 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 13 2016, 12:53 PM 
AAF: No . . . no . . . no "Quexit"!
I suspect it's just the New South Wales Blues fans who want to stop losing to our Queensland Mighty Maroons (note: pronounced "marowns", not "maroons').
They managed to stop the whitewash in the third match tonight, although our conversion attempt, although spectacular, didn't help.
Ufonaut99: [Szczepan] nicknamed "The Central Pole" ... standing on top of the central NONrotating 4metre pole that the merrygoround goes round.
AAF: Because, if the observer does not adjust periodically the direction of the camera to keep pointing to S_1, then the camera will go with S_2 and will point towards S_1 just once per rotation.
No, it doesn't.
On our merrygoround, as our A0 camera (on it's plain, static tripod, with no adjustments by the observer) is carried around by the merrygoround's rotation, then obviously it remains always pointing straight at the merrygoround's central pole (and so equally obviously, at S_1 Szczepan standing on it) throughout the rotation.
Sorry, but I have no idea how or why you could possibly disagree with something that is so utterly basic, plain and straightforward.

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 14 2016, 12:00 AM 
Ah; but they are still charging for this:
a little bit too much;
https://www.facebook.com/qldmaroons/
right?
…......................................................................................................................................
"And still just another bland repetition of the claim, with NOTHING given in support of it. It doesn't even need to be at/above the North Pole. As before: If I am standing due south of Brisbane CBD, then by definition Brisbane CBD is due North of me. Throughout Earth's rotation, with me not moving a muscle, Brisbane CBD will ALWAYS remain due North of me  and ALWAYS fixed stationary at THE EXACT SAME PLACE in my field of view. NO CIRCLE !. That remains so clearly, obviously, blatantly, selfevidently true for ANY object that is due North of ANY northwardfacing observer (including objects at/above the North Pole), I have no idea how or why you could possibly question it. Sorry, but your claim remains ludicrous, no matter how many times you repeat it. "AAF: take our hats off, for the first time, to the great Aristarchus" Love that Star; Also made me think of that song Putting on the Ritz"
Of course, if you're standing due south of Brisbane CBD, then, by definition,
Brisbane CBD is always due North of you.
But that is only true with respect to the reference frame, in which Brisbane CBD
is at rest; correct?
As seen, from S_1, you & Brisbane CBD are constantly moving in a circle around
the rotational axis of S_2.
Surely, you & Brisbane CBD  if like to do so and have the required energy
for it  can go gyroscopic like the rotational axis of the earth, and keep pointing
to the same direction in space all the time:
http://spaceplace.nasa.gov/seasons/en/
But, nonetheless, your direction, with respect to the rotational axis of S_2,
must keep changing constantly, because you & Brisbane CBD, by definition,
are in circular motion around the rotational axis of S_2.

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 16 2016, 12:00 AM 
""AAF: Since Szczepan is stationary at the center of the spinning 3metre radius merrygoround, then Alice ... looking inwards towards him, must always see Szczepan ... constantly rotating in a circle around Alice herself." Nonsense. Since Alice is ALWAYS looking "inwards towards him", then BY DEFINITION Szczepan is ALWAYS in front of her, in her field of view. I do not understand how can you disagree with something so straightforward. But OK, you claim that he is making a "big circle" around her, which means sometimes he will be positioned behind her  yes? So please describe  clearly and with maths  when you claim that Szczepan will be behind Alice, and so out of Alice's field of view (bearing in mind we've both agreed that he is ALWAYS in front of her)."
Is the physical statement that: "Since Szczepan is stationary at the center
of the spinning 3metre radius merrygoround, then Alice ... looking inwards towards
him, must always see Szczepan ... constantly rotating in a circle
around Alice herself" nonsense?
Absolutely NOT!
Provided, of course, that the following two basic conditions
are fulfilled:
A. Alice assumes herself to be at rest.
B. And Alice sees nothing else beside
Szczepan & the spinning 3metre radius merrygoround.
Now, the BIG question is this:
How can one explain the process of reversing something so straightforward like
the spinning 3metre radius merrygoround, in which Szczepan is stationary
at its center, and Alice is standing on its edge?
Quite simple!
All kinds of motion are reflectable;
right?
That is to say, whenever you move, you will see your movement reflected
back to you by surrounding objects.
For example, if you drive your car at 75 miles per hour between two rows of trees,
you will see the two rows of trees traveling at 75 miles per hour
in the opposite direction.
And likewise, if you fly a warplane at 455 miles per hour over the Australian Outback,
you will see the Australian Outback traveling at 455 miles per hour
in the opposite direction.
And the same, of course, applies to Alice, Szczepan, & the spinning 3metre
radius merrygoround.
Alice, in this scenario, is rotating at a specific angular velocity around
Szczepan who is, by definition, is stationary at the center.
And therefore, Alice will see her circular motion reflected back to her
by the stationary Szczepan.
It's that simple.

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 17 2016, 8:36 AM 
but they are still charging for this: a little bit too much;
Hmmm, but cheaper than some leather jackets AAF: Is the physical statement that: "Since Szczepan is stationary at the center of the spinning 3metre radius merrygoround, then Alice ... looking inwards towards him, must always see Szczepan ... constantly rotating in a circle around Alice herself" nonsense?
Absolutely NOT!
Of course it is. It flies blatantly against basic, everyday, common sense and experience.
Look, if a picture is worth a thousand words, a video must be worth a million  yes? So let's look at an actual video of a merrygoround so we can directly see what happens.
S_1 Szczepan (in Red) is the clearly visible top of the nonrotating central pole of the merrygoround, and we can mark the position of "Brisbane in the East" with a yellow spot at a fixed spot on the merrygoround  easily marked by the girl above it.
In addition, we also have a central "hub" (in purple) that is physically attached to the rotating S_2 merrygoround by the spokes.
To recap, your claim is that " if the observer does not adjust periodically the direction of the camera to keep pointing to S_1, then the camera will go with S_2 and will point towards S_1 just once per rotation. "
In this video, for the first 14 seconds (3 rotations) the observer (let's call that person "Alice") is doing a good job of keeping still (not moving a muscle, so stationaryrelativetoS_2)  no counterrotation nor adjusting "the direction of the camera".
So let's roll the tape and see what happens :
http://footage.framepool.com/mov/183112869.mp4
So : does "the camera ... point towards S_1 just once per rotation" ??? Of course not.
Does Alice see S_1 " constantly rotating in a circle around Alice" ? Wrong again.
How about S_1 " over Brisbane in the East"? Never happens.
And of course, let's not forget the " big difference between the behavior of objects physically attached to the rotating object of S_2 and the behavior of external objects like S_1. "
You claim that S_1 must be "rotating in a circle around Alice", whereas the Hub (being "physically attached to the rotating object of S_2") would therefore not make such a "big circle"  utterly ridiculous, since that would mean they must separate.
A reminder, In contrast to all those claims, quite simply Ufonaut99 (June15): I claim that : As Alice is rotating on the merrygoround, she will always see the central pole ("S_1"  Szczepan ... and, of course, the metal thing he's on) straightahead in her line of sight. He remains always fixed at exactly the same place in her view throughout the rotation, moving neither to the left nor right, and never leaving her view. As the merrygoround spins, she see his front, then his leftside, then his back, then his rightside, then front again  in clear apparent rotation around his central axis.
Bingo  spot on with what the video shows. 
 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 18 2016, 12:00 AM 
Yep . . .
Einstein's old & smelly leather jacket:
http://www.cnet.com/news/geniusbuyeinsteinssmellyleatherjacketsellsfor146000/
is quite expensive!
…..............................................................................................................................................................
"Briefly, your whole claim seems to be based on wanting BOTH that she is always looking inwards (as per our graphics), AND that she is seeing the roundabout rotate around her (which is only possible if she is rotating onthespot herself), and ignoring the fact that those statements are in CONTRADCTION. "AAF: And that is because the above 'stationary Szczepan  rotating Alice' system is very similar to the 'stationary Sun – rotating Earth' system, which had fooled very smart astronomers into deducing the exact opposite what is really going on, for hundreds of years". As I have said before : The Sun is NOT  repeat NOT  a valid substitution for S_1 nor Szczepan. S_1 and Szczepan are along the axis of our observer's rotation. The Sun is NOT (of Earth's daily rotation, which is our analogue for S_2's rotation). So YET AGAIN : Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis."
The Sun might not be a valid substitution for S_1, since the Sun is located at the center
of the solar system; while S_1, as described by Einstein, is located along the rotational
axis of S_2.
But, for sure,the Sun is a valid substitution for Szczepan who, as described earlier
by Colleague Ufonaut99, is located at the center of the spinning 3metre
radius merrygoround.
There can be no little bit of doubt about it!
However, the difference between the two above scenarios is not very big:
Objects, located at the center, can have only an unchanging circle whose angular
size does not change with distance from the observer.
But objects, located along the rotational axis, can have a changing circle whose
angular size varies with distance from the observer as well as with the angular
diameter of the geographical latitude of the observer.

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 19 2016, 4:35 AM 
AAF: The Sun might not be a valid substitution for S_1, since the Sun is located at the center of the solar system; while S_1, as described by Einstein, is located along the rotational axis of S_2.
Right AAF: But, for sure,the Sun is a valid substitution for Szczepan who, as described earlier by Colleague Ufonaut99, is located at the center of the spinning 3metre radius merrygoround.
Wait, WHAT ???? NO !!
I first introduced the Merrygoround saying The central "spindle" represents the common axis of rotation, so the top of the spindle represents S_1/Polaris.
The circular platform represents a slice through the earth at a given latitude  could even be through the equator.
So "the rotational axis of S_2" IS the axis through "the center of the spinning 3metre radius merrygoround"; The Sun is still not a valid substitution.
I subsequently introduced Szczepan by saying Szczepan ... also affectionately known as "S_1" ...
Now, imagine someone  "Alice"  standing on the outer edge of a spinning 3metre radius merrygoround in the middle of a grassy park. She is not moving a muscle, but looking inwards (so always looking inwards ) towards Szczepan, who is (also without moving a muscle) standing on top of the central NONrotating 4metre pole that the merrygoround goes round.
Someone's painted a couple of words on the merrygoround  "Brisbane in the east" to Alice's right, "Oslo in the west" to her left.
I claim that :
As Alice is rotating on the merrygoround, she will always see the central pole ("S_1"  Szczepan .....
So from all that, you've chosen to interpret that merrygoround  marked with "Brisbane in the east" and "Oslo in the west"  as representing earth's orbit around the sun ???
And that " "S_1"  Szczepan" is not S_1, but the Sun ???
And so also totally missed that "3metre radius merrygoround ... on top of the 4metre pole" is a scale model of our 3,000km radius, 4,000km height graphic of S_1 that we've been posting ???
Let's recap :
Einstein's experiment ONLY talks about S_1 and S_2, which are EXPLICTLY defined as bodies aligned along a common axis of rotation.
Clearly, only the DAILY rotation about the axis fits that bill, NOT the annual rotation in orbit.
ANYTHING not talking about THAT EXPLICIT SCENARIO (for example, anything talking about the Sun) is NOT talking about Einstein's experiment  ie. it is IRRELEVANT.
EVERYTHING that I have written (including "S_1  Szczepan" and "Stanisława .. known simply by .. S_1") IS about that scenario (and is NOT about The Sun) .
So please drop talking about the Sun, and instead stick solely to Einstein's experiment  Yes ? 
 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 20 2016, 12:00 AM 
"Thus far, over both these threads, you have NEVER followed through and applied anything about the sun nor Parallax to S_1, S_2; You have simply referenced them, but never applied the maths TO OUR SCENARIO. Case in point: "AAF: While the solar and stellar parallax are caused by the linear displacement due to the observer's linear velocity; that is say, if the observer is traveling at a linear velocity equal to v, then, in every interval of time equal to delta_t, the celestial bodies will appear to make an angular displacement equal to theta = arctan([v x delta_t]/d); where d is the distance between the observer and the celestial body in question." Yes, that equation theta = arctan([v x delta_t]/d) does give the change in line of sight for parallax  for an observer in STRAIGHTLINE motion. S_2 is NOT straightline motion; It is CIRCULAR."
So, you agree that "equation theta = arctan([v x delta_t]/d) does give
the change in line of sight for parallax  for an observer
in STRAIGHTLINE motion".
Good!
The BIG question, therefore, is, now, this:
Does circular motion have, in the inside of it, any form
of linear velocity?
Absolutely!
Circular motion, always, has, in the inside of it, a special form of linear velocity
labeled as 'tangential velocity'.
Generally, if an object is rotating at an angular velocity equal to omega,
along the periphery of a circle with a radius r around a fixed axis of rotation,
then the direction of its linear velocity, v, is always along the tangent to
that circle; and the magnitude of that linear velocity is always obtained
by using this equation:
v = omega x r.
What is the total value of the linear displacement,
in this particular case?
Well; it's quite obvious!
The total value of the linear displacement, in every rotational period,
is always equal to 2r; nothing more and nothing less.
For example, the geographical latitude of the city of Brisbane is equal
to 27.4698° S.
And since the equatorial radius of the earth is equal to 6,371 km, the radius
of Brisbane's geographical latitude is equal to:
6,371 km x cos(27.4698°) = 5652.696 km.
And as a result, the city of Brisbane must make a linear displacement equal
to about 11305.392 km around the South Pole everyday.

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 22 2016, 12:00 AM 
Actually,the city of Brisbane must make a linear displacement equal
to about 11305.392 km around the South Pole
every 11 hours & 58 minutes.
........................................................................................
"Look, imagine a car driving in a straight line lefttoright past our merrygoround, camera out the side window. One moment the central pole is to the right of the camera, next it's to the left. That's Parallax, as given by that equation. Fine. However, that does NOT describe Alice, who is rotating round with the merrygoround always looking straight at the central pole. That pole is ALWAYS straight in front of her, not to her left nor right. That is exactly what we expect from commonsense and see by looking at both merrygorounds and Polaris. Your claims that the central pole in that merrygoround scenario would swing to her left and to her right remain ludicrous despite the mathematics of straightline motion."
There is always a linear displacement in every sort of circular motion.
As mentioned earlier in this thread, the geographical latitude of the city of Brisbane
is equal to 27.4698^{o} S.
And since the equatorial radius of the earth is equal to 6,371 km, the radius
of Brisbane's geographical latitude is equal to:
6,371 km x cos(27.4698°) = 5652.696 km.
And accordingly, the city of Brisbane makes a linear displacement, D, equal to about
11305.392 km every 1/2 sidereal day around the South Pole.
So, what is the value of the Sun's parallax, theta, as measured
from the city of Brisbane?
Very simple!
As observed from Brisbane Downtown, the Sun has a daily parallax, as one single unit,
along the periphery of a small circle, centered on the South Pole,
with a total diameter equal to:
theta = arctan(D/AU) = arctan(11305.392/149597870.7) = 15.588 seconds of arc.
Of course, in this case, the Brisbane's observers can only see one half
of the above circle.
But if the Sun was located along the axis of the South Pole, the whole circle
of the Sun's parallax would be visible from the city of Brisbane.

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 24 2016, 12:00 AM 
"" AAF: The Sun, too, has daily parallax … Utterly irrelevant to "S_1 will always appear to be moving, as one single unit, in a circle over the North Pole" ... or, indeed, to anything in OUR SPECIFIC SCENARIO. AGAIN : THE SUN IS NOT  REPEAT NOT – S_1. Look, We are SUPPOSED to be discussing Einstein's Scenario. S_1 and S_2 are clearly defined in Einstein's scenario as being positioned along a common central axis of rotation. Discussions about bodies that are NOT so positioned along a common central axis of rotation, are therefore NOT discussions about Einstein's scenario. If you think that there is some issue about the Sun, Parallax, etc that is applicable to Einstein's scenario, then you should be able to APPLY it TO Einstein's scenario, referencing only S_1 and S_2. If you cannot apply that issue to Einstein's scenario, then by definition that issue is IRRELEVANT to Einstein's scenario, and therefore to our discussion."
Being positioned along a common central axis of rotation does not imply directly
or indirectly that Einstein's S_1 cannot have parallax.
PERIOD!
Is the star Polaris very close to being positioned along Earth's central
axis of rotation?
Certainly, it's very close:
Polaris' celestial latitude is equal to +89° 15′ 50.8″.
Does the star Polaris have stellar parallax?
Absolutely!
Take a look:
https://en.wikipedia.org/wiki/Polaris
The star Polaris has an amount of stellar parallax equal to about
7.54 ± 0.11 mas.
It follows, therefore, that no matter where the location of S_1 on the celestial sphere
of S_2 is, it must show a certain amount of parallax that can be measured
by any observer on S_2.
And that, for sure, destroys Einstein's main supposition
of relative axial rotation.

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 24 2016, 7:03 AM 
AAF: Does the star Polaris have stellar parallax? Absolutely!
You are STILL refusing to apply parallax to THIS scenario. Unfortunately, it looks like we'll have to wait months for you to catch up before we can even begin to discuss it
After all, your past THREE posts are all in response to the SAME ONE of mine, from all the way back to June 29th.
Any reason those past three posts could not have been combined into just one, posted in the days following June 29th ?
In every other forum I've participated in, people simply respond to the previous post, answering all the salient points that post raised. I make post #1. You make a post #2 in response. I respond to your your #2 with post #3. You respond to that post #3 with post #4. Postbypost, makes sense; that way, it's actually a discussion that moves forward. Be nice to have something like that 
 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 26 2016, 12:00 AM 
""AAF: Does the star Polaris have stellar parallax? Absolutely!" You are STILL refusing to apply parallax to THIS scenario. Unfortunately, it looks like we'll have to wait months for you to catch up before we can even begin to discuss it. After all, your past THREE posts are all in response to the SAME ONE of mine, from all the way back to June 29th. Any reason those past three posts could not have been combined into just one, posted in the days following June 29th? In every other forum I've participated in, people simply respond to the previous post, answering all the salient points that post raised. I make post #1. You make a post #2 in response. I respond to your your #2 with post #3. You respond to that post #3 with post #4. Postbypost, makes sense; that way, it's actually a discussion that moves forward. Be nice to have something like that."
Tell me how far Einstein wanted to place his S_1 from his S_2;
and I shall calculate the parallax of his S_1 and show it to you!
Surely, in every other forum you've participated in, people simply respond
to the previous post, answering all the salient points that post raised.
You make post #1.
They make a post #2 in response.
You respond to their #2 with post #3.
They respond to that post #3 with post #4.
And it goes on . . . and on; postbypost.
But, please, notice this Huge difference between
these two situations:
Those people, most of the time, are discussing things that have been known
for long time and discussed, by many other people before them, countless times;
and hence, they're traveling along major highways and city roads.
While we, in our current discussion, by contrast, are traveling along the dusty roads
of the Australian Outback or something like it; i.e., we're doing something
quite different!
This new and very difficult topic has never been discussed by anybody before us.
There are, really, no ready and welltested answers to most of the posed questions
and potential objections in this regard.
In other words, they are doing formal and informal discussions; but we are, right here,
doing some risky sort of scientific research and making somewhat hazardous incursions
and trips into the realm of the unknown.
And I may add that it's very risky, because, as you know, the biggest genius
of the physics community ( e.g. Albert Einstein) had said, time and time
again, throughout most of his professional life, that axial rotation is relative;
and most people believe him.
But their 'SUPER GENIUS' was baldly wrong.
I'm very sure about that.
And I shall prove it . . .
Just relax and enjoy the cool & nice Australian winter!

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 27 2016, 12:13 PM 
AAF: While we, in our current discussion, by contrast, are traveling along the dusty roads of the Australian Outback or something like it; i.e., we're doing something quite different! ... Just relax and enjoy the cool & nice Australian winter!
Not quite; Haven't been out to the Australian Outback recently  but rather to the top of Mount Haleakala, where I took this photo of Orion :
Not too bad (though I say so myself) for a first attempt at a longexposure  30 seconds. I looked for Polaris, but it wasn't in view In it's absence, I'm claiming that the "smudged" stars are the beginning of star trails (rather than a camera knock !)
Not quite, also, since our discusion IMHO is hardly a dusty outback of mathematics or physics. After all, all we're talking about are the mathematics of rotation and of Parallax, both of which are very wellworn paths.
AAF: Tell me how far Einstein wanted to place his S_1 from his S_2; and I shall calculate the parallax of his S_1 and show it to you!
And that's your problem.
You've found an equation : "theta = arctan([v x delta_t]/d)" (or more generally "theta = arctan(R/d)")
and decided that that's ALL there is to parallax regardless of any other factors
Wrong
and that therefore that equation can be blindly applied to ANY scenario, including Einstein's
Wrong
which means you can just hit Einstein over the head with it
Wrong
For example :
AAF: Does the star Polaris have stellar parallax? Absolutely!
Ufonaut99: You are STILL refusing to apply parallax to THIS scenario.
Earth has two rotations :
 Daily rotation around it's axis
 Annual orbital rotation around the Sun
Einstein's scenario is explicitly about rotation about the axis (the DAILY one) ONLY  not the annual orbital one (*)
So yes, Polaris has STELLAR parallax  that is, parallax as measured at opposite ends of earth's annual orbit. In fact, I already made that exact point earlier:
Ufonaut99 (dec 26): Remember also that this [Polaris will remain motionless in frame] does not apply to observations of parallax months apart, using the earth's orbit around the sun. For these, Polaris is not the centre of rotation, and so will display measurable parallax
but annual orbital rotation is NOT this (Einstein's) scenario (hence my comment), and therefore does NOT discredit Einstein. There is more to it than simply shouting "Parallax" !
So no more STELLAR or Sun based arguments, OK ?
And guess what : DailyRotation vs AnnualOrbitalRotation is not the ONLY factor affecting Parallax and this scenario AAF: But their 'SUPER GENIUS' was baldly wrong. I'm very sure about that. And I shall prove it . . .
Go for it But just remember  Einstein was precise in specifying his scenario, so to prove it you have to stay within those specialcase conditions  so basically :
a) the centre of S_1 is directly along the axis of rotation of S_2 (ie. directly above the north pole)
b) AAF (July 16): Alice sees nothing else beside Szczepan & the spinning 3metre radius merrygoround.  ie. we're concerned ONLY with S_1 and S_2.
c) AAF (Feb 11): Einstein preferred to make the actual rotation the BIGGEST secret of his thought experiment; i.e. his two observers do not know which of the two bodies, S_1 and S_2, is actually rotating around its geometrical axis.
Agreed ?
And that's before we even begin to discuss your model ......
(*) And again, it's way past time to leave behind comments like "it's very close", when we can and should be dealing with the EXACT maths.

 Anonymous
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 27 2016, 12:55 PM 
 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 28 2016, 12:00 AM 
Hi; Anonym!
Yep . . .
Bla . . . bla . . . bla . . .
....................................................................................
Hello; Ufonaut99:
Is that 'dark cloud' coming from an active volcano close to Mount Haleakala?
https://www.nps.gov/hale/index.htm
I presume not!
Anyway, this is a great picture of Orion:
….......................................................................................
"I still honestly have absolutely no idea how you reckon you have ANY sort of workable model here. The options are: 1) The everyday commonsense view, true to everybody's memories, that the central pole (around which, of course, the merrygoround is rotating), always remains at the same place straight in front in Alice's field of view (as I claim) 2) Alice sees Szczepan rotate from in front, to her right, to behind her, etc because she is counterrotating (ie. spinningonthespot) herself (which in addition to being contradicted by the explicit setup underlined above, also does not produce the "over brisbane in the east" effect your claims are based on) 3) Alice sees the park rotating around her, and also sees Szczepan against that backdrop, so reckons Szczepan is also rotating around her (which, although being a derivative of (1) and so still supporting my claim that Szczepan remains stationary in Alice's view, is not what we are discussing, nor will it produce that "over brisbane in the east" effect) 4) Alice sees the central pole  around which the merrygoround is rotating  physically leave the merrygoround, and go over the grass of the park in rotation around her. (4) is the only option that has Szczepan being seen over "brisbane to the east". (4) is also the only one that is utterly insane. So still after all this time : WHAT IS YOUR MODEL ??? Please say which, of (1) to (4)  Thanks"
There is no background, outside of the rotating merrygoround, for Alice!
That is one of the basic conditions.
Alice, also, does not know she's standing on the edge of the rotating merrygoround.
Moreover, Alice thinks that Szczepan is standing on the edge of the rotating merrygoround.
And finally, Alice believes that she herself is standing at the center of the rotating merrygoround.
Got the picture?
Alice, in this situation, is just like Claudius Ptolemy:
http://wwwgroups.dcs.stand.ac.uk/~history/Biographies/Ptolemy.html
who honestly thought that his earth and he himself were at the center of the universe;
and the Sun, the planets, and the star were rotating around them.
However, Claudius Ptolemy was correct with regard to the Moon.

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 29 2016, 10:23 PM 
Hi Anonymous,
This topic is supposed to be about Einstein's scenario, so as far as I'm concerned, anything not about that (or anything contrary to known physics) is bla, bla  and yes, there has been an endless amount AAF: Is that 'dark cloud' coming from an active volcano close to Mount Haleakala? I presume not!
Fortunately not (it's just a cloud  came out lovely when the sun rose, though )
AAF: There is no background, outside of the rotating merrygoround, for Alice!
Great, we agree AAF: And finally, Alice believes that she herself is standing at the center of the rotating merrygoround. Got the picture? Alice, in this situation, is just like Claudius Ptolemy:
Whoa  got it, but don't quite believe it ! So let's get this straight, translating it back from merrygoround to Earth/S_2 ......
We both agree that an observer (say "Fred") at Earth's North Pole sees Polaris ^{(*)} fixed in the sky directly above them, and all other stars circle "horizontally" (and so parallel, or a constant height) above the horizon.
You're claiming that the sheer fact that Alice considers herself stationary, means that she considers her current position to be the north pole, and therefore that the axis of revolution of Earth/S_2 runs not from the NorthPoleSouthPole, but through her.
This new axis of rotation results in the concrete physical effect that she sees that star's motions the same as Fred sees them  ie. the stars to her appear to rotate around a spot above her instead of around Polaris, and consequently therefore, she will see Polaris/S_1 rotate around her at a constant height above the horizon.
Is that seriously your model ?????
(if so, no, nothing like Ptolemy )
(*) Usual caveat about ideal Polaris

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  July 30 2016, 12:00 AM 
Thank you very much . . .
I thought it was coming from the East Maui Volcano:
https://thedestinationcenter.com/images/tourimages/96842600_1208481555.jpg
Or from this other volcano:
...............................................................................
"So let's go back to Alice on the merrygoround (and also, since you're currently responding to posts from 2 weeks ago, it's about due for your next response) Ufonaut99: Alice standing on the outer edge of a spinning 3metre radius merrygoround in the middle of a grassy park. She is not moving a muscle, but looking inwards (so always looking inwards) towards Szczepan, who is (also without moving a muscle) standing on top of the central NONrotating 4metre pole that the merrygoround goes round. Someone's painted a couple of words on the merrygoround  "Brisbane in the east" to Alice's right, "Oslo in the west" to her left. "AAF: Since Szczepan is stationary at the center of the spinning 3metre radius merrygoround, then Alice ... looking inwards towards him, must always see Szczepan ... constantly rotating in a circle around Alice herself." Ufonaut99: Nonsense. Since Alice is ALWAYS looking "inwards towards him", then BY DEFINITION Szczepan is ALWAYS in front of her, in her field of view. I do not understand how can you disagree with something so straightforward."
So straightforward?
Oh; yeah!
Let's look at the case of stationary Szczepan first.
How does stationary Szczepan see Alice in this situation?
Well; it's very obvious!
Stationary Szczepan sees Alice to the east of him.
And after exactly half of the rotational period, stationary Szczepan sees Alice
to the west of him.
And after a quarter of the rotational period, stationary Szczepan
sees Alice in the south.
And after exactly half of the rotational period, stationary Szczepan
sees Alice in the north.
And so on . . . and so on . . .
Assuming these two basic conditions are fulfilled:
A. Alice sees nothing outside of the rotating merrygoround.
B. And Alice assumes herself to be stationary.
Alice, therefore, must perceive the entire situation in reverse:
When Szczepan sees Alice in the west, she sees him in the east.
And when Szczepan sees Alice in the east, she sees him in the west.
And when Szczepan sees Alice in the north, she sees him in the south.
And when Szczepan sees Alice in the south, she sees him in the north.
And so on . . . and so on . . .

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  August 1 2016, 12:00 AM 
""AAF: There is no background, outside of the rotating merrygoround, for Alice"! Great, we agree. "AAF: And finally, Alice believes that she herself is standing at the center of the rotating merrygoround. Got the picture? Alice, in this situation, is just like Claudius Ptolemy." Whoa  got it, but don't quite believe it ! So let's get this straight, translating it back from merrygoround to Earth/S_2 ...... We both agree that an observer (say "Fred") at Earth's North Pole sees Polaris(*) fixed in the sky directly above them, and all other stars circle "horizontally" (and so parallel, or a constant height) above the horizon."
Does Fred, at Earth's North Pole, see the star Polaris fixed
in the sky directly above him?
Not exactly . . .
Actually, it depends on how sophisticated & smart the polar observer
Fred really is!
If Fred is a casual observer or just one of Scandinavia's Sami
Reindeer Herders:
http://ngm.nationalgeographic.com/2011/11/samireindeerherders/benkotext
THEN, surely, he will see the star Polaris fixed in the sky
directly above him.
However, even in this very simple case, Albert Einstein loses and badly so
I may add!
And that is, of course, because Einstein believed that Fred would observe the star
Polaris rotating around its geometrical axis due to the rotation of the earth around
its geometrical axis.
That is on one hand.
On the other hand, if Fred is a highly sophisticated observer and equipped with very
sophisticated instruments, THEN Fred may well be able to notice and to measure
as well the tiny amount of Polaris' parallax due to Earth's daily rotation around
its geometrical axis.
And once again, Albert Einstein loses really bad and very big here!
That is because, in this case, the star Polaris appears to move as one single unit
along the periphery of a very minute circle; with no apparent axial rotation
at all.

 Ufonaut99
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  August 1 2016, 7:51 PM 
Hi AAF AAF: On the other hand, if Fred is a highly sophisticated observer and equipped with very sophisticated instruments, THEN Fred may well be able to notice and to measure as well the tiny amount of Polaris' parallax due to Earth's daily rotation around its geometrical axis.
OK, this is a nice simple test case  Can you calculate precisely how much parallax Fred will detect ?
Go on, you can even use your linear equations for Parallax AAF: the celestial bodies will appear to make an angular displacement equal to theta = arctan([v x delta_t]/d); where d is the distance between the observer and the celestial body in question.
... its linear velocity, v ... is always obtained by using this equation: v = omega x r.
Let's say :
 omega = 1 revolution per 24 hours
 d = 3900km (North Pole is just a bit above Oslo's latitude)
 delta_t = 12 hours (half a rotation, so maximum parallax)
 and of course, since Fred is at actually on the North Pole on the axis of rotation, his displacement radius "r" = 0
So, can you calculate Fred's parallax angle theta ? 
 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  August 3 2016, 12:00 AM 
Hi; Ufonaut99:
Tell me how much the cross section of Fred's body is; and I will tell you how much
parallax Fred will detect!
…............................................................................................................................................................
"You're claiming that the sheer fact that Alice considers herself stationary, means that she considers her current position to be the north pole, and therefore that the axis of revolution of Earth/S_2 runs not from the North PoleSouth Pole, but through her. This new axis of rotation results in the concrete physical effect that she sees that star's motions the same as Fred sees them  ie. the stars to her appear to rotate around a spot above her instead of around Polaris, and consequently therefore, she will see Polaris/S_1 rotate around her at a constant height above the horizon. Is that seriously your model ????? (if so, no, nothing like Ptolemy) (*) Usual caveat about ideal Polaris"
That is not my model.
That is Einstein's model about relative rotation generally!
And so, it just happens in this special case, that I personally agree with him that
if Alice considers herself stationary, although she is actually standing on the outer
edge of the spinning 3metre radius merrygoround, then she must perceive herself
to be at the center of the same the spinning 3metre radius merrygoround.
So, now, what is my own model?
Well; I told you many times that, according to my model,
parallax is extremely important.
And because Albert Einstein neglected, in all of his published papers on this subject,
the effect of parallax, his main conclusion about relative axial rotation
is absolutely wrong.
To put it differently, due to the presence of parallax, apparent axial rotation
can never be seen or observed in the physical universe.

 AAF
 Re: You were right: Rotational motion is relative, too, Mr. Einstein!  August 5 2016, 12:00 AM 
""AAF: On the other hand, if Fred is a highly sophisticated observer and equipped with very sophisticated instruments, THEN Fred may well be able to notice and to measure as well the tiny amount of Polaris' parallax due to Earth's daily rotation around its geometrical axis." OK, this is a nice simple test case  Can you calculate precisely how much parallax Fred will detect? Go on, you can even use your linear equations for Parallax. "AAF: the celestial bodies will appear to make an angular displacement equal to theta = arctan([v x delta_t]/d); where d is the distance between the observer and the celestial body in question." ... its linear velocity, v ... is always obtained by using this equation: v = omega x r. Let's say:  omega = 1 revolution per 24 hours  d = 3900km (North Pole is just a bit above Oslo's latitude)  delta_t = 12 hours (half a rotation, so maximum parallax)  and of course, since Fred is at actually on the North Pole on the axis of rotation, his displacement radius "r" = 0. So, can you calculate Fred's parallax angle theta?"
Certainly, if Fred were a geometrical line, then no parallax, due to Earth's daily
rotation, would be detected by him at Earth's North Pole.
However, Fred's cross section is much wider than a geometrical line.
For the sake of argument, let's assume that the radius of Fred's cross section
is equal to 0.4 meter.
The distance of the star Polaris, from Earth, is estimated to be between 323
lightyears and 433 light years.
So, let's just assume that the distance of Polaris, from Earth, is equal
to 350 lightyears.
Into the following general equation for computing parallax:
P = 2*[arctan(r/d)]
we insert these two values:
r = 0.4 m;
d = (350)*(31536000)*(299792458) = 3308989234420800000 m.
And it follows, therefore, that if Fred is standing exactly at Earth's North Pole,
then he should be able to detect an amount of parallax, for the star Polaris, equal
to about 4.987 x 10^{11} mas.
That very tiny amount can be measured, of course, only if Fred is a highly
sophisticated observer and equipped with very sophisticated instruments!
 
   

