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**New Theories / Re: how gravity works**

« **on:**17/03/2019 09:56:08 »

And i dont understand SR & GR.It shows.

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And i dont understand SR & GR.It shows.

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I have a explanation why people are so keen to dismiss GR and SR , while on an engineering at Siemens Hell in Kiel I quoted Einstein to the "ex" Nazi instructor his reply "but he was a Jew"lAwkward...

But it does raise an interesting point.

There was a conference where a bunch of nazi scientists (I use the term loosely) under the direction of their government all pointed out "problems" with Einstein's theories.

He was asked what he thought about such a powerful and prestigious group attacking his ideas.

His response was that, if they were actually correct, then a single undergraduate student pointing it out would be all that was needed.

So, we had a situation where the political forces went the other way- the nazi government was determined to undermine Einstein and his ideas.

Great blessings of a powerful elite would rain down on anyone who could do it.

And yet, the Germans continued to try to build an atom bomb that could only work if he was correct.

So the idea that Einstein's theories are only supported by a political conspiracy is nonsense.

We know that when a political conspiracy tried to oppose them, it failed.

All that stuff about

"

owadays big money hurts science, especially in theusofa.

And politics, especially in theusofa.

And Christians, especially in theusofa.

is nonsense.

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So, you accept you were talking nonsense about the Moon's rotation.

So, it would be a bit silly to choose that nickname.

Incidentally, the phrase is "nit pick".

in your honor, we should change nit pick to something you do well, how about nose pick!

I guess that's a start.

my mistake, I have withdrawn most of my posts to this thread. my apology to you and kyptid and to the community on a whole

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This is an actual , practical , form of the Compton Effect-type of engines .So "practical" that it starts with unobtanium.

Mount a dense-plasma container (X-ray transparent)

These should generate strong shockwaves , which hit the pad & wall as hammer blows .Yes, it will create a shockwave.

And,as I keep pointing out (and you keep trying to ignore) the shockwave will spread out in all directions.

It will hammer the tank equally in all directions.

The net force will be practically zero (apart from photon pressure).

THAT is the first working Reactionless Drive !No.

It's not even a credible start on the way to one.

If the ejected plasma hits a reciever , within a closed system , and is recycled , it ...... produces no net force..

thus justifying the name "Phaser" , or Phase-change laser .OK, so that shows that you don't know what a laser is...

No shock there.

Nobody has seen anything except you posting gibberish.

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What exactly is their logic?It worked.

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Yes.

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That the "lunar tidal wave" is moving from east to west, at a speed of 1600 km per hour, I do not claim. This is written in the lunar theory of tides.

Is there any way to get you to understand the difference between the tidal bulge and a tidal wave?

If not, there is no way that this thread will ever make progress.

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Divide a dot of energy by a piece of string,Srsly?

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This is like a child saying 'you can't speak to my imaginary friend because they use a language that only I can understand'.Which proves that you are a mathematically illiterate fool.

I don't mean to sound rude, but obviously you do not understand if you think the maths is meaningless and illiterate. The maths works with my conceptual envision I have explained with theory . The NFEU model does not ignore present Physics, to the contrary the model is based on some of the present physics such as Coulombs law, space-time and Dirac. I think you are possibly not considering Alpha or Beta point energy particle manifestations , popping into and out of existence , the dissipate resulting in dispersion to 0 value.

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If my notion is so wrong, then why isn't there a single attempt to falsify my notions?Because people have learned by past experience it is just pigeon chess.

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To:KryptidAs already noted the Earth Moon barycenter doesn't shift due to the tidal acceleration of the Moon.

Must have logic , live by it !

Speaking of planetary systems , how do you like that Earth-Moon system ? Much of Earth's initial angular momentum has been transferred to Luna , BUT much of it was converted to heat in the process . Were it a perfect transfer , the E.-M. system would be in a different location , with a different momentum . If you now spun the Earth in the opposite direction , the opposite effect would occur . Bottom line , destroying energy of momentum works as well as removing mass. Sounds like a propellantless rocket to me !

Hokay , next victim ! P.M.

P.S.-" l'm freakin' Bobby Fisher ! "

If we consider the angular momentum of the Earth -Moon system:

The Moon's orbital radius grows by some 4 cm per year, and if we use 384,000 km as its average orbital radius, We find that moving out by 4 meters ( the expected distance in one century) puts it into an orbit with an orbital velocity of 1018.9608624m/sec , vs its starting velocity of 1018.9608677 m/s with the Moon's mass, that works out to a gain of angular momentum of 1.348e26 kgm^2/s ( the Moon will also gain orbital energy by the relationship of E_0 = mv^2/2-u/r = -u/2a where u is the gravitational parameter for the Earth, GM= 3.987e14 m^3/s^2, and a is the semi-major axis of the orbit.)

The Earth, in that same century, increases its period of rotation by 1.7 milliseconds. Using the Earth mass, and assuming it is a sphere of uniform density in order to calculate its moment of inertia, we get a loss of 1.39e26 kgm^2/s

Pretty damn close considering we took a number of shortcuts along the way. (treating the Moon as a point mass, and assuming uniform density for the Earth, etc)

If we were to accurately account for all the factors we would get agreement in the numbers to whatever level of accuracy we had for those factors.

Now the Earth does heat up in this exchange, so some part of its remaining angular momentum will be due to the "effective" mass increase due to that thermal energy, And the Earth will radiate that energy. But this does not mean that there is any net change in total angular momentum. Because, as I already mentioned earlier Electromagnetic radiation exhibits the property of momentum and this "lost" angular momentum is being carried off by this radiation. Thus, if you take the Moon, Earth and the radiation emitted by the system all into account, there is no net gain or loss of momentum, angular or otherwise.

There is nothing "reaction-less" about this at all.

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You answered only two questionsLearn to count, or stop being silly.

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I can't help thinking that, if SR had failed, I'd have heard about it on the national news, rather than some web page here.

Would anyone like to explain how it's plausible that SR has been refuted, but nobody bothered to mention it?

Would anyone like to explain how it's plausible that SR has been refuted, but nobody bothered to mention it?

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I understand electric fields quite well, there isn't much I don't know about QFT.LOL

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and nobody in his right mind considers death to be a bad thing.

You are saying death is a good thing?

Without it, the world would be packed full of animals fighting for food, air and pensions. You would have been born into a world dominated by dinosaurs or possibly their predecessors who lived in an atmosphere that could not support humans. No cell in your body could die, so you would actually be a malignant tumor, not a human being, as would every other living thing. There being no effective means of killing anything, whatever minor infection you might have would develop into an infinite torture. Or you might live forever whilst being digested by a carnivore. Death is what makes life tolerable, and the finite nature of life is what makes every day exciting.

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Good.The first major problem is that you don't know what a Tesla coil does.That called spamming dude.

There's no point to any further discussion until that's sorted out.

In your political correct terms Tesla coil

noun

a form of induction coil for producing high-frequency alternating currents.

Your just being stupid again, are you drinking? being a bot? or just always this much of a fool?

Now that you have some idea what a Tesla coil does- something you should have found out before posting, you will realise that it doesn't produce a force and won't push a spaceship.

This thread is therefor founded on nonsense.

There's nothing further to discuss.

It's not spamming to continue to raise the same point until it's actually answered.

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Why very often as I've just drifted off to sleep do I awake feeling like something has bumped my bed ...

I feel there must be a scientific explanation for this as opposed to some energy force disturbing me.

If you're being startled just after you've fallen asleep, without an external trigger, could be hypnic jerk.

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Well, I'll kick this off:

X is the existence of 1 observable universe sized volume containing no antiprotons

Probability of X is (# of outcomes in which X is true)/(# outcomes possible)

Using these parameters:

• cubic universe with side length of 10^{1010122} parsecs. Let's call this length "L"

• which is a cubic array of cubes of side length <10^{–12} meters. Let's call this length "S"

• 3 possible states for each small box, proton, antiproton or nothing.

• Recall that there are 24 equivalent orientations a cube can have (think of it as each of 6 faces can be "down" and then there are 4 choices for which face is "North").

Thus the # of possible states = 3^{(L/S)3}/24

To address the # of outcomes in which X is true we need to take account of a few more parameters:

• probability of small cube containing proton is "p" and is the same as the probability of containing an antiproton, but the cases are mutually exclusive. so we are looking for "not p" or 1–p. (you don't care whether this space contains protons are not, just as long as no matter. Choice of value for p will play a huge role in what the final probability of X is, and can be anywhere between 0 and 0.5.)

• the observable universe is a cubic region of space with side length M (S << M << L) (choice of M will also have a very significant role in the final probability of X)

For X to be true, there has to be at least 1 contiguous cubic volume of M^{3} that has 0 antiprotons in it.

We need to know the expression for the probability of a cubic volume of M^{3} having 0 antiprotons in it, which is:

(1–p)^{(M/S)3}/24

and we need to know the expression for how many cubic volumes of M^{3} fit in the cubic volume of L^{3}, which is:

((L–M)/S)^{3}

Putting these together, # outcomes in which at least one (not none) of the M^{3} sized cubes has no antiprotons in it:

(1–((1–p)^{(M/S)3}/24))^{((L–M)/S)3}

Now, putting this all together:

(1–((1–p)^{(M/S)3}/24))^{((L–M)/S)3}

P(X) = ————————————

3^{(L/S)3}/24

I leave the reader to experiment with values of L, M, S and p

X is the existence of 1 observable universe sized volume containing no antiprotons

Probability of X is (# of outcomes in which X is true)/(# outcomes possible)

Using these parameters:

• cubic universe with side length of 10

• which is a cubic array of cubes of side length <10

• 3 possible states for each small box, proton, antiproton or nothing.

• Recall that there are 24 equivalent orientations a cube can have (think of it as each of 6 faces can be "down" and then there are 4 choices for which face is "North").

Thus the # of possible states = 3

To address the # of outcomes in which X is true we need to take account of a few more parameters:

• probability of small cube containing proton is "p" and is the same as the probability of containing an antiproton, but the cases are mutually exclusive. so we are looking for "not p" or 1–p. (you don't care whether this space contains protons are not, just as long as no matter. Choice of value for p will play a huge role in what the final probability of X is, and can be anywhere between 0 and 0.5.)

• the observable universe is a cubic region of space with side length M (S << M << L) (choice of M will also have a very significant role in the final probability of X)

For X to be true, there has to be at least 1 contiguous cubic volume of M

We need to know the expression for the probability of a cubic volume of M

(1–p)

and we need to know the expression for how many cubic volumes of M

((L–M)/S)

Putting these together, # outcomes in which at least one (not none) of the M

(1–((1–p)

Now, putting this all together:

(1–((1–p)

P(X) = ————————————

3

I leave the reader to experiment with values of L, M, S and p

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My N-field theoryIt's not a theory.

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Whereas you stated on your N Field thread that:attached is a pdf presentation proposing a hypothesis that tries to quantize gravity

1+1=2

Couldn't be more simpler

A rain drop fell into my cup, a second raindrop fell into my cupAt least try to to keep your attention seeking consistent.

1+1 = 1

Only in your dreams are you smarter than me

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