Billzilla Posted August 14, 2002 Share Posted August 14, 2002 Is this it? Spin gravity = K dI sine 0 / r^ 2 K = constant dI = induced electric vector created by magnetism and spin acceleration sine 0 = spin vector r = distance from centre of spin .. to which the first reponse was - http://www2b.abc.net.au/science/k2/stn/posts/post133858.shtm ... then http://www2b.abc.net.au/science/k2/stn/posts/post133883.shtm The professional opinion was totally unsurprising. Link to comment Share on other sites More sharing options...
Zarkov Posted August 14, 2002 Author Share Posted August 14, 2002 That needs to be integrated into a spherical shape, but yes that equation is basically correct. > Link to comment Share on other sites More sharing options...
Radical Edward Posted August 15, 2002 Share Posted August 15, 2002 right. now we have an underived formula that is entirely devoid of units. Link to comment Share on other sites More sharing options...
Zarkov Posted August 15, 2002 Author Share Posted August 15, 2002 Oh, well Rad E, it is better than last month...no I have more just keeping it shy, for a while :) Link to comment Share on other sites More sharing options...
Radical Edward Posted August 15, 2002 Share Posted August 15, 2002 it isnt better at all. Link to comment Share on other sites More sharing options...
Zarkov Posted August 15, 2002 Author Share Posted August 15, 2002 Rsd E, I think your ship is heading to rocks, you had concentrate on another course! Link to comment Share on other sites More sharing options...
Radical Edward Posted August 15, 2002 Share Posted August 15, 2002 oh I'm fine. I am merely trying to point out to you that a physical formula with no units, ore clearly defined quantities is about as useful as no formula at all. could you clearly define specifically what dI and sin0 are? ... Link to comment Share on other sites More sharing options...
Zarkov Posted August 15, 2002 Author Share Posted August 15, 2002 dI sin(theta) is the induced electric emf, eddy current! Link to comment Share on other sites More sharing options...
Radical Edward Posted August 15, 2002 Share Posted August 15, 2002 uh? you said sin(theta) was a vector though. 1) what is dI and how do you calculate it? 2) what is theta an agle with respect to? and does mass really have nothing to play in this whatsoever? Link to comment Share on other sites More sharing options...
Zarkov Posted August 15, 2002 Author Share Posted August 15, 2002 Mass has no part other than creating a magnetic field, so the permeability constant has to be included, so does the rate of acceleration of spin. A negative rate produces a push in, towards the centre. There are not many positive rates of acceleration in Nature, other than at ejection time, but if this was the case the emf would push away from the centre. That formula was for the force at a point. It has been taken out of context, but it is really basically correct! I have gone furthur, but I am not ready to be critised, since nobody wants to help Link to comment Share on other sites More sharing options...
Radical Edward Posted August 15, 2002 Share Posted August 15, 2002 Originally posted by Zarkov I have gone furthur, but I am not ready to be critised, since nobody wants to help go further. there's nothing worse than half an idea. Link to comment Share on other sites More sharing options...
Zarkov Posted August 16, 2002 Author Share Posted August 16, 2002 Thanks, for the encouragement RadE Link to comment Share on other sites More sharing options...
Radical Edward Posted August 16, 2002 Share Posted August 16, 2002 I mean, tell us the whole formula. I want to digure some predictions out from it, and with this scrap that you have posted I can't figure anything out, as it's impossible to know what numbers to put in. Link to comment Share on other sites More sharing options...
Zarkov Posted August 18, 2002 Author Share Posted August 18, 2002 MATHEMATICS E(electromotive) = d(phi)/dt so if we extend that into a spherical field then = [1/4*pi*r^2]*[d(phi)/dt] = K*[d(phi)/dt] / r^2 = K a I / r^2 Where d(phi)/dt = integral of emf produced by electric field I and acceleration a (+ - ) and so M (magnetomotive force)= K a B / r^2 Spin gravity force = torque set up by these two vectors E and M generates a dipolar eddy spin that spirals into the centre of spin, if a is negative and out from the centre of spin, if a is positive. In both cases K = 1/ 4 pi a constant r = distance from centre of spin Link to comment Share on other sites More sharing options...
Zarkov Posted August 18, 2002 Author Share Posted August 18, 2002 Note..... I do not know how to post the integral sign so I have used a differential, maybe some one will correct my deficiency!! Link to comment Share on other sites More sharing options...
fafalone Posted August 18, 2002 Share Posted August 18, 2002 E(electromotive) = d(phi)/dt -Faraday's Law, I've never seen this applied outside of circuits. so if we extend that into a spherical field then = [1/4*pi*r^2]*[d(phi)/dt] = K*[d(phi)/dt] / r^2 = K a I / r^2 Where d(phi)/dt = integral of emf produced by electric field I and acceleration a (+ - ) -emf is a derivative, not an integral. -if K = 1/4pi, Kr^2 * dphi/dt isn't not equivalent to K dphi/dt/r^2, basic algebra -acceleration times an electric field is magnetic flux? not quite. -I is Ohm's Law, E/R, it represents ; dphi/dt does not equal R, or E/R, , or aE/R and so M (magnetomotive force)= K a B / r^2 -B? magnetic flux density? -howd you calculate mmf without reluctance? Spin gravity force = torque set up by these two vectors -this doesn't make sense E and M generates a dipolar eddy spin that spirals into the centre of spin, if a is negative and out from the centre of spin, if a is positive. -and how does this happen? dipolar eddies require a stratified solution In both cases K = 1/ 4 pi a constant r = distance from centre of spin -this is pretty much the only unflawed part of this post Link to comment Share on other sites More sharing options...
Zarkov Posted August 18, 2002 Author Share Posted August 18, 2002 At least Fafalone, you are getting some idea what I am trying to achieve, the formulae are crude I know, but a little help here, a criticism there, and soon it will make sense. At least I am now delving into the maths of the phenomena! And the answer lies in all this somewhere!! Thanks for your input Link to comment Share on other sites More sharing options...
Zarkov Posted August 19, 2002 Author Share Posted August 19, 2002 Here is decisive proof that the solar system is a vortex that does not rely on mass. mean orbital velocity (km/sec) / mean distance from the Sun / period yrs / K= rv^2 :- Mercury 47.89 / 0.39 / 0.24 / 895 Venus 35.03 / 0.7 / 0.62 / 882 Earth 29.79 / 1 / 1 / 888 Mars 24.13 / 1.52 / 1.88 / 871 Jupiter 13.06 / 5.2 / 11.86 / 878 Saturn 9.64 / 9.54 / 29.46 / 875 Uranus 6.81 / 19.81 / 84.01 / 925 Neptune 5.43 / 30.06 / 164.8 / 903 Pluto 4.74 / 39.44 / 247.7 / 870 Mean K = 887 Graph of distance against velocity is exponential so rv^2 / 4 pi^2 = 887 / 40 = 22. The 22 is a characteristic of the spiral in the vortex. This vortex is independant of matter, and is spiralling out from the centre, if current observations are correct! Link to comment Share on other sites More sharing options...
Zarkov Posted August 19, 2002 Author Share Posted August 19, 2002 The rv^2 is Keplers 3 rd law in another form. period p^2 is proportional to distance from centre r^3 but v = r / p using this v in rv^2 rv^2 = (r / p ) ^2 * r = r^3 / p^2 I used this though because it calculates the angular velocity an object should have at a point at distance from the centre of a vortex. Since they all equal K ( 887) the planets are all where they should be, and verify that they are actually in a vortex, that is independant of mass! Link to comment Share on other sites More sharing options...
Radical Edward Posted August 19, 2002 Share Posted August 19, 2002 I see no relation to a vortex, whatsoever. and yes, the orbital periods are independent of their masses, but they are not related to the axial spins of the bodies in any way. Link to comment Share on other sites More sharing options...
fafalone Posted August 19, 2002 Share Posted August 19, 2002 You could always get a telescope and look at the planets, since they're all on pretty much the same plane that's not characteristic of a vortex. Angular velocity is v/r, not rv2, or the others. Link to comment Share on other sites More sharing options...
Zarkov Posted August 19, 2002 Author Share Posted August 19, 2002 I am talking of a plane vortex, a spiral. All the planets are in a relation, Bode recognised this, and Kepler formalised it a bit more rigoursly. Centripetal force in an open system is dependant upon rv^2, in a closed system it is v^2 /r . Link to comment Share on other sites More sharing options...
fafalone Posted August 19, 2002 Share Posted August 19, 2002 Also, you neglected units. You gave velocities in km/sec and distance in AUs. Furthermore, Kepler's Third Law is T2 = r3, and cannot be written as rv2. Not only is this not a documented form of the law, it is in no way shape or form equivalent. Link to comment Share on other sites More sharing options...
fafalone Posted August 19, 2002 Share Posted August 19, 2002 Centripetal force also depends on mass. (mv2)/r Link to comment Share on other sites More sharing options...
Zarkov Posted August 19, 2002 Author Share Posted August 19, 2002 Ah but theres the rub, Fafalone, the position of the planets, and their speed does NOT rely on it's mass, as I have shown. The implications of this are simple, the whole motion of the whole system is in a spiral vortex, objects are not moving independant of each other, implying IMO that the planets are ejected from the Sun. Data that has bee TOLD to me by an astronomer indicates the Moon is moving away from the Earth, and the Earth is moving away from the Sun. The system can not be static, it is either going in or out! Most people think it is going in, and the theory of "attractive" gravity would imply a spiralling in. Link to comment Share on other sites More sharing options...
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