Prepared by: JOHN A. KOSSEY
Editor: HERMAN L. HOEH
FIRST EDITION
AMBASSADOR COLLEGE PRESS
Pasadena, California
1971, 1974 Edition
----------###----------
PROGRAM I
USING THE TIME UNITS OF THE HEBREW CALENDAR
INTRODUCTION
Why should YOU study the Hebrew calendar?
One of the major identifying signs of the
Church of God is the
observance of the Sacred Festivals. As you study the twenty-third
chapter of Leviticus, you will notice that God employed a calendar
to
indicate when each holy day must be properly kept during the year.
The
Jews were given the responsibility of preserving that calendar
for the
rest of the world.
Since it is the responsibility of the Church
to announce the time
of each festival to the congregations, detailed understanding of
the
Hebrew calendar is not even necessary for a lay church member.
When a
holy day is to be kept is not for the individual Christian to decide.
On the other hand, the education you are
privileged to receive as
an Ambassador College student equips you with a special depth of
biblical understanding. Shallow or sketchy knowledge of basic
background areas would lessen one's effectiveness. But working
out the
calendar principles yourself is going to widen your perspective.
You already know that the holy days portray
God's master plan of
salvation for mankind. Shouldn't you also have a working knowledge
of
the very calendar which houses God's Sacred Festivals?
This is why a study of the Hebrew calendar
is included in
Theological Research I-II.
THE PURPOSE OF THESE LEARNING PROGRAMS
Your study of the Hebrew calendar in this
course has two major
facets. One is the historical development of the calendar. This
is the
primary function of the class lectures. The other is for you to
achieve
needed computational facility with the calendar itself.
Fortunately, the Hebrew calendar requires
surprisingly little
mathematical sophistication. A fifth grade background in arithmetic
will suffice! Nevertheless, a certain number of skills and concepts
must be learned for you to become adept at working with the Hebrew
calendar. These programs are designed to provide you with that
understanding and practice.
Just what will you be able to accomplish
when you complete this
series of learning programs?
For any year, such as 4 BC, 31 AD, 1520
AD, and 1979 AD, you will
correctly determine the dates on a common Roman calendar of the
holy
days listed in Leviticus 23.
How long will this operation require? With
nothing but a pencil
and a blank sheet of paper, you might need anywhere from thirty
to
forty-five minutes. If you use a table of reduced numbers (which
is
included in one of the programs), it might take you only ten to
fifteen
minutes.
That skill is the OVERALL GOAL of this
series of programs. Another
less tangible aim is to give you the confidence that you can actually
SUCCEED in working a calendar problem!
To that end, each learning program takes
a necessary part of the
main goal, and gives you the practice needed to become adept at
it.
Success will breed success as you progress!
The first page of each learning program
has a clear statement of
what you must be doing by the time you complete the program. You
might
think of each program as a "checkpoint" on route to your destination.
Be sure you can accomplish each program goal before going on to
the
next one.
One word of caution. Have you ever learned
mathematics simply by
glancing over the textbook or watching someone else work through
a
problem? No, you can't! The exercises in each program are entirely
for
your benefit. In most cases, these exercises will be worked out
in
detail later in the program. This is for you to have a model with
which
to compare your own procedures and to check your work immediately
for
errors.
But WORK you must! Proverbs 4:13 says to
"take fast hold of
instruction; let her not go." Learning the mathematical operations
of
the Hebrew calendar will take ACTIVE EFFORT. With this diligence,
you
will achieve the exhilaration only success can bring.
----------###----------
Contents
Program
Introduction
i
Using the time units of the Hebrew calendar
1
Calculating the day of the week of the molad Tishri
2
Calculating the day of the month of the molad Tishri
3
Using tables to find the molad Tishri
4
Making Roman leap year and Julian/Gregorian corrections
5
Applying the postponement rules to find Tishri one
6
Counting the days of the week and the days of the month
7
Determining the dates of the annual festivals
8
----------###----------
PERFORMANCE GOAL 1A:
Without hesitation or uncertainty you will
write (or recite) from
memory in any order the following time relationships:
1 part (chalyek)
= 76 moments (regaim)
1 hour
= 1080 parts (chalakim)
1 day
= 24 hours
1 week
= 7 days
1 lunar month
= 29 days 12 hours 793 parts
1 common year
= 12 lunar months
1 intercalary year =
13 lunar months
1 nineteen year cycle = 235 lunar
months, or 12 common
years & 7 intercalary years
PERFORMANCE GOAL 1B:
With only paper and pencil (or pen), you
will accurately add,
subtract, multiply, and divide the time relationships listed in
lA as
necessary to compute specified problems. Large numbers will be
"reduced
to lowest terms" when requested. For example,
28 hours reduces to 1 day, 4 hours
31 hours 650 parts reduces to 1 day, 7
hours, 650 parts.
PERFORMANCE GOAL 1A
In order for you to become adept with calendar
calculations, you
must become very familiar with a minimal number of time relationships.
Some of these you already know; others require memorization. These
numbers will occur so frequently that you simply must know them.
Otherwise, the lessons will take much longer to understand and
you will
feel frustrated in the process. Learn them now -- for your own
benefit.
For a detailed yet concise description
of the time elements of the
Hebrew calendar, consult either of the following references:
"The Jewish
Encyclopedia", volume 3, "Calendar" (either new
or old edition).
Burnaby,
Sherrard Beaumont, "Elements of the Jewish and
Muhammadan Calendars". London: George Bell & Sons, 1901. 554
pages.
(See chapter II, page 21.)
Here is a brief explanation of some of
these time elements to
assist your memorization.
DAY: Genesis 1:5 shows that the day begins
in the evening. "And
the evening and the morning were the first day. " Although each
day
begins at sunset, 6 PM is the arbitrary commencement of a new day
for
CALENDAR CALCULATIONS.
Christ put his divine approval upon dividing
the day into
twenty-four hours. See John 11:9, which states, "Are there not
twelve
hours in the day?" Context shows that this verse refers to the
daylight
portion of a twenty-four hour period. For any given day the periods
of
darkness and light are usually unequal. The total length of a day,
however, is always 24 hours--except for Divine intervention!
HOUR: Instead of being divided into minutes
and seconds, the hour
is divided into parts, or chalakim. One hour consists of 1080 parts,
or
3600 seconds. Using parts instead of minutes and seconds has the
advantage of eliminating fractions. (The smaller unit, the moment
or
rega, is seldom needed for fundamental calculations.) Both the
hour and
the part are considered fixed units anywhere on earth, just as
the
minute and the second are non-varying time elements.
MONTH: A lunar month is the time needed
for the moon to revolve
around the earth. Even though this period varies from month to
month,
29 days 12 hours 793 parts is the traditional average used for
calculation. Actual calendars cannot be based upon 29 1/2 days,
so the
Hebrew calendar incorporates months of 29 and 30 days.
YEAR: The Hebrew calendar has two basic
types of years, common and
intercalary. (The latter term is also called "embolismic.") An
intercalary Hebrew year will have 30 additional days, so it can
also be
called a leap year. By contrast, recall that the Roman leap year
has
366 days instead of 365. You should also remember that the following
terms are interchangeable for the Hebrew calendar:
LEAP (year or month) = embolismic (year
or month) =
INTERCALARY(year or month)
Common years will have 353, 354, or 355
days. Leap years may have
383, 384, or 385 days.
19 YEAR CYCLE: The Western world is accustomed
to a solar year of
365 1/4 days, since the Roman calendar in common use is solar.
This
means that a given month of the year will always occur during the
same
season. January, for example, is invariably a winter month.
On the other hand, the Hebrew year by itself
does not closely
match the length of a solar year. Twelve months which are each
approximately 29 1/2 days results in a year which has only 354
days --
about eleven days less than a solar year of 365 1/4 days. A common
Hebrew year is thus SHORTER than a Roman year.
What about a Hebrew leap year? Thirteen
months of 29 1/2 days is
383 1/2 days, which is LONGER than a solar year.
God has ordained that the holy days must
be kept "in their
seasons" (Lev. 23:4). He also appointed BOTH the sun and the moon
"for
signs, and for seasons, and for days, and years" (Genesis 1:14).
This
means that the calendar which God designed to house His sacred
festivals would be luni-solar. The months must occur at the proper
times for the holy days to fall within the proper season of the
year.
How, then, are the Hebrew lunar months
related to the solar year?
Every 19 solar years (of 365 1/4 days), the moon revolves around
the
earth 235 times, each "lunation" being on the average 29 days,
12
hours, 793 parts. This remarkable astronomical relationship makes
it
possible to combine common years and leap years together within
a
fundamental pattern that repeats itself every nineteen years:
12 common years (12 months each)is 144
months (each month is
7 leap years (13 months each)
is 91 months 29d, 12 h,
-----------------------------------------------
793p)
19 Hebrew years is 235 months =
19 solar years
You should be aware, however, that 235
lunar months is about an
hour and a half less than 19 Julian years. (To be precise, 235
lunations is 1 hour, 485 parts LESS than 19 Julian years.)
The 19-year cycle is also known as the
cycle of Meton, or the
Metonic cycle.
Be sure you have MEMORIZED the time relationships
listed at the
beginning of this section! Don't assume that you know them "pretty
well."
PERFORMANCE GOAL 1B
The source of many errors in calendar calculations
is faulty
arithmetic. Most occur in the basic addition, subtraction, and
multiplication operations. All of these skills you learned before
high
school. Consequently, some review and practice are absolutely
necessary. WORK the problems; don't be deluded into thinking that
glancing over them will suffice.
ADDITION: The key to successful computation rests in one simple rule:
add only likes together
Just as 3 apples and 4 oranges don't equal 7 apples, neither does
adding 11 hours to 4 days equal 15 hours. Never put a number down
without being positively certain that you know what it represents.
The
best way is to label every number: 671 p is 671 parts. Another
way is
to keep like quantities in clearly defined columns:
days hours
parts
------ ------- -------
1 12
103
3 15
1186
This is the same as: 1 d
12 h 103 p
3 d 15 h 1186
p
Select a format you like and use it consistently. Here are some
sample
problems:
5 d
15 h 175 p
+ 1 d
7 h 801 p
------------------------------
6 d
22 h 976 p
So long as you add only
like quantities, no
outstanding difficulties
will happen.
6 d
237 h 1189 p
+24 d
107 h 2159 p
------------------------------
30 d
344 h 3348 p
The numbers can become
very large, as you can
see. Later on you will
learn how to reduce these
to lowest terms.
8 d
13 h 198 p
18 d
23 h 432 p
13 d
45 h 1835 p
+ 7 d
96 h 103 p
-------------------------------
46 d
177 h 2568 p
Adding a whole string of
numbers together will
often occur. Don't be
afraid to double and
triple-check your
arithmetic!
Solve the following problems below. If you aren't "comfortable"
by the
time you complete them, try to make up a few of your own for additional
practice.
1.1) 6 d
17 h
879 p
+
3 d 14 h
198 p
--------------------------------------------
1.2) 67 d
95 h
777 p
+ 275 d
777 h 2589
p
--------------------------------------------
1.3) 55 d
178 h 976 p
31 d 1 h
134 p
178 d 23 h
1937 p
225 d 9 h
11581 p
308 d 768 h
649 p
+
29 d 12 h
793 p
----------------------------------------------------------------
1.4) 131 d
29 h 433 p
227 d 8 h
191 p
0 d 1 h
485 p
+ 130 d
12 h 883 p
----------------------------------------------------------------
The answers to these problems appears below:
1.1) 9 d
31 h 1077 p
1.2) 342 d
872 h 3366 p
1.3) 826 d
991 h 16070 p
1.4) 488 d
50 h 1992 p
REDUCTION: Before attempting to subtract or multiply these numbers,
it
will be helpful for you to understand how they are reduced.
You have learned how to reduce quantities such as inches to feet
and
yards, or seconds to hours and minutes, back in junior high
mathematics:
45 inches reduces to
1 yard 0 feet 9 inches
49 inches reduces to
1 yard 1 foot 1 inch
436 seconds "
" 0 hours 7 min. 16 seconds
3600 seconds "
" 1 hour 0 min. 0
seconds
Reduction of large numbers is not difficult. An EQUIVALENT way of
saying exactly the same thing is all that you are doing. Both
quantities are equal.
The primary units in your calculations will be days, hours, and
parts.
Each of these quantities are related to each other, as you have
already
learned from performance goal 1A.
1. 5) Complete the following:
________ parts = 1 hour
________ hours = 1 day
The answers, of course, are 1080 parts in an hour, and 24 hours
in a
day. This means that 1080 parts can be written as:
0 days 1 hour 0 parts
In like manner, 24 hours can be shifted to the days column:
1 day 0 hours 0 parts.
Take a quantity like 3 d, 49 h, 1400 p. How do you reduce it? What
you
want to do is transfer all EXCESS WHOLE HOURS contained in the
parts
column to the hours column. Then you will take out all the WHOLE
days
contained in the hours column and move them to the days column.
Just
like reading Hebrew, you work from right to left!
To reduce numbers, apply the following operations:
1) Divide the parts in the
parts column by 1080. The whole
number in the quotient represents hours.
2) Add the whole number in
the quotient to the hours column.
This becomes the "revised" hours. Place the remainder in the "reduced"
parts column. If remainder is 0, put this number in that column.
3) Take the revised number
of hours and divide this by 24. The
whole number in the quotient is the number of whole days.
4) Add the whole number of
days to the days column, and place
the remainder in the "reduced" hours column.
-----------------------------------------------------------------------
Reduced numbers for days, hours, parts will have:
any number
of days 23 or less hours 1079 or less parts
-----------------------------------------------------------------------
Here's how to apply these rules to reduce
3 d, 49 h, 1400 p:
1)
1 h
1 h 320 p
---------
1080 / 1400 p
1080
---------
320
2)
1 h 320 p
+ 3 d 49 h
--------------------
3 d 50 h
320 p
3) 2
2 d 2 h
--------
24 / 50 h
48
--------
2
4) 2 d 2 h 320 p
+ 3 d
--------------------
5 d 2 h
320 p
REDUCED
Problems with large numbers are handled the same way. The answer
to 1.
2 is 342 d 872 h 3366 p.
1)
3 h
3 h 126 p
------------
1080 / 3366 p
3240
---------
126
2)
3 h 126 p
+ 342 d 872 h
-----------------------
342 d 875 h
126 p
3) 36
d
36 d 11 h
---------
24 / 875 h
72
---------
155
144
---------
11
4) 36 d 11 h 126 p
+ 342 d
----------------------
378 d 11 h 126
p
Reducing numbers isn't difficult. Your accuracy will be enhanced
if you
consistently stick to a single format. Slopping numbers down
haphazardly on the page is inviting computational errors.
1.6) Reduce 5, 796 parts.
1.7) Reduce 579, 600 parts.
1.8) Reduce 85 d 91 h 150,000 p.
1.9) Reduce 567 d 5228 h 254,404 p.
The answers to 1.6 - 1.9 are on the next pages. Work these problems
on
separate paper, then compare your calculations with the complete
arithmetical details supplied.
1.6
5 h
0 d 5 h 396 p answer
---------
1080 / 5796 p
5400
---------
396
1.7
536 h
536 h 720 p; 536 hours
---------------
must be reduced.
1080 / 579600 p
5400
---------
3960
3240
--------
7200
6480
22 d 22 d 8 h 720 p answer
------- --------
720 24 / 536 h
48
-------
56
48
------
8
1.8 85 d 91 h 150,000 p
(1)
138 h
138 h 960 p; add this to
------------
85 d 91 h.
1080 / 150,000 p
108 0
-----------
42 00
32 40
---------
9600
8640
---------
960
(2)
138 h 960 p
85 d
91 h
-------------------
85 d
229 h 960 p
(3)
9 d
9 d 13 h; add together the
--------
reduced 960 p,
24 / 229
h
9 d 13 h, and
216
85 d.
--------
13 h
(4)
13 h 960 p
85 d
------------------
94 d
13 h 960 p
answer
1.9 567 d 5228 h
254,404 p
(1)
235 h
235 h 604 p
-----------
1080 / 254404
p
2160
---------
3840
3240
---------
6004
5400
---------
604 p
(2)
235 h 604 p
567 d
5228 h
---------------------
567 d
5463 h 604 p
(3)
227 d
227 d 15 h
--------
24 / 5463
h
48
--------
66
48
--------
183
168
--------
15 h
(4) 227 d 15 h 604 p
567 d
--------------------
794 d
15 h 604 p
MULTIPLICATION: After adding days, hours, and parts, you will find
that
multiplication is quite easy. The important key to correct
multiplication is that ALL TERMS must be multiplied! Forgetting
to
multiply days while you complete the operation for the hours and
parts
can easily happen if you aren't wary.
Look at a sample problem:
2 d 16 h 595 p
x 3
-----------------
6 d 48 h 1785 p
Notice that the multiplier operates
on the days, the hours, and the parts.
Multiplication can help you solve many problems that would take
much
longer by straight addition. Here is an example:
How many days, hours, parts are in an AVERAGE common year?
You should easily remember:
The number
of lunar months in a common year;
The number
of days hours and parts in a lunar month.
If you cannot recall these numbers, return immediately to the first
page of this section and drill intensely on the time relations
listed
in 1A.
A common year has 12 months. Each month has 29 d, 12 h, 793 p.
Therefore an average common year has:
29 d 12 h
793 p
x 12 months
--------------------
58
24 1586
29
12 793
----- ----- ------
348 d 144 h 9516 p
If this number were to be used in a series of computations, it could
stand "as is." On the other hand, when it is the final answer,
can't
you see the need to reduce such a number? You may want to verify
that
an average common year has 354 d 8 h 876 p. (This is
worth
remembering.)
1.10 How many days, hours, and parts are in an average intercalary
year?
1.11 What number of d, h, p are in the 19 year cycle?
1.12 7 x (5 d 21 h 589 p) = _____________
1.13 (4 d 8 h 876 p) x 12 = _____________
1.14 (18 d 15 h 589 p) x 6 = _____________
Problems 1.10 - 1.14 are worked out for you below.
1.10 29 d 12 h 793
p
x 13 months
----------------------
87
36 2,379
29
12 793
----------------------
377 d 156 h 10,309 p
This reduces to 383 d 21 h 589 p.
1.11 You may calculate this problem by two methods:
a) add the lengths of 12 common years
and 7 leap years together,
using the example and 3.10.
b) Find the length of 235 lunations.
Here's how method b) is solved:
29 d
12 h 793 p
x 235 months
---------------------
145
60 3965
87
36 2379
58
24 1586
----- -----
------
6815 d 2820 h 186355
p
This reduces to
6939 d 16 h 595 p.
1.12 5 d 21 h 589 p
x 7
----------------------
35 d 147 h 4123
p
This reduces to
41 d 6 h 883 p.
1.13 4 d 8 h 876
p
x 12
----------------------
8
16 1752
4
8 876
----------------------
48 d 96 h 10512
p
This reduces to
52 d 9 h 792 p.
1.14 18 d 15 h 589 p
x 6
----------------------------------
108 d 90 h 3534 p
This reduces to
111 d 21 h 294 p.
SUBTRACTION: This is the most error-prone operation for many students.
Two facets of subtraction are particularly troublesome:
a) borrowing
b) negative numbers
First look at a straight-forward subtraction problem:
5 d
21 h 589 p
- (4 d 19 h
98 p)
-----------------------
1 d
2 h 491 p
Notice that the minus sign affects all the quantities in the second
line. Obviously, adding one of the units and subtracting the others
isn't kosher! The first principle to keep in mind:
SUBTRACT ALL TERMS called for.
Practice on two of these problems, comparing your answer with that
given.
1.15 6 d 6 h 883 p
- (1 d 5 h 785 p
---------------------
answer: 5 d 1 h 98 p
1.16 - ( 74 d 14 h 45 p)
111 d
21 h 294 p
-------------------------
answer: 37 d 7 h 249 p
(You CAN subtract "upside
down.")
In the next example, "borrowing" will be necessary:
8 d
23 h 403 p
- (6 d 22 h 528
p)
-----------------------
Since 1080 parts are in an hour, 8 d 23 h 403 p can
be changed to
8 d (23-1) h (403 + 1080) p, or 8 d 22 h
1483 p. Be sure you
understand that neither expression is different. Borrowing merely
places the number in a more convenient form for subtraction.
By borrowing quantities as needed, the problem becomes ordinary
subtraction:
8 d
22 h 1483 p
- (6 d 22 h
528 p)
------------------------
2 d
0 h 955 p
In order to firm up the concept of borrowing in your mind, examine
another illustration:
3 d
3 h 0 p
- (1 d 18 h 6000 p)
-----------------------
On this problem, you cannot simply take one hour and transfer 1080
parts. A quick estimate will tell you that at least 6 hours need
to be
changed to parts. Only 3 are in the hours column. Where do you
get
them?
First, make up for the lack of hours in the second column by converting
a day (or more!) into hours. 3 d 3 h 0 p transfers
to
2 d (24 + 3) h 0 parts, or 2 d 27 h 0 p.
Next borrow 6 hours and change them to parts:
2 d (27-6) h (6480 + 0) p or,
2 d 21 h
6480 p
Finally, subtract as required by the original problem:
2 d 21 h
6480 p
- (1 d 18 h 6000 p)
-----------------------
1 d
3 h 480 p
Once in a while, you will find that in borrowing parts, you loose
too
many hours for subtracting the hours. (Instead of 18 hours, suppose
you
needed to subtract at least 22 hours in the above example.) When
that
happens, simply borrow again from the next column to the left.
Borrowing is a reverse of reduction. Remember that both quantities
are
equivalent -- before reduction and after reduction, or before and
after
borrowing. A "golden rule" for these operations:
----------------------------------------------------------------
WHAT YOU ADD TO ONE
COLUMN MUST BE SUBTRACTED FROM ANOTHER.
----------------------------------------------------------------
1.17 40 d 6 h 104 p
- (35 d 2 h 1141 p)
----------------------
answer: 5 d 3 h 43 p
1.18 136 d 123 h 5291 p
- ( 89 d 29 h
7679 p)
--------------------------
answer: 47 d 91 h 852 p
1.19 21 d 30 h 811 p
- (17 d 69 h 2050 p)
------------------------
answer: 2 d 7 h 921 p
The last complication to hurdle with subtraction is negative numbers.
By borrowing, you were able to keep numbers positive. But sometimes
the
arithmetic is simplified by allowing numbers to become negative-without
borrowing.
There is nothing mysterious about negative time. All that it means
is
time BEFORE a prescribed reference point. Such a reference point
is the
blast-off of a moon launch. Events before lift-off are considered
negative times, as you heard on many telecasts: "t MINUS eight
minutes
and holding." After the rocket lifts off, time becomes positive
-- with
reference to the ZERO lift-off.
An expression like days, hours, parts refers to a particular instant
in
time. When you are computing a problem, some of these numbers will
occasionally become negative. Nonetheless, the expression is still
identifying a precise moment. For example look at:
5 d -2 h -810 p
Does this specified time occur on the fifth day or on the fourth?
The
answer is the fourth day. This is immediately clear if you change
the
expression to all positive signs by "borrowing. " These are the
steps:
5
d -2 h
-810 p
(5-1) d (-2 +
24) h -810
p
4
d 22 h
-810 p
4
d (22 - 1) h (-810 + 1080) p
4
d 21 h
270 p
So 5 days -2 hours -810 parts is the same as
4 days 21 hours 207
parts.
You should become familiar enough with these negative expressions
that
you accurately complete computations involving them. In ordinary
subtraction, all numbers were positive. Now you will find that
subtraction and addition of numbers of either sign frequently occurring
when you eventually calculate the days, hours, parts, of the month
for
a conjunction of the moon.
The next example illustrates how this type of problem is worked:
36 d 22 h 1284 p
-17
d + 3 h - 541 p
------------------------
19 d 25 h 743 p or
20 d 1 h 743 p
When a number is not preceded by a sign, it is taken to be positive.
A
few rules apply for these problems:
adding two negatives together (e.g.
-5 + -3) results in a
negative number larger than either (-8).
adding a negative number and a positive
number together is the
same as subtracting. The sign of the larger value
(called the
absolute value in mathematics) is the sign of the final number.
(-30
+ 50 = +20)
(-48
+ 38 = -10)
When you subtract a negative number from
another negative number,
you change the sign of the number you are subtracting, and then
add as
above. (-5 - -3 = -5 + +3
= -2)
Adding two negative numbers together results in a larger negative
number. Subtracting two negative numbers results in a smaller negative
number -- closer to zero. Any time you are working with negative
numbers, be sure that you carry the signs with you. Although you
need
not specifically mark positive numbers, this may be advisable for
you
initially in order to keep the signs clear in your mind. Whenever
you
are subtracting, remember that the subtraction sign affects every
number (term) in the expression:
45 d
23 h 150 p
- ( -8 d 15 h -125 p)
------------------------
Notice how the numbers subtracted change signs:
45 d
23 h 150 p
+8 d
-15 h + 125 p
----------------------
53 d
8 h 275 p
Try the following problem:
1.20 - 18 d - 2 h - 780
p
- 0 d
- 1 h - 485 p
- 97 d -
2 h - 756 p
+ 74 d + 14 h
+ 196 p
The answer appears below.
+ 13 d
+ 36 d + 22 h
+ 1284 p
-----------------------
1.20 The problem becomes easier if you total the negative
quantities
separate from the positive, and then combine.
- 18 d
- 2 h - 780 p
74 d 14 h 196 p
0 d - 1 h - 485 p
13 d
- 97 d
- 22 h - 756 p
36 d 22 h 1284 p
------------------------
-------------------------------
-115 d
- 25 h - 2021 p
123 d 36 h 1480 p
123
d 36 h 1480 p
-115 d
- 25 h - 2021 p
------------------------
8 d 11 h - 541 p
This reduces (by taking one
hour and changing it to
parts: + 1080 - 541) to
8 d 10 h 539 p.
1.21 Work the next problem without reducing until the final step:
210
d - 15 h - 971 p
- ( 37 d
19 h - 1185 p )
--------------------------
Change signs and subtract.
Your result should be
173 d -34 h +214 p.
This reduces to
171 d 14 h 214 p.
1.22 Add 92 d 581 h 471 p to eight times
(-10 d -21 h -204 p).
Then subtract -152 d - 589 h 188 p from
the sum. Reduce at the
final answer. How many days over a full number of weeks is this?
First multiply: - 10 d - 21
h - 204 p
x 8
--------------------------
- 80 d - 168 h - 1632 p
Add:
92 d 581 h 471 p
---------------------------
+ 12 d + 413 h - 1161 p
Subtracting -152 d -589 h 188 p, you change the signs and add:
12 d + 413 h - 1161 p
+ 152 d + 589 h - 188 p
----------------------------
164 d + 1002 h - 1349 p
Borrow 2 hours, and then divide 1000 hours
by 24 to convert the
hours column to a number less than 24. This gives:
205 d 16 h 811 p
To find how many days this is over a full number of weeks, you divide
by 7 the REDUCED number of days. 205 / 7 leaves a remainder of
2. Two
days plus 16 hours and 811 p is the time over a full number of
weeks.
* * *
After working the problems in this section,
you should feel
confident about your ability to successfully add, subtract, multiply,
and reduce days, hours, and parts. Some may still require additional
drill on these operations. Design your own problems for more practice.
Have another student check your work, or see your instructor for
assistance.
----------###----------
PROGRAM II
CALCULATING THE DAY OF WEEK FOR MOLAD TISHRI
PERFORMANCE GOAL 2:
Given a required Roman year, you will correctly
calculate the day
of the week, hours, and parts for the molad Tishri of that Roman
year.
This will be accomplished WITHOUT CHARTS AND TABLES.
Why do you need to be concerned with the
molad of Tishri? The
answer is that you must know when it occurs before you can determine
the date of the Festival of Trumpets. And all other holy days within
a
Roman year (January-December) are ultimately referenced to that
holy
day. The molad of Tishri is prerequisite to most calculations involving
the Hebrew calendar. Correct calculation of the molad of Tishri
is thus
essential for determining what dates on the Roman calendar we commonly
use for God's Sacred Festivals to occur.
Certain definitions and concepts need to
become crystallized in
your mind:
What is a molad?
When is Tishri?
What is a bench mark?
How is time reckoned in the calculations?
What is meant by the "advancement" of
the molad?
Let's find the answers to these questions.
What is the molad Tishri?
TISHRI is the seventh month of the sacred
calendar. The computed
time for the conjunction of the sun, moon, and the earth is called
a
MOLAD, from the Hebrew MOLED (plural, MOLEDOTH). This word means
renewal, or rejuvenescence.
Molad of Tishri is the computed time of
the new moon of the month
of Tishri, which corresponds to September/October. As Tishri is
also
the first month of the civil Hebrew year, the molad Tishri is also
the
calculated astronomical commencement of the year.
Another term which you will be using is
bench mark. All this means
is a point of reference from which measurements can be made. Any
known
molad (expressed as day of the month, day of the week, hours, and
parts, e.g., October 6, Sunday, 23 h, 204 p in 3761 BC) can serve
as a
bench mark. The most practical choice for a bench mark, however,
will
be the molad Tishri of year one in a 19 year cycle. 3761 BC is
such a
year.
In order to avoid a mental mix-up later,
let's clarify how we
reckon time. Just what does an expression like 4 d 7 h
503 p mean?
Such an expression can be taken either of two ways: a) as an interval
of time b) a time in the week.
Consider case a). Here, when you are speaking
of a length of time,
you assume that you are starting from a reference point of 0 days,
0
hours, 0 parts. This is just the same as clocking a track runner
on a
stop watch that starts ticking at the sound of the gun. A certain
time
span is involved, figuring from a bench mark of 0 days, 0 hours,
0
parts.
With case b), you are still reckoning time,
but from a DIFFERENT
reference point. In the calendar calculations you are working in
Theological Research, the week begins at Sunday midnight. Sunday
is
regarded as the first day of the week; midnight the zeroth hour,
zero
parts:
The week begins: Sunday: 1 d 0 h 0 p.
An expression like 1 day 13 hours 0 parts
MUST BY DEFINITION OF
THE STARTING POINT refer to Sunday, the 13th hour after midnight,
or
Sunday 1 PM.
Likewise, 1 day 13 hours 179 parts refers
to a time slightly later
than 1 PM on Sunday.
The reason why we arbitrarily begin Sunday
as 1 day 0 hours 0
parts is so that there will be an exact coincidence with the days
of
the week: 1 d is Sunday; 2 d is Monday; 4 d is Wednesday; 7 d is
Sabbath. So long as all the numbers are reduced, you merely look
at the
number of days, and these correspond to the day of the week. (For
the
same reason of SIMPLICITY, we begin the day with midnight instead
of 6
PM so far as the CALCULATIONS are concerned. Since the bench mark
is
expressed by this same reckoning, the final results will be exactly
the
same.)
By contrast, if we decided to define the
beginning of the week,
Sunday, as the zero day, and Saturday sunset (say 6 PM) as the
start of
Sunday, we would have the expression 0 d 0 h
0 p = 6 PM Saturday
evening. Then 1 d 13 h 179 p would have an entirely
different
meaning, if it referred to a time in the week. 1 d would then be
Monday; 13 h 179 p would be slightly after 7 AM. Do you see
how much
more complicated your work would then become?
Here is another distinction that can help
you understand the
difference in meaning between an interval of time, and a time in
the
week:
a) an interval of time:
parts are added to the hours are added to the days
b) a time in the week:
parts are added to the hours WITHIN the day of the week.
If 4 days 7 hours 503 parts refers to an
interval of time, then
503 parts plus 7 hours is added to 4 days. Therefore the interval
of
time goes as far as the 7th hour(and 503 parts) of the fifth day.
If 4 days 7 hours 503 parts is to be taken
as a point in the week,
then 503 parts plus 7 hours is WITHIN the 4th day of the week.
Remember this difference in what is meant
comes about by the way
the starting point is DEFINED, and for no other reason.
During your calculation of the advancement
of the molad over a
full number of weeks, you may come up with a number like 0 days
4 hours
and 71 parts. Be comforted by the fact that the expression is simply
an
interval of time by the way Sunday midnight is defined.
How can you relate a time interval to an
expression of time during
a week?
YOU MUST ADD AN INTERVAL OF TIME to the
BENCH MARK before you can
determine a real day of the week. The bench mark will implicitly
tell
you where the week begins. The bench mark for the year 3761 BC
is
Sunday, the 23rd hour 204 parts. (We could have called that bench
mark
Monday if we started the day at 6PM instead of midnight. But then
we
would be confronted with a more complicated interpretation of what
the
time expressions mean.)
Now transform each of the following expressions
to a) a time
interval b) a time during the week. Use this format:
a: _______ days, plus _______ hours _______ parts of the _______ day
b: _______ (day of the week), between _______ & _______ AM or PM
2.1 4 d 3 h 191 p
2.2 1 d 16 h 304 p
2.3 6 d 7 h 8 p
2.4 7 d 22 h 5 p
2.5 2 d 13 h 871 p
The answers are given below.
2.1 a) 4 days plus 3 hours 191 parts of the 5th
day
b) Wednesday, between 3 &
4 AM
2.2 a) 1 day plus 16 hours 304 parts of the 2nd
day
b) Sunday, between 4 &
5 PM
2.3 a) 6 days plus 7 hours 8 parts of the 7th
day
b) Friday, between 7 &
8 AM
2.4 a) 7 days plus 22 hours 5 parts of the 8th
day
b) Sabbath, between 10 &
11 PM
2.5 a) 2 days plus 13 hours 871 parts of the 3rd
day
b) Monday, between 1 &
2 PM
One particularly important time interval
will occur throughout
your experience with the Hebrew calendar. In order to calculate
the
molad Tishri, you will be working with two molads:
* a known
molad, such as 3761 BC -- a bench mark
* the molad
of the Roman year which you are determining
The time interval between these two molads
is the ELAPSED TIME.
Since you are dealing with the molad of Tishri in both the required
year and the bench mark, the elapsed time will be a whole number
of
years, such as 4520 years, 1503 years, 38 years.
Let's investigate another feature of the
Hebrew calendar which we
can call MOLAD "ADVANCEMENT." Now you know that the MOLAD itself
doesn't really move, since the term is defined as the calculated
conjunction of the sun, moon, and the earth. We are using a figure
of
speech very similar to sun "rise."
Here's an illustration of what is meant
by the advancement of the
molad. In 3761 BC, the molad Tishri was on Sunday 1 d
23 h 204 p. In
1980 AD the molad Tishri will be Tuesday 3 d 23 h
206 p. Although
thousands of years have elapsed over the time span, the APPARENT
"advancement" in the week of the second molad is only 2 days 0
hours
and 2 parts.
Why does the molad occur on different days
of the week? The length
of an average lunar month is 29 d 12 h 793 p.
How much greater than
four full weeks is this?
29 d 12 h
793 p
-(28 d 0 h
0 p)
-----------------------------
1 d 12 h
793 p; about a day and a half.
The molads of two successive months cannot
occur on the same day
of the week because of this EXCESS over 28 days. In one month,
if the
molad were Monday 8 AM (2 d 8 h 0 p), the molad of
the next month
would be:
2 d
8 h 0p
+(1 d 12 h 793
p)
-----------------------------
3 d 20 h
793 p, or Tuesday, between 8 and 9 PM.
Although the second molad occurred
29 d 12 h 793 p after the
first one, the second molad was "displaced" WITH REFERENCE TO THE
WEEK
by 1 d 12 h 793 p. Merely as a convenient label,
we will refer to
that apparent shift as MOLAD ADVANCEMENT. But the molad doesn't
move;
only the time of its occurrence IN THE WEEK apparently advances.
The TOTAL MOLAD ADVANCEMENT is simply the
EXCESS over the number
of full weeks in the elapsed time from the bench mark to the molad
Tishri of the desired Roman year.
Just as monthly molads will occur on different
days, the molad of
Tishri will advance in the week over the previous molad of Tishri.
If
in three years the total advance were 13 days, the Molad would
be 13
days later. But in terms of the day of the week, this would be
13-7, or
6 days later in the week.
Now you will learn how to calculate the
day of the week for the
molad Tishri.
You will find it easier to understand how
to determine the day of
the week for a given molad if the steps are explained first without
the
mathematical details:
In order to determine the day of the week
for the molad of Tishri,
you must find the TOTAL ADVANCEMENT of the molad that occurs within
the
time span involved from a known molad, or bench mark.
What will make up that time span? From
that bench mark, a certain
number of years will elapse to the particular year in question:
molad of Tishri
molad of Tishri
(bench mark)
of required year
(elapsed time)
*________________________________________________*
Later on in this program, you will learn
how to express elapsed
time as:
a whole multiple of 19 year cycles,
plus the excess number of
common years,
plus the excess number of
leap years.
Logically, the total advancement of the
molad, or excess over full
weeks in the elapsed time, for the year in question can be found
by
adding:
the advancement
due to whole multiples of 19 year cycles,
+ the advancement due
to the number of common years,
+ the advancement due
to the number of leap years.
An example will clarify this. If the elapsed
time is 156 years,
there are 8, 19 year cycles, 4 common years, and 2 leap years (you'll
see how this is done later). The total advancement of the molad
will
be:
8 times the
advancement of one 19 year cycle
+ 4 times the advancement
of one common year,
+ 2 times the advancement
of one leap year.
The final step is simply adding the total
excess over full weeks
to whatever bench mark you started with. This figure will give
you the
day of the week of the molad Tishri in question.
Be sure that you understand the qualitative
elements connected
with the advancement of the molad. A molad "advances" with respect
to a
known molad because of the excess time in one average lunar month
over
a full number of weeks. All you are doing is adding the total
advancement that occurs within the time interval from the bench
mark to
the molad Tishri of the year in question. You are actually finding
the
excess over the full weeks from the bench mark to the molad of
the
required year.
Your success at determining the day of
the week of a required
molad impinges upon:
a)
Correctly finding the ELAPSED TIME from a bench mark to
the required year.
b)
Expressing the elapsed time in terms of multiples of 19
year cycles
+ common years in the remainder
+ leap years in the remainder
c)
Calculating the molad "advancement" attributable to each
element of b), then adding these together, reducing as necessary.
d)
Adding the reduced advance of the molad c) to the bench
mark.
ELAPSED TIME, 2A
Three possibilities exist for the elapsed
time from the bench mark
to the required year:
a1)
Both years are AD dates.
a2)
Both years are BC dates.
a3)
One year is BC and the other is AD
(Although you might work backwards in time, this program will only
consider problems in which the bench mark is the EARLIER of the
two
years.)
a1: Bench mark and required year both AD
dates. The elapsed time
is simply the difference between the two years. If the bench mark
is
1845 AD and the required year is 1931, the elapsed time is:
1931 - 1845 = 86 years
If the bench mark is 895 AD and the required year is 1751, the elapsed
time is:
1751 - 895 = 856 years
a2: Both years are BC dates. Many calculations
use 3761 BC as a
bench mark. To be mathematically consistent, it is helpful to place
a
negative sign ( - ) before all BC years. Since you will still be
subtracting in order to find the difference, the second number
will
become positive - ( - ) = +. As you are primarily
concerned with
years after 3761 BC, or -3761, the year in question will always
be more
positive (closer to zero). As a check when both years are BC, expect
the elapsed time to be SMALLER than 3761 years. (Use only the "absolute
value" for the elapsed time, which means you can disregard the
final
negative sign.)
Using 3761 BC as a bench mark, what is the elapsed time to 585 BC?
-3761 - (-585) = -3761 + 585 = -3176
The elapsed time is 3176 years.
What is the elapsed time to 1486 BC?
-3761 - (-1486) = -3761 + 1486 = -2275
The elapsed time is 2275 years.
a3: Bench mark is BC and the required year
is AD. 3761 BC is
frequently used as bench mark for AD years. Only one thing is really
different here. There is no year allotted for 0 AD or 0 BC.
Mathematically, the number system does have a 0. What do you do?
Graphically, the two systems look like:
Calendar: 5 BC 4 BC
3 BC 2 BC 1 BC 1 AD 2 AD
3 AD
Number Scale: -5 -4
-3 -2 -1
0 1 2
The number scale has one more place, a zero, than the calendar.
Therefore in time intervals that cross AD/BC, you must SUBTRACT
ONE
from the answer you compute arithmetically. (The time span from
4 BC to
2 AD is only five years.) BE SURE YOU REMEMBER TO SUBTRACT ONE
FROM THE
MATHEMATICAL COMPUTATION! But only when you cross over from BC
to AD.
What is the time elapsed from 3761 BC to 1000 AD?
-3761 - +1000 = -3761 -1000 = -4761
The elapsed time, however, is 4761 - 1 year, or 4760 years.
What is the time span (elapsed time) to 1974 AD from 3761 BC?
-3761 BC - (+1974) = -3761 - 1974 = -5735
The elapsed time is 5735 - 1, or 5734 years. As a check,
you should
be aware that in going from BC to AD, the elapsed time will be
a larger
number (taken as an absolute value) than either the bench mark
or the
required year.
Drill yourself on elapsed time calculations with the following
problems. Remember that if you are wrong here, all the rest of
your
calculations w ill be off!
Bench mark Required year Elapsed time
2a.1
1845 AD 1971 AD
2a.2
1883 AD 1945 AD
2a.3
3761 BC 1442 BC
2a.4
3761 BC
4 BC
2a.5
3761 BC 721 BC
2a.6
1861 BC 604 BC
2a.7
3761 BC 31
AD
2a.8
3761 BC 1859 AD
2a.9
3761 BC 1982 AD
See problems 2b.1 to 2b.9 for the elapsed time.
EXPRESSING ELAPSED TIME, 2B
Once you've determined the number of years
between the bench mark
and the required year, you need to express that time in terms of
the
Hebrew calendar.
On a practical basis, 19 year cycles are
convenient. Find the
number of 19 year cycles in the elapsed time, divide the elapsed
time
by 19. The QUOTIENT is the number of 19 year cycles.
Usually, however, this division will result
in a remainder, a
number from 1 to 18. This remainder will tell you the number of
years
elapsed in the next cycle.
As you already know each 19 year cycle
consists of 12 common years
and 7 leap years.
=====================================================================
SINCE 142 AD (see footnote 1), the years
in a cycle that are leap
years are: 3 6
8 11 14 17
19
=====================================================================
Be sure to memorize these numbers! (Of course, the years 1, 2, 4,
5, 7,
9, 10, 12, 13, 15, 16, and 18 are common.)
======================================================================
Before 142 AD, the leap years were:
2 5 7 10 13 16 18
======================================================================
For any remainder you acquire after dividing the elapsed time by
19,
you:
1) Decide whether the required
year is before 142 AD, or 142 AD
and after.
2) Count the number of leap
years that fit in the remainder (or
you could count the common years).
3) The number of common years
will be the remainder minus the
number of leap years.
For example, the elapsed time from 1845 AD to 1975 AD is 130 years.
This is 130/19 time cycles, or six 19 year time cycles plus 16
years
remainder.
The leap years for 1975 (after 142 AD) are 3, 6, 8, 11, 14. Five
leap
years altogether. Since 1975 has 16 elapsed years and there are
five
leap years thus far, there must be 16 - 5 or 11 common years.
--------------------------------------------------------------------
(footnote 1) There is some evidence that
an adjustment to the
Hebrew calendar may have taken place during the patriarchate of
Simon
III (140-163). See Cyrus Adler, "Calendar, History of," in "The
Jewish
Encyclopedia" (New York: Funk and Wagnalls, 1907), Vol. 3, p. 500.
--------------------------------------------------------------------
Here is another example. If the bench mark is 3761 BC and the required
year is 27 AD, the elapsed time is:
-3761 - (+27) =
-3788 years; the elapsed time is 3787 years,
since you go from BC to AD.
3787 / 19 = 199 19 year cycles, plus 6 elapsed years.
For 27 AD (before 142 AD), the leap years of the cycle are 2 and
5.
With two leap years, there must be 6 - 2, or
4 common years.
Practice expressing elapsed time in terms of 19 year cycles, the
number
of leap years, and the number of common years. See problems 2a.1
to
2a.9.
Elapsed
19 year # of leap # common
time
cycles years
years
2b.1 126 yrs
2b.2 62
2b.3 2319
2b.4 3757
2b.5 3040
2b.6 1257
2b.7 3791
2b.8 5619
2b.9 5742
Check answers against those below.
2b.1 6 cycles
4 leap 8 common
2b.2 3 "
1 " 4 "
2b.3 122
0 1
2b.4 197
5 9
2b.5 160
0 0
2b.6 66
1 2
2b.7 199
4 6
2b.8 295
5 9
2b.9 302
1 3
"ADVANCEMENT" OF THE MOLAD. 2c
From your previous program (1A & 1B),
you are equipped to find out
the required information regarding the advancement of the molad.
Since
a lunar month has 29 days, 12 hours 793 parts, it is 1 day, 12
hours,
and 793 parts in excess of a full number of weeks. As you saw before,
a
monthly molad (two successive months) advances 1 d 12 h
793 p.
Knowing this, you can easily determine
how much a molad "advances"
in a common year of 12 months, in a leap year of 13 months, or
in a 19
year cycle of 235 months.
How much does the time of a molad "advance" in the week during a
common
year?
1 d 12 h 793 p
(the monthly "advancement",
or excess over 4 weeks)
x 12 (number of lunar
months in a common year)
-----------------
2 24
1586
1 12
793
-----------------
6 days
8 hours
12 d 144h 9516 p
------- ---------
24 / 152 h 1080 / 9516 p
144
8640
-------
-------
18 d 8 h 876 p
reduced 8 h
876 parts
After 12 months, the molad "advancement"
is 18 d 8 h 876 p.
Since full weeks will not affect the days of the week, you can
divide
the reduced number of days by 7. So with respect to the week the
molad
"advancement" for a common year is:
4 d 8 h 876 p.
You should notice an alternative way of
arriving at the same
number. How much does a common year exceed the number of full weeks
in
the year? Multiply the length of an average lunar month by 12 months:
12 x (29 d 12 h 793 p) = 348 d 144 h 9516 p
Reduce this number: 354 d 8 h 876 p
Divide the reduced number of days by 7 to eliminate full weeks:
4 d 8 h 876 p.
2c.1 Verify that an average leap year exceeds a full number
of weeks
by 5 d 21 h 589 p.
2c.2 Verify that a 19 year cycle exceeds a full number of
weeks by 2
d 16 h 595 p.
>From here on, you already are competent to handle the details of
multiplying the advancement of the molad, and then adding. For
4 BC (as
in 2a.4 and 2b.4) you will do the following:
197 x (2 d 16 h 595 p)
molad advancement of 19 year cycles
5 x (5 d 21 h
589 p) molad advancement of leap years
9 x (4 d
8 h 876 p) molad advancement of common years
Multiplying and adding you will discover that this is:
567 d 5228 h 254,404 p.
When reduced, and divided by seven (only the full reduced days are
divided by 7), the advancement of the molad over a week is:
3 d
15 h 604 p.
ADDING REDUCED ADVANCEMENT OF MOLAD TO THE BENCH MARK, 2d
The bench mark for 3761 BC is Sunday
23 h 204 p. This is
reckoning from midnight. As Sunday is the first day of the week,
the
bench mark can be expressed as:
1 d 23 h 204 p.
>From here on, you merely add the reduced advancement of the molad
to
the bench mark. For 4 BC, this is:
3 d 15 h 604 p
+
1 d 23 h 204 p
----------------------------
4 d 38 h 808 p
or
5 d 14 h 808 p
The fifth day of the week is Thursday,
so the molad occurred on
Thursday, the 14th hour (2 PM), 808 parts.
So long as the final d, h, p, are reduced, you need only be concerned
(for now) with the first column.
2d.1 Verify that the molad for 721 BC is 5 d 7 h 364 p.
2d.2 Verify that the molad for 31 AD is 5 d 23 h 941 p.
2d.3 At this point, you should test yourself to be sure you
can
fulfill the goal of program 2. Calculate the DAY OF THE WEEK of
the
molad Tishri for 1996 AD without consulting this program. You'll
need
to have memorized (or else work out again) the molad "advancement"
for
a 19 year cycle, a common year, and a leap year, as well as remember
a
bench mark.
Compare your calculations to the answer below, which is worked out
in
detail.
Calculate the DAY OF THE WEEK of the molad Tishri for 1996 AD.
-3761
-1996
-----------
5757 yrs
-1
302 cycles
-----------
----------
5756 yrs
19 / 5756
57
--------
56
38
------
18 elapsed yrs; 6 leap yrs; 12 common yrs
2 d 16 h
595 p
5 d 21h 589 p
x
302 (19 yr
x 6 (leap
-------------------- cycles)
------------------ yrs)
4 d 32 h
1190
30 d 126 h 3534 p
60 480
17850
-----------------------
604 d 4832 h 179690 p
4 d 8 h 876 p
x 12 (common
------------------ years)
48 d 96 h 10,512p
604 d
4832 h 179690 p
+ 30 d
126 h 3534
p
+ 48 d
96 h 10512 p
---------------------------------------
682 d
5054 h 193736 p
total advancement
---------------------------------------
+ 218 d
+179 h
128 weeks
218 d
179 h
-------
--------
-----------
7 / 900 d 24 / 5233 h
1080 / 193736 p (reducing)
xxx
xxx
xxx
------
--------
----------
4 d
1 h
416 p reduced advance-
ment of molad
+ (1 d
23 h
204 p) bench mark
-----------------------------------------------------
5 d
24 h
620 p
6 d
0 h
620 p
=====================================================
The day of the week of molad Tishri in 1996 AD is Friday.
----------###----------
PROGRAM III
CALCULATING THE
DAY OF THE MONTH OF THE MOLAD TISHRI
PERFORMANCE GOAL 3
Without tables, you will correctly determine
the day of the month
of the molad Tishri for the Roman years specified in this program.
The procedure used to find the day of the
month of the molad
Tishri parallels that of the day of the week calculation you already
learned in Program 2. Part of the work needed for the day of the
month
calculation is ALREADY accomplished in the day of the week calculation!
Let's review part of the day of the week
calculation to see just
what operations are in common. You must have a required year and
the
bench mark (assumed to be 3761 BC unless otherwise specified).
From
these two years, you can determine elapsed time between them (read
over
2a again if this is hazy). Be sure you remember to subtract 1 from
the
total whenever you go from BC to AD. This is a common mistake!
Next, you express the elapsed time in terms
of 19 year cycles,
number leap years, and the number of common years (2b):
* Divide the elapsed time by 19.
* The remainder is the
number of elapsed years in the 19 year
cycle of the required year. If the required year is 142 AD or after,
the intercalary years are 3, 6, 8, 11, 14, 17 and 19. (Before 142
AD, just subtract 1 from each of these numbers to obtain the
intercalary years of that cycle.)
* Count up the number
of leap years that have occurred within
the cycle, including the last number. For example, if the remainder
is
11, there are 4 leap years in the cycle.
* Subtract the number
of leap years from the remainder, and you
will have the number of common years in that cycle.
You have already done this much of the
calculation in order to
determine the day of the week for the molad of Tishri. And you
will use
this SAME information to find the day of the month. How is it applied?
The answer rests in the basic difference
between the Hebrew year
and the Roman (Julian) year. Common years in the Hebrew calendar
have
353, 354, or 355 days, whereas a Roman year has 365 1/4 days. Compared
to the Julian year, the Hebrew common year falls short. A Hebrew
intercalary year (383, 384, or 385 days) is longer than the Roman
year.
Exactly how much will the Hebrew calendar
trail the Roman calendar
in a common year of 12 months? The average common year is 12 times
(29
d 12 h 793 p). This is 354 days, 8 hours and 876 parts.
365 d 6 h 0 parts (average Julian year)
- ( 354 d
8 h 876 p (average common year)
--------------------------------
10 d 21 h 204 p
In other words, the average common year of the Hebrew calendar is
10 d
21 h 204 p LESS than an average Julian year. For calculation
purposes,
remember the number as -10 d -21 h -204 p. This
is the same as -(10
d 21 h 204 p). If you cannot recall this number under
the pressures
of a test, you should remember HOW it is found.
How much LONGER is an average intercalary
(or leap) year than the
average Julian year? To find the length of an average intercalary
year,
multiply 13 months times (29 d 12 h 793 p).
383
d 21 h 589 p (average leap year)
- (365 d
6 h 0 p) (average Julian year)
-----------------------------
18 d 15 h 589 p
In an intercalary year, the Hebrew calendar EXCEEDS the Julian calendar
by +18 d +15 h +589 p. The plus signs are carried
in order to
minimize confusion. As you can see, this number is very easy to
find.
What about the 19 year cycle and the two calendars?
29 d 12 h 793 p
(lunar months) 365 d 6 h 0 p (Roman year)
x 235 months per cycle
x 19 years
-------------------
----------------
6939 d 16 h 595 p
6939 d 18 h 0 p
(length of 235 lunar months)
(length of 19 Roman years)
How much shorter is the 19 year cycle of 235 lunar months than 19
Julian years?
6939
d 18 h 0 p
19 Julian years
- (6939 d 16 h
595 p) 235 average lunations
----------------------
1 h 485 p
Express this as -1 h -485 p. A 19 year cycle is one
hour and 485
parts LESS than 19 Julian years.
Let's summarize these three important numbers.
Be sure you
understand exactly what each means!
---------------------------------------------------------------------
0 d - 1 h - 485 p
19 year cycle is SHORTER than 19 Julian
years.
- 10 d - 21 h - 204 p
average common year is SHORTER than an
average Julian year.
+18 d + 15 h + 589 p
average leap year is LONGER than an
average Julian year.
---------------------------------------------------------------------
You use these three numbers quite like
the three other numbers you
worked with in calculating the day of the week for the molad Tishri.
The only complication is that some of these numbers are negative,
and
you must be certain that you do not overlook a negative number
by
assuming it is positive. It's safer and surer to carry the signs
along
through every step.
Now let's illustrate a calculation of the
day of the month for the
molad Tishri. Be aware of the fact that the years called for in
this
program avoid certain complications, which will be explained in
program
5.
What is the day of the month of the molad
Tishri in 1520 AD?
Proceed as you would for finding the day of the week.
-3761 BC
-(1520) AD
277 cycles elapsed
-------------
-------
-5281; 5281 - 1 = 5280 years elapsed
19 / 5280 years
xxx
-------
17; 3, 6, 8, 11,
14, 17 are
leap years.
6 leap years;
17 - 6 = 11
common years.
For the day of the month, you are determining how far the Hebrew
calendar lags or leads the Roman calendar:
- 1 h 485
p (lag per 19 yr cycle) -10 d -21 h
-204 p
x
277 cycles
x 11 com. yrs.
--------------
------------------------
3395
10 21 204
3395
10 21 204
970
------------------------
--------------
-110 d -231 h -2244 p
-277 h - 134345 p
+ 18 d +15 h + 589 p
x 6 leap years
-------------------------------
+ 108 d +90 h +3534 p
The amount the Hebrew calendar will LAG behind the Roman during
the
elapsed time is the sum of the amounts trailed in the common years
and
the 19-year cycles:
-110 d -231 h
- 2244 p
-277 h -134345 p
----------------------------------
-110 d -508 h
-136589 p Amt. Hebrew calendar LAGS Roman
calendar during elapsed time.
+108 d + 90 h
+ 3534 p Amt. Hebrew calendar leads Roman
calendar during elapsed time.
---------------------------------
-2 d -418
h -133055 p
-22 d
-123 h
--------
------------
24 / -541 h 1080 / -133,055
p
xxx
xxx
--------
------------
-24 d
-13 h
-215 p
-24 d -13 h -215 p represents
the TOTAL AMOUNT that the Hebrew
calendar trails the Roman calendar. (Just reduce the number to
a
convenient negative form; full reduction isn't necessary.) NEVER
divide
this number by 7! To find the day of the month, add the lag you
have
calculated to the bench mark. Since September has 30 days, you
can
express Oct. 6 as Sept. 36.
September 36 23 h 204 p
September 36 22 h 1284 p
+(24 d -13 h -215
p) or
-24 d -13 h -215 p
----------------------------- ------------------------------
September 12 + 9 h +1069 p
The molad Tishri in 1520 AD was on September 12, just before 10 AM.
The procedure for finding the day of the
month for the molad
Tishri is almost exactly parallel to that for finding the day of
the
week. To give you an overview of both calculations, here in schematic
form are the steps. Read this chart from left to right, as well
as down
the page.
-----------------------------------------------------------------------
Molad Tishri
Day of the week Day
of the month
-----------------------------------------------------------------------
Bench mark & required 3761 BC - required
yr 3761 BC - required yr
year. Going from BC (-1)?
(-1?)
to AD? Elapsed time ---------------------
---------------------
in years: (divide by
19)
number of 19 yr cycles times (2d 16h 595p)
times (-1h -485p)
number of common years times (4d 8h 876p)
" (-10d -21h -204p)
number of leap years times (5d 21h 589p)
" (+18d +15h +589p)
------------------- ---------------------
Sum is the molad "ad- Sum is the amount
vancement," or excess that the Hebrew
over full weeks, in the calendar lags (-) or
elapsed time. (Divide leads (+) the Roman
the REDUCED days by 7.) calendar. (Reduce as
necessary.)
Add to the bench mark +(1 d 23 h 204 p) Oct. 6 (Sept 36) 23h 204p
[Corrections, in program 5]
Answer:
(* day of week) (*
day of month)
---------------------- ---------------------
-----------------------------------------------------------------------
In most cases, you will determine the day
of the week first, as
there is less chance for computational errors. This calculation
will
give you a certain number of hours and parts besides the day of
the
week.
Is there any way of knowing that your calculations
are correct?
Yes! The hours and parts for the day of the week of the molad Tishri
must be identical to the hours and parts for the day of the month.
For
1520 AD, the example above, the hours and parts for the day of
the week
are also 9h 1069p. Remember this rule for checking your work.
Since the format involved in calculating
the day of the month is
so very similar to the day of the week, a large amount of additional
practice will not be necessary. You should, however, work out a
few
problems.
3.1 What is the day of the month of the molad Tishri for 1492 AD?
3.2 What is the day of the month of the molad Tishri for 28 AD?
The problems are partially worked out for you below.
3.1 1492 AD is 5252 elapsed years. This is 276 19 year cycles,
and 8
elapsed years. Therefore there are 3 leap years (3, 6, 8) and 5
common
years. The molad was on September 21, 19 hours and 1011 parts.
(This
was a Friday, in case you want to check that calculation, too.)
3.2 28 AD is 3788 elapsed years. 3788 years is 199 19 year
cycles and
7 years. In 7 elapsed years, there are 3 leap years (2, 5, 7) and
four
common years. (If you noticed that these multipliers for the leap
and
common years are the same as in 3.1, you saved yourself some work!)
The
molad was on October 7, 8 hours 760 parts. (This was a Thursday.)
Now that you have almost completed this program, you should feel
confident of your ability to calculate the day of the month for
the
molad Tishri. If you experienced a bit of difficulty with problems
3.1
and 3.2, you may want to work out 1520 AD again, and then compare
each
of your steps with the example already worked above.
3.3 As a final self-test, without consulting the program or
your
notes, determine the day of the month of the molad Tishri for 1448
AD.
The answer is worked out in detail below.
Calculate the day of the month of the molad Tishri for 1448 AD:
-3761
-1448
274 cycles
-------
-------
5209; 5208 elapsed years
19 / 5208
xxx
-------
2 years; no leap years
2 common
-10 d 21 h -204 p
x 2 common
--------------------
-1 h -485 p -20 d
42 h -408 p
274 cycles
----------------
4 1940
7 3395
2 970
-------------------
-274 h -132890 p
-20 d -42 h
-408 p
------------------------
-20 d -316 h -133298 p (no
leap years)
18 h
-123 h
--------
------------
-20 d
24 / -439 h 1080 / -133298 p
-18 d
xxx
xxx
------
------
------------
-38 d
-7 h
- 458 p
Sept 36
+ 23 h
+ 204 p (bench mark)
----------------------------------------------------------
Sept -2
+ 16 h
- 254 p
Sept -2
+ 15 h
826 p
==========
Sept 0
= August 31
Sept - 1 =
August 30
Sept - 2 =
AUGUST 29 day of the month
=========
----------###----------
PROGRAM IV
USING TABLES TO FIND THE MOLAD TISHRI
PROGRAM GOAL 4
Given a table of reduced days; hours, and
parts, for the Hebrew
calendar, you will correctly calculate the day of the week and
the day
of the Roman month for the molad Tishri of selected years. Use
of the
table will considerably shorten the time needed.
Having worked out the problems in the first
three programs of this
series, you may have decided that Hebrew calendar calculations
are more
tedious than difficult.
Yes, some of the arithmetical operations
tend to be time
consuming. Perhaps you feel that you can calculate the molad Tishri
quite well, but you'd prefer having some kind of a desk top computer
just to save time and frustration! Not that all the steps are too
involved for you to do yourself; only the lengthy multiplication
and
division.
Finding the elapsed time is just a quick
subtraction or addition
operation. Expressing the elapsed time in terms of 19 year cycles
still
isn't demanding. The real hang-up comes in the multiplication of
the 19
year cycles, the leap years, and the common years, right?
Most of that step, so far as the multiplication
is concerned, can
be eliminated by using a table of reduced days, hours, and parts.
You
will be able to calculate the molad Tishri in less than half the
time
previously required!
Examine part I of the chart included in
this program. On the left
side of the page as you read it is a table marked "19 YEAR TIME
CYCLES." Toward the bottom of the page are two other tables
"INTERCALARY YEARS" and "COMMON YEARS." (We'll return to the other
table on part I a little later.)
As you suspect, each of these three tables
will replace the
multiplication operations you did before in order to find the excess
over a number of full weeks, or the molad "advancement." Now you
can
read a reduced number from the table very conveniently!
Suppose you have a problem in which the
elapsed time is 200 19
year cycles, three leap (intercalary) years, and five common years.
What will be the excess over full weeks?
200 19 year cycles: 5 d 22 h 200 p
3 leap years: 3 d
16 h 687 p
5 common years: 0 d 20
h 60 p
----------------------------------------
Total:
8 d 48 h 947 p
Excess over full
weeks:
2 d 0 h 947 p
Of course, most of your calculations of
elapsed time will involve
intermediate values of 19 year cycles, which are not directly on
the
first table. What do you do with 189 cycles, 315 cycles, and the
like?
Just add them with the units you already have. If, in the example
above, we had 276 19 year time cycles instead of 200, you would
add the
molad "advancement" for 70 cycles and for 6 cycles:
200 19 year cycles: 5 d 22 h 200 p
276
70 19 year cycles: 6 d 6 h 610 p
cycles
6 19 year cycles: 2 d 3 h 330 p
3 leap years: 3 d 16 h
687 p
5 common years: 0 d 20 h 60 p
---------------------------------------
Total:
16 d 67 h 1887 p
You can add these numbers in less time
than it takes to multiply
276 x (2 d 16 h 595 p). Notice that you can perform all your addition
in one step.
For the 19 year cycles, it may be easier
in some problems to do a
subtraction of two numbers in the table rather than an addition
of
three. The excess over full weeks of 297 19 year cycles can be
found
either way:
add 200: 5 d 22 h
200 p
90: 4 d 1 h 630 p
7: 4 d 19 h 925 p
-----------------------------------
297 cycles: 13 d 42 h 1755 p
reduced: 0 d 19 h
675 p
subtract 300: 1 d 21 h 300 p (1 d 20 h 1380
p)
-3: -1 d -1 h -705 p
-----------------------------------
0 d 19 h 655 p
Now practice using the table to find the
day of the week of the
molad Tishri for:
4.1 1520 AD. The problem is worked out below.
3761 BC
1520 AD
277 cycles
-----------
-------
5281;
5280 elapsed years 19 / 5280
xxx
-------
17 yrs: 3, 6, 8,
11, 14, 17
6 leap
11 common
Using the table:
200 cycles: 5 d 22 h 200 p
70 "
6 d 6 h 610 p
7 " 4
d 19 h 925 p
6 leap: 0 d 9 h
294 p
11 common: 6 d 0 h 996
p
-----------------------------------
Total: 21 d
56 h 3025 p
Excess over full
weeks: 2 d 10 h 865 p
bench mark +( 1 d 23 h 204 p)
------------------------------------
Reduced 3 d
33 h 1069 p
4 d 9 h 1069 p
Molad Tishri: Wednesday 1520 AD
The first time through a calculation with
the chart took more time
than you will need later. But we can even go through a problem
faster
by using the table on the first page called "ELAPSED YEARS IN ONE
19
YEAR CYCLE."
This table combines the leap years and
the common years of a 19
year cycle together, both for dates before 142 AD and for those
after.
Once you divide the elapsed time by 19, you can match the remainder
directly without figuring leap years and common years separately.
If
your remainder is 12 (for a year after 142 AD), the excess over
full
weeks would be 2 d 12 h 724 p.
4.2 Re-calculate the day of the week for the molad Tishri
for 1520 AD
using the combined table. The answer is below.
3761 BC
1520 AD
277 cycles
----------
-------
5281; 5280
years
19 / 5280 years
xxx
-------
17 elapsed years
>From the chart, part I:
200
cycles: 5 d
22 h 200 p
70
" 6 d
6 h 610 p
7 "
4 d 19 h 925 p
17
elapsed years: 6 d 10 h
210 p
-------------------------------------------
Total
21 d 57 h 1945 p
bench mark
1 d 23 h 204 p
----------------------------
22 d 80 h 2149 p
Reduced:
4 d 9 h 1069 p
Molad Tishri: Wednesday.
Notice that you need not reduce until the
last step. If you want,
it's even possible to add the benchmark after the elapsed years
of a
cycle-without sub-totaling first. Once you thoroughly understand
the
concepts involved in calculating the molad Tishri, you can make
these
economies in your work.
(Incidentally, if you choose to add the
bench mark along with the
numbers from the table, your answer -- after reduction -- may take
the
form of 0d xx h xxx p. What day of the
week is 0 d? Just add 7 to 0
d, and you will have the real day -- the sabbath. When you are
working
with the days of the week, "7" is the "additive complement" of
numbers
from -6 to 0. There's nothing mysterious here, just a mathematical
"law".)
Now look at part II of the Hebrew calendar
chart. This section
will assist you in calculating the day of the month. The tables
are
very much like those of part I. The negative numbers represent
the
amount the Hebrew calendar lags behind the Roman calendar for a
given
unit of time. The positive numbers indicate that the Hebrew calendar
leads the Roman.
Because of the different signs involved,
you'll find it easier to
total the negative numbers together before combining the positive.
For
example, how much does the Hebrew calendar trail the Roman after
276
19 year cycles plus 8 elapsed years?
200 cycles:
-12 d - 1 h - 880
p
70 cycles:
- 4 d - 5 h - 470
p
6 cycles:
- 0 d - 8 h - 750
p
--------------------------------------------
276 cycles:
-16 d - 14 h - 2100 p
8 elapsed
yrs. + 2 d - 12 h + 747
p
--------------------------------------------
Total:
-14 d - 26 h - 1353 p
All you need do now is add the bench mark.
The Hebrew calendar is
usually behind the Roman, so the total will be mostly negative
numbers.
The parts or hours COULD be positive in some problems. Watch your
SCRIBAL ACCURACY!
Sept 36 23
h 204 p (bench mark)
-14 d
-26 h -1353 p
-----------------------------
Borrowing:
Sept 35 45
h 2364 p
-14 d
-26 h 1353 p
-----------------------------
Sept 21
19 h 1011 p
Whenever you are making a calculation,
it will be faster to use
the combined table instead of the individual tables of intercalary
years and elapsed years. But you can take either option.
---------------------------------------------------------------------
If you find that calculations of the day of the month are different
from those of the day of the week, be sure that you used the right
table for the right number!
---------------------------------------------------------------------
4.3 Calculate with the use of the chart the day of the month
of molad
Tishri for 1520 AD. The answer is below.
1520 AD
3761 BC
277 cycles
------------
-------
5281; 5280
years
19 / 5280
xxx
-------
17 elapsed years
>From the chart, part II:
200 cycles:
-12 d - 1 h - 880 p
70 cycles:
- 4 d - 5 h - 470 p
7 cycles:
- 0 d -10 h - 155 p
--------------------------------------------
- 16 d -16 h -1505 p
17 years
- 7 d -20 h + 210 p
---------------------------------------------
- 23 d -36 h -1295 p
bench mark Sept. 35
45 h 2364 p (borrowing)
---------------------------------------------
Sept. 12 d 9 h 1069 p
You can set up both the day of the week
and the day of the month
calculations side by side to save space if you like.
4.4 Test yourself on a complete calculation of the molad Tishri,
day
of the week and the Roman date, for 1000 AD. Consult the chart
in an
efficient manner.
1000 AD
3761 BC
250 cycles
-----------
-------
4761; 4760
years
19 / 4760
xxx
-------
10 elapsed years
Day of the week Day of the Roman month
200 cycles: 5 d
22 h 200 p -12 d
- 1 h - 880 p
50 cycles: 1 d
11 h 290 p - 3 d
- 0 h - 490 p
-------------------------
10 elapsed yrs. 6 d 6 h
339 p -15 d - 1 h
- 1370 p
bench mark: 1 d
23 h 204 p -20 d
- 6 h + 339 p
-----------------------------------
-------------------------
Total:
13 d 62 h 1333 p
-35 d - 7 h - 931 p
Reduced:
1 d 15 h 253 p Sept 36
+ 22 h + 1284 p
-------------------------
1 d 15 h 253 p
The hours and the parts agree.
In 1000 AD the molad Tishri was Sunday,
Sept. 1, 15 h 253 p.
By using the Hebrew calendar chart, notice
how much less space on
the page a complete calculation of the molad Tishri now takes.
And
easier, isn't it!
-----------------------------------------------------------------------
THE HEBREW CALENDAR
Tables of Reduced Days,
Hours, and Parts for 19 Year
Time Cycles, Intercalary
Years, and Common Years.
I. Advancement of the Molad Over a Full Number of Weeks
19 YEAR TIME CYCLES
Elapsed Excess over full weeks
Cycles
------- ----------------------
1 2d
16h 595p
2 5
9 110
3 1
1 705
4 3
18 220
5 6
10 815
6 2
3 330
7 4
19 925
8 0
12 440
9 3
4 1035
10 5
21 550
20 4
19 20
30 3
16 570
40 2
14 40
50 1
11 590
60 0
9 60
70 6
6 610
80 5
4 80
90 4
1 630
100 2
23 100
200 5
22 200
300 1
21 300
----------###----------
INTERCALARY YEARS
Elapsed Yrs. Excess over full weeks
------------ ----------------------
1 5d
21h 589p
2 4
19 98
3 3
16 687
4 2
14 196
5 1
11 785
6 0
9 294
7 6
6 883
----------###----------
ELAPSED YEARS IN ONE 19 YEAR TIME CYCLE
Elapsed
Before 142 AD | 142 AD and after
Elapsed
Years Excess over
full weeks | Excess over full weeks Years
------- ----------------------
| ---------------------- -------
1
4d 8h 876p |
4d 8h 876p
1
2
3 6 385 |
1 17 672
2
3
0 15 181 |
0 15 181
3
4
4 23 1057 |
4 23 1057
4
5
3 21 566 |
2 8 853
5
6
1 6 362 |
1 6 362
6
7
0 3 951 |
5 15 158
7
8
4 12 747 |
4 12 747
8
9
1 21 543 |
1 21 543
9
10
0 19 52 |
6 6 339
10
11
5 3 928 |
5 3 928
11
12
2 12 724 |
2 12 724
12
13
1 10 233 |
6 21 520
13
14
5 19 29 |
5 19 29
14
15
3 3 905 |
3 3 905
15
16
2 1 414 |
0 12 701
16
17
6 10 210 |
6 10 210
17
18
5 7 799 |
3 19 6
18
----------###----------
COMMON YEARS
Elapsed Yrs. Excess over full weeks
------------ ----------------------
1
4d 8h 876p
2
1 17 672
3
6 2 468
4
3 11 264
5
0 20 60
6
5 4 936
7
2 13 732
8
6 22 528
9
4 7 324
10
1 16 120
11
6 0 996
12
3 9 792
-----------------------------------------------------------------------
II. Time Differences
19 YEAR TIME CYCLES
Elapsed Cycles Time
Difference
-------------- -----------------------
1
- 0d - 1h - 485p
2
- 0 - 2 - 970
3
- 0 - 4 - 375
4
- 0 - 5 - 860
5
- 0 - 7 - 265
6
- 0 - 8 - 750
7
- 0 -10 - 155
8
- 0 -11 - 640
9
- 0 -13 - 45
10
- 0 -14 - 530
20
- 1 - 4 -1060
30
- 1 -19 - 510
40
- 2 - 9 -1040
50
- 3 - 0 - 490
60
- 3 -14 -1020
70
- 4 - 5 - 470
80
- 4 -19 -1000
90
- 5 -10 - 450
100
- 6 - 0 - 980
200
-12 - 1 - 880
300
-18 - 2 - 780
----------###----------
INTERCALARY YEARS
Elapsed Inc.
Years Time Difference
------------ --------------------
1 + 18d +15h
+589p
2 + 37
+ 7 + 98
3 + 55
+22 +687
4 + 74
+14 +196
5 + 93
+ 5 +785
6 +111
+21 +294
7 +130
+12 +883
----------###----------
ELAPSED YEARS IN ONE 19 YEAR TIME CYCLE
Elapsed Before 142 AD
142 AD and after Elapsed
Years Time Difference
Time Difference Years
------- ---------------
---------------- -------
1
-10d -21h -204p -10d -21h -204p
1
2
+ 8 - 6 +385 -21 -18
-408 2
3
- 3 - 3 +181 - 3 - 3
+181 3
4
-14 0 - 23 -14
0 - 23 4
5
+ 5 - 9 +566 -24 -21
-227 5
6
- 6 - 6 +362 - 6 - 6
+362 6
7
+13 -15 +951 -17 - 3
+158 7
8
+ 2 -12 +747 + 2 -12
+747 8
9
- 9 - 9 +543 - 9 - 9
+543 9
10
+10 -17 + 52 -20 - 6
+339 10
11
- 1 -15 +928 - 1 -15
+928 11
12
-12 -12 +724 -12 -12
+724 12
13
+ 7 -20 +233 -23 - 9
+520 13
14
- 4 -17 + 29 - 4 -17
+ 29 14
15
-15 -15 +905 -15 -15
+905 15
16
+ 4 -23 +414 -26 -12
+701 16
17
- 7 -20 +210 - 7 -20
+210 17
18
+11 - 5 +799 -18 -17
+ 6 18
----------###----------
COMMON YEARS
Elapsed Comm.
Years
Time Difference
------------- -------------------
1
- 10d -21h - 204p
2
- 21 -18 - 408
3
- 32 -15 - 612
4
- 43 -12 - 816
5
- 54 - 9 -1020
6
- 65 - 7 - 144
7
- 76 - 4 - 348
8
- 87 - 1 - 552
9
- 97 -22 - 756
10
-108 -19 - 960
11
-119 -17 - 84
12
-130 -14 - 288
----------###----------
as they are often different; e. g. 1 positive and 2 negative.
- signifies that the Hebrew calendar is behind,
or trails, the Julian calendar.
+ signifies
that the Hebrew calendar is ahead of,
or leads, the Julian calendar.
----------------------------------------------------------------------
PROGRAM V
MAKING ROMAN LEAP YEAR AND JULIAN / GREGORIAN CORRECTIONS
PERFORMANCE GOAL 5
For any required Roman year, AD or BC,
you will correctly make the
necessary adjustments for:
a) the Roman
leap year / common year pattern
b) the conversion
from the Julian to the Gregorian calendar
In your day of the month calculations of the molad Tishri.
By actively working through the first four
programs, you've
learned how to calculate the molad Tishri for the specified years.
Now
you will apply two corrections to the DAY OF THE MONTH calculations.
This will extend your ability to find the molad of Tishri to virtually
any Roman year required, whether AD or BC. Keep in mind that these
corrections affect ONLY the day of the month calculations! The
day of
the week part of your work needs no adjustment.
The first correction involves the length
of a Roman year. On a
calendar, a Roman year has either 365 days (common year) or 366
days
(leap year). In your calculations for the day of the month, you
used
the average length of a Roman year, 365 1/4 days. How do you make
allowance for the difference?
You add six hours to the molad (day of
the month) for every year
after a Roman leap year:
If the required
Roman year is leap, add 0 hours.
If the required
Roman year is one year after a leap year, add
6 hours.
If the required
Roman year is two years after a leap year,
add 12 hours.
If the required
Roman year is three years after a leap year,
add 18 hours.
(Don't become confused by what a "leap
year" means. A Hebrew leap
year has 13 months; a Roman leap year has 366 days. The context
will
tell you what applies.)
How can you know if the required Roman
year is leap or common?
Simply divide the year by four and note the remainder: 10 AD divided
by
4 gives a remainder of two; 51 BC divided by 4 gives a remainder
of
three; 1977 AD divided by 4 leaves a remainder of one.
For AD years, the remainder numerically
corresponds to the Roman
leap year/common year pattern:
A remainder of 0 signifies that the required
Roman year is leap.
"
1 "
one after a leap.
"
2 "
two after a leap.
"
3 "
three after a leap.
BC years have a different pattern of remainders.
Since 4 AD is a
leap year, 1 AD is three years before a leap year. 1 BC is a leap
year,
too, being four years before 4 AD. Four years before 1 BC is 5
BC, a
leap year. And 9 BC is a leap year. Then 8 BC is one year after
a Leap
year; 7 BC is two years after; 6 BC is three after. If you divide
these
years by 4, you can find the remainder that corresponds to a leap
year
(9 divided by 4 leaves a remainder of 1), a year after a leap year,
etc. This pattern of remainders from 9 BC to 6 BC will be valid
for
years further back into antiquity.
The chart below summarizes the remainder
patterns for both BC and
AD Roman years.
Roman leap year corrections (dividing the
required year by 4)
=======================================================================
AD:
If the remainder is 0, add no hours to the day of month
calculations
"
1 6 hours
"
2 12 hours
"
3 18 hours
BC:
If the remainder is 1, add no hours to the day of month
calculations
"
0 6 hours
"
3 12 hours
"
2 18 hours
-----------------------------------------------------------------------
To insure that you can apply this correction
to the day of the
month, indicate how many hours you would add for each of the following
years.
5.1 1520 AD 5.10 1917 BC
5.2 1974 AD 5.11 4 BC
5.3 1983 AD 5.12 971 BC
5.4 1871 AD 5.13 1181 BC
5.5 1699 AD 5.14 1020 BC
5.6 31 AD 5.15 1984 AD
5.7 33 AD 5.16 1020 AD
5.8 142 AD 5.17 2001 BC
5.9 1486 BC 5.18 2001 AD
The answers appear below.
5.1 0 hours
5.7 6 hours
5.13 0 hours
5.2 12 hours
5.8 12 hours
5.14 6 hours
5.3 18 hours
5.9 18 hours
5.15 0 hours
5.4 18 hours
5.10 0 hours
5.16 0 hours
5.5 18 hours
5.11 6 hours
5.17 0 hours
5.6 18 hours
5.12 12 hours
5.18 6 hours
When you are checking your calculations
for the molad of Tishri,
you may find that the day of the month calculation differs from
the day
of the week by six hours, twelve hours, or eighteen hours. Here's
what
has happened: You forgot to make the Roman leap year correction
in the
day of the month calculation! Divide the Roman year by 4 and inspect
the remainder on EVERY calculation involving the day of the month.
----------###----------
The second correction to the day of the
month calculation of the
molad Tishri concerns the corresponding Gregorian calendar date
of a
day on the Julian calendar. By understanding the reasons for the
change
from the Julian calendar to the Gregorian, you will have no difficulty
in performing this correction.
The Julian calendar was named after Julius
Caesar; the Gregorian
calendar after Pope Gregory XIII. Both calendars are "Roman." The
Gregorian calendar is what you use every day.
When did the Roman calendar, based upon
an average yearly length
of 365 1/4 days, come into use? The answer is 45 BC. With the aid
of
the Egyptian astronomer Sosigenes, Julius Caesar completely revised
the
previous lunisolar calendar, which had drifted badly with respect
to
the seasons. To effect the reform, "46 BC" had 445 days assigned
to it
in order to correct for all the previous errors. That year was
very
appropriately called the "year of confusion!" In 45 BC the vernal
(spring) equinox occurred on March 25.
The Julian calendar wasn't without its
faults, however. The
average Julian year was eleven minutes and fourteen seconds LONGER
than
a "tropical year." (A tropical year is measured from one vernal
equinox
to the next.) After 128 years the Julian calendar had an extra
day,
compared to an equal number of tropical years.
By the time of the famous Council of Nicea
in 325 AD, the Julian
calendar was about three days behind the tropical year. This meant
that
the vernal equinox was several days "early" on the Julian calendar.
Accordingly, the churchmen based their rules for the date of Easter
on
the spring equinox falling on March 21 of the Julian calendar.
It had
been March 25 in 45 BC.
Towards the end of the Counter-Reformation
in the sixteenth
century, the spring equinox had "crept back" to about March 11.
To
alleviate the problem, Pope Gregory reformed the Julian calendar
by
deleting ten days from the month of October 1582: The day after
October
4 officially became October 15. So after 1582 the spring equinox
shifted back to March 21--where it had been during the time of
Constantine the Great. (Of course, the equinox never shifted; the
calendars did the moving!)
Pope Gregory also invoked a new rule concerning
Roman leap years.
The Julian calendar considers every fourth year as having 366 days.
To
shorten the new Roman calendar by three days in 400 years, the
Pope
declared that century years NOT EVENLY DIVISIBLE BY 400 would remain
COMMON (no February 29).
Therefore, 1700 AD, 1800 AD, and 1900 AD,
which are not evenly
divisible by 400, were common years according to the Gregorian
calendar
rules. By the old Julian calendar, they would have been leap years.
Instead of the eleven minute error in the
average Julian year, the
average Gregorian year is only 26 seconds too long.
How do you convert a Julian date into a
day on the Gregorian
calendar? Just add the TOTAL DAYS that have been dropped from the
Julian! Before 1582 AD no correction is needed, as all the Roman
dates
are understood to be Julian. But after 1582 AD, you must add at
least
ten days to the Julian day of the month calculation for the molad
Tishri.
Bear in mind that 1600 AD was a leap year
in both the Julian and
the Gregorian calendars. The difference between the calendars remained
ten days until 1700. The Gregorian calendar omitted February 29
that
year (because the year wasn't evenly divisible by 400), while the
Julian retained the extra day. From 1700 to 1799, the Julian calendar
was eleven days behind the Gregorian since that was the total number
of
days dropped.
In the same manner, from 1800 to 1899 you
must add twelve days to
the Julian date to find the corresponding Gregorian day. During
the
1900's and 2000's, you add thirteen days.
How many days will you add to your day
of the month calculations
of the molad of Tishri to convert from the Julian calendar to the
Gregorian calendar for each of the following years?
(Before 1582 AD,
simply respond "0 d".)
5.19 1520 AD
5.20 1601 AD
5.21 1798 AD
5.22 1914 AD
5.23 1851 AD
5.24 1666 AD
5.25 1583 AD
5.26 1984 AD
5.27 2001 AD
5.28 2145 AD
5.29 1979 AD
5.30 1712 AD
The answers appear below.
5.19 + 0 d 5.25 +10 d
5.20 +10 d 5.26 +13 d
5.21 +11 d 5.27 +13 d
5.22 +13 d 5.28 +14 d
5.23 +12 d 5.29 +13 d
5.24 +10 d 5.30 +11 d
* * *
Neither of the corrections to the day of
the month calculation of
the molad Tishri is complicated. However, you must remember to
perform
them whenever the Roman year calls for such adjustments. The last
problem in this program will challenge you to put together all
the
calendar skills you have learned thus far.
5.31 What is the day of the week and the date of the
Roman month for
the molad Tishri in 2055 AD? (You may use the chart of reduced
numbers.) This problem is worked out in detail for you on the next
page.
What is the day of the week and the date of the Roman month for
the
molad of Tishri in 2055 AD?
A. The ELAPSED TIME is: - 3761
- 2055
-------
- 5816 - subtract 1 going from BC to AD:
5815 years.
B. 5815 years is 5815 / 19 nineteen year cycles.
306
-------
19 / 5815 306 cycles and the first year of
the
57 next cycle have elapsed.
------
115
114
------
1
C. THE DAY OF THE WEEK FOR THE MOLAD TISHRI:
From the tables, find the advancement of the molad.
300 19-year cycles:
1 d 21 h 300 p
6 19-year cycles:
2 d 3 h 330 p
1 year of next cycle:
4 d 8 h 876 p
------------------
7 d 32 h 1506 p
Reducing
this:
+1 h -1080 p
------------------
7 d 33 h 426 p
+ 1 d -24 h
------------------
The advancement
is: 1 d 9 h 426 p
over full number
of weeks.
Add the
advancement to
the bench
mark: +(1 d 23 h
204 p)
------------------
2 d 32 h 630 p
This reduces to: 3 d 8 h 630 p
The molad was on the third day of the week, Tuesday.
D. THE DAY OF THE ROMAN MONTH:
>From the tables, find how far behind the Roman calendar the Hebrew
calendar is:
300 19-year cycles:
-18 d - 2 h - 780 p
6 19-year cycles:
- 0 d - 8 h - 750 p
1 year of next cycle:
-10 d -21 h - 204 p
--------------------------
-28 d -31 h -1734 p
The benchmark is:
Sept. 36 23 h 204 p
"Borrow" 2 x 1080 parts, and one day:
Sept. 35 45 h 2364 p
Add the time difference in the calendars to the benchmark:
+(-28 d -31 h -1734 p)
--------------------------------
Sept. 7 14 h 630 p
Two corrections must be made, that of the ROMAN leap year, and the
Julian / Gregorian:
513
-------
4 / 2055
20
-------
5
3 years after a Roman leap year means
4
you must add 18 HOURS.
---
15
12
---
3
During the 1900's, the correction to the Julian calendar is 13 days.
Since 2000 is evenly divisible by 400, the correction during the
21st
century is STILL 13 days.
Sept. 7
14 h 630 p
+18 h Roman
leap year correction
13 d
Julian / Gregorian correction
--------------------------
Sept. 20
32 h 630 p
Molad
Tishri: Sept. 21 8 h 630 p
parts and hours agree
with day of week
----------###----------
PROGRAM VI
APPLYING THE POSTPONEMENT RULES TO FIND TISHRI ONE
PERFORMANCE GOAL 6A: Given the molad Tishri
for a required year,
you will correctly determine the day of the week and the day of
the
month of the Feast of Trumpets, Tishri 1, by applying from memory
the
four postponement rules explained in this program.
Example: In 1987, the molad of Tishri will be:
September 23 Wednesday (4 d) 3 h 77 p
What is the date for the Feast of Trumpets?
Answer: September 24 Thursday (by rule two)
PERFORMANCE GOAL 6B: Without consulting
notes, you will correctly
explain (in a brief paragraph for each rule) why each postponement
for
Tishri 1 is important for the Hebrew calendar.
PERFORMANCE GOAL 6A: After correctly calculating
both the day of
the week and the day of the month for the molad of Tishri the next
step
is to find the date of Tishri 1, the Feast of Trumpets. You will
analyze the time of the molad to see how four postponement rules
affect
the Feast of Trumpets.
At the outset, you should understand that
a conjunction of the
earth, moon, and sun takes place completely apart from man's doings.
The MOLAD CANNOT BE POSTPONED BY HUMAN ENACTMENTS! Tishri 1, the
civil
New Year in the Hebrew calendar, is what is postponed.
Here are the four postponement rules, prefaced
by a general
statement of the case where there is no postponement:
When the molad Tishri occurs at a time
of the week that's
unaffected by the four postponement rules, the Feast of Trumpets
is on
the SAME DAY as the molad.
Rule one: When the molad of Tishri occurs
AT NOON OR LATER (12 h
0 p or more in your calculations), the Feast of Trumpets is postponed
until the next day.
Rule two: When the molad of Tishri OR a
postponement occurs on a
Sunday, Wednesday, or Friday, the Feast of Trumpets is postponed
one
day, to a Monday, Thursday, or Sabbath, respectively.
Rule three: When the molad of Tishri of
a COMMON YEAR is on a
Tuesday, at or after (3 d) 3 h 204 p the Feast of Trumpets
is
postponed to a Wednesday, and by Rule two, further postponed
to a Thursday.
RULE FOUR: When the molad of Tishri of
a COMMON year IMMEDIATELY
FOLLOWING an intercalary year occurs on a Monday, at or after (2
d) 9
h 589 p the Feast of Trumpets is postponed to a TUESDAY.
You should take notice that the maximum
postponement of Tishri 1
is two days. Also realize that the order of these rules is significant.
They are easiest to apply in the numerical order above. Rule one
governs all afternoon time periods, so rules three and four ONLY
AFFECT
A PORTION OF THE MORNINGS.
You will quickly discover that rules one
and two are by far the
most frequently used, both individually and together -- one, two,
or
one and two. And years where no postponement of Tishri 1 is required
are rather common, too.
This means, of course, that rules three
and four are quite
infrequent, especially the last one. Rule three has rule two "built
in", since it is a very specific case. Therefore, rule three and
rule
four are always separately used. Whenever rules one and two apply,
either separately or in combination, you know in advance that rules
three and four will not be involved. All this speeds up your use
of
these rules considerably.
Before you see examples of each rule applied,
one more thing needs
to be explained. How do you know when the year in question is common,
and when does it immediately follow a leap year?
Remember how you determined elapsed time
by adding (or
subtracting) the required year from the bench mark? Well, the year
in
question is the year AFTER the number of elapsed years. The distinction
between the terms "elapsed years" and "required year" is quite
similar
to your age. All through your 21st year of life, your age is twenty!
What do you do to find which year of the
cycle is a required year?
Add 1 to the remainder you already found when you divided the elapsed
years by 19. This new remainder will be the number of the year
in the
cycle. If the remainder for elapsed years is 10, the required year
is
year 11 of the l9-year cycle.
You will recall that for the years 142
AD (handwritten note: 256
AD -- Dr. Hoeh correct date) and after, the intercalary years
are 3,
6, 8, 11, 14, 17, 19. The years immediately following these years
are
the ones you are concerned with for rule four: 4, 7, 9, 12, etc.
All
the years in the cycle which aren't intercalary are common.
Whenever you are working with years earlier
than 142 AD, be sure
to use the proper intercalary years, which are one year earlier
than
the present cycle: 2, 5, 7, 10, 13, 16, 18. The year after these
leap
years will then apply for rule four.
Now you will
learn how to analyze the molad Tishri to apply
the postponement rules.
1492 AD: Molad Tishri was September 21, Friday 6 d 9 h 1011 p.
19 h 1011 p is after 12 h 0
p; RULE ONE postpones to the
Sabbath. The Sabbath is a permissible day; no further postponement.
Feast of Trumpets was September 22, Sabbath.
1584 AD: molad Tishri was September 5, Wednesday 4 d 2 h 852 p.
2 h 852 p is before 12 h 0
p; rule one not involved. Wednesday
is a forbidden day; RULE TWO applies.
Feast of Trumpets was September 6, Thursday.
1615 AD: molad Tishri was September 22, Tuesday 3 d 20 h 804 p.
20 h 804 p is after 12 h 0
p; RULE ONE postpones to Wednesday.
Wednesday is a forbidden day; RULE TWO re-postpones to Thursday.
Trumpets was September 24, Thursday.
1632 AD: molad Tishri was September 14, Tuesday, 3 d 6 h 1014 p.
6 h 1014 p is before 12 h 0
p. Rule one doesn't apply. Tuesday
is a permissible day. Rule two doesn't apply. 6 h 1014 p
is after 3 h
204 p. Rule three could be involved. Is 1632 a common year?
The elapsed time from 3761 BC is 5392
years to 1632 AD. Then the
year in question is the 5393rd year, or the 16th year of a cycle
after
283 full cycles. This is a common year. RULE THREE does apply.
Tishri 1 is September 16, Thursday.
841 AD: molad Tishri was September 19, Monday, 2 d 11 h 735 p.
Rules one, two, and three, don't apply.
The molad is on a Monday,
and 11 h 735 p is after 9 h 589 p. Is 841 AD a common
year
immediately after a leap year? From 3761 BC to 841 AD is 4601 years;
the required year is the 4602 year. This is 242 19-year cycles
plus 4
years of the next. This is a common year immediately following
a leap
year. RULE FOUR applies.
Trumpets was September 20, Tuesday.
1910 AD: molad Tishri was October 4, Tuesday, 3 d 0 h 61 p.
0 h 61 p is before 12 h 0 p;
rule one doesn't apply. Tuesday is
a permissible day; rule two doesn't apply. 0 h 61 p
is before 3 h
204 p; rule three doesn't apply. Rule four can't apply to Tuesday.
No
postponement.
Trumpets is October 4, Tuesday.
As you can see, only a few seconds are
needed to check a molad of
Tishri for applicable postponements. Don't under-estimate the
importance of doing this operation correctly, however! All of the
other
Holy Days are derived from the date of Tishri one!
Practice applying the postponement rules
to the following molads.
The correct dates for the Feast of Trumpets are listed on the next
page, along with the postponement rules which are needed.
6a.1 1264 AD September 22, Monday 2 d 13 h 351 p
6a.2 1255 AD September 3, Friday 6 d 3 h 95 p
6a.3 1259 AD September 18, Thurs. 5 d 15 h 865 p
6a.4 2001 AD September 17, Monday 2 d 22 h 106 p
6a.5 2008 AD September 30, Tuesday 3 d 1 h 1057 p
6a.6 2014 AD September 24, Wednesday 4 d 8 h 339 p
6a.7 1984 AD September 25, Tuesday 3 d 11 h 976 p
6a.8 1985 AD September 14, Sabbath 7 d 20 h 772 p
6a.9 462 AD September 10, Monday 2 d 1 h 511 p
6a.10 496 AD September 23, Monday 2 d 10 h 644 p
6a.11 134 AD October 5, Monday 2 d 23 h 343 p
6a.12 118 AD October 2, Sabbath 7 d 21 h 1009 p
6a.13 588 BC September 27, Tuesday 3 d 3 h 209 p
6a.141 953 BC September 12, Sunday 1 d 12 h 4 p
Answers on following page:
Answers: Rule
6a.1 September 23, 1264 AD; Tuesday 1
6a.2 September 4, 1255 AD; Sabbath 2
6a.3 September 20, 1259 AD; Sabbath 1, 2
6a.4 September 18, 2001 AD; Tuesday 1
6a.5 September 30, 2008 AD; Tuesday ---
6a.6 September 25, 2014 AD; Thursday 2
6a.7 September 27, 1984 AD; Thursday 3
6a.8 September 16, 1985 AD; Monday 1, 2
6a.9 September 10, 462 AD; Monday ---
6a.10 September 24, 496 AD; Tuesday 4
6a.11 October 6, 134 AD; Tuesday 1
6a.12 October 4, 118 AD; Monday 1, 2
6a.13 September 29, 588 BC; Thursday 3
6a.14 September
13, 953 BC; Monday 1
* * * *
* *
PERFORMANCE GOAL 6B: You do not need to understand the reasons behind
the postponement rules in order to correctly apply them. However,
your
appreciation of the Sacred calendar will be enhanced by grasping
the
purpose of each rule.
WHY RULE ONE, NOON OR AFTER POSTPONEMENTS?
Whenever the molad of Tishri occurs at
noon or after (12 h 0 p),
Tishri one is postponed to the next day (at least). But how did
the
molad come to be associated with noon? Rule one points back to
the
initial formulation of the Hebrew calendar -- in the days of Seth,
according to Josephus. God intended that the heavenly bodies would
intrigue man to study their movements carefully. "Let there be
lights
in the firmament of the heaven to divide the day from the night;
and
let them be for signs, and for seasons, and for days, and for years"
(Gen. 1:14).
As you may have observed, the sun, moon
and stars rise and set
each day. But the sun rises each day about four minutes later with
respect to the stars. It has a proper motion of its own, independent
of
the stars. This motion is not due to the daily rotation of the
earth on
its axis, which gives the illusion that the sun moves westward
each
day. The apparent path of the eastward journey of the sun through
the
stars, known as the "ecliptic", is due to the earth's annual orbital
revolution.
The earth is tilted 232 degrees on its
axis. By studying the
diagram below, you can see that the plane of the earth's equator
is
inclined 232 degrees from the plane of the earth's orbit. Astronomers
call the projection of the earth's equator into space the "celestial
equator".
---------------------
(NOTE: To view the diagram mentioned above, see the file HEB-CAL1.TIF
in the Images\OtherWCG directory.)
---------------------
Seasons occur because the earth's orbital
plane (the plane of the
earth's orbit) is not coincident with the plane of the earth's
equator.
The vernal equinox takes place when the apparent northward moving
sun
crosses the celestial equator. It again intersects the celestial
equator when the sun passes below the celestial equator moving
southward--at the time of the autumnal equinox. On the days of
the
equinoxes, the day and night periods are equal.
Consider the next diagram, which is a closeup
of the vernal
equinox. The moment of the equinox occurs when the center of the
sun
crosses the celestial equator. But the sun's apparent diameter
is 1/2
degree; therefore the TRAILING EDGE of the sun is still 1/4 degree
behind the equinoctial point.
---------------------
(NOTE: To view the diagram mentioned above, see the file HEB-CAL2.TIF
in the Images\OtherWCG directory.)
---------------------
How long will it take for the trailing
edge of the sun to pass the
point of the vernal equinox? The sun's eastward progress along
the
ecliptic is about 1 degree per day (since the earth revolves 360
degrees around the sun in about 365 days). In 24 hours the sun
moves 1
degree; in 12 hours it moves 1/2 degree; in 6 HOURS it moves 1/4
degree.
Six hours before sunset is 12 noon. Unless
the equinox occurs
BEFORE noon, the sun's trailing edge will not have passed the celestial
equator by sunset! When the equinox occurs at noon or later, Sol
is
still a winter sun. The first day of spring is therefore assigned
to
the following day.
This is one reason why noon became a logical
demarcation point for
time in astronomical matters. Noon is a stable observation point
for
time, irrespective of the observer's latitude. (The time of sunrise
and
sunset, but not noon, vary considerably during the course of a
year,
especially in the extreme latitude.) From earliest known times
until
1925, astronomers had traditionally used noon as a reference point
for
the day. Thus noon became in antiquity a limit point not only for
the
equinox, but also for the molad.
Just as noon arbitrated the first day of
spring and the first day
of fall, it served a similar function with the assigned day of
the
molad -- that is, the molad or conjunction of the moon had to have
a
natural and arbitrary limit, in this case noon.
WHY RULE TWO, FORBIDDEN DAYS?
This postponement rule prevents Holy Days from falling on a Sunday
(during the fall) and the Passover occurring at awkward times:
If Trumpets could occur on a Wednesday,
the Day of Atonement,
Tishri 10, would fall on a Friday, the preparation day for the
Sabbath!
And the Passover would also be observed Saturday night, a most
difficult time.
If Trumpets could occur on a Friday, the
Day of Atonement would be
on a Sunday, the day after the Sabbath.
If Trumpets could occur on a Sunday, then
the first day of the
Feast of Tabernacles, as well as the last Great Day would be on
a
Sunday.
WHY RULE THREE, THE TUESDAY-COMMON YEAR POSTPONEMENT?
As you recall, the maximum length of a common year in the Hebrew
calendar is 355 days. Without this rule, a common year might have
356
days!
Anytime in the morning of a Tuesday, 3h 204 p is affected
by this
rule. Why this particular moment?
Start with Tuesday
3 d 3 h 204 p
Add an average
+ ( 4 d 8 h 876 p) the excess over full
common year
----------------------- number of weeks
7th 11 h 1080 p
This number is the same as the seventh
day 12 hours and no parts.
This means that the next molad Tishri would occur on a Sabbath
at noon.
But rule one says you must postpone to a Sunday, and rule two says
to
re-postpone until a Monday.
From Tuesday until Monday is six days.
The full number of weeks in
a common year (50) gives 7 x 50 = 350 days. So that year would
have 356
days -- if rule three didn't intervene.
WHY RULE FOUR, THE MONDAY-COMMON YEAR FOLLOWING LEAP YEAR POSTPONEMENT?
The minimum number of days an intercalary
year can have is 383
days. If rule four weren't in effect, some leap years would have
only
382 days. An average leap year has 5 d 21 h 589 p over
a full number
of weeks, which is 54.
Suppose that a common year just after a
leap year began on the
same day as the molad, Monday 2 d 9 h 589 p. When would
the preceding
leap year have begun?
Monday
2 d 9 h 589 p
average leap yr. -(5
d 21 h 589 p) excess over full number of
------------------- weeks.
- 3 d -12 h
-3 d -12 h is the same as -4 d
+12 h. Whenever you are working
with negative days of the week, you merely add 7 d to find out
what day
of the week you are actually on. -4 d +7 d is the 3 d of
the week, or
Tuesday. The molad of the leap year thus occurred on a Tuesday
at noon.
But rule one says that this must be postponed to a Wednesday, and
by
rule two, re-postponed to a Thursday.
If the leap year began on a Thursday and
ended on a Monday, there
would be four days over the full number of weeks. 7 x 54 weeks
is 378
days. 378 days plus 4 days is only 382 days; less than the minimum.
Without these postponement laws, the sacred
calendar would be in a
perpetual state of confusion. Holy Days would fall on a Sunday.
The
lengths of years would be irregular. Calendar reformers would be
tempted to tamper with the sacred calendar more often. Picture
the
difficulty of a deacon trying to keep a Sabbath Day holy while
frantically making last minute preparations for the Passover ceremony!
But all that turmoil is avoided by four
very simple and easily
applied postponement rules. Instead of the Sacred Festivals being
subordinate to the Hebrew calendar the latter serves the Holy Days.
----------###----------
PROGRAM VII
COUNTING THE DAYS OF THE WEEK AND THE DAYS OF THE MONTH
PERFORMANCE GOAL 7A:
You will count forward or backward a specified
length of time from
a given day of the week and correctly determine the new day of
the
week.
For example:
What day of the week is 164 days before
Thursday?
Answer: Monday.
PERFORMANCE GOAL 7B:
Given the date and the day of the week
for any one day in a Roman
or Hebrew month, you will correctly indicate:
a) The day of the week for any other given
date in that month, or
b) the dates for any specified day of
the week in that month.
For example:
a) If the 13th of a month is on a Friday,
what day of the week
would the 4th be?
Answer: Wednesday.
b) What are the dates for Tuesday in a
month when a Monday in that
month is on the 9th?
Answer: 3, 10, 17, 24, (31).
COUNTING THE DAYS OF THE WEEKS 7A
Counting the days of the week is so simple
that you're probably
inclined to use "finger-calculations" quite often! Despite the
apparent
lack of complexity, however, mistakes can easily creep in. Wouldn't
you
prefer a less embarrassing technique than finger-counting to explain
why a Holy Day occurs when it does? A more dignified demonstration
is
possible!
All of the Sacred Festivals of God are
anchored to Tishri one. For
example, the Day of Atonement is ten days after Trumpets. Passover
is
164 days before. Accurate determination of the day of the week
of the
Festivals depends on correct counting procedures. Basic arithmetic
is
all that's needed.
As you recall from Program 3, Calculating
the day of the week of
the molad Tishri, multiples of full weeks do not affect the final
day
of the week involved. Only the excess, or remainder, is significant.
Just as seven days from Monday is still a Monday, 147 days (twenty
one
weeks) before or after Monday is still a Monday.
In the same manner as Program 3, equate
the days of the week with
the numbers 1 to 7. Sunday is the first day, Wednesday is the fourth
day, the Sabbath is the seventh day.
Days AFTER a reference day are ADDED to
the numerical equivalent
of the reference day. Days BEFORE the reference day are SUBTRACTED
from
the numerical equivalent. You will find it much easier if you take
out
full multiples of weeks before you add or subtract.
Here are some examples. On what day of
the week will Atonement
occur when Trumpets is on a Monday? Atonement (Tishri 10) is nine
days
after Trumpets, or one week and two days later. Monday is the second
day of the week.
2 + 2 = 4 The fourth day of the week is Wednesday.
On what day of the week will Passover occur
when Trumpets is on a
Sabbath? Passover is 164 days before Trumpets. Divide 164 by 7,
and you
can express the time interval as 23 weeks and 3 days.
7 -
3 = 4 The fourth day is Wednesday. (It's observed
Tuesday night.)
What happens when you want to find the
day of Passover if Trumpets
were on a Monday? Arithmetically, the problem is simply 2
- 3 = -1.
What day of the week does this negative number indicate? Simply
ADD 7
to any negative number in these calculations of the day of the
week:
-1 + 7 = 6 The sixth day of the week is Friday.
Here is an alternative way of expressing
the same reasoning. First
indicate the total time in terms of full weeks and days over a
full
week As an illustration, Passover is always 164 days before Trumpets.
This is the same as 23 weeks and 3 days before. If Trumpets occurs
on
Tuesday, 23 weeks before the Tuesday of Trumpets is another Tuesday.
What is three days before Tuesday? One day before is Monday; two
days
before is Sunday; three days before Tuesday is the Sabbath. Passover
is
on a Sabbath, observed Friday evening.
Pentecost always occurs on a Sunday because
of inclusive
reckoning. Leviticus 23:15 states that Pentecost is counted "from
[beginning with] the morrow after the Sabbath, from the day that
ye
brought the sheaf of the wave offering ..." The wavesheaf offering
was
performed on Sunday, the day after the weekly Sabbath. With a Sunday
the first day, seven full weeks (or forty-nine days) takes one
to the
fiftieth day, again on a Sunday.
By counting inclusively, 50 days is seven
weeks. If you divide
seven weeks by seven, you have no remainder. Sunday is the first
day of
the week, so arithmetically you have 1 + 0 =
1, or Sunday, as you
started.
Practice finding the day of the week in
the following problems.
Set them up in terms of a simple arithmetical expression.
7a.1 What is 17 days after a Monday?
7a.2 What day is 6 days before Tuesday?
7a.3 What is 50 days after a Saturday.
7a.4 On what day does Passover occur if Trumpets is on a Tuesday?
7a.5 What day is 53 days before Thursday?
7a.6 If Passover is on a Friday (Thursday evening), when does
Atonement occur?
The answers are on the next page.
7a.1 17 days = 2 weeks, 3 days. Monday is the second day of
the week.
2 + 3 = 5; Thursday.
7a.2 Tuesday is the third day of the week. 3 - 6 = -3. Add
7 to change
this to a real day: -3 + 7 = 4; Wednesday
7a.3 50 days = 7 weeks, one day. Sabbath is the seventh day.
7 + 1 =
8. Divide by seven and look at the remainder: 1;
Sunday.
7a.4 164 days = 23 weeks, three days. Tuesday is the third
day. 3 - 3
= 0. Convert this to a real day by adding seven:
0 + 7 = 7; Sabbath.
7a.5 53 days is 7 weeks, 4 days. Thursday is the fifth day.
5 - 4 = 1;
Sunday.
7a.6 Passover and Trumpets are separated by 23 weeks, 3 days.
Atonement is nine days after Trumpets, or one week, two days later.
Passover is on the sixth day. Then Trumpets is on:
6 + 3 = 9.
Eliminate the seven (9-7): 2; Monday.
When Trumpets is on Monday, Atonement
will be:
2 + 2 =
4; Wednesday.
COUNTING THE DAYS OF THE MONTH, 7B
Hebrew months have either 29 or 30 days.
Most months on a Roman
calendar contain 30 or 31 days. As you recall from the form of
most
printed calendars, the same days of the week are placed over top
one
another.
For example, take a month of 30 days whose
first day is Sunday. It
will appear:
Sun Mon Tue Wed Thu Fri Sat
1 2 3 4
5 6 7
8 9 10 11 12
13 14
15 16 17 18 19
20 21
22 23 24 25 26
27 28
29 30
Take particular notice of the sequence
in the columns (up and down
the page). The Tuesday column contains all the days of the month
which
fall on Tuesday. They are, in this case, 3, 10, 17, 24.
What number separates these days of the
month? Seven. One week
from the third day of the month is 7 days later, or 3 + 7 days
into the
month. One week after the 10th is 10 + 7, or the 17th, etc. Learn
to
count by sevens for numbers up to 31.
Not every month will begin with the first
day of the week, of
course. You still can utilize the sequence of numbers separated
by
seven no matter what day of the week the month commences.
7b.1 What is the sequence of days of the month which occur on the
Sabbath when the first Sabbath is on the
2nd?
7b.2 What is the sequence of days of the month that fall on Monday
when
the first day of the month is a Friday?
7b.3 If one of the Tuesdays of the month is the 22nd, what would
be the
other days of the month for Tuesdays?
7b.4 What are the days of the month for the Sabbath in a month where
one of them is on the 16th?
7b.5 The month of AB has 30 days. The 10th of Ab is a Thursday.
On what
days of the month will the Mondays occur?
7b.6 The month of Thammuz has 29 days. The 6th of Ab is a Sabbath.
On
what days of the month will the Wednesdays
occur?
7b.7 If the 14th of a month is on Wednesday, what day of the week
would
the 26th be?
7b.8 If the 22nd of the month is on Sunday, what day of the week
would
the 10th be?
The answers to these problems are below.
7b.1 2, 9, 16, 23, 30 are Sabbaths.
7b.2 Friday, the 6th day of the week, is the first of
the month. Then
Monday,
three days after Friday, is the fourth. The Mondays
are: 4,
11, 18, 25.
7b.3 The 29th is the last Tuesday in the month. To find
the others,
count backward
"by sevens" from the 22nd: 15, 8, 1.
7b.4 16, 23, 30; 9, 2. These are the Sabbaths.
7b.5 10th = Thursday; Monday is 4 days later, or the
14th. 7, 14, 21,
28 are the
Mondays.
7b.6 Sabbath = 6th of the month. Wednesday is three
days earlier in
the week,
or the 3rd: The Wednesdays in Thammuz (in this
case) are
3, 10, 17, 24.
7b.7 Two weeks after the 14th is the 28th, a Wednesday.
The 26th is
two days
before the 28, so it is a Monday.
7b.8 Two weeks before the 22nd is the 8th, a Sunday.
The 10th will be
two days
later, or Tuesday.
----------###----------
PROGRAM VIII
DETERMINING THE DATES OF THE ANNUAL FESTIVALS
PERFORMANCE GOAL 8:
Given the day of the week and the Roman
date for the Feast of
Trumpets, you will correctly determine the day of the week and
the day
of the Roman month for each of the other annual Festivals in the
Roman
year. These Festivals are listed in Leviticus 23.
One of the most practical aspects of studying
the Hebrew calendar
is understanding the layout of Sacred Festivals. Obviously, if
the
Western World were using the Hebrew calendar for routine business
affairs, there would be no need for transforming the dates of Trumpets,
Atonement, Passover, etc., to the Roman calendar. But since this
isn't
the case, one must know what day of the Roman calendar each Holy
Day
corresponds.
Program 8 builds upon the calculation skills
you have learned in
program 7. Take time to review the previous program as is necessary
for
you.
In order to understand the relationship
of the dates of the Holy
Days, let's summarize the calendrical information given in Leviticus
23
Reference Festival
Hebrew Date
=======================================================================
Lev. 23:5 Passover
Nisan 14; 164 days before Tishri 1
:6-7 1st
day of
Unleavened Bread Nisan 15
:8
7th day of
Unleavened Bread Nisan 21
:15-16 Pentecost
50 days beginning with the Sunday of
the wavesheaf offering, (which is
the day after the [a] regular weekly
Sabbath) [ NOT AN ANNUAL SABBATH ]
[ THE SUNDAY WHICH IS ] during the
Days of Unleavened Bread. (Always a
Sunday.)
[ See the WWN 5-11-87 & GN 6-74 for
comments in brackets ]
:24
Trumpets
Tishri 1
:27
Atonement Tishri
10 (9 days after Trumpets)
:34-35 1st day of
Tabernacles Tishri 15
:36
Last Great Day Tishri 22; (the "8th day"
of the
Festival, or 7 days after Tishri 15)
-----------------------------------------------------------------------
One of the first questions you will have
is this: How do you know
that Nisan 14, Passover, is 164 days before Tishri 1? The answer
is
quite simple: the intervening months are always the same length,
no
matter whether the year is leap or common. Nisan has 30 days; Zif
(Iyar) 29; Sivan 30; Tammuz 29; Ab 30; Elul 29. Then comes Tishri.
After the 14th day, Nisan has 16 days remaining in the month. The
next
five months, Iyar through Elul have 147 days (29 + 30 + 29 + 30
+ 29).
One more day to Tishri 1. Add 16 + 147 + 1, and you have 164 days.
Remember the "count by seven" pattern that
separates the same day
of the week in any month? You learned in Program 7 that days 1,
8,15,
22, and 29 of a month are the same day of the week. Notice that
three
of the four Holy Days in the fall season occur on the SAME DAY
OF THE
WEEK!
All the fall Holy Days occur in the month
of Tishri, on the 1st,
10th, 15th, and 22nd. Once you determine the day of the week of
Trumpets, you know immediately the day of the week of the first
day of
Tabernacles and the Last Great Day. They're all the same! And Atonement
is simply two days later in the week than Trumpets.
Another convenient feature of the Holy
Days is that Tishri 1 can
fall only on a Monday, Tuesday, Thursday or Sabbath. (Recall
Postponement Rule 2 in Program 6?) This means that in any Roman
year,
the Holy Days will have one of these four patterns based upon what
day
of the week the Feast of Trumpets occurs. You have a "Monday pattern",
a "Tuesday pattern", a "Thursday pattern", and a "Sabbath pattern".
Once you understand these four patterns -- and they need NOT be
memorized, as you will see! -- you can list the day of the week
of ALL
the Sacred Festivals for any given year in about 30 seconds. Yes,
it's
that simple!
Here is a brief diagram to show you how
all the Holy Days in a
Roman year hinge upon the day of Tishri 1. The Hebrew civil year
is
reckoned from Tishri to Tishri, so in that sense, two Hebrew years
are
involved when you specify all the Holy Days in one Roman year.
How the Annual Holydays Are Governed By Tishri 1
HEBREW YEAR
HEBREW YEAR
Tishri 1
Tishri 1
Tishri 1
(Trumpets)
(Trumpets)
(Trumpets)
|
|
|
|
|
|
--//--|
|
|
|-------------------------|------------------------|
|
| |
| |
|---
Passover |---
Passover |---
| |
| |
| |
| Atonement
| Atonement
| Atonement
| Tabernacles
| Tabernacles
| Tabernacles
======================================================================
|
|
|
January
January
January
| ROMAN YEAR
| ROMAN YEAR
|
|
|
|
How do you lay the Holy Days out on the
Roman Calendar? Since you
determine Tishri l first, it's a good policy to work with the fall
festivals first, then the Passover and Days of Unleavened Bread,
and
finally Pentecost. Of course, the day of the week for Pentecost
is
always on a Sunday because of the way God defined the time for
observing that Holy Day. Nevertheless, you still need to find the
date
for Pentecost.
Take the "Sabbath pattern" as an illustration
of the days of the
week. Your reasoning will go something like this:
For this year, Trumpets (Tishri 1) is on
a SABBATH
Atonement is 9 days later, or two days
later in the week: MONDAY
Feast of Tabernacles begins on the 15th
of Tishri: SABBATH
The Last Great Day is Tishri 22: SABBATH
(Or, in more abbreviated fashion, think
of Tishri 1, 10, 15, 22
corresponding to Sabbath, Monday, Sabbath, Sabbath.)
Passover is 164 days before Trumpets, or
three days earlier in the
week: WEDNESDAY
The first day of Unleavened Bread is the
next day: THURSDAY
The last day of Unleavened Bread is Nisan
21, one week after
Passover: WEDNESDAY
(In a less verbose manner, think Nisan
14 is 3 days earlier in the
week than Trumpets. Then Nisan 14, 15, 21 correspond to Wednesday,
Thursday, Wednesday.)
Pentecost is a SUNDAY.
If you are still uncertain about how this
pattern of days is
thought out, please read Program 7 again. It will become much simpler
once you understand the basic concepts of counting days that are
explained in that program.
Now practice on the four patterns yourself.
You might want to do
them more than once just to impress the thinking process more fully
in
your mind. The four patterns of Holy Days are extremely important!
Work
each pattern independently from the others -- for your own benefit.
8.1 What is the "Thursday pattern" of Holy Days? In other
words, what
are the days of the week for each of the annual festivals when
Trumpets
is on a Thursday?
8.2 What is the "Monday pattern" of Holy Days?
8.3 What is the "Tuesday pattern" of Holy Days?
8.4 What is the "Sabbath pattern" of Holy Days?
The answers are given below. (Don't try
to memorize them UNLESS
you can first figure them out yourself!)
8.1 Tishri 1 is on Thursday. Tishri 1, 10, 15, 22 correspond
to
Thursday,
Sabbath, Thursday, Thursday.
Passover is 3 days earlier in the week
than Trumpets: Monday.
Nisan 14, 15, 21 correspond to Monday,
Tuesday, Monday.
Pentecost is on Sunday.
On an examination, of course, you should
actually list these dates
along with the Holy Day:
Passover, (Nisan 14 ), Monday
1st day of Unleavened Bread, (Nisan 15),
Tuesday
Last day of Unleavened Bread, (Nisan 21),
Monday
Pentecost, Sunday Trumpets, (Tishri 1),
Thursday
Atonement, (Tishri 10), Sabbath 1st day
of Tabernacles, (Tishri
15), Thursday
Last Great Day, (Tishri 22),Thursday
8.2 Tishri 1 is on Monday. Tishri 1, 10, 15, 22 correspond
to Monday,
Wednesday,
Monday, Monday.
Passover is three days earlier in the
week: Friday.
Nisan 14, 15, 21 correspond to Friday,
Sabbath, Friday.
Pentecost is on Sunday.
8.3 Tishri 1 is on Tuesday. Tishri 1, 10, 15, 22 correspond
to
Tuesday,
Thursday, Tuesday, Tuesday.
Passover is three days earlier in the
week than Trumpets: Sabbath.
Nisan 14, 15, 21 correspond to Sabbath,
Sunday, Sabbath.
Pentecost is on Sunday.
8.4 See the example given previously in this program for the answer.
Now that you can easily determine the day
of the week for each of
the annual Holy Days, the final task is to specify the day of the
month
on a Roman calendar for each Festival.
The Roman dates of the fall festivals are
very easy to determine
if you simply work with the "seven pattern" of days in a month
-- that
you learned in Program 7. All you are concerned with are the equivalent
Roman dates to Tishri 1, 10, 15, and 22.
You will either be given the day of the
week and the day of the
Roman month for Trumpets, or else you will calculate the molad
Tishri
for the required year yourself. The known correspondence between
the
Hebrew calendar and the Roman calendar for one day provides the
key to
determining the other dates.
Imagine a Hebrew calendar for the month
of Tishri, but with most
of the days except for 1, 10, 15, and 22 deleted.
TISHRI
1
(8) (9) 10
15
22
Suppose that Tishri 1 in a particular year
occurred on September
7. To find the dates of the other fall Holy Days, lay out an
abbreviated calendar for the month of September (and October, if
necessary):
Tishri 1
September: 7
(8)
(9) 10
(14) (15) 16
15
21
22
28
Essentially, all you are doing in either
calendar is counting by
sevens from your starting day, the Feast of Trumpets. Atonement
is nine
days later in the month, so it might help you to think 1, 8-9-10,
15,
22 and correspondingly for this example, 7, 14-15-16, 21, 28.
As another illustration for the fall festivals,
what happens when
Trumpets is on September 25? Here you must remember to change months,
but that's no horrendous problem!
Tishri: 1
September : 25
- - 10
(32) (33) (34) = Oct. (2) (3)
15
39 =
Oct. 9
22
46 =
Oct. 16
September has 30 days, so you can simply
call October 1, September
"31" for calculation purposes. September 46 is simply 16 days after
September 30, or October 16.
What happens when Trumpets is on August
28? The other fall
Festivals will occur in the month of September. In order not to
get
involved with negative numbers, go ahead and add seven days, which
is
equivalent to finding Tishri 8. What day is August 35? September
4,
because August has 31 days. September 1 is the same as August 32;
September 2 the same as August 33, etc.
Your abbreviated calendar will appear:
Tishri 1
August 28
(8) (9) 10
Sept. (4) (5) 6
15
11
22
18
Observe that you are making a direct correspondence
between the
Hebrew calendar and the Roman calendar. Counting by full weeks
-- "by
sevens" -- enables you to avoid other computational confusions
that can
creep into your work if you just add days. You can add 9 days to
the
Roman date of Tishri 1 and find the date for Atonement; 14 days
to
Trumpets for the first day of Tabernacles, and 21 days for the
Last
Great Day. Counting by sevens with an abbreviated calendar is less
error-prone, once you understand the principle.
Here are a few problems for you to practice
determining the Roman
dates for the annual Festivals. Just indicate the dates for Atonement,
the first day of Tabernacles, and the Last Great Day, but include
the
day of the week.
8.5 Trumpets is September 5, Monday
8.6 Trumpets is September 12, Thursday
8.7 Trumpets is September 23, Tuesday
8.8 Trumpets is August 25, Tuesday
8.9 Trumpets is September 16, Thursday
8.10 Trumpets is August 26, Sabbath
The problems are worked for you on the next page.
8.5 September '5'
Monday (Feast of Trumpets)
12 - '14' Wednesday Atonement, September 14
'19' Monday
Tabernacles, Sept, 19 (first day)
'26' Monday
Last Great Day, Sept. 26
The single quotation mark dates are the answers, following the same
format as above:
8.6 September '12'
Thursday (Trumpets)
19 - '21' Sabbath
Atonement
'26' Thursday
First day of Tabernacles
'33' (Oct. 3) Thursday Last Great
Day
8.7 September '23'
Tuesday
30 - '2' October 2
Thursday
'7' October
7 Tuesday
'14' October
14 Tuesday
8.8 August '25'
Tuesday
32 - '34', or September 1 - '3' Thursday
'8' Tuesday
'15' Tuesday
8.9 September '16'
Thursday
23 - '25'
Sabbath
'30'
Thursday
'37' which is October '7' Thursday
8.10 August '26'
Sabbath
33 - 34, or September 2 - '4' Monday
'9' Sabbath
'16' Sabbath
Take a look at the spring season. Once
you find the date for
Passover (which you'll learn how to do very shortly), all you need
to
do is find the equivalent dates for Nisan 15 and Nisan 21. An
abbreviated calendar for Nisan looks like this:
Nisan 14 15
21
If Passover were April 18, you would have:
April 18 19
25
The procedure is the same as for the fall
season. Just remember
that March has 31 days, so March 32 is the same as April 1. (You
handled the 31 days in August the same way.)
Now indicate the Roman dates and day of
the week for the first day
of Unleavened Bread and the last day of Unleavened Bread in the
following problems:
8.11 Passover is April 10, Monday
8.12 Passover is March 25, Friday
The problems are
8.13 Passover is March 29, Friday
worked for yo on
8.14 Passover is March 23, Wednesday
the next page.
8.11 Passover is April 10 '11' Monday
Tuesday
'17' Monday
To be sure you understand the pattern of
this answer,
Passover
is Monday, April 10
1st day
of Unleavened Bread is Tuesday, April 11
7th day
of Unleavened Bread is Monday, April 17
8.12 Passover is March 25 '26'
Friday Sabbath
'32' which is APRIL 1 Friday
8.13 Passover is March 29 '30'
Friday Sabbath
'36' which is APRIL 5 Friday
8.14 Passover is March 23 '24'
Wednesday Thursday
'30'
Wednesday
Now turn your attention to finding the
date of Passover, once you
know the Roman date for Trumpets. As you learned earlier in this
program, Passover is 164 days before the Feast of Trumpets. What
you do
is count by whole Roman months until you obtain a number slightly
larger than 164. (You can also find a number of days below 164.)
An example will clarify the procedure.
Suppose Trumpets is on
September 13. You want a number bigger than 164 days, so you mentally
keep track of the months you are adding:
September: 13 days
August: 31 days
July:
31 days
June:
30 days
May:
31 days
April: 30 days
--------------------------------
166 days
166 days brings you to the last day of
March. How many days past
164 did you go? 166 - 164 = 2. Two days
into April is April 2 --
Passover.
Doing this the other way, had you only
counted as far as May, you
would have come up with 136 days. 136 days brings you to April
30. How
many days into April must you go? 164 - 136 =
28 days. April 30
minus 28 days is April 2 -- Passover.
Take another illustration, Trumpets being on August 30. You have:
August: 30 days
July:
31 days
June:
30 days
May:
31 days
April: 30 days
March: 31 days
--------------------------
183 days
183 days takes you to the last day of February,
How many days into
March is Passover? 183 - 164 = 19 days. Passover is March 19.
Once you practice on a few problems, you
will have no difficulty
in determining the Roman date for Passover. In these problems,
also
indicate the day of the week:
8.15 Trumpets is September 8, Thursday
8.16 Trumpets is September 11, Sabbath
8.17 Trumpets is September 5, Monday
8.18 Trumpets is August 25, Sabbath
The answers are listed for you below.
8.15 Passover is March 28, Monday
8.16 Passover is March 31, Wednesday
8.17 Passover is March 25, Friday
8.18 Passover is March 14, Wednesday
You've now learned how to find the day
of the week and the Roman
day of the month for all the Holy Days except Pentecost. Of course,
you
already know the day of the week. But how about the Roman date?
Before you can count fifty days to Pentecost,
you need the
starting point, which is the Sunday of the wavesheaf offering.
The
table below will give you an overview of the relationship of Passover
and the wavesheaf offering:
Days after
Passover
(Nisan 14) Wavesheaf offering Passover
========================================================
Friday (Thurs.
eve) Nisan 16 (Sunday)
2
Wednesday
Nisan 18
4
Monday
Nisan 20
6
Sabbath
Nisan 15
1
There's really no need to memorize this
table, because you can
easily determine the number of days from the Passover to the wavesheaf
offering once you find the day of the week Passover occurs. Just
count
from the Passover to the Sunday of the wavesheaf offering.
When the Passover is on Monday, March 28,
the Sunday of the
wavesheaf offering is 6 days later, March 34 = April 3. With April
3 as
the first day (counting inclusively), Pentecost, the fiftieth day,
will
be 49 days later. There are 30 - 3 days left in April, or
27 days.
Pentecost is 49 - 27 days into May, or on May 22. As another way
for
determining the date of Pentecost, count seven full weeks from
April 3.
Here you are counting by sevens: April 10, 17, 24, May 1, 8, 15,
22.
Again you have arrived at May 22 for the date of Pentecost.
Here's another example for determining
the date of Pentecost. If
Passover is on April 8, a Wednesday, the wavesheaf offering will
be
April 12, Sunday. Count seven full weeks from April 12: 19, 26,
May 3,
10, 17, 24, 31. Pentecost is May 31.
Work out a few problems in order to firm
up the process in your
mind. Determine Pentecost for each of the following cases
8.19 Passover is Friday, March 30
8.20 Passover is Wednesday, March 16
8.21 Passover is Sabbath, April 2
8.22 Passover is Monday, April 20
The answers are on the next page.
8.19 Pentecost is Sunday, May 20
8.20 Pentecost is Sunday, May 8
8.21 Pentecost is Sunday, May 22
8.22 Pentecost is Sunday, June 14
You have learned how to work with the Holy
Days in separate steps,
fall festivals, Passover and Pentecost. Now integrate those skills
and
find for all the Holy Days in a Roman year the respective date
and day
of the week.
8.23 Trumpets is Monday, September 16
8.24 Trumpets is Sabbath, October 3
8.25 As the last problem in this series of programs, determine the
day
of the week
and the Roman date for all the annual festivals
in the year
2055 AD. You may use the chart included in
Program
4.
The answers are below.
8.23 Passover
Friday, April 5
1st day
of Unleavened Bread Sabbath, April 6
7th day
of Unleavened Bread Friday, April 12
Pentecost
Sunday, May 26
(Trumpets
Monday, September 16)
Atonement
Wednesday, September 25
1st day
of Tabernacles Monday, September
30
Last Great
Day
Monday, October 7
8.24 Passover
Wednesday, April 22
1st day
of Unleavened Bread Thursday, April 23
7th day
of Unleavened Bread Wednesday, April 29
Pentecost
Sunday, June 14
(Trumpets
Sabbath, October 3)
Atonement
Monday, October 12
1st day
of Tabernacles Sabbath, October
17
Last Great
Day
Sabbath, October 24
8.25 See Program 5, pages 44 and 45 for the determination of Tishri 1
Passover
Monday, April 12
1st day
of Unleavened Bread Tuesday, April 13
7th day
of Unleavened Bread Monday, April 19
Pentecost
Sunday, June 6
Trumpets
Thursday, September 23
Atonement
Sabbath, October 2
1st day
of Tabernacles Thursday, October
7
Last Great
Day
Thursday, October 14
----------###----------
SAMPLE TEST QUESTIONS
1. How many lunar months are in a 19-year cycle?
2. How many 19-year cycles, leap years, and common years
are between
the bench
mark and 171 AD?
3. Without using your chart, determine the REDUCED number
of days,
hours, and
parts in TWO common Hebrew years.
4. Demonstrate (or derive) without charts the amount
that the Hebrew
common year
leads or trails the Roman calendar.
5. What corrections (Roman leap year and Julian / Gregorian)
are
applicable
for the following years:
971 BC 1716 AD 1291 AD 1900 AD 2155 AD
6. In 1984 AD the molad Tishri is September 25, Tuesday
11 h 976 p.
What is
the day of the week and the day of the month for
Trumpets?
7. If the 29th of Elul is a Wednesday, what day of the
week is the
15th of
Elul?
8. What day of the week is 313 days before a Friday?
9. In 1981 AD Trumpets is on September 28, Monday. List
the day of
the week
and the day of the Roman month for each of the
annual festivals
given in Leviticus 23.
10. List the day of the week and the day of the Roman month
for all
the annual festivals in the year 29 AD. You may use your chart.
HEBREW CALENDAR
Name of Month*
#Sacred #Civil Begins with new moon of
--------------
------- ------ -----------------------
Aviv, or Nisan
1st 7th
March-April
Ziw 2nd 8th April-May
Siwan 3rd 9th May-June
Tammuz 4th 10th June-July
Av 5th 11th July-August
Elul 6th 12th August-September
Tishri, or Ethanim 7th 1st September-October
Marcheshwan 8th 2nd October-November
Kislew 9th 3rd November -December
Teveth 10th 4th December-January
Shevat 11th 5th January-February
Adar 12th 6th February-March
(*) Modern Hebrew transliteration. Please compare with that below:
The spelling given in W. M. Feldman's,
"Rabbinical Mathematics and
Astronomy," is as follows:
Nisan Tishri
Iyar Marcheshvan
Sivan Kislev
Tammuz Tebeth
Ab
Sh'bat
Elul Adar
V'Adar
Use this as a guide for pronunciation.
THE SIX TYPES OF HEBREW YEARS
COMMON LEAP
Month Deficient Regular Perfect Deficient Regular Perfect
Tishri 30 30 30 30 30 30
Marcheshwan
(Heshwan) 29
29 30
29 29
30
Kislew 29 30 30 29 30 30
Teveth 29 29 29 29 29 29
Shevat 30 30 30 30 30 30
Adar 29 29 29 30 30 30
V'Adar __ __ __ 29 29 29
Nisan (Aviv) 30 30 30 30 30 30
Ziw (Iyar) 29 29 29 29 29 29
Siwan 30 30 30 30 30 30
Tammuz 29 29 29 29 29 29
Av 30 30 30 30 30 30
Elul 29
29 29
29 29
29
----------------------- -------------------------
353 354 355
383 384
385
======================= =========================
(*) Modern Hebrew transliteration. See the top of this page for
alternate spelling and pronunciation.