Large Hadron Collider

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16 years 9 months ago #17968 by Stoat
Replied by Stoat on topic Reply from Robert Turner
I made a silly mistake there, when I thought that the universe would whiz past the node. We have three quarks rotating about this fourth "electron like" node, at the speed of light. At zero the node will want the universe to rotate about it.

Remember that our chassis potential is set to the speed of light. When the fourth wire goes negative, it's only negative in respect of this chassis datum. When the fourth wire is at negative equipotential i.e. two protons near to each other, the node can draw on gravitational energy from the quarks. The two protons are violently pushed apart.

Don't know if this is important but this set up wouldn't show any signs of gravitational shielding, the quarks are way smaller than h. The fourth wire gives us our inertial frame but it's "lazy." it doesn't want to pull gravitational "current."

We need to move our datum line up and reduce the amount of "clipping" we have.

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16 years 9 months ago #17979 by Stoat
Replied by Stoat on topic Reply from Robert Turner
In one post i suggested that we could set up an experiment with a laser to see what a star delta connected tetrahedron of three quarks and a node would do. Now that's something that could be done but the bearings have to be extremely good.

For someone with a lab. Take three piezoelectric crystals and arrange them on a card as an equaltateral triangle of sixty degrees. Star delta connect these and then spin the card. Of course it will need some slip rings. I reckon that we'll get a gyroscopic pump engine. So add a variable tank circuit to find resonance effects.

I suppose we could try the same idea with plastic tubes and ball bearings. Drop ball bearings into a star connection, then they will spray out to the corners of a triangular tube. They wil bunch but start to rotate round the rotating triangle. Some will fall into a second star connection, that is at sixty degrees. Because of the angle they have to slow down. The supply tank of ball bearings will be at some potential but won't be able to pump against gravity in this case.

(Edited) To sum up, as this is not getting any posts. At the centre of a proton we have a tetrahedron spinning at near light speed. On one face, there is a triangular shadow tube, in which ftl particles counter rotate. The fourth point is the inertial node, a star connection.

But this is a bose einstein condensate, so this is gravitational energy and electromagnetic energy and its mass equivalent is not present near the tetrahedron. Gravity is the strong force in this region.

Around about these shadow tubes is the higgs field. Wave like, in that we have wave like entropy. As we move out from the quark shadow tubes we arrive at a shell of higgs particles. We have a tiny little dyson sphere which is full of a BEC ftl liquid.

If we push this BEC proton upto the speed of light, its internal energy starts to fall. The rotating tetrahedron at its core "wobbles" and looks like an icosahedron. Its shaows beome "fuzzier." It must conserve charge, so it star connects and pulls electromagnetic power from the inertial node, which is at the speed of light in "normal" space. (This is not a mass increase, just the expected momentum increase.) The outer shell of higgs particles thickens, it "cools" the higgs field in order to conserve its charge. Any secondary shadows i.e. those connecting protons to neutrons in the nucleus, will also become fuzzier. Note that these shadows "quantum tunnel" as a matter of course.

(Edited again) Oops, that's pushing the proton to the speed of gravity and not the speed of light.[:I][:D]

(Edited again, already!) For the speed of gravity, using Mach's limit, I get 17.4 billion times c. Though as I've said earlier, I think the circuit acting as a 3 phase star delta wired full rectifier, knocks this down to a "d.c" component of 5.5 billion c.

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16 years 9 months ago #19763 by Stoat
Replied by Stoat on topic Reply from Robert Turner
A little on Mach's limit, it's the basis of the idea that gravitational energy and inertial energy are equivalent. GM^2 / r = MC^2 where M is the mass of the entire universe. lose the M and we have GM /r = C^2 Now let's make GM = 1. A rather small universe. So what's r when the speed of light is 20 billion times c? It makes it 23 thousand times smaller than h. About the interest level of a quark. Now I wanted to tie in the idea of the mass ratio of the electron and proton times pi. This is because I'm after the full wave rectifier effect that I think quarks can do. Hence I get 17.4 billion c for the speed of gravity.



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