First, I calculated that y = x2 / 2r if it is to approximate the effect of a round Earth at short distances (r being the radius of the RE). This means that dy / dx = x / r and therefore dy / dt = (x / r) * (dx / dt). Substituting x = f(t) and y = g(t), you get g'(t) = f(t) * f'(t) / r.And the solution:
Next, if the speed is to remain constant, then by Pythagoras' theorem f'(t)2 + g'(t)2 = c2. Squaring the above equation, c2 - f'(t)2 = f(t)2 * f'(t)2 / r2. Rearranging this gives:
(http://i33.tinypic.com/fmjpeq.png)
f2 (df/dt)2 = r2 (c2 - (df/dt)2)
f2 = (rc)2 (dt/df)2 - r2
(f2 + r2)0.5 = +/- rc (dt/df)
Integrate (http://integrals.wolfram.com/index.jsp?expr=sqrt(x^2+%2B+r^2)&random=false) w.r.t. f to get the equation in blue.
f * (f2-r2)0.5 - log[ f + (f2-r2)0.5] = +/- 2 r c t
I used x(t) and y(t) instead of f(t) and g(t), just because it is more descriptive.
We know (if I understand EAT correctly):
y'(t)2 + x'(t)2 = c2 (speed of light is constant)
y''(t) = a (light accelerates upwards constantly with acceleration a)
y(0) = 0, x(0) = 0, y'(0) = 0 (initial conditions)
y''(t) = a => y'(t) = at => y(t) = at2/2 This is assuming the EA is constant and directed upwards.
So, by pythagoras: (at)2 + x'(t)2 = c2
Re-arrange for x'(t) to get a difficult integral, type into the online integrator (http://integrals.wolfram.com/index.jsp?expr=sqrt(c^2-(ax)^2)&random=false) to get:
x(t) = (c2/2a) arctan (at/{c2-(at)2}0.5) + (t/2)(c2-(at)2)0.5
There is already a thread for this. (http://theflatearthsociety.org/forum/index.php?topic=21912.0) It is locked temporarily because it was being trolled. You are officially FES's most redundant citizen. To summarise: in b4 teh lock!
Robosteve hasn't really figured out how the EA works, so atm doing any proper maths is impossible.
Here are some speculative calculations:First, I calculated that y = x2 / 2r if it is to approximate the effect of a round Earth at short distances (r being the radius of the RE). This means that dy / dx = x / r and therefore dy / dt = (x / r) * (dx / dt). Substituting x = f(t) and y = g(t), you get g'(t) = f(t) * f'(t) / r.And the solution:
Next, if the speed is to remain constant, then by Pythagoras' theorem f'(t)2 + g'(t)2 = c2. Squaring the above equation, c2 - f'(t)2 = f(t)2 * f'(t)2 / r2. Rearranging this gives:
(http://i33.tinypic.com/fmjpeq.png)f2 (df/dt)2 = r2 (c2 - (df/dt)2)
f2 = (rc)2 (dt/df)2 - r2
(f2 + r2)0.5 = +/- rc (dt/df)
Integrate (http://integrals.wolfram.com/index.jsp?expr=sqrt(x^2+%2B+r^2)&random=false) w.r.t. f to get the equation in blue.
f * (f2-r2)0.5 - log[ f + (f2-r2)0.5] = +/- 2 r c t
Using a different set of postulates, it works out like this:I used x(t) and y(t) instead of f(t) and g(t), just because it is more descriptive.
We know (if I understand EAT correctly):
y'(t)2 + x'(t)2 = c2 (speed of light is constant)
y''(t) = a (light accelerates upwards constantly with acceleration a)
y(0) = 0, x(0) = 0, y'(0) = 0 (initial conditions)
y''(t) = a => y'(t) = at => y(t) = at2/2 This is assuming the EA is constant and directed upwards.
So, by pythagoras: (at)2 + x'(t)2 = c2
Re-arrange for x'(t) to get a difficult integral, type into the online integrator (http://integrals.wolfram.com/index.jsp?expr=sqrt(c^2-(ax)^2)&random=false) to get:
x(t) = (c2/2a) arctan (at/{c2-(at)2}0.5) + (t/2)(c2-(at)2)0.5
The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?
I see no maths...
The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?
I see no maths...
When there is maths to post, it will be posted in that thread. Again, redundant thread is redundant.
There is already a thread for this. (http://theflatearthsociety.org/forum/index.php?topic=21912.0) It is locked temporarily because it was being trolled. You are officially FES's most redundant citizen. To summarise: in b4 teh lock!
The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?
First off, I don't believe FE or EAT, I just find this an interesting theory.
You say the mineshaft was straight. How was this straightness measured? Spirit level? A beam of light? Looking down the length of it and seeing it is straight?
Over 1 mile light wouldn't deviate enough to be significant.
Over 1 mile light wouldn't deviate enough to be significant.
OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.
(http://i12.photobucket.com/albums/a239/ghazwozza/EAT-test.png)
A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).
Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.
FE+EAT predicts A will arrive first. RET predicts B will arrive first.
OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.
A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).
Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.
FE+EAT predicts A will arrive first. RET predicts B will arrive first.
Of course, just setting up the experiment would give you the answer about which one was longer. Just measure the amount of cable on each "leg" of the experiment as you string it up.
Here's another approach regarding the OP. Why don't you just Lorentz transform the straight-line path of light in an inertial frame to an accelerating frame?
Here's another approach regarding the OP. Why don't you just Lorentz transform the straight-line path of light in an inertial frame to an accelerating frame?
I do believe you may have simplified the necessary calculations quite considerably, good sir! :o
Using the conditions of the height of the sun, ~4800km
OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.
(http://i12.photobucket.com/albums/a239/ghazwozza/EAT-test.png)
A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).
Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.
FE+EAT predicts A will arrive first. RET predicts B will arrive first.
I don't think this will work. If FE+EAT is correct, then the light in A will keep getting bent upwards and reflected back down away from the edge of the optic fibre. This will result in a path that zigzags through the fibre optic. Of course, the same effect will be observed if RET is true and the light tries to follow a straight line through a curved optic fibre. This zigzag effect will probably be of greater significance to the time it takes the light to go from one end to the other than will the shape of the Earth.
Umm...the concept of fiberoptic cables is that the difference in refractive index between the cable and the medium is such that light will bounce off the edges of the inside of the cable as it travels down. All light in fiberoptic cables travels in a zigzag fashion. In fact, if light were to bend, long-distance fiberoptic cables could not work as eventually the light would be too angled to continue reflecting. It would reach a point where its vector relative to the edge of the fiber is so direct that it would not reflect. Thank goodness light doesn't bend. Otherwise we wouldn't have fiberoptic technology.
Honestly, the calculations haven't simplified much at all. I've been messing around with the equations, and I obtained an expression for the path of light in an accelerated reference frame. The path is not parabolic and the expression is very complicated.
Using the conditions of the height of the sun, ~4800km, and the radius of the spotlight, ~10000km, I'm trying to solve for the required acceleration. It's going to require some numerical approximations, and I don't know if I get a precise value or not. So far it, just by playing with numbers, it seems that an acceleration of around 10^10 m/s^2 will be required.
Umm...the concept of fiberoptic cables is that the difference in refractive index between the cable and the medium is such that light will bounce off the edges of the inside of the cable as it travels down. All light in fiberoptic cables travels in a zigzag fashion. In fact, if light were to bend, long-distance fiberoptic cables could not work as eventually the light would be too angled to continue reflecting. It would reach a point where its vector relative to the edge of the fiber is so direct that it would not reflect. Thank goodness light doesn't bend. Otherwise we wouldn't have fiberoptic technology.
You misunderstand the EA, and you fail to acknowledge that in a perfectly straight optical fibre light (if it travels in a straight line) would not need to zigzag to get to the other end.
It is only theoretically possible for a photon to enter a fiberoptic cable in perfect parallel to the cable (or perfectly orthogonal to the beginning of the cable). Because it is not practically possible, all light will zigzag inside of a fiberoptic cable - even a perfectly straight one. Suppose, however, that you do manage to get a photon to do this. It would be undetectable in relation to the numerous photons that will follow the "zigzag" pattern. The next argument would be LASER light, as all the photons should propagate parallel to each other. However, we do not have the technology to ensure perfectly ideal orthogonality. As such, even the LASER photons would be refracted, even if only slightly, upon entering the optic cable. Even supposing you managed a test where you did get perfect orthogonality, it would be virtually impossible to repeat. You simply cannot achieve such perfect conditions that this could work. Even vibrations of the earth or a single dust particle or even a handful of large molecules would cause light scattering so that perfect orthogonality could not be achieved. This would occur even if the LASER was in contact with the cable inside of the best vacuum achievable with our technology. Not to mention the variance in the synchronization of the LASER or tolerance level for non-simultaneous emission.
It is only theoretically possible for a photon to enter a fiberoptic cable in perfect parallel to the cable (or perfectly orthogonal to the beginning of the cable). Because it is not practically possible, all light will zigzag inside of a fiberoptic cable - even a perfectly straight one. Suppose, however, that you do manage to get a photon to do this. It would be undetectable in relation to the numerous photons that will follow the "zigzag" pattern. The next argument would be LASER light, as all the photons should propagate parallel to each other. However, we do not have the technology to ensure perfectly ideal orthogonality. As such, even the LASER photons would be refracted, even if only slightly, upon entering the optic cable. Even supposing you managed a test where you did get perfect orthogonality, it would be virtually impossible to repeat. You simply cannot achieve such perfect conditions that this could work. Even vibrations of the earth or a single dust particle or even a handful of large molecules would cause light scattering so that perfect orthogonality could not be achieved. This would occur even if the LASER was in contact with the cable inside of the best vacuum achievable with our technology. Not to mention the variance in the synchronization of the LASER or tolerance level for non-simultaneous emission.
Okay, fair point. Still, the zigzag effect would be more pronounced in a curved optical fibre than a straight one.
Congratulations on your first concession (at least that I have ever witnessed). As to the more pronounced effect, that seems intuitively correct, but I seem to recall that one of the nifty features of a fiberoptic cable is that it would actually average out so that the amount of time it takes light to travel a severely coiled fiberoptic cable (or any random pattern) is exactly the same as light traveling another cable of the exact same length (worked out mathematically, not experimentally). I cannot remember the details, as I am nearly certain the phrae "averaging out" conveys the wrong concept. I did this research about 4 years ago while preparing a patent. I was considering the use of fiberoptic cable in the design of an instrument. The design was abandoned for something that put the light source and detector in nearly direct contact with the sample, eliminating the need for fiberoptics, so my study of fiberoptics was only tangential to my work and not done with vigor.
Congratulations on your first concession (at least that I have ever witnessed). As to the more pronounced effect, that seems intuitively correct, but I seem to recall that one of the nifty features of a fiberoptic cable is that it would actually average out so that the amount of time it takes light to travel a severely coiled fiberoptic cable (or any random pattern) is exactly the same as light traveling another cable of the exact same length (worked out mathematically, not experimentally). I cannot remember the details, as I am nearly certain the phrae "averaging out" conveys the wrong concept. I did this research about 4 years ago while preparing a patent. I was considering the use of fiberoptic cable in the design of an instrument. The design was abandoned for something that put the light source and detector in nearly direct contact with the sample, eliminating the need for fiberoptics, so my study of fiberoptics was only tangential to my work and not done with vigor.
I see. Well, if this is true, then the experiment would certainly work. Is there anyone here with more experience dealing with optic fibre?
This zigzag effect will probably be of greater significance to the time it takes the light to go from one end to the other than will the shape of the Earth.
So I have a bending light math question. In the FE, light makes it from the sun to the earth in as little as .01 seconds to around .04 seconds. How do we not notice the EA as it would have to be super strong?
So I have a bending light math question. In the FE, light makes it from the sun to the earth in as little as .01 seconds to around .04 seconds. How do we not notice the EA as it would have to be super strong?
We have nothing to compare it to. We have never seen the sun unaffected by the EA.
So I have a bending light math question. In the FE, light makes it from the sun to the earth in as little as .01 seconds to around .04 seconds. How do we not notice the EA as it would have to be super strong?
We have nothing to compare it to. We have never seen the sun unaffected by the EA.
But I thought you claim vertical light is unaffected.
Also why did you not acknowledge the fact that the EA would have to be super strong?
But I thought you claim vertical light is unaffected.
Also why did you not acknowledge the fact that the EA would have to be super strong?
Vertical light is unaffected. And the EA is quite probably super strong, in terms of the acceleration it causes as a function of time.
Well, if the EA can bend light enough to cause a noticeable change at 10 km then it has to be super strong. It would take light around .000033 seconds to go 10 km. If I did any math close to correctly, its around a 3030 m/s^2 acceleration.
Well, if the EA can bend light enough to cause a noticeable change at 10 km then it has to be super strong. It would take light around .000033 seconds to go 10 km. If I did any math close to correctly, its around a 3030 m/s^2 acceleration.
That is horizontal light.
Yes.
noYes.
So we are in agreement that it can work.
no
no
Then what is your point?
Your theory is unsound.
Your theory is unsound.
Making unjustified claims is no way to debate.
It was justified.
It was justified.
Show me your justification, then.
3030 m/s^2 would be noticeable. There will also be more arguments to come.
3030 m/s^2 would be noticeable. There will also be more arguments to come.
It is noticeable, when the light is horizontal.
You know it would also lead to a non linear horizon distance right?
I guess that could be an answer to my question. A crappy answer though.
It is noticeable, when the light is horizontal.
Except it isn't. Do I have to post the picture of the interferometer again?
when I create the official document outlining the EA.
Except it isn't. Do I have to post the picture of the interferometer again?
I don't consider that adequate evidence against it. I shall expand on my reasoning when I create the official document outlining the EA.
I guess that could be an answer to my question. A crappy answer though.
Non linear with respect to what? Time? Angle at which horizon is being observed? Elevation of observer? The mass of the observer's genitals?
The distance to the horizon would depend on the angle of the light.
All this arguing is completely pointless, because Robosteve still fails to deliver.
I mean, he's saying all this stuff about what observations his EA can account for, but he still hasn't shown how it can even work.
Untill some sort of equation is presented, then everything he says with regard to the EA is just meaning less fluff.
y = x2 / 2r
y = x2 / 2r
How was this derived?
y = x2 / 2r
How was this derived?
By calculating the concavity of the expected secant curve traced out by a light ray tangential to the surface of the round Earth at the point where it meets the Earth, and then using that to calculate the equation for the parabolic arc traced out by a light ray in FET with the EA, such that it will have the same concavity when horizontal. r is the radius of the round Earth.
You're assuming light travels in parabolic arcs. I see no reason that this should be true.
The distance to the horizon would depend on the angle of the light.
As it does in RET.All this arguing is completely pointless, because Robosteve still fails to deliver.
I mean, he's saying all this stuff about what observations his EA can account for, but he still hasn't shown how it can even work.
Untill some sort of equation is presented, then everything he says with regard to the EA is just meaning less fluff.
Here's an equation:
y = x2 / 2r
And another:
(dx/dt)2 + (dy/dt)2 = c2
What shape would a ray of light trace out in a uniform gravitational field in GR? I think this would be the most logical shape to assume. Maybe it is a parabola.
You are still not getting it.
Here's an equation:
y = x2 / 2r
And another:
(dx/dt)2 + (dy/dt)2 = c2
Here's an equation:
y = x2 / 2r
And another:
(dx/dt)2 + (dy/dt)2 = c2
Ok, let's go with it.
Now for an experiment to show if it's correct or not.
My educated guess would be some sort of conic section, depending on various factors. Of course, one possible conic section is indeed a parabola.
The more I think about it, the more a parabola makes sense, but it's by no means a dead cert.
OK, let me give you guys my approach to the problem; it's long so be patient...
OK, let me give you guys my approach to the problem; it's long so be patient...
and the answer is... four?
Yup....that's what I got. Definitely 4.
Now, here is the interesting part. The density of the quintessence increases with height, so presumably so does its index of refraction towards light, possibly in a nonlinear way. The increase of refractive index with height is exactly what is needed to bend light to cause sunrises/sunsets, sinking ship effect, etc. Keep in mind it was the negative pressure property of dark energy that made this possible.
Thoughts, criticisms?
Consider the bendy light pictures shown here so far. All the rays appear tohitemanate from the sun at what I'm calling an "90 degree" angle. Furthermore all those curves have no inflection point. (This will be important.)
Now, consider what happens to a beam of light emanating from the sun at "89 degrees" or... lower angles if necessary.
Either all light that hits the earth from the sun comes from that "90 degree corridor" and consequently no light at even 89.9999 degrees hits the earth, or some light emanating at less than 90 degrees hits the earth.
I strongly believe that the models you and robosteve are kicking around imply that light emanating from the sun at something other than 90 degrees takes a path that does have an inflection point in it. And that just seems crazy to me. The only way that can happen is if the sun exerts its own component of EA and where the solar EA and the terrestrial EA are "equal" in strength is where this inflection point occurs.
So, there currently is no math to explain it, therefore you make it up because it doesn't actually exist/happen right?
The more intriguing properties of the quintessence (or whatever) are:
Now, here is the interesting part. The density of the quintessence increases with height, so presumably so does its index of refraction towards light, possibly in a nonlinear way. The increase of refractive index with height is exactly what is needed to bend light to cause sunrises/sunsets, sinking ship effect, etc. Keep in mind it was the negative pressure property of dark energy that made this possible.
^ Can we get this guy (Candleja) banned please?
Does bendy light affect all things EM?
Magnetic fields?
X-rays?
Gamma rays?
What about particles?
Neutron radiation?
Electrons?
What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight
These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!
These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!
You have no idea what you are talking about. There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider. It is real so stop making shit up.
What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight
These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!
You have no idea what you are talking about. There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider. It is real so stop making shit up.
So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
I have some pictures of the particle accelerator at Cornell University that I personally took. If I were to post them, would that be credible evidence that particle accelerators exist?
What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight
QuoteWhat about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight
What makes you think that they're build flat? Particle Accelerators are several miles long and allegedly bend along the curvature of the earth.
So yes, a photon emitted from one end would arrive at a higher altitude at the other end in both RE and FE + EA. The physicists just compensate by placing the detector where the particles arrive.
They're circular, the particles travel in a beam and travel around the accelerator several hundred or thousand times before emerging. This wouldn't be possible if light were constantly accelerating upwards.
On the basis of circular particle accelerators, I think we can declare EAT dead.
I have some pictures of the particle accelerator at Cornell University that I personally took. If I were to post them, would that be credible evidence that particle accelerators exist?
Oh, there are structures that are supposed to resemble particle accelerators. They're duds; they don't perform any functions whatsoever. So no, pictures will not prove anything.
Of course, it makes you wonder why the international community, which "knows" about the conspiracy ever approved funding for the super-collider in Switzerland.
I have some pictures of the particle accelerator at Cornell University that I personally took. If I were to post them, would that be credible evidence that particle accelerators exist?
Oh, there are structures that are supposed to resemble particle accelerators. They're duds; they don't perform any functions whatsoever. So no, pictures will not prove anything.
QuoteThey're circular, the particles travel in a beam and travel around the accelerator several hundred or thousand times before emerging. This wouldn't be possible if light were constantly accelerating upwards.
Those particles are traveling through something similar to fiber optic cables, where if the beam or particle deviates from its course, it just bounces back into its correct path.
Those particles aren't exactly traveling in a free open environment to the detector.
QuoteOn the basis of circular particle accelerators, I think we can declare EAT dead.
Obviously the particles are bouncing against the sides of something to follow the layout of the particle accelerator. Did you think that particles naturally traveled in big circles?
There goes my argument.
So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
What kind of cable could contain a beam of high energy particles? This is madness!
The reason particles travel in circular paths is because they are being directed by powerful magnetic fields. Ever noticed particle accelerators only accelerate ions?
QuoteWhat kind of cable could contain a beam of high energy particles? This is madness!
The reason particles travel in circular paths is because they are being directed by powerful magnetic fields. Ever noticed particle accelerators only accelerate ions?
Well there's your answer. The particles in a particle accelerator don't travel upwards and escape the facility by EA because they're being constantly pulled to follow the layout of the facility by strong magnetic fields all around them.
Tom, isn't this whole upward bendy effect completely incompatible with you own observations? You have said that you can see children playing on a beach 30 miles away.
Surely your observations falsify this bendy hypotheses? Or are your own observations now in doubt?
QuoteTom, isn't this whole upward bendy effect completely incompatible with you own observations? You have said that you can see children playing on a beach 30 miles away.
Surely your observations falsify this bendy hypotheses? Or are your own observations now in doubt?
If EA is correct, then it just makes such observations of being able to see past the curvature of the earth nothing more than a curiosity.
Perhaps refraction did it, as I'm so very often told.
In other words, you don't know and don't care what did it. You'll just go with whatever FE nonsense is first on your cut and paste list.
QuoteTom, isn't this whole upward bendy effect completely incompatible with you own observations? You have said that you can see children playing on a beach 30 miles away.
Surely your observations falsify this bendy hypotheses? Or are your own observations now in doubt?
If EA is correct, then it just makes such observations of being able to see past the curvature of the earth nothing more than a curiosity.
Perhaps refraction did it, as I'm so very often told.
No crisis, just change.
QuoteWhat kind of cable could contain a beam of high energy particles? This is madness!
The reason particles travel in circular paths is because they are being directed by powerful magnetic fields. Ever noticed particle accelerators only accelerate ions?
Well there's your answer. The particles in a particle accelerator don't travel upwards and escape the facility by EA because they're being constantly pulled to follow the layout of the facility by strong magnetic fields all around them.
QuoteSo a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
(http://www.physik.tu-muenchen.de/einrichtungen/department/luftbild03.jpg)
I happen to be there at the TUM. It looks quite real to me ;)
P.S. did i mention i have just been granted a new BMW 535d for saying that!? Yay me :D
So Robosteve... if EA doesn't affect PA......why are they fake?
If they wanted money then a casino would be an easier way to do it. Open enough of them. You don't even have to fight for funding.
How do any of us make money? There are no ads or donations or anything, and hosting costs mean poor Daniel has to pay hundreds of quid for you to spam with this shit.
How can you prove that what is shown in that picture performs any sort of function whatsoever?
You have no idea what you are talking about. There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider. It is real so stop making shit up.
So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
You have no idea what you are talking about. There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider. It is real so stop making shit up.
So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
I have no primary evidence for its existence.
You have no idea what you are talking about. There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider. It is real so stop making shit up.
So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
I have no primary evidence for its existence.
That's all you needed to say.
I can link to unpublished photos or I can say I am in direct contact with a conspirator.
Like I said, I am not in direct contact with a conspirator as she would of told her brother.I can link to unpublished photos or I can say I am in direct contact with a conspirator.
I don't think you understand what is meant by "primary evidence".
Like I said, I am not in direct contact with a conspirator as she would of told her brother.
I don't think you understand what is meant by "primary evidence".
Also where is your evidence that particle colliders don't work?
I'm not the one making unsubstantiated claims here. I have yet to see a shred of direct evidence that they do work.
So the world has to revolve around you? It's not their job to make sure you do research. Everyone already knows you can't. All the information you need to know they exist is out there. Go find it or stop making false claims.
So you're all out of evidence? Good, then I can stop wasting my time here.
All the information you need to know they exist is out there.
I'm not the one making unsubstantiated claims here. I have yet to see a shred of direct evidence that they do work.
If you dont believe they work just talk to a laboratory which has one that you CRAVE to stand right in the ray of particles.
And then you will have your proof by enjoying your own little demise :)
I was born in South Africa, now shut the fuck up.
And shooting somebody and standing in the ray of a particle accelerator is quite a differene even at the eyes of a non professional forensic guy ;)
And if they pay him three times his usual rate to keep quiet, do you really think the information will leak?
And really hope that you never piss the guy off. And there is also the question of what happens around retirement time, do you keep paying the guy huge sums in addition to now having to "bribe" his replacement?
Not to mention having to pay people to generate all of that "fake" data that is being analyzed by numerous independent groups.
And really hope that you never piss the guy off. And there is also the question of what happens around retirement time, do you keep paying the guy huge sums in addition to now having to "bribe" his replacement?
No, he just becomes his replacement's first job.
So how is the bent light theory (BLT hehe) going? How many FE-followers?
I just wanted to remind everyone that if you do believe in BLT, then you also admit that earth looks like a sphere. So if you support BLT, which I think is a great theory btw, be careful with your basic arguments like "earth looks flat, therefore its probably flat".
What is the fundamental property of the UA that accelerates the FE matter, penetrates and then affects light without affecting other matter?
How does one establish what is perfectly flat in this world of bendy light? How can you attempt to measure something on a level plane when there is no agreement on how to determine if the plane is flat?
With bendy light a slightly upwardly curved tunnel would appear dead straight.
I was wondering about a particle beam vs laser test to establish a flat line.
Has anyone come back to answer the Michelson-Morley inferometer yet?
There should be some reservoirs long enough for this purpose. Since they are normally situated in valleys they should offer a measure of protection from wind.
A particle beam would measure a geodesic in spacetime, not in space.
Interresting. How would the laser meassure?
QuoteTom, isn't this whole upward bendy effect completely incompatible with you own observations? You have said that you can see children playing on a beach 30 miles away.
Surely your observations falsify this bendy hypotheses? Or are your own observations now in doubt?
If EA is correct, then it just makes such observations of being able to see past the curvature of the earth nothing more than a curiosity.
Perhaps refraction did it, as I'm so very often told.
Interresting. How would the laser meassure?
In RET, they would measure a geodesic in spacetime. In FET with the EA, they would trace out a parabolic path in space. I am not certain what geometric shape this would correspond to in spacetime.
Anyway, getting off topic.
How long before we see some refined bendy light theories from the FEs.
Quantifiable predictions. When can we see some?
Well with the images i took in the sinking ship experiment
The distance was 20.0km from a height of 1m.
The lighthouse has an elevation of 80.5m and is 38.7m tall.
http://www.lighthouse.net.au/lights/WA/Rottnest%20Main/Rottnest%20Main.htm (http://www.lighthouse.net.au/lights/WA/Rottnest%20Main/Rottnest%20Main.htm)
There is something like 40m hidden behind the curve of the Earth.
Or the light has bent 40m over 20km.
Thats 2m per kilometer.
Cmon FEs. Put something together. It's been weeks.
Resurrection
http://bovitz.com/photo/traditional/jpgphotos/2005/Rays-of-light-through-cloud.jpg
What about sunbeams like these? There should be noticeable curvature of the light beams.
FE explanation please.
Well with the images i took in the sinking ship experiment
The distance was 20.0km from a height of 1m.
The lighthouse has an elevation of 80.5m and is 38.7m tall.
http://www.lighthouse.net.au/lights/WA/Rottnest%20Main/Rottnest%20Main.htm (http://www.lighthouse.net.au/lights/WA/Rottnest%20Main/Rottnest%20Main.htm)
There is something like 40m hidden behind the curve of the Earth.
Or the light has bent 40m over 20km.
Thats 2m per kilometer.
Cmon FEs. Put something together. It's been weeks.
Converting units from my earlier calculation of the curve in inches per mile per mile, the light should have bent by about 15.7 metres over a distance of 20 kilometres.
Resurrection
http://bovitz.com/photo/traditional/jpgphotos/2005/Rays-of-light-through-cloud.jpg
What about sunbeams like these? There should be noticeable curvature of the light beams.
FE explanation please.
To head off possible Tom Bishop explanations...
No it isn't glare in the camera lens. Here is a picture that I took showing the same effect.
http://img376.imageshack.us/img376/4479/godsfingersza7.jpg
It is definitely something that you can see without holding a camera to your eye.