"Equator" problem

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Re: "Equator" problem
« Reply #150 on: October 25, 2014, 07:47:33 PM »
The only factor determining the daily rate of rotation wrt the stars is the rotation of the earth. Period. This is a constant. You're getting tangled up in your  "daily rate of annual rotation", which is a convoluted way of describing the difference between the lengths of the sidereal day (constant) and apparent solar day (not quite constant) and thinking the length of the mean solar day (which is irrelevant here) is somehow mixed in here. It's not.

Well, this is not an answer, and you know it. Now, i have to ask you this: If you are aware that what you are offering as an explanation doesn't explain anything, why then you do this?

The whole thing is very simple:

24 hours (mean solar day) MINUS 23h56m4s = 3m56s= 0,986 degree = daily rate of annual rotation of the stars above the Earth

Now, if you want to proceed to defend HC theory the only possible logical way out of this is this:

1. Either you claim that 0,986 degree is not equal to daily rate of annual roatation of the stars

2. Or you claim that it is constant indeed

If 1 then you disagree not just with FET, but with basic HC rules also, and you have to prove what you claim which will be very difficult...

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?

It is very strong argument against fraudulent HC theory, and there is no place to hide from it.

Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

It seems we agree on the basic premises (if you disagree with any of those first three statements in the paragraph above, please say clearly which one(s) you think is wrong, and, if possible, why), but your interpretation of them goes off the rails.

As I asked, and you ignored, how would you explain the varying length of the apparent solar day in a Geocentric model?
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #151 on: October 26, 2014, 01:43:17 AM »
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Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

I see, you vote for the second option, wise!

Well, now the only thing that remains to be solved is this question (i repeat):

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?

You partially answered it:

If you subtract a constant from a constant, you get a constant.

Bravo! That is the correct answer!

But the essence of my question wasn't about if we can get the constant by substracting one constant from another constant, since this is a self-evident logical inference, my question was about this:

Mean Solar Time doesn't happen every day, it happens only during equinoxes!

So, if every day of the year was an equinox then there would be no flaw in this matter.

But, the core of this flaw is in the fact that everyday is not an equinox!

So, if everyday is not an equinox, how can HC theory be defended by doing astronomical calculations as if everyday was an equinox indeed?

24 hours is a constant but an artificial one, and why this constant is an artificial one you know very well, don't pretend that you don't know.

It is an artificial one at any other day except at those equinox days.

You cannot just say: "here we have got three constants and everything stays in perfect order", and proceed as if nothing of great importance hasn't just happened (been changed), because something (replacement of identity) which makes big difference has happened and has been changed (replaced):

One out of these three constants doesn't present anything that exist in reality (except during equinox) and you still pretend that you are not aware of it!!!

And i repeat: Everyday is not an equinox!!!

So how can stars shift EVERYDAY EXACTLY 0,986 degrees if the time between sun culminations vary from day to day?

There you go: when you say "the time between sun culminations vary from day to day" what you really say is: One of these constants is not a REAL-ACTUAL constant at all!!!

Once again: How can you use an artificaial constant as if it were something that fits everyday reality?

Exit heliocentrism and join us, why would you support an obvious lie, you are too smart to be a heliocentrist (or a cop ), aren't you?
« Last Edit: October 26, 2014, 02:13:30 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #152 on: October 26, 2014, 08:30:35 AM »
So many words. So little understanding.

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Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

I see, you vote for the second option, wise!

Well, now the only thing that remains to be solved is this question (i repeat):

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?
It depends on your definition of day, as in "constant daily rate of annual rotation". If, by "day" you mean exactly 24 hours, then the shift will be a constant. This appears to be the case here since that angle represents the "overshoot" of 360 degrees of rotation in exactly 24 hours. If your definition of "day" is local apparent noon to local apparent noon (solar culmination to solar culmination) and thus slightly variable, then the shift will be slightly variable. How hard is this to understand?

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You partially answered it:

If you subtract a constant from a constant, you get a constant.

Bravo! That is the correct answer!

But the essence of my question wasn't about if we can get the constant by substracting one constant from another constant, since this is a self-evident logical inference, my question was about this:

Mean Solar Time doesn't happen every day, it happens only during equinoxes!
No, the difference between mean and apparent solar days is largest near the equinoxes and solstices. It would be largest exactly on those dates (longest apparent solar days on the solstices, shortest on the equinoxes)  if it weren't for the effect of the elliptical orbit, which shifts them a few days. The length of the apparent solar day is exactly 24 hours four times a year - roughly midway between the equinoxes and solstices. You either didn't look at the link for EoT, didn't understand it, or both.

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So, if every day of the year was an equinox then there would be no flaw in this matter.

But, the core of this flaw is in the fact that everyday is not an equinox!

So, if everyday is not an equinox, how can HC theory be defended by doing astronomical calculations as if everyday was an equinox indeed?

24 hours is a constant but an artificial one, and why this constant is an artificial one you know very well, don't pretend that you don't know.

It is an artificial one at any other day except at those equinox days.
Ignoring your error about when those days occur, the core of the flaw is still your reasoning. You're the one insisting on using the mean solar day by insisting on a constant daily rate of change; the HC model doesn't require this. As I said earlier, the mean solar day has no astronomical meaning; it's used for everyday timekeeping as a convenience.

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You cannot just say: "here we have got three constants and everything stays in perfect order", and proceed as if nothing of great importance hasn't just happened (been changed), because something (replacement of identity) which makes big difference has happened and has been changed (replaced):

One out of these three constants doesn't present anything that exist in reality (except during equinox) and you still pretend that you are not aware of it!!!
As I said earlier, the mean solar day has no astronomical meaning; it's used for everyday timekeeping as a convenience.

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And i repeat: Everyday is not an equinox!!!

So how can stars shift EVERYDAY EXACTLY 0,986 degrees if the time between sun culminations vary from day to day?
It depends on your definition of day. If, by "day" as in "EVERYDAY" you mean exactly 24 hours, then the shift will be a constant. If your definition of "day" is local apparent noon to local apparent noon (solar culmination to solar culmination), then the shift will be slightly variable. How hard is this to understand?

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There you go: when you say "the time between sun culminations vary from day to day" what you really say is: One of these constants is not a REAL-ACTUAL constant at all!!!

Once again: How can you use an artificaial constant as if it were something that fits everyday reality?

Exit heliocentrism and join us, why would you support an obvious lie, you are too smart to be a heliocentrist (or a cop ), aren't you?

Dude... will you lose the "lie" stuff, already? It's just a gratuitous insult that adds nothing to your argument (in fact, it weakens it, and your argument doesn't need that kind of "help"). It also makes it that much harder for you to back down once you recognize your mistake.

Why would I want to join you? You still haven't explained the Equation of Time in your model at all. I already have one that explains it well, and in detail, as well as other measured phenomena such as parallax, well, and in detail.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #153 on: October 26, 2014, 09:59:40 AM »
What determines a daily rate of annual rotation of the stars according to HC?

Answer: A daily rate of annual rotation of the stars around the Earth is determined by the the speed of the Earth's rotation on it's axis (which is a constant, in reality it is not a constant because it doesn't exist in the first place, but let's pretend that it exists and that it is a constant indeed), and by the orbital speed of the Earth which is not a constant (and which doesn't exist also)!

Mean time follows the "mean sun", best described by Meeus:

    "Consider a first fictitious Sun travelling along the ecliptic with a constant speed and coinciding with the true sun at the perigee and apogee (when the Earth is in perihelion and aphelion, respectively). Then consider a second fictitious Sun travelling along the celestial equator at a constant speed and coinciding with the first fictitious Sun at the equinoxes. This second fictitious sun is the mean Sun..."

Now, let's say that the Earth's moderate speed is around equinoxes (and it is so) and let's pretend that there is no axial tilt of the Earth (as it is in reality indeed), in order to simplify the whole thing a little bit.

During equinoxes the alleged orbital speed of the Earth is moderate, so let's call it The Mean Speed.
During northern solstice the alleged orbital speed of the Earth is lowest, so let's call it The Low Speed.
During southern solstice the alleged orbital speed of the Earth is highest, so let's call it The High Speed.

Apparent solar day depends on the current spatial orientation of the Earth due to Earth's alleged tilt and current orbital position with respect to the position of the Sun, and all this depends on the Earth's alleged orbital speed.

Now, for the sake of simplicity let's say that the longest apparent day coincides with The Low Speed (of the Earth), that the shortest apparent day coincides with The High Speed (of the Earth), and that the mean sun coincides with equinox a.k.a. The Mean Speed.

So, now you should finally understand what is the deal here.

As the speed of the Earth changes through the year, so the daily rate of an annual motion of the Stars has to change too, BUT IT DOESN'T CHANGE!!!

What i really meant by saying "Equinox is not everyday!" was: The orbital speed of the Earth is not the same through the year, and that is why a daily rate of an annual motion of the stars wouldn't be a constant if it were really depend on irregular orbital speed of the Earth.

But it doesn't depend on that, since the orbital speed of the Earth doesn't exist in the first place!!!
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #154 on: October 26, 2014, 11:04:00 AM »
Meanwhile for FET, the sun moves overhead twice as fast at different times of the year and different latitudes than others, but somehow this isn't noticed on the ground.
« Last Edit: October 26, 2014, 07:48:27 PM by 29silhouette »

Re: "Equator" problem
« Reply #155 on: October 26, 2014, 03:45:32 PM »
What determines a daily rate of annual rotation of the stars according to HC?

Answer: A daily rate of annual rotation of the stars around the Earth is determined by the the speed of the Earth's rotation on it's axis (which is a constant

This is correct so far.

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, in reality it is not a constant because it doesn't exist in the first place

This is your opinion.

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, but let's pretend that it exists and that it is a constant indeed)

There's no need to pretend, but, as they say, for the sake of the argument...

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, and [also] by the orbital speed of the Earth which is not a constant

This depends on what you mean by "daily". If "daily" means apparent solar days, then yes; if "daily" means mean solar days, then no. Which do you prefer?

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(and which doesn't exist also)!

More opinion.

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Mean time follows the "mean sun", best described by Meeus:

    "Consider a first fictitious Sun travelling along the ecliptic with a constant speed and coinciding with the true sun at the perigee and apogee (when the Earth is in perihelion and aphelion, respectively). Then consider a second fictitious Sun travelling along the celestial equator at a constant speed and coinciding with the first fictitious Sun at the equinoxes. This second fictitious sun is the mean Sun..."

OK. Note that Meeus is not saying that the length of the apparent solar day equals the length of the mean solar day at the equinoxes. He is only setting his second fictitious Sun to coincide with the first fictitious Sun at those points.

You still haven't looked at a plot of the Equation of Time, have you? Yes or no, please? If yes, do you realize that the points on that plot where the curves level out (the four peaks and valleys on the graph are the times when apparent solar days are the same length as mean solar days and the steepest parts of the curve are when the lengths differ the most?

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Now, let's say that the Earth's moderate speed is around equinoxes (and it is so)

Are you referring to orbital speed? If so, then this is approximately correct because the line of apsides happens to very nearly coincide with the solstices.

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and let's pretend that there is no axial tilt of the Earth (as it is in reality indeed), in order to simplify the whole thing a little bit.

More opinion, but, for the sake of the argument, OK, no tilt.

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During equinoxes the alleged orbital speed of the Earth is moderate, so let's call it The Mean Speed.
During northern solstice the alleged orbital speed of the Earth is lowest, so let's call it The Low Speed.
During southern solstice the alleged orbital speed of the Earth is highest, so let's call it The High Speed.

A couple of points here. If there's no tilt, then there will be no discrete equinoxes and no solstices; every day will be an equinox. Can we restate these as:

At aphelion orbital speed of the Earth is lowest, so let's call it The Low Speed.
At perihelion orbital speed of the Earth is highest, so let's call it The High Speed.
Midway between these, orbital speed of the Earth is moderate, so let's call it The Mean Speed.

It happens that aphelion is near our (current) actual northern solstice, and perihelion is near our actual southern solstice, so your original statements can be thought of as fairly close, although the language used is imprecise since the solstices don't necessarily have to coincide with the apsides, and in our simplified model the solstices have been abolished, anyway.

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Apparent solar day depends on the current spatial orientation of the Earth due to Earth's alleged tilt and current orbital position with respect to the position of the Sun, and all this depends on the Earth's alleged orbital speed.

Wait... I thought we were simplifying by removing axial tilt. Now it's back, without ever using the simplification.

And that paragraph is kind of a hodge-podge of confused terminology and extraneous qualifiers. The length of the apparent solar day is affected by where the Earth is in its elliptical orbit, and thus its speed, and on how its axis of rotation is aligned wrt the Sun. The axial alignment does not depend on its orbital speed; in fact, the angle between the line of apsides and equinoxes is slowly changing due to precession.

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Now, for the sake of simplicity let's say that the longest apparent day coincides with The Low Speed (of the Earth), that the shortest apparent day coincides with The High Speed (of the Earth), and that the mean sun coincides with equinox a.k.a. The Mean Speed.

You have longest and shortest backwards, but, with that correction, OK. So are we back to no tilt? It looks like we are, but I'm losing track.

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So, now you should finally understand what is the deal here.

Absolutely! You are confusing yourself. That's been apparent all along.

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As the speed of the Earth changes through the year, so the daily rate of an annual motion of the Stars has to change too, BUT IT DOESN'T CHANGE!!!

I really think that the term "daily rate of an annual motion of the Stars" is a poor one and is causing you unneeded confusion. One part of the problem is that "daily" is ambiguous. "Daily" can refer to mean days or apparent days. PICK ONE! If you pick mean solar days, then, no the daily change of Right Ascension at the meridian at, say local noon each (mean solar) day does not change. If you pick apparent solar days, then the rate of change of the Sun's RA at (apparent solar) noon does change. You are mixing these definitions and claiming that proves something. It only proves you can't keep your terms straight.

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What i really meant by saying "Equinox is not everyday!" was: The orbital speed of the Earth is not the same through the year, and that is why a daily rate of an annual motion of the stars wouldn't be a constant if it were really depend on irregular orbital speed of the Earth.

The varying orbital speed is actually the lesser component in the varying length of the apparent solar day. The larger component is due to the obliquity of the ecliptic (since you're quoting Jean Meeus, I presume you know what that means.)

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But it doesn't depend on that, since the orbital speed of the Earth doesn't exist in the first place!!!

More opinion asserted as fact. So how's that non-rotating Geocentric explanation of the Equation of Time coming? And have you looked at stellar and planetary parallax yet?
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #156 on: October 26, 2014, 04:58:22 PM »
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Alleged orbital speed of the Earth (a main factor) is not a constant!

So, daily rate of annual rotation (a function) can't be a constant too!

You are busted! And you know that!

In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Time for retraction!

Good night!

P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!
"I can't breathe" George Floyd RIP

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sokarul

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Re: "Equator" problem
« Reply #157 on: October 26, 2014, 06:04:21 PM »
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Alleged orbital speed of the Earth (a main factor) is not a constant!

So, daily rate of annual rotation (a function) can't be a constant too!

You are busted! And you know that!

In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Time for retraction!

Good night!

P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!
Can you teach us the "when things get further away they appear smaller hoax" as well?
« Last Edit: October 26, 2014, 06:08:30 PM by sokarul »
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It's no slur if it's fact.

Re: "Equator" problem
« Reply #158 on: October 26, 2014, 06:40:54 PM »
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

I think I understand what you're trying to argue but recognize errors in it.

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Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Not true.

If:
f(x) = x + 2
g(x) = 1 - x
h(x) = f(x) + g(x)

Then h(x) is a constant for any x even though f(x) and g(x) are not. Sure, this is a contrived example, but, remember, you said any. I only have to find one counter example to prove your assertion wong.

Here's another:
f(x) = 3x
g(x) = 6(x + 1) - 6x - 6
h(x) = f(x)g(x)

f(x) is not constant for any x, but h(x), which has f(x) as a factor is a constant for any x.

"If any of the determining factors is not the constant the result (function) can't be a constant!" is proven false by counterexample(s).

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Alleged orbital speed of the Earth (a main factor) is not a constant!

We've never disagreed on this. The bigger factor in the variation of the length of the apparent solar day, obliquity, is, I think, also accepted.

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So, daily rate of annual rotation (a function) can't be a constant too!

Why not? You still haven't shown that "daily rate of annual rotation" isn't a constant because you haven't defined what you mean by "daily". This means that "daily rate of annual rotation" may or may not depend on orbital speed of the Earth and obliquity of the ecliptic, so it may indeed be a constant. Even if you use a definition for "daily" that does depend on orbital speed of earth, you still haven't disproved the Heliocentric model, which was your original assertion, since, if you use that definition, you can no longer say "daily rate of annual rotation" is a constant. You gotta be consistent with your "days" or your argument is invalid.

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You are busted! And you know that!

Do you really still believe this, or are you just unwilling to admit you know you're wrong?

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In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Wait...  what? Don't you understand what your quote from Meeus was saying?

First he posits a "first fictitious Sun" that travels along the ecliptic at a constant rate, coinciding with the real sun at the solstices. Sometimes this fictitious Sun will lead the real Sun and sometimes it will trail, but completes a circuit in exactly the same time as the real Sun, "smoothing out" the variations due to the elliptical orbit.

Next he posits a "second fictitious Sun"  that travels along the equator at a uniform rate, sometimes leading the first fictitious sun in RA and sometimes trailing it, but coinciding with it at the equinoxes. This smooths out the variations in change in RA per unit time due to obliquity of the ecliptic.

Are you suggesting there is a difference in how changes in the length of the apparent solar day due to eccentricity and obliquity affects the length of the apparent solar day or the apparent position of the stars relative to the Sun's RA? Why? Would these affect "daily rate of annual rotation" differently (if "daily" means apparent, not mean, solar days)? Please amplify this.

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Time for retraction!

Great! I've been waiting for days. It's tough to do, so I admire you for owning up that you were wrong.

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P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!

Ohboy! I can't wait! I trust that, unlike sceptimatic, you will actually deliver something as promised.

Still nothing on the Equation of Time? Not even to deny it exists?
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #159 on: October 27, 2014, 05:22:04 AM »
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree. The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

As for the equation of time: what that notion really presents?

It is about lining up ecliptic (irregular yearly motion of the Sun) with a steady-regular of the stars!

That is the manner in which we get our Mean Sun.

A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

http://www.energeticforum.com/264541-post421.html

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...
« Last Edit: October 27, 2014, 05:34:36 AM by cikljamas »
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Rama Set

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Re: "Equator" problem
« Reply #160 on: October 27, 2014, 05:39:16 AM »
You know that the 3:56 seconds you are fixated on is not constant but rather becoming shorter as the rotation of the Earth slows down?  So it is in fact, not a constant at all.
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Re: "Equator" problem
« Reply #161 on: October 27, 2014, 07:09:21 AM »
You know that the 3:56 seconds you are fixated on is not constant but rather becoming shorter as the rotation of the Earth slows down?  So it is in fact, not a constant at all.

He has much bigger issues than that, like a basic misunderstanding of what he's arguing about. I'm not sure he knows about this and I've been neglecting this very small variable as a needless complication. There are other small factors, such as precession and nutation, that also can be safely neglected in the discussion at hand (precession was touched on briefly in a late post in reference to the angle between apsides and the line of equinoxes slowly changing).

Also, the slowing spin will increase the difference between sidereal and solar days.

[Edit to add] And, actually, it's closer to 3:55.9084 according to Wikipedia, but 3:56 is close enough for the discussion.
« Last Edit: October 27, 2014, 07:13:14 AM by Alpha2Omega »
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cikljamas

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Re: "Equator" problem
« Reply #162 on: October 27, 2014, 07:24:35 AM »
1. If we accept the Copernican viewpoint and its unavoidable extrapolations with regard to the structure of the universe, we have to accept the consequences. Then we cannot hold on to the picture of a simple sun- centered cosmos, of which not even Newton was fully convinced, but which Bradley and Molyneux took for granted. Today the astronomers assure us that our Great Light is only an insignificant member of a spiral Milky Way galaxy, containing billions of stars. Our sun flies at a speed of about 250 km/sec around the center of this system. And that is not all, the ruling cosmology also tells us how the Milky Way itself whirls at 360,000 km/hr through the space occupied by the local group of galaxies. Now all these imposing particulars are theoretically gathered from observations assuming the speed of light to be 300,000 km/sec, at least, everywhere through our spatial neighborhood. But if this cosmological panorama is put through its paces, there is a hitch somewhere. The astronomical theorists cannot have their cake and eat it. If they accept— as all the textbooks still do!—Bradley's “proof” of the Copernican truth, then their cosmological extrapolations of that truth clash with a not-yet developed simple heliocentrism; that is to say, with the model of an earth orbiting a spatially unmoved sun.

The other way around, when holding on to their galactic conjectures, they are at a loss how to account for a steady 20”.5 stellar aberration. For in that scheme our earth, dragged along by the sun, joins in this minor star's 250 km/sec revolution around the center of the Milky Way. If, for instance, in March we indeed would be moving parallel to the sun's motion, our velocity would become 250+30 = 280 km/sec, and in September 250-30 = 220 km/sec. The “aberration of starlight,” according to post- Copernican doctrine, depends on the ratio of the velocity of the earth to the speed of light. As that velocity changes the ratio changes. Hence Bradley's 20”.496 should change, too. But it does not. Therefore, there is truly a fly in this astronomical ointment, paraded and promoted as a truth.

2.For those who maybe don't know:

1 sec. of an arc = An angle subtended by a U.S. dime coin at a distance of 4 km!!!
1 mas = 1/1000000 sec. of an arc
Tycho Brahe was able to measure angles about 0,3 minutes of an arc.
Casini was able to measure angles about 3,6 seconds of an arc.

The North Star, also known as Polaris, is known to stay fixed in our sky. It marks the location of the sky’s north pole, the point around which the whole sky turns. That’s why you can always use Polaris to find the direction north.

Just look at this insanity:

In a recent letter to the Astrophysical Journal, Turner et al.(2013) (TKUG from here on) suggested that the parallax as measured by Hipparcos (van Leeuwen 2007a,b) for Polaris (HIP 11767, HD 8890) is signi?cantly lower than it should be. The distance of 99 ± 2 pc suggested by TKUG on the basis of he assumed pulsation mode of Polaris is equivalent to a parallax of 10, 1 ± 0 2 mas, very di?erent from the parallax as measured by Hipparcos, 7, 54 ± 0 11 mas. Consequently, I have recently frequently been asked if it is at all possible for the Hipparcos parallax measurement to be so far o?. This letter shows the Hipparcos astrometric solution for Polaris in all detail as a means to assess the robustness of that solution, to assess whether its measurement of the parallax could be o?set by 23 times its standard error. It also brie?y discusses other arguments that have been used to suggest a signi?cantly shorter distance for Polaris than what has been measured by Hipparcos. READ MORE: http://arxiv.org/pdf/1301.0890v1.pdf

All that this "stars parallax" nonsense is about is this : throwing sand in the eyes of a sincere thinkers, nothing more than that!

3. http://www.realityreviewed.com/Negative%20parallax.htm

4 .Again and again have their theories been combated and exposed, but as often have the majority, who do not think for themselves, accepted the popular thing. No less an authority in his time than the celebrated Danish astronomer, Tycho Brahe, argued that if the earth revolves in an orbit round the sun, the change in the relative position of the stars thus necessarily occasioned, could not fail to be noticed. In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.

If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

5. This is not about the parallax but it gives us additional insight into a sanity of the heliocentric theory.: http://www.energeticforum.com/264118-post355.html

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markjo

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Re: "Equator" problem
« Reply #163 on: October 27, 2014, 09:42:54 AM »
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice.
Which fits quite nicely with what one would expect if the earth has an elliptical orbit around the sun.
Science is what happens when preconception meets verification.
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Besides, perhaps FET is a conspiracy too.
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Re: "Equator" problem
« Reply #164 on: October 27, 2014, 10:20:08 AM »
...
In a recent letter to the Astrophysical Journal, Turner et al.(2013) (TKUG from here on) suggested that the parallax as measured by Hipparcos (van Leeuwen 2007a,b) for Polaris (HIP 11767, HD 8890) is signi?cantly lower than it should be. The distance of 99 ± 2 pc suggested by TKUG on the basis of he assumed pulsation mode of Polaris is equivalent to a parallax of 10, 1 ± 0 2 mas, very di?erent from the parallax as measured by Hipparcos, 7, 54 ± 0 11 mas. Consequently, I have recently frequently been asked if it is at all possible for the Hipparcos parallax measurement to be so far o?. This letter shows the Hipparcos astrometric solution for Polaris in all detail as a means to assess the robustness of that solution, to assess whether its measurement of the parallax could be o?set by 23 times its standard error. It also brie?y discusses other arguments that have been used to suggest a signi?cantly shorter distance for Polaris than what has been measured by Hipparcos. READ MORE: http://arxiv.org/pdf/1301.0890v1.pdf
...
Sorry about the time you spent putting that tome together. It doesn't address the question I asked.  Repeating it here:

How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #165 on: October 27, 2014, 12:26:39 PM »
Sorry about the time you spent putting that tome together.
What tome? " class="bbc_link" target="_blank" rel="noopener noreferrer">

How come you skipped one whole post? Repeating it here:

Quote
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree. The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

As for the equation of time: what that notion really presents?

It is about lining up the ecliptic (irregular yearly motion of the Sun) with a steady-regular motion of the stars!

That is the manner in which we get our Mean Sun.

A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

http://www.energeticforum.com/264541-post421.html

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...

How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.


No, i don't think that stellar parallax exists, not even geocentric one. There are some phenomena that HCs and GCs link with that parallax concept, but what these phenomena really are, is a big question. Micro arc seconds, tell me about it...thousands of light years, tell me about it...

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

In which post? Last one (about stellar parallax hoax) or one that i just have repeated here?

« Last Edit: October 27, 2014, 12:33:05 PM by cikljamas »
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Re: "Equator" problem
« Reply #166 on: October 27, 2014, 12:35:53 PM »
This didn't post. Fortunately, I saved a copy. Sorry for the confusion.

Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree.

OK. That's the apparent motion of the sun along the ecliptic, which is due to the eccentricity of the Earth's orbit.

Quote
The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

You can't simply take the arithmetic average of fastest and slowest to get the mean here, but your answer is fairly close.

Note that this is not the only factor in the varying length of the apparent solar day, though. A larger factor is obliquity of the ecliptic, and this works to make the apparent solar days shortest at the equinoxes and longest at solstices, in case "apparent = mean at the equinoxes" is what you're getting at here; it's not.

Quote
We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

Hold it. Your ASTRONOMICAL DAY here is just the mean solar day we've been talking about all along (call it a civil day if you want a shorter name). The term 'ASTRONOMICAL DAY' would be better applied to the sidereal day, but we already have a perfectly good and well-accepted names for both of these, so why introduce a new term?

Quote
After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

Quote
As for the equation of time: what that notion really presents?

It is about lining up ecliptic (irregular yearly motion of the Sun) with a steady-regular of the stars!

That is the manner in which we get our Mean Sun.

Actually, it shows the difference in RA between the steady-regular mean Sun and the sometimes-leading and sometimes-lagging true Sun. The stars aren't involved.

Quote
A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical civil day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

Bingo!

Quote
It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

Not sure how this adds anything to the discussion, but, OK, sure

Quote
WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator." A slightly longer answer is that "Our celestial coordinate system is typically aligned with the Equator and Equinox of J2000, but the origin of this coordinate system is often the barycenter of the solar system, which is usually very close to the center of the Sun."

Quote
<snip random link>

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...

That's not a bad idea. [Oh, well...]
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #167 on: October 27, 2014, 02:52:50 PM »
How come you skipped one whole post? Repeating it here:


Probably user error. Sorry about that.

Quote
How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

No, i don't think that stellar parallax exists, not even geocentric one. There are some phenomena that HCs and GCs link with that parallax concept, but what these phenomena really are, is a big question. Micro arc seconds, tell me about it...thousands of light years, tell me about it...

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

In which post? Last one (about stellar parallax hoax) or one that i just have repeated here?

The one where you cited a study about parallax.

There wouldn't be parallax if the GC model were correct, so I can see why you don't want to believe it exists. I gather from your remarks that you don't think Aberration of Light exists, either. The nice thing about simply ignoring data is it doesn't get in the way of whatever model you want to believe in. This way you can "continue to seek the truth" without being bothered by actual observations and measurements. This won't get you very far discussing ideas with professional astronomers, however; you were asking me to get an opinion of this stuff from one earlier. It certainly won't get you very far in "finding the truth" either, but that's your problem.

Any thoughts about the apparent retrograde motion of the outer planets? This is easy to see and is attributed to parallax in the HC model, which is quite accurate in predicting where the planets will appear in the sky well into the future. I would like to see how you explain this without resorting to parallax, or is this a hoax, too?

Just for the record:

mas is milliarcsecond, not microarcsecond (micro is usually abbreviated with 'u' if the lower-case Greek letter 'mu' isn't available). But, hey, what's three orders of magnitude between friends?

Polaris is not exactly at the North Celestial Pole, and does not "stay fixed" in the sky. It's close, and for casual observation that's a good enough approximation, but the small circle it traces in the sky is evident in "star trail" photographs taken with even simple equipment. I'm not sure why you think this matters for the discussion of parallax; its position relative to the ecliptic pole is what counts in that case.

"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #168 on: October 27, 2014, 03:40:18 PM »
Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

You are very close to the truth, only it doesn't depend only on the rate of rotation of the Earth, it depends only on the rate of rotation of the Stars. But the problem is that it is not that simple in HC theory...Watch: " class="bbc_link" target="_blank" rel="noopener noreferrer">

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator."

So, if we lived on the Sun, which star would be our reference (fixed) star around which all other stars and clasters of stars would rotate in perfectly concentric circles?

Bingo!

What that supposed to mean? You are trying to be a smart guy?

Just for the record: mas is milliarcsecond, not microarcsecond

O.K., thanks for that correction! It is still small enough to be laughable, don't worry...

Apparent retrograde motion of the outer planets...that's another topic...


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sokarul

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Re: "Equator" problem
« Reply #169 on: October 27, 2014, 05:48:54 PM »
You shotgun attacks on RE in this thread but won't answer one topic that goes against what you preach?  I am disappointed.
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Re: "Equator" problem
« Reply #170 on: October 27, 2014, 07:20:40 PM »
Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

You are very close to the truth, only it doesn't depend only on the rate of rotation of the Earth, it depends only on the rate of rotation of the Stars. But the problem is that it is not that simple in HC theory...Watch: " class="bbc_link" target="_blank" rel="noopener noreferrer">

No, you asked for a Heliocentric explanation so I gave a Heliocentric answer - one where the Earth spins. Here's the answer in context:
Quote
Quote
Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

Since the Earth being fixed and the distant stars rotating about it, or the distant stars being fixed and the Earth rotating under them is simply a change in the frame of reference, they are equivalent. So thanks for acknowledging that my answer from the rotating-earth frame of reference was correct. I'll take that.

Can you summarize what you think that youtube link explains? Sometimes these will provide a visual insight that isn't easily described in words or static diagrams, but much more often they're a complete waste of time.

Quote
This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator."

So, if we lived on the Sun, which star would be our reference (fixed) star around which all other stars and clasters of stars would rotate in perfectly concentric circles?

Here's the whole answer in context:
Quote
Quote
WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator." A slightly longer answer is that "Our celestial coordinate system is typically aligned with the Equator and Equinox of J2000, but the origin of this coordinate system is often the barycenter of the solar system, which is usually very close to the center of the Sun."


If we lived on the sun, I seriously doubt the Earth's equator would play any part in solar-dweller cosmology. Note that modern cosmology doesn't expect perfect circles, nor, since it's based on an inertial universe (one that doesn't rotate), it doesn't have any point that everything revolves around. Assuming we could see any celestial objects at all (remember, if we lived on the Sun it would always be "daytime" because the "ground" would be literally as bright as the Sun), the solar-dweller celestial coordinate system would probably be based on the Sun's equator, but the origin for the RA coordinate would be up for grabs since there would be no significant point like the Vernal Equinox to tie it to. The origin would likely be the SS barycenter or the center of the Sun. What else would it be?

Quote
Bingo!

What that supposed to mean? You are trying to be a smart guy?

It means "we have a winner". In this context it means you got that part right. It may be idiomatic enough that it didn't make sense to someone not from the US; if that's the case, sorry. It comes from a game of chance called "Bingo", usually played in large groups, and play continues until someone wins. The winner shouts "BINGO!" ending the game.

Quote
Just for the record: mas is milliarcsecond, not microarcsecond

O.K., thanks for that correction! It is still small enough to be laughable, don't worry...

It's not that small. Instruments capable of measuring angles well below the arcsecond level were being built and used in the 19th century and have gotten better since then. Instead of laughing, do some research.

Quote
Apparent retrograde motion of the outer planets...that's another topic...

It's not really, but since you've ruled out parallax, it may be a difficult problem. It's easily explained in a heliocentric solar system with parallax due to the Earth's orbit making the outer planets appear to move backward - and is strong evidence in favor of the Heliocentric model over the Geocentric one.  In a Geocentric system it's not so simple. Surely not impossible, but not so simple, either. I'm really looking forward to seeing your approach.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #171 on: October 28, 2014, 10:07:36 AM »
This is the problem: Astronomical day = Solar day

If we assumed that the Earth rotates but doesn't revolve then we could consider principally almost identical situations when comparing HC & GC. But since the HC supposes Earth's revolution around the Sun also, then you have to expect that you will be faced with unresolvable difficulties.

Whenever the Earth allegedly completes one sidereal rotation what follows right after is Our Famous Constant!

The difference between one Astronomical day and one Sidereal day is equal to Our Constant.

The difference between one Solar day and one Sidereal day is equal to Our Constant AGAIN.

Anything suspicious over here?

Why are these differences equal to the same period (amount) of time which is 3min56sec???

How do we get so perfect synchronization between one Astronomical and one Solar day, although the Earth's orbit is not a perfect circle, and the Earth's orbital speed is not a constant???

To make it even clearer for our audience let's pose this question:

How probable (to be accomplish in reality) do you think is this purely theoretical scenario:

The Earth makes exactly one SLOW (annual) rotation on it's axis after exactly one year, and each day of the year while traveling along the ECCENTRIC orbit around the Sun at different speeds, the Earth aligns with the Sun (Solar day) at the exact same time that takes for a completion of one Astronomical day???

Shall we talk about  the identically miraculous synchronization of the alleged one SLOW (monthly) rotation of the Moon while traveling along the eccentric orbit at different speeds around the Earth?

In addition:

1. What is the miraculous cause for the tilt of the Earth?
2: What maintains it?
3: What is the miraculous cause for the fixed spatial orientation of the Earth's axis?
4. What maintains it?
5. The Earth's axis is spatially fixed with respect to miraculous "what"?
6. Why there are no alternations regarding the relative positions of the Stars (Tycho Brahe stated that had the Earth traveled through space in it's orbit around the Sun he should have been able to observe such alternations.)?
7.Why are the parallaxes perfectly circular? (It shouldn't be so due to combination of the insanely fast journey of our "Solar" System around the "galactic" centre and the alleged Earth's orbit around the Sun, as W.V.D.Kampe had nicely explained)
8. What about the negative parallaxes?
9. The entire universe is centered to the Earth (not to the Sun)! What was your answer on this? Was it something like this: "Because we live on the Earth!" - Hm, the answer is correct, but it belongs to the realm of theology. So, how come that you are bringing in theological arguments into an astronomical discussion?

P.S. I know the meaning of the word "bingo", you missed the point of my question...it was about the irony that i wrongly noticed in that word...but it seems that there was no irony in your usage of that "bingo" either...so the only reasonable solution of this misunderstanding is that i was just paranoid without the reason, my apology!
« Last Edit: October 28, 2014, 10:16:22 AM by cikljamas »
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Rama Set

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Re: "Equator" problem
« Reply #172 on: October 28, 2014, 11:04:45 AM »
You know a solar day is an arbitrary number? Just a sidereal day rounded up to the nearest hour?
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

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cikljamas

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Re: "Equator" problem
« Reply #173 on: October 28, 2014, 11:19:28 AM »
You know a solar day is an arbitrary number? Just a sidereal day rounded up to the nearest hour?

It seems to me that a solar day in HC has to be much more than just an arbitrary number! If it turns out to be a correct inference then my argument becomes a winner of this game! I am going to refrain myself of yelling "Bingo" at this moment...
« Last Edit: October 28, 2014, 11:21:04 AM by cikljamas »
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Re: "Equator" problem
« Reply #174 on: October 28, 2014, 01:53:55 PM »
This is the problem: Astronomical day = Solar day
What is it that defines an "Astronomical day"? You said earlier that its length is 24 hours, which is the length of a mean solar day. So why is this a problem?

Quote
If we assumed that the Earth rotates but doesn't revolve then we could consider principally almost identical situations when comparing HC & GC. But since the HC supposes Earth's revolution around the Sun also, then you have to expect that you will be faced with unresolvable difficulties.
Nope. No difficulties, unresolvable or otherwise.

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Whenever the Earth allegedly completes one sidereal rotation what follows right after is Our Famous Constant!
Whenever the Earth completes one sidereal rotation it has to rotate an additional amount to bring the Sun back to the same position because it's progressed along its orbit. This additional rotation varies slightly from day to day but averages almost exactly 3m 54s over a year.

Quote
The difference between one Astronomical day and one Sidereal day is equal to Our Constant.

The difference between one Solar day and one Sidereal day is equal to Our Constant AGAIN.

Anything suspicious over here?
Nope. Since they are both defined as the same thing, 24 hours, what else would you expect?  It does make one wonder why you think a new term is necessary.

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Why are these differences equal to the same period (amount) of time which is 3min56sec???
Because the length of the two types of "days" you're subtracting the length of the sidereal day from are the same?

Quote
How do we get so perfect synchronization between one Astronomical and one Solar day, although the Earth's orbit is not a perfect circle, and the Earth's orbital speed is not a constant???
Because you defined "one Astronomical Day" as the same length of time as one mean solar day. Recall that the mean solar day has a constant length, unlike apparent solar days, which do depend on the season and orbital speed.

Quote
To make it even clearer for our audience let's pose this question:

How probable (to be accomplish in reality) do you think is this purely theoretical scenario:

The Earth makes exactly one SLOW (annual) rotation on it's axis after exactly one year, and each day of the year while traveling along the ECCENTRIC orbit around the Sun at different speeds, the Earth aligns with the Sun (Solar day) at the exact same time that takes for a completion of one Astronomical day???
Since it's purely theoretical, it doesn't have to be probable at all. But your scenario doesn't make sense as written; you posit one rotation per year, and then talk about "each day of the year" and aligning with the sun each day. If earth rotates exactly one time per year, there will be zero "days"; the Sun will trace the Analemma in the sky and won't set at all in some locations, and will never rise at all in some others. Can you restate the question?

Quote
Shall we talk about  the identically miraculous synchronization of the alleged one SLOW (monthly) rotation of the Moon while traveling along the eccentric orbit at different speeds around the Earth?
It's neither 'miraculous' nor a coincidence. The moon's period of rotation is equal to its period of orbit because it's tidally locked.   

Because the Moon's orbit is slightly eccentric, it appears to "wobble" a little as its rotation is alternately slightly faster and slower than its orbital speed. See  Libration.

Rather than lots of quote blocks, answers below will be in italics.

Quote
In addition:

1. What is the miraculous cause for the tilt of the Earth?
I don't know. There's no reason to expect the Earth's axis of rotation to be exactly perpendicular to the plane of the orbit, though. It is what it is; why does what caused it matter?

2: What maintains it?
The Earth's spin.

3: What is the miraculous cause for the fixed spatial orientation of the Earth's axis?
See the answer to 2: above. Note that is isn't truly fixed, though; it precesses, completing a full circuit in about 26,000 years, and nutates (wobbles) by a few seconds of arc roughly every 19 years.

4. What maintains it?
See the answer to 2: above.

5. The Earth's axis is spatially fixed with respect to miraculous "what"?
The plane of the orbit is our most obvious reference for its orientation. But, actually, it's its orientation is fixed (neglecting precession and nutation) with respect to the inertial reference frame - that is, the universe as a whole

6. Why there are no alternations regarding the relative positions of the Stars (Tycho Brahe stated that had the Earth traveled through space in it's orbit around the Sun he should have been able to observe such alternations.)?
There are. It's called parallax. Tycho was mistaken about how far away the nearest stars were, so parallax was much smaller than he expected.

7.Why are the parallaxes perfectly circular? (It shouldn't be so due to combination of the insanely fast journey of our "Solar" System around the "galactic" centre and the alleged Earth's orbit around the Sun, as W.V.D.Kampe had nicely explained)
They're not. With no proper motion, parallax would be perfectly circular only at the ecliptic poles, a straight line on the ecliptic, and elliptical elsewhere. Most of the stars we can measure parallax for are moving in about the same direction at about the same speed as we are, so, combined with the great distances, their proper motion is low. Those speeds only seem "insanely fast" if you think the universe is small.

8. What about the negative parallaxes?
Binary systems can produce motion unrelated to, but mistaken for, parallax. Stellar parallax can be measured only by comparing positions wrt other stars assumed to be much further away; sometimes that assumption is not correct. Error margins when measuring small values of parallax can exceed the actual value itself and sometimes land on the wrong side of zero. This paper is a good analysis.

9. The entire universe is centered to the Earth (not to the Sun)! What was your answer on this? Was it something like this: "Because we live on the Earth!" - Hm, the answer is correct, but it belongs to the realm of theology. So, how come that you are bringing in theological arguments into an astronomical discussion?
If you reread my answer, I describe "references" and "coordinate systems" which are based on what is convenient to us. The universe is inertial (not spinning), and has no center, so any point (or direction) is as good as any other to use as a reference; we choose what makes things easiest for us.

[Edit] Clarification in answer to 5.
« Last Edit: October 28, 2014, 02:05:31 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #175 on: October 28, 2014, 09:49:02 PM »
To be sure I wasn't missing something obvious without waiting for your answer to:
This is the problem: Astronomical day = Solar day
What is it that defines an "Astronomical day"?

I found the post where the term was introduced. Here it is:
...
We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec
...

Let's use AD for "ASTRONOMICAL DAY", SD for "SIDEREAL DAY", C for "OUR CONSTANT", and MS for Mean Solar day, which is exactly 24 hours.

Now

MS - SD = C   (I think we can agree on this)

So

MS = SD + C   (simple algebra)

Your statement 1, above, says

AD = SD + C

Now, since

MS = SD + C

and

AD = SD + C

then

MS = AD.

According to your own definition of "ASTRONOMICAL DAY", it's identical to Mean Solar Day.

Further, you use "solar day", "mean solar day", "mean day", and "24 hours" interchangeably later in the same post
A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

So, again, why, exactly, is

"Astronomical day = Solar day" a problem? You have defined them to be exactly the same thing: the length of a sidereal day plus the difference between mean solar day and sidereal day.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #176 on: October 29, 2014, 06:42:17 AM »
Astronomical day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Stars only!

Solar day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Sun only!

That is the difference!

Quote
An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

If the Earth's orbit were a perfect circle and if the Earth's orbital speed were a constant,  one additional (slow-annual)  rotation of the Earth (assuming that the Earth is placed at the perfect distance from the Sun) would perfectly match the Earth's annual sidereal rotation. That would be some synchronization...

Quote
The problem is that in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December; 24 "hours" of solar time can be 21 seconds less or 29 seconds more than 24 hours of clock time. As explained in the equation of time article, this is due to the ellipticity of the Earth's orbit and the fact that the Earth's axis is not perpendicular to the plane of its orbit.

What is even greater miracle is that although the Earth allegedly travels in the eccentric orbit at a different speeds, one additional annual rotation of the Earth still perfectly matches the Earth's annual sidereal rotation.

The apparent sun is the true sun as seen by an observer on Earth. Apparent solar time or true solar time is based on the apparent motion of the actual Sun. It is based on the apparent solar day, the interval between two successive returns of the Sun to the local meridian.

The question is this:

If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???

Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?


In addition:

Quote
Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"

Does this make any difference?
« Last Edit: October 29, 2014, 06:46:25 AM by cikljamas »
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Re: "Equator" problem
« Reply #177 on: October 29, 2014, 11:44:44 AM »
Astronomical day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Stars only!
Why do you think "Our Constant" is related to the stars? It's related only to the mean sun. Sidereal days alone are related to the distant stars.

Quote
Solar day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Sun only!
This is true, but the "in this case" qualifier is unnecessary. This constant only applies to the mean solar day.

Quote
That is the difference!
What difference? When you add the same two constants together you get the same answer, thus "Astronomical day" = Solar day. Didn't we already establish this? Why, again, do we need a new term, and a confusing one since it isn't really "astronomical"?

Quote
Quote
An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

If the Earth's orbit were a perfect circle and if the Earth's orbital speed were a constant,  one additional (slow-annual)  rotation of the Earth (assuming that the Earth is placed at the perfect distance from the Sun) would perfectly match the Earth's annual sidereal rotation. That would be some synchronization...
If you're contriving your example to be perfectly synchronized, why would you expect it to be other than synchronized?

Quote
Quote
The problem is that in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December; 24 "hours" of solar time can be 21 seconds less or 29 seconds more than 24 hours of clock time. As explained in the equation of time article, this is due to the ellipticity of the Earth's orbit and the fact that the Earth's axis is not perpendicular to the plane of its orbit.
Exactly like this:


The displacement of the solid black line above or below zero is the time the apparent sun leads (above zero) or lags (below zero) the mean sun. The slope of the solid black line is the difference between the lengths of the mean solar day and apparent solar day. When upward, apparent days are shorter (the apparent sun catches up with and passes the mean sun); when downward, apparent days are longer. At the four minima and maxima, where the slope momentarily flattens out, apparent and mean days are the same length. These descriptions also apply to the two component sine waves; note that the seasonal component (red dash-dot line) is somewhat larger than the orbital component (green dashed line) and has much higher maximum slopes. The latter means that the changes in the length of apparent solar days are more strongly affected by the seasonal component than the orbital component.

From the quote above, "in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December". Note that the black line is sloping upward in September and even more steeply downward in December.

Quote
What is even greater miracle is that although the Earth allegedly travels in the eccentric orbit at a different speeds, one additional annual rotation of the Earth still perfectly matches the Earth's annual sidereal rotation.
It's neither a miracle nor even mysterious. The Sun makes exactly one complete circuit of the ecliptic in exactly one year (that's what defines the year) even though it appeared to "speed up" and "slow down" a little along the way. That complete circuit of the Sun around the sky "uses up" exactly one complete [sidereal] rotation of the Earth and means that there will be exactly one less solar day in a year than sidereal days - regardless of the number of sidereal days in the year, or whether the motion was completely uniform or not. This is simple geometry and not magical or coincidental in any way. It really is that simple.

Quote
The apparent sun is the true sun as seen by an observer on Earth. Apparent solar time or true solar time is based on the apparent motion of the actual Sun. It is based on the apparent solar day, the interval between two successive returns of the Sun to the local meridian.

The question is this:

If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???
Are you sure that's right? I think you have it backward and apparent solar days will be longest near the southern solstice, where the perihelion of the orbit (earth moves faster, which lengthens the apparent solar days) nearly coincides with a solstice (which also lengthen the apparent solar days). 

Quote
Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

Quote
In addition:

Quote
Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"
Citation needed.

Quote
Does this make any difference?
Only if true.

[Edit] Typo.
« Last Edit: October 29, 2014, 11:47:27 AM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #178 on: October 29, 2014, 02:09:48 PM »

 

Quote
Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

Correct!

That is why this curve should look like this:



When the Earth allegedly speeds up the above curve should go downwards, and vice versa...

Isn't that obvious to you?
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Re: "Equator" problem
« Reply #179 on: October 29, 2014, 03:15:36 PM »
Quote
Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

Correct!

That is why this curve should look like this:



When the Earth allegedly speeds up the above curve should go downwards, and vice versa...

Isn't that obvious to you?
Um... yes. When is the Earth at perihelion and therefore going fastest? Early January, right? [Hint: the answer is yes.]

Yeah... you said it right here:
If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???
You got the length-of-day effect backward in that, but the timing of perihelion and aphelion were correct. You corrected the length-of-day effect of orbital speed later in your post and confirmed it at the beginning of your reply quoted here.

Do you see the green dashed line in the drawing? That one represents the component of the Equation of Time due to the Earth's elliptical orbit. It's going down in early January as you say it should.

The dash-dot red line represents the component of the Equation of Time due to the obliquity of the ecliptic. It is independent of the green curve, but happens to also cross the zero line going downward in late December close to when the green curve crosses going downward. Since the obliquity causes the apparent days to be longest at the solstices and shortest at the equinoxes, we would expect this. The black curve is the sum of the other two, so it also crosses zero going downward in late December - early January.

So why did you think it necessary to butcher the original drawing? It was showing exactly what you said it should.

When the Earth allegedly speeds up the above curve should go downwards, and vice versa...
This is what the original plot showed.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan