Dear Joe and Calendar People,
Your comment is exhaustive explanation. There are the mean day = 1 MEAN day, a MEAN tropical solar year = 365,242195 MEAN days. I hope Karl will correct his web-pages and Wikipedia JOE SAYS: Your solar year is the mean tropical year, which uses an imaginary or fictitious Sun moving along the ecliptic at a constant speed, that is, an average or mean Sun. In contrast, the true tropical year uses the true or visible Sun moving along the ecliptic at a variable speed. The length of the true tropical year depends on its point of origin. The four equinoxes and solstices result in four different kinds of true tropical years: the vernal equinox year, the summer solstice year, the autumnal equinox year, and the winter solstice year. Their lengths depend on the movement of the perihelion (fast Sun) and aphelion (slow Sun) relative to the equinoxes and solstices. The mean Sun of the mean tropical year does not even have a perihelion or aphelion. Although the vernal equinox is a single point on the celestial sphere, where the ecliptic crosses the celestial equator, the mean Sun and the visible Sun reach it at different times. The time/date of the vernal equinox given in almanacs is when the visible Sun reaches it. MIKHAIL SAYS: - Mean solar year (tropical) = 365,242195 mean days - mean duration of the time interval between two consecutive passages of the centre of the Sun, for example, through the Eastern equinox point, KARL SAYS: The definition "mean duration of the time interval between two consecutive passages of the centre of the Sun through the Northward equinox point" refers to what we call the Vernal equinox year, but the value 365.242195 days refers to what we call the mean tropical year (as defined below). These two tropical years are different. This failure to differentiate the two is the well established ERROR IN THE STATEMENT OF THE TROPICAL YEAR. http://www.hermetic.ch/cal_stud/cassidy/err_trop.htm A tropical year is the length of time that the Sun, as viewed from the Earth, takes to return to the same position along the ecliptic (its path among the stars on the celestial sphere) measured relative to the equinoxes and solstices (not the fixed stars). The mean duration of this period depends on where on the ecliptic that 'same point' is. If this is averaged for every 'same point' you get what we call the Mean Tropical Year which is about 365.2422 days. However a simpler definition in which the 'same point' is the March equinox is often used. The correct value for this tropical year is about 365.2424 days. This we call the Vernal Equinox Year. Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
Mikhail wrote:
> Your comment is exhaustive explanation. > There are the mean day = 1 MEAN day, a MEAN tropical solar year > = 365,242195 MEAN days. > I hope Karl will correct his web-pages and Wikipedia I assume your 'mean day' is the average length of the solar day over an entire year. The true solar day can be as much as 30 seconds longer or shorter than the mean solar day at certain times of the year. The summation or integral of all those long and short days over the course of a year is the "equation of time", which can reach +17 minutes or -14 minutes. But all astronomical days have been mean solar days since about 1830. The 'mean' in the mean tropical year is something else entirely. The length of the tropical year in mean solar days varies from year to year due to perturbations of Earth's elliptical orbit by the other planets. Its average over many years is called the mean tropical year. Its J2000.0 value is 365.242190 mean solar days and is gradually decreasing. Newcomb's mean tropical year was 365.242199 days at 1900.0. Your 365.242195 days is midway between them, maybe its B1950.0 value. You might be using a value from an old reference book. My earlier explanation was wrong. I'll try again. For this explanation, assume that Earth's elliptical orbit is not perturbed by any other planet. If so, it is fixed in inertial space, and the time to complete one orbit (360°) is always the same, regardless of its starting point. With this restriction, this fixed time is either the anomalistic year or the sidereal year, depending on your point of view. Precession is 50" per year. For Earth to complete its orbit, it needs to move an extra 50" beyond one tropical year. The time it takes the Earth to traverse most of its orbit, 360°-50", its partial tropical-year orbit, varies depending on the relation of its starting point (an equinox or solstice) to the apsides (perihelion and aphelion) of the ellipse. Earth's fast speed near perihelion at the winter solstice causes Earth to traverse that extra 50" in an unusually short time. When subtracted from the fixed anomalistic/sidereal year, the time for the partial tropical-year orbit is lengthened to 365.242740 days, the winter solstice tropical year. Near aphelion the summer solstice tropical year in shortened to 365.241626 days. The equinoctial years are midway--the vernal equinox tropical year is 365.242374 days and the autumnal equinox tropical year is 365.242018 days. The average of these four tropical years is the mean tropical year. If Earth's orbit was circular, it would always traverse the partial tropical-year orbit in 365.242190 days, the mean tropical year. Joe Kress |
In reply to this post by Mikhail Petin
Dear Joe and Calendar People,
JOE SAYS that Mikhail wrote: > Your comment is exhaustive explanation. > There are the mean day = 1 MEAN day, a MEAN tropical solar year > = 365,242195 MEAN days. > I hope Karl will correct his web-pages and Wikipedia JOE SAYS: I assume your 'mean day' is the average length of the solar day over an entire year. The true solar day can be as much as 30 seconds longer or shorter than the mean solar day at certain times of the year. The summation or integral of all those long and short days over the course of a year is the "equation of time", which can reach +17 minutes or -14 minutes. But all astronomical days have been mean solar days since about 1830. The 'mean' in the mean tropical year is something else entirely. The length of the tropical year in mean solar days varies from year to year due to perturbations of Earth's elliptical orbit by the other planets. Its average over many years is called the mean tropical year. Its J2000.0 value is 365.242190 mean solar days and is gradually decreasing. Newcomb's mean tropical year was 365.242199 days at 1900.0. Your 365.242195 days is midway between them, maybe its B1950.0 value. You might be using a value from an old reference book. MIKHAIL SAYS: You say truly. JOE SAYS: My earlier explanation was wrong. MIKHAIL SAYS: It is impossible. JOE SAYS: I'll try again. For this explanation, assume that Earth's elliptical orbit is not perturbed by any other planet. If so, it is fixed in inertial space, and the time to complete one orbit (360.) is always the same, regardless of its starting point. With this restriction, this fixed time is either the anomalistic year or the sidereal year, depending on your point of view. MIKHAIL SAYS: On my view the explanation given below is wrong. JOE SAYS: Precession is 50" per year. For Earth to complete its orbit, it needs to move an extra 50" beyond one tropical year. The time it takes the Earth to traverse most of its orbit, 360.-50", its partial tropical-year orbit, varies depending on the relation of its starting point (an equinox or solstice) to the apsides (perihelion and aphelion) of the ellipse. Earth's fast speed near perihelion at the winter solstice causes Earth to traverse that extra 50" in an unusually short time. When subtracted from the fixed anomalistic/sidereal year, the time for the partial tropical-year orbit is lengthened to 365.242740 days, the winter solstice tropical year. Near aphelion the summer solstice tropical year in shortened to 365.241626 days. The equinoctial years are midway--the vernal equinox tropical year is 365.242374 days and the autumnal equinox tropical year is 365.242018 days. The average of these four tropical years is the mean tropical year. If Earth's orbit was circular, it would always traverse the partial tropical-year orbit in 365.242190 days, the mean tropical year. MIKHAIL SAYS: 1. The tropical year = 365,242190 MEAN days should be the only thing for any points of the ecliptic. This value shows an angular way made by the Earth during the year. this angular way is the same for any points of the ecliptic. Values: 365,242740 days, 365,241626 days, 365,242374 days, 365,242018 days, don.t show an angular way made by the Earth during the year and, therefore, they are a nonsense. It is possible a Karl.s delusion. 2. Precession has a constant value and constant mark during the year and, therefore, is not the reason of changing of an angular Earth.s velocity during the year. 3. The reason of changing of an angular Earth.s velocity during the year is 2-nd law of Keppler. Joe, you are an experienced calendar-man. I have a favour to ask of you come back to previous your objective true positions. Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
In reply to this post by Palmen, KEV (Karl)
Am 24.02.2006 um 13:56 schrieb Palmen, KEV (Karl):
>> KARL SAYS: > A tropical year is the length of time that the Sun, as viewed from > the Earth, takes to > return to the same position along the ecliptic (its path among the > stars on the celestial > sphere) measured relative to the equinoxes and solstices (not the > fixed stars). > Hi Karl, i miss the word AVERAGE in the length of the year ... Sepp |
In reply to this post by Mikhail Petin
Dear Mikhail and Calendar People
-----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]]On Behalf Of Mikhail Petin Sent: 25 February 2006 10:58 To: [hidden email] Subject: Re: The New World Calendar Dear Joe and Calendar People, Your comment is exhaustive explanation. There are the mean day = 1 MEAN day, a MEAN tropical solar year = 365,242195 MEAN days. I hope Karl will correct his web-pages and Wikipedia KARL SAYS: My web-pages and wikipedia are correct and in accordance with the explanations provide by Joe and myself. The MEAN sun passes along the ecliptic at a constant speed (unlike the real sun), so gives rise to the mean tropical year, regardless of start point (unlike the real sun). Karl 07(17(29 till noon >JOE SAYS: Your solar year is the mean tropical year, which uses an imaginary or fictitious Sun moving along the ecliptic at a constant speed, that is, an average or mean Sun. In contrast, the true tropical year uses the true or visible Sun moving along the ecliptic at a variable speed. The length of the true tropical year depends on its point of origin. The four equinoxes and solstices result in four different kinds of true tropical years: the vernal equinox year, the summer solstice year, the autumnal equinox year, and the winter solstice year. Their lengths depend on the movement of the perihelion (fast Sun) and aphelion (slow Sun) relative to the equinoxes and solstices. The mean Sun of the mean tropical year does not even have a perihelion or aphelion. Although the vernal equinox is a single point on the celestial sphere, where the ecliptic crosses the celestial equator, the mean Sun and the visible Sun reach it at different times. The time/date of the vernal equinox given in almanacs is when the visible Sun reaches it. >MIKHAIL SAYS: - Mean solar year (tropical) = 365,242195 mean days - mean duration of the time interval between two consecutive passages of the centre of the Sun, for example, through the Eastern equinox point, >KARL SAYS: The definition "mean duration of the time interval between two consecutive passages of the centre of the Sun through the Northward equinox point" refers to what we call the Vernal equinox year, but the value 365.242195 days refers to what we call the mean tropical year (as defined below). These two tropical years are different. This failure to differentiate the two is the well established ERROR IN THE STATEMENT OF THE TROPICAL YEAR. http://www.hermetic.ch/cal_stud/cassidy/err_trop.htm A tropical year is the length of time that the Sun, as viewed from the Earth, takes to return to the same position along the ecliptic (its path among the stars on the celestial sphere) measured relative to the equinoxes and solstices (not the fixed stars). The mean duration of this period depends on where on the ecliptic that 'same point' is. If this is averaged for every 'same point' you get what we call the Mean Tropical Year which is about 365.2422 days. However a simpler definition in which the 'same point' is the March equinox is often used. The correct value for this tropical year is about 365.2424 days. This we call the Vernal Equinox Year. Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
In reply to this post by Mikhail Petin
Dear Karl and Calendar People,
KARL SAYS: ..If this is averaged for every 'same point' you get what we call the Mean Tropical Year which is about 365.2422 days. However a simpler definition in which the 'same point' is the March equinox is often used. The correct value for this tropical year is about 365.2424 days. This we call the Vernal Equinox Year. MIKHAIL SAYS: The last sentence is wrong. 1. The tropical year = 365,242190 MEAN days should be the only thing for any points of the ecliptic. This value shows an angular way made by the Earth during the year. This angular way is the same for any points of the ecliptic. Values: 365,242740 days, 365,241626 days, 365,242374 days, 365,242018 days, don.t show an angular way made by the Earth during the year and, therefore, they are a nonsense. Karl, this is your delusion. 2. It is necessary to specify each calendar postulate for the members of List. I offer 2 postulate for the beginning: - day (day + night) - mean duration of a revolution of the Earth around the its axis during the annual moving of Sun along the ecliptic, - solar year (tropical) = 365,242195 days - mean duration of the time interval between two consecutive passages of the centre of the Sun, for example, through the Eastern equinox point, Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
In reply to this post by Sepp Rothwangl
On 2006-02-24, Sepp Rothwangl wrote:
> I miss the word AVERAGE in the length of the year ... A tropical year is not defined as an average in modern astronomy, nor does it seem to be an average in any mathematical sense. Astronomers have used the notion of "tropical year" for centuries, and its definition has been refined correspondingly. Meeus and Savoie give an easily readable survey at [http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1992JBAA..102...40M &db_key=AST&data_type=HTML&format=] The most exact definition was applied in 1955 when the "tropical year" was used to define the ephemeris second. For that occasion, the tropical year was taken as (2 pi rad)/( angular speed of the geometric mean longitude of the Sun with respect to the mean equinox on the ecliptic of date ) where this angular speed is measured relative to ephemeris time (ET). This tropical year changes with time; the value used for the definition of the ephemeris second was the one when ET was 1899 December 31 + 12 h (noon) exactly. Some terms in this definition merit an explanation: "mean longitude" is not an average in any sense but the "secular" part of the longitude of the Sun. In order to define this precisely, one needs a theory of the movement of the Sun (giving the difference between the observable positions of the Sun and the mean Sun); Newcomb's theory was used in 1955. The series expansion of the geometric longitude that is used in such theories is a sum of the following types of terms: - secular terms, ie, a polynomial function of ET; - periodic terms, ie, products of sines (and cosines) of angles that are linear functions of ET; - other Poisson terms, ie, products of terms of the preceding two types. The last kind of terms represents periodic summands with a varying amplitude -- these would not cancel by taking averages. Also, some of the strictly periodic terms have long periods (even longer than the interval of validity of the ephemeris) -- they would not average out either. Thus, the secular part is not an average in the mathematical sense but just the part that does not involve any periodic functions. It is in this sense that the tropical year gives the long term trend in the longitude of the Sun. "geometric longitude" refers to the longitude of a theoretical point on the mean ecliptic that is obtained by subtracting the yearly aberration, as computed with a nominal value of 20.47 arcs for the aberration constant. For modern, relativistic dynamical thoeries of the solar system, "geometric longitude" does not make much sense; but the effect of switching to geocentric longitude should be negligible on the value of the tropical year. The "mean equinox" is a nominal point on the celestial sphere that approximates the secular movement of the equinox (it is not an average either). The "mean equinox of date" is its position at the instant of observation (rather than at a standard epoch, as common in numerical ephemerides). The definition of the ecliptic and the mean equinox changed in 1984. It probably will be changed again in the near future. (Some people say "mean mean longitude" for "mean longitude referred to the mean equinox".) A tropical year as used in modern astronomy: - represents the instantaneous angular speed of a fictituous point that moves on the ecliptic with a speed that is a polynomial function (ie, without periodic components) of the time scale that is taken as the basis of the ephemeris (ET or TDB or Teph). - is not defined as any average time interval between two solar phenomena. - has nothing to do with the declination of the Sun ("crossing the celestial equator") but only with the Sun's ecliptical longitude - depends to a certain degree on astronomical conventions (mean equinox, choice of timescale) and on an analytical theory of the motion of the Sun. For subsecond precision, these dependencies become relevant. Michael Deckers |
In reply to this post by Mikhail Petin
What you don't seem to understand, is that in an inertial frame -- in which
it can be assumed that the ellipse of the Earth's orbit is fixed -- there is no such thing as *the* equinox point. An equinox occurs when the Earth's axis is perpendicular to the Sun-Earth line; because of precession, if you extend this line towards the Zodiac, it will point to a different point each year, and that difference depends on where on the ellipse it happens. BTW, an equinox can be called either Spring (Northwards) or Autumn (Southward); I don't know what you mean by "Eastern". And another point: if you insist on calling anything you don't understand or agree with "nonsense" or "delusion", you're well on your way to Aristeanism... >From: Mikhail Petin <[hidden email]> >Date: Mon, 27 Feb 2006 12:57:31 +0300 > >>I offer 2 postulate for the beginning: > >- day (day + night) - mean duration of a revolution of the Earth around the >its axis >during the annual moving of Sun along the ecliptic, >- solar year (tropical) = 365,242195 days - mean duration of the time >interval >between two consecutive passages of the centre of the Sun, for example, >through the >Eastern equinox point, > _________________________________________________________________ FREE pop-up blocking with the new MSN Toolbar - get it now! http://toolbar.msn.click-url.com/go/onm00200415ave/direct/01/ |
In reply to this post by Mikhail Petin
Dear Amos and Calendar People,
AMOS SAYS: BTW, an equinox can be called either Spring (Northwards) or Autumn (Southward); I don't know what you mean by "Eastern".. .. And another point: if you insist on calling anything you don't understand or agree with "nonsense" or "delusion", you're well on your way to Aristeanism... MIKHAIL SAYS: The Eastern equinox is a point of intersection of the ecliptic Plane with the plane of a Earth.s equator. The Eastern equinox is always to the right of the projection N-S line if see from a North Pole of the Earth. Had Aristeo already said about this? >From: Mikhail Petin <[hidden email]> >Date: Mon, 27 Feb 2006 12:57:31 +0300 > >> I am offering 2 postulates: > >- day (day + night) - mean duration of a revolution of the Earth around the its axis during the annual moving of Sun along the ecliptic, >- solar year (tropical) = 365,242195 days - mean duration of the time interval between two consecutive passages of the centre of the Sun, for example, through the Eastern equinox point, > Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm > Amos Shapir <[hidden email]>: > What you don't seem to understand, is that in an inertial frame -- in which > it can be assumed that > the ellipse of the Earth's orbit is fixed -- there is no such thing as > *the* > equinox point. An equinox > occurs when the Earth's axis is perpendicular to the Sun-Earth line; > because > of precession, if you > extend this line towards the Zodiac, it will point to a different point > each > year, and that difference > depends on where on the ellipse it happens. > > BTW, an equinox can be called either Spring (Northwards) or Autumn > (Southward); I don't know > what you mean by "Eastern". > > And another point: if you insist on calling anything you don't understand > or > agree with "nonsense" > or "delusion", you're well on your way to Aristeanism... > > > >From: Mikhail Petin <[hidden email]> > >Date: Mon, 27 Feb 2006 12:57:31 +0300 > > > >>I offer 2 postulate for the beginning: > > > >- day (day + night) - mean duration of a revolution of the Earth around > the > >its axis > >during the annual moving of Sun along the ecliptic, > >- solar year (tropical) = 365,242195 days - mean duration of the time > >interval > >between two consecutive passages of the centre of the Sun, for example, > >through the > >Eastern equinox point, > > > > _________________________________________________________________ > FREE pop-up blocking with the new MSN Toolbar - get it now! > http://toolbar.msn.click-url.com/go/onm00200415ave/direct/01/ > Best regards Mikhail Petin |
In reply to this post by Mikhail Petin
> AMOS SAYS:
> BTW, an equinox can be called either Spring (Northwards) > or Autumn (Southward); I don't know what you mean by "Eastern".. > > .. And another point: if you insist on calling anything > you don't understand or agree with "nonsense" or "delusion", > you're well on your way to Aristeanism... > > MIKHAIL SAYS: > The Eastern equinox is a point of intersection of the ecliptic > Plane with the plane of a Earth.s equator. > The Eastern equinox is always to the right of > the projection N-S line if see from a North Pole of the Earth. I'm having trouble visualizing this. I have no trouble with the intersection of the ecliptic with the plance of Earth's equator. But, if you are on the North Pole, which way is right? The answer depends upon which direction you are looking. Also, if you're on the North Pole, aren't you also on the "projection N-S line"? If not, what do you mean by that? Victor |
In reply to this post by Mikhail Petin
But there is no single "point of intersection"! Two (non parallel) planes
always intersect along a line, and this line intersects the earth's surface at the equator on two points, one on the right and the other on the left (or East and West, or front and rear, depending on where you are looking from). >From: Mikhail Petin <[hidden email]> >Date: Mon, 27 Feb 2006 17:39:17 +0300 > >Dear Amos and Calendar People, > >AMOS SAYS: >BTW, an equinox can be called either Spring (Northwards) >or Autumn (Southward); I don't know what you mean by "Eastern".. > >.. And another point: if you insist on calling anything >you don't understand or agree with "nonsense" or "delusion", >you're well on your way to Aristeanism... > >MIKHAIL SAYS: >The Eastern equinox is a point of intersection of the ecliptic >Plane with the plane of a Earth.s equator. >The Eastern equinox is always to the right of >the projection N-S line if see from a North Pole of the Earth. _________________________________________________________________ Express yourself instantly with MSN Messenger! Download today it's FREE! http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/ |
In reply to this post by Mikhail Petin
Dear Amos, Mikhail and Calendar People
Actually Shriramana and I had a considerable discussion on CALNDR-L over Easter/Western Equinox etc stating from 15 March 2005. We eventually agreed that any naming of the equinoxes Eastern or Western is in fact hemisphere biased, thereby being no better than Spring/Autumn Equinox. I show after the note to which I'm replying an early note from this thread. Mikhail was involved in the discussion at this early stage. Karl 07(17(30 -----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]]On Behalf Of Amos Shapir Sent: 27 February 2006 16:30 To: [hidden email] Subject: Re: The New World Calendar But there is no single "point of intersection"! Two (non parallel) planes always intersect along a line, and this line intersects the earth's surface at the equator on two points, one on the right and the other on the left (or East and West, or front and rear, depending on where you are looking from). >From: Mikhail Petin <[hidden email]> >Date: Mon, 27 Feb 2006 17:39:17 +0300 > >Dear Amos and Calendar People, > >AMOS SAYS: >BTW, an equinox can be called either Spring (Northwards) >or Autumn (Southward); I don't know what you mean by "Eastern".. > >.. And another point: if you insist on calling anything >you don't understand or agree with "nonsense" or "delusion", >you're well on your way to Aristeanism... > >MIKHAIL SAYS: >The Eastern equinox is a point of intersection of the ecliptic >Plane with the plane of a Earth.s equator. >The Eastern equinox is always to the right of >the projection N-S line if see from a North Pole of the Earth. ----------- Note of 15 March 2005 ------- Dear Mikhail, Shriramana and Calendar People -----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]]On Behalf Of Mikhail Petin Sent: 12 March 2005 13:26 To: [hidden email] Subject: Re: Western (Eastern) equinox??! Dear Shriramana, Karl and Calendar People, KARL SAYS: > > On the Arctic circle, the horizontal ecliptic would have the March equinox due East (sun moves leftwards on the ecliptic). This suggests calling it the Eastern equinox. On the Antarctic circle, the horizontal ecliptic would have the March equinox due West (sun moves rightwards on the ecliptic). This suggests calling it the Western equinox. > > I SAY: Excuse me for my misprints in previous message. It is necessary to read the textby the following way: The Sun during the year moves against clock hand on ecliptic if we see from the Pole-star. Therefore, the March equinox will be the Eastern equinox and the September equinox will be the Western equinox on Arctic and Antarctic circles. KARL SAYS: Mikhail is almost correct, except that it does not apply to the Antarctic circle. The Pole-star (which I took to be the north Pole-star) CANNOT be seen at the Antarctic circle. The reason why I said Mikhail was almost correct is because the (north) Pole-star is not the relevant star. The relevant star is situated about 23.5 degrees away from the Pole-star and exactly 90 degrees from the ecliptic (not the equator). This makes no difference at either of the polar circles or anywhere else that is not tropical. Karl 07(06(05 > > Shriramana Sharma <[hidden email]>: > > > Palmen, KEV (Karl) wrote: > > > > > On the Arctic circle, the horizontal ecliptic would have the March equinox due East (sun moves leftwards on the ecliptic). This suggests calling it the Eastern equinox. On the Antarctic circle, the horizontal ecliptic would have the March equinox due West (sun moves rightwards on the ecliptic). This suggests calling it the Western quinox. > > > > Actually Petin names the two the other way round. > > > > -- > > > > 2005-03-06-Sym454 UTC+0530 > > http://samvit.org/calendar > > > |
In reply to this post by Mikhail Petin
Dear Michael and Calendar People,
MICHAEL SAYS: A tropical year is not defined as an average in modern astronomy, nor does it seem to be an average in any mathematical sense. MIKHAIL SAYS: It is necessary to correct quickly. MICHAEL SAYS: Astronomers have used the notion of "tropical year" for centuries, and its definition has been refined correspondingly. Meeus and Savoie give an easily readable survey at [http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1992JBAA..102...40M &db_key=AST&data_type=HTML&format=] MIKHAIL SAYS: I don.t see the refined definition of .tropical year.. MICHAEL SAYS: The most exact definition was applied in 1955 when the "tropical year" was used to define the ephemeris second. For that occasion, the tropical year was taken as (2 pi rad)/( angular speed of the geometric mean longitude of the Sun with respect to the mean equinox on the ecliptic of date ) where this angular speed is measured relative to ephemeris time (ET). This tropical year changes with time; the value used for the definition of the ephemeris second was the one when ET was 1899 December 31 + 12 h (noon) exactly. MIKHAIL SAYS: Formula: (2 pi rad)/( angular speed of the geometric mean longitude of the Sun with respect to the mean equinox on the ecliptic of date ) shows that was measured the time interval between two consecutive passages of the centre of the Sun through the Eastern equinox point (t=365,242195 days, i.e. a tropical year, which is equal to a tropical years in any other points of an ecliptic). After this was defined a MEAN angular speed of the geometric mean longitude of the Sun with respect to the mean equinox on the ecliptic of date. It is understood. MICHAEL SAYS: A tropical year as used in modern astronomy: - represents the instantaneous angular speed of a fictituous point that moves on the ecliptic with a speed that is a polynomial function (ie, without periodic components) of the time scale that is taken as the basis of the ephemeris (ET or TDB or Teph).. MIKHAIL SAYS: The value 365,242195 days can.t represent the instantaneous angular speed of a fictituous point. This is a time. MICHAEL SAYS: - is not defined as any average time interval between two solar phenomena. - has nothing to do with the declination of the Sun ("crossing the celestial equator") but only with the Sun's ecliptical longitude - depends to a certain degree on astronomical conventions (mean equinox, choice of timescale) and on an analytical theory of the motion of the Sun. For subsecond precision, these dependencies become relevant. MIKHAIL SAYS: The last indention need in details Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
In reply to this post by Mikhail Petin
Dear Victor and Calendar people,
VICTOR SAYS: I'm having trouble visualizing this. I have no trouble with the intersection of the ecliptic with the plane of Earth's equator. But, if you are on the North Pole, which way is right? The answer depends upon which direction you are looking. Also, if you're on the North Pole, aren't you also on the "projection N-S line"? If not, what do you mean by that? MIKHAIL SAYS: It is necessary to project an the Earth.s axis on the ecliptic plane and do a perpendicular to N-S line. AMOS SAYS: > BTW, an equinox can be called either Spring (Northwards) > or Autumn (Southward); I don't know what you mean by "Eastern".. > > .. And another point: if you insist on calling anything > you don't understand or agree with "nonsense" or "delusion", > you're well on your way to Aristeanism... > > MIKHAIL SAYS: > The Eastern equinox is a point of intersection of the ecliptic > plane with the plane of a Earth.s equator. > The Eastern equinox is always to the right of > the projection N-S line if see from a North Pole of the Earth. Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
In reply to this post by Deckers, Michael
Am 27.02.2006 um 11:49 schrieb Michael Deckers: > On 2006-02-24, Sepp Rothwangl wrote: > >> I miss the word AVERAGE in the length of the year ... > > A tropical year is not defined as an average in modern astronomy, > nor > does it seem to be an average in any mathematical sense. Michael, OK! You can determine a year with many different values and to many aims. Briefly: Calendrically it is determined by the secular average days and no more by seconds. Astronomically you can determine it by angular expansion and no days Temporally by atomic seconds or by the beats of a pulsar ... Servus Sepp > > Astronomers have used the notion of "tropical year" for centuries, > and > its definition has been refined correspondingly. Meeus and Savoie > give > an easily readable survey at > > [http://adsabs.harvard.edu/cgi-bin/nph-bib_query? > bibcode=1992JBAA..102...40M > &db_key=AST&data_type=HTML&format=] > > The most exact definition was applied in 1955 when the "tropical > year" > was used to define the ephemeris second. For that occasion, the > tropical > year was taken as > (2 pi rad)/( angular speed of the geometric mean longitude > of the Sun with respect to the mean equinox on the > ecliptic of date ) > where this angular speed is measured relative to ephemeris time > (ET). > This tropical year changes with time; the value used for the > definition > of the ephemeris second was the one when ET was 1899 December 31 > + 12 h (noon) exactly. > > Some terms in this definition merit an explanation: > > "mean longitude" is not an average in any sense but the > "secular" part of the longitude of the Sun. In order to define this > precisely, one needs a theory of the movement of the Sun (giving > the difference between the observable positions of the Sun and > the mean Sun); Newcomb's theory was used in 1955. > > The series expansion of the geometric longitude that is used in > such theories is a sum of the following types of terms: > - secular terms, ie, a polynomial function of ET; > - periodic terms, ie, products of sines (and cosines) of > angles that are linear functions of ET; > - other Poisson terms, ie, products of terms of the preceding two > types. > The last kind of terms represents periodic summands with a varying > amplitude -- these would not cancel by taking averages. Also, > some of the strictly periodic terms have long periods (even longer > than the interval of validity of the ephemeris) -- they would not > average out either. Thus, the secular part is not an average in > the mathematical sense but just the part that does not involve any > periodic functions. It is in this sense that the tropical year > gives the long term trend in the longitude of the Sun. > > "geometric longitude" refers to the longitude of a theoretical point > on the mean ecliptic that is obtained by subtracting the yearly > aberration, as computed with a nominal value of 20.47 arcs for the > aberration constant. For modern, relativistic dynamical thoeries of > the solar system, "geometric longitude" does not make much sense; > but the effect of switching to geocentric longitude should be > negligible on the value of the tropical year. > > The "mean equinox" is a nominal point on the celestial sphere that > approximates the secular movement of the equinox (it is not an > average > either). The "mean equinox of date" is its position at the instant > of observation (rather than at a standard epoch, as common in > numerical ephemerides). The definition of the ecliptic and the mean > equinox changed in 1984. It probably will be changed again in the > near future. (Some people say "mean mean longitude" for "mean > longitude > referred to the mean equinox".) > > A tropical year as used in modern astronomy: > > - represents the instantaneous angular speed of a > fictituous point that moves on the ecliptic with a speed that > is a polynomial function (ie, without periodic components) of > the time scale that is taken as the basis of the ephemeris (ET > or TDB or Teph). > > - is not defined as any average time interval between two > solar phenomena. > > - has nothing to do with the declination of the Sun ("crossing > the celestial equator") but only with the Sun's ecliptical > longitude > > - depends to a certain degree on astronomical conventions > (mean equinox, choice of timescale) and on an analytical > theory of the motion of the Sun. For subsecond precision, > these dependencies become relevant. > > Michael Deckers > > www.calendersign.com Y -669; CEP - 244368 ======================= Anno-Domini-hoax 2006 Since we should state not only the truth, but also the cause of error... Aristotle, Nicomachean Ethics VII 14 ********************************* Why organize the world's timekeeping by religious superstition? The web-net is the best device to catch the ICHTHYS: I.WWW.WWW.WWI======================= I.W <')/I/H/S/)<< W.I I.WWW.WWW.WWI |
In reply to this post by Mikhail Petin
Dear Mikhail,
OK. So what we're really looking at, then, is the plane of the ecliptic. We project Earth's axis down to this plane and view this shadow to see which way the shadow is pointing. Right? Are we viwing the shadow from the north or from the south? Victor > -----Original Message----- > From: East Carolina University Calendar discussion List > [mailto:[hidden email]]On Behalf Of Mikhail Petin > Sent: Monday, February 27, 2006 1:12 PM > To: [hidden email] > Subject: Re: The New World Calendar > > > Dear Victor and Calendar people, > > VICTOR SAYS: > I'm having trouble visualizing this. I have no trouble > with the intersection of the ecliptic with the plane > of Earth's equator. But, if you are on the North Pole, > which way is right? > The answer depends upon which direction you > are looking. Also, if you're on the North Pole, > aren't you also on the "projection N-S line"? > If not, what do you mean by that? > > MIKHAIL SAYS: > It is necessary to project an the Earth.s axis > on the ecliptic plane and do a perpendicular > to N-S line. > > AMOS SAYS: > > BTW, an equinox can be called either Spring (Northwards) > > or Autumn (Southward); I don't know what you mean by "Eastern".. > > > > .. And another point: if you insist on calling anything > > you don't understand or agree with "nonsense" or "delusion", > > you're well on your way to Aristeanism... > > > > MIKHAIL SAYS: > > The Eastern equinox is a point of intersection of the ecliptic > > plane with the plane of a Earth.s equator. > > The Eastern equinox is always to the right of > > the projection N-S line if see from a North Pole of the Earth. > > Best regards > Mikhail Petin > http://WorldCalendarPetin.narod.ru/Bible.htm > http://WorldCalendarPetin.narod.ru/index.htm > http://Petin1Mikhail.narod.ru/index.htm > http://NewWorldCalendar.narod.ru/index.htm > |
In reply to this post by Mikhail Petin
Dear Victor,
If look at the ecliptic plane from a South Pole of the Earth then the directions of revolutions and lines will be contrary (opposed). Best regards Mikhail Petin |
In reply to this post by Mikhail Petin
On 2006-02-27, Mikhail Petin wrote:
> MICHAEL SAYS: > A tropical year is not defined as an average in modern astronomy, > nor does it seem to be an average in any mathematical sense. > > MIKHAIL SAYS: > It is necessary to correct quickly. It is the task of astronomers to define their notions as they need them. For the background and rationale of these definitions you may consult a textbook on spherical astronomy or on celestial mechanics. It does not help communication (except for Humpty Dumpty) if everybody uses their private definitions. > MICHAEL SAYS: > ............................................. Meeus and Savoie give > an easily readable survey at > [http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1992JBAA..102...40M > &db_key=AST&data_type=HTML&format=] > > MIKHAIL SAYS: > I don't see the refined definition of "tropical year". Meeus and Savoie describe in detail how Hipparcos, Copernicus, and Leverrier (among others) determined the tropical year. These methods measure different things, and therefore correspond to different definitions of "tropical year". > MIKHAIL SAYS: > Formula: > (2 pi rad)/( angular speed of the geometric mean longitude > of the Sun with respect to the mean equinox > on the ecliptic of date ) > shows that was measured the time interval between two consecutive > passages of the centre of the Sun through the Eastern equinox point > (t=365,242195 days, i.e. a tropical year, which is equal to > a tropical years in any other points of an ecliptic). You seem to misunderstand. Speed (angular or any other) has nothing to do with a time interval. You can drive 55 miles per hour without driving for an hour. Taking the derivative to determine instantaneous angular speed is described in the article cited above. > After this was defined a MEAN angular speed of the geometric > mean longitude of the Sun with respect to the mean equinox > on the ecliptic of date. No, not a "mean" speed of geometric longitude is meant; the instantaneous speed of mean geometric longitude is meant. (Actually, taking the derivative does commmute with dropping the non-secular terms of a series, but it does not commute with taking an average in whatever sense -- but that is probably too much math for the occasion.) > MIKHAIL SAYS: > The value 365,242195 days can't represent the instantaneous angular speed > of a fictituous point. This is a time. It is a time all right, and it also represents a speed -- in the same way as the fact that a car needs 60 seconds per kilometer can represent an instantaneous speed. Michael Deckers |
In reply to this post by Sepp Rothwangl
On 2006-02-27, Sepp Rothwangl wrote:
> OK! You can determine a year with many different values and to many > aims. > Briefly: Calendrically it is determined by the secular average days > and no more by seconds. > > Astronomically you can determine it by angular expansion and no days > Temporally by atomic seconds or by the beats of a pulsar ... This is not a question of time units (day or second or Julian year) but a question of timescale. If you determine the angular speed of longitude you take the first derivative with respect to a timescale. There are many different timescales in astronomy to choose from, and all are measured and denoted with standard time units (s, d, min, h) and the standard date notations (Gregorian calendar, Julian date, MJD). You probably want to allude to the choice of timescale, not time unit. For the definition of the tropical year, ephemeris time (ET) was used until it was abolished in 1984. The successor of ET was meant to be TDB ("barycentric dynamical time", which also has been abolished in the meantime) and properly is Teph, the timescale used in the numerical integration of modern ephemerides. However, Teph is not a coordinate timescale in the sense of physics, nor is it a proper timescale. The timescales TAI (proper time on the geoid) and TT (a scaled form of a coordinate timescale) agree very well with ET over long ranges, so that both can substitute ET in the definition of the tropical year. An easy introduction into the maze of astronomical timekeeping is in chapter 2 of [aa.usno.navy.mil/kaplan/Circular.pdf]. Of course, for civil time (which is based on UTC), the angular speed should be taken relative to UT1 (universal time, the modern form of mean solar time). Actually, scientists are currently reconsidering the definition of UTC -- the proposal is to make it a proper timescale (TAI minus an offset of some 33 plus seconds). If this is adopted, the target values (eg, synodic month/calendar day) for calendars intended to represent astronomical phenomena would change. Michael Deckers |
In reply to this post by Mikhail Petin
Dear Michael and Calendar People,
> MICHAEL SAYS: > A tropical year is not defined as an average in modern astronomy, > nor does it seem to be an average in any mathematical sense. > > MIKHAIL SAYS: > It is necessary to correct quickly. MICHAEL SAYS: It is the task of astronomers to define their notions as they need them. For the background and rationale of these definitions you may consult a textbook on spherical astronomy or on celestial mechanics. It does not help communication (except for Humpty Dumpty) if everybody uses their private definitions. MIKHAIL SAYS: Thanks. As I am basing solely on a textbook on spherical astronomy or on celestial mechanics I recommend you these textbooks too (without fall - SI System International, 1960) instead of your Humpty Dumpty and your private definitions mentioned below > The value 365,242195 days can't represent the instantaneous > angular speed of a fictituous point. This is a time. MICHAEL SAYS: It is a time all right, and it also represents a speed . in the same way as the fact that a car needs 60 seconds per kilometer can represent an instantaneous speed. MIKHAIL SAYS: This is wrong (Humpty Dumpty). The instantaneous angular speed in given time moment is a function of time t, not a function (t, t+Dt). The instantaneous angular speed in given time moment is an abstraction because the mean angular speed may be measured only, but not the instantaneous angular speed. For information (from a textbook on spherical astronomy or on celestial mechanics): - solar year (tropical) = 365,242190 days - mean duration of the time interval between two consecutive passages of the centre of the Sun, for example, through the Eastern equinox point, Year Duration,ephemeris Change for 100 year, days (1950,0 г.) days Star 365,256360 +0,11.10-6 Тропический 365,242196 - 6,16.10-6 Anomalist 365.259641 + 3,04.10-6 Draconist 346,620047 + 32.10-6 Lunar(12months) 354,3670 - 2,4.10-6 Julian 365,25 . Gregorian 365,2425 . (mean duration) Best regards Mikhail Petin http://WorldCalendarPetin.narod.ru/Bible.htm http://WorldCalendarPetin.narod.ru/index.htm http://Petin1Mikhail.narod.ru/index.htm http://NewWorldCalendar.narod.ru/index.htm |
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