Bjarne on: Dark Energy, only an illusion.

[PiedPiper]

I'm still waiting on Bjarnes promised explanation as to why all those arrows were necessary; I'm sure it will be the keystone to understanding his overall theory.

 

[Bjarne]

I'm still waiting on Bjarnes promised explanation as to why all those arrows were necessary; I'm sure it will be the keystone to understanding his overall theory.

If all the arrows not were there you would have no question

 

[MRC_Hans]

One of the reasons I hang in here, apart from masochism, is because you keep learning something. Not being a' priori a specialist in gravity, I need to dig up facts to counter all Bjarne's weird claims. I don't think Bjarne learns much, but that's his loss.

One claim Bjarne recently made is that the gravity acceleration on the surface of an object must be degraded if the object is not a sphere. On the face of it, this seems logical, but it is actually false. ... Let me explain:

If you take a ball of dough and roll it into a long stick, then, of course, the gravity attraction from the dough, measured on the end of the stick gets lower. But this is because the distance to the center of gravity (COG) becomes larger.

However, if we take a building, like the Domus Vista, which is tall and flat, and want to calculate its gravity attraction, say, on the roof, what then?

The formula for this is g=GM/r²
But that is the formula for a sphere. Will the gravity attraction for a tall, flat box be higher or lower?

Well in the dough-ball example above, we changed the shape of a given mass, but we retained the same density, so the resulting form became much longer; the r in the formula increased.

However, the building has the mass it has. If we were to convert it to a sphere, that sphere would have a much lower density. This would mean that much of the mass would be farther away and at a greater angle than in the box. So most of the mass is actually concentrated right 'below' us, giving a higher attraction.

If we were to make the building into a long thin rod, it would have to be incredibly dense to have the same mass, and its surface gravity would be even higher.

Hans

 

[MRC_Hans]

Another claim form Bjarne, related to the above, is that the gravity I derive from his measurements is far too high. How this is not a problem for his own thesis is unclear, but I'll let that rest.

However, using the above formula on the Domus Vista, and assuming the 50,000 ton that is the only weight that can be found in the documentation, we get:

G*50,000,000/50² = 1.3484*10^-6m/s² = 0.13484 mGal

The value derived from the measurements is 0.273 mGal. Not nearly as great a discrepancy as Bjarne claims, and if we assume the the box shape gives an increase of 1.5 (see my previous post), and the mass of the building is really 75,000,000kg, then the figure fits.

So, again Bjarne's measurements debunk his thesis and confirm conventional gravity.

Hans

 

[MRC_Hans]

Finally, we have a curious logical short-circuit in Bjarne's 'argumentation':

Bjarne recognizes the formula for surface gravity ( g=GM/r² ), and uses it himself. Obviously, the same formula is extensively used by astronomers and has been shown to consistently fit observations.

However, that formula implies that the square rule also applies inside an object, but Bjarne denies this. According to Bjarne's idea, the gravity pull is constant throughout the object, but then we would have 'r' in the formula and not 'r²'.

Hans

 

[MRC_Hans]

Isn't the ground floor by definition 0 meters above the ground?

Actually not. The basement comes above the ground. In fact, I use the absolute elevations (over standard sea level) from the drawing. This is 9.35m. The advantage of this is that we can use normalized base gravity values from measurements in the general area, without bothering about the local elevation.

Hans

 

[Bjarne]

However, using the above formula on the Domus Vista, and assuming the 50,000 ton that is the only weight that can be found in the documentation, we get:

G*50,000,000/502 = 1.3484*10-6m/s² = 0.13484 mGal

The value derived from the measurements is 0.273 mGal. Not nearly as great a discrepancy as Bjarne claims, and if we assume the the box shape gives an increase of 1.5 (see my previous post), and the mass of the building is really 75,000,000kg, then the figure fits.

This is in no way what the pattern shows, and also not the way it normally is calculated.

Bjarne recognizes the formula for surface gravity ( g=GM/r² ), and uses it himself.
I have wrote several times that the deviation from a ball shape is crucial to understand, and include in yours calculations / estimations…

However, that formula implies that the square rule also applies inside an object, but Bjarne denies this.

Off course I deny that,- also the prevailing theory deny that.

- /r² can only be use to find the ADG at the surface at (homogeneous) ball shaped body, and further out in space.
NOT inside the body.

According to Bjarne ‘s idea, the gravity pull is constant throughout the object, but then we would have 'r' in the formula and not 'r²'.
This is NOT what I claim, the "picture” is much more complex like that.


However, if we take a building, like the Domus Vista, which is tall and flat, and want to calculate its gravity attraction, say, on the roof, what then?

The formula for this is g=GM/r²

In no way….
As I wrote; the more an object deviates from a ball shape, the less gravity can it produce.
This is no surprise according to the new theory, because a ball shapes is the best possible object to “trap ” - “encircle” central space, and this is what gravity is about.
Only to that degree that central space is trapped, - the strenght of ADG at the surface is a result.

Hans we are talking about creating “a depression" in the space density.
Compare it to create a vacuum in a container, the vacuum cannot be stronger as the walls of the container..
It is the exact same story with gravity.

Mass encircling central space, is that; - "Wall of the container"

This is also what happens when ball shaped astronomic bodies are created…

If gravity always could decide the shape, - all shapes would be ball shape and you or Domus Vista would not exist, but only a complete ball shaped Earth.

If central space not is trapped to a maximum degree, due to the shape of the object, - then matter / gravity is more or less “spread out” of a large area, - in this process it get more and more ineffective to create strong gravity.

But still atoms create gravity, but just not so much as they could in a ball shaped body.

The more a body deviate from a ball shape, is certainly significant reducing ADG,- and not only due to that the distance to the COG get’s larger.

If Domus Vista would be a ball shape (and homogenous mass) and did had radius 50 meter to the COG, - YES we could use g=GM/r² - But since this is not the case, the overall gravity pattern from that building is MUCH less .
So gravity is not how you are used to think. - It is always cheating you.

Do you remember the history lessons how Napoleons Soldiers was relocating all the time to cheat the English admiral visiting France, - so that he believed that France had a huge army ?

Gravity is playing the same game; so long it can repeat itself in ball shaped phenomena’s.

• First like atoms
• then like ball shaped astronomic bodies,
• then like the center of galaxies (black holes, and “dark matter”)
• then in the center of the cluster of galaxies,
• and my friend I can guarantee you that in the center of the Universe, there is a similar area of gradually huge stretched space towards the center.

A ball shaped galaxy would created much larger gravity as the one you live in.

Domus Vista measurement project confirms that there is really weak gravity due to that building.

Shortly; - the idea that a certain amount of matter / mass must give a certain amount of gravity, - is scrap, - it all depends on the shape of the bodies this matter is part of - locally as well as globally.

There is no law that tells gravity always to give max result.
There is also no law to tell you that you must participate in 37 julefrokost ever year, - a ½ julefrokost is also OK.

And that’s it my friend

Whether you believe in that or not is not decisive..

Simple measurement and simple logic, clearly tells you, so just trust these.

 

[MRC_Hans]

This is in no way what the pattern shows, adn also not the way the way it normally is calculated.

You have shown this formula yourself. g=GM/r²
And it is the normal way to calculate this. Otherwise explain exactly what is wrong.

Off course I deny that,- also the prevailing theory deny that.
- /r² can only be use to find the ADG at the surface at (homogeneous) ball shaped body, and further out in space.
NOT inside the body.

It works everywhere, but inside a body, not all mass is below you, so you get pull in different directions. I'm sure you agree that if Earth was larger, but with the same mass, the surface gravity would be smaller. Since the gravity reaches us from inside Earth, this shows that the square rule is in power.

Maybe this is over your head, however.

As I wrote; the more an object deviates from a ball shape, the less gravity can it produce.

You wrote it, but it is wrong. The amount of 'gravity produced' ONLY depends on the mass. If you get far enough from a body, its shape makes no difference at all.

However, the gravity on any spot on the surface depends on the shape. For a given density, the shortest r is obtained in a ball shape. So you can get closest to the COG.

However, in the Domus Vista case, both the 'r' and the mass are given. So the only thing that can change withthe shape is the density, and in this case the ball shape is actually the one with the lowest density.

Hans we are talking about creating “ a depression in the space density”

No, you are talking about it. And it is nonsense. As shown by your own experiment.

The more a body deviate from a ball shape is certainly a significant reducing ADG,- and not only due to the distance to the COG get’s larger.

If Domus Vista would be a ball shape (and homogenous mass) and did had radius 50 meter to the COG, - YES we could use g=GM/r2 - But since this is not the case, the overall gravity pattern from that building is MUCH less .

Perhaps according to your idea, but not in reality. And this is confirmed by the data.

Hans

 

[Paulhoff]

You wrote it, but it is wrong. The amount of 'gravity produced' ONLY depends on the mass. If you get far enough from a body, its shape makes no difference at all.

So damn correct.

Paul

 

[Bjarne]

You have shown this formula yourself. g=GM/r²
The equation is not the one you can use, also not according to the prevailing thesis.
I don't know which one you must use, but I am sure it is pretty complex…

I have got the equation to calculate ADG different places on the surface of a container (from DTU). Such structure is much more simple..
But lost these, and could not find anything at the internet.

A container is close to a ball shape, but still there is a pretty huge “lost” compared to if it was a ball shape..

And it is the normal way to calculate this. Otherwise explain exactly what is wrong.

As i wrote above, - I have my intuitive understanding based on both the new theory, and measurement at MDmus Vista as well as the Ullerup measurement.

It works everywhere, but inside a body, not all mass is below you, so you get pull in different directions. I'm sure you agree that if Earth was larger, but with the same mass, the surface gravity would be smaller. Since the gravity reaches us from inside Earth, this shows that the square rule is in power.

If the Earth would be double so big, ADG would be 4 times weaker.
Inside the Earth and towards the center ADG would also only be 4 times less.
The reason to that is ; the less space there is available to the “same amount of mass”

Each time the radius of a circle doubles, then the area will quadruple.
However the area of the outer circle is only 3 times
that of the inner circle (75% larger).

Each time the square increases 75%, then the ADG is decreases 75%
This shows perfect proportionality between space / volume and ADG.

This proportionality cannot simply be a coincidence.​

It reveal that gravity is a space vacuum that is decreases
as a result of volume increase.

So already 300 years old mathematic reveals the cause/nature of gravity.

The way you measure ADG is both at a point but at the same time in a 3D perspective, because the result is the same everywhere at the surface of a ball shape (with same radius and of a homogeneous mass)

If the square is increasing 75 % ADG at the surface-ADG is decreasing 75%

As you can see does in fact reveal the principles between the “pressure on space” – the space consumption/vacuum and available space (inside a body).

You wrote it, but it is wrong. The amount of 'gravity produced' ONLY depends on the mass. If you get far enough from a body, its shape makes no difference at all.

It depends on the perspective.
If a building was reaching the whole way around the Earth, the buildings COG would be inside the Earth, and seen from outside the earth that mass would be parts of the Earth and therefore the effect almost 100% since the earth is almost a 100 % ball shape.

But if the building was floating in space, it would NOT produce that much gravity, as it would if it was a ball shape.
My friend this is also surprising me, BUT this is the consequences of understanding how gravity is created.

The more you can encircle space the more you can create gravity “once again”

Gravity is no longer a sum of the mass, - this old thinking only applies for 100%ball shaped bodies.

If the Earth had the shape of Domus Vista, gravity would only be some percent of what you are used to.

Imaging the Earth would be stretching to a million km long steel wire with the same mass (6E24Kg) and diameter 1 meter. - Imaging you was living at that “planet” and exactly at the COG.

This mean at the middle at the steel wire. – Do you believe that gravity here would be 6E24*6,67E-11/ 1 meter = 400000000000000 m/s – or how would you calculate ADG here ?

Do you believe that a moon would be orbit that "planet" ?

Do you belive gravity would be the same seen from a long distance ?

I believe we both agree ADG at that spot would be insignificant.
Hans attemp to understand what gravity does, - without understanding the basic nature of how gravity is created, - is doomed to fail.

However, the gravity on any spot on the surface depends on the shape. For a given density, the shortest r is obtained in a ball shape. So you can get closest to the COG.
However, in the Domus Vista case, both the 'r' and the mass are given. So the only thing that can change with the shape is the density, and in this case the ball shape is actually the one with the lowest density.

No the shape is extremely important, and so long you cannot understand this, you have totally failed to understand the nature of gravity.

 

[Paulhoff]

If one were R2 out from the earth's center, the gravity there would be the same even if the earth was the size of a black hole to any size up to R2.

Paul

 

[MRC_Hans]

You have shown this formula yourself. g=GM/r²
The equation is not the one you can use, also not according to the prevailing thesis.
I don't know which one you must use, but I am sure it is pretty complex…

I think you are being intentionally obtuse, now. I know it is the formula for a sphere, and I said as much. Since the Domus Vista has all its mass concentrated in a box much smaller than a 50m radius sphere, with that mass placed right under your measurement stations, the measured gravity will be be stronger. That is very simple.

As i wrote above, - I have my intuitive understanding based on both the new theory, and measurement at MDmus Vista as well as the Ullerup measurement.

Yeah, I know it's all in your imagination. Nice of you to admit it.

If the Earth would be double so big, ADG would be 4 times weaker.
Inside the Earth and towards the center ADG would also only be 4 times less.

I guess he can't grasp it.

This shows perfect proportionality between space / volume and ADG.

This proportionality cannot simply be a coincidence.​

It isn't a coincidence. It follows from the square law.

Gravity is no longer a sum of the mass, - this old thinking only applies for 100%ball shaped bodies.

Wrong.

If the Earth had the shape of Domus Vista, gravity would only be some percent of what you are used to.

Depends on the density.

Bjarne, I have shown you the numbers, based on your own data. I have shown you the math. Quite simple math. I can't help you any more than that.

Hans

 

[MRC_Hans]

If one were R2 out from the earth's center, the gravity there would be the same even if the earth was the size of a black hole to any size up to R2.

Paul

I don't think he wants to understand.

Hans

 

[Paulhoff]

I don't think he wants to understand.

Hans

And he still doesn't understand why GPS still works without his so-called understanding of gravity. If It wasn't understood to the level that it is, that system would fail within a day.

Paul

 

[MRC_Hans]

And he still doesn't understand why GPS still works without his so-called understanding of gravity. If It wasn't understood to the level that it is, that system would fail within a day.

Paul

Part of Bjarnes private fantasy is to imagine that everyone is as ignorant as himself.

He has to; otherwise it would dawn on even him that if a million scientists, after centuries of research, say he is wrong, then, quite likely, he is ...... wrong.

Hans

 

[Bjarne]

I think you are being intentionally obtuse, now. I know it is the formula for a sphere, and I said as much. Since the Domus Vista has all its mass concentrated in a box much smaller than a 50m radius sphere, with that mass placed right under your measurement stations, the measured gravity will be stronger. That is very simple.

You cannot use g=GM/r² in this case....

Not according to the prevailing thesis and not according to the new theory.

In thread 2459 you wrote;

That is actually wrong. The strength of the gravity field is determined by the mass, and only by the mass. The ADG at a given distance from the COG (outside the mass) is determined by the square of the distance (r²).
So, the surface gravity of an object will be higher if the density is higher, but ONLY because the surface is closer to the COG.

http://www.internationalskeptics.com/forums/showpost.php?p=6610616&postcount=2459

Hans these 2 statements are so wrong these possible can be.

Both according to the prevailing thesis, - as well as according to the new theory.

What you are trying to say must mean;
If you imaging you was living at a looong heavy wire with the same mass as the earth, say just few meter thick, and a HUGES length, - you would (as a consequence of your own statement ) be affected by a ADG larger as the speed of light, (if you was living near the COG, at the center of the wire).

Tell me honestly,, do you really believe in this ?

According to my opinion ADG everywhere on this wire would be insignificant, and probably even not possible to measure, doesn’t matter where on it you would live.
And there would also not be any significant ADG to measure far away from any angel of that wire.

The example is extreme, - but only to force you to face the simple point; that the more a body deviate from a ball shape, the weaker ADG you’ll get.

And this effect Domus Vista as well.

If you get far enough from a body, its shape makes no difference at all.

And also this statement emphasizes how poor you understanding really is.

The shape of a body is crucial for how strong ADG you'll get.

Off course the; “1 million km wire”, is only a nexample.
Doesn’t matter if you are close or far away from the wire, a such wire would “produce” insignificant gravity.

We can say that a wire like that has a extreme deviation from a ball shape. And therefore the ADG result is extremely reduces.

Also when looking / measuring the ADG of the wire from long distance...

The principal is the exact the same with Domus Vista, - if the Earth had that shape, and if we would say it is only 10% of a ball shape, - you would due to that only get 10% ADG of that reason.

But there is off course much more to take into consideration.

If you fail to understand gravity already at the basic level, you are not prepared to step 2, and even not at all to understand the limitation of the simple equation; g=GM/r² and why this ONLY applies for ball shape bodies.

So Hans, now pack together the loooooong wire, into a ball shape, - the more you’ll succeed to pack it to a ball shape, the more you will discover you are creating STRONG gravity, - both on short as well on long distance.

But so long the wire deviate A LOT form a ball shape, the strength of gravity will do the same.

After to have finish this basic lesson, go back to Domus Vista and ADMIT that we are dealing with gravity NOT MORE THAN 15µgal

OK amigo ?

 

[Paulhoff]

Part of Bjarnes private fantasy is to imagine that everyone is as ignorant as himself.

He has to; otherwise it would dawn on even him that if a million scientists, after centuries of research, say he is wrong, then, quite likely, he is ...... wrong.

Hans

The thing is, the rest of us should just exchange information that we have learned about the subject and not bother to respond to him anymore.

Paul

 

[MRC_Hans]

You cannot use g=GM/r² in this case....

Not according to the prevailing thesis and not according to the new theory.

You should stop making statements about real gravity, because you obviously don'r understand it. You are fre to make statements about your own fantasies, of course, although I doubt you understand that.

What you are trying to say must mean;
If you imaging you was living at a looong heavy wire with the same mass as the earth, say just few meter thick, and a HUGES length, - you would (as a consequence of your own statement ) be affected by a ADG larger as the speed of light, (if you was living near the COG, at the center of the wire).

Tell me honestly,, do you really believe in this ?

Of course not. Stop misinterpreting me. I will try once more, and if you don't get it, I'll conclude that you don't want to understand:

Take a given mass, say 50,000 tons. Imagine this is the mass of a ball with 50m radius. The surface gravity will be g=GM/r² = 1.35mGal.

NOW, squeeze this mass together to a flat box 100m long. It still has a mass of 50,000 tons, but now it is a narrow column, and you are standing on top of it, 50m from the COG as before. In this case, the surface gravity will be somewhat higher than for the ball. ... This is true for real gravity. I have no idea how it will work in your fantasy gravity.

According to my opinion ADG everywhere on this wire would be insignificant, and probably even not possible to measure, doesn’t matter where on it you would live.

Since you have here spread the mass out over a huge amount of space, yes, the gravity at any spot on the surface will be low.

And there would also not be any significant ADG to measure far away from any angel of that wire.

That, however, is wrong. All the mass is still there, and even if the wire is long, if you get, say, as far away as Pluto, the gravity field will be the same. Again, I'm talking about real gravity.

And this effect Domus Vista as well.

If you compare the building with a ball with the same mass and volume, then yes, the surface gravity is lower, but I don't; I compare it with a ball with the same 'radius' as the building.

If you don't get it now, you never will.

After to have finish this basic lesson, go back to Domus Vista and ADMIT that we are dealing with gravity NOT MORE THAN 15µgal

OK amigo ?

If that is your opinion, then please explain how you have measured 274µgal?

You have this litle problem: You OWN measurments contradict you, evey bit of the way.

Hans

 

[Bjarne]

Take a given mass, say 50,000 tons. Imagine this is the mass of a ball with 50m radius. The surface gravity will be g=GM/r2 = 1.35mGal.

NOW, squeeze this mass together to a flat box 100m long. It still has a mass of 50,000 tons, but now it is a narrow column, and you are standing on top of it, 50m from the COG as before. In this case, the surface gravity will be somewhat higher than for the ball. ... This is true for real gravity. I have no idea how it will work in your fantasy gravity.

I understand your point, but you do not understand that that building deviate “a lot” from a ball shape, and this simply mean less gravity (ADG).

Since you have here spread the mass out over a huge amount of space, yes, the gravity at any spot on the surface will be low.

The building is also spread over a larger distance.
The principal is the exact same as the wire example.
The more a body deviate from a ball shape = the less gravity.
A huge long steel wire, is only a extreme example that “should” turn on the light for you, to be able to discover the law / principle behind this.
The steel wire deviate more that 1 billion time, and the building “only” about 90 % (sorry I am too lazy to calculate exactly.).

Bjarne
And there would also not be any significant ADG to measure far away from any angel of that wire.
Hans
That, however, is wrong. All the mass is still there, and even if the wire is long, if you get, say, as far away as Pluto, the gravity field will be the same. Again, I'm talking about real gravity.

No there would be insignificant AGD from that wire, simply because a wire does not encircle space in a significant way.
This is so simple.
If shape of a body did not matter (as you stated) then ADG should be huge at the middle at the wire. – This is not the case.
Also not the case in the two ends of the wire either.

If you compare the building with a ball with the same mass and volume, then yes, the surface gravity is lower, but I don't; I compare it with a ball with the same 'radius' as the building.
If you don't get it now, you never will.

Off course I will NEVER accept it, because it is totally wrong.
The wire experiment should tell you that.
Gravity on that wire stretching to Pluto would in no way, and at no place could “carry” (attract) a cow, or even a mosquito.

If that is your opinion, then please explain how you have measured 274µgal?

What 274µgal?

Hans , honestly the door to understand the cause of gravity is widely open, all it require is only that you walk in of that door.
Do you self a favor try to understand why a ball shape is so unique and why the math-law of understating it is so nice and simple g=GM/r²
Doesn’t matter how big or small the Earth would be, the ADG at the surface would always follows that beautiful equation.

What it (indirectly) tells you is how much space is available inside the earth, - and the proportional affection by matter.

Try to understand the proportionality between “available inner space” and ADG, and you are on the right track.

Each time the square increases 75%, then the ADG is decreases 75%

This shows perfect proportionality between space / volume and ADG

And it happens so nice ONLY because space inside a ball shaped body is perfectly nice trapped by matter.

Just open you mind, and you eyes, and accept the big nothing (space) is a big “something” and most likely a big "elastic something".

It is hard for matter fixated people to imaging that everything are build of a substance you brain tells you is nothing, but as rather is “something”….beyond you imagination.

Matter does not only affect space. Matter is made of that space.

The more compact matter is , the stronger it is encircling space, - the space cumsumption is the same, - THIS is what the propotional stronger effect on that inner space shows you.
THIS is what the old beautiful equation of Sir Isaac Newton tells you.

So the mathematical evidence has been between us in 300 year.

Unfortunately only cows have really understood the full and true meaning of that beautiful equation (you now tries to disgrace) .

You know humans are so clumsy slow understanding, - cows are much more intuitive, elegant, logical and open minded.

 

[MRC_Hans]

The thing is, the rest of us should just exchange information that we have learned about the subject and not bother to respond to him anymore.

Paul

That might be an idea.

For instance, I now have a nice spreadsheet for calculating the surface gravity of objects. I have been speculating, in a Sci-Fi vein, how large a planet could be, and still have roughly Earth gravity, and actually I had expected that it would require some pretty exotic properties, but that is far from the case: A planet twice the radius of Earth, and nearly the same density, has 1G of gravity. But obviously, this function cannot be linear.

Hans