I'm Looking For Special Research Group

Discussion in 'Pseudoscience Archive' started by ABV, Sep 4, 2010.

  1. D H Some other guy Valued Senior Member

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    2,257
    I gave you three answers.
    1. Stop tilting at windmills.
    2. Stop with the stupid Rube Goldberg contraptions, and
    3. Go learn some physics.
     
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  3. ABV Registered Senior Member

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    Thanks. Very useful

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    Back again to my question. To explain force difference, a few people from other forum think the frame of reference of spring is not the same as from of reference of initial point. Therefore, the forces of both sides of spring have a different values.
    It's not quite rights. The forces difference is proportional to mass of spring and spring acceleration. However, the weightless spring has same forces on both sides.
    Actually, the effect is very small. I'm not surprise if none din't find practical difference on experiment before. It's easily discard to standard deviation on experiment.
     
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  5. D H Some other guy Valued Senior Member

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    Alex, in this new silly example, your text, your diagrams, and your equations are all in conflict with one another.

    BTW, what happened to all of the other silly examples that I and others have shot down? Why don't you try to fix those instead of creating yet another nonsense device?
     
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  7. ABV Registered Senior Member

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    The natural experiment first

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  8. D H Some other guy Valued Senior Member

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    Alex,

    You have made multiple novice mistakes in this problem. Had you not made these novice mistakes you would have realized that you cannot make the platforms massless. What you can do is account for this platform mass and then take the limit as this platform mass goes to zero.

    Set the problem up correctly, and do not double-book things as is your wont.


    Ben: Alex has a penchant for creating silly, overly complex problems. As soon as a hole is poked in one of them, he abandons that one only to create yet another silly, overly complex problem. This is the hallmark of a crackpot. I suggest this thread by sent to pseudoscience, or even better, the cesspool.
     
  9. ABV Registered Senior Member

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    79
    I removed masses from platforms and spring especially, because I showing asymmetric force which came from modern classical mechanics calculation.
    If add masses to these objects then effect will be smaller. However it will present anyway. The lab equipment is sensitive enough to catch up this effect. Need experiment.



    The problem cannot be set up incorrectly. Base on problem a physics misunderstanding may happen, because substitution primary laws with their consequences gives an adequate meaning. In this particular instance, third newtons law and consequence law of momentum conservation. The key is the consequence law of momentum conservation in simple form covers just simple objects interaction. For complex objects interaction needs a law of momentum conservation in complex form.


    So, you don't have any arguments to explain an asymmetrical force for this problem.
    http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18#
    Sorry

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  10. ABV Registered Senior Member

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    One more time for everybody why I excluded mass of platforms and spring.
    This is a normal practice for physics ideal model.

    For instance, if repulse two objects using weightless spring in simple translational motion case on a line theirs center of masses then these objects are taking same momentums by value and opposite by direction where
    \(\vec P = - \vec P \) Correct?
    No matter if they have different masses the law of momentum conservation must work in simple form here. The spring is weightless and we don't care what acceleration is spring getting and where it goes to, we don't count it in ideal model and mention the total momentum of isolated system is constant and equal zero.
    \(\vec P +( - \vec P)=0 \)
    Here is we don't include momentum of spring because mass of it zero
    \(\vec P_(_s_p_r_i_n_g_) = m \vec v_(_s_p_r_i_n_g_)\\ 0=0 \vec v_(_s_p_r_i_n_g_)\)

    Here's same thing. As it shown before, for ideal model I'm excluding mass of platforms and spring and compare momentums of rolling objects. In trivial case, objects on one platform are not rotating. On this trivial case easy to see one force which induce translational motion only for one side is equal to net force on the other platform where each component of this net force induce only own type of motion.
    Following by modern classical mechanics with using law of momentum conservation in simple form the component of net force for rotational motion must be excluded. However, this force exclusion gives a conflict with third newtons law of symmetrical interaction where \(\vec F = - \vec F\). Therefore, I think it's impossible use law of momentum conservation in simple form for complex interaction. The law of momentum conservation for complex interaction must be in complex form too.

    P.S.
    Probably I'll make another figure for trivial case soon. I would be easy to see difference between force for translational motion on one side and net force for translational and rotational motion on other side.
     
  11. D H Some other guy Valued Senior Member

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    2,257
    Alex, you always set things up incorrectly. Part of the setup is the equations of motion. You get those wrong. When you get the math wrong you will almost always get the wrong answer. Just because you get the wrong answer does not mean physics is wrong. It just means you are wrong.


    Sigh. Since you appear to need me to waste my time on this crap, ....


    Let's do this last Rube Goldberg device of yours correctly. I'll look at the subsystem on the right; what happens on the left is mirror image of what happens on the right. First, I'll define some terms.
    • Subscripts u, l, w, and p respectively designate the the upper wheel on the right, the lower wheel on the right, the wheels (where no distinction is needed), and the platform on the right.
    • m[sub]u[/sub], m[sub]l[/sub], and m[sub]p[/sub] designate the masses of the wheels and the platform.
    • F[sub]s[/sub] is the force exerted on the platform by the spring.
    • F[sub]u[/sub], F[sub]l[/sub], and F[sub]p[/sub] designate the forces on the wheels and and on the platform.
    • v[sub]u[/sub], v[sub]l[/sub], and v[sub]p[/sub] designate the magnitudes of the velocities of the centers of mass of the two wheels and of the velocity of the platform. These quantities are zero at time t=0
    • a[sub]u[/sub], a[sub]l[/sub], and a[sub]p[/sub] designate magnitudes of the accelerations of the centers of mass of the two wheels and of the acceleration of the platform.
    • I[sub]u[/sub] and I[sub]l[/sub] designate the moments of inertia of the upper and lower wheels.
    • r[sub]u[/sub] and r[sub]l[/sub] designate the radii of the upper and lower wheels.
    • γ[sub]u[/sub] and γ[sub]l[/sub] are unitless parameters that relate the mass, radii, and moments of inertia of the wheels: γ[sub]u[/sub]=m[sub]u[/sub]r[sub]u[/sub][sup]2[/sup]/I[sub]u[/sub], and similarly for the lower wheel.
    • ω[sub]u[/sub] and ω[sub]l[/sub] designate the magnitudes of the angular velocities of the upper and lower wheels.
    • α[sub]u[/sub] and α[sub]l[/sub] designate the magnitudes of the angular accelerations of the upper and lower wheels.
    • x is the position of the right edge of the spring (left edge of the right platform).
    • x[sub]i[/sub] is the initial value of x at time t=0, at which time the platform and the wheels are at rest.
    • x[sub]0[/sub] is the final uncompressed length of the right half of the spring.

    Per your setup, the upper and lower wheels are constrained to move with one another and both wheels are rolling without slipping. The first constraint means that

    \((1)\qquad v_u = v_l \equiv v_w\)
    \((2)\qquad a_u = a_l \equiv a_w\)

    where v[sub]w[/sub] and a[sub]w[/sub] designate the common wheel center of mass velocity and acceleration. The second constraint means that

    \((3)\qquad v_u+\omega_u r_u = v_p\)
    \((4)\qquad v_l+\omega_l r_l = v_p\)

    Differentiating yields

    \((5)\qquad a_u+\alpha_u r_u = a_p\)
    \((6)\qquad a_l+\alpha_l r_l = a_p\)

    The sole force acting on the upper (lower) wheel is F[sub]u[/sub] (F[sub]l[/sub]), leading to the equations of motion for the wheels,

    \((7)\qquad a_u = \frac{F_u}{m_u}\)
    \((8)\qquad \alpha u = \frac{F_u r_u}{I_u} = \frac{\gamma_u}{r_u} a_u\)
    \((9)\qquad a_l = \frac{F_l}{m_l}\)
    \((10)\qquad \alpha l = \frac{F_l r_l}{I_l} = \frac{\gamma_l}{r_l} a_l\)

    Apply the above to equations (5) and (6) yields

    \((11)\qquad (1+\gamma_u)a_u = a_p = (1+\gamma_l)a_l\)

    By equation (2), a[sub]u[/sub]=a[sub]l[/sub]=a[sub]w[/sub]. Equation (11) thus says that γ[sub]u[/sub] and γ[sub]l[/sub] must be equal to one another (and hence I'll just use γ from now on) to have the system that you described. The parameter γ describes the mass distribution of the wheel. For example, a value of 1 results when all the mass is concentrated at the rims while a value of 2 arises for a solid disk of uniform thickness and density. If the two wheels have different gamma values it is impossible for both wheels to be rolling without slipping.

    Thus

    \((12)\qquad a_w = a_u = a_l = \frac{a_p}{1+\gamma}\)


    The net force on the platform is given by

    \((13)\qquad F_p = F_s - (F_u + F_m)\)

    or

    \((14)\qquad F_u + F_m+ F_p = m_u a_u + m_l a_l + m_p a_p = F_s\)

    Applying equation (12) yields

    \((14)\qquad \left(m_p + \frac{m_u+m_l}{1+\gamma}\right) a_p = F_s\)

    The spring force is given by Hooke's law,

    \((15)\qquad F_s = k(x_0-x)\)

    Thus

    \((16)\qquad \frac{d^2 x}{dt^2} = \frac{(1+\gamma)k}{(1+\gamma)m_p + (m_u+m_l)} (x_0-x)\)

    or

    \((17)\frac{d^2 x}{dt^2} = \Omega^2 (x_0-x)\)

    where

    \((18)\qquad \Omega^2 \equiv \frac{(1+\gamma)k}{(1+\gamma)m_p + (m_u+m_l)}\)

    With the initial conditions x(0)=x_i and dx/dt|[sub]t=0[/sub] = 0, the solution of equation (18) is

    \((19)\qquad x(t) = x_0 - (x_0-x_i) \cos(\Omega t)\)

    At time Ωt[sub]rel[/sub]=π/2 the platform+wheels mechanism releases from the spring with the platform having a velocity

    \((20)\qquad v_p(t_{\mbox{rel}}) = (x_0-x_i)\Omega\)

    The translational and angular velocities of the wheels at this release time are given by

    \((21)\qquad v_w(t_{\mbox{rel}}) = \frac{v_p}{1+\gamma}\)
    \((22)\qquad \omega_u(t_{\mbox{rel}}) = \frac{\gamma}{r_u}v_w\)
    \((23)\qquad \omega_l(t_{\mbox{rel}}) = \frac{\gamma}{r_l}v_w\)

    The kinetic energy of the system at this time is

    \((24)\qquad T(t_{\mbox{rel}}) = \frac 1 2 \left(m_p v_p^2 + m_u v_u^2 + I_u \omega_u^2 + m_l v_l^2 + I_l \omega_l^2\right) = \frac 1 2 k (x_0-x_i)^2\)

    which, not surprisingly, is identically equal to the potential energy in the spring at time t=0. Add in the symmetric left hand side subsystem, and the total linear and total angular momentum are always identically equal to zero and the total energy is always equal to the potential energy initially stored in the spring.


    This is the last time I waste my time on this crap. I'll just ask the moderator to send your nonsense to its rightful home in the sciforums sewer.
     
  12. ABV Registered Senior Member

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    79
    DH thank you for solution. But equations 3,4,5,6.
    What did you mean?
    The forces are applying to platforms from spring. The wheels are rotating in inverse mode. If translational acceleration is trying move rolling object away from spring, however the angular acceleration is returning rolling objects to spring. The wheels are rolling on inversion mode back to spring.
    Did you included this wheels rotation inversion into your equations?
    I don't see it.
     
  13. D H Some other guy Valued Senior Member

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    2,257
    Let's look at the upper right wheel. Similar arguments will dictate the behaviors of the other three wheels. Designate the +x axis pointing to the right, the +y axis pointing to the top of the screen, and the +z axis as pointing out of the screen toward your eyes. The platform, accelerating to the right, exerts a rightward force on the wheel with the point of application of the force being the contact point between the platform and the wheel. This force makes the wheel accelerate to the right and exerts a torque on the wheel that makes the wheel rotate about the +z axis. The bottom of the upper right wheel (the contact point with the platform) has a greater velocity than the wheel's center of mass.

    That is exactly what equations 3 to 6 state, mathematically.
     
  14. ABV Registered Senior Member

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    79
    DH.

    The mathematically the angular acceleration of wheels depends on platform translational acceleration and wheel translational acceleration. The difference between these acceleration through radius of wheel gives an angular acceleration of the wheel.

    Anyway. Is you mentioned before and I agree with that, here's no possibility to use mechanic spring and platform are weightless. For ideal model it's easy, but for real model. The effect will be too small and it's hard to see it.
    However, what if change mechanic spring to something real weightless which will gives a momentum? Is it possible? Yes! Photons!

    Here some example of it.
    http://knol.google.com/k/paradox-of-classical-mechanics-2#Repulsion_objects_using_laser(2E)

    Repulsion objects using laser.

    All explanation of this effect has a little problem. The spring should have very little weight.The ideal spring may have mass zero, but it's just an ideal model Is it possible make the spring is weightless on real world?
    The answer is - YES. it just need replace mechanic spring by photons. The photons have and impulse, however the stationary mass of them is zero.
    This diagram shows this repulsion using laser ray.

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    The laser itself has two rays with same intensive in opposite detections to each other. The net momentum for for laser is equal zero. However, these rays hit two cylinders with different alignment relatively to rays position.
    Is it was described before these cylinders with same mass will have a different translational velocities.
     
  15. D H Some other guy Valued Senior Member

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    2,257
    Alex, Why do you persist? You are not going to find a flaw in Newtonian mechanics, at least not in the areas in which you are looking.

    Flaws most certainly do exist in Newtonian mechanics. It does not apply to the domains of the very small, the very massive, or the very fast. One can set up configurations of point masses in which objects will, due to gravitation, go to infinity in a finite time. One can set up configurations of objects such that Newtonian mechanics is not deterministic.

    You haven't done any of those. You are instead trying, in vain, to attack Newtonian mechanics at its very heart. That isn't going to work, Alex. In the domains in which Newtonian mechanics does apply, it has been tested and retested, re-retested and ... re-re-...-retested ad infinitum since Newton's day.
     
  16. ABV Registered Senior Member

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    79
    DH, Thank you for your point of view. I'm continue posting my articles with related to my idea.

    This article came form this site
    http://knol.google.com/k/alex-belov/repulsing-objects-by-photons/1xmqm1l0s4ys/23#

    Repulsing objects by photons

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    Base radiation pressure or light pressure phenomenon, the the particle photon with stationary mass zero has a momentum which can be transfered to an object. The good example is a light mill.
    The light radiation and objects repulsion must be follow by law of momentum conservation. The figure1 shows repulsing two identical objects with mass m. After a period of time the objects are utilizing photons momentums and start conduct a translational motion. If laser system has identical ray intensity in both directions then objects will have translational momentums P with same value.

    Other words, the objects with same mass m will have same velocities v1.

    The laser system with identical ray intensity in both directions has zero net momentum.

    However, the objects will utilize less photon momentums when these objects increase it's own velocities relatively to laser. The speed of light is constant. However, base on Doppler effect the photons will increase wavelength which reduce photons momentum.


    Other words, the objects linear velocities will gain in non-linear mode.

    How photons will repulse objects if one of them is rotating?
    The figure2 shows repulsing objects by photons where one of them conducts rotation around it's own center of mass.

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    In this case, this objects will utilize photon momentum differently. After some period of time, the non-rotated object will utilize photons momentums on velocity v. However, the rotated object will utilize photons momentums on velocity v+wR. Base on Doppler effect the rotated object will utilize photons momentums less then non-rotated object. Therefore, after some period of time the rotated and non-rotated objects will have different translational velocities v1 and v2.

    Will it work into air environment?
    The goal of this article is transfer translational momentum without mass transfer. Would it possible to do it into air environment? What if transfer momentums to objects through air wave? The Doppler effect for rotated object will be a plus.


    Conclusion
    Base on Doppler effect and particles(photons) with zero stationary mass, the repulsed objects may have different translational velocities by value inside isolated system.
     
  17. D H Some other guy Valued Senior Member

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    2,257
    Alex, yet again you are jumping to a new subject replete with a new Rube Goldberg device. Yet again you have failed to learn anything from your previous endeavor and its Rube Goldberg device.

    Go learn some physics.
     
  18. ABV Registered Senior Member

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    79
    DH,

    I'm not jumping. I'm continue.
    I'm trying to say the physics world can safe a world from global disasters as consequence of world economy recession if aggressive starts research and build new technologies.
    This topic is one of them. The mother nature has a lot of examples which show center of mass of isolated system position instability. We just ignore them.

    Here's possible propulsion system:

    http://knol.google.com/k/alex-belov/the-concept-of-wave-propulsion-system/1xmqm1l0s4ys/24#

    This propulsion concept use ideal models like:
    1. The wave source produce pure directional waves which transfer momentum only in particular direction.
    2. The wave receivers completely utilize directional wave and don't reflect secondary waves back.
    3. The known wave which can transfer momentum without mass - photon.

    The figure1 shows first concept of wave propulsion system.

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    The wave doesn't transfer mass. However, the wave may transfer momentum. Base on Doppler effect, the rotated wave receivers utilize waves with different wavelength. Base on this wavelength difference the receivers will utilize momentum of wave with different values. This momentum difference let system move in particular direction.

    The figure 2 shows second concept of wave propulsion system.

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    Here's wave path through a curve channel and hits wave receivers into one direction.
    The utilized by receivers momentums of waves let system moves into particular direction.
     
    Last edited: Mar 10, 2011
  19. ABV Registered Senior Member

    Messages:
    79
    DH, back to mechanic.
    During platform repulsion the net force will equal to wheel translational motion plus force for rotating this wheels around it's own center of mass.

    Same for momentums. The momentum platform with rotated wheels will be a sum of translational momentum and momentum which spend for wheels rotation.
    If hook back the weightless platform with rotated wheels then need momentum to stop wheels and momentum to stop these wheels rotation.

    It's possible to complicate experiment and make cover for platform with rotated wheels and set up their axis on it.
    If repulse this platform with cover and solid object then momentum for solid object will be equal to net of translational momentum of platform with cover plus momentum for platform motion relatively to cover. On the end of experiment platform will collide to the end of cover and cancel this second part of net momentum. That's all.
     
  20. D H Some other guy Valued Senior Member

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    2,257
    Ben, cess this nonsense please.
     
  21. ABV Registered Senior Member

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    79
    DH, what wrong?

    a few questions.
    What weight will show scale if holding thread let spool roll down with acceleration?
    If person standing on scale lifts up with acceleration some object, what weight is showing there?
    I hope you remember classics examples for accelerated elevator from physics book.
    Here's same thing. The platform relatively to wheels moving with acceleration. This acceleration will depend on wheels moment inertia. The wheels acceleration will be equal to difference of spring force and force for accelerating platform relatively to wheels divided to mass wheels.

    Any questions? Fell free to ask

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  22. ABV Registered Senior Member

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    One more time.
    http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18#
    Base on trivial case of problem, the figure 4 shows a model for experiment where wheels and solid block covers by another platforms on wheels.

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    The wheel1_1 and wheel1_2 connect with cover by axis.
    For easies calculation, all doted elements, platforms and spring are weightless.

    Let's assume the law of momentum conservation always works in simplest form. In this case, the forces on both sides of spring are equal by value and induce translational motions for solid block and rolling objects on both platforms. However, which force induce rotational motion for rolling objects on one platform? The solid block doesn't conduct rotational motion on other platform. On one platform, the rolling objects rotate on opposite direction for each other and have same angular momentum by value and opposite direction. However, need a force to induce this rotation. The asymmetrical force cannot exist during interaction. Therefore, the assumption where law of momentum conservation exist in simple form during complex interaction is wrong.
     
  23. Tach Banned Banned

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    If you think that you proved the above, you have some very serious mental problems, dude, you should get checked up
     

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