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solar year range
by Irv Bromberg
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Dear Calendar People:
I have added "Longest Solar Year" (grey) and "Shortest Solar Year" (black) curves to my equinoctial and solstitial year length plot, attached as a PDF (36 KB). These curves are almost tangent to the helical curves that they contain, a surprisingly good fit despite the fact that the helical curves are based on numerical integration (SOLEX 9.1) whereas the solar year range curves are based on astronomical algorithms (Meeus' mean aphelion polynomial to find the target solar longitude of aphelion and perihelion, and the Reingold/Dershowitz/Meeus truncated VSOP87 for the solar longitude). I am especially surprised to see no evidence that the polynomial goes "wild" within this relatively wide range of years. The longest solar year is calculated as follows: For each 200-year interval, find the ecliptic longitude of mean perihelion. Find the moment when Sun was/will be at same ecliptic longitude (not the longitude of perihelion) 100 years before and after, and calculate the average year length by subtracting them and dividing by 200 years. Express the year length as the time in excess of 365 days 5 hours by subtracting 365 days 5 hours. This generates an array of points follow along the top edge of the helical curves with some scatter, so I fit a 3rd order polynomial (least squares regression) to that, which is the grey plotted trendline, then I hid the scattered points to get them out of the way. I found that higher order polynomials didn't improve the fit. The shortest solar year is calculated the same way, but substitute aphelion for perihelion, and fit a 3rd order polynomial to the array of points following along the bottom edge of the helical curves, and assigned it a black color. Stable calendar seasons can only exist between the longest and shortest solar year curves as appropriate to the target era. Note that these year lengths are very different from the mean anomalistic year, the time required for Earth to revolve from perihelion to perihelion or from aphelion to aphelion, which is presently about 365 days 6 hours 13 minutes and 52 seconds, slightly longer than the mean sidereal year because of the advance of perihelion and aphelion. It would be nice at some point (when I figure out how) to redo this chart using SOLEX 10.1, in "planets only" numerical integration mode to generate the data for plotting all of the curves according to the Earth-Moon barycenter, taking advantage of several of the new SOLEX features. (Note that SOLEX doesn't numerically integrate the Earth axial tilt = obliquity of the ecliptic, but simply uses Williams' polynomial for ±10000 years from the present era.) -- Irv Bromberg, Toronto, Canada |
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Re: solar year range
by Tom Peters-6
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Re: solar year range
by Irv Bromberg
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Symmetry Statement and Divide-by-Six RE: using Primary cycles of 11,15,19&33-years RE:
by Karl Palmen
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Dear Brij, Irv, Tom and Calendar People Brij said (quoting me in bold type): >If a leap cycle is arranged such that the list
of leap years is symmetrical, so that year n of each cycle has the same
leap status as >the symmetrical year occurring n years before the
first year of the next cycle, then the start of the first year of every cycle
will >always be at the average for that cycle. Brij refers to his idea of having a leap week on each year whose
number is divisible by 6 plus some additional years referred to as Kepler’s
Leap Week years. No such cycle can have the symmetry to which I refer to in the
statement I made and Brij quoted and so the statement does not apply to any
such cycle (i.e. the first year of such a cycle need not have an average start). However the 834-year cycle may have one or two years that do
have an average start, but they are not easy to find. The 896-year cycle, has no year with an average start no matter
how the 159 leap weeks are arranged, because each year has a start that is an
odd multiple of 1/256 days from average. The same applies to the 400-year cycle
with 71 leap weeks, because each year has a start that is an odd multiple of
1/800 days from average. Karl 10(07(28 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Irv,
Tom Peters, Karl & CC: (MJD 2454944)/1361+D-122W17-02 (G. Tuesday, 2009 April 21H16:49
(decimal) EST
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Re: solar year range
by Irv Bromberg
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Re: Symmetry Statement and Divide-by-Six RE: using Primary cycles of 11,15,19&33-years RE:
by Karl Palmen
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Dear Brij, Irv and Calendar People What I says applies to Irv’s statement too. I thought there might be one or two years in the 834-year divide-by-six
cycle that have an average start, but this is not the case. I show this next. The 834 -year cycles has every year whose number is divisible by
six has a leap week (139 years) along with nine Kepler years 87, 183,
273, 369, 459, 555, 645, 741, 831, whose intervals alternate between 90 and 96
years, except for year 831, which is 90 years from both its neighbours. The
cycle is symmetrical about year 831, but this symmetry does not allow any year
to be the year after its mirror image so be the first year of a symmetrical
cycle to which the statement applies. Because the cycle is symmetrical about year 831, its middle
coincides with the middle of the mean years and so its start is 3.5*(53 -
(52 + 148/834)) = 2401/834 days earlier than average. Now we can work out how
early or late other years start compared with average. 831: 2401/834 days early 832: 2401/834 days late 833: 1365/834 days late 834: 329/834 days late 001: 5131/834 days late 002: 4095/834 days late etc.. Because steps are of 4802/834 for a 53-week year or 1036/834,
every year must start an odd multiple of 1/834 days from average and so no year
starts exactly on average. I now realise that if the 148 leap years were arranged in two
identical cycles of 417 years and 79 leap years, then an odd number of moves
would be needed to change this into the 834-year cycle described, where one
move is a shift of one leap week by one year. Karl 10(07(30 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
sir: Date: Thu, 23 Apr 2009 08:57:51 +0100 Dear Brij, Irv, Tom and Calendar People Brij said (quoting me in bold type): >If a leap cycle is arranged such that the list of leap years is
symmetrical, so that year n of each cycle has the same leap status as
>the symmetrical year occurring n years before the first year of the
next cycle, then the start of the first year of every cycle will >always be
at the average for that cycle. Brij refers to his idea of having a leap week on each year whose
number is divisible by 6 plus some additional years referred to as
Kepler’s Leap Week years. No such cycle can have the symmetry to which I
refer to in the statement I made and Brij quoted and so the statement does not
apply to any such cycle (i.e. the first year of such a cycle need not have an
average start). However the 834-year cycle may have one or two years that do
have an average start, but they are not easy to find. The 896-year cycle, has no year with an average start no matter
how the 159 leap weeks are arranged, because each year has a start that is an
odd multiple of 1/256 days from average. The same applies to the 400-year cycle
with 71 leap weeks, because each year has a start that is an odd multiple of
1/800 days from average. Karl 10(07(28 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij Bhushan
Vij Irv,
Tom Peters, Karl & CC: (MJD 2454944)/1361+D-122W17-02 (G. Tuesday, 2009 April 21H16:49
(decimal) EST
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Re: 400-yrs,834-yrs & 896-yrs RE: Symmetry Statement and Divide-by-Six RE:
by Karl Palmen
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Dear Brij and Calendar People The Kepler’s leap week years of the 834-year cycle can be
arranged symmetrically thus: 45th, 141st, 231st, 327th
417th, 507th, 603rd, 693rd, 789th
years. The intervals between the successive Kepler's leap weeks are either 90
or 96 years, including the interval over the end of the cycle. This is much better than Brij’s suggestion of 81st, 165th, 249th, 333rd,
417th, 501st, 855th, 669th 753rd
years, which suffers from a huge gap of 162 years over the end of the
cycle. Even the simple 57th, 147th, 237th, 327th,
417th, 507th, 597th, 687th, 777th
is better with a gap of 114 years. Brij seems to be unaware of the importance of having the all the
intervals between Kepler's leap weeks with either 90 or 96 years to avoid
excessive jitter. Also the intervals of 90 and 96 years need to be roughly
equal in proportion with slightly more 90s than 96s. The cycles listed although symmetrical do not have the same
symmetry in the symmetry statement, which requires the first year to be of the
same type as the last year. In this case year 1 has no leap week, but year 834
does has a leap week. So the statement that Irv and I have made does not apply
to them. What the symmetry of the 834-year cycle does tell you is that
the middle of year 417 and 834 is time at the average of middles
of all years. About the 895-year cycle, Brij stated
I
hope this clarify my approach, to place Kepler's Leap Weeks in 400-year,
834-year & 896-years cycle. Multiples of these cycles could be used, in
necessary to achieve better placing/results. The 896-year cycle as stated is here erroneous. If he were to
add 10 Kepler’s leap week years like this every 896-year cycle, then in
three 896-year cycles, he have 3*896/6 = 448 leap weeks in years divisible by
six and 30 Kepler’s leap weeks adding to 478 in total. But three 896-year
cycle require just 3*159=477 leap weeks and so such a 896-year cycle would be
one week out every 3 cycles of 2688 years. Perhaps, Brij intends to omit
one of these 10 Kepler’s leap weeks once every third 896-year cycle. This
would correct the number of leap weeks, but produce extra jitter by leaving a
gap of 180 or 186 years between two of the Kepler’s leap weeks. Instead one can use a 2688-year cycle (equal to three 896-year
cycles) made up of two 834-year cycles and one 1020-year cycle. The Kepler’s
leap years can be as stated above on the 45th, 141st, 231st,
327th 417th, 507th, 603rd, 693rd
and 789th year of each 834-year cycle and also the 879th
and 975th years of each 1020-year cycle. This gives rise to
9+9+11=29 Kepler’s leap weeks in a whole 2688-year cycle, which added to
the 448 divide-by-six leap weeks gives the required 477=3*159 leap weeks. The
1020-year cycle cannot be made symmetrical in the same way as the 834-year
cycle, because the number of years is not twice an odd number. The same
applies to the entire 2688-year cycle. However, the 462-year cycle of 5 Kepler’s leap weeks and
same mean year as 33-year cycle can have this symmetry with Kepler’s leap
weeks on the 45th, 141st, 231st, 321st
and 417th year of each 462-year cycle. Karl 10(08(03 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
Irv & CC:
Distribution
shown now in http://www.brijvij.com/bb_harappaTithi-Cycles.pdf
is aslo (Y1920+417 =Y2337) as the mid year for symmetry.
I
hope this clarify my approach, to place Keplers' Leap Weeks in 400-year,
834-year & 896-years cycle. Multiples of these cycles could be used, in
necessary to achieve better placing/results. Date: Fri, 24 Apr 2009 14:38:15 +0100 Dear Brij, Irv and Calendar People What I says applies to Irv’s statement too. I thought there might be one or two years in the 834-year
divide-by-six cycle that have an average start, but this is not the case. I
show this next. The 834 -year cycles has every year whose number is divisible by
six has a leap week (139 years) along with nine Kepler years 87, 183,
273, 369, 459, 555, 645, 741, 831, whose intervals alternate between 90 and 96
years, except for year 831, which is 90 years from both its neighbours. The
cycle is symmetrical about year 831, but this symmetry does not allow any year
to be the year after its mirror image so be the first year of a symmetrical
cycle to which the statement applies. Because the cycle is symmetrical about year 831, its middle
coincides with the middle of the mean years and so its start is 3.5*(53 -
(52 + 148/834)) = 2401/834 days earlier than average. Now we can work out how
early or late other years start compared with average. 831: 2401/834 days early 832: 2401/834 days late 833: 1365/834 days late 834: 329/834 days late 001: 5131/834 days late 002: 4095/834 days late etc.. Because steps are of 4802/834 for a 53-week year or 1036/834,
every year must start an odd multiple of 1/834 days from average and so no year
starts exactly on average. I now realise that if the 148 leap years were arranged in two
identical cycles of 417 years and 79 leap years, then an odd number of moves
would be needed to change this into the 834-year cycle described, where one
move is a shift of one leap week by one year. Karl 10(07(30 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
sir: Date: Thu, 23 Apr 2009 08:57:51 +0100 Dear Brij, Irv, Tom and Calendar People Brij said (quoting me in bold type): >If a leap cycle is arranged such that the list of leap years is
symmetrical, so that year n of each cycle has the same leap status as
>the symmetrical year occurring n years before the first year of the
next cycle, then the start of the first year of every cycle will >always be
at the average for that cycle. Brij refers to his idea of having a leap week on each year whose
number is divisible by 6 plus some additional years referred to as
Kepler’s Leap Week years. No such cycle can have the symmetry to which I
refer to in the statement I made and Brij quoted and so the statement does not
apply to any such cycle (i.e. the first year of such a cycle need not have an
average start). However the 834-year cycle may have one or two years that do
have an average start, but they are not easy to find. The 896-year cycle, has no year with an average start no matter
how the 159 leap weeks are arranged, because each year has a start that is an
odd multiple of 1/256 days from average. The same applies to the 400-year cycle
with 71 leap weeks, because each year has a start that is an odd multiple of
1/800 days from average. Karl 10(07(28 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Irv,
Tom Peters, Karl & CC: (MJD 2454944)/1361+D-122W17-02 (G. Tuesday, 2009 April 21H16:49
(decimal) EST
Windows
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it out.
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Re: Never Before RE: 400-yrs,834-yrs & 896-yrs RE: Symmetry Statement and Divide-by-Six RE:
by Karl Palmen
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Dear Brij and Calendar People From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
sir & CC: >….. The
intervals between the successive Kepler's leap weeks are either 90 or 96 years,
including >the interval over the end
of the cycle….. Placing of Keplers’ Leap
Weeks is left best to judgments of ‘astronomers, mathematicians &
software engineers’ like yourself. I have tried to be as symmetrical as I
could possibly be. My point of suggesting *CENTRAL YEAR* (Y2000-80)+417 =Y2337,
in separation of 84-years all through; as 0081, 0165, 0249, 0333, 0417,
0501, 0585, 0669, 0753.... that I had been shown long ago. Does
extending the gaps to 90 & 96-yr intervals present EXTRA benefits. Yes. Less jitter. Each date
varies no more than 2 weeks with respect to the seasons (rather than almost 3
weeks for Brij’s suggestion). Note that ALL the gaps including the gap
across the end of the cycle are 90 or 96 years usually alternating. It is the cyclic
repetition of 834-year cycle that I considered i.e.
in cycle (1920 thro 2754), 1st KLWk to be at Y2001 (2004 being
divisible by six) and then every 84-years to have Keplers’ Leap Week as
shown. Why do we want to count distance between TWO KLWks (unless there is a
specific advantage), while 1st cycle ends at Y2754th year.
ERA start at YEAR ZERO could be formed, as shown at: http://www.brijvij.com/bbv_417-year-div.6.pdf 0000, 0834, 1668, 2502, 3336,
4170, 5004…..and repeating. Having the Kepler’s
leap weeks once every 84 years is equivalent to having a Julian
Calendar mean year of 365.25 days. In the suggested 834-year
cycle its like running the Julian calendar for 834 years before correcting it. It has a huge gap of
162 years between year 0753 and year 834+81=0915. Brij’s suggestion
would have the dates moving back and forth almost three weeks with respect to
the seasons rather than two weeks that is possible by a divide-by-six rule. I feel, as shown, central year
417th i.e. *Y0417th or Y2337th is of
significance* with repetition at 834-year cyclic interval. This shall be more
conducive to software designers! I am not allergic to placing
extra NINE Keplers’ LWks as Karl has shown: 45th,
141st, 231st, 327th 417th, 507th,
603rd, 693rd, 789th years, BUT why
‘wider & alternate placing’ 96, 90 years interval and call them
symmetrical? It is the start that has been delayed/enhanced by 45 years. I see
this as: 45-96-90-96-90-90-96-90-96-45 =834-years; whereas I place KLWk every
84-years, starting 3-yaers earlier for 1st KLWk i.e. Y0081 or Y2001. Brij must pay
attention to the interval between the last Kepler’s leap week of one
cycle and the first Kepler’s leap week of the next cycle and so the two
45s add together to form an interval of 90 years. If the middle year (417) is a
Kepler’s leap week year and there are an odd number (9) of Kepler’s
leap weeks, then the last year (834) must occur half way between two
Kepler’s leap week years. This is possible for an interval of 90 (but not
96) making the first Kepler’s leap week year the 45th year. The cycle is symmetrical
thus: 045+789=834 141+693=834 231+603=834 327+507=834 417+417=834. Brij’s suggestion is
also symmetrical, but the two 81s in his 81-84-84-84-84-84-84-84-84-81 add to
form a huge interval of 162 years between Kepler’s leap weeks. > “Even
the simple 57th, 147th, 237th, 327th,
417th, 507th, 597th, 687th, 777th
is better with a gap of 114 >years” This is 90-years from
417th…..shifts the start of next cycle from 57th to 33rd
year [(777+90=(867 - 834) =33] for next cycle.. I therefore leave THIS for
software designers to choose, so long as there are 9 KLWks between 834-years
& repeating every 834-years. I see that Brij did pay
attention to this interval across the end of the cycle! If this interval were 90
years, then the next Kepler’s year would be the 33rd year of
the next cycle, but it is a 114-year (twice 57) interval to make it into a
symmetrical 834-year cycle. For a symmetrical cycle the interval over the end
is always twice the first Kepler’s year. That why the 45th
year is a good choice. If all intervals were 90
years we’d have a mean year of 365.2444444444 days. >
Perhaps, Brij intends to omit one of these 10 Kepler’s leap weeks once
every third 896-year cycle. >This would correct the
number of leap weeks, but produce extra jitter by leaving a gap of 180 or 186 >years between two of the
Kepler’s leap weeks. YES sir, this was an exercise
that I undertook to see if I could achieve cyclic placing of 896-year cycles
around 448th year (CENTRAL) to fulfill: Mean Year
=7*(52+159/896), since there are EXACTLY 159 LWks in 896-year cycle. I agree,
as earlier discussed 3*896 =2688-years make better sense, to get MY
=7*[(52+1/6+29/2688)] days - since 896 is NOT divisible by six BUT 10 KLWks are
needed every 896-year cycle. Wrong! I’ve shown
this many times before! BASICALLY, the point I make is
that DIVIDE six(6) plan that had NEVER been attempted has a solution;
additionally resolving the ZERO YEAR riddle. Likewise, the 400-year
cycle, using div. six(6) for LWks has Mean Year =7*[52+1/6+13/1200)] =365.2524
days. The 400-year cycle does not
work with divide-by-six for the same reason that 896-year cycle does not work.
The number of years must be divisible by six. The 400-year becomes a
1200-year cycle with 71*3-200=13 Kepler’s leap weeks. Because 1200 is not
twice an odd number, the Kepler’s leap weeks cannot be arranged symmetrically
like the 834-year cycle. Making the first Kepler’s leap week year the 45th
year and using intervals of 96 and 90 years, we can get an almost symmetrical
cycle of 45th, 141st
231st, 321st, 417th, 507th,
597th, 693rd, 783rd, 879th,
969th, 1059th and 1155th years. This
cycle would be symmetrical if the 597th year were moved to the 600th
year. It has eight intervals of 90 years (including the interval between the
1155th year and the 1200+45th year) and five intervals of
96 years. Karl 10(08(04 Regards, (MJD 2454950)/1361+D-128W18-01 (G. Monday, 2009 April 27H15:36
(decimal) EST Date: Mon, 27 Apr 2009 12:42:29 +0100 Dear Brij and Calendar People The Kepler’s leap week years of the 834-year cycle can be
arranged symmetrically thus: 45th, 141st, 231st, 327th
417th, 507th, 603rd, 693rd, 789th
years. The intervals between the successive Kepler's leap weeks are either 90
or 96 years, including the interval over the end of the cycle. This is much better than Brij’s suggestion of 81st, 165th, 249th, 333rd,
417th, 501st, 855th, 669th 753rd
years, which suffers from a huge gap of 162 years over the end of the
cycle. Even the simple 57th, 147th, 237th, 327th,
417th, 507th, 597th, 687th, 777th
is better with a gap of 114 years. Brij seems to be unaware of the importance of having the all the
intervals between Kepler's leap weeks with either 90 or 96 years to avoid
excessive jitter. Also the intervals of 90 and 96 years need to be roughly
equal in proportion with slightly more 90s than 96s. The cycles listed although symmetrical do not have the same symmetry
in the symmetry statement, which requires the first year to be of the same type
as the last year. In this case year 1 has no leap week, but year 834 does has a
leap week. So the statement that Irv and I have made does not apply to them. What the symmetry of the 834-year cycle does tell you is that
the middle of year 417 and 834 is time at the average of middles
of all years. About the 895-year cycle, Brij stated
I hope this clarify my approach, to place
Kepler's Leap Weeks in 400-year, 834-year & 896-years cycle. Multiples of
these cycles could be used, in necessary to achieve better placing/results. The 896-year cycle as stated is here erroneous. If he were to
add 10 Kepler’s leap week years like this every 896-year cycle, then in three
896-year cycles, he have 3*896/6 = 448 leap weeks in years divisible by six and
30 Kepler’s leap weeks adding to 478 in total. But three 896-year cycle
require just 3*159=477 leap weeks and so such a 896-year cycle would be one
week out every 3 cycles of 2688 years. Perhaps, Brij intends to omit one
of these 10 Kepler’s leap weeks once every third 896-year cycle. This
would correct the number of leap weeks, but produce extra jitter by leaving a
gap of 180 or 186 years between two of the Kepler’s leap weeks. Instead one can use a 2688-year cycle (equal to three 896-year
cycles) made up of two 834-year cycles and one 1020-year cycle. The
Kepler’s leap years can be as stated above on the 45th, 141st,
231st, 327th 417th, 507th, 603rd,
693rd and 789th year of each 834-year cycle and also the
879th and 975th years of each 1020-year cycle. This gives
rise to 9+9+11=29 Kepler’s leap weeks in a whole 2688-year cycle, which
added to the 448 divide-by-six leap weeks gives the required 477=3*159 leap
weeks. The 1020-year cycle cannot be made symmetrical in the same way as the
834-year cycle, because the number of years is not twice an odd number.
The same applies to the entire 2688-year cycle. However, the 462-year cycle of 5 Kepler’s leap weeks and
same mean year as 33-year cycle can have this symmetry with Kepler’s leap
weeks on the 45th, 141st, 231st, 321st
and 417th year of each 462-year cycle. Karl 10(08(03 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
Irv & CC:
Distribution
shown now in http://www.brijvij.com/bb_harappaTithi-Cycles.pdf
is aslo (Y1920+417 =Y2337) as the mid year for symmetry.
I
hope this clarify my approach, to place Keplers' Leap Weeks in 400-year,
834-year & 896-years cycle. Multiples of these cycles could be used, in
necessary to achieve better placing/results. Date: Fri, 24 Apr 2009 14:38:15 +0100 Dear Brij, Irv and Calendar People What I says applies to Irv’s statement too. I thought there might be one or two years in the 834-year
divide-by-six cycle that have an average start, but this is not the case. I
show this next. The 834 -year cycles has every year whose number is divisible by
six has a leap week (139 years) along with nine Kepler years 87, 183,
273, 369, 459, 555, 645, 741, 831, whose intervals alternate between 90 and 96
years, except for year 831, which is 90 years from both its neighbours. The
cycle is symmetrical about year 831, but this symmetry does not allow any year
to be the year after its mirror image so be the first year of a symmetrical
cycle to which the statement applies. Because the cycle is symmetrical about year 831, its middle
coincides with the middle of the mean years and so its start is 3.5*(53 -
(52 + 148/834)) = 2401/834 days earlier than average. Now we can work out how
early or late other years start compared with average. 831: 2401/834 days early 832: 2401/834 days late 833: 1365/834 days late 834: 329/834 days late 001: 5131/834 days late 002: 4095/834 days late etc.. Because steps are of 4802/834 for a 53-week year or 1036/834, every
year must start an odd multiple of 1/834 days from average and so no year
starts exactly on average. I now realise that if the 148 leap years were arranged in two
identical cycles of 417 years and 79 leap years, then an odd number of moves
would be needed to change this into the 834-year cycle described, where one
move is a shift of one leap week by one year. Karl 10(07(30 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Karl,
sir: Date: Thu, 23 Apr 2009 08:57:51 +0100 Dear Brij, Irv, Tom and Calendar People Brij said (quoting me in bold type): >If a leap cycle is arranged such that the list of leap years is
symmetrical, so that year n of each cycle has the same leap status as
>the symmetrical year occurring n years before the first year of the
next cycle, then the start of the first year of every cycle will >always be
at the average for that cycle. Brij refers to his idea of having a leap week on each year whose
number is divisible by 6 plus some additional years referred to as
Kepler’s Leap Week years. No such cycle can have the symmetry to which I
refer to in the statement I made and Brij quoted and so the statement does not
apply to any such cycle (i.e. the first year of such a cycle need not have an
average start). However the 834-year cycle may have one or two years that do
have an average start, but they are not easy to find. The 896-year cycle, has no year with an average start no matter
how the 159 leap weeks are arranged, because each year has a start that is an
odd multiple of 1/256 days from average. The same applies to the 400-year cycle
with 71 leap weeks, because each year has a start that is an odd multiple of
1/800 days from average. Karl 10(07(28 From: East Carolina University Calendar
discussion List [mailto:CALNDR-L@...] On Behalf Of Brij
Bhushan Vij Irv,
Tom Peters, Karl & CC: (MJD 2454944)/1361+D-122W17-02 (G. Tuesday, 2009 April 21H16:49
(decimal) EST
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Re: solar year range
by Irv Bromberg
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