Res Ipsa Loquitor-- Disproved:The Impossiblity of absolute motion detection.

Take your amateurish professorial bullshit and go fuck yourself with it James R.

 

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I was commenting on your "frame definition" bullshit. No I am not touchy, I just abhor phony incompetence. Regular incompetence I can undersatand and live with in peace, it is your kind of incompetence that generates the fuck you James R.

 

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LMAO. You finally realize why your threads had no point, and you get all upset.

 

Einstein was the premier myth buster, as he busted the biggest myth of them all: the universal spacetime metric, the aether. Now, people like poltergeist there are green with envy over Einstein's veneration. He/they want to get in on all the glorious myth busting, but since Einstein won the Grand Prize of Questioning & Debunking Authority, others can only hope for scraps... and so they get vicious and attack Relativity. All in vain, utter vanity, alas!

 

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I have noticed that you never have any physics to offer us, so in violation of a deeply sincere promise I nade to a wise advisor of mine, and who, flatly speaking 'as an 'erring cndishn', though is is never in error, I am going to simply tell you Mr. Dristam, to go fuck off.

ADDENDUM TO POST: [Note, I decded to offer this that was posted in another thread for Mr. Dristam to comment on. I trust he technically up to the task.]

This is a continuation of a discussion of an absolute motion measuring device. It is based on the fundamental postulate that the motion of the emitted photon is independent of the motion of the source of the photon. The schematic is shown at three instantaneous times as the photon continues downward, while the frame continues to the right where constant velocity is assumed.

Just in case there is any misunderstanding about where I have been coming from here is a brief sketch of the peap machine. The two horizontal lines show a grid (one dimension of a xy plane grid) where all activity is in the x direction. Here the photon has just been emitted from a peap unit UA10 (upper A # 10)

Code:

T1
UA 1  2   3   4   5   6   7   8   9  10   11   12
 -||--||--||--||--||--||--||--||--||--||--||--||-
                                      o


DA 1  2   3   4   5   6   7   8   9  10   11   12
 -||--||--||--||--||--||--||--||--||--||--||--||-

Code:

T2
        UA 1  2   3   4   5   6   7   8   9  10   11   12
         -||--||--||--||--||--||--||--||--||--||--||--||-
                                       
                                      o 

        DA 1  2   3   4   5   6   7   8   9  10   11   12
         -||--||--||--||--||--||--||--||--||--||--||--||-

Code:

T3
                UA 1  2   3   4   5   6   7   8   9  10   11   12
                 -||--||--||--||--||--||--||--||--||--||--||--||-
                                      

                                      o
                DA 1  2   3   4   5   6   7   8   9  10   11   12
                 -||--||--||--||--||--||--||--||--||--||--||--||-

els of the three dimensional two tiered planar grid (3DTTPG).

 

You're the scum of the Earth, geist, so why would I bother commenting on your rubbish notions. If they have merit, then you can take them to the publication, Nature.

 

The source of the photons is merely a point in space, but it is invariant. It may be difficult to determine at times, but thr reaity of the invariance of the photon trajectory is cast in steel.

You didn't answer the question.

How do you determine that point in space?

And now another question arises...

Why is your method (yet to be relinqusihed) "difficult to determine at times?"

Are you saying that it may not be entirely possible to determine the sources' position using only space as your reference?

If you were on Earth and I was on a ship 5 light years away - what coordinates could you send to me that will allow me to go to that source point in space?

 

The system measures frame velocity with respect to the position of the photon that is emitted. The device, the scematic is two or three post back, has an emitted photon from the frame. The photons are directed directly at the absorber located a known distance from the emitter. T rest the photon always is absorbed bits mirror image absorber/emitter, the up/down designations being the only difference in the location of he emitter/absorbers.

If the frame is moving, the "mirror image" has moved before the arrival of the photon, ergo the absorber/emitter that absorbes the photon will have a different identiofication and our ordinary lapto computer calulates the measured velocit of the frame wrt the invariant trajectory of the photon. As thge frame contionues to move the left side will eventuall catch up to the emitted phpotns, but before this happens the computer orders the photons to be emitted from an emitter/absorber located in a direct line as far from the on coming wall, before the photons are interferred with.

Methinks, by the nature of your questions that you have put more into what the device is designed to accomplish than I had intended, an what the device is limited to provide, namely cvelocity of he frame with a known straightline trajectoery.

I was thinking about technical problems of calibration of the device starting from an unknown velocity, direction and absolute speed, whwich now that I have thunk about it, is not a real problem of any substance.

I probably wouldn't. Using EM transmission would put an enormous error into the problem, 10 light years, as a minimum, assuming you transmitted your position relative to a commonly known point, like Ve the earth.

I'm not being a smart ass, I'm just reminding you that the device design is no more a road map than is the speedometer in your automobile. I suppose iI would at least send an estimate of our mutual positions wrt some known stellar object and provide the best astronomical map available and let you determine your own direction.

 

But the direction of the emitted photon is different for different observers, hence your system doesn't work.

 

Yeah, Einstein thunk that too: it's not a significant problem! You start with an unknown and you end up with a deeper unknown, and your device goes into the nearest dumpster. The only known will ever be C.

 

The system measures frame velocity with respect to the position of the photon that is emitted.

So, the postition of the photon when it was emitted is located in respect to where the photon was emitted - brilliant.

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The photons are directed directly at the absorber located a known distance from the emitter.

In other words, the distance the emitter is in relation to the absorver is 'relative.'

T rest the photon always is absorbed bits mirror image absorber/emitter, the up/down designations being the only difference in the location of he emitter/absorbers.

Wtf?

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If the frame is moving, the "mirror image" has moved before the arrival of the photon...

Again, wtf?

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the device is limited to provide, namely cvelocity of he frame with a known straightline trajectoery.

Which is little more than a device that might measure the speed of light. So what?

It does nothing to tell us the position of the source of the emitter, as I had asked.

is not a real problem of any substance.

Au contraire, mon fraire. It is a huge problem, one that your system fails to solve.

I probably wouldn't. Using EM transmission would put an enormous error into the problem, 10 light years, as a minimum, assuming you transmitted your position relative to a commonly known point, like Ve the earth.

Exactly - in other words, your system is pointless without the use of relativity.

Thank you.

 

You may characterize the system anyway you choose, but be aware that the postulate of light that tells us the motion of a photon is independent of the motion of the sources, in this case the peap frame attached to the "inertial frame", and therefore is an independent entity in the universe. No one goes bonkers over the lack of locating the sun, for instance, wrt to the center of mass of the universe. The sun is just there. Of course some make measurements of relative motion wrt galaxies, stars in the Milky Way and so on, but what else can one do in such a huige place we find ourselves?

Q, the trajectory of the photon is invariant, wrt to the moving frame. There is "no after emission history" physically linking the photon and frame. The photon is located in a unique position in space. It doesn't require any external "refrerence point" and certainly for the purposes of measurement of the frame velocity. The exact location iof the trajectory with respect to any other stellar object, or any onject is thoroughly insignificant, irrelevant. The photon only has to be measured in the realtively small volume element of each of three coplanar peap grids, only.

The device measures velocity. It isn't a road may to Betelguese .

The photon when emitted is directed at the opposite peap unit. As the photon is in flight and the frame is moving the ultimate arrival point of the photon will be so many peap units from the last recorded by the computer system, a dx/dt is easily calculated.

I wouldn't describe it as such, but if this blows your skirt up, go for it. I would say that the location of peaps are rigidly fixed in the frame. The planar grid is two tiered wiith the physical distance between all peaps exactly the same.

I live 2.23 miles from my friend's house. Do I live 2.23 miles relative to my friend. I don't talk that way do you?

OK, confused, me too. Here is the translation. The difference in location of a peap, the upper and lower unit are identified by location. The only difference in the location is that distance between the peaps. One is the upper, the other the lower. Let's say the UM27 and DM27 are pairs. The Computer only measures the time the absorber detected a photon. The emission time of the emitter, say the upper, is also known from the time of arrival of the photon from the down direction. If there is a constant exchange of photons between the U/D M27 pair, obviously the frame is at rest. If however, the UM27 emitted a photon at T1 and the D R14 was the arrival point of the photon, then this delta T is not different from the "at rest T", for motion along the X axis

Anyway, the computer is a slave. It only calculates delta arrival times and then publishes the Dx (distant between pairs looking down the trajectory of the photon. The delta T is just another number to crunch by the calculating algorythm.

Lets say you are on a bridge over a rail system. A Train of small boxcars is at rest under the bridge, and as you drop little rocks into the boxcar beneath you all the stones end up in the same box car, Now the train begins to move, but you aren't moving. You did not accelerate. Now that stone you released just as the train begain it motion, does not end up in the same bioxcar as when the train was at rest. The next, or the 14th next boxcar, depending on the speed along the tracks, will get your stone. The 1st boxcar moved and was replaced by a cousin boxcar following behind.

The analogy is direct.

It already knows the speed of light. The devce is merely comparing arrival times by any absorber absorber relative tot he time of emission. So what?

The known velocity of light allows the algorythm to assume a known delta T of travel time for photon between emitter and absorber (actually it doesn't assume anything. The devices are the source of data the computer uses to make calculations).

Even though the frame is moving, and assume no variation off the x-axis, the delta T will be constant. It is the newly arriving peap that provides the measure of frame velocity. But because the algirythm knows or anticipates off x-axis motion, the variation in dT, while small is real. In any event there are two other grid panels designed to measure X and Z motion.

D = VT. and V = D/T. Here D is the linear distance the frame has traveled, or the distance between peaps, calibrated in meters, or km. And T the time between emission and absorbtion.

Well you know where the source of the emitter is located .

There is an xy planar grid, where the up and down planes are parallel. The location of each peap unit on the planar grid is known exactly. The system merely measures DT and DX. The DT in uniform motion is a slam dunk trivial task as one direction along a planar grid can be arbitrarily selected as X. If the motion is uniform then all distances of the yz-plane are constant, or the device can be oriented such hat the yz-palne is perpendicul;ar to the motion and hence no variation in the DY direction will be mesured as the photons are moving in the same direction as the motion of the grid. The DT of this frame will be different as the lower plane is moving into the direction of the oncoming photons, and the upper plane is moving away from the oncoming return photons. The zx-plane is moving parllal to the x-axis and functions as the X-axis as the photon DT does not exhibit the "collision" with oncoming photons and is "catching up" of the return photons.

Here is a simple round trip of photons moving in a plane perpendiculalry to the frame motion.
D/___________<-/ U -> v. A photon arrives at the D absorber after traveling a distance ct. That point of arrival is VT short of the known distance between the emitter-absorber. On the return trip the frame has moved VT by the time the photon just leaves the D absorber (now emitter). When the distance ct is covered in the X direction it arrives at the original invariant location of the emitted photon, and the U absorber is now 2VT ahead of the photon. Counting from here the photon must cover a distance cT' = 2VT + VT'. The VT' of course, is the distance the frame has moved when he photon covers the 2VT. Hence, after some algebra, T' = T(2V)/(C - V).

You see Mr Q, this T' is the mysterious lost time period that special relativity says is covering for us by frrame contracvion and/or time dilation, when these esoteric calculations are not required.

This T' is also the lost time giving us loss of simultaneity, which SR says is recovered by SRT, which when one considers the resulting SRT demanding time dilation, frame contraction, loss of simultaneity, loss opf absolute time and space, is a substantially and totally unnecessary deflection of physical law. The resulting confusion perpetrated on an innocent general population is magnified tremendously.

I added your question (in red) to which I responded in blue). You inadvertently ommitted this question when you published this post. Also there seems to be a typo and the question on its face is ambiguous , buit I understand what you are asking.

As I said, the device measures frame speed rela'tive to the invariant straight-line trajectory motion of photons. The device is not a road map to Betelguese.

Here is the same answer I published before:

"I probably wouldn't. Using EM transmission would put an enormous error into the problem, 10 light years, as a minimum, assuming you transmitted your position relative to a commonly known point, like Ve the earth [frame].

I'm not being a smart ass, I'm just reminding you that the device design is no more a road map than is the speedometer in your automobile. I suppose I would at least send an estimate of our mutual positions wrt some known stellar object(s) and provide the best astronomical map available and let you determine your own direction."

 

No one goes bonkers over the lack of locating the sun, for instance, wrt to the center of mass of the universe. The sun is just there.

That would be a very good answer - if it came from an elementary school kid.

Of course some make measurements of relative motion wrt galaxies, stars in the Milky Way and so on, but what else can one do in such a huige place we find ourselves?

We make use of relativity - that's what everyone is telling you.

The device measures velocity. It isn't a road may to Betelguese .

Then it is entirely useless, as I've already said. We can already meaure velocity, thanks.

I wouldn't describe it as such, but if this blows your skirt up, go for it.

It's not a matter of 'going for it,' its a matter of what works, and your so-called system does not work.

I live 2.23 miles from my friend's house. Do I live 2.23 miles relative to my friend. I don't talk that way do you?

Yes, I do - the two statements are the same.

One is the upper, the other the lower.

In other words, they are 'relative' to one another.

The analogy is direct.

It is an analogy of relativity.

The devce is merely comparing arrival times by any absorber absorber relative tot he time of emission. So what?

Then what purpose does it serve?

Here D is the linear distance the frame has traveled, or the distance between peaps, calibrated in meters, or km.

Again, one is 'relative' to the other.

Well you know where the source of the emitter is located .

I could never know if space is my only reference. I need to know the location 'relative' to other objects in space.

The location of each peap unit on the planar grid is known exactly.

Yes, because they are 'relative' to each other.

direction will be mesured as the photons are moving in the same direction as the motion of the grid

The photons are moving 'relative' to the grid.

The resulting confusion perpetrated on an innocent general population is magnified tremendously.

The confusion is on your part. The 'innocent general population' is safe.

As I said, the device measures frame speed rela'tive to the invariant straight-line trajectory motion of photons. The device is not a road map to Betelguese.

Again, it measures velocity, which is already a known. Other than that, your system is pointless, hence useless.

Everything you've tried to explain thus far has required the use of relativity, and so far you have failed miserably to attempt to explain absolute motion. You've done little more than explain a method of measuring light speed.

Back to the drawing board.

 

[

You say a few sentences below that you don't need my machine as the "velocity" is already known.

Why are ypu involved in this post anyway, just to keep me busy answering your useless posts?

How do you measure velocity in a closed container in free space? You know the postulate of SR that says it is impossible to detect or measure translaory motion of unaccelerated bodies in free space? This is SR, basic stuff., so how do you measue velocity?
I told you how I measured velocity, now you tell me how you do it.

Yes, they are relative to each other..
What is so magic about the use of the word "relative"?


I mean the roof is over my head. You seem to place a different meaning on the situation by insisting that the roof is "relative" to my head. Have you added anything to understanding anything?

The purpose of the device is to measure the velocity of a material frame with respect to absolute zero velocity.

There is nothing in SR that can come near to providing this kind of velocity information the peaps provide. SR theory rejects the concept of absolute anything including velocity.

yes they afre "relative" to each other, Why do you persist in asking such questions? So what? When you say one s traveliong relative to the other? One what is traveling wrt to the other what? The peaps are rigidly fixed in the planar grid. The relative velocity of the peap units is zero. The relative velocity is the motion of the peaps with reszpect to the invariant photon trajectory.

.

There could be as many as a few million double tiered planar arrays where the exact ralative location of each unit is known and the device is in space also.

Further, the iunits are connected to the input port of a computer where each data bit is used to calculate the velocity of the light trajectory that is unvariably located in space. What other kind of information do you need, what kind of information do you need outside the device? When you look at the speedometer of your automobile do you ask for any other external information to tell you how freaking fast your are traveling?

OK, they are relative toeach other , so what? .

I have to correct the statement above, Direction is measured as the grid frame is moving with respect to an invariantly located photon trajectory
located in a unique spot in absolute space and it is the frame that is moving with respect to a phuysically invariant frame at absolte velocity = 0.. Generally the frame is moving pependicualr with respect to the absolutely located photon trajectory.

What do you mean about the velocity that is already known? Why does your speedomenter keep out putting your speed of your automobile if the velocity is already known?

The velocity of the spacecraft is noiwn by whom? Where did yiu get that statement? What does you mean about that statement? What in he hell are you talking about man.

You've got it backwards Q man. The concepts of the device I designed are contrary an anethema, to the fundamental postulates of SR and you know it. I suppose you feel threatened.?
Well relax man. MacDonalds is hiring , and they train you too.

 

The purpose of the device is to measure the velocity of a material frame with respect to absolute zero velocity.

Failed. You were unable to show absolute zero velocity and you have admitted SR is required to explain the positions of your test equipment.

You've got it backwards Q man. The concepts of the device I designed are contrary an anethema, to the fundamental postulates of SR and you know it.

Sorry, your concepts are pointless - that has already been confirmed.

I suppose you feel threatened.?

Only by your lack of syntax.

Well relax man. MacDonalds is hiring , and they train you too.

No thanks, and besides, you probably wouldn't want me as your boss.

 

The trajectory of the photon is invariant. The trajectory is straight lince sine the sinstant th ephototn was emitted. Thevariation in straighline motion means the photon drift is zero velocity.

confirmed by you, that's what I'm betting on?

What is this syntax of which you speak? You mean the ladies of the night are supposed to file some proof of income with the IRS? So if I provided yopu with some syntax you wouldn't feel threatened? Sorry, I ain't a gummint man.

I love the positivity you express for the future. You're going into management in the food service industry!

Geistkiesel

 

The trajectory is straight lince sine the sinstant th ephototn was emitted. Thevariation in straighline motion means the photon drift is zero velocity.

hehe - you really do love making stuff up as you go along.

Baffle them with bullshit.

 

Yet again, wrong. We've been pointing this out since you've started this little diatribe.

The trajectory changes.