AAF

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

May 29 2016, 12:44 AM
 "...I can still remember being surprised at how inquisitive and friendly this thing was - it was actively looking around, and seeing what was going on. Interesting creatures.... An observer who moves has never been in question; I am disputing your claim that an observer who does NOT move will see see an up/down motion of S_1. THAT claim, since travel “along the periphery of a big circle” includes an up/down component. Let’s say the radius of your supposed circle is X degrees. It is clear from the illustration that S_1 is ALWAYS 3,000km straight ahead of the camera, and ALWAYS 4,000km above the camera. It’s line of sight, by simple trigonometry, is therefore ALWAYS at an angle of 53 degrees. Except you claim it’s not. You claim that that 3,000km 4,000km right-angled triangle sometimes has an angle of 53+X degrees, and sometimes has an angle of 53-X degrees, even though (by definition from the illustration) the lengths of the triangle never change. THAT is the claim I am asking you to justify." Of course, the carpet python is very inquisitive and friendly, when it's young and tiny like this: Even Jack the Ripper was so lovely and friendly, when he was a kid: https://en.wikipedia.org/wiki/Jack_the_Ripper I presume! But we have to wait, until its size becomes slightly gigantic like this:: O.K. . . The observer on S_2, as depicted in the above illustration, is always at rest in the same location. Now, S_2 is either rotating or at rest: [1.] If S_2 is at rest, then S_1 will appear, to the observer, to be at rest as well. [2.] But if S_2 is rotating around its geometrical axis, then S_1 will appear, to the same observer, to be in orbital motion around absolutely nothing. What is the justification for the last one? Well; it's obvious . . . The observer, in the latter case, thinks he's at rest; but he's, actually, constantly moving along the circumference of a huge circle whose diameter is equal to 6,000 km. And furthermore, the angle of 53 degrees,between the observer and S_1, is not a 2-dimensional, plane angle, but a 3-dimensional, solid angle, like this one: And therefore, S_1 has no other scenario beside appearing, to the observer on S_2, to be traveling constantly along the circumference of a humongous circle on the sky and centered around the North Pole.
Ufonaut99

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

May 30 2016, 12:32 AM
 Actually, young jungle pythons can be quite aggressive and snappy - maybe because they are so small and vulnerable. My son’s also tend to stay out of their hides more than others, which probably also contributes. As they get older, they mellow (a little!) - maybe because they’re bigger and/or maybe because they get used to being handled. ￼AAF: The observer on S_2, as depicted in the above illustration, is always at rest in the same location. …. The observer, in the latter case, thinks he's at rest; but he's, actually, constantly moving along the circumference of a huge circle whose diameter is equal to 6,000 km. And furthermore, the angle of 53 degrees,between the observer and S_1, is not a 2-dimensional, plane angle, but a 3-dimensional, solid angle, like this one: Yes. So in that picture, our camera is travelling around the upper slice, and S_1 is the point at the centre of the sphere, so forming a downward-pointing cone. ￼ AAF: And therefore, S_1 has no other scenario beside appearing, to the observer on S_2, to be traveling constantly along the circumference of a humongous circle on the sky and centered around the North Pole. And no. Again, just another unsubstantiated repetition of the claim. If I regard myself as stationary, then by definition I regard everything else that is stationary-relative-to-me as also being stationary. So, where is S_1 relative to me? IT IS STATIONARY. It is always at exactly 3,000km straight in front and 4,000km up from me. It is at that location regardless of whether I am stationary or whether I am moving - there is no change. To take the cone picture, my camera is whizzing around the circle at the base of the cone looking up (or down!) at the tip. If my camera is stationary, then I regard the cone as rotating, but that does not translate into my seeing the tip moving. (*) Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You are actually arguing the general case of bodies at OTHER locations, but repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis. For example, taking the cone again, if we were to be looking at the general case of a star that is away from the axis, then that cone would be lopsided. The camera being stationary and the cone rotating would then certainly result in our seeing the tip moving. For our special case of a body positioned along the axis, however, the cone is symmetrical, and so the tip does not move. So, leave those words, and instead concentrate on the plain simple maths that is clearly shown in the illustration : - S_1 is always 3,000km straight ahead and 4,000km up from the camera - That results in the line-of-sight always being at a 53 degree angle - You claim that, at say the bottom of the supposed circle, the line of sight will be less than 53 degrees Please describe in concise mathematical detail how you reckon this is possible. (*) For completeness sake, it’s worth mentioning that an alternate picture - that of the cone rotating “around” (from in-front, to behind, to in front again) the camera - is not applicable for this scenario, as it is in stark contradiction to the illustration where S_1 is always in front of the camera
AAF

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

May 31 2016, 12:00 AM
 "Actually, young jungle pythons can be quite aggressive and snappy . . . . Yes. So in that picture, our camera is travelling around the upper slice, and S_1 is the point at the centre of the sphere, so forming a downward-pointing cone. "And therefore, S_1 has no other scenario beside appearing, to the observer on S_2, to be traveling constantly along the circumference of a humongous circle on the sky and centered around the North Pole". And no. Again, just another unsubstantiated repetition of the claim. If I regard myself as stationary, then by definition I regard everything else that is stationary-relative-to-me as also being stationary. So, where is S_1 relative to me? IT IS STATIONARY. It is always at exactly 3,000km straight in front and 4,000km up from me. It is at that location regardless of whether I am stationary or whether I am moving - there is no change. To take the cone picture, my camera is whizzing around the circle at the base of the cone looking up (or down!) at the tip. If my camera is stationary, then I regard the cone as rotating, but that does not translate into my seeing the tip moving. (*) Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You are actually arguing the general case of bodies at OTHER locations, but repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis. For example, taking the cone again, if we were to be looking at the general case of a star that is away from the axis, then that cone would be lopsided. The camera being stationary and the cone rotating would then certainly result in our seeing the tip moving. For our special case of a body positioned along the axis, however, the cone is symmetrical, and so the tip does not move. So, leave those words, and instead concentrate on the plain simple maths that is clearly shown in the illustration: - S_1 is always 3,000km straight ahead and 4,000km up from the camera - That results in the line-of-sight always being at a 53 degree angle - You claim that, at say the bottom of the supposed circle, the line of sight will be less than 53 degrees. Please describe in concise mathematical detail how you reckon this is possible. (*) For completeness sake, it’s worth mentioning that an alternate picture - that of the cone rotating “around” (from in-front, to behind, to in front again) the camera - is not applicable for this scenario, as it is in stark contradiction to the illustration where S_1 is always in front of the camera." YEP . . . They can be quite snappy: http://www.acreptiles.com/main/index.php?option=com_content&view=article&id=81&Itemid=140 In this illustration: S_1 is only 4,000 km over the North Pole. And so, if the observer, on the rotating S_2, looks at S_1 from the east, S_1 will appear in the west. And if the observer, on the rotating S_2, looks at S_1 from the west, S_1 will appear in the east. And because the observer, on the rotating S_2, is traveling along a big circle, the end result of changing perspective, in this particular case, is a big circle, on the sky, along which S_1 appears to travel around the North Pole. But why doesn't S_1, in this scenario, behave like the star Polaris and stay stationary over the North Pole? It's because S_1 is only 4,000 km over the North Pole. While the star Polaris, by contrast, is more than 323 light-years over the same North Pole. And therefore, the angle between the observer and the star Polaris is much greater than 53 degrees; and almost equal to 90 degrees. In short, in order to make S_1 appear stationary, the observer must stand right smack on the exact geometrical point of the North Pole. Right?
Ufonaut99

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

May 31 2016, 8:59 AM
 ￼AAF: YEP . . . They can be quite snappy: Heh, he hasn’t got snapped like that. They do have teeth and can draw blood - so that guy would have had an interesting story to dine out on Mind you, it could have been worse .. AAF: S_1 is only 4,000 km over the North Pole. ONLY 4,000 km ???? Hey, I think it’s a long way down to the chemists …….. AAF: And so, if the observer, on the rotating S_2, looks at S_1 from the east, S_1 will appear in the west. And if the observer, on the rotating S_2, looks at S_1 from the west, S_1 will appear in the east. I don’t see what that’s got to do with up/down motion that we were discussing, but OK, its you want to add Left/Right motion (which, of course, is equally implicit in circular motion) to the question …. - when the observer is to the West, he is looking due East, and S_1 all also appear due East - ie. S_1 appears exactly in front of him. - when the observer is to the South, he is looking due North, and S_1 all also appear due North - ie. S_1 appears exactly in front of him. - when the observer is to the East, he is looking due West, and S_1 all also appear due West - ie. S_1 appears exactly in front of him. - when the observer is to the North, he is looking due South, and S_1 all also appear due South - ie. S_1 appears exactly in front of him. So, in addition to when the observer sees S_1 move Up/Down, we can now add the question : when and how you reckon he sees S_1 move Left/Right ? AAF: But why doesn't S_1, in this scenario, behave like the star Polaris and stay stationary over the North Pole? It's because S_1 is only 4,000 km over the North Pole. While the star Polaris, by contrast, is more than 323 light-years over the same North Pole. And therefore, the angle between the observer and the star Polaris is much greater than 53 degrees; and almost equal to 90 degrees. Utterly irrelevant. The maths remains the same, whatever the value for the height. Let’s replace that 4,000km with a variable H km. Consequently, our line of sight angle of 53 degrees is simply arctan(H/3,000) - let’s call it D. So, make H however large you like, the question remains : when and how you reckon he sees S_1 move above/below D ? AAF: In short, in order to make S_1 appear stationary, the observer must stand right smack on the exact geometrical point of the North Pole. Right? Blatantly wrong. Not according to the maths (nor the illustration) he doesn’t. So again, you are just making claims some “Big Circle” with words instead of maths - and those words are blatantly wrong. As I said before : don’t just claim it …. SHOW IT, yes? SHOW how something that is always 3000Km straight ahead, and 4000Km up, can be at anything other than that 53 degree angle in the sky. SHOW how something that is always straight ahead of the camera, can appear to move to the Left or Right. SHOW WITH MATHS, not just word claims. Right ? As an aside, it’s worthwhile considering what an effect a 4,000km height instead of a 323 light year height would have. The 4,000km will appear lower in the sky than Polaris. The other existing far stars, of course (assuming they could be seen ), would of course be entirely unaffected by this, and continue to produce their existing circular star trails. This would mean that those circles would not be centred on our 4,000km object. That’s irrelevant of course, since we are discussing S_1 staying at the same point in the sky, not staying at the centre of star-trails. S_1 will appear motionless, just as the centre pole of a merry-go-round that I’m riding appears motionless to me, even though that’s nowhere near even the 4,000km high
AAF

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

June 2 2016, 12:00 AM
 "ONLY 4,000 km ???? Hey, I think it’s a long way down to the chemists …….. I don’t see what that’s got to do with up/down motion that we were discussing, but OK, its you want to add Left/Right motion (which, of course, is equally implicit in circular motion) to the question …. - when the observer is to the West, he is looking due East, and S_1 all also appear due East - ie. S_1 appears exactly in front of him. - when the observer is to the South, he is looking due North, and S_1 all also appear due North - ie. S_1 appears exactly in front of him. - when the observer is to the East, he is looking due West, and S_1 all also appear due West - i.e. S_1 appears exactly in front of him. - when the observer is to the North, he is looking due South, and S_1 all also appear due South - ie. S_1 appears exactly in front of him. So, in addition to when the observer sees S_1 move Up/Down, we can now add the question : when and how you reckon he sees S_1 move Left/Right?" YEP! It's June 2 – Yell "Fudge" at the Cobras in North America Day: http://every-day-is-special.blogspot.com/2013/06/june-2-yell-fudge-at-cobras-in-north.html We are not discussing 'up/down' motion anymore, because you said earlier that the observer is not allowed to approach or recede from the North Pole! And so, the observer, on the rotating S_2, can only travel along a circle with a radius equal to 3,000 km. In short, our current scenario is just like as if you tie one end of a very long rope to the top of this Uluru (Ayers Rock): http://amazingdubaigu.blogspot.com/2013/01/50-natural-wonders-ultimate-list-of.html#sthash.DT3fOMhh.4GCpMJMK.dpbs and then you stretch it and hold its other end and begin moving in a big circle around that giant rock. Now, if you assume that you are stationary, then you must accept that Uluru (Ayers Rock) is really moving in a big circle around you; right? And the MATH of it, of course, is next time ...
Ufonaut99

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

June 2 2016, 7:52 AM
 AAF: It's June 2 – Yell "Fudge" at the Cobras in North America Day: Mmmm, Fudge I wonder if that’s how St Patrick got all the snakes out of Ireland ?? Wish it would work against Cane Toads, though But not the maroon cane toads - they’re great at crushing the blue cockroaches AAF: We are not discussing 'up/down' motion anymore, because you said earlier that the observer is not allowed to approach or recede from the North Pole! We most certainly ARE still discussing up/down motion; As I clearly stated : “I am disputing your claim that an observer who does NOT move will see an up/down motion of S_1” From the illustration and basic trigonometry, S_1 at height H km will always appear in the sky at angle arctan(H/3,000) - let’s call it D degrees. But you claim that our observer stationary on S_2 will see S_1 make a circular orbit in the sky - let’s say with radius R degrees. That would mean that our stationary observer sometimes sees S_1 at D+R degrees, and sometimes at D-R degrees. So yet again : don’t just claim it …. SHOW IT, yes? SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW how a right-angled-triangle with base 3,000km and height H km, can have that angle being D-R degrees. Take our H=4,000km straight from the illustration if you like - SHOW how and when our observer sees S_1 at less than 53 degrees SHOW WITH MATHS, not just word claims. Next post, yes?
AAF

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

June 4 2016, 12:00 AM
 Hi, Ufonaut99: It is not just Ireland; but, also, New Zealand, Hawaii, Greenland, Iceland, and Antarctica don’t have snakes: http://www.popsci.com/why-doesnt-ireland-have-snakes Just take a closer look at this thing: It's called the 'hooded malpolon'. And although harmless, it's playing the role of the cobra so well, that Indian and Australian soldiers in Montgomery's army, fighting in the Sahara desert in 1942, were more scared of it than of Erwin Rommel himself and his tanks! ….................................................................................................................................. "Utterly irrelevant. The maths remains the same, whatever the value for the height. Let’s replace that 4,000km with a variable H km. Consequently, our line of sight angle of 53 degrees is simply arctan(H/3,000) - let’s call it D. So, make H however large you like, the question remains : when and how you reckon he sees S_1 move above/below D? Blatantly wrong. Not according to the maths (nor the illustration) he doesn’t. So again, you are just making claims some “Big Circle” with words instead of maths - and those words are blatantly wrong. As I said before : don’t just claim it …. SHOW IT, yes? SHOW how something that is always 3000Km straight ahead, and 4000Km up, can be at anything other than that 53 degree angle in the sky. SHOW how something that is always straight ahead of the camera, can appear to move to the Left or Right. SHOW WITH MATHS, not just word claims. Right? As an aside, it’s worthwhile considering what an effect a 4,000km height instead of a 323 light year height would have. The 4,000km will appear lower in the sky than Polaris. The other existing far stars, of course (assuming they could be seen), would of course be entirely unaffected by this, and continue to produce their existing circular star trails. This would mean that those circles would not be centred on our 4,000km object. That’s irrelevant of course, since we are discussing S_1 staying at the same point in the sky, not staying at the centre of star-trails. S_1 will appear motionless, just as the centre pole of a merry-go-round that I’m riding appears motionless to me, even though that’s nowhere near even the 4,000km high". The height of S_1, over the North Pole, is very important. Because the diameter of the projected circle of the observer's latitude, on the sky, is inversely proportional to the height of S_1 over the North Pole; i.e., the larger the height; the smaller the subtended circle; and vice versa. As for the math, here, it's exactly the same as the math of parallax: http://www.esa.int/Our_Activities/Space_Science/Gaia/Parallax But whether S_1 is assumed to be too close or faraway, Einstein's assumption of relative rotation is always false. And that is because, if S_1 is nearby, then it must appear to travel along a huge circle, because its parallax is big. And if S_1 is faraway, then it must appear to be non-rotating and stationary, because its parallax is so tiny and very close to nil.
Ufonaut99

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

June 4 2016, 7:30 PM
 Hey AAF AAF: It is not just Ireland; but, also, New Zealand, Hawaii, Greenland, Iceland, and Antarctica don’t have snakes: http://www.investigatemagazine.co.nz/Investigate/12827/snakes-drain-snake-free-nz-found-aussie-serpents/”>Not native snakes, no That hooded malpolon looks fierce - I can certainly understand why even brave soldiers got jittery Equally fierce looking is this one - love that array of bulging teeth : When I first arrived in Australia, I thought I’d take up Scuba diving. After getting my licence, I thought “What’s a good first dive to do?”, and it just so happened a local dice shop was having a Shark-Dive So we drove up to Seal Rocks, and dived to see these (that photo’s nothing to do with me, though ! ) These are “Grey Nurse” sharks, and are actually quite docile - beautiful creatures. Unfortunately, their appearance led to them getting a bad reputation - even as man eaters - and they were hunted, so they’re still on the endangered list today Let’s clarify the last point first : AAF: if S_1 is faraway, then it must appear to be non-rotating and stationary, because its parallax is so tiny and very close to nilSorry, but I couldn’t care less about “very close to nil” I’m talking about the maths - and maths is not dependent on the resolving power of our telescope. Either S_1 is in EXACTLY the same position in the sky, or it is not. So let’s leave the fudge factor of “so tiny” out of it; that is not what we are talking about - right ? And that, in turn, means The height of S_1, over the North Pole, is still totally unimportant. Which brings us to : AAF: As for the math, here, it's exactly the same as the math of parallax: Really ??? Because that page you linked to, that has that graphic, describes Parallax as : “Now open the other eye, and close the first. Your finger appears to move against the more distant objects, even though you haven’t moved it. This is parallax!” …. but our scenario explicitly does not have any “more distant objects”, remember ? So no, it can’t be the same as that. What about the “close the first [eye]” bit? Again no, that’s a different scenario from ours, since “your finger” is not on any axis of rotation (unless you have very weird eyes !!), so the maths can’t be the same as that one either. So what I said earlier still holds : 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. I am asking you, to make a clear case, in your own words, on OUR explicit scenario. That’s actually one of the main reasons I’ve been repeatedly asking “SHOW WITH MATHS, not just word claims”, because that it’s only by the maths - APPLIED TO THIS SCENARIO - that your argument becomes clear and unambiguous. So again, for OUR scenario where S_1 is positioned along the axis of rotation : D = arctan( H / 3000 ) is the equation, and for a given H and a stationary observer, that angle D does NOT change during S_2’s revolution …… except, of course, you claim that it does. SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW how a right-angled-triangle with base 3,000km and height H km, can have that angle being D-R degrees. Take our H=4,000km straight from the illustration if you like - SHOW how and when our observer sees S_1 at less than 53 degrees SHOW WITH MATHS, not just word claims - Yes ?
AAF

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

June 6 2016, 12:00 AM
 YEP! It looks as fiercesome & fearsome as its cousin, the great white shark: https://en.wikipedia.org/wiki/Sand_tiger_shark …....................................................................... "We most certainly ARE still discussing up/down motion; As I clearly stated : “I am disputing your claim that an observer who does NOT move will see an up/down motion of S_1”. From the illustration and basic trigonometry, S_1 at height H km will always appear in the sky at angle arctan(H/3,000) - let’s call it D degrees. But you claim that our observer stationary on S_2 will see S_1 make a circular orbit in the sky - let’s say with radius R degrees. That would mean that our stationary observer sometimes sees S_1 at D+R degrees, and sometimes at D-R degrees. So yet again : don’t just claim it …. SHOW IT, yes? SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW how a right-angled-triangle with base 3,000km and height H km, can have that angle being D-R degrees. Take our H=4,000km straight from the illustration if you like - SHOW how and when our observer sees S_1 at less than 53 degrees. SHOW WITH MATHS, not just word claims." An observer who does not move, can see nothing: No up/down motion & no circular motion! But the rotating observer, on S_2, can see that S_1 is making a circle, on the sky, directly proportional to the circle, along which he/she is moving; and inversely proportional to the distance between S_1 & S_2. Surely, S_1 is always '3000Km straight ahead, and H km up'. Nonetheless, the straight line between the observer & S_1 is constantly sweeping the sky and drawing a huge circle, along which S_1 appears to be continuously and tirelessly moving around the North Pole. What about the math of it? Well; the math has already been worked out by the astronomers: http://www.math.nus.edu.sg/aslaksen/gem-projects/hm/0506-1-16-Parallax.pdf
Ufonaut99

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

June 7 2016, 12:06 AM
AAF

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

June 8 2016, 12:00 AM
 "Let’s clarify the last point first: "if S_1 is faraway, then it must appear to be non-rotating and stationary, because its parallax is so tiny and very close to nil". Sorry, but I couldn’t care less about “very close to nil”. I’m talking about the maths - and maths is not dependent on the resolving power of our telescope. Either S_1 is in EXACTLY the same position in the sky, or it is not. So let’s leave the fudge factor of “so tiny” out of it; that is not what we are talking about - right? And that, in turn, means The height of S_1, over the North Pole, is still totally unimportant. Which brings us to: "As for the math, here, it's exactly the same as the math of parallax". Really ??? Because that page you linked to, that has that graphic, describes Parallax as: “Now open the other eye, and close the first. Your finger appears to move against the more distant objects, even though you haven’t moved it. This is parallax!” …. but our scenario explicitly does not have any “more distant objects”, remember? So no, it can’t be the same as that. What about the “close the first [eye]” bit? Again no, that’s a different scenario from ours, since “your finger” is not on any axis of rotation (unless you have very weird eyes !!), so the maths can’t be the same as that one either. So what I said earlier still holds : 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. I am asking you, to make a clear case, in your own words, on OUR explicit scenario. That’s actually one of the main reasons I’ve been repeatedly asking “SHOW WITH MATHS, not just word claims”, because that it’s only by the maths - APPLIED TO THIS SCENARIO - that your argument becomes clear and unambiguous. So again, for OUR scenario where S_1 is positioned along the axis of rotation: D = arctan( H / 3000 ) is the equation, and for a given H and a stationary observer, that angle D does NOT change during S_2’s revolution …… except, of course, you claim that it does. SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW how a right-angled-triangle with base 3,000km and height H km, can have that angle being D-R degrees. Take our H=4,000km straight from the illustration if you like - SHOW how and when our observer sees S_1 at less than 53 degrees. SHOW WITH MATHS, not just word claims - Yes?" Do I have to do the math? Well; why bother? I don't have to do that . . . You already have shown the math of it, yourself: D = arctan(H / 3000) And so, according to the above equation, if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90o. And hence, the projected circle, on the sky, becomes smaller and smaller until it finally vanishes even on the viewing screens of the Hubble Space Telescope. And conversely, if the numerical value of H approaches zero, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 0o as well. And as a result, the projected circle, on the sky, becomes bigger and bigger, until even an old rhino can see it: http://www.lasikmd.com/blog/5-animals-worst-vision/ Consequently, the numerical values of H are extremely important. But, no matter how big or how small is the area of the projected circle, on the height of S_1, the object S_1 must always appear, to the rotating observer, to be moving, as one single unit, along the periphery of that circle. And that, certainly, contradicts, sharply, Einstein's main supposition, according to which the rotating observer, on S_2, must see S_1 rotating, around S_2's axis of rotation, if S_1 is located along S_2's axis of rotation.
AAF

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

June 10 2016, 12:00 AM
 "No kidding, if I saw a great white while I was diving, I’d be jumping out of that water faster than a stingray! Grey Nurses are nice, though. I remember one dive, we turned up where there were 4 or 5, so they decided just to go away ... except one came back just to have a look at us (not to eat us! Great - thought I'd start with this one, as it's always nice to start with an agreement. Now, ’3000Km straight ahead, and H km up' is ALWAYS the same triangle, as shown in our graphic - and therefore by definition, S_1 is always at angle D = arctan( H / 3000 ) straight ahead (neither left nor right) to our observer. Not D + R nor D - R; ALWAYS just plain D. By definition. THEN SHOW IT ! Don’t just keep claiming it - show it with maths, and not just words – yes?" Actually, swimming with the sharks is far less dangerous than swimming above the sharks. I presume! And also having a shark shield, like this one, at hand, is a very good idea: https://en.wikipedia.org/wiki/Shark_Shield O.K. . . . So, S_1 is always at angle D = arctan( H / 3000 ) straight ahead (neither left nor right) to our observer. Not D + R nor D - R; ALWAYS just plain D. By definition. Now, let's consider the geographical terrain within the circle of the horizon around the rotating observer, on S_2. In relation to that circle, which, by definition, is always at rest relative to the rotating observer, on S_2, the celestial object named 'S_1' is always moving along a huge circle on the sky. And so, 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. But how & why? Well; it's because, the straight line between the observer and S_1 is NOT rigidly connected to S_1; ; right? Instead, the straight line between the observer and S_1 is actually rotating around S_1, because the observer is rotating around the North Pole of S_2. And therefore, S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west.
Ufonaut99

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

June 10 2016, 11:29 PM
 AAF: Actually, swimming with the sharks is far less dangerous than swimming above the sharks. I presume! You presume correctly Sharks like nice big juicy seals; humans are scrawny by comparison, so we're not on their favourites menu. When we're below the waves, it's obvious we're not seals - but if they sense something splashing around on the surface .... When I saw your picture with the diver, my first thought was that you were praising the spear-gun; Spear guns would probably do nothing to a shark except really cheese it off - so not too good an idea ! SharkShield looks ideal - but an Aussie with a good right-hook could be even better ! AAF: Do I have to do the math? Well; why bother? I don't have to do that . . . You already have shown the math of it, yourself: D = arctan(H / 3000) True ..... although you would have to do a bit more math if you are claiming something different from that equation - say, for example, if you reckoned the line of sight to S_1 sometimes had a different angle AAF: And so, according to the above equation, if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90o. Scary .... we're still agreeing (*)AAF: And hence, the projected circle, on the sky And then you go and spoil it ! WHAT "projected circle" ??? Again, two simple and obvious facts : 1) D is the angle of the line-of-sight from the observer to S_1; For our illustration with a given H, D is constant, meaning a camera fixed at that angle keeps S_1 in view. 2) If S_1 is moving around a circle in the sky, it will be continuously changing it's angle of line-of-sight : you'd have to move your head - and camera - up/down and left/right to follow it Those statements are CONTRADICTORY - so you have to choose And if you choose option (2), then you have to SHOW IT with maths, not just words - yes? AAF: And therefore, S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west. Since our observer is always facing north, then east is to his right, and west is to his left. OK, explain to me how S_1, which we've agreed is ALWAYS straight in front of our observer, sometimes appears off to his left, and sometimes to his right??? That just makes no sense. If I had to guess what you mean (and I shouldn't have to guess, because you should be spelling it out clearly, yes? ) - but if I had to guess ..... Taking a low value for H first: Let's say our camera such a really wide-angle lens it can take in both Brisbane and Oslo. At 1am it just so happens that the camera films at low H (so D is just above the horizon) exactly due north (ie. directly across the north pole / axis of rotation) a distant star - say Sirius. Now, Sirius will behave exactly as you've been saying - the camera will film it rotating around at a constant angle D(**) in the sky, and indeed will film it sometimes over Brisbane, and sometimes over Oslo - and in fact, it's sometimes out-of-shot behind the camera. In short, yes, it certainly does travel a "Humungous circle" in the sky. If instead we tilt the camera so choosing a high H - say we find another star again directly north at 1am but higher in the sky - call it Mizar - then Mizar will make a much smaller circle in the sky. Thing is : I couldn't care less about Sirius or Mizar; They are NOT what we're talking about. We are suposed to be talking about S_1. Neither Sirius nor Mizar are candidates for S_1. S_1 is EXPLICITLY positioned directly up exactly along the axis of rotation. Sirius and Mizar are not. Let's go back to our view of Sirius at 1am. Suppose someone at Santa's workshop puts a yellow ball suspended directly above the exact North Pole, at a height that precisely blocks our camera's view of Sirius at that time. So, during S_2's rotation, our camera will film Sirius moving out from behind that ball - but since our camera is always facing North, and that ball is (by definition since it's at the north pole) always due north of it, then the camera will always have the ball in view. S_1 is THAT BALL, not Sirius, since it is that ball that is positioned along our axis of rotation, above our north pole. Which brings us back again to what I said earlier : 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. So let's fix that now - yes? Explain clearly (with maths ) to me how how our camera (that is always facing due north) can ever film that BALL (that is always positioned straight ahead, since it is always due north of the camera) way off to the right, over Brisbane in the east. (*) pretty much. The graphic should have had the right-angle inside the earth at the latitude of the camera, so the angle we're talking about is really between that point rather than the north pole. Trivial detail that doesn't change anything, though (**) Actually, D in this case would not be constant. For any distant body that is NOT positioned directly along the axis of rotation, then the distance between our camera and that body will vary as the rotation brings the camera closer then further to that body. That means that that 3000km figure will be changing, so D will be changing.
AAF

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

June 12 2016, 12:00 AM
 It's absolutely correct . . . The sharks are doing horrible things to 'nice big juicy' mammals! But let's NOT overlook what the ORCAS of the mammals are doing to the sharks: https://www.youtube.com/watch?v=GvTmrieZXBY ........................................................................ "As I said before : 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. Speaking of neglecting to take into account S_1 being on the axis of rotation ..... Yes - and that maths in that paper is full of scenarios like: - “your finger” in front of “your” eye - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Calculating distances using Parallax - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Simultaneous measurements from two locations - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Appearance shifting against background of stars - NOT APPLICABLE TO THIS EXPLICIT SCENARIO. Those scenarios (and more) are all applicable and incredibly useful with real-life Earth-based Parallax calculations - but that does NOT make them applicable to THIS EXPLICIT SCENARIO. Seriously, I asked you to make a clear case, in your own words, on OUR explicit scenario. That’s actually one of the main reasons I’ve been repeatedly asking “SHOW WITH MATHS, not just word claims”, because that it’s only by the maths - APPLIED TO THIS SCENARIO - that your argument becomes clear and unambiguous. Instead, you just point to some 50 page document, leaving all readers to guess which - if any - of the maths you think is applicable. So no more links to random pages on Parallax, which will be full of those scenarios not applicable to us - OK ? If you continue to think that Parallax is somehow applicable, then you should be able to SHOW it here - right? After all, ours is not a complex scenario, so easy to describe with no reason to link off anywhere else. So once again, PLEASE: MAKE A CLEAR CASE IN YOUR OWN WORDS ON THIS EXPLICIT SCENARIO. SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW WITH MATHS HERE, not just word claims, nor links talking about other unrelated scenarios - Yes?" It's true that S_1 is located along the axis of rotation over the North Pole. Does that make S_1 appear to be rotating around its geometrical axis? The answer is no! The rotation of the observer, on S_2, can only make S_1 appear to be moving, as one single unit, along a specific circle, on the sky. What does the math say about it? The math says, here, that there can be no apparent rotation of S_1 around its geometrical axis. Consequently, Albert Einstein got it all wrong. Period! How exactly does the mathematics of this situation say that there can be no apparent rotation of S_1 around its geometrical axis? The math says that the angle between S_1, the observer, and the North Pole varies with the value of the height H according to this equation: D = arctan(H / 3000) And that means: 1. If S_1 is too close above the North Pole, then S_1 will appear to make, periodically, a huge circle on the sky above the North Pole of S_2, along the circumference of which S_1 will always appear to move as one single unit. 2. And if S_1 is too faraway above the North Pole, then S_1 will appear to make, periodically, an extremely tiny circle on the sky above the North Pole of S_2, along the circumference of which S_1, also, will always appear to move as one single unit. And therefore, there can be no apparent axial rotation of S_1; none whatsoever.
Ufonaut99

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

June 12 2016, 5:51 AM
 AAF: But let's NOT overlook what the ORCAS of the mammals are doing to the sharks: Heh, loved the fact the guy said "", and he's still got over 1.5 million views ! (plus one ... but must admit, wish I had that time back !) AAF: The math says that the angle between S_1, the observer, and the North Pole varies with the value of the height H according to this equation: D = arctan(H / 3000) Look at the illustration again - H is CONSTANT throughout S_2's revolution. That means D is also CONSTANT throughout S_2's revolution. And that means S_1 has NO up/down motion. And that means S_1 is NOT moving in any circle. That's what the maths says But OK, let's take the simplest possible test case : Say hello to Stanisława from Krakow : 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. AAF: 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 ????
AAF

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

June 14 2016, 12:00 AM
 "True ..... although you would have to do a bit more math if you are claiming something different from that equation - say, for example, if you reckoned the line of sight to S_1 sometimes had a different angle. "And so, according to the above equation, if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90^o". Scary .... we're still agreeing (*). "And hence, the projected circle, on the sky". And then you go and spoil it! WHAT "projected circle" ??? Again, two simple and obvious facts: 1) D is the angle of the line-of-sight from the observer to S_1; For our illustration with a given H, D is constant, meaning a camera fixed at that angle keeps S_1 in view. 2) If S_1 is moving around a circle in the sky, it will be continuously changing it's angle of line-of-sight : you'd have to move your head - and camera - up/down and left/right to follow it. Those statements are CONTRADICTORY - so you have to choose. And if you choose option (2), then you have to SHOW IT with maths, not just words – yes?" I'm so glad that we've finally agreed that " according to the equation: D = arctan(H / 3000), if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90o". But the projected circle is extremely important within the current context of debunking, once and for all, Einstein's baseless conjecture about the relativity of axial rotation. What is the projected circle? 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. Generally, if the observer moves in a straight line, then he/she will see the external object moving in a straight line. And if the observer moves in a circle, then he/she will see the external object moving in a circle. And if the observer moves in a square, then he/she will see the external object moving in a square. And if the observer moves in an ellipse, then he/she will see the external object moving in an ellipse. And if the observer moves in hexagon, then he/she will see the external object moving in a hexagon. And if the observer moves in polygon, then he/she will see the external object moving in a polygon. And so on . . . and so on. In all cases, the basic geometrical properties of the observer's path of movement are conserved,upon projection on external objects; but their relative sizes vary inversely with increasing distance. Does the observer have to keep continuously changing the direction of the angle of the line-of-sight and to move his/her head left/right in order to follow the external object? You bet! Does the observer have to keep continuously changing the size of the angle of the line-of-sight and to move his/her head up/down in order to follow the external object? Absolutely not! And that is because the observer has assumed him/herself to be at rest; and therefore, he/she is always at the straight line that passes right smack through the center of the circle along which the external object is moving.
Ufonaut99

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

June 15 2016, 7:42 PM
 AAF: I'm so glad that we've finally agreed .... "finally agreed" ????? I never disagreed (beyond the earlier correction) ; I said it was irrelevant to the effect you were claiming (which it is). AAF: 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. Straightforward, simple, and consistent with maths and with everyone's (well, except maybe yours ) memories and expectations of merry-go-rounds. S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west. In contrast, based on spurious talk of "projected circles" and the like, you have convinced yourself that : As Alice is rotating on the merry-go-round, she sees the central pole move around her in a "big circle", from straight ahead, over the "Brisbane in the east", out behind her over the grass, back over the "Oslo in the west", and back to centre again. Sorry, I don't know what types of merry-go-rounds you've been on, but seriously, surely you must see that is utter nonsense. It is still nonsense if the merry go round is transplanted to the North Pole. And it remains nonsense if the 3-metre-radius circular merry-go-round is scaled up to our illustration of the 3,000km-radius circular slice through a line of latitude.
AAF

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

June 16 2016, 12:00 AM
 ""And therefore, S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west". Since our observer is always facing north, then east is to his right, and west is to his left. OK, explain to me how S_1, which we've agreed is ALWAYS straight in front of our observer, sometimes appears off to his left, and sometimes to his right??? That just makes no sense. If I had to guess what you mean (and I shouldn't have to guess, because you should be spelling it out clearly, yes?) - but if I had to guess ..... Taking a low value for H first: Let's say our camera such a really wide-angle lens it can take in both Brisbane and Oslo. At 1am it just so happens that the camera films at low H (so D is just above the horizon) exactly due north (ie. directly across the north pole / axis of rotation) a distant star - say Sirius. Now, Sirius will behave exactly as you've been saying - the camera will film it rotating around at a constant angle D(**) in the sky, and indeed will film it sometimes over Brisbane, and sometimes over Oslo - and in fact, it's sometimes out-of-shot behind the camera. In short, yes, it certainly does travel a "Humongous circle" in the sky. If instead we tilt the camera so choosing a high H - say we find another star again directly north at 1am but higher in the sky - call it Mizar - then Mizar will make a much smaller circle in the sky." "That just makes no sense"? Oh, yeah . . . That is what the Sun is doing everyday! And furthermore, the star Sirius: is located way south at the celestial latitude of −16° 42′ 58.0171; and therefore, it can see the South Pole; but it cannot see the North Pole: So, now, let's just forget Sirius and go back to Einstein's S_1! Because S_1 is over the North Pole of S_2; and because the observer is rotating around the North Pole of S_2, the observer, in this particular case, must see S_1 constantly moving in a circle over the North Pole of S_2, and centered around him/her personally. Does that make a perfect sense? You bet! It's just like this:
AAF

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

June 18 2016, 12:00 AM
 "Thing is: I couldn't care less about Sirius or Mizar; They are NOT what we're talking about. We are supposed to be talking about S_1. Neither Sirius nor Mizar are candidates for S_1. S_1 is EXPLICITLY positioned directly up exactly along the axis of rotation. Sirius and Mizar are not. Let's go back to our view of Sirius at 1am. Suppose someone at Santa's workshop puts a yellow ball suspended directly above the exact North Pole, at a height that precisely blocks our camera's view of Sirius at that time. So, during S_2's rotation, our camera will film Sirius moving out from behind that ball - but since our camera is always facing North, and that ball is (by definition since it's at the north pole) always due north of it, then the camera will always have the ball in view. S_1 is THAT BALL, not Sirius, since it is that ball that is positioned along our axis of rotation, above our north pole. Which brings us back again to what I said earlier : 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. So let's fix that now - yes? Explain clearly (with maths.) to me how how our camera (that is always facing due north) can ever film that BALL (that is always positioned straight ahead, since it is always due north of the camera) way off to the right, over Brisbane in the east. (*) pretty much. The graphic should have had the right-angle inside the earth at the latitude of the camera, so the angle we're talking about is really between that point rather than the north pole. Trivial detail that doesn't change anything, though. (**) Actually, D in this case would not be constant. For any distant body that is NOT positioned directly along the axis of rotation, then the distance between our camera and that body will vary as the rotation brings the camera closer then further to that body. That means that that 3000km figure will be changing, so D will be changing." Surely, the two stars Sirius & Mizar are not candidates; but the star Polaris certainly is: Well; the star Polaris, too, is not exactly over the North Pole. So, let's just go back to Einstein's S_1. And let's assume, as Einstein did in his 1916 paper, that S_1 is EXPLICITLY positioned directly up exactly along the axis of rotation; i.e., over the North Pole. And therefore, according to this equation: D = arctan(H / 3000) If the value of H is so large, then S_1 will always appear to be stationary over the North Pole and without any apparent motion or rotation whatsoever. That is on one hand. 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. But regardless of whether S_1 appears to be stationary or to be moving in a circle, this object labeled S_1 will never ever show any sort of axial rotation caused by the axial rotation of S_2. S_1 might show axial rotation, if it's actually rotating around its geometrical axis. But nonetheless, S_1 can never ever show any degree of axial rotation due to the axial rotation of S_2
Ufonaut99

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

June 18 2016, 8:50 PM
 Ufonaut99: That just makes no sense AAF: 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. 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???? 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. 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

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