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AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 20 2016, 12:00 AM 









"But OK, let's take the simplest possible test case : Say hello to Stanisława from Krakow:

[linked image]

She's the oldest of her sisters, so often known simply by the abbreviation "S_1". In addition, Stanisława is spending a couple of days standing on the the northern surface of S_2 exactly on the axis of rotation - which, of course, leads to her other common nick-name. That's right - Stanisława IS "The North Pole". Now that, of course, means that Stanisława has a constant and low value of H."






Do you mean Stanisława Leszczyńska?

https://en.wikipedia.org/wiki/Stanisława_Leszczyńska


Anyway, you can't see the one, in the picture, from Oslo, if she stands at the North Pole,
because this Stanisława is too short!


wink.gif


Let's assume, for the sake of argument, that Stanisława is 1.75 m tall.


By inserting the above value into this equation:

D = arctan(H/3000),

therefore, we obtain this value of D:

D = arctan(0.00175/3000) = 1.11 x 10-8 degrees.


That is almost equal to zero!


And accordingly, if S_2 is assumed to be spherical, no one can see Stanisława from Oslo.


But let's once again, for the sake of argument, assume that the spherical cap between
Oslo and the North Pole is completely removed; and hence,the flat circle
of Oslo's geometrical latitude is totally exposed.


It's absolutely certain, in this hypothetical case, that a rotating observer,
located in Oslo, will see Stanisława projected on the opposite arc
of the flat circle of Oslo's geometrical latitude, and NOT standing
at the center of the circle on the North Pole as she actually is.


And furthermore, if the rotating observer, located in Oslo, assumes
that he/she is stationary, then he/she will see Stanisława doing all
the circular motion in a big circle centered around Oslo.


Does this well-known optical illusion of circular motion become,
now, very clear?



happy.gif














 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 21 2016, 10:57 AM 

AAF: Do you mean Stanisława Leszczyńska? https://en.wikipedia.org/wiki/Stanisława_Leszczyńska

Wow, what a lady. To see someone standing for goodness, in the face of such evil, really shows our own petty troubles in context.
AAF: a rotating observer, located in Oslo, will see Stanisława projected on the opposite arc .... Does this well-known optical illusion of circular motion become, now, very clear?
No - it remains blatantly false. The "opposite arc" of Oslo is Alaska, so the rotating observer will see Stanisława ALWAYS fixed stationary against Alaska throughout S_2's entire rotation (though again, we're supposed to be concerned solely with S_1 and S_2, not any background).
AAF: if the city of Brisbane is located on the eastern horizon .. S_1, sometime, seems to be over Brisbane ... Surely, S_1 is always '3000Km straight ahead

From where I am, real-life Brisbane CBD is due north, and the Pacific is to the east. While I am standing (not moving a muscle) facing due North :
- There is NO optical illusion (let alone any "well-known" one (*)) that will put Brisbane CBD over the Pacific.
- There is NO optical illusion (let alone any "well-known" one (*)) that will result in the Pacific ever being straight ahead of me.

Any reader could substitute their own landmarks, to see that these will always be self-evidently true - and would remain true even if Brisbane was actually at the north pole (Brrrr, please no !)

These have been long threads, and IMHO, this claim is the central core point of difference between us, upon which your remaining arguments are hinged.
I have repeatedly asked for clear justification of this claim, but we're still not progressing from the above statements, so it's time to call a halt.

In summary, I see our respective positions, as stated above and what I said in my previous post :
I go instead with the commonsense straightforward view that anything that remains on (or directly above) the north pole remains due north of our observer. Since our observer is always facing due north, then he sees that object fixed in front of him. THAT is what makes perfect sense.

In contrast, you claim that an object positioned at/above the north pole travels until it's due EAST of our observer (.... and so continues until that NORTH pole object is due SOUTH of him, yes?) However, even though the observer remains facing north, you reckon this object (that is now East of him) is not off to his right - No, you say it's still straight ahead of him !!

I'm happy to leave any other readers to decide the relative credibility of these positions, and consequently the merit of all your arguments against Einstein in these threads happy.gif

(*) Going incredibly cross-eyed doesn't count ! wink.gif


 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 22 2016, 12:00 AM 









Oh, so you've decided suddenly to suspend it; just like
Ted Cruz after losing the Indiana primary:



[linked image]



Well; speaking for myself, it's too late,
for me, to quit now . . .

I still have a huge file full of unpublished posts related
to this THREAD!


[linked image]


…..................................................................................................................................................................


""if the city of Brisbane is located on the eastern horizon and the city of Oslo is located on the western horizon of the rotating observer, then S_1 will, sometimes, appear over the city of Brisbane; and it will appear, at some other times, over the city of Oslo." Surely it's commonsense that the North Pole will always be due North - yes? .... so please explain clearly, with maths, exactly: how and why you reckon our camera sees The North Pole over at Brisbane due .... East ???? And then, of course, how you reckon that The North Pole swings over to Oslo due West ????"




The North Pole, by definition, does not swing over to Oslo due west.


But the external object S_1 does swing over to Oslo due west.


And also, the external object S_1 does swing over to Brisbane due east.


How do we explain it exactly with math?


Very simple!


happy.gif


Firstly, we calculate the angle D between the S_1, the rotating observer, and the North Pole,
by using this equation: D = arctan(H/3000).


And since the height of S_1 over the North Pole is equal to 4,000 km,
then D is equal to about 53 degrees.


And secondly, we rotate the straight line between the observer and S_1 at
a 53-degree angle across the sky.


And finally, by simply assuming that observer is stationary, we make S_1 appear to travel
in a circle at about 53 degrees over the stationary circle of the observer's horizon.


It's that simple!



happy.gif












 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 23 2016, 11:48 AM 

AAF: Oh, so you've decided suddenly to suspend it; just like Ted Cruz after losing the Indiana primary:

Oh no !!! If I'm Ted Cruz, that means you regard yourself as Donald Trump Shocked eyes

OK, you put a bit more detail into that post, so lets give it another shot happy.gif

[linked image]

So, from what you described in the final section of your post, seemed to imply the following :

1) Let's say that, as per the graphic, at 1am we point our camera directly at S_1, that is positioned fixed at a constant height H above the north pole.
2) Let's also say Mizar happens to also be in the camera's frame (although ignored by our observers wink.gif ).

3) Now, we agree that S_2 has actual rotation about it's axis (although again, not known by our observers wink.gif )
4) However, assuming our camera-observer is stationary, then it will remain stationary throughout the rotation
5) (and so, consequently, always remain pointing towards Mizar)
6) while S_2's rotation means S_2 will be rotating about it.
7) That is, although at 1am the north pole is directly ahead of the camera, at other times the pole will be rotating around (and at 1pm, be behind) it.
8) All of which means, whatever is above S_2's north pole will also be effectively rotating around the camera
9) which results in S_1 being seen travelling by our observer "in a big circle in the sky", such that Parallax can be applied.

So is that a fair summary of your position ?

Pending your confirmation, it's worth pointing out that that position is totally inconsistent with the idea that "the external object S_1 does swing over to Brisbane due east" (Actually, your repetition of those "S_1 over Brisbane" claims is one of the reasons I've been thinking you couldn't have that model in mind.)
but that's minor issue happy.gif

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 24 2016, 12:00 AM 











I hope we all will, someday, be as good at making BIG money
as Mr. Trump:


[linked image]



[linked image]



….............................................................................................................................................



"finally agreed"????? I never disagreed (beyond the earlier correction); I said it was irrelevant to the effect you were claiming (which it is). "And if the observer moves in a circle, then he/she will see the external object moving in a circle." Let's take a simple every day example - but first give a warm welcome to Stanisława's brother Szczepan. He's the oldest son, so also affectionately known as "S_1" He has a habit of climbing on top of things that are in the middle of something - so yes, he's nicknamed "The Central Pole". Now, imagine someone - "Alice" - standing on the outer edge of a spinning 3-metre radius merry-go-round 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 NON-rotating 4-metre pole that the merry-go-round goes round. Someone's painted a couple of words on the merry-go-round - "Brisbane in the east" to Alice's right, "Oslo in the west" to her left. I claim that: As Alice is rotating on the merry-go-round, she will always see the central pole ("S_1" - Szczepan ... and, of course, the metal thing he's on) straight-ahead 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 merry-go-round spins, she see his front, then his left-side, then his back, then his right-side, then front again - in clear apparent rotation around his central axis."






The consequences of the equation:

D = arctan(H/3000) are relevant and very important!


So, Alice is standing on the outer edge of a spinning 3-metre radius merry-go-round
looking inwards towards Szczepan; right?


Of course, I got it right?


wink.gif


Since Szczepan is stationary at the center of the spinning 3-metre radius merry-go-round,
then Alice, standing on the outer edge of the spinning 3-metre radius merry-go-round and
looking inwards towards him, must always see Szczepan, NOT stationary as he actually
is, but projected on the other outer edge of the spinning 3-metre radius merry-go-round
and constantly rotating in a circle around Alice herself.


Do you agree with this result?


Of course, everyone must agree with this result.


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.



happy.gif


















 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 25 2016, 11:26 AM 

AAF: I hope we all will, someday, be as good at making BIG money as Mr. Trump:

I once made a $1,000,000,000 note printed on my printer - does that count ? wink.gif

AAF: Since Szczepan is stationary at the center of the spinning 3-metre radius merry-go-round, 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).

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 possble if she is rotating on-the-spot 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.

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 26 2016, 12:00 AM 











YEP . . .


Printing a $1,000,000,000 note is half the job . . .


The other half of the job is to convince the Reserve Bank of Australia:


[linked image]


http://www.rba.gov.au

that the printed $1,000,000,000 note is genuine!



[linked image]



…...........................................................................................................................................



"In contrast, you claim that an object positioned at/above the north pole travels until it's due EAST of our observer (.... and so continues until that NORTH pole object is due SOUTH of him, yes?) However, even though the observer remains facing north, you reckon this object (that is now East of him) is not off to his right - No, you say it's still straight ahead of him!! All based on some idea that The projected circle is the circle of the rotating observer's latitude as seen by the same observer projected on an external object like S_1. In other words, rather than simply calculating line-of-sight directly with maths, you're basing the whole thing on some flawed idea that "if he sees me moving in a "big circle", then I must see him moving in one as well" (*) - an absurd proposition that is clearly contradicted by the everyday experience of riders viewing central poles of merry-go-rounds. And you reckon all that "makes perfect sense" ????? (*) Heh, reminded me of The Ravenous Bug-Blatter Beast of Traal, that believes that if you can't see it, then it can't see you - so the best way to escape it is to wrap a towel around your head."





An object positioned at/above the north pole must travel in a circle,
if the observer is rotating in a circle.


And if the observer decides to divide and segment that circle to east, west,
south, and north, then he/she can do it in any way he/she likes it to be.


happy.gif


But the fact of the matter is that, come hell or high water, an object positioned
at/above the north pole must travel in a circle, if the observer
is rotating in a circle.


And so, now, let's, please, take our hats off, for the first time,
to the great Aristarchus, the original discoverer
of the solar parallax:


http://www.astro.cornell.edu/academics/courses/astro201/aristarchus.htm


Because ancient Aristarchus of Samos, for certain, was, way more scientific,
visionary, and extremely smarter than modern Albert Einstein!



[linked image]














 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 27 2016, 3:43 AM 

AAF: An object positioned at/above the north pole must travel in a circle, if the observer is rotating in a circle.

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, self-evidently true for ANY object that is due North of ANY northward-facing 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 [linked image]


Love that Star; Also made me think of that song Putting on the Ritz wink.gif

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 28 2016, 12:00 AM 









What are EXACTLY the odds of convincing the Reserve Bank of Australia that
OUR printed $1,000,000,000 note is genuine?


Although I haven't done any math for it, I shall assume, at once, that the CHANCES,
here, are slightly better than those of selecting & sending UKIP Leader, Nigel Farage,
as an EU chief ambassador to Civilization #2 on Planet #7
around the star Alpha Centauri:


[linked image]

http://www.ukip.org/statement_from_ukip_leader_nigel_farage_on_eu_referendum_designation



[linked image]



….................................................................................................................................



""AFF: Oh, yeah . . . That is what the Sun is doing everyday"! Utterly irrelevant. The Sun is NOT a valid candidate for S_1. 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. "AAF: On the other hand, if the value of H is NOT so large, then S_1 will always appear to be moving, as one single unit, in a circle over the North Pole." And STILL just another empty repetition of the claim, with absolutely no basis or support (let alone any maths) given. Again: Don’t just claim it …. SHOW IT, yes? SHOW WITH MATHS, not just word claims."




The Sun, too, has daily parallax, which is the angle subtended by the mean equatorial
radius of the Earth at a distance of one astronomical unit:


http://www.thefreedictionary.com/solar+parallax


[linked image]


http://www.didaktik.physik.uni-due.de/~backhaus/Venusproject/Quarks/Idea.htm



So, now, let have your close attention, please!


The daily rising and setting of the Sun is NOT the solar parallax.


I repeat once again:


The daily rising and setting of the stars is NOT the stellar parallax.


happy.gif



The vast majority of people, who have ever lived on the face of the earth,
have and must have noticed the daily rising and setting
of celestial bodies very well.


But, at the same time, those people could have never noticed or even suspected
the presence of the solar and stellar parallax.


It's just unbelievably strange and too abstract for them . . .


Even the illustrious genius called 'Albert Einstein' never suspected,
in all his entire life, the existence of the S_1 parallax.


And therefore, take, please, your hat off to Aristarchus,
the discoverer of the solar parallax:

http://www.astro.cornell.edu/academics/courses/astro201/aristarchus.htm


Because Aristarchus, for sure, is so many light-years ahead of Einstein!


[linked image]


Anyway, the daily rising and setting of celestial bodies are caused by the angular
displacement due to the observer's angular velocity; that is to say, if the observer
is rotating at an angular velocity omega, then, in every interval of time equal
to delta_t, the celestial bodies will appear to rotate by an angle
equal to omega times delta_t.


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.























 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 29 2016, 12:52 PM 

AAF: What are EXACTLY the odds of convincing the Reserve Bank of Australia that OUR printed $1,000,000,000 note is genuine?

Probably better than the odds that people were giving Iceland !

AAF: Nigel Farage, as an EU chief ambassador to Civilization #2 on Planet #7 around the star Alpha Centauri:

I can picture another Civilization that would vie against that one
Ufonaut99: Utterly irrelevant. The Sun is NOT a valid candidate for S_1. 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.

AAF: if the value of H is NOT so large, then S_1 will always appear to be moving, as one single unit, in a circle over the North Pole.

Ufonaut99: And STILL just another empty repetition of the claim, with absolutely no basis or support (let alone any maths) given. Again: Don’t just claim it …. SHOW IT, yes? SHOW WITH MATHS, not just word claims."

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.

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 STRAIGHT-LINE motion.

S_2 is NOT straight-line motion; It is CIRCULAR.

Look, imagine a car driving in a straight line left-to-right past our merry-go-round, 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 merry-go-round 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 common-sense and see by looking at both merry-go-rounds and Polaris. Your claims that the central pole in that merry-go-round scenario would swing to her left and to her right remain ludicrous despite the mathematics of straight-line motion.

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.

PLEASE: apply any points and maths to S_1 and S_2, preferably referencing our graphic. There was nothing in your previous post that did.


Now, let's put some actual figures into your claim - eg. say that our camera has a field of view of a square 1 degree (so nice big telephoto lens happy.gif ).
We agree that :
- D = arctan(H / 3000) , where D is the angle of line of sight from our camera to the centre of S_1
- H is constant throughout S_2's revolution - let's take our graphic, with H = 4,000km
- Consequently :
D = arctan(H / 3000) = arctan( 0.75 ) = 53 degrees (-ish!)

However, you claim that it will be observed travelling in "big circles in the sky" - say 5 degrees radius, OK? happy.gif So at the top of the circle it's at 58 degrees, at the bottom at 48 degrees.

So what we have is (as seen by a human behind our camera) is the following - our camera's field of view being the bit in Yellow :
[linked image]

So I say S_1 is always at that D=53 degrees line, by definition.

You claim that, even though neither the H nor the 3,000 values change, that the line of sight to S_1 - that D = arctan( 0.75 ) sometimes = 58, and sometimes = 48 !!!

THAT is what I am still waiting for you to come up with some justification for.

Height in the sky is the angle D by defintion - and also by definition, D is constant in our scenario. Your wordy descriptions of "projected circles", etc are wrong.

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

June 30 2016, 12:00 AM 








Is it Iceland:
Where 1 in 10 people will publish a book?


http://www.bbc.com/news/magazine-24399599


[linked image]


…....................................................................................................................................................



"So you seem to be claiming that our observer 3,000 km from the pole, will see Polaris - or let's make it a low H value: will see Stanisława - making a "big circle" in the sky, always at the same angle D. Except that's impossible - the only way "D" could be constant would be for Stanisława to remain at the same height above the horizon, which is NOT true for star trails (nor parallax) - as is plainly evident in photos of star trails from those locations. Further, that would also mean Stanisława goes around (so sometimes behind) our observer, which clearly contradicts our graphic that shows that our observer is always directly facing Stanisława. So still, after all this time, just WHAT are you claiming????"





That is not only my claim; but it's also what the equation:

D = arctan(H/3000)

is claiming as well!


happy.gif


For example, if the height of S_1, over the north pole of S_2, is equal
to 4,000 km, then D is equal to about 53 degrees.


And that means, the rotating observer, on S_2, will see S_1 making a circle,
with an angular radius equal to 36.87 degrees, on the sky,
around the north pole of S_2.


And surely, a circle whose diameter, on the sky, is equal
to 73.74 degrees is very huge indeed.


There can be no doubt about it!



wink.gif










 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 2 2016, 12:00 AM 










""AAF: Surely, S_1 is always '3000Km straight ahead, and H km up'. .... if the value of H is NOT so large, then S_1 will always appear to be moving, as one single unit, in a circle over the North Pole. ... Does that make a perfect sense?" Hardly! I go instead with the commonsense straightforward view that anything that remains on (or directly above) the north pole remains due north of our observer. Since our observer is always facing due north, then he sees that object fixed in front of him. THAT is what makes perfect sense."






According to this illustration:


[linked image]


H is equal to 4,000 km.


And consequently:


D = arctan(4000/3000) = 53.13 degrees.


As a result, S_1 is always at 53.13 degrees over the horizon
of the observer on S_2.


What is the angular diameter of the circle, along which S_1 is traveling,
on the sky?


Well; the angular radius of that circle, R, can be obtained by using
this equation:


R = arctan(3000/4000) = 36.87 degrees.


And therefore, the angular diameter of the S_1 circle is equal to twice
the angular radius; i.e., it's equal to about 73.74 degrees


Let's, now, take our hats off to Eratosthenes,
for 'his remarkably accurate calculation of the
Earth’s circumference'
:

https://www.khanacademy.org/partner-content/big-history-project/solar-system-and-earth/knowing-solar-system-earth/a/eratosthenes-of-cyrene

[linked image]


That is because Eratosthenes, like Aristarchus before him, is, obviously,
at least, two thousands lightyears more modern than Einstein, Eddington,
&, perhaps, even Hawking!





[linked image]












 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 3 2016, 5:55 AM 

AAF: Is it Iceland: Where 1 in 10 people will publish a book?

Could be a certain UK supermarket happy.gif
[linked image]


AAF: That is not only my claim; but it's also what the equation: D = arctan(H/3000) is claiming as well!

Nonsense. That equation ONLY says that S_1 is at a constant height above the horizon; S_1 being fixed at one spot in the sky also fully consistent with that description.

OK, your claim is that "the big circle" is one at a constant height above the horizon - meaning the picture in my previous post (being my guess at your model) was not your model (although in my defence, you were the first to post it, and with the comment "S_1 making one of these circles" wink.gif )
Still, that answers that Up/Down question happy.gif ... but opens others ...

For example, say our observer is in Oslo, facing due north towards Lillehammer to the North, and with Gothenburg behind him due south. We agree S_1 is always due north above the North Pole, but you claim that S_1 will sometimes be due south over Gothenburg ???? That is just plainly ridiculous; the ONLY way that could happen would be for the Earth's axis of rotation to shift from through-the-North-Pole to through-Oslo instead, which is obviously just nuts.

In contrast, my claim is simply that what is above the North Pole (like Polaris) remains fixed at the same spot in the sky throughout the revolution (like Polaris so plainly does, and as plainly illustrated by our graphic).

AAF: What is the angular diameter of the circle, along which S_1 is traveling, on the sky? Well; the angular radius of that circle, R, can be obtained by using
this equation: R = arctan(3000/4000) = 36.87 degrees.

No - that's the radius of the circle that an observer at the centre of S_1 would see the camera making. Your idea does not agree wth the graphic, that clearly shows the observer at ALL times FACING S_1 - which would not be the case with S_1 being over Gothenburg (using the previous example)
So again: you're basing the whole thing on some flawed idea that "if he sees me moving in a "big circle", then I must see him moving in one as well" - an absurd proposition that is clearly contradicted by the everyday experience of riders viewing central poles of merry-go-rounds.

(oh, and that Parallax stuff about Aristarchus and Eratosthenes ... observers on both S_1 and S_2 can calculate H, so those posts are totally irrelevant to Einstein's scenario)

So let's go back to Alice on the merry-go-round (and also, since you're currently responding to posts from 2 weeks ago, it's about due for your next response happy.gif )

Ufonaut99: Alice standing on the outer edge of a spinning 3-metre radius merry-go-round in the middle of a grassy park. She is not moving a muscle, but looking inwards (so always looking inwards happy.gif ) towards Szczepan, who is (also without moving a muscle) standing on top of the central NON-rotating 4-metre pole that the merry-go-round goes round.
Someone's painted a couple of words on the merry-go-round - "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 3-metre radius merry-go-round, 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.

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 merry-go-round 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 counter-rotating (ie. spinning-on-the-spot) 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 merry-go-round is rotating - physically leave the merry-go-round, 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 happy.gif

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 4 2016, 12:00 AM 









This Iceland is way cozier than the other one:


[linked image]



I presume . . .



[linked image]



…...................................................................................................................



""AAF: a rotating observer, located in Oslo, will see Stanisława projected on the opposite arc .... Does this well-known optical illusion of circular motion become, now, very clear?" No - it remains blatantly false. The "opposite arc" of Oslo is Alaska, so the rotating observer will see Stanisława ALWAYS fixed stationary against Alaska throughout S_2's entire rotation (though again, we're supposed to be concerned solely with S_1 and S_2, not any background)."




Noway . . .


That is impossible . . .


Alaska & Siberia are supposed to have been removed along with the polar spherical cap
between Oslo & the North Pole!



happy.gif



And so, on the opposite arc of the circle of Oslo's geographical latitude,
there is only Stanisława along with the blue sky behind here.


The simple fact of the matter is that, as observed from the city of Oslo,
Stanisława is going to be projected on the other side of the latitudinal
circle regardless of any other objects in the background.


And this simply means that, in the previous hypothetical scenario, the polar lady,
Stanisława, will be seen, from Oslo, to travel in a big circle,
in which the city of Oslo is at the center.


But, for the sake argument, let's bring back Alaska!


Let's suppose furthermore that the state of Alaska is exactly at the other
side of Oslo's geographical latitude.


What will happen to the projection of the polar Stanisława
in this case?


Well; it's geometrically obvious,
Ufonaut99!


wink.gif


The polar Stanisława, in this case, will be projected on the state of Alaska
on the other side of Oslo's geographical latitude.


And therefore, Stanisława & Alaska will be seen, from Oslo, to travel together
along the periphery of a huge circle centered on the city of Oslo.


And that is it.














 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 4 2016, 10:37 AM 

Sigh ... just been through a long election campaign .... only to have it STILL dragging on sad.gif

The election was Soooo boring, one betting company was laying odds on what colour tie one of the commentators would be wearing on TV on election night !
It gave 6 options people could bet on ..... so the commentator in question swapped ties through the night to wear all 6 happy.gif

AAF: This Iceland is way cozier than the other one:

Yep - and not quite as many volcanoes wink.gif Apparently, Iceland has more volcanoes than it has professional football players .... err, I meant the OTHER iceland, not the food-store !!

AAF: Well; it's geometrically obvious, Ufonaut99!
The polar Stanisława, in this case, will be projected on the state of Alaska on the other side of Oslo's geographical latitude.
And therefore, Stanisława & Alaska will be seen, from Oslo, to travel together

YES ! EXACTLY !! Stop right there and stick with that ! happy.gif(*)

S_1 and Alaska will ALWAYS be seen from Oslo to travel together.

Therefore, our observer in Oslo will ALWAYS see S_1 against Alaska.

Therefore, our observer in Oslo will NEVER see S_1 "swing over Brisbane in the East" !

And, what about when S_1 is higher up? Same thing.

Let's rebuild poor old S_2 to restore it's north pole, but then get a HUMUNGOUS knife to slice a narrow slit down the Oslo/Alaska circle of longitude, so our observer can still see through to Alaska.
The slit, of course, marks due North from Oslo on the surface of S_2, and is a fixed feature on the surface of S_2 - It runs through Lillehammer to the north of Oslo, but not through Brisbane to the East, for example.
That means our observer can no longer see Brisbane in the East, for the same reason you've highlighted in the past when I've mentioned the north pole - yes? happy.gif

So, throughout S_2's rotation, our observer will always see S_1 against Alaska no problem - and never finds his view blocked of S_1 blocked (as it would be against Brisbane in the east).

What if Stanisława is standing on a bridge right at the tip of the north pole? No problem - our Oslo observer can still see her fixed at that location on S_2, always directly in line with Lillehammer along the slit (never over Brisbane nor Gothenburg).

What if Stanisława is raised, say, 4,000 km ? (let's be kind, and give her a spacesuit happy.gif )

STILL THE SAME THING. S_1 is still directly above that slit - that fixed point on S_2's surface - throughout S_2's rotation. If you doubt that, just imagine that they captured a few asteroids, and built up S_2 (with slit) all the way up to S_1. End result is the same - our observer would STILL always be able to see S_1 along the slit (through Lillehammer), which he wouldn't if S_1 was hovering over Brisbane in the East.

In summary, our observer in Oslo will always see S_1 over the SAME FIXED spot on Earth/S_2's surface (just the same as our Earth-based observers in Oslo see Polaris (**) )- always along the slit (and so always in line with Lillehammer) - and that means:

our observer is NOT seeing S_1 travel in any "big circle in the sky" around him.

You're absolutely right - it IS geometrically obvious happy.gif



(*) What about the rest of the sentence "along the periphery of a huge circle centered on the city of Oslo." ? That doesn't change the relative positions between Oslo, Stanislawa, Alaska and Brisbane - and therefore doesn't change the argument above

(**) Usual caveat of "idealised" polaris happy.gif

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 6 2016, 12:00 AM 











I see . . .


Queensland used to be yellow:



[linked image]




wink.gif



…...........................................................................................................................................................



""AAF: if the city of Brisbane is located on the eastern horizon .. S_1, sometime, seems to be over Brisbane ... Surely, S_1 is always '3000Km straight ahead". From where I am, real-life Brisbane CBD is due north, and the Pacific is to the east. While I am standing (not moving a muscle) facing due North: - There is NO optical illusion (let alone any "well-known" one (*)) that will put Brisbane CBD over the Pacific. - There is NO optical illusion (let alone any "well-known" one (*)) that will result in the Pacific ever being straight ahead of me. Any reader could substitute their own landmarks, to see that these will always be self-evidently true - and would remain true even if Brisbane was actually at the north pole (Brrrr, please no !)"





There is a 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.


Oslo & Brisbane are firmly attached to S_2; and hence the external object S_1 will
have no trouble at all to be over Oslo in the late afternoon, and to be over
Brisbane in the early morning.


That is one one hand.


On the other hand, attached Stanisława & Alaska will always be seen, from Oslo,
to travel together along the periphery of a huge circle centered
on the city of Oslo.


In other words, Stanisława cannot be be over Alaska in the afternoon, and over
Oslo in the morning; because she is firmly attached to S_2.


It's very clear and simple!



happy.gif















 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 8 2016, 12:00 AM 










"OK, you put a bit more detail into that post, so lets give it another shot.
[linked image]
So, from what you described in the final section of your post, seemed to imply the following: 1) Let's say that, as per the graphic, at 1am we point our camera directly at S_1, that is positioned fixed at a constant height H above the north pole. 2) Let's also say Mizar happens to also be in the camera's frame (although ignored by our observers). 3) Now, we agree that S_2 has actual rotation about it's axis (although again, not known by our observers) 4) However, assuming our camera-observer is stationary, then it will remain stationary throughout the rotation 5) (and so, consequently, always remain pointing towards Mizar)"





All the points, from Point #1 to Point #4
are O.K.



happy.gif



But Point #5 is NOT!


And that is clearly because, if the observer does not adjust periodically the direction
of the camera to keep pointing to S_1, then the camera will just keep rotating at the same
angular velocity of S_2 and will make 360 degrees relative to the sky every day.


Surely, S_1 will come back again to the same point 'at 1am' the next
day and everyday.


But that is it:
Just a few minutes per day.


How to make the camera keep pointing to S_1?


Quite simple!



wink.gif



If S_2 is rotating with an angular velocity equal to omega in a clockwise direction,
then simply make the camera rotate with an angular velocity equal to omega
but in a counterclockwise direction:


[linked image]


















 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 10 2016, 12:00 AM 











" 6) while S_2's rotation means S_2 will be rotating about it. 7) That is, although at 1am the north pole is directly ahead of the camera, at other times the pole will be rotating around (and at 1pm, be behind) it. 8) All of which means, whatever is above S_2's north pole will also be effectively rotating around the camera 9) which results in S_1 being seen travelling by our observer "in a big circle in the sky", such that Parallax can be applied."





Certainly, S_2's rotation means S_2 will be rotating around
the North Pole.


happy.gif


But, even though the North Pole is always pointing to the same point on the sky,
the line between the observer and the North Pole continues to
change direction periodically.


For example, 'at 1 am' each morning, the line between the observer
and the North Pole points towards the star Mizar.


While, 'at 8 pm' each evening, the line between the observer
and the North Pole points towards the star Vega.


And of course, the line between the observer and the North Pole makes
an angle of 360 degrees in every period.












 
 
Ufonaut99

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 11 2016, 7:57 PM 

AAF: I see . . . Queensland used to be yellow:

... And now it's blue ! After Queensland voted one particular candidate into the Senate, some people in other states started calling for a "Quexit" !

35E7665D00000578-3672142-image-a-28_1467


OK, I've got a few points wink.gif But I'll wait until your posts catch up before we get the discussion going again.

But just to further clarify your position :

AAF: But Point #5 is NOT! "[always remain pointing towards Mizar]"
If S_2 is rotating with an angular velocity equal to omega in a clockwise direction, then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction:

Aww, don't like Mizar ?? OK, let's go back to our merry-go-round, but add some extra details in :
- The park is surrounded by buildings - "Lillehammer Mall" to the North, "Brisbane Tourism" to the east, "Gothenburg Garage" to the south, and "London Towers" to the west.
- The Merry-Go-Round is rotating at the breakneck pace of one revolution per day clockwise.
- In keeping with our graphic, let's replace Alice with her nice simple objective cameras - Sony Alphas wink.gif - instead happy.gif
She has two of them, which we'll call "A0" and "A1", mounted one above the other, both with a 180 degree field of view.

The difference between them is that A1 has a motorised panning mount as below (so it'll counter-rotate as you've detailed above), whereas A0 is a plain, static tripod (so zero rotation of its own - zero - geddit wink.gif ).
Sevenoak-Automatic-Motorized-Rotating-Ti

So we start at midnight with the cameras due south, and Alice sets A1's motor to 1 revolution per day counter-clockwise.

This means we have :

* At 12am, cameras are due south, so both seeing the central pole directly ahead, against Lillehammer mall

* At 6am, cameras are due west, so see the central pole against Brisbane Tourism. However, while A0 is still pointing at the central pole, A1 has counter-rotated, so the pole is to it's right (and will be on the right edge of it's picture)

* At noon, cameras are due north, so see the central pole directly against Gothenburg Garage. However, A1 has counter-rotated 180 degrees, so the central pole is behind (and so out of shot) of that camera.

* At 6pm, cameras are due east, so see the central pole directly against London Towers. The central pole is now to the left of A1 (and so will be on the left edge of it's picture).

So to confirm that I understand you correctly : Your model is like that of A1, and the "big circle in the sky" that you say our observer sees S_1 make is that "straight-ahead" / "to it's right" / "behind" / "to the left" / "straight-ahead" circle that A1 sees the central pole travelling.

Is that your position ?

 
 
AAF

Re: You were right: Rotational motion is relative, too, Mr. Einstein!

July 12 2016, 12:00 AM 









No . . . no . . . no "Quexit"!


Just let the Australian Treasury dig a little deeper into its packets
and send this fine lady:


[linked image]


along with her voters to spend one, two, or three nights overseas with
very nice gentlemen like these:


[linked image]

http://www.smithsonianmag.com/travel/sleeping-with-cannibals-128958913/?no-ist



[linked image]


…......................................................................................................................................................................



"So is that a fair summary of your position? Pending your confirmation, it's worth pointing out that that position is totally inconsistent with the idea that "the external object S_1 does swing over to Brisbane due east" (Actually, your repetition of those "S_1 over Brisbane" claims is one of the reasons I've been thinking you couldn't have that model in mind.) but that's minor issue."




As I pointed out earlier, all points are O.K, except Point #5!


happy.gif


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.


Is it totally inconsistent with the idea that "the external object S_1 does swing
over to Brisbane due east?


Of course, NOT . . .


wink.gif


Let me explain:

Every value of the angle, between S_1, the observer and the North Pole, must make 360 degrees
around the observer in each and every period of S_2's rotation.


Except, of course, the value of 90 degrees between S_1, the observer
and the North Pole, where in this particular case, the object S_1 will
remain always stationary exactly overhead with respect to an observer
standing still at the S_2's axis of rotation.
















 
 
 
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