Unbalanced (action/reaction) elastic vs inelastic collisions the secret behind inertial propulsion
JT
of billiards, the player gives kinetic energy to the cue ball by
striking it with the cue stick. If the cue ball collides with another
ball, it will slow down dramatically and the ball it collided with
will accelerate to a speed as the kinetic energy is passed on to it.
Collisions in billiards are effectively elastic collisions, where (by
definition) kinetic energy is preserved.
OBSERVE********************************************************************
***************
In inelastic collisions, kinetic energy is dissipated as: heat,
sound,
binding energy (breaking bound structures), or other kinds of energy.
***************************************************************************
*********************
Can you now explain where the extra energy in these inelastic
collisons come from, because you and rest of parrot farm claim
momentum conserved.
I said you are wrong only energy conserved, you want elastic
collision
from action and a inelastic collsions transfer heat,light, radiation
from the reaction.
And you will have forces that are bigger for the recoil then for the
thrust action.
So here is an idea about an inertial propulsion design.
The system needs a cooler and an evaporator and a comressor/pump (and
of course an engine to drive it) mounted within a closed system
between two plates "basically a closed two membran with a cold side
and a warm
side.
The thrust from the top mounted pump is directed downwards the lower
plate is a plate evaporator so no waterthrust actually reach the
plate. The vaporised steam raise the pressure inside membran and heat
to level by raising upwards upperplate. Above the pump is a
pipecooler
connected to the upperplate and with a vessel connected to the pump
inlet valve.
You get basicly a waterpiston directed downwards, with a right angle
recoil force upwards but the thrust force is leveled out into a
chaotic force by the vaporizer, where molecules pushing in every
direction and the plate will have a bigger force upwards to
counteract
escape gravity, levitate accelerate and gain momentum.
Maybe you should study the following video and explain the the
radiation exchange around the rim.
http://www.youtube.com/watch?v=1Utr0lt6XR8
JT
Cwatters
"JT" <jonas.t...@hotmail.com> wrote in message
news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
> Kinetic energy can be passed from one object to another. In the game
> of billiards, the player gives kinetic energy to the cue ball by
> striking it with the cue stick. If the cue ball collides with another
> ball, it will slow down dramatically and the ball it collided with
> will accelerate to a speed as the kinetic energy is passed on to it.
> Collisions in billiards are effectively elastic collisions, where (by
> definition) kinetic energy is preserved.
> OBSERVE********************************************************************
> ***************
> In inelastic collisions, kinetic energy is dissipated as: heat,
> sound,
> binding energy (breaking bound structures), or other kinds of energy.
> ***************************************************************************
> *********************
> Can you now explain where the extra energy in these inelastic
> collisons come from, because you and rest of parrot farm claim
> momentum conserved.
If the collision between the balls were inelastic their KE after impact is
less than before. That's where the KE comes from.
JT
we try again.
Kinetic energy can be passed from one object to another. In the game
of billiards, the player gives kinetic energy to the cue ball by
striking it with the cue stick. If the cue ball collides with another
ball, it will slow down dramatically and the ball it collided with
will accelerate to a speed as the kinetic energy is passed on to it.
Collisions in billiards are effectively elastic collisions, where (by
definition) kinetic energy is preserved.
OBSERVE********************************************************************
***************
In inelastic collisions, kinetic energy is dissipated as: heat,
sound,
binding energy (breaking bound structures), or other kinds of energy.
***************************************************************************
*********************
Can you now explain where the extra energy in these inelastic
collisons come from, because you and rest of parrot farm claim
momentum conserved.
I said you are wrong only energy conserved, you want elastic
collision
from action and a inelastic collsions transfer heat,light, radiation
from the reaction.
And you will have forces that are bigger for the recoil then for the
thrust action.
I've been thinking some more about a inertial propulsion design.
The system needs a cooler and an evaporator and a comressor/pump (and
of course an engine to drive it) mounted within a closed system
between two connected plates "basically a membran with a cold side and
a warm
side.
The thrust from the top mounted pump is directed downwards the lower
plate is a plate evaporator so no waterthrust actually reach the
plate. The vaporised steam raise the pressure inside membran and heat
to level by raising upwards upperplate. Above the pump is a
pipecooler
Maybe you should study the following video and explain the the
radiation exchange around the rim.
http://www.youtube.com/watch?v=1Utr0lt6XR8
JT
connected to the upperplate and with a vessel connected to the pump
inlet valve.
You get basicly a waterpiston directed downwards, with a right angle
recoil force upwards but the thrust force is leveled out into a
chaotic force by the vaporizer, where molecules pushing in every
direction and the plate will have a bigger force upwards to
counteract gravity, accelerate andgain momentum
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> "JT" <jonas.thornv...@hotmail.com> wrote in message
Well that pretty much the point, isn't it?
You want the reaction force without a action/thrust, basically you
should a rubber ball within a closed vessel/system but it never reach
the other end.
You will have a recoil force and only a recoil force because the
rubber ball explodes halfway. Your vessel will move in the recoil
direction.
JT
JT
The linear bulk momentum of watermolecules be extinguised or less and
redirected into chaotic oscillation. Your linear action force of
watermolecules with linear momentum, will transform/level into chaotic
oscillation expanding radially between the two plates due to pressure
and temperature differences.
The KE is just the exchange/transform of relative momentum into
radiating energy.
So you could as well written that the linear momentum transforms into
oscillation by chaotic interaction in a media. Thus the gained linear
action of bulk watermolecules is extinguished and your left with a
reaction/recoil force in the vessel.
JT
Sam Wormley
>
> You want the reaction force without a action/thrust, basically you
> should a rubber ball within a closed vessel/system but it never reach
> the other end.
>
> You will have a recoil force and only a recoil force because the
> rubber ball explodes halfway. Your vessel will move in the recoil
> direction.
>
> JT
Bzzzt!
Conservation of Momentum!
Sam Wormley
>
> The KE is just the exchange/transform of relative momentum into
> radiating energy.
Fortune: Self-education in physics reduces embarrassment.
JT
Mr glueball when your balls travel space how do you think they carry
KE. Because as soon they hit an obstacle they will transfer some KE,
Between your glueballs and the new object and some will dissapate as
radiation.
Now go learn something about elastic VS inelastic collision, your
glueballs is an idealisation it do not exist in macro realworld.
Perfect inelastic collisions only between atoms. Not even molecules
smashing into eachother have conservation of linear momentum.
JT
Sam Wormley
>
> Mr glueball when your balls travel space how do you think they carry
> KE. Because as soon they hit an obstacle they will transfer some KE,
> Between your glueballs and the new object and some will dissapate as
> radiation.
>
> Now go learn something about elastic VS inelastic collision, your
> glueballs is an idealisation it do not exist in macro realworld.
> Perfect inelastic collisions only between atoms. Not even molecules
> smashing into eachother have conservation of linear momentum.
>
> JT
A perfectly elastic collision is defined as one in which
there is no loss of kinetic energy in the collision. An
inelastic collision is one in which part of the kinetic
energy is changed to some other form of energy in the
collision.
The assumption of conservation of momentum as well as the
conservation of kinetic energy makes possible the calculation
of the final velocities in two-body collisions.
JT
Typo of me above perfect elastic collision.....
However as i said they do not really exist in macroworld.
Your crashbag in car tell you so, your waterboiler, your cars
deformation zone any action will tell you so heat is dissapated.
Your slow thoughts even proof of the noneexistens of perfect elastic
collisions.
You just started a thread about quantum computers, your CPU if any
should tell you that there is something very wrong with your line of
thought Sam.
JT
Sam Wormley
Can you tell me in detail where I have gone wrong, JT?
JT
Yes you think inelastic collisions conserve momentum. Only perfect
elastic does.
JT
JT
>
> > Mr glueball when your balls travel space how do you think they carry
> > KE. Because as soon they hit an obstacle they will transfer some KE,
> > Between your glueballs and the new object and some will dissapate as
> > radiation.
>
> > Now go learn something about elastic VS inelastic collision, your
> > glueballs is an idealisation it do not exist in macro realworld.
> > Perfect inelastic collisions only between atoms. Not even molecules
> > smashing into eachother have conservation of linear momentum.
>
> > JT
>
> A perfectly elastic collision is defined as one in which
> there is no loss of kinetic energy in the collision. An
Energy can not be lost according to conservation laws, but it can be
transformed.
> inelastic collision is one in which part of the kinetic
> energy is changed to some other form of energy in the
> collision.
Yes the relative momentum of gas/waterthrust will transform to KI at
collision, the reaction collision will be elastic with minimal
potential momentum/KI transformed into radiation. An object that
travel with relative constant velocity thru space, do not carry any
KI, the KI is the energy potential relative another object at
collision.
A vessel can achieve reaction by causing an action accelerating an
object/objects giving them momemtum, to get an elastic collision this
by using an upper enclosing with rubberdampers or other elastic
compound.
This reaction of vessel can be achived without the vessel having an
equal action force the action force can be dissipated as chaotic
oscilation within a gasmedia before the action of the accelerated
objects reach the opposite interior vessel hull.
JT
JT
momentum and redirected within the gas or liquid media to.
Sam Wormley
The law of conservation of linear momentum is a fundamental law of nature,
and it states that the total momentum of a closed system of objects (which
has no interactions with external agents) is constant. within that closed
system Kinetic Energy can be converted to other forms of energy. One of
the consequences of this is that the center of mass of any system of
objects will always continue with the same velocity unless acted on by a
force from outside the system.
Owen Jacobson
> Kinetic energy can be passed from one object to another. In the game
> of billiards, the player gives kinetic energy to the cue ball by
> striking it with the cue stick. If the cue ball collides with another
> ball, it will slow down dramatically and the ball it collided with
> will accelerate to a speed as the kinetic energy is passed on to it.
> Collisions in billiards are effectively elastic collisions, where (by
> definition) kinetic energy is preserved.
> OBSERVE********************************************************************
> ***************
> In inelastic collisions, kinetic energy is dissipated as: heat,
> sound,
> binding energy (breaking bound structures), or other kinds of energy.
> ***************************************************************************
> *********************
> Can you now explain where the extra energy in these inelastic
> collisons come from, because you and rest of parrot farm claim
> momentum conserved.
Actually, you can derive the principle that momentum is conserved from
Newton's other laws of motion. For a trivial example, have a look at
message
<eeffd8cc-bfeb-44ba...@a1g2000hsb.googlegroups.com>
(http://groups.google.ca/group/sci.physics/msg/989e1fb5a2c702af), which
works through the simple case of a ball being launched from one end of
a closed container and stopping inelastically at the other end.
-o
Sam Wormley
If the vessel is only a part of the closed system of object, mass
(or energy) ejected from the vessel can change the vessel's momentum,
however, the center of mass of any system of objects will always
continue with the same velocity unless acted on by a force from
outside the system.
The law of conservation of linear momentum is a fundamental law of
Sam Wormley
> Kinetic energy can be passed from one object to another. In the game
> of billiards, the player gives kinetic energy to the cue ball by
> striking it with the cue stick. If the cue ball collides with another
> ball, it will slow down dramatically and the ball it collided with
> will accelerate to a speed as the kinetic energy is passed on to it.
> Collisions in billiards are effectively elastic collisions, where (by
> definition) kinetic energy is preserved.
>
> JT
In an isolated system with only two objects, the change in momentum
of one object must be equal and opposite to the change in momentum of
the other object. Mathematically,
∆p1 = -∆p2
Momentum has the special property that, in a closed system, it is always
conserved, even in collisions and separations caused by explosive forces.
Kinetic energy, on the other hand, is not conserved in collisions if they
are inelastic. Since momentum is conserved it can be used to calculate an
unknown velocity following a collision or a separation if all the other
masses and velocities are known.
PD
> Kinetic energy can be passed from one object to another. In the game
> of billiards, the player gives kinetic energy to the cue ball by
> striking it with the cue stick. If the cue ball collides with another
> ball, it will slow down dramatically and the ball it collided with
> will accelerate to a speed as the kinetic energy is passed on to it.
> Collisions in billiards are effectively elastic collisions, where (by
> definition) kinetic energy is preserved.
> OBSERVE********************************************************************
> ***************
> In inelastic collisions, kinetic energy is dissipated as: heat,
> sound,
> binding energy (breaking bound structures), or other kinds of energy.
> ***************************************************************************
> *********************
> Can you now explain where the extra energy in these inelastic
> collisons come from, because you and rest of parrot farm claim
> momentum conserved.
This seems to surprise you.
If you set up an experiment where you fire an 30 gram (0.030 kg)
bullet at 900 m/s toward a 6.0 kg block of wood sitting on a ice
surface (say, in a hockey rink), you will *observe* (that is, in real
measurement) that the bullet will become embedded in the block of
wood, and the block of wood will coast on the ice at a speed of 4.48 m/
s.
Now it's a simple matter from these *measurements* to see what has
happened to the momentum and kinetic energy.
Momentum:
Before the collision = (0.030 kg)(900 m/s) + (6.0 kg)(0 m/s) = 27 kg*m/
s.
After the collision = (6.030 kg)(4.48 m/s) = 27 kg*m/s.
Kinetic energy:
Before the collision = (1/2)(0.030 kg)(900 m/s)^2 + (1/2)(6.0 kg)(0 m/
s)^2 = 12200 J.
After the collision = (1.2)(6.030 kg)(4.48 m/s)^2 = 60.45 J.
Now, from the above measurements, it is plain that momentum has been
conserved but kinetic energy has not.
I realize that you find it hard to understand how this is possible.
But the point is that these conservation rules are what are found in
nature by making *measurements*.
Autymn D. C.
my momentum lab with a sled on airtrack and timing gates shew the
inelastic collisions breachd conservation of momentum.
JT
>
> - Visa citerad text -
Well you are confused, the vessel have an internal bulk acceleration
of gas or watermolecules. There are alot of internal agents
interacting that will lead to radiation, cause is pressure and heat.
There is no outside force needed to dissipate or change the vector of
the action force, this is gas/watermolecules only heat and pressure
will vaporize the bulk movement into chaotic oscillation that expands
radially within membran. It is that easy, you can bitch about
conservation of momentum to hell freeze over. But it is still your
missunderstanding of your conception of conservation of momentum. It
is an idealisation of noneexistent microworld objects atoms that smash
into eachother within a system free of external agents magnetics and
gratvity.
The ideal environment do not exist the ideal collision do not exist in
macro world, keep bitching about santa he still do not exist. You want
elastic collisions for reaction to keep momenum and inelastic to
transform bulk momentum of gas/liquid into heat end of story.
JT
JT
>
> - Visa citerad text -
No the vessel is the system there is no ejection except for radiation,
there is a recycling process of bulk movement going on inside vessel
however. No the center of mass is not static within an enclosed pump
it change with pressure you are dreaming and you are confused, the
JT
>
> - Visa citerad text -- Dölj citerad text -
>
> - Visa citerad text -
Bullshit there is not conservation of momentum in inelastic
collisions.
You are confused, the vessel have an internal bulk acceleration of gas
or watermolecules. There are alot of internal agents interacting that
will lead to radiation, what cause it is pressure and heat.
Cwatters
"JT" <jonas.t...@hotmail.com> wrote in message
news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
> but the thrust force is leveled out into a
> chaotic force by the vaporizer,
That's the bit I have trouble with. .
How does the water impact the vaporiser without exerting a force on it? I
mean even if this was a microwave oven cavity that turned the water to steam
without contact - the resulting steam would still have the momentum of the
water and exert a pressure on the cavity in the direction the water was
travelling. Trace the path and momentum of an individual water molecule and
explain where the momentum goes.
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
>
> > but the thrust force is leveled out into a
> > chaotic force by the vaporizer,
>
> That's the bit I have trouble with. .
I do not see why, where does the forces go when you drop a stone in a
pound.
> How does the water impact the vaporiser without exerting a force on it? I
> mean even if this was a microwave oven cavity that turned the water to steam
> without contact - the resulting steam would still have the momentum of the
> water and exert a pressure on the cavity in the direction the water was
> travelling. Trace the path and momentum of an individual water molecule and
> explain where the momentum goes.
Well i think you know that a pressure boiler excert a force the
pressure is highest around the evaporator. The water will try to level
out within the closed system, that's why you need a cooler.
JT
Sam Wormley
In an isolated system with only two objects, the change in momentum
of one object must be equal and opposite to the change in momentum of
the other object. Mathematically,
∆p1(internal to the vessel) = -∆p2(stuff escaping the vessel
Sam Wormley
> No the vessel is the system there is no ejection except for radiation,
> there is a recycling process of bulk movement going on inside vessel
> however. No the center of mass is not static within an enclosed pump
> it change with pressure you are dreaming and you are confused, the
> vessel have an internal bulk acceleration of gas or watermolecules.
> There are alot of internal agents interacting that will lead to
> radiation, cause is pressure and heat.
>
> There is no outside force needed to dissipate or change the vector of
> the action force, this is gas/watermolecules only heat and pressure
> will vaporize the bulk movement into chaotic oscillation that expands
> radially within membran. It is that easy, you can bitch about
> conservation of momentum to hell freeze over. But it is still your
> missunderstanding of your conception of conservation of momentum. It
> is an idealisation of noneexistent microworld objects atoms that smash
> into eachother within a system free of external agents magnetics and
> gratvity.
>
> The ideal environment do not exist the ideal collision do not exist in
> macro world, keep bitching about santa he still do not exist. You want
> elastic collisions for reaction to keep momenum and inelastic to
> transform bulk momentum of gas/liquid into heat end of story.
>
> JT
In an isolated system with only two objects, the change in momentum
of one object must be equal and opposite to the change in momentum of
the other object. Mathematically,
∆p1(vessel) = -∆p2(radiation)
G=EMC^2 Glazier
with one ball and only the top ball goes out. hit it with two balls and
two balls move out. Here you see equal action creating an equal action.
The balls in the middle do not move but pass the energy through. its
some what amazing Bert
Sam Wormley
Conservation of Momentum.
If the middle balls didn't move, the initiating ball(s) would bounce
back. http://en.wikipedia.org/wiki/Newton's_cradle
PD
What I just gave you is a measurable example of an inelastic
collision.
Set it up at a hockey rink, watch it happen in front of your eyes,
make the measurements, and THEN tell me there's no such thing.
>
> You are confused, the vessel have an internal bulk acceleration of gas
> or watermolecules.
What gas or water molecules in the bullet or block of wood?
jbriggs444
> On 25 Nov, 10:54, "Cwatters"
>
> <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> > "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> >news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
>
> > > but the thrust force is leveled out into a
> > > chaotic force by the vaporizer,
>
> > That's the bit I have trouble with. .
>
> I do not see why, where does the forces go when you drop a stone in a
> pound.
Into the water. The stone is decelerated (upward force) and the water
is accelerated downward (downward force).
This induces sound waves and ordinary water waves in the water. The
pond is supported by the muck at the bottom. The downward motion of
the water results in a transient increase in supporting force from the
muck on the bottom of the pond.
And you end up with a splash and ripples move through the water and
you could even argue the the downward impulse has actually moved the
water upward. Yes, this is chaotic. But that doesn't mean there
wasn't a net impulse delivered to the water. And And it doesn't mean
the resulting momentum wasn't further transferred from the water to
the muck.
You seem to operate on the premise:
"If I can't figure out where the force goes and if it's small
compared to other forces that I can see then it must just disappear".
This is the fallacy of argument by failure of imagination.
You're like the fellow who thinks that the "round down the pennies and
put the excess in my account" scheme popularized in Office Space not
only works, but actually creates money out of nothing.
It ain't so. If money is conserved at each transaction and you end up
with more in your account, somebody else winds up with less. Do the
math.
If momentum is conserved at each transaction and you end up with more
on your object then some other object winds up with less.
Wiggling your fingers and chanting "leveled out into a chaotic force"
doesn't change this.
Androcles
"jbriggs444" <jbrig...@gmail.com> wrote in message
news:5bba0708-0b7f-4e19...@x5g2000prf.googlegroups.com...
On Nov 25, 6:07 am, JT <jonas.thornv...@hotmail.com> wrote:
> On 25 Nov, 10:54, "Cwatters"
>
> <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> > "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> >news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
>
> > > but the thrust force is leveled out into a
> > > chaotic force by the vaporizer,
>
> > That's the bit I have trouble with. .
>
> I do not see why, where does the forces go when you drop a stone in a
> pound.
Into the water. The stone is decelerated (upward force) and the water
is accelerated downward (downward force).
===========================================
There are 14 pounds in a stone - ask your mother.
http://www.weightlossresources.co.uk/body_weight/weight_converter.htm
Cwatters
> On 25 Nov, 10:54, "Cwatters"
> <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
>> "JT" <jonas.thornv...@hotmail.com> wrote in message
>>
>> news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
>>
>> > but the thrust force is leveled out into a
>> > chaotic force by the vaporizer,
>>
>> That's the bit I have trouble with. .
>
> I do not see why, where does the forces go when you drop a stone in a
> pound.
Well waves travel out in all directions AND the planet moves. Momentum is
still conserved.
>> How does the water impact the vaporiser without exerting a force on it? I
>> mean even if this was a microwave oven cavity that turned the water to
>> steam
>> without contact - the resulting steam would still have the momentum of
>> the
>> water and exert a pressure on the cavity in the direction the water was
>> travelling. Trace the path and momentum of an individual water molecule
>> and
>> explain where the momentum goes.
>
> Well i think you know that a pressure boiler excert a force the
> pressure is highest around the evaporator. The water will try to level
> out within the closed system, that's why you need a cooler.
>
> JT
Sorry. Not I'm not familiar with pressure boilers.
jbriggs444
> There are 14 pounds in a stone - ask your mother.
>
> http://www.weightlossresources.co.uk/body_weight/weight_converter.htm
Ask the guy at animal control with the high pitched voice. Maybe he
dropped a stone in the pound.
Owen Jacobson
I wouldn't even call it a "special property". The conservation of
momentum is a mathematical consequence of Newton's laws of motion: you
can derive it from p = mv and Newton's second and third laws.
Arguing that conservation of momentum can be violated is equivalent to
arguing that at least one of Newton's laws of motion is incorrect. The
questions that raises are:
- Which one(s)?
- What would you replace it (them) with?
- What experiments can be performed to differentiate between Newton's
laws and the proposed new theory?
Of course, we already know Newton's laws as normally taught in grade
school are inaccurate at relativistic velocities, but the corrected
versions also imply that momentum is conserved.
-o
JT
I think i told you ten times Sam that even in a system of three
objects spring, rubberball, vessel that the linear momentum of the
accelerated ball is not transfered into linear motion of the closed
vessel.
You just let it bounce sideways zigzag by but a simple obstacle
changing the direction of ball. I do not mean to be offensive but
surely a 5 year old person can see that there will be forces making
the vess rock sideways.
It really do not have to bounce zigzag since it is a rubberball not
your glueball. It will not stuck to the vesselwall it will get a
bounce acceleration back.
You seem to think that the momentum will be extinguished and all
momentum will be transfered to the vessel linear when it reach the
hull, you are quite funny.
JT
JT
> On Nov 25, 6:07 am,JT<jonas.thornv...@hotmail.com> wrote:
>
> > On 25 Nov, 10:54, "Cwatters"
>
> > <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> > > "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> > >news:e4fcaffc-fc76-4190...@g23g2000vbr.googlegroups.com...
>
> > > > but the thrust force is leveled out into a
> > > > chaotic force by the vaporizer,
>
> > > That's the bit I have trouble with. .
>
> > I do not see why, where does the forces go when you drop a stone in a
> > pound.
>
> Into the water. The stone is decelerated (upward force) and the water
> is accelerated downward (downward force).
You are quite funny you think a meteor impact hitting water have a
force downwards trancends linear momentum thru earth and you get a
water piston on other side of earth funny....
No there will be a radial wave of watermolecules that moves outwards
from impact zone, as the meteor push the water molecules aside they
ineract with other watermolecules pushing them sideways.
The watermolecules are not glued to the meteor like some linear
piston, you tell your silly ideas to anyone that been in a tsunami or
earthquake that the force epicentrum that was directed upwards, only
created a force that travelled conserving linear momentum upwards thru
the atmosphere. And they punch you on the nose with a linear force
that breaks your nosebone spreading body fluids expanding radially
from impact point.
> This induces sound waves and ordinary water waves in the water. The
> pond is supported by the muck at the bottom. The downward motion of
> the water results in a transient increase in supporting force from the
> muck on the bottom of the pond.
No you are stupid the forces is extinguished long gone before the
stone reach bottom of the pond. There is no downward force hitting the
bottom of pond a 5 year old now this so either you have somekind of
comphrehension or intellectual deficency or you just had a bad day.
> And you end up with a splash and ripples move through the water and
> you could even argue the the downward impulse has actually moved the
> water upward. Yes, this is chaotic. But that doesn't mean there
> wasn't a net impulse delivered to the water. And And it doesn't mean
> the resulting momentum wasn't further transferred from the water to
> the muck.
You are just plain silly or stupid the stone momentum do not transform
to linear net momentum it transforms to radial momentum a force that
travel outwards from impact point.
> You seem to operate on the premise:
>
> "If I can't figure out where the force goes and if it's small
> compared to other forces that I can see then it must just disappear".
What dissapear? It change direction and finally dissipate as
oscillation, radiation, heat. Now your IQ fastly decrease or just
wasn't there to begin with. The force vector relative an accelerated
body and the vessel isn't glued, the momentum of an accelerated body
or bulk of bodies do not travel linear within a gas or liquidated
medium.
> This is the fallacy of argument by failure of imagination.
More a fallacy argument due to lack of IQ on your part.
> You're like the fellow who thinks that the "round down the pennies and
> put the excess in my account" scheme popularized in Office Space not
> only works, but actually creates money out of nothing.
No you and Sam think there can be created radiating energy without
loss of momentum, Utterly stupid i would say a 5 year old knows it,
you are the ones with the energy out of nothing scheme if your ideas
about conservation of momentum was right we would had free energy
devices a couple of hundred years ago.
> It ain't so. If money is conserved at each transaction and you end up
> with more in your account, somebody else winds up with less. Do the
> math.
I did and the winner was the radiation and the loser the momentum.
> If momentum is conserved at each transaction and you end up with more
> on your object then some other object winds up with less.
But it is not stupid there is less do you have a comprehension
deficiency, the reaction/recoil will only creat a small loss in
momentum when colide with hull, due to radiation of vessel while the
action will dissipate as radiation before reach other hull. And the
vessel will move.
> Wiggling your fingers and chanting "leveled out into a chaotic force"
> doesn't change this.
Your IQ is limited and i have a feeling that will not change
conservation of momentum.
JT
Sam Wormley
Are the spring and rubber ball inside the vessel?
>
> You just let it bounce sideways zigzag by but a simple obstacle
> changing the direction of ball. I do not mean to be offensive but
> surely a 5 year old person can see that there will be forces making
> the vess rock sideways.
>
> It really do not have to bounce zigzag since it is a rubberball not
> your glueball. It will not stuck to the vesselwall it will get a
> bounce acceleration back.
>
> You seem to think that the momentum will be extinguished and all
> momentum will be transfered to the vessel linear when it reach the
> hull, you are quite funny.
>
> JT
Laugh now, but learn the physics, JT.
Momentum is conserved in a closed system, i.e., your vessel containing
a spring and a ball.
JT
>
> - Visa citerad text -
Well Sam you could do this yourself with a spring a papertube and a
rubberball, inside tube is an obstacle that make the ball bounce
sideways down the tube.
All is rigged on a floating piece of styrofoam, the spring have to be
quite powerful when you release spring you will see styrofoam rocking
sideways creating waves while moving in the direction of recoil.
JT
Sam Wormley
Visualize in free space. The center of mass does not change its state
of uniform motion (or rest) depending on who's looking.
Conservation of momentum applies.
Cwatters
>You just let it bounce sideways zigzag by but a simple obstacle
>changing the direction of ball.
..and the ball changing the direction of the obstacle.
Cwatters
us know how you get on.
Owen Jacobson
What's your procedure for compressing and releasing that spring? Are
you sure that that procedure doesn't introduce momentum to the spring-
tube-ball system?
There's a fairly simple intuitive argument for why momentum should be
conserved that you probably haven't considered.
Remember that momentum (p = mv) is a vector quantity -- that is, it
has a direction and a magnitude. In fact, momentum depends on your
frame of reference (since velocity also depends on your frame of
reference). I'm going to assume that your argument is that the
*magnitude* of the system's momentum vector decreases over time as
momentum is "lost to friction and chaotic forces." If that's so, then
there exists some frame of reference where |p_initial| > |p_final|.
However, because the laws of physics are (so far as we can tell) frame-
invariant even under simple (Gallilean) relativity, there must also be
a frame where the magnitude of the system's momentum vector
*increases* and |p'_initial| < |p'_final|.
Consider the frame of reference frame_r where the system is initially
at rest (|p_initial_r| = 0) and the frame of reference frame_m where
the system is initially not at rest (|p_initial_m| > 0). The frame of
reference frame_r must necessarily be moving with respect to frame_m
at the same velocity that the system is moving in frame_m. Follow?
If |p_final_m| = 0 and frame_m is moving with respect to frame_r, then
|p_final_r| > 0: in fact, with a little work you can show that
p_final_r = -1 * p_initial_m.
So, in frame_r, the system appears to have *gained* momentum - it was
initially at rest, and now it's not.
If momentum is conserved, on the other hand, then in every frame_x,
p_initial_x = p_final_x so in no frame does momentum appear to grow.
Of course, this is the hard way. The easy way would be to work through
the post I referred you to that works through the math for a simple
example and discusses the difference between conservation of momentum
and conservation of energy.
-o
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> "JT" <jonas.thornv...@hotmail.com> wrote in message
Ehhh yes of course the obstacle is not exactly freefloating it is part
of the tube/vessel. It will change the vector of the ball so it is
bouncing the tubes interior walls.
That was the whole point with the story...the action momentum of the
ball is not an opposite force to the recoil, it will change direction.
You have a action force that make the tube rock oscillate.
So spring reaction is linear acceleration recoil for tube while the
action is mainly sideways force that make the tube rock from side to
side. That mean the reaction force for the closed system will be
bigger then the action force into the opposite direction. That is
inertial propulsion in a recycle process.
JT
JT
>
> - Visa citerad text -
No it does not the vessel will move forward in the recoil direction
while rocing sideways.
JT
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> The best way to convince someone who doesn't understand is to build it. Let
> us know how you get on.
I'm sure Sam will keep you updated.
JT
JT
>
> - Visa citerad text -
As i said the only point i have with my post is to show that an action
force do not necessarily have a reaction ***in oposite direction*** of
same magnitude.
The action reaction forces is unbalanced and the system will move, the
object/vessel will move respective the initial rest system.
This is done by transfer maximum momentum for the reaction force to
the vessel hull, but the accelerated molecules bulk momentum is
transformed to chaotic oscillation radiation before the action force
reach the opposite hull.
JT
Sam Wormley
>
> No it does not the vessel will move forward in the recoil direction
> while rocing sideways.
>
> JT
JT -- Why do you deny the real physics?
Sam Wormley
I will update you JT on other also attempting to skirt the
physics conservation laws. See: http://en.wikipedia.org/wiki/Reactionless_drive.
Welcome to the world of kooks and cranks.
JT
>
> - Visa citerad text -
Why do you promote objects with momentum only to have linear
interaction.
JT
Sam Wormley
What happens inside the vessel, is of no concern to me, it being
a closed system. The momentum of that closed system is conserved
unless acted upon by an external force.
F = dp/dt
Owen Jacobson
> On 27 Nov, 23:05, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On Nov 27, 3:14�pm, JT <jonas.thornv...@hotmail.com> wrote:
>>
>>> Well Sam you could do this yourself with a spring a papertube and a
>>> rubberball, inside tube is an obstacle that make the ball bounce
>>> sideways down the tube.
>>
>>> All is rigged on a floating piece of styrofoam, the spring have to be
>>> quite powerful when you release spring you will see styrofoam rocking
>>> sideways creating waves while moving in the direction of recoil.
>>
>> What's your procedure for compressing and releasing that spring? Are
>> you sure that that procedure doesn't introduce momentum to the spring-
>> tube-ball system?
Could you address this, please?
>> There's a fairly simple intuitive argument for why momentum should be
>> conserved that you probably haven't considered.
>>
>> Remember that momentum (p = mv) is a vector quantity -- that is, it
>> has a direction and a magnitude. In fact, momentum depends on your
>> frame of reference (since velocity also depends on your frame of
>> reference). I'm going to assume that your argument is that the
>> *magnitude* of the system's momentum vector decreases over time as
>> momentum is "lost to friction and chaotic forces." If that's so, then
>> there exists some frame of reference where |p_initial| > |p_final|.
>> However, because the laws of physics are (so far as we can tell) frame-
>> invariant even under simple (Gallilean) relativity, there must also be
>> a frame where the magnitude of the system's momentum vector
>> *increases* and |p'_initial| < |p'_final|.
>
> As i said the only point i have with my post is to show that an action
> force do not necessarily have a reaction ***in oposite direction*** of
> same magnitude.
Direction doesn't matter. If momentum is not *exactly* conserved, then
there exists a frame of reference where the system's final momentum has
a larger magnitude than the system's initial momentum. Momentum cannot
be "lost" globally; a "loss" of momentum in one frame is a "gain" of
momentum in another. Do the math yourself, if you don't believe me.
More to the point, do the experiment. You'll discover that 1. total
energy is conserved, 2. momentum is conserved, and 3. energy is
converted momentarily into kinetic energy before being lost as heat:
kinetic energy is not conserved. At best, your experimental system will
oscillate back and forth for a while to keep its center of mass
stationary.
> The action reaction forces is unbalanced
No experiment has ever falsified Newton's third law.
-o
JT
>
> - Visa citerad text -
Inelastic collisions do not preserve momentum end of story, it is
transformed to radiation heat. Of course the total energy is conserved
no one ever doubted that, but some morons think there can be heat
without loss of momentum. There is no such thing as a perfect
collision without loss of momentum.
P=mv at the collision some of the velocity is lost into radiation if
not we would have systems that could radiate forever *free energy*
devices just by accelerating a body and let them endlessly crash into
eachother to produce free energy.
Because some moron state that in a closed system there can be free
energy radiation wihtout loss of momentum. So let the damn
watermolecules oscillate and bounce into eachother endlessly within a
vessel to create radiation exchanged to heat within our homes.
It is a moron theory conservation of momentum is an idealisation of
microscopic collision without exchange of kinetic energy only momentum
transfered between object it do not exist things oscillate radiat when
they collide. The perfect elastic macroscopic collison is a figment of
a broken mind.
I do not know what moron started to interpret Newtons idealisation
into macroscopic collsiions i just know that within inelastic
collision there is no conservation of momentum because if it was a
colided body would never stop to deform/wobble/radiate but is always
stop to move relative the rest system. The oscillation of object is
transformed to radiating heat within an endless chain of collision
even boiling water in a vessel in space will cool of finally it is due
to kinetic losses into radiation after each collision the oscillation
get weaker to finally dissipate no more momentum, end of story.
JT
JT
I started to get tired to explain there is heat exchange within
inelastic collisions but you fucking do not listen. That is why the
reaction "*elastic collision*"(there is no macroscopic perfect
collisions though) transfer more momentum to the vessel then the
inelastic oscillating bulk of collisions transfer momentum to the
vessel hull in opositie direction.
The object move in the reaction direction end of story.
> More to the point, do the experiment. You'll discover that 1. total
> energy is conserved,
Sure it is conserverd *****into radiation**** + minimal action
momentum but conserved
> 2. momentum is conserved, and
Total bullshit and dreaming on your part it will never happen the the
collision that transfer action vs reaction is unbalanced within the
closed system. The action force on is far weaker then the reaction
force on the interior hull because energy is lost into radiation for
the inelastic collisions. End of story end of chain of lies.
3. energy is
> converted momentarily into kinetic energy before being lost as heat:
> kinetic energy is not conserved.
Correct that is what happens with the bulk watermolecules that is
collided into chaotic oscillation.
>At best, your experimental system will
> oscillate back and forth for a while to keep its center of mass
> stationary.
Bullshit the reaction force vs the action force is unbalanced the
action force transfer momentum the reaction force dissapate as heat
*********energy conserved******* there is no bulk momentum to the
acting as an action force on the vessel hull and transfer momentum,
that force is depleted into a chaotic oscillation radiating system.
> > The action reaction forces is unbalanced
>
> No experiment has ever falsified Newton's third law.
Maybe there is a fucking good reason for it.
Most people can be fooled into think that Newtons idealisation of a
microscopic event of elastic collision can be transfered to into
macroscopic events.
JT
Sam Wormley
> Inelastic collisions do not preserve momentum end of story, it is
> transformed to radiation heat.
What makes you that "radiation heat" doesn't have momentum?
What makes you think momentum is not conserved?
Conservation of momentum applies.
∆p1(vessel) = -∆p2(radiation)
∆p1(internal to the vessel) = -∆p2(stuff escaping the vessel
Sam Wormley
Conservation of momentum applies.
∆p1(vessel) = -∆p2(radiation)
JT
>
> - Visa citerad text -
Yes the vessel moves and radiates. There is no linear conserved action
momentum though, only the reaction force is linear the action force is
chaotic oscillation radiation and the vessel will move. Radiant energy
oscillation can be stored within a medium.
JT
Sam Wormley
>
> Total bullshit and dreaming on your part it will never happen the the
> collision that transfer action vs reaction is unbalanced within the
> closed system. The action force on is far weaker then the reaction
> force on the interior hull because energy is lost into radiation for
> the inelastic collisions. End of story end of chain of lies.
>
Conservation of momentum applies.
∆p1(vessel) = -∆p2(radiation)
Note that p = h/λ , photon Momentum as the vector
p_ph = h/2π k
where k is the wave number vector, with |k| = 2π/λ such that |p_ph| = h/λ
The photon momentum p_ph is directed and have a magnitude of h/λ
Sam Wormley
>> Conservation of momentum applies.
>>
>> ∆p1(vessel) = -∆p2(radiation)
>>
>> - Visa citerad text -
>
> Yes the vessel moves and radiates. There is no linear conserved action
> momentum though, only the reaction force is linear the action force is
> chaotic oscillation radiation and the vessel will move. Radiant energy
> oscillation can be stored within a medium.
>
> JT
In that case, ∆p1(vessel) = -∆p2(radiation) = 0
F = dp/dt = 0
No change in momentum. Momentum is conserved.
JT
You are tiresome Sam p2 do now have oposite vector/direction to p1,
you know very well that heat oscillation will try to level out in a
homogenious media. p2 will escape the media where you let is transfer
radiation that is up to the structural design of vessel where you let
it radiate or is you chose to using it as a giant laser or store it,
some material can store thermal energy quite good.
JT
Cwatters
If the ball doesn't leave the tube, some time later the ball must end up
travelling at the same speed as the tube.
Where does the force come from to accelerate the ball so that it's
travelling along at the same speed as the tube ?
JT
>
> - Visa citerad text -
When the zigzag action force is depleted it will stop be at rest
within tube.
You could easily calculate the exact force needed to make the ball
travel forth and back the tube to the initial position. And yet the
system moved from the initial rest system point in a kinetic
gravitational environment. In space the tube would have gained
momentum.
JT
PD
Perhaps there is a misunderstanding here -- a quite common one.
Newton's third law says that if a body A exerts a force F on body B,
then body B exerts a force -F on body A.
Beginning students commonly mistake the third law to say that if there
is a force F on body B, then there is also a force -F on body B. This
is obviously wrong, because it would imply that there is a net force
of 0 on any body at any given time, and therefore no body can ever
accelerate.
PD
Momentum is not energy. Kinetic energy gets transformed into heat, but
momentum does not get transformed into ANY kind of energy.
> Of course the total energy is conserved
> no one ever doubted that, but some morons think there can be heat
> without loss of momentum. There is no such thing as a perfect
> collision without loss of momentum.
No, that's not right. Momentum is TRANSFERRED across the boundary of a
system that is not isolated. But no momentum is lost.
JT
>
> - Visa citerad text -- Dölj citerad text -
>
> - Visa citerad text -
Bullshit F is shirinking when the mass is deaccelerated thru collision
your iinitial mv is transform and you get a new relative velocity to
vessel v(2). So your new F(2)=mv(2).
The v-v(2) is the lost into radiation, and your new relative force
with vessel is mv(2)
JT
PD
What? What makes you say so?
> your iinitial mv is transform and you get a new relative velocity to
> vessel v(2). So your new F(2)=mv(2).
F is not momentum. They don't even have the same units.
Owen Jacobson
I heard you just fine. You believe that there is a mechanism through
which momentum (a vector quantity described by p = mv) can be converted
into energy (any of a number of scalar quantities). The problem is that
this violates conservation of energy -- any change in momentum without
a corresponding change in the momentum of some other part of the system
would change the total energy of the system, and there always exists a
frame of reference where that change in total energy takes the form of
an increase in total energy.
Collisions can convert kinetic energy into heat, but only until the
entire system is moving at the same velocity, at which point no more
collisions occur and the system keeps any remaining kinetic energy
until acted on by an outside force. Momentum, on the other hand, is not
equivalent to any form of energy and cannot be converted into energy
(and thus lost as heat, which is a form of energy) through any known
means -- and most certainly not through collisions. This is really easy
to verify experimentally.
As far as I can tell your proposed inertial drive amounts to injecting
mass in the direction of thrust at one end of a closed container and
removing mass from the container in a direction not parallel to the
direction of travel, then cycling that mass to the end of the container
to be re-injected. If such a setup produced net thrust, the ISS would
fall out of orbit or fly off into space: that's essentially the
arrangement of its cooling and ventilation system.
-o
Sam Wormley
> On 28 Nov, 15:36, Sam Wormley <sworml...@mchsi.com> wrote:
>> JT wrote:
>>> Inelastic collisions do not preserve momentum end of story, it is
>>> transformed to radiation heat.
>> What makes you that "radiation heat" doesn't have momentum?
>> What makes you think momentum is not conserved?
>>
>> Conservation of momentum applies.
>>
>> ∆p1(vessel) = -∆p2(radiation)
>>
>> ∆p1(internal to the vessel) = -∆p2(stuff escaping the vessel
>
> You are tiresome Sam p2 do now have oposite vector/direction to p1,
> you know very well that heat oscillation will try to level out in a
> homogenious media.
You certainly like to ignore conservation of momentum.
If -∆p2(radiation) is net zero, as you are implying, then
∆p1(vessel) = -∆p2(radiation) = 0
F = dp/dt(vessel) = 0, without some external force.
JT
And it told you at least 10 times that oscillation is momentum of a
system, do you deny that?
And i told you that oscillation will deplete due to loss of energy
(radiation), do you deny that?
> Collisions can convert kinetic energy into heat, but only until the
> entire system is moving at the same velocity, at which point no more
> collisions occur and the system keeps any remaining kinetic energy
> until acted on by an outside force. Momentum, on the other hand, is not
> equivalent to any form of energy and cannot be converted into energy
> (and thus lost as heat, which is a form of energy) through any known
> means -- and most certainly not through collisions. This is really easy
> to verify experimentally.
Well you talk bullshit a linear travelling object have ***relative***
momentum that is not oscillating if no external forces acting on it.
All rotating systems have momentum but also oscillation without any
external force and will eventually stop, that is fact.
A vessel can have internal action and reaction forces when moving in a
vector these action forces includes accelerating a bulk of molecules
inside the object the acceleration will if correct designed result in
giving the molecules relative bulk momentum to the vessel an action
force within the vessel.
However the action force do not need any oposite reaction force within
vessel a bulk of accelerated molecules can be forced to oscillate by a
pressure change. Thus the action force is depleted within the vessel
into radiation, thus the reaction force have no inverse force.
End of story.
> As far as I can tell your proposed inertial drive amounts to injecting
> mass in the direction of thrust at one end of a closed container and
> removing mass from the container in a direction not parallel to the
> direction of travel,
There is no mass loss in radiation m is not transformed into exotic
matter, the v in mv though is depleted you lose momentum the force F
acting on oposite hull is gone.
>then cycling that mass to the end of the container
> to be re-injected. If such a setup produced net thrust, the ISS would
> fall out of orbit or fly off into space: that's essentially the
> arrangement of its cooling and ventilation system.
Certainly ISS would have trouble in only let to radiate high energy
from a single point in one vector, now the whole system is radiating
so no problem.
I am quite sure ISS radiate spherical around the radiator, and there
is no accelerated thrust inside the ISS just thermal heat from
electronics and crew.
JT
>
> - Visa citerad text -- Dölj citerad text -
>
> - Visa citerad text -
No F is the relative force of the object with momentum.
JT
JT
> JT wrote:
> > On 28 Nov, 15:36, Sam Wormley <sworml...@mchsi.com> wrote:
> >> JT wrote:
> >>> Inelastic collisions do not preserve momentum end of story, it is
> >>> transformed to radiation heat.
> >> What makes you that "radiation heat" doesn't have momentum?
> >> What makes you think momentum is not conserved?
>
> >> Conservation of momentum applies.
>
> >> ∆p1(vessel) = -∆p2(radiation)
>
> >> ∆p1(internal to the vessel) = -∆p2(stuff escaping the vessel
>
> > You are tiresome Sam p2 do now have oposite vector/direction to p1,
> > you know very well that heat oscillation will try to level out in a
> > homogenious media.
>
> You certainly like to ignore conservation of momentum.
>
> If -∆p2(radiation) is net zero, as you are implying, then
>
> ∆p1(vessel) = -∆p2(radiation) = 0
No there is no perfect elastic collisions and no perfect inelastic
collision as i told you at least 10 times however F reaction < F
action if the action is radiating energy.
> F = dp/dt(vessel) = 0, without some external force.
>
> p2 will escape the media where you let is transfer
>
True, that is my main point although i have no idea what p represent
but the v in mv representing the relative force will deplete in a
radiating system of bulk collision.
Each collision will result in a diminishing v for the bulk objects.
the momentum is shrinking with as the radiation goes on. The radiation
is not part of the bulk momentum in a system. Radiating system do not
lose any mass only bulk velocity and momentum when oscillating.
Now there is probably some morons who say that the bulk oscillation do
not carry relative momentum to the vessel system.
But oscillation is not necessarily radiation as you can see within my
system with a rubberball, an oscillation system is any system with a
period like a pendulum or a bouncing rubberball between two points. It
is not possible to make the rubberball
bounce infinitly between two points within a vessel even in space.
Since there is no perfect elastic collisions without loss of momentum
v in mv will diminish with each bounce.
Each bump against vessel will result in an inelastic collision where
the total ***oscillating system*** lose momentum due to kinetics i
guess that always where the kinetics result in heat or radiation.
JT
>
> > radiation that is up to the structural design of vessel where you let
> > it radiate or is you chose to using it as a giant laser or store it,
> > some material can store thermal energy quite good.
>
> > JT- Dölj citerad text -
>
> - Visa citerad text -- Dölj citerad text -
>
> - Visa citerad text -
No recoil(p2)=action(p1) where p2 is momentum/force acted on vessel
and p1 is force/momenutm depleted as radiation leaving the vessel
hull.
The hull experience only(p2) and of course minor(p1), in ideal world
p1 is either depleted inside or let out radiating around the rim. You
have no oposite bulk momentum to p2, there is no force p1 equal to p2
acting on oposite of hull.
JT
Sam Wormley
> On 28 Nov, 20:26, Sam Wormley <sworml...@mchsi.com> wrote:
>> JT wrote:
>>> On 28 Nov, 15:36, Sam Wormley <sworml...@mchsi.com> wrote:
>>>> JT wrote:
>>>>> Inelastic collisions do not preserve momentum end of story, it is
>>>>> transformed to radiation heat.
>>>> What makes you that "radiation heat" doesn't have momentum?
>>>> What makes you think momentum is not conserved?
>>>> Conservation of momentum applies.
>>>> ∆p1(vessel) = -∆p2(radiation)
>>>> ∆p1(internal to the vessel) = -∆p2(stuff escaping the vessel
>>> You are tiresome Sam p2 do now have oposite vector/direction to p1,
>>> you know very well that heat oscillation will try to level out in a
>>> homogenious media.
>> You certainly like to ignore conservation of momentum.
>>
>> If -∆p2(radiation) is net zero, as you are implying, then
>>
>> ∆p1(vessel) = -∆p2(radiation) = 0
>
> No there is no perfect elastic collisions and no perfect inelastic
> collision as i told you at least 10 times however F reaction < F
> action if the action is radiating energy.
Did not PD point out to you that you cannot equate force and energy?
>
>> F = dp/dt(vessel) = 0, without some external force.
>>
>> p2 will escape the media where you let is transfer
>>
> True, that is my main point although i have no idea what p represent
> but the v in mv representing the relative force will deplete in a
> radiating system of bulk collision.
p represent momentum, as in p = mv
>
> Each collision will result in a diminishing v for the bulk objects.
> the momentum is shrinking with as the radiation goes on. The radiation
> is not part of the bulk momentum in a system. Radiating system do not
> lose any mass only bulk velocity and momentum when oscillating.
Momentum must be conserved. You it is not going to change unless
there is an external force, F = dp/dt ≠ 0
>
> Now there is probably some morons who say that the bulk oscillation do
> not carry relative momentum to the vessel system.
Momentum is defined as the product of velocity (for an observer) and
mass. Momentum is a conserved quantity.
> But oscillation is not necessarily radiation as you can see within my
> system with a rubberball, an oscillation system is any system with a
> period like a pendulum or a bouncing rubberball between two points. It
> is not possible to make the rubberball
> bounce infinitly between two points within a vessel even in space.
> Since there is no perfect elastic collisions without loss of momentum
> v in mv will diminish with each bounce.
You are neglecting the momentum of radiation p = h/λ
>
> Each bump against vessel will result in an inelastic collision where
> the total ***oscillating system*** lose momentum due to kinetics i
> guess that always where the kinetics result in heat or radiation.
The bottom line, JT, is the fact that your vessel is going nowhere
unless you expel some mass or energy
∆p1(vessel) = -∆p2(mass and/or radiation)
where p1 and p2 are momenta of the vessel and ejecta respectively.
JT
Your confused radiation have no mass unless it is X-rays that
something radiating do not mean it is losing mass. It means that there
is radiant energy created in collision of molecules.
But even if you except this fact you are wrong there is no need to
expel any energy from ship, thermal heat can nowadays be stored
without expel any radiation.
It will be a hot cake though.
JT
> where p1 and p2 are momenta of the vessel and ejecta respectively.- Dölj citerad text -
JT
>
> - Visa citerad text -- Dölj citerad text -
>
> - Visa citerad text -
Well there is no perfect radiation free thermal energy source.
But i've seen it mentioned molten salt is good to store thermal
energy.
From Wikipedia:
+*****************************************************************************************
Molten salt can be employed as a heat store to retain heat collected
by a solar tower or solar trough so that it can be used to generate
electricity in bad weather or at night. It was demonstrated in the
Solar Two project from 1995-1999. The system is predicted to have an
annual efficiency of 99%, although it is not clear if this is the
second law efficiency.[3][4] The molten salt is a mixture of 60
percent sodium nitrate and 40 percent potassium nitrate, commonly
called saltpetre. It is non-flammable and nontoxic, and has already
been used in the chemical and metals industries as a heat-transport
fluid, so experience with such systems exists in non-solar
applications.
The salt melts at 221 °C (430 °F). It is kept liquid at 288 °C (550
°F) in an insulated "cold" storage tank. The liquid salt is pumped
through panels in a solar collector where the focused sun heats it to
566 °C (1,051 °F). It is then sent to a hot storage tank. This is so
well insulated that the thermal energy can be usefully stored for up
to a week.
+*********************************************************************************************
One week it will be a very hot cake but there probably even more
exotic materials that store thermal heat even better.
JT
Sam Wormley
All electromagnetic photons (including x-rays) have momentum
p = hv/c = h/λ
and energy
E = hv
When I wrote
∆p1(vessel) = -∆p2(mass and/or radiation)
The "mass" refers to atoms or other particles with rest mass
greater than zero and radiation, refers to photons.
Both have momentum. In the case of mass, it's
p = mv
In the case pg photons, it's
p = hv/c = h/λ
I'll repeat an indisputable fact, without an external force, your
vessel is going nowhere unless you expel mass and or radiation, such
that:
∆p1(vessel) = -∆p2(mass and/or radiation)
And that is the end of the story.
JT
Well energy can be stored momentum conserved.
End of story
JT
JT
Oh and i forgot the photon that leaves the vessel is not part of the
closed system, the mass remains intact.
You should really have a look at what a closed system and inertial
propulsion is about. It is about a vessel gain momentum without
expelsion of mass.
Nowhere in litterature there is any mentioning about energy can not
leave the vehicle as photon energy. The mass is not changing.
And as i said you are simply wrong about photons have to leave the
vehicle thermal energy is easier to store in the void of vaccua. My
bet is that liquidised salt crystals store energy better in the void
of space then in a molecular environment.
PD
There is no such thing as "relative force", JT.
I'm sure it's crossed your mind now that you're just babbling.
PD
Oh dear. Wherever did you get these foolish notions?
JT
You are the confused one, a linear moving object A always have
momentum P=mv relative a restframe.
If v=0 for the object then P=0 . So in objects A restframe P always 0
momentum is zero. There is a frame where a linear moving object do not
carry momentum it is always relative another object or rest system.
This is also true for forces, if P=0 between two objects there can
be no collision no force exchanged so F=0 or the relative force
potential is zero.
A linear moving objects energy potential is always relative a
restsystem or object thus the force it can exchange with a system is
always relative a specific rest frame.
P and ****F**** for object A is variant between restsystem B and C.
JT
JT
The force it can transfer from A to B, is dependent of B's relative
restframe.
We can conclude that the vessel B have the restframe and that A carry
the momentum since molecule A was accelerated within vessel B.
JT
From wikipedia:
****************************************************************************************
A reactionless drive or inertial propulsion engine (also reactionless
thruster, reactionless engine, bootstrap drive, and inertia drive) is
any form of propulsion not based around expulsion of fuel or reaction
mass.
*****************************************************************************************
If the mass stay intact but the object gain momentum you have a
reaction less drive, inertial propulsion. But i would rather call it
*unbalanced action/reaction propulsion* since it is based on the
reaction force being greater against the closed system hull then the
oposite action force on the hull.
JT
Sam Wormley
>
> From wikipedia:
> ****************************************************************************************
> A reactionless drive or inertial propulsion engine (also reactionless
> thruster, reactionless engine, bootstrap drive, and inertia drive) is
> any form of propulsion not based around expulsion of fuel or reaction
> mass.
http://en.wikipedia.org/wiki/Reactionless_drive
"A reactionless drive or inertial propulsion engine (also reactionless thruster,
reactionless engine, bootstrap drive, and inertia drive) is any form of propulsion not
based around expulsion of fuel or reaction mass.
"The name comes from Newton's Third Law of Motion, usually expressed as: "For every
action, there is an equal and opposite reaction." Such a drive would use a hypothetical
form of thrust that does not require any outside force or net momentum exchange to produce
linear motion, and therefore necessarily violates the conservation of momentum, a
fundamental principle of all current understandings of physics.
"In spite of their physical impossibility, such devices have been often proposed in recent
history.
"The fundamental scientific problem is one of momentum transfer. If there is no momentum
transfer, the postulated device is classified as a "reactionless" drive and labelled a
fraud. If there is a mechanism for momentum transfer, then the device is classed as a
reaction drive and is therefore by definition not a "reactionless" drive".
Sam Wormley
Energy and momentum have different units, JT.
One requires a force to change momentum.
F = dp/dt
Owen Jacobson
> On 29 Nov, 02:07, PD <thedraperfam...@gmail.com> wrote:
>> On Nov 28, 2:50�pm, JT <jonas.thornv...@hotmail.com> wrote:
>>
>>> No F is the relative force of the object with momentum.
>>
>> There is no such thing as "relative force", JT.
>> I'm sure it's crossed your mind now that you're just babbling.
>
> [A] linear moving object A always have momentum P=mv ...
This part is okay.
> ... relative a restframe.
This part is poorly expressed.
> If v=0 for the object then P=0 .
True.
> So in objects A restframe P always 0 momentum is zero.
Your writing is atrocious, but this is also true.
> There is a frame where a linear moving object do not carry momentum it
> is always relative another object or rest system.
This sentence is meaningless. The closest thing I can come up with that
has meaning is that for every object that is not accelerating, there is
a frame where it is at rest. Which is true, but you already said that.
> This is also true for forces, if P=0 between two objects ...
There is no momentum "between two objects". However:
> there can be no collision
As best as I can tell, this is false, and I'll show you why in a couple
of posts. But first, I have some questions for you. I am very sincerely
interested in your answers, so please take some time to work them out
and show your math. It will help me understand your reactionless
propulsion system.
Feel free to add any additional assumptions you feel are necessary, but
please include them in your answers.
1.
a) You have a 1 g mass at rest, and a 1 g mass approaching it from the
left at 2 m/s. What is the initial momentum of this system?
b) These two masses collide inelastically and become one 2-gram mass.
What is the velocity of this mass after the collision?
2.
a) You have a 1 g mass at rest, and a 1 g mass approaching it from the
right at 2 m/s. What is the initial momentum of this system?
b) These two masses collide inelastically and become one 2-gram mass.
What is the velocity of this 2 g mass after the collision?
3.
a) You have a 1 g mass approaching from the left at 1 m/s, and a 1 g
mass approaching from the right at 1 m/s. What is the initial momentum
of this system?
b) These two masses collide inelastically and become one 2-gram mass.
What is the velocity of this 2 g mass after the collision?
I know these are trivial questions, but I'm not trying to waste your
time. Your answers and your explanations for your answers will likely
make it much easier to understand your proposed propulsion mechanism.
-o
Cwatters
> > On 27 Nov, 21:30, "Cwatters"
> > <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> If the ball doesn't leave the tube, some time later the ball must end up
> travelling at the same speed as the tube.
>
> Where does the force come from to accelerate the ball so that it's
> travelling along at the same speed as the tube ?- D�lj citerad text -
>
> - Visa citerad text -
>When the zigzag action force is depleted it will stop be at rest
>within tube.
That's not what I asked.
JT
> JTwrote:
>
> > From wikipedia:
> > ****************************************************************************************
I am not sure the last part of post is correct i think that a gyro
that would exhibit a linear net momentum would be classified as a
reactionless drive would it not although it has angular momentum?
JT
JT
> >>JTwrote:
> >>> On 28 Nov, 22:54, Sam Wormley <sworml...@mchsi.com> wrote:
>
> - Visa citerad text -
Yes i was wrong a force is defined in this case as the push that act
on the object and make the object accelerate F=ma.
So there is no such thing defined as potential force although it
certainly can be calculated if you know the objects relative speed
mass and elasticity you could calculate F how much the object will
accelerate.
I guess that *a* is the acceleration relative the other object, so
both objects in a collision experience the same relative force
relative eachothers rest system.
And your right P=mv is the momenum gained after acceleration the
object with mass m reach velocity v and carry that momentum.
You said
*********************************************************
I'll repeat an indisputable fact, without an external force, your
vessel is going nowhere unless you expel mass and or radiation,
such
that:
∆p1(vessel) = -∆p2(mass and/or radiation)
**********************************************************
Which i also beleive to be correct, but i would think that radiation
in this case is basicly molecular oscillation and heat can be stored.
Do not need to be expelled directly although at some point will have
stored up so much energy that it will radiate regardless storing
medium.
If the vessel store or radiate photon i would say it is still inertial
propulsion there is no mass loss or exelsion of propulsion outside
system only radiant energy transfer. I can not see how it would be
possible to build any mechanic moving device without kinetic exchange.
JT
JT
JT
F=ma=0.002
> b) These two masses collide inelastically and become one 2-gram mass.
> What is the velocity of this mass after the collision?
I guess you mean velocity relative stationary system A, i would say
that do depend on how much kinetic energy is developed at the
collision. In a perfect inelastic collision all momentum is transfered
to chaotic oscillation, and oscillation of molecules lead to heat
being radiated. A perfect inelastic collision do not exist, and not
the perfect macroscopic elastic either.
The velocity is undeterminable wihtout knowing how much energy leaving
the system as radiation if you gave the (viscosity and the
conductivity?) of the two bodies someone probably could calculate how
much energy leaving system thru radiation and calculate how much
kinetic energy that is transfered into momentum, i also think that
there must be a unit for thermal sensitivity of a material, how easy
it is to get it oscillating and radiating.
According to Sam the momentum also includes the radiation from your
collision between two objects which i find intriguing the momentum of
the two objects is undeterminable without knowing the bulk energy
transmitted of the leaving photons. So your velocity calculation can
not be performed without know how much energy leaving the system.
How anyone can call this conservation of momentum is beyond my
understanding of the english language. The velocity of the objects is
deaccelerated and leaving the system as radiation. Actually what is
conserved here i can see energy, but your P=mv and v deaccelerate
something is clearly wrong with the definition or my language
interpretation.
How much energy leaving the system probably have todo something with
the molecular bindings in material i do not know the unit though maybe
you could tell me maybe should it be viscosity or conductivity?
Is viscosity and conductivity dependent on eachother.
I answer your other question in another post, i am interested in your
answers
to this.
JT
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> news:db7e4c74-bbc6-45a7...@s31g2000yqs.googlegroups.com...
>
> > > On 27 Nov, 21:30, "Cwatters"
> > > <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> > If the ball doesn't leave the tube, some time later the ball must end up
> > travelling at the same speed as the tube.
>
> > Where does the force come from to accelerate the ball so that it's
>
> > - Visa citerad text -
> >When the zigzag action force is depleted it will stop be at rest
> >within tube.
>
> That's not what I asked.
If we assume that the ball come to stop before it reach end of tube,
after acceleration relative tube you will have an obstacle changing
direction of the ball acting as a force upon the ball where kinetic
energy leaves system through molecular oscillation.
And each bump into the wall will act as a force upon ball where
momentum is transfered into kinetic energy, oscillation of molecules.
The ball lose momentum *AND VESSEL GET MOMENTUM FROM SIDE TO SIDE* i
do not think it would be wrong to say that the system have oscillation
sideway.
JT
JT
I guess that *a* is the acceleration relative the other object, so
both objects in a collision experience the same relative force
relative eachothers rest system.
*************************************
If they have same mass
Owen Jacobson
> On 29 Nov, 08:19, Owen Jacobson <angrybald...@gmail.com> wrote:
>> As best as I can tell, this is false, and I'll show you why in a couple
>> of posts. But first, I have some questions for you. I am very sincerely
>> interested in your answers, so please take some time to work them out
>> and show your math. It will help me understand your reactionless
>> propulsion system.
>>
>> Feel free to add any additional assumptions you feel are necessary, but
>> please include them in your answers.
>>
>> 1.
>> a) You have a 1 g mass at rest, and a 1 g mass approaching it from the
>> left at 2 m/s. What is the initial momentum of this system?
>
> F=ma=0.002
Interesting for two reasons. First, since neither object is
accelerating, a = 0 and F = ma = 0. However, the question was not "how
much force is acting on the system" but "what is the system's
momentum", described by p = mv. We can calculate the system's momentum
by adding up the momenta of the parts of the system:
The momentum of the mass approaching from the left is...
p_left = m_left * v_left
= 1 g * 2 m/s to the right
= 2 g * m/s to the right
(Yes, g * m / s is the right units.)
And the stationary mass is...
p_right = m_right * v_right
= 1 g * 0 m/s
= 0 g * m/s
So the system's total momentum is...
p = p_left + p_right
= 2 g * m/s to the right + 0 g * m/s
= 2 g * m/s to the right
>> b) These two masses collide inelastically and become one 2-gram mass.
>> What is the velocity of this mass after the collision?
>
> I guess you mean velocity relative stationary system A
Relative to the system in which the objects had velocities of 2 m/s and
0 m/s respectively, yes.
> i would say that do depend on how much kinetic energy is developed at the
> collision. In a perfect inelastic collision all momentum is transfered
> to chaotic oscillation, and oscillation of molecules lead to heat
> being radiated.
Given this, I'd be interested in your answer to question 3. I wrote:
>> 3.
>> a) You have a 1 g mass approaching from the left at 1 m/s, and a 1 g
>> mass approaching from the right at 1 m/s. What is the initial momentum
>> of this system?
>> b) These two masses collide inelastically and become one 2-gram mass.
>> What is the velocity of this 2 g mass after the collision?
If it helps, assume both objects are identical in every measurable way
except velocity.
> A perfect inelastic collision do not exist, and not the perfect
> macroscopic elastic either.
Consider the collision between your foot and a tile floor while you
walk: your foot does not rebound off the floor, and the floor does not
rebound off your foot, so this collision is close enough to perfectly
inelastic that you can treat it as perfect. Similarly the collision
between an arrow and an archery target, and a number of other examples.
> According to Sam the momentum also includes the radiation from your
> collision between two objects which i find intriguing the momentum of
> the two objects is undeterminable without knowing the bulk energy
> transmitted of the leaving photons. So your velocity calculation can
> not be performed without know how much energy leaving the system.
Feel free to assume that both objects are at a uniform temperature (and
thus radiating uniformly in all directions, for a net radiation
pressure of zero) and that they are small enough and thermally
conductive enough that any heat generated in the collision heats both
objects uniformly and effectively instantly. 1 gram balls of hot
copper, for example.
You don't actually need this assumption to answer the questions, though.
-o
PD
But the forces are equal. It's just that you're only counting the
contribution from expelled mass to that force.
JT
> > A reactionless drive or inertial propulsion engine (also reactionless
> > thruster, reactionless engine, bootstrap drive, and inertia drive) is
> > any form of propulsion not based around expulsion of fuel or reaction
> > mass.
>
> > If the mass stay intact but the object gain momentum you have a
> > reaction less drive, inertial propulsion. But i would rather call it
> > *unbalanced action/reaction propulsion* since it is based on the
> > reaction force being greater against the closed system hull then the
> > oposite action force on the hull.
>
> > JT
>
> But the forces are equal. It's just that you're only counting the
>
> - Visa citerad text -
No they are not equal in oposite direction.
JT
Cwatters
On 29 Nov, 11:59, "Cwatters"
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
>> "JT" <jonas.thornv...@hotmail.com> wrote in message
>>
>> news:db7e4c74-bbc6-45a7...@s31g2000yqs.googlegroups.com...
>>
>> > > On 27 Nov, 21:30, "Cwatters"
>> > > <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
>> > If the ball doesn't leave the tube, some time later the ball must end
>> > up
>> > travelling at the same speed as the tube.
>>
>> > Where does the force come from to accelerate the ball so that it's
>>
>> > - Visa citerad text -
>> >When the zigzag action force is depleted it will stop be at rest
>> >within tube.
>>
>> That's not what I asked.
>
>If we assume that the ball come to stop before it reach end of tube,
That implies friction between the ball and the walls of the tube.
JT
> > inelastic collisions but you fucking do not listen. That is why the
> > reaction "*elastic collision*"(there is no macroscopic perfect
> > collisions though) transfer more momentum to the vessel then the
> > inelastic oscillating bulk of collisions transfer momentum to the
> > vessel hull in opositie direction.
>
> I heard you just fine. You believe that there is a mechanism through
> which momentum (a vector quantity described by p = mv) can be converted
> into energy (any of a number of scalar quantities). The problem is that
> this violates conservation of energy -- any change in momentum without
> a corresponding change in the momentum of some other part of the system
> would change the total energy of the system, and there always exists a
> frame of reference where that change in total energy takes the form of
> an increase in total energy.
I never doubted energy conservation in my attempt to make inertial
propulsion on the contrary, it will be a very hot cake if you do not
let the thing radiate.
I doubted conservation of ********linear momentum******** within a
chaotic gasmedia or vaporised water. We talk about an accelerated
compressed thrust of water here that is vaporised in a gasmedia.
Within a strong electromagnetic field you can even store plasma see
our sun.
The inertial propulsion idea is basicly a recycling steam engine or
turbine with a conected pipecooler, pump and plate vaporiser. The heat
from pipecooler is redirected into a highly conductive media that can
store the thermal heat.
So there is not even need for any energy to leave the system,
constantly.
But any material probably have a limit for how much heat it can store.
The steam engine is used to pump water/thrust water into an
oscillating hot gasmedia above a vaporizer (if such needed). The
pressure within system will probaby be very high. But the pressure
within the closed system, is lowest around the cooler and highest
around vaporiser. So it recycles the vaporised water into a tank
around the cooler, it is a natural cycle between cooler and vaporiser,
the molecules try to level out heat and pressure within system.
There is no violation of energy conservation, you can store thermal
heat between layer of nonconductive materials and conductive materials
and the system will gain momentum even without radiating heat.
>
> Collisions can convert kinetic energy into heat, but only until the
> entire system is moving at the same velocity, at which point no more
> collisions occur and the system keeps any remaining kinetic energy
> until acted on by an outside force.
You are simply wrong the there is an unbalance in pressure created
within system you have a cycle between high and low pressure between
the cooler and vaporiser.
The whole system will never move at same velocity the flow of gases
even help the system gain a net lift and momentum.
You do not seem to realise that there is an steam engine that thrust
against a vaporiser. How can you think the system is moving at the
same velocity?
> Momentum, on the other hand, is not
> equivalent to any form of energy and cannot be converted into energy
Bullshit linear momentum can be transformed into oscillation in a
media, a stone droped in a waterpound prove you wrong it lose momentum
into heat.
> (and thus lost as heat, which is a form of energy) through any known
> means -- and most certainly not through collisions. This is really easy
> to verify experimentally.
Totally more bullshit as anyone with half a brain can verify, if you
blast the water surface with any material that hold same temperature.
The system will temperature will raise water get hot.
Yes it is easy to verify experimentally the collisions transfer
momentum to heat.
I start to think you lie intentionally, this so far out of any logic
to beleive that linear momentum is conserved if you have a directed
thrust into a pound you create forces sideways. Where do you beleive
the sideway momentum come from?
Well i can tell you it comes from the collision where your linear
momentum of molecules interact with other watermolecules transfer your
linear momentum of watermolecules into radially expanding momentum.
Your watermolecules do not transfer all momentum linear they press
aside other watermolecules and interact chaotic but the bulk pressure
will create a radially expanding force in media your watermolecules
get momentum sideways from impact point the most momentum the surface
water will gain because there is the lowest pressure in media. Very
little momentum is carried downward because of the higher pressure.
I think even you noticed the pressure change at two meter deep water.
I simply think you lye intentionally now, prove me wrong.
>
> As far as I can tell your proposed inertial drive amounts to injecting
> mass in the direction of thrust at one end of a closed container and
> removing mass from the container in a direction not parallel to the
> direction of travel
The simplest form of craft i can think of i describe here.
There is no inject and remove of mass only thermal energy into thermal
mass inside a noneconductive ring tube you ever seen "stargate" ;)...
So the two plates/membrans is welded together preferably discshaped.
Between plates is a round tube of noneconductive material filled with
a material that can store thermal heat a thermal mass a ring
refrigerator.
That thermal material within tube is the battery of the recycle
process heating water for the steamengine. So the material in ring
must be heated before the steamengine and recycle process can work.
The noneconductive tubematerial prevent heat transfer between plates.
they do not easily level if in vaccua you could heat one without the
other getting heated.
There is water in a topmounted tank with a tube cooler inside
thattransport the heat to the thermal mass refrigerator within the
noneconductive tube
The heat led to and stored within noneconductive tube is then used to
drive a steamengine that is in center of the lower copper/aluminia
membran and part of lower membran itself is the steamdrum acting as an
vaporiser.
So the steam led into the steamdrum drive a waterpump/thruster mounted
at the watertank directed downwards the vaporiser.
That is the whole recycle process. Once the thermal mass storage is
loaded the steamengine start working and drive vaporizer, cooler and
pump.
>, then cycling that mass to the end of the container
> to be re-injected. If such a setup produced net thrust, the ISS would
> fall out of orbit or fly off into space: that's essentially the
> arrangement of its cooling and ventilation system.
No the heat is radially spherical dissapated around the iss try a ISS
with a hot and cold side and you would be in trouble, and there is no
strong thrust inside directected one way only ISS have a small action/
reaction momentum inside..
JT
> -o
JT
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> news:089039e3-4d22-4886...@g27g2000yqn.googlegroups.com...
> On 29 Nov, 11:59, "Cwatters"
>
>
>
>
>
> <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> >> "JT" <jonas.thornv...@hotmail.com> wrote in message
>
> >>news:db7e4c74-bbc6-45a7...@s31g2000yqs.googlegroups.com...
>
> >> > > On 27 Nov, 21:30, "Cwatters"
> >> > > <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
> >> > If the ball doesn't leave the tube, some time later the ball must end
> >> > up
> >> > travelling at the same speed as the tube.
>
> >> > Where does the force come from to accelerate the ball so that it's
>
> >> > - Visa citerad text -
> >> >When the zigzag action force is depleted it will stop be at rest
> >> >within tube.
>
> >> That's not what I asked.
>
> >If we assume that the ball come to stop before it reach end of tube,
>
> That implies friction between the ball and the walls of the tube.
Yes that would be the *normal* case wouldn't it? Did you have another
idea that conserve the balls linerar momentum.....
JT
Sam Wormley
>
> Bullshit linear momentum can be transformed into oscillation in a
> media, a stone droped in a waterpound prove you wrong it lose momentum
> into heat.
>
This statement indicate you do not understand fundamental physics.
Linear Momentum is conserved for closed system. Conserved, not
transformed into anything else. Conserved.
JT
You are so tiresome,trying to falsify physic.
The ***linar momentum*** P=mv for each watermolecule is not conserved
within a chaotic environment. If you fill your mouth with water and
spit it out "the water molecules gain momentum" it just happens to be
into boiling kittel. The kittel contain chaotic oscillating
molecules. You see water and gases under pressure do not have linear
momentum as soon your sprout of water hit the surface it will lose all
linear momentum and become part of the chaotic system.
There is no sprout of water going dow to the bottom of the boiling
pot, end of story end of conservation of momentum.
JT
Sam Wormley
> On 30 Nov, 15:31, Sam Wormley <sworml...@mchsi.com> wrote:
>> JT wrote:
>>
>>> Bullshit linear momentum can be transformed into oscillation in a
>>> media, a stone droped in a waterpound prove you wrong it lose momentum
>>> into heat.
>> This statement indicate you do not understand fundamental physics.
>> Linear Momentum is conserved for closed system. Conserved, not
>> transformed into anything else. Conserved.
>
>
> The ***linar momentum*** P=mv for each watermolecule is not conserved
> within a chaotic environment. If you fill your mouth with water and
> spit it out "the water molecules gain momentum" it just happens to be
> into boiling kittel. The kittel contain chaotic oscillating
> molecules. You see water and gases under pressure do not have linear
> momentum as soon your sprout of water hit the surface it will lose all
> linear momentum and become part of the chaotic system.
Any one water molecule is not in a CLOSED SYSTEM, in your scenario.
However the momentum of a CLOSED SYSTEM of water molecules IS conserved.
Cwatters
On 30 Nov, 10:21, "Cwatters"
<colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote:
>> "JT" <jonas.thornv...@hotmail.com> wrote in message
>>
>>
>> That implies friction between the ball and the walls of the tube.
>
>Yes that would be the *normal* case wouldn't it?
If it's frictionless the ball will eventually be left behind and hit the end
of the tube.
If it's not frictionless contact with the walls at an angle will impart spin
on the ball and the balls moment of inertia will slow the tube at each
bounce.
If you are saying the ball hits the wall at right angles then it must be
travelling at the same speed as the tube eg have bounced off the obstruction
in a way that slowed the tube.
If these balls are molecules of gas then there is still no way to slow them
down, turn them around and accelerate them up to tube speed without there
being consequences for the speed of the tube.
At the end of the day it's a closed system regardless of what complicated
collisions you arrange inside the tube.
PD
??