THE WASP-18b & Space Probes Mysteries

[Bjarne]

You gave MRC_Hans the amount of energy lost by every particle every turn due to your claimed kinetic resistance.
Your own words and calculations clearly state that a massive amount of energy should be lost every turn….

“turn” is not my words. But all the time deceleration = m/s

It takes less than a millisecond for all the particles to lose all their energy.

Really
Why ?

You back-pedalling fast now

I Never did, I am talking about deceleration .

Ok, so the 600MJ is pr turn not pr second…

It’s the KE the particles have archived after long time of acceleration.

Well, Bjarne is clearly not talking about turning off all the power, but only the power necessary to prevent the protons from being braked by his imaginary resistance to motion.

Finally one that understood.

I was hoping you'd say that!
Tell me - if the kinetic energy is changed to heat, what happens to a single proton travelling through space (or a particle accelerator)?

The strong force is responsible for atomic heat releas, - ( because this force keeps space contracted) = (the strong force keeps matter "together/contracted ").
The cause of gravity increase that emerges due to motion is different from how gravity normally works. I am not sure if this also realises heat.

I seems quite difficult to convince Bjarne that since his theory have giganormus effect on accelerators and it have not been noticed, there is most likely something wrong with his theory.

I am not convinced, and I still do not believe a accelerator is build to make such a scientifically test.

Why should it require energy to get a object to orbit?

The is not the point - but continuing keeping a object in a circula motion.

We can see here that when gravity radius of a orbiting satellites change from factor 4 to 1, then ADG increases 16 times. But the velocity only 2 times.

This mean that ADG is 8 times so strong compared to the satellites orbit velocity in orbit with radius 4..

Answer 1: Your ignorance of physics
F = ma = MmG/r^2. Thus a = MG/r^2. Thus ADG varies as the inverse radius squared

I can’t see you point either not the answer

F = ma – Why mention this, for what purpose ?
MmG/r^2 = Equation for calculating mass attraction, - it have nothing with the question to do , Why mention this, for what purpose ?
MG/r^2. = Equation for calculating ADG this is already calculated (factor 1 and 16) – there is no reason for : quote “Thus” between these two equation.

Why do you only confuse others ?
Why not calculated it, - when you really have the answer ?

 

[Toke]

You don't believe an accelerator is built to test your theory, ok.
But your theory would mean that it required a power supply several orders of magnitude higher.

You are claiming a resistance of 600MJ pr second or turn.

If we just use the pr second that is a whole power plant of continuous effect (600MW is a coal fired power plant)
The 700MJ of kinetic energy in the beam is achieved over days, do you see the difference between that and 600MJ/s?

And you have not explained where all that energy goes, 600MW does not simply disappear.

I have once tried that comparing KW and A on the switchboard gave 1500A unaccounted for. There were no fire alarms* so we assumed there were something wrong with cos phi on a regulator. Then 2. engineer came running in and reported smoke from generator 2. It was as assumed the regulator that had burned off.

*That kind of power does not go missing without some sparks.

 

[Reality Check]

F = ma – Why mention this, for what purpose ?
MmG/r^2 = Equation for calculating mass attraction, - it have nothing with the question to do , Why mention this, for what purpose ?
MG/r^2. = Equation for calculating ADG this is already calculated (factor 1 and 16) – there is no reason for : quote “Thus” between these two equation.?

FYI Bjarne:
F = ma is the law relating mass and acceleration to force in general.
F = GMm/r^2 is Newton's law of gravitational force.

Notice that F in both equations?
That means that you can put
F = ma = F = GMm/r^2
or
ma = GMm/r^2
Thus
a = GM/r^2 where a is what you call ADG.

That is the answer to your question: Gravitational force varies as 1/r^2. Thus ADG varies as 1/r^2.

 

[steenkh]

“turn” is not my words. But all the time deceleration = m/s

No. m/s is a constant speed.

(Hint: meters per second per second is acceleration or deceleration)

I Never did, I am talking about deceleration .

You were told to calculate the resistance, but now you are claiming that that is not what you calculated? Why did you specify it in Joules?

It’s the KE the particles have archived after long time of acceleration.

You were told to calculate the resistance to constant movement for one proton, and when multiplied by the number of protons, this is the force that is needed to prevent the protons from braking.

Finally one that understood.

Yep. Your theory is thoroughly busted!

 

[Reality Check]

Bjarne's "space wind"

We already know that Bjarne's "kinetic resistance" cannot exist.
The first reason is that his attempt to define it mathematically is pathetic. He defines it as the Lorentz factor (a number) minus 1 (a number) and is so ignorant of physics that he thinks that he can assign this number the units of acceleration just because he want it to have the units of acceleration.
Does Bjarne know basic physics (unit-less quantities cannot be arbitrarily assigned units) First asked 23 October 2009

The second reason is that the fact that particle accelerators work as deigned using the known laws of physics means that "kinetic resistance" cannot exist. For example: when beams are stored in particle rings there should be extra energy needed to keep them there (initially the power bill should be double the expected amount).
Experimental proof that Bjarne "kinetic resistance" is wrong
First asked 2 November 2009
Why particle accelerators disprove Bjarne's impossible "kinetic resistance" (the above post expanded).

Now lets look at "space wind":

A = RM/Qr^2

R = The strength of the rotating SW (Example: The circumference of the Earth's orbit (around the Sun) divided by the number of second one rotation of the Sun takes) => SW velocity in the Suns orbit. (m/s) (angel velocity)
M = Mass (primary) of the rotating object's).
Q = Rotation Force Constant 6.52* 10^21 (temperary')
r^2 = Distance Square

The first point is his continued ignorance of basic physics. Plug in the units for his variables (R = m/s, M = kg, Q = none, r = m) and you get kg/ms. This is not an acceleration.

Bjarne,
Do you know that kg/ms is not an acceleration?

The second point is that he is quite willing to plug in an constant (Q) into his space wind equation - but not into his "kinetic resistance" equation.

Bjarne,
Where is your constant for "kinetic resistance"?

 

[Reality Check]

Bjarne's "space wind"

duplicate

 

[Bjarne]

------

There is a Force. There is Energy. No Energy is required to bend the paths of objects into orbits. A force is required.

According to GR gravity is caused to curvature of space, not to a force..
Or an object is not following an orbit because of a force or because of energy, but because of a property of space.

According to Newtonian gravity, you are right. Gravity is a force. = F
You are not only ridding two horses, but a bull and a horse at the same time.
How is this possible?

1. Is gravity a force or is gravity not a force?
2. And if gravity is a force, how can a force pop up from space do you think ? - I mean without energy ?

If we just use the pr second that is a whole power plant….

No this is not what I am saying
Based on the Lorenz equation you can calculated how much KE must increased to get a gradually diminishing increase of speed. – This is a process in time.
Deceleration is according to my idea, the exact opposite, still a gradually deceleration in time.
To prevent that to happen you only need to replace the KE lose per second. - Which mean - per the first second.- That’s the point.

And you have not explained where all that energy goes, 600MW does not simply disappear.

I have. All motion set space into motion.
This means you affect the whole Universe (with the speed of light) also even though the motion it self is much slower. The energy (disturbance of space) continues to continue through space, - like a magnetic field does.

Where is your constant for "kinetic resistance"?

You don’t need one, the question is answered above…

Do you know that kg/ms is not an acceleration?

The idea is that MG/r^2 ( or M/Q/r^2) is a elastic space “connection” – that you can compare with a gliding clutch, or spinning wheels.

R = the speed of rotating deformed space.. Like towing a car in a elastic rubber...

This equation is so fare only a: "working drawing", or a line on the blackboard.
It will be approved, don’t worry

FYI Bjarne:
F = ma is the law relating mass and acceleration to force in general.
F = GMm/r^2 is Newton's law of gravitational force.

Notice that F in both equations?
That means that you can put
F = ma = F = GMm/r^2
or
ma = GMm/r^2
Thus
a = GM/r^2 where a is what you call ADG.

I can only agree to that
But I still do not think you have got the question right?

At radius 1 ADG is 16 times as strong as at radius 4.
It’s not a question why. I know why.

You can also see that the objects continue to orbit at both radius 1 and radius 4.
This is also not so strange, because object at radius 1 has 4 times so much KE and 4 times so little time to change the course (into a continues circular orbit) hence ADG must be 4*4 times so strong, = ADG 16 at radius 1,- to keep the object in circula motion.

BUT the point is:that the speed of the object at radius 1 is only *2 - even though that gravity ADG is *16 – WHY.

It seems that there is a high price to pay for keeping the object in orbit, and that price is lower orbit speed. – Right ? - Or what's going on here ?

------

 

[Reality Check]

According to GR gravity is caused to curvature of space, not to a force.

And we interpret that curvature of the path of the object as a force.

BUT the point is:that the speed of the object at radius 1 is only *2 - even though that gravity ADG is *16 – WHY.

BUT the point is: that the speed of the object is a speed and the ADG of the object is an acceleration.
Can you tell the difference?

It seems that there is a high price to pay for keeping the object in orbit, and that price is lower orbit speed. – Right ? - Or what's going on here ?-

Wrong.
There is no price at all to pay for keeping the object in orbit.
Putting an object in orbit has a price.
Changing the orbit of an orbiting object has a price.

 

[Reality Check]

The growing list of the flaws and basic ignorance in Bjarne's ideas

Here is the growing list of the flaws and basic ignorance in Bjarne's ideas.

1. "Bended space is Contracted Space"
   Curved space is not contracted space
   First Asked 11 November 2009
2. Physical quantities can be arbitrarily assigned units of measurement, e.g. an equation that returns numbers (no units) can be given units of m/s^2 or coulombs or kilograms just because Bjarne wants it to return acceleration or charge or mass.
   • Does Bjarne know basic physics (unit-less quantities cannot be arbitrarily assigned units)
     First asked 23 October 2009
   • Bjarne's "space wind" has units of kg/ms
     It is as badly defined as his "kinetic resistance".
3. Anything Bjarne states is right even if there is experimental evidence that shows that it is wrong.
   • Experimental proof that Bjarne "kinetic resistance" is wrong
     First asked 2 November 2009
   • Why particle accelerators disprove Bjarne's impossible "kinetic resistance" (the above post expanded).
   • Bjarne's idea is proven wrong (acceleration due to gravity is constant inside a sphere).
     First asked 2 November 2009
4. Energy is not conserved in Bjarne's universe!
   First asked 26 November 2009
5. Can Bjarne understand the physics behind orbital resonances?
   First asked 3 December
   So far the answer is no !
6. Does Bjarne understand orbital mechanics?
   So far the answer is no - see the latest posts !

And the questions he he cannot answer (basically about units of measurement):

• Does Bjarne know that the Lorentz factor does not have any units of measurement?
  First asked 22 November 2009
• Does Bjarne think that the Lorentz factor is measured in kilograms?
  First asked 23 November 2009
• Why does relativistic mass have the units of mass
  First asked 23 November 2009

 

[MRC_Hans]

“turn” is not my words. But all the time deceleration = m/s

No, 'turn' was part of the question. I specifically asked you the calculate the loss of energy per protron per turn. I even mentioned that the lenght is 27km. If you did not give an answer that reflected the question, it is only your own fault.

But, let us take it again, then: What is the energy loss due to your hypothetical 'resistance against motion' in the Hadron Collider, per protron? You can state the answer in J per distance or in J per time (since we know the speed, they are convertible).

The necessary data for the Hadron Collider can be found here: http://en.wikipedia.org/wiki/Large_Hadron_Collider

It’s the KE the particles have archived after long time of acceleration.

We can look that up. It is the putative loss you need to find.

The strong force is responsible for atomic heat releas, - ( because this force keeps space contracted) = (the strong force keeps matter "together/contracted ").
The cause of gravity increase that emerges due to motion is different from how gravity normally works. I am not sure if this also realises heat.

A particle cannot have a temperature. If a particle looses energy, it will give the energy off in the form of radiation.

I am not convinced, and I still do not believe a accelerator is build to make such a scientifically test.

It may or may not be built to make the test, but the point is that according to your own figures (and this seems to be the case no matter how the numbers are massaged), the effect would be impossible to overlook. Not only would it lead to massive energy losses, but that energy would bleed off in the form of some kind of radiation, and since the place is full of various radiation detectors, it is virtually impossible that such an unexpected radiation would not be discovered.

Why do you only confuse others ?

Bjarne, at this point, the only confused one appears to be you.

Why not calculated it, - when you really have the answer ?

There is nothing to calculate, it has all been done already. Where do you think the figures you use for orbital speeds come from? OK, you looked them up somewhere, but how do you think they were created? I'll tell you how: Someone calculated them using the formulas RC showed you.

Hans

 

[MRC_Hans]

You don't believe an accelerator is built to test your theory, OK.
But your theory would mean that it required a power supply several orders of magnitude higher.

You are claiming a resistance of 600MJ pr second or turn.

If we just use the pr second that is a whole power plant of continuous effect (600MW is a coal fired power plant)

Unfortunately, it is much worse: The particles go about 11,000 revolutions per second, so the continuous power is to the tune of 6,6TW!

That is exactly the reason I brought up the Hadron Collider in the first place, in this debate: Almost no matter how infinitesimal Bjarne's 'resistance to movement' is claimed to be, it would have to show up big time in such an instrument. As you can see, Bjarne has to get to out somewhere around the 20th decimal before he might claim that the loss is not noticeable, by which time it would be unable to account for even marginal effects like the Pioneer anomaly.

So the purpose of this exercise has been to prove that Bjarne's idea lacks coherence. I think we have thoroughly succeeded.

The 700MJ of kinetic energy in the beam is achieved over days, do you see the difference between that and 600MJ/s?

Well, over hours, still....

And you have not explained where all that energy goes, 600MW does not simply disappear.

Especially since a particle in flight can only give off energy in the form of radiation.
So, 600MW (or more) of radiation...

... And Bjarne claims nobody noticed it.

Hans

 

[Bjarne]

---

No, 'turn' was part of the question.

You was asking : “Per omgang” - = per round.
If first the top-speed is reached one round doesn’t add more KE, such question is therefore meaningless.

I understand “round” as the total process of acceleration to reach top speed, to me only such question makes sense.

Hence my answer was that the proton at 2.99999997E8 m/s, - have total resistance is KE = 0,000001064J per proton, - but the KE lost due to free deceleration is much less.

Before continuing this part:

1. Let us finish the question regarding the Moon and its KE orbit etc.
2. Answer the question what was the 150 MJ used for
3. What does that ,mean: Quote: - I've already told you what will happen. It takes less than a millisecond for all the particles to lose all their energy.
4. How many Joules does it require to get X number (you can decide) of particles for the accellerator to circulate with the mention speed? (99.9999991% af c = 2.99999997E8) included all the lost ?
5. Mention all kind of KE lost in the accelerator.

And please mention sources.

Unless we have so much fact as possible on the table, it’s meaningless to discuss something what is difficult for none specialist to distinguish what is fact and bogus.

I mean even if you ask specialist in CERN, they properly would answer you what they believes:. I mean if I today would ask 1000 specialist in CERN if the Higgs exists, I would properly get a average answer that this is 99,1 % sure.
But 3 years after (if still not is found) the answer would be 0,00001% sure.
Like if fusion is posisble? - 50 years ago everyone belived it could happen almost in a a wash machine. Today peoples believe differently.
Believes is too often “certain facts” today, - but certain bogus tomorrow.
Hence lets now only get the cool facts on the table, as also mean answer the question above..
And please let's finish the question conserning the Moons KE first.....

BUT the point is: that the speed of the object is a speed and the ADG of the object is an acceleration.
Can you tell the difference?

It’s not a question about differences between the equations.
It’s an question of KE lose: I will get back to that point.

Changing the orbit of an orbiting object has a price.

Imaging you was living on the moon, and suddenly the moon would not orbit around the Earth,.

This mean the moon would move away from the Earth if nothing would be done.
Now imaging you should force the moon to continue to change its orbit in order to continue to orbit around the Earth? You would use a huge racket engine to that purpose.

Let’s say that the Moons speed is 1000 m/s and its KE is 3E29 J
How much energy (J) should you use per second to keep the Moon in Orbit?
---

 

[steenkh]

You was asking : “Per omgang” - = per round.
If first the top-speed is reached one round doesn’t add more KE, such question is therefore meaningless.

No. According to you there is a continuous resistance to motion, not just on the first round. This is why you asked to calculate how much energy was lost per round, so that we could know how much would be needed to be added per round, so keep the particles going.

I understand “round” as the total process of acceleration to reach top speed, to me only such question makes sense.

Well, you are wrong. The particles are brought to full speed before they enter the ring. Your claim is about resistance to constant motion: why are you so interested in knowing how much it costs to bring the particles up to speed? I thought your "theory" and real physics would agree on that part.

Hence my answer was that the proton at 2.99999997E8 m/s, - have total resistance is KE = 0,000001064J per proton, - but the KE lost due to free deceleration is much less.

Why all the mumbo-jumbo? Just state how much kinetic energy a proton moving at close to the speed of light is losing to your mysterious resistance. But be careful: if you set it too low, you will no longer be able to explain all the other effects of your "theory" that you are so fond of!

Let us finish the question regarding the Moon and its KE orbit etc.

No. Let us finish this simple example of protons. It is very easy to calculate, and the space wind cannot cancel out the effect.

 

[MRC_Hans]

-
You was asking : “Per omgang” - = per round.
If first the top-speed is reached one round doesn’t add more KE, such question is therefore meaningless.

I understand “round” as the total process of acceleration to reach top speed, to me only such question makes sense.

Hence my answer was that the proton at 2.99999997E8 m/s, - have total resistance is KE = 0,000001064J per proton, - but the KE lost due to free deceleration is much less.

Now you are outright ridiculous: The possible difference in the interpretation of the word 'round' doesn't come into it AT ALL; because we had that conversation on the Danish forum, and 'omgang' is not ambiguous. Also, I added the information that one 'omgang' is 27km, so you should not be in doubt.

However, you are now saying that your first figure was in error. Whatever the reason, fine! Now please calculate the correct number. Per metre, per 27km round, per second, per nanosecond, or whatever you want.

Let us finish the question regarding the Moon and its KE orbit etc.

What question?

Answer the question what was the 150 MJ used for

Which 150MJ?

1. What does that ,mean: Quote: - I've already told you what will happen. It takes less than a millisecond for all the particles to lose all their energy.
2. How many Joules does it require to get X number (you can decide) of particles for the accellerator to circulate with the mention speed? (99.9999991% af c = 2.99999997E8) included all the lost ?
3. Mention all kind of KE lost in the accelerator.

And please mention sources.

Nice try, but unfortunately soo plain: You are the one who have to provide answers here. Look in any recent RC post for the list of questions.

Your questions about the moon are completely nonsensical.

Hans

 

[Bjarne]

You are the one who have to provide answers here.

You are the one who have to provide answers here.

The accelerator is not my business, and so long one answers the question I will not make any further comments regarding this, before the questions are replied.

I mean just read above how much a answers can be misunderstood, and turned opposite, also after I have explained how I did understand the question.
My reply is very clear, and it's also clear that the last 5 month I have only speak about deceleration. - Not that all the KE of e.g; the Earth or the Moon should be consider like resistance, but rather between 1/10.000.000.000 and 1/400.000.000.000 in the first second.
It "should" not be possible to misunderstand that basic claim today, for any of you Danes.
I mean I ask if I you can you: please calculate the KE per round of one Planet, and the next I do is to drop all that KE above you head like an atom bomb. – This is not fair...

Now it is my turn to ask question.

My attention is now: to prove what I am saying in my way.
Decisive new view on all that is prepared and coming soon.

 

[steenkh]

You are the one who have to provide answers here.

Really? You are claiming something that is contrary to everything that is known in physics, and you think that it physics that has got to provide answers? Think again!

The accelerator is not my business, and so long one answers the question I will not make any further comments regarding this, before the questions are replied.

Loser!

 

[Cuddles]

“turn” is not my words.

This has already been covered by others, but I'll address it anyway. You may not have used the word "turn" (which is again rather telling since that's what anyone involved in accelerator physics will call it), but regardless of what you choose to call it, that is what you were talking about. You were asked to calculate the energy lost by a single particle due to your kinetic resistance in one turn. You did that, or at least gave a number in answer to the question. Your answer clearly states that the energy required to keep particles in motion in an accelerator is at least an order of magnitude higher than is actually observed. Either your answer was wrong, or your entire claim was wrong. Neither would reflect well on you. Personally I rather suspect it is both.

But all the time deceleration = m/s

Firstly, deceleration is not measured in m/s. Secondly, as I have already noted a couple of times, deceleration is not measured at all in particle accelerators, because at the relativistic velocities we are dealing with the velocity is always indistinguishable from the speed of light. It's constant. There is no deceleration. What is important is the loss of energy.

Really
Why ?

Because that's how fast they lose their energy. Seriously, you really need to stop talking about things that you not only don't understand, but have clearly never made any attempt to learn anything about. Misunderstanding science and pretending to understand it is bad enough, but you haven't even bothered to gain the misunderstanding first, you just make nonsense up and appear to be surprised when your imagination doesn't have anything in common with reality.

It’s the KE the particles have archived after long time of acceleration.

No it isn't. It's the number you claimed was the loss per turn, multiplied by the number of particles. If you disagree with that, why did you give the wrong number in the first place?

The strong force is responsible for atomic heat releas

No it isn't.

I am not convinced, and I still do not believe a accelerator is build to make such a scientifically test.

This is because you are, quite simply, delusional. You understand exactly nothing about accelerators, and you simply ignore everything you are told by those who do. If you are not interested in a discussion, please stop pretending that you are. You cannot simply dismiss the answers you get and pretend they were never given. Your nonsense is disproved very easily, and that won't change just because you stick your fingers in your ears.

Why do you only confuse others ?

He doesn't. You are the only person that is confused.

 

[Toke]

Thanks Hans,
6,6TW not 600MW.
Anyway, I noticed that the discussion of whether the accelerator instruments would pick up Bjarne's resistance were focused on 600MJ not 600MJ/turn or second.
That makes details of instrumentation rather irrelevant compared to the question of power supply.

 

[Bjarne]

It's the number you claimed was the loss per turn

Here is my answer:
Everyone can see I have not one answer but 4 answers.

1.) Proton Mass 1.6726211154E-27
2.) Hastighed = (speed) 99.9999991% af c = 2.99999997E8
3.) KE = 0,000001064J
4.) Resistance 7070 m/s

Answer 3.
Everybody that know just a little about what we are talking about can immediately see that the KE relates to the (top speed) mentioned

I did not mention:

• turn / round omgang or what so ever.
• acceleration
• time
• or anything also that could justify a calculation related to that.

Answer 4.
The answer was how much resistance there is per "omgang"
Here my answer start with the word resistance, but as m/s
This mean that a proton will loos 7070 Meter there first second, - of course with reference to the Lorenz Transformation, that we have discuss several weeks here and several Month together with the Dane.

If someone really should believe that I have change my mind now, and believes that resistance of the moons orbit speed (that I guess the most of you fellows have discovered is calculated to 5.77* 10-12 m/ the first second.) correspond to 20.000.000.000.000.0000.000.000 Mega ton atom bombs per second, - then I am off course terrible sorry, - This was really not my intention.

But I rather can see that some like to misunderstand, and to make a huge point of that (nothing), and also to continue to do so , instead just to ask, does this mean this or that..’

Left to say is only: thanks for coffee and for hairsplitting.

Have a nice day fellows.

---

 

[Toke]

Resistance to movement as m/s ?