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