THE WASP-18b & Space Probes Mysteries
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Reality Check
Penultimate Amazing
Are you still on about your idea that computers let the beams hit the walls of the ring and then secretly adjust parameters to let the beams circulate?
This is easily seen to be dumb by considering the fact that the beams have enough energy (total beam energy at top energy, nominal beam, 362 MJ) to melt nearly 1 tonne of copper. This means that they have to have a special area to dump the beams after they are finished with them so that the ring is not destroyed.
If you want details about the LHC energy consumption then: LHC Energy consumption.
The 28 km ring is one of the rings in the LHC. It is used both to accelerate the particles and to store them for 10-24 hours. There is noly oone point that I know of in the ring where there is RF equipment to do the acceleration. The power consumption of the LHC is "between 220 and 300 megawatts of electricity".
The Wikipedia page supplies more detail about how the LHC accelerates and stores the 2 beams
Reality Check
Penultimate Amazing
I have some time to waste so I may as well address your ignorance.
A watt is the SI unit of power (remenber your high school physics?). It is Joules per second.
But there are other things that happen in a real system, e.g.
If it is to say that when you add gravity you still need the energy then you are wrong.
But the answer is zero - just attach the Moon with a rope to the Earth. Then there will be no energy needed to keep the Moon in its orbit.
I mean that the LHC web site states between 220 or 300 MW. Between means staring at the first vaule and going to the second value.
A watt is the SI unit of power (remenber your high school physics?). It is Joules per second.
There is no power needed in an ideal system to keep charged particles orbiting in a magnetic field.
But there are other things that happen in a real system, e.g.
- Synchrotron radiation is significant in the LHC system and the ring has been designed to absorb about 3.7 kW (this is in kilo watts, i.e. 1000 watts, i.e. 1000 joules per second). This means that the cooling system does not just cool the magnets down - it has to continue to cool them not only for ordinary heating but also to compensate for this extra heat.
P.S. I was wrong about the RF system having to compensate for this - that is applicable to electron-positron accelerators. - There are beam losses, e.g. a portion of the beam is lost in the collisions; particles diffuse toward the beam pipe wall and are lost. This means that a mechanism is needed to catch theses particles before they damage the ring. That is the various bean collimator systems. Thes need energy to run.
Magnetic fields continue to the end of the Earth (and further)
It IS a stupid question until you show your purpose.
If it is to say that when you add gravity you still need the energy then you are wrong.
But the answer is zero - just attach the Moon with a rope to the Earth. Then there will be no energy needed to keep the Moon in its orbit.
I have.
Reality Check can and has.
Excellent. I'm not sure how to describe this without resorting to a face and a palm.
No he didn't. From what I've read he just sort of assumed that once someone had unified the two long range forces the other two would be obvious.
Bjarne
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I am a bit busy today and have only time to ask one more question
How much energy (J) does it require to achieve a pull (in the cow) of 10 Newton (per second).
This must be basic physic that the big physicist: Reality Check certainly must be able to answer I hope.
If not who can crack the code ?
How much energy (J) does it require to achieve a pull (in the cow) of 10 Newton (per second).
This must be basic physic that the big physicist: Reality Check certainly must be able to answer I hope.
If not who can crack the code ?
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Zero Joules.
Basic physics, Bjarne. If your cow isn't moving, it doesn't matter how much force you're applying, no work has been done.
Reality Check
Penultimate Amazing
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.
For example:
Here is the growing list of the flaws and basic ignorance in Bjarne's ideas.
- "Bended space is Contracted Space"
Curved space is not contracted space
First Asked 11 November 2009 - 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
So far the answer is no
! - Bjarne's "space wind" has units of kg/ms
It is as badly defined as his "kinetic resistance".
- Does Bjarne know basic physics (unit-less quantities cannot be arbitrarily assigned units)
- 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
- Experimental proof that Bjarne "kinetic resistance" is wrong
- Energy is not conserved in Bjarne's universe!
First asked 26 November 2009 - Can Bjarne understand the physics behind orbital resonances?
First asked 3 December
So far the answer is no! - Does Bjarne understand orbital mechanics?
First asked 3 December 2009.
So far the answer is no - see the latest posts!
- Does Bjarne know that the Lorentz factor does not have any units of measurement?
First asked 22 November 2009
So far the answer is no! - Does Bjarne think that the Lorentz factor is measured in kilograms?
First asked 23 November 2009
So far the answer is no! - Why does relativistic mass have the units of mass
First asked 23 November 2009
For example:
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The dipoles and supply cables are superconducting. There are no resistive losses in them at all.
Whether you like it or not, the answer is that it's between 220-300MW. The exact amount will depend very much on the energy the machine is currently running at, the current it's running at, the particles it's using (the LHC will collide heavy ions as well as protons eventually), and many other things. Yes, these all can, and have been, calculated, but there isn't a single simple answer. More to the point, no-one here actually works there, as far as I know, so none of us know the exact answer.
Given that we're using absolutely standard SI units, you really shouldn't be guessing here. And no, Watts per second is not what anyone has meant, ever.
Once again, you appear to be rather severely confused. As others have already told you, it does not take any energy at all to keep an uncharged particle in orbit. Despite you nonsensical protestations to the contrary, no energy is spent keeping the Moon in orbit. That's 0W. 0J/s. Zero. None.
However, that is for an uncharged particle. Charged particle are slightly more awkward, since when they are accelerated they give off light - synchrotron radiation (cyclotron radiation in general, but we're dealing with highly relativistic particles here). It is this that causes the particles in an accelerator to lose energy as they pass through the dipole magnets. This would happen even in a linear accelerator (although obviously there would be much less since there would only be small corrections rather than large bending magnets), since it is nothing at all do with orbits.
As for where the energy ends, it ends, of course, as heat. As does pretty much all energy everywhere. In this particular case, it will end as heating in the accelerator vessel where it is hit by the synchrotron radiation.
Yes, he does. Unfortunately, you do not know enough of basic physics or maths, let alone accelerator physics, to understand the answers. Or apparently to count to 6.
Yes, someone else can answer as well. You won't understand my answers either.
No, you weren't wrong. All accelerators have an RF system to continuously provide power. The page you link to means the effects are unimportant as far as setting limits are concerned - the power required is easily achievable at the energies involved, but it still needs to be there.
The losses due to collisions are actually pretty much insignificant. The main contributions to losses are due to Tousheck scattering within bunches, scattering off residual gas (since you can never get a perfect vacuum), and effects due to the quantised nature of the synchrotron radiation. Diffusion is not relevant at all.
Actually, collimators are one of the few things that really don't need any energy at all. They're essentially just big lumps of metal that stick into the beam pipe for the particles to hit. Basically, you know that particles are going to get lost somewhere, so you make sure the most likely place for them to be lost is something that doesn't mind being hit. They will require cooling, and will have motors for adjustment, but the actual power requirements relative to everything else are pretty minuscule.
Bjarne
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Thanks for all the replies…
Who was saying it wasn’t move?.
The cow is dead, and the Landover is pulling it home.
It will take exactly 100 second to pull it home, and the resistance is all the time 10 Newton
So how many Joules does is takes for that mission?
I’ll come back to the other replies, but the one above is the most important question.
Who was saying it wasn’t move?.
The cow is dead, and the Landover is pulling it home.
It will take exactly 100 second to pull it home, and the resistance is all the time 10 Newton
So how many Joules does is takes for that mission?
I’ll come back to the other replies, but the one above is the most important question.
Then your question is unanswerable, and if you knew basic physics you'd know that.
W = F.d
where W is the work done (energy), F is the force, and d is the displacement. Time and speed are irrelevant unless the force is dependent on them.
Bjarne, if I have to explain such a basic equation to you, you shouldn't really be working with something as complex as gravity and spacetime.
MRC_Hans
Penultimate Amazing
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Bjarne, are you aware that asking such questions puts you in the league with a guy who wants to tell a mason how to build a house and then asks "what is mortar?"
However, energy can be expressed as force * distance * time. BTW a rather small cow you have there. One of your toy cows, perhaps?
Hans
Bjarne
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Off course it cost energy (Joule) to pull the cow home, - even a dead cow would not deny that, - But because no one of you can calculate it doesn’t mean I am stupid.
Let's look at an everyday’s example even a every blonde would agree too.
The condition bikes have a resistance of 10 Newton
How much many calorie / joule would it take to use it 100 seconds…
dafydd
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What is a condition bike? Use it for what? You are not making sense.
You didn't read what I wrote, did you?
We can't answer the question unless you tell us how far.
The time is irrelevant.
Read the "Work (physics)" article on Wikipedia.
If you're so sure there is an answer, why don't you tell us what you think it is?
EDIT (because apparently you're talking about an exercise bike): The machine might be stationary, but the wheel isn't. We need the radius at which the brakes are applied, and the angular speed of the wheel. The force is effectively moving along the surface of the wheel, and so the distance for the work equation is just the number of revolutions multiplied by the radius.
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It is a literal translation of the Danish word for an exercise bicycle.
Dancing David
Penultimate Amazing
Darn, it sounds like some philosophy term, condition bike.
dafydd
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He can't do the maths,I don't think that he can even add 2+2.