Thornhill's gravity model

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Re: Thornhill's gravity model

Unread postby Solar » Thu Mar 22, 2018 5:27 pm

celeste wrote:A little more here:
You may be familiar with early ideas on “ pushing gravity”. The idea of these LeSage type gravity models did not address this point: What caused the small scale particles themselves to be “slamming into” matter in order to push it together?


Gravitation as emission is the only complement to “Pushing gravity”. This perspective is from ‘relativity of forces’. Not ‘relativity of forces’ from today’s approach using grandiose mathematical formalisms; but ‘relative’ in terms of its actual practical understanding which means ‘to compare’, ‘in comparison to’, or ‘from the perspective of’.

Such that: depending on one’s relative point of view, and for the sake of comparison, gravity, when considered as an emanation - would then bear that relationship of being “small scale particles slamming into something” and thereby “pushing’ that something along. For example, in reference to the gravitational relationship between the Sun and Earth this approach was also analogously used by Wilhelm Reich wherein the Earth were like ‘a ball rolling on water waves more slowly than the waves’.

Gravitational energy possessing emission-absorption qualities between objects would be the impetus behind “pushing gravity”. How else can it occur? That perspective makes for interesting point of view considering other thoughts expressing the “speed of gravity” as being far in excess of that of light. This is also the point of view that I take i.e. gravity as ‘substantive’ emanation-absorption phenomena; although I do not use the “pushing” term. It is more so that all gravitating bodies constitute an intervening “resistance” to the superluminal flow of the gravitating emissions via (what might be called) their unique ‘rate of absorption’. Such as with:

We propose a modified Weber's potential for gravitation that takes into account the influence of intervening matter. Then we obtain equations of motion similar to Newton's first and second laws, and derive the proportionality between inertial and gravitational masses. We conclude that the gravitational absorption coefficient should be proportional to the square root of the density of the intervening medium, and that for solids its value is approximately 10-11 m-1 . All of this is accomplished supposing a limitless, homogeneous and stationary universe. – On the Absorption of Gravity: A.K.T Assis


Then Shift: the principle, or concept, and one finds this expression through the work of say; Reginald Cahill with “3-Space Inflow Theory of Gravity” such that… gravity is assessed as an actual three-dimensional ‘substantive’ quality that may be absorbed.

Then Shift again: the principle, or concept and one finds the same expression through those works having to do with gravity as a kind of Cosmological “Superfluid” that may be absorbed.

Therefore, one does not find “turtles all the way down” but one does find cycles, or circulating (!) gravitational emission-absorption resonant relationships such that the gravitational circulation of the electron(s) gives way to the gravitational circulations of the atom(s), which then gives way to the gravitational circulations of the molecule etc. And yet, they each remain inextricably integrated to their Cosmic Gravitational Circulations as well as to the temporary constructs of "matter" that they compose all at once. There is the individual relationship as well as the group relationships. The work of Ray Tommes comes to mind here.

It is only from the ‘relative position’ (or point of view) that is taken up when trying to compare these qualities that one says “push” or “pull”. The position is only the angle, or point of view, from which the assessment becomes prioritized. It could have been interpreted from the position of the rates of absorption of an ever-present cosmic emanation. Probably best to stop there. Nonetheless, gravitation as ‘substantive’ emission-absorption is also out there as opposed to just the geometry of disembodied forces. There can be no "push" without an emission.
"Our laws of force tend to be applied in the Newtonian sense in that for every action there is an equal reaction, and yet, in the real world, where many-body gravitational effects or electrodynamic actions prevail, we do not have every action paired with an equal reaction." — Harold Aspden
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Re: Thornhill's gravity model

Unread postby Solar » Thu Mar 22, 2018 5:38 pm

With all due respect, for those interested in this idea, imho this aspect of EU is a mess. I don’t understand why interpretations of Sansbury’s pocket cell experiment, which apparently prompted the ellipsoidal dipole gravity inducing approach, aren’t being discussed, refuted, elaborated on, contrasted, compared etc:

The Speed of Light: Cumulative Instantaneous Forces at a Distance: Sansbury

The above presentation was so jumbled that I can’t really make heads or tails of it. Basically, the Sansbury/Thornhill gravity “model” desperately needs a cohesive presentation specifically explaining how it was derived from Sansbury’s work, the data of course, as well as references to other official (or not) experiments and source material that may support same. Normally I enjoy researching topics but for some reason I don’t fully understand the Sansbury/Thornhill approach. I think I read some of this once; do some of you have Sansbury's original source material for this??
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Re: Thornhill's gravity model

Unread postby Webbman » Fri Mar 23, 2018 11:45 am

just put a boat in a stream and youll then understand dipole gravity and gravity in general imo. Maybe ythe stream isnt in the form you expected. Our molten core and 20 degree c climate tells you its there.
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Re: Thornhill's gravity model

Unread postby Zyxzevn » Fri Mar 23, 2018 1:08 pm

Solar wrote:
We present a generalized Weber’s Law for electromagnetism including terms of fourth and higher order in 1/c.


As I understand it, Weber replaces the linear movement relativity of Einstein with a
circular movement relativity.
So: according to Weber, all forces stay the same, independent of how you are rotating.
I have not seen any evidence of this idea, but if we follow this logic,
you may get gravity of a planet rotating around the sun,
because it is the same to Weber as if the planet is standing still.
It is really a weird system.
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Re: Thornhill's gravity model

Unread postby Zyxzevn » Fri Mar 23, 2018 1:25 pm

Siggy_G wrote:
Zyxzevn wrote:And because electrical fields are additive and 1/R^2 the total effect of the electrical field of the sphere is ZERO.

The electrical field of the inside negates the electrical field of the outside. This works even at close distances, because the 1/R^2 field is exactly the opposite of the area that the charge is distributed over. As long the charges are evenly distributed.


You now seem to claim what you just opposed, by saying that the inside zeros out the field of the outside (although that wasn't my point). The important part is that there is a difference in distance. The more distant opposite charges (of equal amount) of the inside won't subtract the closer charges at the surface down to zero or below. The absolute E-field value will always be above zero. This is why it is called a charged sphere, because the exterior has a measured E-field.


Ok, lets get the model right before we do the calculation:
The charge is distributed (on the planet) with positive on the outside, and negative on the inside.
Can I replace this with two hollow spheres?
Where one is positive charged and a smaller negative charged sphere is in the centre?

Quote: The important part is that there is a difference in distance
The charge on the outside is distributed over a larger area than the inside:
The area outside: Ao= 4*PI*Ro*Ro, so charge density= Q/(4*PI*Ro*Ro)
The area inside: Ai= 4*PI*Ri*Ri, charge density= Q/(4*PI*Ri*Ri)

So we have a charge density that is reduced by a factor of 1/R^2,
which is the exactly same reduction as the distance.
This means that due to the charge distribution,
the difference in distance is negated.
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Re: Thornhill's gravity model

Unread postby Siggy_G » Fri Mar 23, 2018 6:18 pm

Zyxzevn wrote:The charge on the outside is distributed over a larger area than the inside:
The area outside: Ao= 4*PI*Ro*Ro, so charge density= Q/(4*PI*Ro*Ro)
The area inside: Ai= 4*PI*Ri*Ri, charge density= Q/(4*PI*Ri*Ri)

So we have a charge density that is reduced by a factor of 1/R^2, which is the exactly same reduction as the distance. This means that due to the charge distribution, the difference in distance is negated.


Charge density describes the E-field intensity and affects the surface area of a test object depending on its location within this field. The distances and directions of two such fields would need to be defined.

I made an illustration below in an attempt to show that the equal amount of charges of the inside shell don't cancel the charges of the outer shell (surface). Each of the particles have a corresponding opposite charge, but they are separated by a distance L (the green radial lines), which means an inverse square reduction over this distance. At the surface the initial value of a charge B is 1, while it is reduced a little by the corresponding opposite charge A / L^2. If one instead considers the other side of the inside shell, the distance is even longer. It is true though, that a relatively strong repulsive/expansive force would be going on within the inner shell, due to the charge density there.

Image

Note: this is a simplified scenario for a charged sphere. Atomic dipoles in a planetary model do likely not work as direct A-to-B dipole chains over such distances, but rather as collective interactions between each and all atomic dipoles. As mentioned before, electrodynamic Laplace forces are additional components.
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Re: Thornhill's gravity model

Unread postby Webbman » Sun Mar 25, 2018 7:13 am

if you simplify charge as a surplus or deficiency of energy in relation to equilibrium, which is what it is, and abandon the deceptive +/- ideas it really does become a much simpler matter of where the energy is going.
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Re: Thornhill's gravity model

Unread postby Zyxzevn » Mon Mar 26, 2018 5:02 pm

Siggy_G wrote:I made an illustration below in an attempt to show that the equal amount of charges of the inside shell don't cancel the charges of the outer shell (surface).


Great work.
Maybe I was mistaken, but lets do a calculations:

Now, if I may simplify your image to focus on only 2 dipoles.
For that I draw a line right through the centre.

(+)<---L--->(-)<---C-->(-)<----L---->(+)
1 2 3 4

Which is very similar to one of the calculations that I did above,
where I found:

Let me use coloumb's law:

E= k*q/(R*R), where R is the distance to the charge.

And we have 4 charges, the first is on distance H,
where H is the height from earth's surface.
E= k*q( 1/H^2 - 1/(H+L)^2 - 1/(H+L+C)^2 + 1/(H+2*L+C)

Wolfram Alpha

E=k*q*Factor
Factor= (C^4 (2 H L + L^2) + 2 C^3 L (4 H^2 + 8 H L + 3 L^2) + C^2 (12 H^3 L + 42 H^2 L^2 + 44 H L^3 + 13 L^4) + 2 C L (H + L) (3 H^3 + 15 H^2 L + 19 H L^2 + 6 L^3) + 2 L^2 (H + L)^2 (3 H^2 + 6 H L + 2 L^2))/(H^2 (H + L)^2 (C + H + L)^2 (C + H + 2 L)^2)

And for the moon or planets we can assume that H>>C and H>>L
that means we can reduce the equation a lot:
Factor = ( CL(6 H^4 ) + L^2(6 H^4 ) )/(H^8)
= 6(CL + L^2)/(H^4)

And as expected we see a huge reduction of the 1/R^2 relationship of the normal electrical field
to a 1/R^4 relationship that is common for a dual-dipole field.

Now this is only for one line through the centre of the sphere.
We still have to add more lines for every charge around the sphere.
It will not improve our 1/R^4 relationship, but worsen it,
because the we will need to multiply C and L with the cosine of the angle
with our initial line.

With N charges spread around the the sphere we will have
have LESS than 1/H^4
E < N*k*q*6( CL + L^2 )/(H^4)

The integral is a bit difficult, but it is already clear that
the 1/H^4 relationship is nowhere near the 1/R^2 relationship that we needed for gravity.
This is on distances that planets and moons are orbiting from their central bodies.

I did seem to make an error in assuming that there is no field, but also
I did not yet take into account the distribution of charge around the sphere.

I just assumed we have N lines of dipoles distributed as shown in the picture,
and that already proved that the electrical field must be less than 1/R^4
on normal planetary distances.
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Re: Thornhill's gravity model

Unread postby Bengt Nyman » Tue Mar 27, 2018 2:48 am

An approximated dipole to dipole equation with 1/r^4 is obviously not the formula for gravity. Calculations should be done including ALL interacting pairs of charges using Coulombs law with 1/r^2. No shortcuts and no approximations please.
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Re: Thornhill's gravity model

Unread postby Zyxzevn » Tue Mar 27, 2018 8:16 am

Bengt Nyman wrote:An approximated dipole to dipole equation with 1/r^4 is obviously not the formula for gravity.

That is exactly what I concluded too.

The 1/R^3 is caused by the addition of ( a/R^2- b/R^2 ), for just one dipole with 2 charges.
And it gets 1/R^4 when 4 charges are involved with 4 charges..

alculations should be done including ALL interacting pairs of charges using Coulombs law with 1/r^2. No shortcuts and no approximations please.


Yes. If you add them together you get even LESS, because I only showed the BEST possible case.

Assume that the moon (M) is orbiting earth.
Then this is this line that I calculated:

M<-------H-------->(+)<-----L----->(-)<----C---->(-)<-----L----->(+)

This is a line straight through the center, with maximum distance L.
With trillions of trillions of dipoles (T), you can only get the T times
the same factor. It still does not get better with distance.
It does not affect the factor 1/R^4

I made the equation a bit easier to assume that distance H (distance from moon)
is much larger than the diameter of earth (2L+C)
While the equation seems to start with 1/R^2 it quickly shifts to 1/R^4 on realistic distances.
Just fill in the data if you like.
Moon distance= 362600 km
Earth's radius= 6371 km
The distance H makes the factors L and C diminish.

The dipoles are also in a different direction, which makes the effective distance
L and C smaller. And this decreases the effect of the electric field, for these dipoles.

Now let's show you the WORST possible case: a line perpendicular to the best case.
On earth you would just see the moon past the horizon.

M
|
|
Y = H +L+1/2C
|
|
(+)<-----L----->(-)<----C---->(-)<-----L----->(+)

Now the (-) are nearer to the moon (M) than the (+)
We can calculate the distances with Pythagoras (smart greek guy)

This gives us 2 (-) charges on a distance of M(-)= SQRT( (C/2)^2 + Y^2 )
And 2 (+) charges on a distance of M(+)= SQRT( (L+C/2)^2 + Y^2 )

The SQRT is the square root.

So the electric field of the negative charges is:
E(--)= 2*k*q/( R^2) = 2*k*q/( (C/2)^2 + Y^2 )
E(++)= 2*k*q/(R^2) = 2*k*q/( (L+C/2)^2 + Y^2 )

The total electrical field is E(++) - E(--)=
Wolfram alpha

E(Total)= 2*k*q*Factor

Factor= -(16 L (C + L))/((C^2 + 4 Y^2) (C^2 + 4 C L + 4 L^2 + 4 Y^2))

You can already see a -1 in the factor. This means that the worst-case force is opposite
of the best-case force. That is because the negative charges (-) are now closer to M
than the positive charges (+).
So this can even negate the force that we got from the best case.

Which of course reduces to 1/Y^4 again for distances where Y>>(L+C).

But to make this easy, let's first assume that C=0.
That does not make so much difference.
E(Total)= 2*k*q*Factor
Factor= -( L^2 ) / (( L^2 * Y^2 + Y^4))

With the moon we get:
Y= 362600 km L= 6371 km
Which gives: Factor= -2.3473*10^-15
link

If I reduce the equation by assuming that L<<Y I get:
Factor= -( L^2 ) / (Y^4)
Factor= -2.3480*10^-15
link

So my assumption that L<<Y is very much justified, and
we can use the 1/Y^4 factor as a good approximation for the electrical force at these distances.

So in total we have both repelling force and attracting electrical forces of
1/R^4 from all dipoles.
So we can indeed conclude that the total electrical force is less than 1/R^4

Which will lead us to the only conclusion that electric dipoles are not
the way to create a 1/R^2 gravity field.
And we need such a field for the orbits of the Moon and for
the orbits of the planets.

Image

On close distances the dipole electric field is still stronger than gravity.
That is why the balls stick to the cat.
On long distances, the electric dipole field diminishes with 1/R^4,
and this makes too small to replace gravity.
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Re: Thornhill's gravity model

Unread postby Bengt Nyman » Tue Mar 27, 2018 8:40 am

Hi Zyxzevn,
Unless you have something new to add I suggest that we stop repeating ourselves. Thank you.
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Re: Thornhill's gravity model

Unread postby Zyxzevn » Tue Mar 27, 2018 9:53 am

I presented a very optimistic situation, easy to calculate.
The real situation is even worse.

http://www.phys.uri.edu/gerhard/PHY204/tsl56.pdf

The electrical field of a uniformly charged Solid sphere.
For any place outside the sphere, the electrical field is:

E(r)= k*Q/r^2
Where Q is the total charge on the sphere,
and r is the distance from the centre.

If we have two spheres the charges subtract from each other.

So we have for any dipole sphere:

E(total) = k*( Q(+) - Q(-) )/r^2

For any dipole we have Q(+) = Q(-)= Q

So we have
E(total) = k*( Q-Q )/r^2 = k*( 0) /r^2 = 0

Image
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Re: Thornhill's gravity model

Unread postby neilwilkes » Tue Mar 27, 2018 10:37 am

Bengt Nyman wrote:
neilwilkes wrote: ...

Because of a lack of clarity in Wikipedia, which is spreading to the readers, let us agree on not using names which can be misassociated.
1. Ionic winds have nothing to do with Dipole gravity.
2. The large, standing capacitor experiment performed by Brown in vacuum as well as in oil showing a change in the "weight" of the capacitor has everything to do with Dipole gravity.
Neil, please feel free to comment.


Sorry for the silence - I have been ill recently so am catching up with the thread right now Bengt.
My apologies.
You are absolutely correct in (2) above and I will dig out (and scan from my books when I cannot find a PDF) the relevant sources to support this. Brown thought of that one & disproved the idea (this I do have, just need to dig them out)
Back to finishing off the thread first though
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Re: Thornhill's gravity model

Unread postby Aardwolf » Wed Mar 28, 2018 10:28 am

willendure wrote:
Aardwolf wrote:We know what G is as measured on Earth. That's all we have. Plus the fact that it oscillates in an 11-year cycle over time. An undisputed and predictable fact, even though you want to desperately believe it can't be true.


You have a handful of data points over about 2.5 such cycles, I hardly call that indisputable. It is very suggestive, yes, but you are really exaggerating here.
It's published and without any valid published criticism, so by definition it's undisputed. You can personally dispute it because you don't like it if you want but that carries no weight. You need to either produce some evidence it's been disputed or critique it yourself, and it's 13 points over 4-5 cycles.

willendure wrote:The mass in newtons formula (GMm/r^2), when taken into GR is replaced with the relativistic mass. If the earth is changing velocity over a 5.9 year cycle, could this be enough to account for this effect?
Velocity in what direction? At what point of the year? At which point on Earth? You're grasping at straws. The only 5.9 year cycle is Jupiter's apogee/perigee and those years correlate exactly with the troughs in the measurement of G.
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Re: Thornhill's gravity model

Unread postby Aardwolf » Wed Mar 28, 2018 6:21 pm

querious wrote:
Aardwolf wrote:will the experiment give the same result for G on the moon without any adjustments to the apparatus?


Absolutely. Why wouldn't it?
Well maybe we have finally got to the bottom of why you misunderstand the experiment (Although we have discussed this extensively here and on another thread so possibly the misunderstanding is wilful).

As explained on numerous occasions, the apparatus works by comparing the vertical attractive force from the earth with the horizontal attractive force from the static weights. The torsion material is specifically chosen so that the resistance creates a degree of oscillating movement in the subject spheres that can be measured. And this is what determines the comparison between the weight of Earth and the weight of the sphere, so that Cavendish, as per the name of his paper, could measure the density of the Earth.

Now your position is that if the downward force is reduced to 1/6 it will still be possible to take readings from the apparatus without any other changes. I’m sure most members here can clearly see this is a ridiculous proposition but you carry on with your delusional beliefs.
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