SMOKING GUN

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Re: SMOKING GUN
« Reply #240 on: April 24, 2019, 04:04:38 AM »
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Now, we have one another issue :

Wiki quote :
Viewed from the same location, a star seen at one position in the sky will be seen at the same position on another night at the same sidereal time.

The question :
If sidereal time = the time it takes for the earth to rotate once on it's axis then how we can see a star at the same sidereal time at the same position given the fact that due to the precession of earth's axis, that same position (at which we should see the same star) after every single rotation should be off for 3,2 seconds (20 min = 1200 sec / 365 days = 3,2 seconds), that is to say : our star should lag (arrive at that same position) every sidereal day for 3,2 seconds???
No, you might "have one another issue" but the Heliocentric Solar System and Astronomers do not have any such problem.
  • The short answer is to be found by asking you, "What is the definition of a sidereal day? Answer that and you have answered your own question.
    The 20 or so minutes has nothing to do with the case. That is related to the time the sun crosses (the sub-solar point on TimeAndDate.com) the equator.

  • The long answer is that the axial precession, with a period of 25,772 years, does slightly affect the length of the sidereal day.
    The stellar day is the true rotational period of the earth and the "sidereal day" is the rotational period of the earth relative to the stars and the one affected by precession.
    Astronomers are primarily interested in the "sidereal day".

    The following values are from: International Earth rotation and Reference systems Service (IERS): USEFUL CONSTANTS. Refer to that reference for more details.
    The stellar day (the true period of rotation of the earth) is currently 86164.098 903 691 seconds and
    the sidereal day (rotation of the earth relative to the "so-called fixed stars") is currently 86164.090 530 832 88 seconds.

There is another point where confusion might arise when nit-pickers question slight differences in these values.
This is that most of the values, such as mean solar day, sidereal day, stellar day (and similar values for a year) change very slowly.
Because these slowly changing values would cause serious issues when it comes to precise time-keeping certain values are defined as of a certain date or period.
Hence the actual values will vary slightly from these defined values.

The solar day definition of 86,400 SI seconds has the most serious implications because the SI second was originally defined from average mean solar days taken before 1900. As a result, solar time does get out of step with these 86,400 SI second days and this "patched" with leap seconds.
This isn't a bad reference on time-keeping, What Is International Atomic Time (TAI)?

These "defined values" can be identified in the above "USEFUL CONSTANTS" because all are labelled as "exact" in the "Relative uncertainty in 10-6 (ppm)"

Are JB and Alpha2Omega included in those "others"?

Rabinoz, are you serious?

Stellar time is longer than sidereal time?

That is very much akin to saying that synodic time is (or could be, or maybe even should be) shorter than sidereal time. This is impossibility in HC Solar System!!!

But even if you confused stellar and solar times this slight difference between them is too small to be accounted for your "20 min" conundrum.

You said :

"The 20 or so minutes has nothing to do with the case."

However, the truth is that it has everything to do with the case.

In HC system 20 min before entirely accomplishing it's singe orbital cycle the earth's axis is (due to the precession of 50'' = 50 arcseconds) slightly tipped in CW direction so that the sun comes 20 min earlier at the equator. If there were no precession then sun's arrival at the equator would coincide with the very end of one sidereal cycle, so that there would be no 20 min discrepancy between tropical and sidereal year, that is to say : tropical and sidereal year would be of equal length.

Now, that you grasped this part, we can go on to the second part of this story.

Forget about the sun, since we have solved this issue (what is the cause for 20 min discrepancy between tropical and sidereal year).

All that you have to pay attention to, now is the amount of the alleged shift of earth's axis (which is the consequence of HC explanation for 20 min difference between tropical and sidereal year) with respect to the MOTIONLESS stars.

You see : the stars are MOTIONLESS, and the earth's spatial orientation is different after one sidereal cycle. However, according to HC theory after each sidereal time (86164 seconds) from the same location on the earth, you should see the same star at the same spot.

Wiki quote :

Viewed from the same location, a star seen at one position in the sky will be seen at the same position on another night at the same sidereal time.

It means that after 366 cycles you should also see the same star at the same spot from the same location on the earth, shouldn't you?

But Houston, we got the problem : since the stars are motionless, and earth's axis precessed, AND ONE ROTATIONAL PERIOD IS EQUAL TO ONE SIDEREAL PERIOD then we can't see the same star from the same location on the earth at the same spot in the sky after 366 rotational periods, since in the meantime, earth's axis precessed for 50'' which has to be the reason (besides 20 min difference between tropical and sidereal year) for waiting additional 20 min for the particular star to come at the same spot in the sky at which we saw it 366 sidereal times (31 536 024 seconds) before, as well.

But it isn't!

How so?

The motion  of the fixed  stars due  to the precession  of the equinoxes 
was a particularly  difficult  topic
since "it  is not reasonable  that  the inequality 
be at all noticeable  in the stars  in the life of one man"  and Tycho conceded  to the
Landgrave  of Hesse  in 1587:  "I  believe  that  an  exact  knowledge  of 
the apparent  motion of the eighth  sphere  is hardly  attainable  for any mortal"
  (VI, 73).

Re: SMOKING GUN
« Reply #241 on: April 24, 2019, 04:55:09 AM »
Stellar time is longer than sidereal time?
Note what he is saying, a stellar day not a solar day. Just because they might sound similar doesn't mean they are.
The exact definitions used can very between groups and context.

However, the truth is that it has everything to do with the case.
I thought we went over this and you already admitted you were wrong.
This is the same as you claiming the tropical year and sidereal year had to be the same for Earth.

Again, going back to my extreme example of a 2 sidereal year precession period, after 1 sidereal year, you are off by half a tropical year.
If what you are saying is true, then you would need to wait half a day for the stars to come back to their original position. But that is looking in the complete opposite direction.

i.e. looking from above the orbit of Earth you see Earth orbiting counter clockwise. Earth is on the right, with you on the right side at night and facing off to the right you see a star. You are claiming that after a year, when you are back on the right, due to the precession of the equinox you instead need to be on the left side of Earth, facing towards the left to see a star that is off to the right.
This makes no sense at all.

Again, you repeatedly ignoring these massive issues is not going to magically make them go away.

It isn't Earth falling behind so the stars aren't in the right position. It isn't simply rotating the celestial sphere. It is the Earth's axis changing direction. This does not produce a simple time delay.

However, according to HC theory
No it isn't.
Quoting wiki without understanding wont make your case.
It is a simplified description primarily for laypeople focusing on the difference between the solar and sidereal day.
Again, it doesn't come back to the exact same spot.
The precession means it will be offset and take the entire cycle to come back. Nutation also distorts that as does parallax and the proper motion of the stars.

IT IS AN APPROXIMATION!
Do you understand that?

So no, after sidereal time you do not see the same star in the exact same spot.

But it isn't!
How so?
Because you don't understand what you are talking about.

Re: SMOKING GUN
« Reply #242 on: April 24, 2019, 05:42:31 AM »
Stellar time is longer than sidereal time?
Note what he is saying, a stellar day not a solar day. Just because they might sound similar doesn't mean they are.
The exact definitions used can very between groups and context.
What is your problem? Who said "solar day"?

i.e. looking from above the orbit of Earth you see Earth orbiting counter clockwise. Earth is on the right, with you on the right side at night and facing off to the right you see a star. You are claiming that after a year, when you are back on the right, due to the precession of the equinox you instead need to be on the left side of Earth, facing towards the left to see a star that is off to the right.
This makes no sense at all.
Yes, looking from above the earth rotates CCW, so, as the consequence of our rotation CCW, the stars rotate (looking from above) CW, so, the stars are approaching to you from the left side (looking from above), and from your right side (looking from below, let's say when laying on your back at the north pole), so, since the earth is precessing to the right (looking from above), and the stars are approaching the observer from the left (looking from above), or from the right (from the observer's at the north pole perspective), that means that the stars have to lag behind 3,2 seconds after every sidereal time is accomplished wrt the same spot in the sky in which we saw them the night before, or 20 min wrt the same spot in the sky in which we saw them 366 sidereal cycles before.
This makes perfect sense.

However, according to HC theory
No it isn't.
Quoting wiki without understanding wont make your case.
It is a simplified description primarily for laypeople focusing on the difference between the solar and sidereal day.
Again, it doesn't come back to the exact same spot.
The precession means it will be offset and take the entire cycle to come back. Nutation also distorts that as does parallax and the proper motion of the stars.

IT IS AN APPROXIMATION!
Do you understand that?

So no, after sidereal time you do not see the same star in the exact same spot.

It doesn't even have to be the exact same spot in absolute terms, but it has to be the exact same spot wrt the observer.
What does that mean?
It means that one stellar day must be shorter 3,2 seconds than one sidereal day (because the earth allegedly precesses in the opposite direction of it's rotation) 50''/365,25 days.
If the earth precessed in the same direction (to the same alleged extent) then within HC theory one stellar day (one rotational period of the earth) would be 3,2 seconds longer than one sidereal day.

But it isn't!
How so?
Because you don't understand what you are talking about.
No, you don't understand what you are talking about, or are you deliberately lying again?
« Last Edit: April 24, 2019, 06:58:57 AM by cikljamas »

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rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #243 on: April 24, 2019, 05:53:04 AM »
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Are JB and Alpha2Omega included in those "others"?

Rabinoz, are you serious? Stellar time is longer than sidereal time?
Yes, I am very serious! Go ask some real astronomers. But did you miss this bit?
"The long answer is that the axial precession, with a period of 25,772 years, does slightly affect the length of the sidereal day.
The stellar day is the true rotational period of the earth and the "sidereal day" is the rotational period of the earth relative to the stars and the one affected by precession."

Quote from: cikljamas
That is very much akin to saying that synodic time is (or could be, or maybe even should be) shorter than sidereal time. This is impossibility in HC Solar System!!!
But even if you confused stellar and solar times this slight difference between them is too small to be accounted for your "20 min" conundrum.
I confused nothing!
The "stellar day" is the true rotational period of the earth but the sidereal day is a little shorter because it includes the effect of the precession.

Quote from: cikljamas
You said :

"The 20 or so minutes has nothing to do with the case." However, the truth is that it has everything to do with the case.
Incorrect!
The "20 or so minutes" the difference between a sidereal year and a tropical year but the stellar day and the sidereal day have nothing to do with what the sun appears to do.
 
I guess you failed to read:
This reference gives some detail and more accurate values for these periods.
Quote from: Professor Michael J. White, Arizona State University (Tempe Campus)
Sidereal, tropical, and anomalistic years

The relations among these are considered more fully in Axial precession (astronomy).

Each of these three years can be loosely called an 'astronomical year'.

The sidereal year is the time taken for the Earth to complete one revolution of its orbit, as measured against a fixed
frame of reference (such as the fixed stars, Latin sidera, singular sidus). Its average duration is 365.256363004 mean
solar days (365 d 6 h 9 min 9.76 s) (at the epoch J2000.0 = January 1, 2000, 12:00:00 TT).[3]

The tropical year is the period of time for the ecliptic longitude of the Sun to increase by 360 degrees. Since the
Sun's ecliptic longitude is measured with respect to the equinox, the tropical year comprises a complete cycle of the
seasons; because of the economic importance of the seasons, the tropical year is the basis of most calendars. The tropical year is often defined as the time between southern solstices, or between northward equinoxes. Because of the Earth's axial precession, this year is about 20 minutes shorter than the sidereal year. The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds (= 365.24219 days).

So a Sidereal Year is (at the epoch J2000.0 = January 1, 2000, 12:00:00 TT) 365.256363004 mean solar days or 366.256363004 sidereal days
But a mean Tropical Year is approximately 365.24219 mean solar days or 366.256363004 sidereal days.
They are different periods and the sidereal year has nothing to do with the sun.

Quote from: cikljamas
In HC system 20 min before entirely accomplishing it's singe orbital cycle the earth's axis is (due to the precession of 50'' = 50 arcseconds) slightly tipped in CW direction so that the sun comes 20 min earlier at the equator. If there were no precession then sun's arrival at the equator would coincide with the very end of one sidereal cycle, so that there would be no 20 min discrepancy between tropical and sidereal year, that is to say : tropical and sidereal year would be of equal length.

Now, that you grasped this part, we can go on to the second part of this story.

Forget about the sun, since we have solved this issue (what is the cause for 20 min discrepancy between tropical and sidereal year).

All that you have to pay attention to, now is the amount of the alleged shift of earth's axis (which is the consequence of HC explanation for 20 min difference between tropical and sidereal year) with respect to the MOTIONLESS stars.

You see : the stars are MOTIONLESS, and the earth's spatial orientation is different after one sidereal cycle. However, according to HC theory after each sidereal time (86164 seconds) from the same location on the earth, you should see the same star at the same spot.
The whole point is that all of the stars cannot be in exactly in the same spot because of the very slight precession of the earth's axis of one day.

Quote from: cikljamas
Wiki quote :

Viewed from the same location, a star seen at one position in the sky will be seen at the same position on another night at the same sidereal time.

It means that after 366 cycles you should also see the same star at the same spot from the same location on the earth, shouldn't you?
No, it does not mean that we should "see the same star at the same spot from the same location on the earth" because after one sideral year the earth's axis it pointing to a very slightly different position in the sky.
And you will find that any serious astronomer, amateur or professional, will get a new almanac for each year. You can dpwnlad it for free from the UNSO.
For 2018: ASTRONOMICAL PHENOMENA FOR THE YEAR 2018
For 2019: ASTRONOMICAL PHENOMENA FOR THE YEAR 2019
Both are marked:
Quote
Prepared Jointly by The Nautical Almanac Office
United States Naval Observatory and
Her Majesty’s Nautical Almanac Office United Kingdom Hydrographic Office

Quote from: cikljamas
But Houston, we got the problem : since the stars are motionless,
No, we do not have a problem!
The stars are not quite motionless - the "proper motions" (relative to earth) of many stars can be measured - though that is irrelevant to the problem..

Quote from: cikljamas
and earth's axis precessed, AND ONE ROTATIONAL PERIOD IS EQUAL TO ONE SIDEREAL PERIOD then we can't see the same star from the same location on the earth at the same spot in the sky after 366 rotational periods, since in the meantime, earth's axis precessed for 50'' which has to be the reason (besides 20 min difference between tropical and sidereal year) for waiting additional 20 min for the particular star to come at the same spot in the sky at which we saw it 366 sidereal times (31 536 024 seconds) before, as well.
I've been saying all along that the stars will not come back to exactly the same position.
The precession will slightly alter the declination (equiv to latitude) and Right Ascension (equiv to longitude) of the stars.

Quote from: cikljamas
But it isn't!
How so?

The motion  of the fixed  stars due  to the precession  of the equinoxes  was a particularly  difficult  topic since "it  is not reasonable  that  the inequality 
be at all noticeable  in the stars  in the life of one man"  and Tycho conceded to the Landgrave  of Hesse  in 1587:  "I  believe  that  an  exact  knowledge  of  the apparent  motion of the eighth  sphere  is hardly  attainable  for any mortal"  (VI, 73).
Oh, rubbish! You claim that because "Tycho . . . in 1587" believed it was impossible then it cannot be done with modern astronomical measurements!
Tycho Brahe didn't have any sort of a telescope!

Re: SMOKING GUN
« Reply #244 on: April 24, 2019, 06:57:36 AM »
However, according to HC theory
No it isn't.
Quoting wiki without understanding wont make your case.
It is a simplified description primarily for laypeople focusing on the difference between the solar and sidereal day.
Again, it doesn't come back to the exact same spot.
The precession means it will be offset and take the entire cycle to come back. Nutation also distorts that as does parallax and the proper motion of the stars.

IT IS AN APPROXIMATION!
Do you understand that?

So no, after sidereal time you do not see the same star in the exact same spot.

It doesn't even have to be the exact same spot in absolute terms, but it has to be the exact same spot wrt the observer.
What does that mean?
It means that one stellar day must be shorter 3,2 seconds than one sidereal day (because the earth allegedly precesses in the opposite direction of it's rotation) 50''/365,25 days.
If the earth precessed in the same direction (to the same alleged extent) then within HC theory one stellar day (one rotational period of the earth) would be 3,2 seconds longer than one sidereal day.

Wait a minute, what did i say? It doesn't even have to be the exact same spot in absolute terms?!?!??!
Well, mr Cikljamas, how could you say something like that?
Have you forgotten that the stars are MOTIONLESS?!?!?!?
So, feel free to correct yourself...
O.K., mr Cikljamas, i am going to correct myself right away :

It has to be the exact same spot in absolute terms, because the stars (in HC theory) are MOTIONLESS!!!

What would be the consequence of this preposterous theoretical postulate if HC theory were correct description of our reality???

There would be absolutely terrific consequences :

After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!

Have you ever seen such a big HC smoking gun???

« Last Edit: April 24, 2019, 07:02:59 AM by cikljamas »

Re: SMOKING GUN
« Reply #245 on: April 24, 2019, 09:41:06 AM »
No, within HC theory it is simply the earth is accomplishing it's annual orbit around the sun
No it isn't. Stop lying.
Do you really think repeating this lie again and again will do anything for you?
All it does is show you don't care about the truth.
No it doesn't, and i am going to prove this for the umpteenth time in a row (something that is absolutely strange, unknown, and unattainable to you) :
You are right, i was wrong (on this particular issue)! You see, how easy it is (for me), and how easy it should be for you (in many other cases), as well, to admit that (you are) i am wrong once that i (you) figure out that i (you) was wrong?!?!?! So, i proved once again that i care about the truth, unlike you!!!

I, for one, appreciate how forthcoming you are when you realize you were wrong. I wish others would do the same.
Are JB and Alpha2Omega included in those "others"?

No. Both of us have admitted it when we were wrong here.

Quote
Now, we have one another issue :

Wiki quote :

Viewed from the same location, a star seen at one position in the sky will be seen at the same position on another night at the same sidereal time.

The question :

If sidereal time = the time it takes for the earth to rotate once on it's axis then how we can see a star at the same sidereal time at the same position given the fact that due to the precession of earth's axis, that same position (at which we should see the same star) after every single rotation should be off for 3,2 seconds (20 min = 1200 sec / 365 days = 3,2 seconds), that is to say : our star should lag (arrive at that same position) every sidereal day for 3,2 seconds???

Which wiki? When you quote an online source, a link would be appropriate.

I ask because the context of the statement matters.

3.2 seconds of time is 213 milliarcseconds of earth rotation angle, which is too small to matter for most purposes except the most precise. This means the statement is true for practical purposes, which is likely the intent for which it was stated. In astrometry, an angle of that size would be significant; in more common use it isn't significant at all.

3,2 seconds per day = 20 minutes per year which is not small at all for most purposes not just for the most precise. And even 3,2 seconds per day is pretty significant since it presumes that sidereal day should be 3,2 seconds longer so to be able to see the same star at the same spot very next day (better to say : the very next night) at the same sidereal time.

Just to make sure neither of us had made a glaring blunder in the analysis or math, I used Stellarium to do some simulations because something didn't seem right, but I couldn't quite put my finger on what we were doing wrong.

Here's what I found:

At epoch J2000.0 [2000/01/01 at exactly noon TT (approximately noon UTC)], a magnitude 11.5 star in Gemini is just a few seconds off the ecliptic at ecliptic longitude 89°31'16.1", RA 5h57m54.74s and Dec +23°26'07.1". This star was selected because it is very near the ecliptic and very close to the solstice, so the effect of precession is greatest (ecliptic) and precession is almost entirely in RA with almost none in Dec (solstice).

Here's where it was at J2000.0, J2020.0, and J2020.0 + 1 day, and how far it moved in 20 years, and in 1 day:

Epoch (or span)
RA
Dec
Ecliptic Long.Comment
J2000.05h57m54.74s+23°26'07.1"
89°31'16.1"
2000/01/01 noon TT
2020 Jan 1.05h59m09.04s+23°26'08.0"
89°48'18.6"
2020/01/01 ~noon TT
2020 Jan 2.05h59m09.05s+23°26'08.0"
89°48'18.8"
2020/01/02 ~noon TT
20.0 years0h01m14.30s+00°00'00.9"
00°17'02.5"
Change after 7305 days
1.0 day0h00m00.01s+00°00'00.0"
00°00'00.2"
Change after 1 more day

So, in the 7305 days between the beginning of year 2000 and the beginning of year 2020, the vernal equinox moves 17' 2.5"  (1022.5" of arc) around the ecliptic, which works out to 0.1366 seconds of arc per day. This is consistent with the 0.2 seconds of arc shown for the one day at the beginning of 2020, considering rounding.

That motion manifested itself mostly as a change of 1m14.30s in the star's RA and a very slight change in declination over 20.0 years. Ignoring the change in Dec, that amounts to a change of 74.30s in RA (24h = 1440m = 86400s in a full circle kind of seconds). 1s in RA is the same as 15" of arc (360° = 21600' =  1 296 000"), so the change in RA is 15 ("/s) * 74.30s = 1114.5". That angle is measured from the axis of the earth and not its center, so it must be multiplied by cos(Dec) to convert it to the geocentric angle, which would be 1114.5" * cos(23.436°) = 1022.56" of arc, which is quite close to the change in ecliptic longitude; there is a small amount of slop in there due to effects like nutation that I have ignored but are built into the Stellarium calculations.

One thing that was bothering me is the daily change of location of the RA of the vernal equinox due to precession works out to something on the order of 0.01s per day on average, not the 3.2s we were discussing. So, what does that 3.2s we were looking at mean? That's how much longer the length of the mean solar day would be if the sidereal year were 20 minutes shorter than it actually is. It's not a real effect, it's an artifact of ignoring the lengths of the slightly-different years and considering only their difference.

It does get confusing, and this had me confused for a while. Here's an excellent example of why I enjoy participating in these discussions. I learn things!

Getting back to the question, the definition of a sidereal day is the length of time between meridian crossings of the vernal equinox (or really, any fixed RA, which has the VE as its reference). So the wiki article is only approximately correct (but close enough for most purposes) with their statement:

Viewed from the same location, a star seen at one position in the sky will be seen at the same position on another night at the same sidereal time.

Because it's the RA that will be at the same position on another night, not necessarily the star because the RA of most stars changes very slowly with time. It's slow enough that the difference is almost always insignificant for most practical purposes, even over a year. You never identified the actual source for that quote, so it's not possible to tell if the change from night to night matters for the intended audience or not. I'm betting not, since it's unlikely that an astronomer who needed the more precise definition would be looking at a wiki, so the quote is correct enough for practical purposes, even though it's not precisely correct.
 
Quote
Further, this translates into a difference of angular position in the sky with that magnitude only for objects on the ecliptic. You have to multiply that angle by the cosine of the ecliptic latitude; at the ecliptic poles there is no precession at all since cosine(90°) = cosine(-90°) = zero.

Can you perchance illustrate this with some helpful diagram?



It has taken me a while to examine the issues and compose this post. I see there have been others added in the meantime. I'm posting this as it was written and will look at the new posts later.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: SMOKING GUN
« Reply #246 on: April 24, 2019, 01:56:32 PM »
However, according to HC theory
No it isn't.
Quoting wiki without understanding wont make your case.
It is a simplified description primarily for laypeople focusing on the difference between the solar and sidereal day.
Again, it doesn't come back to the exact same spot.
The precession means it will be offset and take the entire cycle to come back. Nutation also distorts that as does parallax and the proper motion of the stars.

IT IS AN APPROXIMATION!
Do you understand that?

So no, after sidereal time you do not see the same star in the exact same spot.

It doesn't even have to be the exact same spot in absolute terms, but it has to be the exact same spot wrt the observer.
What does that mean?
It means that one stellar day must be shorter 3,2 seconds than one sidereal day (because the earth allegedly precesses in the opposite direction of it's rotation) 50''/365,25 days.
If the earth precessed in the same direction (to the same alleged extent) then within HC theory one stellar day (one rotational period of the earth) would be 3,2 seconds longer than one sidereal day.

Wait a minute, what did i say? It doesn't even have to be the exact same spot in absolute terms?!?!??!
Well, mr Cikljamas, how could you say something like that?
Have you forgotten that the stars are MOTIONLESS?!?!?!?

No one has forgotten that. It was never true in the first place.

Quote
So, feel free to correct yourself...
O.K., mr Cikljamas, i am going to correct myself right away :

It has to be the exact same spot in absolute terms, because the stars (in HC theory) are MOTIONLESS!!!

That is not correct. Not only do stars exhibit proper motion, but they also have parallax and their location changes in relation to celestial coordinates because of nutation and ... wait for it... precession. All of these are so small, slow, or both, that they can often be neglected with no consequence, but they still do exist and nullify your claims of exactness.

Quote
What would be the consequence of this preposterous theoretical postulate if HC theory were correct description of our reality???

There would be absolutely terrific consequences :

After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

This is not correct after all, as shown above. I admit that I was mistaken about that for a while, too.

Quote
So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!

This interpretation is incorrect.

Quote
Have you ever seen such a big HC smoking gun???
<picture of black-powder shotgun being fired>

As I said before, your English is excellent, and your familiarity with English idioms and figures of speech is quite good. However, the term "smoking gun" refers to a gun that has been fired so recently that it still has smoke coming out of it, not one being fired using black powder or other types of non-smokeless gunpowder. Here's an example:

"I didn't witness the murder, but I heard a shot and saw him standing near the body and holding a smoking gun."

Since "smoking gun" is an expression referring to an object or fact that provides conclusive evidence for something, you have shown no smoking gun in this entire thread.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: SMOKING GUN
« Reply #247 on: April 24, 2019, 01:58:37 PM »
What is your problem? Who said "solar day"?
You seemed to be confusing the 2.
I thought that was the basis of your objection.

If you would like to more clearly explain what your objection is, go ahead.

This makes perfect sense
Instead of completely ignoring what I have said just so you can assert pure garbage of it magically making sense, actually address what I have said.
It isn't the case of stars appropaching you or going away of any of that BS. We are simply discussing where the stars are and thus what direction you need to look, with the direction being in that frame looking down from above.

Do you need a picture?
Here, for the extreme precession example, first we have the start of the year:

The sun is in red, Earth is in blue, the direction our axis is pointing is in green, the distant star is off to the right, our line of sight to the star is the dashed grey line with our orbit shown as a solid grey line.
Now, this is the situation after the hypothetical sidereal year is over (i.e. one orbit around the sun):

Notice how while we are half way through the solar year, the sun is still off to the right, and we can see it just fine.

But instead you claim we need to wait an extra half day and have this nonsense:

Notice how now our line of site goes off towards the sun and away from the star?
It is pure nonsense and does match reality at all.
You don't need to wait the extra half day.
Waiting the extra half day means you are looking in the wrong direction.

Now do you understand?
Now are you going to stop repeating the same refuted nonsense?

It doesn't even have to be the exact same spot in absolute terms, but it has to be the exact same spot wrt the observer.
Doesn't matter. IT ISN'T GOING TO BE!
Do you understand that?
Due to all the different processes going on, it will not be the same exact spot.
If you made it so precession was the only thing affecting it, then in general, you would need to wait for the entire precession cycle to be over before it came back to the same spot. It is not a simple case of just waiting a few extra seconds.

So you need some other reference than "the exact same spot".

What does that mean?
It means that one stellar day must be shorter 3,2 seconds than one sidereal day
Again, PURE BS!
Repeatedly asserting the same BS won't magically make it true.

No, you don't understand what you are talking about, or are you deliberately lying again?
Stop projecting your own inadequacies onto others.
I have clearly explained why you are wrong, how you are completely misrepresenting what you are talking about.
You have been completely unable to refute that and instead just resort to repeating the same nonsense.
You haven't even attempted to honestly and rationally attempt to address the issues raised.

So it is quite clear which of us either has no idea what they are talking about or is intentionally lying; and it isn't me.

Have you ever seen such a big HC smoking gun???
Again, all you are doing is repeatedly asserting refuted BS.
You ignoring the refutation wont magically make it disappear.
The only smoking gun here is shooting yourself in the foot and showing just how dishonest you are.

So much for you finding it easy to admit when you were wrong.

Re: SMOKING GUN
« Reply #248 on: April 24, 2019, 02:30:44 PM »
Everybody seems to be forgetting about basic physics... frames of reference! From any object's frame of reference, it is stationary and everything else moves around it! From an Earthling's or the Earth's frame of reference, the Earth is stationary and the planets revolve 'round the Sun, which goes 'round the Earth, just as from a train's frame of reference, it is a machine for turning the universe around the core of the Earth until the part you want is facing you.

*

rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #249 on: April 24, 2019, 02:50:51 PM »
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wait a minute, what did i say? It doesn't even have to be the exact same spot in absolute terms?!?!??!
Well, mr Cikljamas, how could you say something like that?
Have you forgotten that the stars are MOTIONLESS?!?!?!?
So, feel free to correct yourself...
O.K., mr Cikljamas, i am going to correct myself right away :

It has to be the exact same spot in absolute terms, because the stars (in HC theory) are MOTIONLESS!!!
The stars in reality (ie the Heliocentric System) are not MOTIONLESS:

From: Hipparcos, Education & Images, High Proper Motion Stars
But that is somewhat immaterial in this discussion.

Quote from: cikljamas
What would be the consequence of this preposterous theoretical postulate if HC theory were correct description of our reality???

There would be absolutely terrific consequences :

After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!
Rubbish! The earth does not "precess 3,2 seconds per day in CW direction"! It is the rotational axis of the earth that precesses, ie changes direction slightly
The Axial Precession of the earth is not an extra rotation of the earth.

Re: SMOKING GUN
« Reply #250 on: April 24, 2019, 04:26:46 PM »

What would be the consequence of this preposterous theoretical postulate if HC theory were correct description of our reality???

There would be absolutely terrific consequences :

After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

This is not correct after all, as shown above. I admit that I was mistaken about that for a while, too.

You was mistaken about that for a while, and then you came up with the following ingenious solution :

One thing that was bothering me is the daily change of location of the RA of the vernal equinox due to precession works out to something on the order of 0.01s per day on average, not the 3.2s we were discussing. So, what does that 3.2s we were looking at mean? That's how much longer the length of the mean solar day would be if the sidereal year were 20 minutes shorter than it actually is. It's not a real effect, it's an artifact of ignoring the lengths of the slightly-different years and considering only their difference.

For God sake, if the sidereal year were 20 minutes shorter then sidereal year and tropical year would be of equal length, and your HC theory would be exempted from this massive "20 minutes" FLAW!!!

So, feel free to correct yourself...
O.K., mr Cikljamas, i am going to correct myself right away :

It has to be the exact same spot in absolute terms, because the stars (in HC theory) are MOTIONLESS!!!

That is not correct. Not only do stars exhibit proper motion, but they also have parallax and their location changes in relation to celestial coordinates because of nutation and ... wait for it... precession. All of these are so small, slow, or both, that they can often be neglected with no consequence, but they still do exist and nullify your claims of exactness.

So, according to new Copernicus, or better to say Kepler, or better to say Newton, or better to say Einstein ALPHA2OMEGA stars are NOT motionless, or better to say : they change their locations in relation to celestial coordinates because of nutation and ... wait for it ... precession...

...WAIT FOR IT ALPHA....PRECESSION IS THE EXACT REASON (ACCORDING TO YOUR HC THEORY) for 20 min discrepancy between sidereal and tropical year...so...after reading the second time your words it is obvious that you don't claim that the stars are NOT motionless, you claim that they change their locations in relation to celestial coordinates because of nutation etc...

So, can you answer directly (with yes, or no) to this question : are the stars motionless or not, according to you???

The same reason for 20 min discrepancy between the length of tropical and sidereal year must be the reason for 3,2 seconds per day difference between every consecutive sidereal period, so that after one single sidereal year there must be the same 20 min difference between the first sidereal cycle (86164 seconds) and the last (366th) sidereal cycle (86164 seconds + 20 min) in an accomplished sidereal year, also. You can't have you cake and eat it too...

So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!

This interpretation is incorrect.

Why?
Because MOTIONLESS stars are NOT motionless?!?!?!?!?
Or because you would rather die than admit that HC theory is wrong???
You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have...
Janis Joplin sang : Freedom is just another word for nothing left to lose...
You are not free persons, that is a real tragic life story, the situation is as bad as it can get when someone is slave of his own deliberate lies, so i would like to remind you to this :


« Last Edit: April 24, 2019, 04:31:38 PM by cikljamas »

Re: SMOKING GUN
« Reply #251 on: April 24, 2019, 07:45:16 PM »

What would be the consequence of this preposterous theoretical postulate if HC theory were correct description of our reality???

There would be absolutely terrific consequences :

After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

This is not correct after all, as shown above. I admit that I was mistaken about that for a while, too.

You was mistaken about that for a while, and then you came up with the following ingenious solution :

One thing that was bothering me is the daily change of location of the RA of the vernal equinox due to precession works out to something on the order of 0.01s per day on average, not the 3.2s we were discussing. So, what does that 3.2s we were looking at mean? That's how much longer the length of the mean solar day would be if the sidereal year were 20 minutes shorter than it actually is. It's not a real effect, it's an artifact of ignoring the lengths of the slightly-different years and considering only their difference.

For God sake, if the sidereal year were 20 minutes shorter then sidereal year and tropical year would be of equal length

Not if the tropical year were also 20 minutes shorter. My statement says nothing about the length of the tropical year, it only points out that, if you change the length of the year by 20 minutes, to get the same number of days, you have to adjust the rotation rate by 3.2 seconds per day.

If we start by assuming the tropical year is 365.2422 mean solar days, each exactly 1440 minutes of exactly 60 SI seconds, then the tropical year is 525 948.768 minutes long. There will be one more sidereal day than solar days in a tropical year, so 366.2422 sidereal days in a tropical year. The tropical year is used because the length of the sidereal day is defined in terms of the precessing reference frame.

525 948.768 min/yeartr / 366.2422 dayssi/yeartr = 1436.068176 min/daysi = 86164.09054 sec/daysi

Sound familiar?

If we further find that the mean sidereal year is exactly 20 minutes longer than the mean tropical year (it's not exactly that much longer in real life, but let's work the problem as if it is), then the sidereal year is 525 968.768 minutes long, which is 365.25609 mean solar days, or 366.25612 sidereal days long.

At any rate, since the sidereal year is 20 minutes longer than the tropical year, the earth has to travel 20/525 948.768 of a full orbit further to complete the sidereal orbit, which is 1/26 297.44 of the orbit, which is about what you'd expect after a year if the precessing coordinate systems is rotating once in about 26 000 years with respect to the inertial one.

The rotating earth completes 1/720 of a complete rotation in those 20 minutes since it spins a lot faster than it revolves around the sun, but distributing that rotation over the days doesn't have any significance here because the tropical year/sidereal year comparison involves the relationship between the sun and the precessing equinox and the inertial coordinate system, and neither of those is concerned with the rotation angle of the earth.

Quote
and your HC theory would be exempted from this massive "20 minutes" FLAW!!!

It's not a flaw in the heliocentric model, massive or otherwise. It's a flaw in your understanding of what it happening.

Quote
So, feel free to correct yourself...
O.K., mr Cikljamas, i am going to correct myself right away :

It has to be the exact same spot in absolute terms, because the stars (in HC theory) are MOTIONLESS!!!

That is not correct. Not only do stars exhibit proper motion, but they also have parallax and their location changes in relation to celestial coordinates because of nutation and ... wait for it... precession. All of these are so small, slow, or both, that they can often be neglected with no consequence, but they still do exist and nullify your claims of exactness.

So, according to new Copernicus, or better to say Kepler, or better to say Newton, or better to say Einstein ALPHA2OMEGA stars are NOT motionless, or better to say : they change their locations in relation to celestial coordinates because of nutation and ... wait for it ... precession...

...WAIT FOR IT ALPHA....PRECESSION IS THE EXACT REASON (ACCORDING TO YOUR HC THEORY) for 20 min discrepancy between sidereal and tropical year...so...after reading the second time your words it is obvious that you don't claim that the stars are NOT motionless, you claim that they change their locations in relation to celestial coordinates because of nutation etc...

So, can you answer directly (with yes, or no) to this question : are the stars motionless or not, according to you???

No. But it's not according to me. It is according to all competent astronomers, and I agree with them.

Quote
The same reason for 20 min discrepancy between the length of tropical and sidereal year must be the reason for 3,2 seconds per day difference between every consecutive sidereal period, so that after one single sidereal year there must be the same 20 min difference between the first sidereal cycle (86164 seconds) and the last (366th) sidereal cycle (86164 seconds + 20 min) in an accomplished sidereal year, also. You can't have you cake and eat it too...

There is no 3.2 second per day discrepancy.

First of all, you don't add the 20 minutes here. The length of the sidereal day is defined in terms of the precessing reference frame, which is based on the equator and equinox, not the inertial frame. If you define the length of the sidereal day in terms of the inertial reference system, it will be slightly different... one extra rotation per sidereal year instead of tropical year, which works out to a difference of about 7 milliseconds per day, I think.

Quote
So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!

This interpretation is incorrect. Truly motionless stars along the ecliptic would change their lag versus vernal equinox by 20 minutes in a year due to precession. At the ecliptic poles, they wouldn't change at all.

Why?
Because MOTIONLESS stars are NOT motionless?!?!?!?!?

There are no perfectly motionless stars. Many of them appear motionless because their apparent motion is too slow to detect reliably, but there is no reason to expect even those to be truly motionless.

Quote
Or because you would rather die than admit that HC theory is wrong???

So far no one has offered a plausible alternative to the heliocentric model. You certainly haven't. If someone did, I'd consider it.

Don't forget, your model not only has to accurately model the things we see, but it also needs to provide a plausible physical explanation why what we see happens. The first can be accomplished by a simple change of reference, but it becomes very difficult to explain why everything orbits around a stationary earth. The why is where you should be concentrating your effort; the first part is straightforward.

Quote
You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have...
Janis Joplin sang : Freedom is just another word for nothing left to lose...
You are not free persons, that is a real tragic life story, the situation is as bad as it can get when someone is slave of his own deliberate lies, so i would like to remind you to this :

We're constrained by reality. That's true. That's a problem with working in the real world instead of just fantasizing about how we'd like things to be.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

*

rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #252 on: April 24, 2019, 08:03:42 PM »
Why?
Because MOTIONLESS stars are NOT motionless?!?!?!?!?
In the Heliocentric Solar System the stars are not completely MOTIONLESS they appear to shift due the axial precession and really do move due to their proper motion.

Quote from: cikljamas
Or because you would rather die than admit that HC theory is wrong???
And why is that relevant? After I post I think over what I've written and try to think where I might be wrong - maybe I have been but not intentionally.
Where, exactly, have you proved anyone wrong on things that matter?

Quote from: cikljamas
You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have...
So now you are a qualified psychologist licensed to practice on-line ::)?

All you seem to be concerned with is trying to find nit-picky little differences is what is said and written in Wikipedia etc.

Quote from: cikljamas
You are not free persons, that is a real tragic life story, the situation is as bad as it can get when someone is slave of his own deliberate lies,
Now you are accusing us of "deliberate lies" - ever looked in a mirror? But I'm quite free to accept what I believe best explains what I see and know.

Yet it seems that you are tied to an ancient idea that has been proven wrong numerous times.

I assume that you adhere so tightly to this in the mistaken belief that Scripture teaches that the earth must be stationary but
probably more believe that Scripture teaches that the earth must be both stationary and flat -  who is right, the flat earthers, you or us?

You might read what is written on the "Young Earth Creation" site about Geocentrism:
          Creation Ministries International, Geocentrism and Creation
          Creation Ministries International, ‘Nevertheless, it moves!’: Copernicus, Galileo, and the theory of evolution
          Creation Ministries International, The truth about the Galileo affair
          Creation Ministries International, Refuting absolute geocentrism. Refutation of our detractors

I don't necessarily agree with all that is on Creation.com but I just showed those to impress on you that there are many interpretations of Scripture.
Not only that but there are various different ideas on whether Scripture was ever intended to teach anything about science.
Likewise, I do not support Catholicism but what they say is also relevant:
          Science and Western Civilization’s Debt to Catholic Church
Then there is: It could be said that the "big bang" theory was first proposed by a "creationist" :).
And Stanford SOLAR Center: Does the Judeo-Christian Concept of Creation Conflict with The Big Bang? from WELCOME TO THE AMERICAN SCIENTIFIC AFFILIATION: A Network of Christians in the Sciences

I more follow Galileo, referring to the Geocentrism vs Heliocentrism issue, when he argued that God wrote two books:
Quote from: Kelly James Clark
Science and Religion: Two Books
Galileo argued that God has written two books — the Book of Nature and the Book of Scripture — and that these two books do not, because they cannot, contradict. That means that if one has a well-established scientific explanation of the physical world that seems to contradict a passage of Scripture, one has good reason to reconsider one’s interpretation of Scripture. The surface meaning of the Bible may not be its true meaning.
<< see the rest for more detail >>
Galileo's letter can be found at: Letter to the Grand Duchess Christina of Tuscany (1615) (abridged) by Galileo Galilei

So for you to sit in judgement and claim, "I don't have such a serious psychological problem as you, JB, and Rabinoz have" sounds highly pretentious.

Now would you finally care to answer what you believe are the laws of physics, including motion and gravitation, that govern how things move.

If you do not agree with Newton's Laws of Motion and Universal Gravitation just what rule does your Universe follow?
Do you follow Dr. Robert Sungenis and claim that the Geostationary Universe does obey Newton's Laws as well as Mach's Principle.
You might find this interview interesting: Interview with Robert Sungenis on  Phil Plait’s Criticism of Geocentrism though I can guess who you might support.

By the way, you insisted that you are correct and everybody else from Einstein and Newton down are wrong but what about YouTube videos like:
        THE EARTH IS FLAT - GASLIGHT by odiupicku
        THE FLAT EARTH - ANTARCTICA (reupload) by odiupicku
        The Flat Earth The True Law of Perspective by odiupicku
        THE FLAT EARTH - THE SUN by odiupicku
So back there the earth was flat but since then the earth seems to have magically become a Globe but it still won't move.
Maybe by this time next year, you might discard the Geostationary Universe and finally accept the Heliocentric Solar System - who knows?

Re: SMOKING GUN
« Reply #253 on: April 24, 2019, 11:05:12 PM »
your HC theory would be exempted from this massive "20 minutes" FLAW!!!
The only flaw is in your understand and the nonsense you spout. You are yet to show any problem with the HC model

So, according to new Copernicus, or better to say Kepler, or better to say Newton, or better to say Einstein ALPHA2OMEGA stars are NOT motionless, or better to say : they change their locations in relation to celestial coordinates because of nutation and ... wait for it ... precession...
And the fact that they aren't actually motionless.

So what? That in no way backs up your BS claims.

The same reason for 20 min discrepancy between the length of tropical and sidereal year must be the reason for 3,2 seconds per day difference
Again, PURE BS!
Repeatedly asserting the same refuted BS won't magically make it true.
There is no connection between the real 20 minutes difference and your 3.2 seconds of bullshit.
You clearly either have no idea what you are talking about or you are blatantly lying to everyone. Which is it?

You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have...
Searching for the truth isn't a problem. So we don't have a problem, you do.

You repeatedly assert the same refuted nonsense even when it is exposed to make absolutely no sense.
Rather than address these massive issues you just ignore them or blatantly misrepresent them.
You either have no concern for the truth or you have no idea what you are talking about.

Again, going back to my extreme example to show the stupidity of your claims, this is how to find a star after one year according to us (or at least me) without any of your 3.2 seconds of BS:

Notice that I say you look towards the star.

Meanwhile, you claim this bullshit:

Instead of looking towards the star you claim you have to look in the complete opposite direction, as you need to wait half a day for the star to return to the same position.

Are you going to deal with this massive problem of yours this time? Or are you going to ignore it and keep repeating the same childish BS?

Re: SMOKING GUN
« Reply #254 on: April 25, 2019, 03:42:47 AM »
Jack, it seems that you stuck here https://www.theflatearthsociety.org/forum/index.php?topic=80229.msg2167352#msg2167352

First of all, you misconcieved my words, secondly : stop putting words that i have never spoken out in my mouth, thirdly : stop drawing meaningless drawings only because you desperately need someone (yourself) to persistently quarrel with him (yourself).



No matter to how many smoking-guns you can point we will smoke you each time simply because we are louder than you, and we are louder than you because there are three of us, and you are alone!!! Three musketeers
« Last Edit: April 25, 2019, 03:48:25 AM by cikljamas »

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rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #255 on: April 25, 2019, 04:21:35 AM »
<< Not worth reading. >>
If you must resort to:
You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have...
Run away!

Re: SMOKING GUN
« Reply #256 on: April 25, 2019, 05:14:00 AM »
The rotating earth completes 1/720 of a complete rotation in those 20 minutes since it spins a lot faster than it revolves around the sun, but distributing that rotation over the days doesn't have any significance here because the tropical year/sidereal year comparison involves the relationship between the sun and the precessing equinox and the inertial coordinate system, and neither of those is concerned with the rotation angle of the earth.

If you think that this convoluted passage is some kind of sane respond to my following argument ...
Quote
After the first sidereal day of the creation of CCW rotational earth (which precesses 3,2 seconds per day in CW direction) the MOTIONLESS stars would lag 3,2 seconds behind one stellar (rotational) day, after the second sidereal day the MOTIONLESS stars would lag 6,4 seconds behind one stellar day, and so on...

So, after one sidereal year the MOTIONLESS stars would lag 20 minutes behind one stellar (rotational day), because MOTIONLESS stars are MOTIONLESS!!!

...then feel free to go much deeper into details of your convoluted explanation, and then we we are going to see will your convoluted story hold the water...

Re: SMOKING GUN
« Reply #257 on: April 25, 2019, 05:44:20 AM »
<pathetic insults>
No, I'm not stuck at all.
Your pathetic insults are showing just how poor your position is.

My diagram is a valid representation of your claims.
You are claiming that because the tropical year is off by 20 minutes due to the precession of the equinoxes, that this should then manifest in all stars by them appearing to lag behind, and be distributed throughout the year.
This seems to be tied to you still not understanding what is causing this 20 minutes.
All it is, is Earth's axis not pointing in the exact same direction. It doesn't mean the sun has shifted by 20 minutes. This is based primarily upon the declination, e.g. the timing of the solstices and the equinoxes.

Now, if this was true, then with the extreme hypothetical case I presented, the tropical year is longer by 1 sidereal year or half a tropical year.
According to you, this should manifest as needing to wait a bit longer for each part of the year.
So just what is wrong with it?
You can even just focus on where I said the sun should be and ignore the actual tilt.
According to me, you still look to the right. According to you, for the HC system you think you should look straight out 20 minutes earlier, which as Earth rotates roughly 15 degrees per hour that would be 3 degrees off.
You wouldn't be looking at the star then, you would be looking 3 degrees off.

So what will you do now?
Will you act like a rational adult and admit you were wrong?
Will you continue your childish behaviour in this thread?
Or will now be the time you take your typical exit after being refuted so many times it isn't funny?

Re: SMOKING GUN
« Reply #258 on: April 25, 2019, 06:28:09 AM »
FRAMES OF REFERENCE! YOU'RE BOTH RIGHT!!!

Re: SMOKING GUN
« Reply #259 on: April 25, 2019, 07:29:08 AM »
My diagram is a valid representation of your claims.
You are claiming that because the tropical year is off by 20 minutes due to the precession of the equinoxes, that this should then manifest in all stars by them appearing to lag behind, and be distributed throughout the year.
This seems to be tied to you still not understanding what is causing this 20 minutes.
All it is, is Earth's axis not pointing in the exact same direction. It doesn't mean the sun has shifted by 20 minutes. This is based primarily upon the declination, e.g. the timing of the solstices and the equinoxes.
All it is, is Earth's spatial position (spatial orientation of earth's axis) wrt the stars has changed so that earth's axis precessed in CCW direction!!!
Now, if it affects stationary sun, to come to the Equator 20 min earlier than the earth accomplishes it's full circle around the sun, then in the same manner it must affect stationary stars (at least those at the ecliptic) so that they come at the local meridian 20 min later.



Alpha2Omega said :

One thing that was bothering me is the daily change of location of the RA of the vernal equinox due to precession works out to something on the order of 0.01s per day on average, not the 3.2s we were discussing. So, what does that 3.2s we were looking at mean? That's how much longer the length of the mean solar day would be if the sidereal year were 20 minutes shorter than it actually is. It's not a real effect, it's an artifact of ignoring the lengths of the slightly-different years and considering only their difference.


So, even if 3,2 seconds would be the wrong number, and if Alpha2Omega's number (0.01s per day on average) turned out to be the correct number, that would still be significant difference since 0,01 * 365 = 3,65 seconds, wouldn't it?

Now, if this was true, then ...
Since it isn't true, then i will take into consideration the rest of your passage after you prove that it is true what you claimed above...

So what will you do now?
I will wait patiently!!!
« Last Edit: April 25, 2019, 08:49:40 AM by cikljamas »

Re: SMOKING GUN
« Reply #260 on: April 25, 2019, 01:01:01 PM »
My diagram is a valid representation of your claims.
You are claiming that because the tropical year is off by 20 minutes due to the precession of the equinoxes, that this should then manifest in all stars by them appearing to lag behind, and be distributed throughout the year.
This seems to be tied to you still not understanding what is causing this 20 minutes.
All it is, is Earth's axis not pointing in the exact same direction. It doesn't mean the sun has shifted by 20 minutes. This is based primarily upon the declination, e.g. the timing of the solstices and the equinoxes.
All it is, is Earth's spatial position (spatial orientation of earth's axis) wrt the stars has changed so that earth's axis precessed in CCW direction!!!
Now, if it affects stationary sun, to come to the Equator 20 min earlier than the earth accomplishes it's full circle around the sun, then in the same manner it must affect stationary stars (at least those at the ecliptic) so that they come at the local meridian 20 min later.

No. The sun returns to the vernal equinox 20 minutes per year faster than it returns to the same position relative to a distant star after a year. This is because the direction of the vernal equinox changes with respect to the distant stars. That means the VE has moved (20/1440)/365.25 (approximately) of the way around the ecliptic (roughly 1/26 000) in a year. Since the earth rotates once wrt the VE each sidereal day, then that affects transit times of celestial objects relative to the transit of the VE by a similar amount. 1/26 000 of 86 164 seconds is about 3.3 seconds, not 20 minutes. You're conflating 20 minutes in a year with 20 minutes in a day.

Quote
Alpha2Omega said :

One thing that was bothering me is the daily change of location of the RA of the vernal equinox due to precession works out to something on the order of 0.01s per day on average, not the 3.2s we were discussing. So, what does that 3.2s we were looking at mean? That's how much longer the length of the mean solar day would be if the sidereal year were 20 minutes shorter than it actually is. It's not a real effect, it's an artifact of ignoring the lengths of the slightly-different years and considering only their difference.


So, even if 3,2 seconds would be the wrong number, and if Alpha2Omega's number (0.01s per day on average) turned out to be the correct number, that would still be significant difference since 0,01 * 365 = 3,65 seconds, wouldn't it?

I got 3.3 seconds above, so that's about right. 3.3 seconds per year is about 9 ms (0.009 seconds) per day, which is consistent with "on the order of 0.01s per day on average" as stated above.

I, too, was a little surprised the difference was that great, but it appears to be correct. It would take 28 800 years to accumulate 24 hours at 3 seconds per year, which is consistent with the ~26 000 year precession cycle.

Quote
Now, if this was true, then ...
Since it isn't true, then i will take into consideration the rest of your passage after you prove that it is true what you claimed above...

So what will you do now?
I will wait patiently!!!

You have yet to show any real inconsistencies between observations and the heliocentric model of the solar system. You have made several claims that a geocentric model offers "more flexibility" than the heliocentric model does, but that is incorrect if you constrain that geocentric model to match observations. The heliocentric model is more consistent, since the same laws that govern the motion of the earth govern the motion of everything else. If you require the earth to be fixed, the laws governing the motion of everything else become very messy and difficult to reconcile.

Working in an earth-fixed reference frame makes certain analyses much simpler, but doing so is just a choice of reference frames. For instance, if you wanted to draw a plan of the furnishings in a room on a houseboat, you would most likely make all measurements relative to some point in the room. Making them relative to another point on the boat would also work, and would be beneficial if you needed to locate where something in that room is relative to something in another part of the vessel, but it makes handling the relationships between items within the room more complex. You could measure everything in the room relative to some point not on the boat if you wanted, but that would be much more complex, even if the boat's motions relative to the reference point could be accurately known, and adds no value.

[Edit: clarification]
« Last Edit: April 25, 2019, 01:05:22 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: SMOKING GUN
« Reply #261 on: April 25, 2019, 01:05:36 PM »
I've figured out where i was mistaken : I can't use as a reference point spatial position of Earth's axis (in absolute terms = in absolute space) as reference point for each next sidereal period, instead i must use last spatial position of Earth's axis as reference point for each next sidereal period. This is how the difference between stellar and sidereal period is embedded within one sidereal period.

However, if Earth's axis precesses in CCW direction, how in the world can one sidereal period be shorter than one stellar period?

Re: SMOKING GUN
« Reply #262 on: April 25, 2019, 02:57:37 PM »
And you are still completely misrepresenting it.

All it is, is Earth's spatial position (spatial orientation of earth's axis) wrt the stars has changed so that earth's axis precessed in CCW direction!!!
Yes, the orientation has changed. It hasn't simply slowed down or shifted around the axis. The axis itself is pointing in a different direction.

Now, if it affects stationary sun
Again, in the current model of the universe, NOTHING IS STATIONARY!

to come to the Equator 20 min earlier
Again, it doesn't.
It isn't coming earlier. It isn't coming later.
It has a different declination.

then in the same manner it must affect stationary stars (at least those at the ecliptic) so that they come at the local meridian 20 min later.
Do you not notice the 2 vastly different things you are saying here?

Here let me make it easier:
to come to the Equator 20 min earlier
come at the local meridian 20 min later.

Notice how one appeals to the equator, focusing on the declination of the sun and how that is effected by the precession.
Notice how the other one focuses on the meridian instead, something completely different?

Once again, my example is a clear representation of this.
The precession of Earth's axis isn't going to make a star all the way over to the right appear to the left.
You repeatedly ignoring this and instead just asserting the same refuted nonsense will not help your case at all.

So, even if 3,2 seconds would be the wrong number, and if Alpha2Omega's number (0.01s per day on average) turned out to be the correct number, that would still be significant difference since 0,01 * 365 = 3,65 seconds, wouldn't it?
He didn't actually say 0.01 s. He said on the order of. But again, it depends on exactly what you focus on and how you calculate it and exactly how you are defining it.

As for if it is significant, remember that Astronomers don't typically just use a time, they use an angle.
After a year so you get your ~4 seconds, what angle does that correspond to?
It is roughly only 1 minute of arc.

In order to notice this you will need a very good clock and telescope and mount.
It is not something amateur astronomers (who will typically use the stars themselves to align their scope) will be able to do.

Not even levelling will help you here as the direction of "down" can actually vary due to movements under the crust, and movements of the crust.

So depending upon how you are defining the terms, this difference may exist.
After looking at stellar and sidereal days some more, it seems the difference does exist and is small.

after you prove that it is true what you claimed above...
I have proven it. You just ignore the proof. Although their may be confusion over what is meant by each term.

However, if Earth's axis precesses in CCW direction, how in the world can one sidereal period be shorter than one stellar period?
Firstly, wasn't one of the big objections you had the fact that Earth's axis precessed in a direction opposite its rotation? That it rotates CCW while precessing CW? Or are you now viewing it from below?

When viewing from above the north pole, the Earth's axis rotates CCW so the sun rises in the east and sets in the west. In addition it orbits CCW, meaning after one sidereal or stellar day it still needs to turn a bit more to face the sun, making the solar day longer.
But the precession is opposite this, so if you start with the northern hemisphere facing the sun (at the solstice), then after slightly less than one full orbit (i.e. a tropical year, slightly shorter than a sidereal year) we are again at the solstice but need to wait a bit longer to get back to the same point in our orbit.
If precession was CCW as well then the tropical year would be longer.

If I have the terminology correct, from wiki:
Quote
Earth's rotation period relative to the fixed stars, called its stellar day by the International Earth Rotation and Reference Systems Service (IERS), is 86164.098 903 691 seconds of mean solar time (UT1) (23h 56m 4.098 903 691s).[2][3] Earth's rotation period relative to the precessing or moving mean vernal equinox, its sidereal day, is 86164.090 530 832 88 seconds of mean solar time (UT1) (23h 56m 4.090 530 832 88s).
So this means that according to these definitions, the stellar day is the time taken for Earth to rotate about its axis.
The sidereal day is how long it takes for earth to rotate in a rotating reference frame, that keeps the vernal equinox at the same point.

This can cause serious confusion as when I learnt of the sidereal day it was what that refers to as the stellar day. Perhaps the more annoying part of this is that the sidereal day is not connected to the sidereal year. So which did you mean?

Perhaps the simplest way to compare them is with a comparison with the mean solar day and with the sidereal and tropical year.
If I am understanding it correctly:
For a tropical year, there will be 1 additional sidereal day than solar days.
For a sidereal year, there will be 1 additional stellar day than solar days.

First a qualitative way:
Both the sidereal day and stellar day need to be shorter such that over the course of a "year", this amounts to a 1 day difference.
For a sidereal day, the year used is shorter, meaning it has less time to accumulate that day and thus it must be shorter than the mean solar day by more. Conversely the stellar day has more time (an extra 20 minutes) to make up this difference.
This means the sidereal day will be shorter than the stellar day.

And now some calculations.
First, assuming a tropical year is 365.2425 mean solar days (31556952 s) long. (Note: due to these approximations, these numbers will not match up with those from wiki, but will be "close").
This means we need 366.2425 sidereal days to be the same length.
This makes the sidereal day 86164.09073 s long.

The sidereal year is 20 minutes longer, i.e. 31558152 s.
This means there needs to be 365.2563889 mean solar days in this sidereal year.
This means there needs to be 366.2563889 stellar days in this sidereal year.
This means the stellar day needs to be 86164.09968 s long.

This puts this longer by 0.008945967 seconds.

Note that the difference from wiki is 0.0084 seconds. So as I said, close but not perfect.

*

rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #263 on: April 25, 2019, 03:44:40 PM »
I've figured out where i was mistaken : I can't use as a reference point spatial position of Earth's axis (in absolute terms = in absolute space) as reference point for each next sidereal period, instead i must use last spatial position of Earth's axis as reference point for each next sidereal period. This is how the difference between stellar and sidereal period is embedded within one sidereal period.

However, if Earth's axis precesses in CCW direction, how in the world can one sidereal period be shorter than one stellar period?
It is "sidereal day" and "stellar period", not "sidereal period and "stellar period". There is no "stellar year".

The term "stellar day" is simply another name for the "earth's rotational period" or 2π/Ω where Ω is the "angular velocity of the Earth".
"Stellar day" is a confusing name when it is not related directly to stars other than the "fixed stars" being a sort of reference frame.
But it is used in International Earth Rotation and Reference Systems Service (IERS): USEFUL CONSTANTS for values of these and the IERS are the "go to people" for these definitions and values.

The sidereal day is very slightly shorter because the CCW axial precession adds a little (a few milliseconds) to the angle the earth has rotated so a reference star seems to arrive a little sooner.
But the difference is so small that it can usually be neglected.

One big problem if you try to get too fussy over all these values is that the length of an individual year (sidereal or tropical) can vary considerably due to the gravitational influence of the other planets, mainly Jupiter but Venus and Mars have a slight effect.

Look in Sidereal, tropical, and anomalistic years for more detail.

Look at the variation in the following tropical year's duration from the mean value of 365.24219 SI days. The years are from the northern winter solstice:
Date/time Winter Solstice      Deviation
 2007-12-22 06:04:04.2       +10.51 minutes
 2008-12-21 12:03:19.7       -11.86 minutes
 2009-12-21 17:40:13.2      +15.91 minutes
 2010-12-21 23:44:53.2      -11.94 minutes
 2011-12-22 05:21:41.8      +3.58 minutes
 2012-12-21 11:14:01.9      +2.85 minutes
 2013-12-21 17:05:38.3      +0.86 minutes
 2014-12-21 22:55:15.2      +0.48 minutes

The sidereal year is always the 20 or so minutes longer.

If you want reliable up-to-date information you really should be asking astronomers.

Now what about a few apologies all around for your statements like:
         "So, i proved once again that i care about the truth, unlike you!!!"
         "You see, i don't have such a serious psychological problem as you, JB, and Rabinoz have..."

Now, I've done my best to answer your questions so what about finally commenting on this:
Now would you finally care to answer what you believe are the laws of physics, including motion and gravitation, that govern how things move.

If you do not agree with Newton's Laws of Motion and Universal Gravitation just what rule does your Universe follow?
Do you follow Dr. Robert Sungenis and claim that the Geostationary Universe does obey Newton's Laws as well as Mach's Principle.
You might find this interview interesting: Interview with Robert Sungenis on  Phil Plait’s Criticism of Geocentrism though I can guess who you might support.

By the way, you insisted that you are correct and everybody else from Einstein and Newton down are wrong but what about YouTube videos like:
        THE EARTH IS FLAT - GASLIGHT by odiupicku
        THE FLAT EARTH - ANTARCTICA (reupload) by odiupicku
        The Flat Earth The True Law of Perspective by odiupicku
        THE FLAT EARTH - THE SUN by odiupicku
So back there the earth was flat but since then the earth seems to have magically become a Globe but it still won't move.
Maybe by this time next year, you might discard the Geostationary Universe and finally accept the Heliocentric Solar System - who knows?
Some response would be appreciated, thank you!

Re: SMOKING GUN
« Reply #264 on: April 26, 2019, 02:15:33 AM »
Jack, your mum gives you 50 $, your dad gives you 50 $, you buy some item for 97 $
Now, 3 $ has left
1 $ you put in your pocket
1 $ you give back to your mum
1 $ you give back to your dad
You own your mum 49 $
You own your dad 49 $
So, 49 + 49 = 98
98 + 1 (that one $ which is in your pocket) = 99 $
Where the fuck is 1 $ (difference)???

So, Jack, we are looking from above, yes, from above
Earth's axis precesses in CW direction
Earth allegedly rotates in CCW direction
So, Earth's axis "runs away" from the motionless stars, isn't that so?
Running away from the stars means what?
Means that one stellar day must be shorter than one sidereal day, isn't that so?
If Earth's axis precessed in the same direction in which the earth allegedly rotates, then Earth's axis would go in the same direction as particular meridian on the earth (going "in" the stars, not running away from them), and that would yield an opposite result : sidereal period would be shorter than stellar (rotational) period.
Can it be more simple than that?

Now, I've done my best to answer your questions so what about finally commenting on this:
Now would you finally care to answer what you believe are the laws of physics, including motion and gravitation, that govern how things move.

If you do not agree with Newton's Laws of Motion and Universal Gravitation just what rule does your Universe follow?
Do you follow Dr. Robert Sungenis and claim that the Geostationary Universe does obey Newton's Laws as well as Mach's Principle.
You might find this interview interesting: Interview with Robert Sungenis on  Phil Plait’s Criticism of Geocentrism though I can guess who you might support.

The answer is in the last sentence of this video :


The sidereal day is very slightly shorter because the CCW axial precession adds a little (a few milliseconds) to the angle the earth has rotated so a reference star seems to arrive a little sooner.
But the difference is so small that it can usually be neglected.

If you look from above than it is CW axial precession, and the earth rotates in CCW direction.
So, when you say that axial precession adds a little to the angle the earth has rotated, you imply that axial precession goes in the same direction in which the earth rotates, however you know that it is a blatant misinterpretation of HC theory, don't you? See above (my answer to JB)...
« Last Edit: April 26, 2019, 02:19:02 AM by cikljamas »

*

rabinoz

  • 21640
  • Real Earth Believer
Re: SMOKING GUN
« Reply #265 on: April 26, 2019, 04:20:25 AM »
Now, I've done my best to answer your questions so what about finally commenting on this:
Now would you finally care to answer what you believe are the laws of physics, including motion and gravitation, that govern how things move.

If you do not agree with Newton's Laws of Motion and Universal Gravitation just what rule does your Universe follow?
Do you follow Dr. Robert Sungenis and claim that the Geostationary Universe does obey Newton's Laws as well as Mach's Principle.
You might find this interview interesting: Interview with Robert Sungenis on  Phil Plait’s Criticism of Geocentrism though I can guess who you might support.

The answer is in the last sentence of this video :

So, "All the theories collapse when you cannot see the stars in outer space."
Surely even if you don't believe anything can be learned about what is outside the earth you must have some rules governing motion and "why things fall down".
If you do not agree with Newton's Laws of Motion and Universal Gravitation what are those rules?

But when it comes to astronomy your claimed lack of any possibility of learning anything about what is outside the earth means that there's no point carrying on.

I'd say we are done here then. There's no point in debating with nothing when it comes to the Universe.
Especially as you keep claiming deception on our part. I said I was no astronomer and could easily be mistaken.
For my own satisfaction, I'll keep looking at the matter of precession etc, though it's obviously no point trying to debate it with you.

But when it comes to that sort of thing, have you seen this video?

Flat Earth Calling out a liar "odiupicku" by Sly Sparkane

Just wondering what comments you might have.

Bye bye!

PS It seems so significant that you, like all flat earthers, must claim:
             that all astronomers are liars because they claim to observe a rotating earth orbiting the sun and
             that all space are pure deception and all rocket launches into space are faked.



Re: SMOKING GUN
« Reply #266 on: April 26, 2019, 04:47:44 AM »
Bye bye!
Run away, you little soul!

How come you didn't even try to refer to my refutation of your blatant misinterpretation of HC theory?

If a man has reported to you that a certain person speaks ill of you, do not make any defense to what has been told, but say: that man did not know the rest of my faults, for otherwise he would not have mentioned these only.

 ... the idea being not getting upset at what other people think or say about you.

The uneducated person blames others for his own bad condition. He who has made some progress blames himself. And he who has complete understanding blames neither another, nor himself.

 Thank you for your attention (and for a satisfactory amount of applause)


Re: SMOKING GUN
« Reply #267 on: April 26, 2019, 04:51:59 AM »
How come you didn't even try to refer to my refutation of your blatant misinterpretation of HC theory?
He didn't.
Remember, CW is meaningless unless you specify a reference.
When viewed from below, Earth rotates CW and precesses CCW.

He also just used what you said. So why did you say that? Were you hoping to trick people with more blatant dishonesty?

Are you capable of responding in an honest and rational manner and admitting all your claims so far about smoking guns has been nothing more than dishonest BS?

Re: SMOKING GUN
« Reply #268 on: April 26, 2019, 05:19:45 AM »
How come you didn't even try to refer to my refutation of your blatant misinterpretation of HC theory?
He didn't.
Remember, CW is meaningless unless you specify a reference.
When viewed from below, Earth rotates CW and precesses CCW.

He also just used what you said. So why did you say that? Were you hoping to trick people with more blatant dishonesty?

Are you capable of responding in an honest and rational manner and admitting all your claims so far about smoking guns has been nothing more than dishonest BS?

Jack, have you wrapped your head around that 1 missing $?
If you have, then you know that i know what you had done in your extensive post (the one which you have posted before the last one)!
Do i have to explain it to you or do i have to explain it to others, only (since you know what you are doing)?
Here we go :
In the moment when sidereal year is over, how many times earth allegedly rotated on it's axis?
Let's say 366,25 times
How many times the earth REALLY rotated on it's axis after sidereal year is over?
366,25 + 1' (4 seconds)
You see, Earth REALLY rotated more (for 1' = in 4 seconds) in that period.
If the Earth REALLY rotated more than average astronomer is aware of, then what inference we have to draw from it?
There are more STELLAR rotations in the same period (in one sidereal year) than SIDEREAL rotations, because 4 seconds before one sidereal year is over Earth already rotated 366,25 times, but after 4 additional seconds there is going to be some extra rotational motion of the Earth.
So, because the Earth rotated more (for 1') in these 4 last seconds of one sidereal year you can't say that STELLAR year is longer than SIDEREAL year.
After all : define one STELLAR year if you can!
What one STELLAR year actually means is this : How many times the Earth REALLY rotated on it's axis after 365,25 days.

So, if you don't know the answer to 1 missing $ conundrum, i will gladly explain it to you, as well...
« Last Edit: April 26, 2019, 05:22:42 AM by cikljamas »

Re: SMOKING GUN
« Reply #269 on: April 26, 2019, 05:34:18 AM »
Jack, have you wrapped your head around that 1 missing $?
If you have, then you know that i know what you had done in your extensive post
So you know I have found the correct answer and you are pretending that $1 is missing?
So you are finally admitting to lying to everyone?

In the moment when sidereal year is over, how many times earth allegedly rotated on it's axis?
Let's say 366,25 times
How many times the earth REALLY rotated on it's axis after sidereal year is over?
Well if we are assuming that it is 366.25 times, then it is 366.25 times.
It doesn't magically change.

366,25 + 1' (4 seconds)
Where are you pulling this BS from?
You are literally pulling numbers from no where and expecting people to take them seriously.

There are more STELLAR rotations in the same period (in one sidereal year) than SIDEREAL rotations
You are yet to substantiate this in any way.
Again, you either have no idea what you are talking about or are happy to lie to everyone.
Which is it?

you can't say that STELLAR year is longer than SIDEREAL year.
There is no stellar year.
Stop just making up words.
Again, from what I can tell, alpha is using the following definitions:
Tropical year - The time between 2 vernal equinoxes.
Sidereal year - The time actually taken for Earth to orbit the sun.
Mean solar day - The average time taken for Earth to rotate enough for the sun to come back to the same meridian on Earth.
Sidereal day - The time taken for Earth to rotate enough for the 0 point of the moving right ascension to come back to the same meridian on Earth, or to put it another way, the time taken such that after one tropical year there will be one extra sidereal day than solar day.
Stellar day - The actual time taken for Earth to rotate about its axis.

After all : define one STELLAR year if you can!
It doesn't exist.
While you are at it why not define more nonsense, like what is a potato year?
How about a car year?

So, if you don't know the answer to 1 missing $ conundrum, i will gladly explain it to you, as well...
Again, there is no missing $1. You are just lying about it.