer here.
You can't equate linear momentum and angular momentum. Angular momentum is tied to radius and linear momentum isn't. So I've converted angular accelerating force into separate x and y component forces which would give the exact identical effect as angular accelerating force. Were only interested in the x
component of pull on the end chain during acceleration of the ball and the x
component of the linear deceleration of the ball. The integral of ( force as a function of time) for the x components must be equal for acceleration and
deceleration of the ball otherwise momentum is not conserved.
I not using the rod for clarity in this example it's massless anyways so it doesn't matter. The end of the chain follows a pathway from point B to point T at
a 1 m radius from 45 degrees to 135 degrees by adjusting the pull on the end of the chain you can keep the chain in line with the direction of travel of the end of the chain.
The ball starts accelerating at point C at 0 degrees and is accelerated by the chain
in an arc at a 1.414 m radius to 90 degrees and is released from the chain at Q.
where it follows a straight trajectory in the x direction until decelerated linearly
to zero.
The key to this whole thing is the force acting on the end of the chain in the
x direction is never less than force of acceleration acting on the ball which
accelerates it in it's direction of travel. Simple put the force acting on the end of the chain to accelerate the ball feels like it is accelerating a mass 1.414 times
heavier than the actual mass of the ball. Since the work done to the end of the
chain is the same as done to the ball and time is the same, the integral of (force as a function of time) for the x component of the force acting on the end of the chain during acceleration is greater than the integral of (force as a function of time) deceleration of the ball where distances,forces and time of acceleration acting
on the ball are the same for acceleration and deceleration.
If this is any clearer.
Free energy is a buzz word
It encompasses all energy systems not accepted as possible by mainstream science that generate energy at basically at no cost.
Over unity is where there is no loss of mass to account for the energy produced.
Thats the way I see it
By isolated I mean isolated in space physically not electromagnetically.
You can't equate linear momentum and angular momentum. Angular momentum is tied to radius and linear momentum isn't. So I've converted angular accelerating force into separate x and y component forces which would give the exact identical effect as angular accelerating force. Were only interested in the x
component of pull on the end chain during acceleration of the ball and the x
component of the linear deceleration of the ball. The integral of ( force as a function of time) for the x components must be equal for acceleration and
deceleration of the ball otherwise momentum is not conserved.
I not using the rod for clarity in this example it's massless anyways so it doesn't matter. The end of the chain follows a pathway from point B to point T at
a 1 m radius from 45 degrees to 135 degrees by adjusting the pull on the end of the chain you can keep the chain in line with the direction of travel of the end of the chain.
The ball starts accelerating at point C at 0 degrees and is accelerated by the chain
in an arc at a 1.414 m radius to 90 degrees and is released from the chain at Q.
where it follows a straight trajectory in the x direction until decelerated linearly
to zero.
The key to this whole thing is the force acting on the end of the chain in the
x direction is never less than force of acceleration acting on the ball which
accelerates it in it's direction of travel. Simple put the force acting on the end of the chain to accelerate the ball feels like it is accelerating a mass 1.414 times
heavier than the actual mass of the ball. Since the work done to the end of the
chain is the same as done to the ball and time is the same, the integral of (force as a function of time) for the x component of the force acting on the end of the chain during acceleration is greater than the integral of (force as a function of time) deceleration of the ball where distances,forces and time of acceleration acting
on the ball are the same for acceleration and deceleration.
If this is any clearer.
Free energy is a buzz word
It encompasses all energy systems not accepted as possible by mainstream science that generate energy at basically at no cost.
Over unity is where there is no loss of mass to account for the energy produced.
Thats the way I see it
By isolated I mean isolated in space physically not electromagnetically.
QUOTE
I will try and make myself clearer here.
You can't equate linear momentum and angular momentum. Angular momentum is tied to radius and linear momentum isn't. So I've converted angular accelerating force into separate x and y component forces which would give the exact identical effect as angular accelerating force. Were only interested in the x
component of pull on the end chain during acceleration of the ball and the x
component of the linear deceleration of the ball. The integral of ( force as a function of time) for the x components must be equal for acceleration and
deceleration of the ball otherwise momentum is not conserved.
I'm not quite sure what you mean by 'angular accelerating force.'
When a mass is given angular acceleration the components break down to a tangential and radial component. Is this what you are talking about?
QUOTE (->
| QUOTE |
I will try and make myself clearer here. You can't equate linear momentum and angular momentum. Angular momentum is tied to radius and linear momentum isn't. So I've converted angular accelerating force into separate x and y component forces which would give the exact identical effect as angular accelerating force. Were only interested in the x component of pull on the end chain during acceleration of the ball and the x component of the linear deceleration of the ball. The integral of ( force as a function of time) for the x components must be equal for acceleration and deceleration of the ball otherwise momentum is not conserved. |
I'm not quite sure what you mean by 'angular accelerating force.'
When a mass is given angular acceleration the components break down to a tangential and radial component. Is this what you are talking about?
The ball starts accelerating at point C at 0 degrees and is accelerated by the chain
in an arc at a 1.414 m radius to 90 degrees and is released from the chain at Q.
Ok, I must have missed this in your previous posts.
You seem like a smart person, I'm going to ask you at this point to work this out and post the equations. This setup seems fairly simple.
QUOTE
Since the work done to the end of the
chain is the same as done to the ball and time is the same, the integral of (force as a function of time) for the x component of the force acting on the end of the chain during acceleration is greater than the integral of (force as a function of time) deceleration of the ball where distances,forces and time of acceleration acting
on the ball are the same for acceleration and deceleration.
If this is any clearer.
This isn't clear. The way I'm reading it is that the accelerating force is greater than the decelerating force. Well no kidding. You really need to work these equations out and post them. There's a reason why mathematics is the language of physics.
QUOTE (oomchu+Jul 28 2007, 09:51 PM)
There's a reason why mathematics is the language of physics.
Mathematically is possible to derive from constant speed of light in vacuum the pictures, like this one presented on the Nature web site...
Can you guess from such picture, the light speed in vacuum is constant at all? I cannot.
Mathematically is possible to derive from constant speed of light in vacuum the pictures, like this one presented on the Nature web site...
Can you guess from such picture, the light speed in vacuum is constant at all? I cannot.
Actually that guy you linked to Carl don't strike me as a crank, if what he says is true, i also would like to know from where he gets those results. In this link one can find a pdf wherein he presents his setup and results. I can't do that math he has done but i guess a number of you guys (Yes, i'm talking to ...You :) will find no trouble with it. I would be very curious as to if he's really found something new, or if it's explainable some other way. http://www.firstgravitymachine.com/testresults.phtml
QUOTE (Lalbatros+Jul 25 2007, 05:19 AM)
ubavontuba,
Sorry to say that the medieval battle flail will not shake physics.
Engineers are using classical mechanics daily to simulate systems which are much more complicated than the medieval battle flail. Refering to Popper on such a simple question is quite odd. If you have the time, you would better solve the question.
I remember a documentary on TV also claiming strange things about the medieval battle flail and its mechanics. I wonder what the link is with this thread.
I agree. The momentum is simply transferred from the earth. I'd bet a medieval battle flail would be virtually impossible to swing in space without destablizing the flailer.
Sorry to say that the medieval battle flail will not shake physics.
Engineers are using classical mechanics daily to simulate systems which are much more complicated than the medieval battle flail. Refering to Popper on such a simple question is quite odd. If you have the time, you would better solve the question.
I remember a documentary on TV also claiming strange things about the medieval battle flail and its mechanics. I wonder what the link is with this thread.
I agree. The momentum is simply transferred from the earth. I'd bet a medieval battle flail would be virtually impossible to swing in space without destablizing the flailer.
QUOTE (Lalbatros+Jul 25 2007, 08:05 AM)
Momentum conservation and other conservation laws have not only been verified in-depth experimentally, but they have been analysed and understood in extreme-depth theoretically.
Personally, I consider that the Noether theorem is the culminating analysis of conservation laws. This (proven) theorem relates any conservation law with an existing symmetry of physics.
For example, the energy conservation is related to the time-invariance symmetry: doing an experiment now or doing it next year will give the same results, given the same conditions. The theorem says that this implies the conservation of energy.
The monmentum conservation is related to the space-invariance symmetry: performing an experience here or performing the same experiement 1000 km away from here will deliver exactly the same results. The theorem says that this implies the conservation of (linar) momentum.
There are other applications of the Noether theorem.
This theorem is however based on some assumption. But these assumption are also very well verified and very hard to contradict. As a consequence, I think that the weakest point maybe the validity of a symmetry rather than these other assumptions of the Noether theorem. For example, even if the link with Noether is not immediate, the CPT symmetry is known to be imperfect and this implies imperfections in some quantum conservations law, to speak very roughly.
I hope that you might be interrested by this fascinating topics and that you will forget about finding a contradiction of momentum conservation.
Lalbatros,
This was a very nice interpretation of Noether's Theorem. However, I question this theorem's integrity. I do not believe that it is completely accurate. There are a number of symmetry paradoxes which relate to it. For instance, the CP violation.
That one's old of course. We have new ones. The "Dark Energy" paradox for instance. We assume it must be an actual energy of some sort only because it violates conservation. Maybe Noether's Theorem is wrong on the grand scale and it's simply a normal violation of conservation (maybe there is no dark energy).
In any case, I can think of an instance where energy is not conserved (in the traditional sense). In a two body collision, kinetic energy is generally computed using the formula: KE=.5mv^2. In the case of two attractive bodies (gravitational or magnetic), the KE increases with proximity due to the acceleration of the attractive force. In repulsive bodies, the KE decreases with proximity.
Arguably, the acceleration (and subsequent change in the KE potential) is the result of a conserved force which increases (or decreases) the energy potential without cost to itself, the mass/energy in the system, or any other energy source. The change in the energy potential just happens. Therefore, energy isn't always conserved in symmetrical systems.
Personally, I consider that the Noether theorem is the culminating analysis of conservation laws. This (proven) theorem relates any conservation law with an existing symmetry of physics.
For example, the energy conservation is related to the time-invariance symmetry: doing an experiment now or doing it next year will give the same results, given the same conditions. The theorem says that this implies the conservation of energy.
The monmentum conservation is related to the space-invariance symmetry: performing an experience here or performing the same experiement 1000 km away from here will deliver exactly the same results. The theorem says that this implies the conservation of (linar) momentum.
There are other applications of the Noether theorem.
This theorem is however based on some assumption. But these assumption are also very well verified and very hard to contradict. As a consequence, I think that the weakest point maybe the validity of a symmetry rather than these other assumptions of the Noether theorem. For example, even if the link with Noether is not immediate, the CPT symmetry is known to be imperfect and this implies imperfections in some quantum conservations law, to speak very roughly.
I hope that you might be interrested by this fascinating topics and that you will forget about finding a contradiction of momentum conservation.
Lalbatros,
This was a very nice interpretation of Noether's Theorem. However, I question this theorem's integrity. I do not believe that it is completely accurate. There are a number of symmetry paradoxes which relate to it. For instance, the CP violation.
That one's old of course. We have new ones. The "Dark Energy" paradox for instance. We assume it must be an actual energy of some sort only because it violates conservation. Maybe Noether's Theorem is wrong on the grand scale and it's simply a normal violation of conservation (maybe there is no dark energy).
In any case, I can think of an instance where energy is not conserved (in the traditional sense). In a two body collision, kinetic energy is generally computed using the formula: KE=.5mv^2. In the case of two attractive bodies (gravitational or magnetic), the KE increases with proximity due to the acceleration of the attractive force. In repulsive bodies, the KE decreases with proximity.
Arguably, the acceleration (and subsequent change in the KE potential) is the result of a conserved force which increases (or decreases) the energy potential without cost to itself, the mass/energy in the system, or any other energy source. The change in the energy potential just happens. Therefore, energy isn't always conserved in symmetrical systems.
QUOTE (ubavontuba+Jul 29 2007, 05:42 AM)
Maybe Noether's Theorem is wrong on the grand scale and it's simply a normal violation of conservation (maybe there is no dark energy).
No, it isn't, you talk about things that you obviously have no clue what they mean (CP violation, etc).
In any case, I can think of an instance where energy is not conserved (in the traditional sense). In a two body collision, kinetic energy is generally computed using the formula: KE=.5mv^2. In the case of two attractive bodies (gravitational or magnetic), the KE increases with proximity due to the acceleration of the attractive force. In repulsive bodies, the KE decreases with proximity.
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
Arguably, the acceleration (and subsequent change in the KE potential) is the result of a conserved force which increases (or decreases) the energy potential without cost to itself, the mass/energy in the system, or any other energy source. The change in the energy potential just happens. Therefore, energy isn't always conserved in symmetrical systems.
More mumbo-jumbo, I bolded the funnier ones.
You're just twaddling strong words, rhetorical questions and social games here, as usually. This is, what are you good in.
Especially after I got you to falsify your very own AWT, remember? That was fun!
The "accelerating force" is the conserved, attractive or repulsive force I mentioned. Did you miss it? The acceleration (and hence the change in potential energy) arises from this force. I said that too, did you miss it?
Hahahahahaha, no I didn't miss it. Your high school teacher wold have rightfully graded your above statement as an F.
Once again: your 9-th grade book STILL says : don't wander why your energy and momentum is not conserved if you have external forces acting on the system. If you need to retake 9-th grade, by all means. Either way, this forum is for discussing new theories, not your misconceptions about oldtheories.
Can you find a reference in regards to your statements?
Yes, open your 9-th grade physics book.
Did I say that momentum is not conserved in the system in question? Where?
How quickly you "forget" Right here
You did not provide a single proper reference. If you had, you could have quoted the relevant portions. Can you quote any relevant portions?
For the laws of energy conservation , you need to read (and hopefully comprehend) page 113.
If that fails, hunt down for your 9-th grade physics book, it is all there
For the laws of energy conservation , you need to read (and hopefully comprehend) page 113.
If that fails, hunt down for your 9-th grade physics book, it is all there
That's still not a quote. What's it say that you feel is relevant? Do you even know? You certainly didn't know what the discussion was about!
If you cannot provide truly relevant and meaningful materials, then our conversation is at an end.
.... then our conversation is at an end.
...it ended long ago, you just didn't know about that, see if you can buy a 9-th grade physics book, the answers are all in there
...check out that 9-th grade book, your local library might have extra copies. Until you check it out (and you understand it) leave ms. Noether out of it, she might get very upset that you are mangling her theorem.
Obviosly, you can't read a ninth grade textbook! You can't even find a relevant passage!
What did I write?
You wrote that energy is equal to energy and force is equal to mass times acceleration. Sigma (E) cancels out, so all you really wrote is F=ma.
Now, how is it applicable?
No,this is not what the equations say.And you pretend to talk about Noether's theorem?
Congratulations, you managed to get everything wrong.
You failed physics in 9-th grade, you'll need to retake it. It is kind of late for it, Eric
Fine then, what do the equations say?
Again, your mistake. F=ma is the classical (Newtonian) expression. p=mv is another. They just don't include the sum of the masses or time.
Eric,
The only person you are fooling is yourself . Go find those 9-th grade books and try taking over that 9-th grade you flunked some 30+ years ago.
They mean that you don't have a clue when it comes to mechanics in general and conservation of energy and momentum in specific.
It means that you need to take 9-th grade physics over . The stuff that you flunked about 40 years ago.
Nothin' but bluster. Is that all you got? Admit it, you didn't understand my last post.
What a moronic program you are. If you're the leading edge in AI, then useful AI isn't likely to happen in my lifetime!
I could go on, but what's the point. I have demonstrated that I can do Highschool Physics - something you have yet to do, presumably because your too busy with you irrational attacks on your fellow posters - a sure sign of a truely juvenile mind.
But "young" Eric=ubavontuba is about 50 years old. He has been doing this sort of stuff (insulting the people who attempt hopelessly to teach him a little physics) for a loooong time. He's been doing this since he flunked 9-th grade, for about 40 years.
No, it isn't, you talk about things that you obviously have no clue what they mean (CP violation, etc).
QUOTE
In any case, I can think of an instance where energy is not conserved (in the traditional sense). In a two body collision, kinetic energy is generally computed using the formula: KE=.5mv^2. In the case of two attractive bodies (gravitational or magnetic), the KE increases with proximity due to the acceleration of the attractive force. In repulsive bodies, the KE decreases with proximity.
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
QUOTE (->
| QUOTE |
In any case, I can think of an instance where energy is not conserved (in the traditional sense). In a two body collision, kinetic energy is generally computed using the formula: KE=.5mv^2. In the case of two attractive bodies (gravitational or magnetic), the KE increases with proximity due to the acceleration of the attractive force. In repulsive bodies, the KE decreases with proximity. |
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
Arguably, the acceleration (and subsequent change in the KE potential) is the result of a conserved force which increases (or decreases) the energy potential without cost to itself, the mass/energy in the system, or any other energy source. The change in the energy potential just happens. Therefore, energy isn't always conserved in symmetrical systems.
More mumbo-jumbo, I bolded the funnier ones.
QUOTE (Dallas+Jul 29 2007, 01:27 PM)
No, it isn't, you talk about things that you obviously have no clue what they mean (CP violation, etc).
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
More mumbo-jumbo, I bolded the funnier ones.
Dallas,
You just mimicked the the skeptics without adding much of your own. Do you even know anything? Do you know what the CP violation is? Do you know the proposed solution?
The one part you added is in error. You wrote:
...energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
You obviously don't understand the system described. It is a closed system and none of the effects are the result of external forces (I said so). Therefore, the effects are still present even when the external forces are zero. The potential energy changes, without an apparent source.
So do you know what you're talking about, or are you just a funny monkey - imitating the better educated (but still obnoxious) skeptics on this site?
You may have learned in 9-th grade (but you probably forgot ) that energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
More mumbo-jumbo, I bolded the funnier ones.
Dallas,
You just mimicked the the skeptics without adding much of your own. Do you even know anything? Do you know what the CP violation is? Do you know the proposed solution?
The one part you added is in error. You wrote:
...energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
You obviously don't understand the system described. It is a closed system and none of the effects are the result of external forces (I said so). Therefore, the effects are still present even when the external forces are zero. The potential energy changes, without an apparent source.
So do you know what you're talking about, or are you just a funny monkey - imitating the better educated (but still obnoxious) skeptics on this site?
QUOTE (ubavontuba+Jul 29 2007, 08:47 PM)
Do you know what the CP violation is? Do you know the proposed solution?
He cannot know about it, because no satisfactory explanation of CP violation has yet been devised by mainstream physics. The size of the effect, only about two parts per thousand, has prompted a theory that invokes a new force, called the "superweak" force, to explain the phenomenon. Another theory, named the Kobayashi-Maskawa model after its inventors, posits certain quantum mechanical effects in the weak force between quarks as the cause of CP violation.
The explanation of CP violation presented by AWT is closely related to the binary time concept, where the dimensions of time are closely packed in paired surface gradients of Aether foam. But the stability/probability of these gradients is not equal, once they're created. The antiparticles are formed by the outer surface gradients of Aether foam with higher curvature, so they're slightly less stable and the instability even increases toward the Universe history, when the observable matter was created. It means, by AWT the particles and antiparticles could never exist in dynamic equilibrium, because the curvature of Aether foam has decreased with time during inflation.
He cannot know about it, because no satisfactory explanation of CP violation has yet been devised by mainstream physics. The size of the effect, only about two parts per thousand, has prompted a theory that invokes a new force, called the "superweak" force, to explain the phenomenon. Another theory, named the Kobayashi-Maskawa model after its inventors, posits certain quantum mechanical effects in the weak force between quarks as the cause of CP violation.
The explanation of CP violation presented by AWT is closely related to the binary time concept, where the dimensions of time are closely packed in paired surface gradients of Aether foam. But the stability/probability of these gradients is not equal, once they're created. The antiparticles are formed by the outer surface gradients of Aether foam with higher curvature, so they're slightly less stable and the instability even increases toward the Universe history, when the observable matter was created. It means, by AWT the particles and antiparticles could never exist in dynamic equilibrium, because the curvature of Aether foam has decreased with time during inflation.
QUOTE (Zephir+Jul 29 2007, 06:15 PM)
He cannot know about it, because no satisfactory explanation of CP violation has yet been devised by mainstream physics. The size of the effect, only about two parts per thousand, has prompted a theory that invokes a new force, called the "superweak" force, to explain the phenomenon. Another theory, named the Kobayashi-Maskawa model after its inventors, posits certain quantum mechanical effects in the weak force between quarks as the cause of CP violation.
Hence the word, "proposed."
Thanks (sarcasm) for ruining my trap for him. If he knew anything about physics at all, he should have been able to answer this without help. Now, we'll never know.
Next time, MYOB.
Hence the word, "proposed."
Thanks (sarcasm) for ruining my trap for him. If he knew anything about physics at all, he should have been able to answer this without help. Now, we'll never know.
Next time, MYOB.
QUOTE (ubavontuba+Jul 29 2007, 05:47 PM)
Dallas,
You just mimicked the the skeptics without adding much of your own. Do you even know anything? Do you know what the CP violation is? Do you know the proposed solution?
Yes, I do. From what you write it is clear that you don't. If you did, you would have answered. If you did, you wouldn't have written the erroneous stuff about energy-momentum conservation.
Yeah, "you said so"
What about the "accelerating force"? You "forgot" about it, didn't you?
You just mimicked the the skeptics without adding much of your own. Do you even know anything? Do you know what the CP violation is? Do you know the proposed solution?
Yes, I do. From what you write it is clear that you don't. If you did, you would have answered. If you did, you wouldn't have written the erroneous stuff about energy-momentum conservation.
QUOTE
The one part you added is in error. You wrote:
...energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
You obviously don't understand the system described. It is a closed system and none of the effects are the result of external forces (I said so). Therefore, the effects are still present even when the external forces are zero. The potential energy changes, without an apparent source.
...energy (and momentum) are conserved for closed systems. Your counter example is senseless since the energy-momentum are conserved when the sum of external forces is zero.
You obviously don't understand the system described. It is a closed system and none of the effects are the result of external forces (I said so). Therefore, the effects are still present even when the external forces are zero. The potential energy changes, without an apparent source.
Yeah, "you said so"
What about the "accelerating force"? You "forgot" about it, didn't you?
QUOTE (ubavontuba+Jul 29 2007, 09:26 PM)
If he knew anything about physics at all, he should have been able to answer this without help. Now, we'll never know.
Why? Dallas can still propose his own explanation, I mean his own opinion without reference to some randomly choosen article on the net and stupid rhetorical questions, as usually..
And I'm perfectly sure, he doesn't know about physics at all. You an ask him for the explanation of inflation, photon, Lorentz symmetry, renormalization, electroweak interaction, Abelian symmetry, imaginary time, quantum entanglement, number of particle generations, virtual quark formation, ...simply whatever else with the very same result.
Dallas is just another Pupamancur/Manco reincarnation.
Why? Dallas can still propose his own explanation, I mean his own opinion without reference to some randomly choosen article on the net and stupid rhetorical questions, as usually..
And I'm perfectly sure, he doesn't know about physics at all. You an ask him for the explanation of inflation, photon, Lorentz symmetry, renormalization, electroweak interaction, Abelian symmetry, imaginary time, quantum entanglement, number of particle generations, virtual quark formation, ...simply whatever else with the very same result.
Dallas is just another Pupamancur/Manco reincarnation.
QUOTE (Zephir+Jul 29 2007, 06:34 PM)
You an ask him for the explanation of inflation, photon, Lorentz symmetry, renormalization, electroweak interaction, Abelian symmetry, imaginary time, quantum entanglement, number of particle generations, virtual quark formation,
AN and Euler have already shown multiple times that you don't know the meaning of any of the terms you listed, so I will not waste my time proving again the same thing. Especially after I got you to falsify your very own AWT, remember?
AN and Euler have already shown multiple times that you don't know the meaning of any of the terms you listed, so I will not waste my time proving again the same thing. Especially after I got you to falsify your very own AWT, remember?
QUOTE (Dallas+Jul 29 2007, 09:38 PM)
...I got you to falsify your very own AWT, remember?
You're just twaddling strong words, rhetorical questions and social games here, as usually. This is, what are you good in.
You're just twaddling strong words, rhetorical questions and social games here, as usually. This is, what are you good in.
QUOTE (Zephir+Jul 29 2007, 06:42 PM)
You're just twaddling strong words, rhetorical questions and social games here, as usually. This is, what are you good in. Especially after I got you to falsify your very own AWT, remember? That was fun!
QUOTE (Dallas+Jul 29 2007, 06:33 PM)
Yes, I do. From what you write it is clear that you don't. If you did, you would have answered. If you did, you wouldn't have written the erroneous stuff about energy-momentum conservation.
Answered what? You didn't ask a question! Are you really that dumb?
I, however, asked questions. Why haven't you answered them? Can you?
The "accelerating force" is the conserved, attractive or repulsive force I mentioned. Did you miss it? The acceleration (and hence the change in potential energy) arises from this force. I said that too, did you miss it?
You obviously not only can't write and spell at the ninth grade level, you can't comprehend at the ninth grade level either! Who are you, some cocky fourth grader perhaps?
It's obnoxious people like you that give open forums like this a bad reputation.
Answered what? You didn't ask a question! Are you really that dumb?
I, however, asked questions. Why haven't you answered them? Can you?
QUOTE
Yeah, "you said so"
What about the "accelerating force"? You "forgot" about it, didn't you?
What about the "accelerating force"? You "forgot" about it, didn't you?
The "accelerating force" is the conserved, attractive or repulsive force I mentioned. Did you miss it? The acceleration (and hence the change in potential energy) arises from this force. I said that too, did you miss it?
You obviously not only can't write and spell at the ninth grade level, you can't comprehend at the ninth grade level either! Who are you, some cocky fourth grader perhaps?
It's obnoxious people like you that give open forums like this a bad reputation.
QUOTE (Zephir+Jul 29 2007, 06:34 PM)
Dallas is just another Pupamancur/Manco reincarnation.
Looks like you got it right, Zephir
Looks like you got it right, Zephir
QUOTE (ubavontuba+Jul 29 2007, 09:56 PM)
It's obnoxious people like you that give open forums like this a bad reputation.
Only the people, who are disputing with such people like with normal people are giving the forum like this one a bad reputation. Such people must be handled separately. For example, if you know, the Dallas will never answer the question about physics, you can never put him the question from another area, which he can answer.
Only the people, who are disputing with such people like with normal people are giving the forum like this one a bad reputation. Such people must be handled separately. For example, if you know, the Dallas will never answer the question about physics, you can never put him the question from another area, which he can answer.
QUOTE (Zephir+Jul 29 2007, 07:08 PM)
For example, if you know, the Dallas will never answer the question about physics, you can never put him the question from another area, which he can answer.
Looks like I got under your skin, Zephir. Making you refute your beloved AWT must be still stinging. Let's do it again some time in the future, it was fun seeing you falsify your own theory
Looks like I got under your skin, Zephir. Making you refute your beloved AWT must be still stinging. Let's do it again some time in the future, it was fun seeing you falsify your own theory
QUOTE (ubavontuba+Jul 29 2007, 06:56 PM)
The "accelerating force" is the conserved, attractive or repulsive force I mentioned. Did you miss it? The acceleration (and hence the change in potential energy) arises from this force. I said that too, did you miss it?
Hahahahahaha, no I didn't miss it. Your high school teacher wold have rightfully graded your above statement as an F.
Once again: your 9-th grade book STILL says : don't wander why your energy and momentum is not conserved if you have external forces acting on the system. If you need to retake 9-th grade, by all means. Either way, this forum is for discussing new theories, not your misconceptions about oldtheories.
QUOTE (Dallas+Jul 30 2007, 01:04 AM)
Hahahahahaha, no I didn't miss it. Your high school teacher wold have rightfully graded your above statement as an F.
Once again: your 9-th grade book STILL says : don't wander why your energy and momentum is not conserved if you have external forces acting on the system. If you need to retake 9-th grade, by all means. Either way, this forum is for discussing new theories, not your misconceptions about oldtheories.
Wow, your grammar and spelling are atrocious. What grade are you in? I'm thinking fourth, maybe fifth. Why can't you answer simple questions? Are you a mindless chatbot, perhaps?
Anyway, the attractive/repulsive force is always incorporated within the described isolated, symmetrical system. Why are you calling it "external?"
Once again: your 9-th grade book STILL says : don't wander why your energy and momentum is not conserved if you have external forces acting on the system. If you need to retake 9-th grade, by all means. Either way, this forum is for discussing new theories, not your misconceptions about oldtheories.
Wow, your grammar and spelling are atrocious. What grade are you in? I'm thinking fourth, maybe fifth. Why can't you answer simple questions? Are you a mindless chatbot, perhaps?
Anyway, the attractive/repulsive force is always incorporated within the described isolated, symmetrical system. Why are you calling it "external?"
QUOTE (ubavontuba+Jul 30 2007, 05:53 AM)
Wow, your grammar and spelling are atrocious. What grade are you in? I'm thinking fourth, maybe fifth. Why can't you answer simple questions? Are you a mindless chatbot, perhaps?
Why can't you answer to the PHYSICS errors pointed out in your posts? Why is your only answer a collection of personal attacks?
Anyway, the attractive/repulsive force is always incorporated within the described isolated, symmetrical system. Why are you calling it "external?"
Does the fact that your force is producing acceleration tell you anything? Why do you even wonder tha momentum is not conserved? If you have a system of forces with non-zero resultant what happens to the conservation of energy and momentum? It is written in your 9-th grade physics textbook.
Can you find your high school books? Apparently not
Why can't you answer to the PHYSICS errors pointed out in your posts? Why is your only answer a collection of personal attacks?
QUOTE
Anyway, the attractive/repulsive force is always incorporated within the described isolated, symmetrical system. Why are you calling it "external?"
Does the fact that your force is producing acceleration tell you anything? Why do you even wonder tha momentum is not conserved? If you have a system of forces with non-zero resultant what happens to the conservation of energy and momentum? It is written in your 9-th grade physics textbook.
Can you find your high school books? Apparently not
QUOTE (Zephir+Jul 29 2007, 07:08 PM)
Only the people, who are disputing with such people like with normal people are giving the forum like this one a bad reputation. Such people must be handled separately. For example, if you know, the Dallas will never answer the question about physics, you can never put him the question from another area, which he can answer.
No physics in this post, the whole post is just a personal attack. Must be that getting you to refute your own AWT is still stinging. We should do it again sometime, it was fun !
No physics in this post, the whole post is just a personal attack. Must be that getting you to refute your own AWT is still stinging. We should do it again sometime, it was fun !
QUOTE (Dallas+Jul 30 2007, 01:44 PM)
Why can't you answer to the PHYSICS errors pointed out in your posts? Why is your only answer a collection of personal attacks?
Same to you. All you've done is pointlessly attack me without interpreting the specifics of my questions. In fact, you have yet to answer a single proper question with a proper answer. Why is that?
Can you find a reference in regards to your statements?
Did I say that momentum is not conserved in the system in question? Where? In fact, I'll simply declare it: Momentum is conserved in the system I described.
The acceleration is not caused by an external force. It is a property of the isolated, symmetrical system. If Noether is correct, kinetic energy should be conserved within the system. How can that be if the kinetic energy changes with proximity?
Same to you. All you've done is pointlessly attack me without interpreting the specifics of my questions. In fact, you have yet to answer a single proper question with a proper answer. Why is that?
QUOTE
Does the fact that your force is producing acceleration tell you anything? Why do you even wonder tha momentum is not conserved? If you have a system of forces with non-zero resultant what happens to the conservation of energy and momentum? It is written in your 9-th grade physics textbook.
Can you find your high school books? Apparently not
Can you find your high school books? Apparently not
Can you find a reference in regards to your statements?
Did I say that momentum is not conserved in the system in question? Where? In fact, I'll simply declare it: Momentum is conserved in the system I described.
The acceleration is not caused by an external force. It is a property of the isolated, symmetrical system. If Noether is correct, kinetic energy should be conserved within the system. How can that be if the kinetic energy changes with proximity?
QUOTE (ubavontuba+Jul 31 2007, 04:54 AM)
Can you find a reference in regards to your statements?
Yes, open your 9-th grade physics book.
QUOTE
Did I say that momentum is not conserved in the system in question? Where?
How quickly you "forget" Right here
QUOTE (Dallas+Jul 31 2007, 05:14 AM)
Yes, open your 9-th grade physics book.
That's not a proper reference. Can you provide a proper reference?
"Right here," where? Where does it say momentum isn't conserved in a two-body collision? Be specific.
Are you a chatbot? Kann u REED THees sen-tentses? I-f sew, pl-ease Kerect da s-pell-in-g an-d re-prnt dem.
That's not a proper reference. Can you provide a proper reference?
QUOTE
How quickly you "forget" Right here
"Right here," where? Where does it say momentum isn't conserved in a two-body collision? Be specific.
Are you a chatbot? Kann u REED THees sen-tentses? I-f sew, pl-ease Kerect da s-pell-in-g an-d re-prnt dem.
QUOTE (ubavontuba+Aug 2 2007, 03:04 AM)
That's not a proper reference. Can you provide a proper reference?
Yes, your 9-th grade physics book and any freshman college physics book. All of them say the same thing.
Yes, your 9-th grade physics book and any freshman college physics book. All of them say the same thing.
QUOTE (Dallas+Aug 2 2007, 05:06 AM)
Yes, your 9-th grade physics book and any freshman college physics book. All of them say the same thing.
Obviously then, you cannot provide a proper reference. I wonder if you even "know" what the subject of our conversation has been, do you?
You have also revealed yourself to be a chatbot. Is this forum generally maintained for the purpose of testing "AI" programs? There sure seems to be a lot of them here...
Obviously then, you cannot provide a proper reference. I wonder if you even "know" what the subject of our conversation has been, do you?
You have also revealed yourself to be a chatbot. Is this forum generally maintained for the purpose of testing "AI" programs? There sure seems to be a lot of them here...
QUOTE (ubavontuba+Aug 4 2007, 05:10 AM)
Obviously then, you cannot provide a proper reference.
Can't you find your freshman college book?Is this because you never made it into college?
No problem, the theory can be found in your 9-th grade high school book. Try there.
If you get realy desperate buy Spacetime Physics by Taylor and Wheeler and check out page 105, it has some nice pictures.
Can't you find your freshman college book?Is this because you never made it into college?
No problem, the theory can be found in your 9-th grade high school book. Try there.
If you get realy desperate buy Spacetime Physics by Taylor and Wheeler and check out page 105, it has some nice pictures.
QUOTE (Dallas+Aug 4 2007, 06:08 AM)
Can't you find your freshman college book?Is this because you never made it into college?
No problem, the theory can be found in your 9-th grade high school book. Try there.
If you get realy desperate buy Spacetime Physics by Taylor and Wheeler and check out page 105, it has some nice pictures.
And what should I be looking at? How does it relate to our conversation?
Do you know what our discussion is about?
No problem, the theory can be found in your 9-th grade high school book. Try there.
If you get realy desperate buy Spacetime Physics by Taylor and Wheeler and check out page 105, it has some nice pictures.
And what should I be looking at? How does it relate to our conversation?
Do you know what our discussion is about?
QUOTE (ubavontuba+Aug 5 2007, 07:23 AM)
And what should I be looking at? How does it relate to our conversation?
Do you know what our discussion is about?
...it's about you being wrong about conservation of momentum.
....it's about your incessant asking for references and obviously not reading (or not understanding) them
Everyone else got your mistakes, except you.
Do you know what our discussion is about?
...it's about you being wrong about conservation of momentum.
....it's about your incessant asking for references and obviously not reading (or not understanding) them
Everyone else got your mistakes, except you.
QUOTE (Dallas+Aug 5 2007, 02:13 PM)
...it's about you being wrong about conservation of momentum.
....it's about your incessant asking for references and obviously not reading (or not understanding) them
Everyone else got your mistakes, except you.
You really are just a stupid automated chatbot, aren't you?
My topic was on the conservation of energy, not momentum. How wrong were you?
You did not provide a single proper reference. If you had, you could have quoted the relevant portions. Can you quote any relevant portions?
Do you get your own mistakes?
Is anyone monitoring this chatbot's foolishness? When are you going to make one that can actually follow a conversation?
....it's about your incessant asking for references and obviously not reading (or not understanding) them
Everyone else got your mistakes, except you.
You really are just a stupid automated chatbot, aren't you?
My topic was on the conservation of energy, not momentum. How wrong were you?
You did not provide a single proper reference. If you had, you could have quoted the relevant portions. Can you quote any relevant portions?
Do you get your own mistakes?
Is anyone monitoring this chatbot's foolishness? When are you going to make one that can actually follow a conversation?
QUOTE (ubavontuba+Aug 5 2007, 10:10 PM)
You did not provide a single proper reference. If you had, you could have quoted the relevant portions. Can you quote any relevant portions?
For the laws of energy conservation , you need to read (and hopefully comprehend) page 113.
If that fails, hunt down for your 9-th grade physics book, it is all there
QUOTE (Dallas+Aug 5 2007, 11:59 PM)
For the laws of energy conservation , you need to read (and hopefully comprehend) page 113.
If that fails, hunt down for your 9-th grade physics book, it is all there
That's still not a quote. What's it say that you feel is relevant? Do you even know? You certainly didn't know what the discussion was about!
If you cannot provide truly relevant and meaningful materials, then our conversation is at an end.
QUOTE (ubavontuba+Aug 8 2007, 04:50 AM)
.... then our conversation is at an end.
...it ended long ago, you just didn't know about that, see if you can buy a 9-th grade physics book, the answers are all in there
QUOTE (Dallas+Aug 8 2007, 05:06 AM)
...it ended long ago, you just didn't know about that
I didn't know about what? ...that you can't follow a simple conversation?
I didn't know about what? ...that you can't follow a simple conversation?
QUOTE (ubavontuba+Aug 8 2007, 05:08 AM)
I didn't know about what? ...that you can't follow a simple conversation?
...check out that 9-th grade book, your local library might have extra copies. Until you check it out (and you understand it) leave ms. Noether out of it, she might get very upset that you are mangling her theorem.
...check out that 9-th grade book, your local library might have extra copies. Until you check it out (and you understand it) leave ms. Noether out of it, she might get very upset that you are mangling her theorem.
QUOTE (Dallas+Aug 8 2007, 05:16 AM)
...check out that 9-th grade book, your local library might have extra copies. Until you check it out (and you understand it) leave ms. Noether out of it, she might get very upset that you are mangling her theorem.
Obviosly, you can't read a ninth grade textbook! You can't even find a relevant passage!
QUOTE (ubavontuba+Aug 8 2007, 05:23 AM)
Obviosly, you can't read a ninth grade textbook! You can't even find a relevant passage!
...of course , Eric, it is just in your forgotten 9-th grade book, it says right there:
Sigma(E)=const
Sigma(dp/dt)=Sigma(F)
Did you have to take physics over in 9-th grade, Eric?
...of course , Eric, it is just in your forgotten 9-th grade book, it says right there:
Sigma(E)=const
Sigma(dp/dt)=Sigma(F)
Did you have to take physics over in 9-th grade, Eric?
QUOTE (Dallas+Aug 8 2007, 06:16 AM)
...of course , Eric, it is just in your forgotten 9-th grade book, it says right there:
Sigma(E)=const
Sigma(dp/dt)=Sigma(F)
Did you have to take physics over in 9-th grade, Eric?
And how is that applicable to the discussion?
Sigma(E)=const
Sigma(dp/dt)=Sigma(F)
Did you have to take physics over in 9-th grade, Eric?
And how is that applicable to the discussion?
QUOTE (ubavontuba+Aug 9 2007, 02:44 AM)
And how is that applicable to the discussion?
What did I write?
What did I write?
QUOTE (Dallas+Aug 9 2007, 05:07 AM)
What did I write?
You wrote that energy is equal to energy and force is equal to mass times acceleration. Sigma (E) cancels out, so all you really wrote is F=ma.
Now, how is it applicable?
QUOTE (ubavontuba+Aug 10 2007, 05:44 AM)
You wrote that energy is equal to energy and force is equal to mass times acceleration. Sigma (E) cancels out, so all you really wrote is F=ma.
No,this is not what the equations say.And you pretend to talk about Noether's theorem?
Congratulations, you managed to get everything wrong.
You failed physics in 9-th grade, you'll need to retake it. It is kind of late for it, Eric
No,this is not what the equations say.And you pretend to talk about Noether's theorem?
Congratulations, you managed to get everything wrong.
You failed physics in 9-th grade, you'll need to retake it. It is kind of late for it, Eric
QUOTE (Dallas+Aug 10 2007, 06:02 AM)
No,this is not what the equations say.And you pretend to talk about Noether's theorem?
Congratulations, you managed to get everything wrong.
You failed physics in 9-th grade, you'll need to retake it. It is kind of late for it, Eric
Fine then, what do the equations say?
QUOTE (ubavontuba+Aug 10 2007, 06:11 AM)
Fine then, what do the equations say?
Eric,
The equations say that you insulted me for no reason.
The equations say that you flunked physics starting from 9-th grade.
The equations say that you are too lazy to open a book , that you pretend to rtalk about advanced subjects and that you expect the people that you are insulting to explain physics to you such that you could continue to pretend to talk about advanced subjects.
The first equation is the conservation of energy. The second equation is the conservation of momentum. Please do not continue to pretend that you know physics, you aren't fooling anyone but yourself.
Eric,
The equations say that you insulted me for no reason.
The equations say that you flunked physics starting from 9-th grade.
The equations say that you are too lazy to open a book , that you pretend to rtalk about advanced subjects and that you expect the people that you are insulting to explain physics to you such that you could continue to pretend to talk about advanced subjects.
The first equation is the conservation of energy. The second equation is the conservation of momentum. Please do not continue to pretend that you know physics, you aren't fooling anyone but yourself.
It's fun you talked about Noether theorem. By now I'm only four chapters and about 100 pages away from it in my textbook.
( OK, I know , nobody cares )
( OK, I know , nobody cares
)
QUOTE (Dallas+Aug 10 2007, 02:28 PM)
The first equation is the conservation of energy.
Pupamancur,
Conservation is expressed in the verbal equation: "energy equals energy." Duh!
If I said energy equals more (or less) energy, then I would NOT have expressed conservation. I just simplified it (obviously not enough for you).
You're just trying to make it look complicated and impressive (it's not), rather than make it clear and concise.
Again, your mistake. F=ma is the classical (Newtonian) expression. p=mv is another. They just don't include the sum of the masses or time. It's all the same for my purposes. However F=ma is particularly interesting because of the force of gravity present in the system.
Anyway, I'm still waiting for you to explain how this relates to the thought experiment I presented. Can you?
Pupamancur,
Conservation is expressed in the verbal equation: "energy equals energy." Duh!
If I said energy equals more (or less) energy, then I would NOT have expressed conservation. I just simplified it (obviously not enough for you).
You're just trying to make it look complicated and impressive (it's not), rather than make it clear and concise.
QUOTE
The second equation is the conservation of momentum. Please do not continue to pretend that you know physics, you aren't fooling anyone but yourself.
Again, your mistake. F=ma is the classical (Newtonian) expression. p=mv is another. They just don't include the sum of the masses or time. It's all the same for my purposes. However F=ma is particularly interesting because of the force of gravity present in the system.
Anyway, I'm still waiting for you to explain how this relates to the thought experiment I presented. Can you?
QUOTE (Cédric H.+Aug 10 2007, 02:32 PM)
It's fun you talked about Noether theorem. By now I'm only four chapters and about 100 pages away from it in my textbook.
( OK, I know , nobody cares )
I care Cedric. Good for you. Good luck on the adventure!
( OK, I know , nobody cares
) I care Cedric. Good for you. Good luck on the adventure!
QUOTE (ubavontuba+Aug 11 2007, 04:08 AM)
Again, your mistake. F=ma is the classical (Newtonian) expression. p=mv is another. They just don't include the sum of the masses or time.
Eric,
The only person you are fooling is yourself . Go find those 9-th grade books and try taking over that 9-th grade you flunked some 30+ years ago.
QUOTE (Dallas+Aug 11 2007, 06:26 AM)
Eric,
The only person you are fooling is yourself . Go find those 9-th grade books and try taking over that 9-th grade you flunked some 30+ years ago.
Apparently your programming doesn't include classical physics. Do you recognize the formulas F=ma and p=mv? What do they mean?
Again, how do your formulas apply to my thought experiment? Do you even know to which thought experiment I am referring? Can you describe it? Any ninth grader can, can you?
The only person you are fooling is yourself . Go find those 9-th grade books and try taking over that 9-th grade you flunked some 30+ years ago.
Apparently your programming doesn't include classical physics. Do you recognize the formulas F=ma and p=mv? What do they mean?
Again, how do your formulas apply to my thought experiment? Do you even know to which thought experiment I am referring? Can you describe it? Any ninth grader can, can you?
QUOTE (ubavontuba+Aug 11 2007, 09:32 PM)
Apparently your programming doesn't include classical physics. Do you recognize the formulas F=ma and p=mv? What do they mean?
They mean that you don't have a clue when it comes to mechanics in general and conservation of energy and momentum in specific.
It means that you need to take 9-th grade physics over . The stuff that you flunked about 40 years ago.
They mean that you don't have a clue when it comes to mechanics in general and conservation of energy and momentum in specific.
It means that you need to take 9-th grade physics over . The stuff that you flunked about 40 years ago.
QUOTE (Dallas+Aug 11 2007, 09:40 PM)
They mean that you don't have a clue when it comes to mechanics in general and conservation of energy and momentum in specific.
It means that you need to take 9-th grade physics over . The stuff that you flunked about 40 years ago.
Nothin' but bluster. Is that all you got? Admit it, you didn't understand my last post.
What a moronic program you are. If you're the leading edge in AI, then useful AI isn't likely to happen in my lifetime!
Actually, Dallas is right in this case.
The first equation says more then just energy = energy, and the second equation says more then....
You know, I'm not even sure what Ubavontuba thinks the second equation is about.
The first equation says more then just energy = energy, and the second equation says more then....
You know, I'm not even sure what Ubavontuba thinks the second equation is about.
QUOTE (Trippy+Aug 27 2007, 09:40 PM)
Actually, Dallas is right in this case.
The first equation says more then just energy = energy, and the second equation says more then....
You know, I'm not even sure what Ubavontuba thinks the second equation is about.
Your grammar is terrible and you don't know what F=ma or p=mv means? How old are you?
Please elaborate on the first and second equations. Enlighten me. What do they say that's "more then" (sic) I stated?
The first equation says more then just energy = energy, and the second equation says more then....
You know, I'm not even sure what Ubavontuba thinks the second equation is about.
Your grammar is terrible and you don't know what F=ma or p=mv means? How old are you?
Please elaborate on the first and second equations. Enlighten me. What do they say that's "more then" (sic) I stated?
QUOTE (ubavontuba+Aug 28 2007, 05:48 PM)
Your grammar is terrible and you don't know what F=ma or p=mv means? How old are you?
Please elaborate on the first and second equations. Enlighten me. What do they say that's "more then" (sic) I stated?
No.
I'm tired of bashing my head against a brick wall trying to talk to rude ignorami.
Which you clearly are.
You open your post with an assault on my english and physics, and then attempt to discredit me on what are clearly false, and irrational grounds, then expect me to waste my time by engaging in a rational debate?
You so smart.
You figure out for self.
(If you couldn't figure it out I'm being sarcastic in those last two sentences).
And as for your formulae.
f=ma "The force required to accelerate any given mass at a constant rate is the product of the mass of the object and the acceleration of the object."
p=mv "The momentum experienced by any object traveling at a constant velocity is the product of the mass of the object, and the velocity of the object."
I could go on, but what's the point. I have demonstrated that I can do Highschool Physics - something you have yet to do, presumably because your too busy with you irrational attacks on your fellow posters - a sure sign of a truely juvenile mind.
Please elaborate on the first and second equations. Enlighten me. What do they say that's "more then" (sic) I stated?
No.
I'm tired of bashing my head against a brick wall trying to talk to rude ignorami.
Which you clearly are.
You open your post with an assault on my english and physics, and then attempt to discredit me on what are clearly false, and irrational grounds, then expect me to waste my time by engaging in a rational debate?
You so smart.
You figure out for self.
(If you couldn't figure it out I'm being sarcastic in those last two sentences).
And as for your formulae.
f=ma "The force required to accelerate any given mass at a constant rate is the product of the mass of the object and the acceleration of the object."
p=mv "The momentum experienced by any object traveling at a constant velocity is the product of the mass of the object, and the velocity of the object."
I could go on, but what's the point. I have demonstrated that I can do Highschool Physics - something you have yet to do, presumably because your too busy with you irrational attacks on your fellow posters - a sure sign of a truely juvenile mind.
QUOTE (Trippy+Aug 28 2007, 06:49 AM)
I could go on, but what's the point. I have demonstrated that I can do Highschool Physics - something you have yet to do, presumably because your too busy with you irrational attacks on your fellow posters - a sure sign of a truely juvenile mind.
But "young" Eric=ubavontuba is about 50 years old. He has been doing this sort of stuff (insulting the people who attempt hopelessly to teach him a little physics) for a loooong time. He's been doing this since he flunked 9-th grade, for about 40 years.
QUOTE (Trippy+Aug 28 2007, 06:49 AM)
No.
I'm tired of bashing my head against a brick wall trying to talk to rude ignorami.
Which you clearly are.
You open your post with an assault on my english and physics, and then attempt to discredit me on what are clearly false, and irrational grounds, then expect me to waste my time by engaging in a rational debate?
I assaulted YOU? Get it straight. YOU assaulted me first!
That's a copout which simply means you can't really add to my statements.
That's a copout which simply means you can't really add to my statements.
And as for your formulae.
f=ma "The force required to accelerate any given mass at a constant rate is the product of the mass of the object and the acceleration of the object."
p=mv "The momentum experienced by any object traveling at a constant velocity is the product of the mass of the object, and the velocity of the object."
Not bad. But since you put them in quotes, I think I'm safe in assuming you had to look them up. How are they applicable (or not) to my answer to Dallas?
So, how old are you?
I'm tired of bashing my head against a brick wall trying to talk to rude ignorami.
Which you clearly are.
You open your post with an assault on my english and physics, and then attempt to discredit me on what are clearly false, and irrational grounds, then expect me to waste my time by engaging in a rational debate?
I assaulted YOU? Get it straight. YOU assaulted me first!
QUOTE
You so smart.
You figure out for self.
(If you couldn't figure it out I'm being sarcastic in those last two sentences).
You figure out for self.
(If you couldn't figure it out I'm being sarcastic in those last two sentences).
That's a copout which simply means you can't really add to my statements.
QUOTE (->
| QUOTE |
| You so smart. You figure out for self. (If you couldn't figure it out I'm being sarcastic in those last two sentences). |
That's a copout which simply means you can't really add to my statements.
And as for your formulae.
f=ma "The force required to accelerate any given mass at a constant rate is the product of the mass of the object and the acceleration of the object."
p=mv "The momentum experienced by any object traveling at a constant velocity is the product of the mass of the object, and the velocity of the object."
Not bad. But since you put them in quotes, I think I'm safe in assuming you had to look them up. How are they applicable (or not) to my answer to Dallas?
QUOTE
I could go on, but what's the point. I have demonstrated that I can do Highschool Physics - something you have yet to do, presumably because your too busy with you irrational attacks on your fellow posters - a sure sign of a truely juvenile mind.
That's right. YOU started in with the first juvenile attack, so I simply replied with the logical question: "How old are you?"So, how old are you?
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