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 &nbs