Do the longest periods of 10 Hebrew years always begin on Monday?
YES!
The reason why the longest periods of 10 Hebrew years always begin on Monday will be given in the answer to the next Weekly Question.
Do the shortest periods of 10 Hebrew years always end on Monday?
YES!
The smallest number of Hebrew years that can have the maximum 34 day difference between its shortest and longest durations is 10 years.
The shortest period of 10 Hebrew years is 3,630 days and occurs 3,237 times in the full Hebrew calendar cycle of 689,472 years.
The longest period of 10 Hebrew years is 3,664 days and occurs 9,473 times in the full Hebrew calendar cycle of 689,472 years.
The smallest number of Hebrew years that can have the
maximum 34 day difference between its
shortest and longest durations is 10 years.
Therefore, for any period of H Hebrew years, the maximum
possible difference of days between its shortest and longest durations
is 34 days.
The shortest period of 10 Hebrew years is 3,630 days
and occurs 3,237 times in the full Hebrew calendar cycle of
689,472 years.
The longest period of 10 Hebrew years is 3,664 days
and occurs 9,473 times in the full Hebrew calendar cycle of
689,472 years.
Let D~ = the start day of the shortest period
Let D' = the start day of that period 10 years after
Then (D' - D~) Mod 7d = (A - 2) MOD 7d
Since the shortest period takes place when D~ is after a 2 day postponement
D~ MOD 7d = 5d.
Hence, A MOD 7d = (D' - 5d + 2d)
A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d
= {4d, 6d, 0d, 2d}
Since the shortest period takes place when D~ is after a 2 day postponement
D~ MOD 7d = 5d.
Hence, A MOD 7d = (D' - 5d + 2d) MOD 7d
A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d
= {4d, 6d, 0d, 2d}
Now,
Let D~ = the start day of the longest period
Let D' = the start day of that period 10 years after
Then (D' - D~) Mod 7d = (A + 34d) MOD 7d
Since the longest period takes place when D' is after a 2 day postponement
D' MOD 7d = 5d.
A MOD 7d = (5d' - D~ - 34d) MOD 7d
A MOD 7d = (1d - {0d , 2d, 3d, 5d} ) MOD 7d
= {1d, 6d, 5d, 3d}
For A to satisfy both the minimum and the maximum cases it is therefore
necessary that A MOD 7d = 6d, since that is the only value common
to both cases.
For the shortest period of 10 years
(D' - D~) Mod 7d = (A - 2d) MOD 7d
D' Mod 7d = (A - 2d + D~) MOD 7d
= (6d - 2d + 5d) MOD 7d
= 2d
Similarly for the longest period of 10 years
(D' - D~) Mod 7d = (A + 34d) MOD 7d
D~ Mod 7d = (D' - A + 6d ) MOD 7d
= (5d - 2d + 6d) MOD 7d
= 2d
Therefore, the shortest periods of 10 years always end on Monday
while, the longest periods of 10 years always begin on Monday.
For some given period of H Hebrew years
Let its lunar length be A + a
where A is the integral number of days in that length
and a is the outstanding fraction of a day in that length
Then Ls, the shortest possible INTEGRAL length of that
period, is A - 2 days.
If the H years are not a multiple of 19 years, then
the alternate length includes an additional month and the
longer lunar length is A + a + 29d 12h 793p.
Then Lm, the maximum possible INTEGRAL length of that
longer period, is A + 29 + 3 days = A + 32 days.
The difference between the maximum length and the shortest possible
length is Lm - Ls = A + 32 - (A - 2) = 34 days.
The above conclusion holds for any period of Hebrew years that
varies 34 days between its shortest and longest periods!
The Weekly Question will resume after Pesach.
YES!
The smallest number of Hebrew years that can have the maximum 34 day difference between its shortest and longest durations is 10 years.
The shortest period of 10 Hebrew years is 3,630 days and occurs 3,237 times in the full Hebrew calendar cycle of 689,472 years.
The longest period of 10 Hebrew years is 3,664 days and occurs 9,473 times in the full Hebrew calendar cycle of 689,472 years.
The smallest number of Hebrew years that can have the
maximum 34 day difference between its
shortest and longest durations is 10 years.
Therefore, for any period of H Hebrew years, the maximum
possible difference of days between its shortest and longest durations
is 34 days.
The shortest period of 10 Hebrew years is 3,630 days
and occurs 3,237 times in the full Hebrew calendar cycle of
689,472 years.
The longest period of 10 Hebrew years is 3,664 days
and occurs 9,473 times in the full Hebrew calendar cycle of
689,472 years.
Let D~ = the start day of the shortest period
Let D' = the start day of that period 10 years after
Then (D' - D~) Mod 7d = (A - 2) MOD 7d
Since the shortest period takes place when D~ is after a 2 day postponement
D~ MOD 7d = 5d.
Hence, A MOD 7d = (D' - 5d + 2d)
A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d
= {4d, 6d, 0d, 2d}
Since the shortest period takes place when D~ is after a 2 day postponement
D~ MOD 7d = 5d.
Hence, A MOD 7d = (D' - 5d + 2d) MOD 7d
A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d
= {4d, 6d, 0d, 2d}
Now,
Let D~ = the start day of the longest period
Let D' = the start day of that period 10 years after
Then (D' - D~) Mod 7d = (A + 34d) MOD 7d
Since the longest period takes place when D' is after a 2 day postponement
D' MOD 7d = 5d.
A MOD 7d = (5d' - D~ - 34d) MOD 7d
A MOD 7d = (1d - {0d , 2d, 3d, 5d} ) MOD 7d
= {1d, 6d, 5d, 3d}
For A to satisfy both the minimum and the maximum cases it is therefore
necessary that A MOD 7d = 6d, since that is the only value common
to both cases.
For the shortest period of 10 years
(D' - D~) Mod 7d = (A - 2d) MOD 7d
D' Mod 7d = (A - 2d + D~) MOD 7d
= (6d - 2d + 5d) MOD 7d
= 2d
Similarly for the longest period of 10 years
(D' - D~) Mod 7d = (A + 34d) MOD 7d
D~ Mod 7d = (D' - A + 6d ) MOD 7d
= (5d - 2d + 6d) MOD 7d
= 2d
Therefore, the shortest periods of 10 years always end on Monday
while, the longest periods of 10 years always begin on Monday.
For some given period of H Hebrew years
Let its lunar length be A + a
where A is the integral number of days in that length
and a is the outstanding fraction of a day in that length
Then Ls, the shortest possible INTEGRAL length of that
period, is A - 2 days.
If the H years are not a multiple of 19 years, then
the alternate length includes an additional month and the
longer lunar length is A + a + 29d 12h 793p.
Then Lm, the maximum possible INTEGRAL length of that
longer period, is A + 29 + 3 days = A + 32 days.
The difference between the maximum length and the shortest possible
length is Lm - Ls = A + 32 - (A - 2) = 34 days.
The above conclusion holds for any period of Hebrew years that
varies 34 days between its shortest and longest periods!
In the foreseeable future, will there ever be a Gregorian year without a Rosh Hashannah?
YES! Although that future will be quite distant!
Assuming that no changes take place to either the Gregorian or the Hebrew calendars, then Rosh Hashannah will NOT be observed during the year 22,335g.
This will be the first time that a Gregorian year will not host such an important holiday.
Rosh Hashannah 26,095H will fall on
Thu 13 Dec 22,334g while the immediately following
Rosh Hashannah 26,096H will
occur Thu 2 Jan 22,336g.
The topic on The Calendar Drift found in the Additional Notes fully explains why the 1st day of Tishrei eventually lands in the Gregorian month of December.
Correspondent Larry Padwa sent the following response:
I don't have calendars that go that far into the future, but I can say that the situation you describe will definitely happen, and approximately when, as follows:Currently, the latest date for Rosh Hashannah is approximately 5-Oct.
The drift is roughly four days per thousand years, so it should take about 17,000 years for Rosh Hashannah to begin on or about 12-Dec.
If the Rosh Hashannah of a leap year begins between 12-Dec and 31-Dec, then the following Gregorian year will not have a Rosh Hashannah.
Thus, in about 17,000 years there should be a Gregorian year without a Rosh Hashannah.
Without calendars, (or running of a calendar program) I cannot be more precise than that, but I think that my guess is a fair approximation.
Thank you Larry Padwa for sharing an excellent insight into the dynamics which lead to this particular phenomenon of the Hebrew calendar.
In eliminating the 356 day year, what unforeseen problem would have arisen if the period of the molad had been 29d 13h 1p?
IF the molad period had been 29d 13h 1p then the 12 month year would have been 354d 12h 12p.
The 356 day year could then have been eliminated by postponing Rosh Hashannah to Thursday (5d) whenever the molad of Tishrei of a 12 month year fell on Tuesday (3d) past 11h 1067p.
This postponement would then cause the year to be 354d.
However, suppose that in this scenario, the molad of Tishrei of a 12 month year occurred on Thursday (5d) at 23h 1056p.
Then 5d 23h 1056p + 354d 12h 12p = 360d 11h 1068p
Now (360d 11h 1068p) MOD 7d = 3d 11h 1068p
Since, this time of the molad is Tuesday (3d) past 11h 1067p
it is necessary to postpone Rosh Hashannah to Thursday (5d)
thereby adding 2 days to the year.
Consequently, the final sum of days is 362d.
Since the year began at 5d, the length of the year is
362d - 5d = 357 days.
Therefore, if the period of the molad had been 29d 13h 1p instead of 29d 12h 793p, the unexpected problem arising from the elimination of the 356 day year would have been the creation of a 357 day year.
It appears that all of the literature that deals with the fixed Hebrew calendar's method for eliminating the 356 day year gives the impression that no other Hebrew year length is created in that process.
And it appears that noone has actually demonstrated, other than by computer simulation, that this is the case.
Why is the 357 day Hebrew year impossible in the fixed Hebrew calendar?
In the fixed Hebrew calendar,
absent Dehiyyah Molad Zakein,
the 356 day year is eliminated by postponing
Rosh Hashannah to Thursday (5d) whenever the
molad of Tishrei of a
12 month year falls on Tuesday (3d) past 15h 203p.
This postponement then causes the year to be 354d.
However, suppose that in this scenario, the molad of Tishrei of a 12 month year occurred on Thursday (5d) at 23h 1079p.
Then 5d 23h 1079p + 354d 8h 876p = 360d 8h 875p
Now (360d 8h 875p) MOD 7d = 3d 8h 875p.
Since this is the maximum possible time for a 12 month year ending on Tuesday, and does not trigger a postponement rule, the largest 12 month year that can be generated in the fixed Hebrew calendar is 360d - 5d = 355d.
Therefore, it is impossible to create a 357 day year in the fixed Hebrew calendar.
Correspondent Ram Sinclair asked this week's intriguing question which concerns the disappearing days Heshvan 30 and February 29.
Can Heshvan 30 ever coincide with February 29?
YES!
In the fixed Hebrew calendar, Heshvan 30 occurs only when the year is 355 or 385 days long.
In the Gregorian calendar, February 29 occurs only when the Gregorian year is a leap year.
That simple observation made correspondent Ram Sinclair wonder whether or not Heshvan 30 could ever coincide with February 29.
Ram Sinclair then researched the matter.
He correctly deduced that Heshvan 30 could coincide February 29 provided that Rosh Hashannah mapped onto January 1 of a Grgorian leap year.
With that in mind, he noted from the topic
The Rosh Hashannah Drift
that his search
would have to start after Rosh Hashannah 25963H
(Sat 1 Jan 22203g), since that represents the first
coincidence of these dates.
Assuming that the Hebrew and Gregorian calendar rules remain unchanged, Ram Sinclair discovered that 30 Heshvan 28184H would coincide with 29 February 24424g and also that an earlier 30 Heshvan 28032H would coincide with 29 February 242724g.
Thank you Ram Sinclair, not only for asking this very interesting question, but also for sharing with us the results of your excellent research.
Correspondent Ram Sinclair's second discovery was an earlier coincidence of Heshvan 30 and February 29. But was this the earliest possible such coincidence?
When will be the earliest coincidence of Heshvan 30 and February 29?
Once again, correspondent Ram Sinclair provided the correct answer.
I am almost sure that Feb 29, 23680g is the first 30 Heshvan.I realized that in such a year, Rosh Hashannah is on Jan 1.
So I looked in the table in your site and found that 22203g is the first occurrence of Jan 1 on Rosh Hashannah.
So I tried every 4th year starting with 22204g (skipping multiples of 100 which are not multiples of 400).
It was interesting because there were many close calls on the way, like Kislev 1 when there were only 29 days in Heshvan.
Thank you Ram Sinclair, not only for asking this very interesting question, but also for sharing with us the results of your excellent research.
Still on the idea of disappearing dates, correspondent Dennis Kluk
discovered that
Thu 30 Adar 5585H coincided with
29 February 1824g.
Correspondent Dennis Kluk did not know if there were any other coincidences of these dates.
So Weekly Question 127 will deal with this issue.
When was the most recent coincidence of Adar 30 and February 29?
Once again, correspondent Ram Sinclair provided the correct answer.
According to my research 1824g is the most recent occurrence of 29 Feb and 30 Adar.Send my appreciation to Dennis Kluk. To my understanding 30 Adar is less frequent than 30 Heshvan which makes his challenge tougher to meet.
Thank you Ram Sinclair, for once again sharing with us the results of your excellent research.
And also, thank you correspondent Dennis Kluk for passing on your thought provoking question.
Of course, if the most recent coincidence of Thu 30 Adar 5585H was 29 February 1824g then it would be interesting to know how soon the next coincidence will be.
So Weekly Question 128 will deal with this issue.
When will be the next coincidence of Adar 30 and the Gregorian date Feb 29?
Assuming that both the Hebrew and the Gregorian calendars rules do not change, once again, correspondent Ram Sinclair provided the correct answer.
The next occurrence will be about 80,000 years from now.I found a Feb 29 on Adar 30 in 83820g = 87579H.
I am not sure that it is the first.
Correspondent Ram Sinclair did find the next date for this particular coincidence.
In order that Adar 30 coincide with February 29, it is necessary that the Hebrew year be a leap year, and that the next Rosh Hashannah coincide with September 23 of a Gregorian leap year.
All of these conditions are established next in the year 87579H (83819g/83820g).
Thank you Ram Sinclair, for once again sharing with us the results of your excellent research.
Can the Dehiyyah Molad Zakein test be bypassed when calculating for Tishrei 1?
YES!
Dehiyyah Molad Zakein limits the maximum possible calculated
time of the molad to within the first day of any new month.
This purpose is fully explained in
Understanding the Molad Zakein Rule
The Molad Zakein rule requires that 1 Tishrei
be postponed to the next allowable day whenever
f => 18h on an allowable day for Rosh Hashannah.
An unsuccesful attempt to increase that limit to 18h 642p was made by
Aaron Ben Meir in 920g (4680H). For more information on this attempt,
please refer to The Ben Meir Years
When the Hebrew calendar is considered without
Dehiyyah Molad Zakein, and it is desired to maintain the
Keviyyot, then 6 hours must be added to the limits which
govern the postponements that eliminate the 356 and 382 day years.
This is demonstrated in The Keviyyot
Those two limits then become 15h 204p and 21h 589p.
Then, by adding 6 hours to the molad of Tishrei, and also using the limits as modified, there is no further need to include a test for Dehiyyah Molad Zakein.
In this way, the Dehiyyah Molad Zakein test is entirely bypassed.
This technique is interesting to anyone involved in the calculation of the Hebrew calendar since it automatically applies Dehiyyah Molad Zakein without having to test for it.
The technique actually was used by the 18th century
mathematical genius Karl Friedrich Gauss in his well known
Pesach formula which automatically calculates the Julian date
for Pesach given any Hebrew year.The formula is shown in
The Gauss Pesach Formula
The next question will deal with the effect the 6 hour addition to the molad of Tishrei.
Why does adding 6 hours to the molad of Tishrei automatically bypass the Dehiyyah Molad Zakein test when calculating for Tishrei 1?
Dehiyyah Molad Zakein limits the maximum possible calculated time of the molad to within the first day of any new month. This purpose is fully explained in Understanding the Molad Zakein Rule.
The Molad Zakein rule requires that 1 Tishrei
be postponed to the next allowable day whenever
the molad of Tishrei => 18h on an allowable day for
Rosh Hashannah.
By adding 6 hours to the time of the molad of Tishrei, and also to the threshold limits of Dehiyyot GaTaRaD and BeTU'TeKaPoT, there is no further need to include a test for Dehiyyah Molad Zakein.
In this way, the Dehiyyah Molad Zakein test is entirely bypassed.
When the time of the molad of Tishrei => 18h
on either Shabbat (0d), Monday (2d), Tuesday (3d),
or Thursday (5d),
Dehiyyah Molad Zakein requires that the first day of Tishrei
be postponed to
either 2d, 3d, 5d, or 0d.
Adding 6 hours to that time cause the resulting day of the week to be either 1d, 3d, 4d, or 6d.
Dehiyyah LO ADU Rosh will then force the first day of Tishrei to be either 2d, 3d, 5d, or 0d, giving the same result as if the Molad Zakein had been applied prior to the addition of the 6 hours.
In order to be complete, the demonstration also requires consideration of the 6 hour addition in the 3 cases in which the Molad of Tishrei is
1. either on 1d, 4d, or 6d,
2. or past 9h 203p on Tuesday,
3. or past 15h 588p on a post-leap year Monday.
However, the Weekly Question will leave that demonstration to its readers.
The 6 hour addition technique is interesting to anyone involved in the calculation of the Hebrew calendar since it automatically applies Dehiyyah Molad Zakein without having to test for it.
The technique actually was used by the 18th century mathematical genius Karl Friedrich Gauss in his well known Pesach formula which automatically calculates the Julian date for Pesach given any Hebrew year.The formula is shown in The Gauss Pesach Formula.
The next question first appeared as Weekly Question 19.
In the foreseable future can 2 consecutive Rosh Hashannah's ever begin in the same Gregorian year?
Correspondent Larry Padwa sent in the correct answer and said as follows:
The answer is yes, and it follows immediately from your reply to the previous question.RH 26,096H falls on 2-Jan 22336g. Since 26096H (mod 19) = 9, 26096H is not a leap year and hence has at most 355 days. 355 days from 2-Jan falls in December of the same year, so RH 26096H and 26097H fall in the same Gregorian year.
Correspondent Ram Sinclair also gave the correct answer without specific dates.
Thank you correspondents Larry Padwa and Ram Sinclair for sharing with us the results of your excellent research.
Holding constant both the Hebrew and Gregorian calendars, it is possible to discover the occurrence of Gregorian years in which no Rosh Hashannah will occur. It is also possible to discover Gregorian years in which 2 Rosh Hashannot will be observed.
The next question first appeared as Weekly Question 20.
In the foreseeable future, can 2 Gregorian years ever see the same Rosh Hashannah?
Correspondent Larry Padwa once again sent in a correct answer. Thank you Larry.
The Rosh Hashannah Drift shows that the Hebrew year moves into the Gregorian calendar at the rate of about one day in every 216 years.
Left unchecked by any changes to either calendar, it is then possible to predict the event of the first day of Rosh Hashannah on December 31. In that instance, the second day of the festivity will occur in the next Gregorian year.
The event of one Rosh Hashannah spanning 2 Gregorian years will then first be seen to occur for Saturday, 1 Tishrei 25716H (31 Dec 21955g).
Correspondents Ram Sinclair and Winfried Gerum both sent in the dates on which the first day of Rosh Hashannah and the Gregorian New Year would theoretically coincide. That is also a correct answer because the start of Rosh Hashanah would be on the eve of December 31, which of course, is New Year's Eve.
Thank you Ram Sinclair and Winfried Gerum for noticing and sharing with us this very festive coincidence.
The next question first appeared as Weekly Question 21.
Some traditions suggest that the Exodus took place on a
Thursday
either in the year 2448H (-1313g/-1312g)
or in the year 2449H (-1312g/-1311g).
Assuming the fixed calendar rules, could the Exodus have taken place on a Thursday, in either the years 2448H or 2449H?
NO!
According to the fixed calendar, the first day of Pesach would have
occurred on
Tuesday 13 Mar -1312g in the year 2448H.
Similarly, the first
day of Pesach would have taken place on
Sunday 31 Mar -1311g in the year 2449H.
By one of these very strange coincidences, the first day of Pesach would have taken place on THURSDAY 24 Mar -1313g in the year 2447H.
A very quick way to derive the day of the week for the first day of Pesach is to subtract two days from the day of the week of the subsequent Rosh Hashannah.
Consequently, it may be entirely possible that the advocates of a Pesach originating on a Thursday in the year 2448H may have overlooked or forgotten to realize that Thursday was the first day of Pesach for the previous year, namely, 2447H.
Since 17 Tammuz 5761H occurred Sun 8 Jul 2001g, correspondent Ira Walfish asked if the fast of the 17th day of Tammuz occurred most often on a Sunday.
First Begun 21 Jun 1998 First Paged 5 Nov 2004 Next Revised 12 Nov 2004