(Those of you who are fans of my work are really going
to love this one...)
Part 1:
Analysis of some undeciphered Babylonian math problems
And yes indeed, I do believe that I have not only found info
on Babylonian tuning, but also that I have solved a riddle that
has eluded both Assyriologists and mathematicians who have been
trying to decipher Babylonian math. I believe it is evidence
that the Sumerians recorded the earliest surviving example of
the tuning of the so-called
'Pythagorean' scale.
CONVENTIONS:
BM 85194 and BM 85210 refer to the catalog numbers
of the actual tablets. They are housed in the British
Museum, Department of Western Asiatic Antiquities.
Their provenance is Sippar, a city-state a bit north
of Babylon, most likely from the Kassite period
(c.1600-1150 BC) or possibly from the MB [Middle Babylonian]
(c.1150-626 BC) or NB [Neo-Babylonian] (625-539 BC) periods.
By this time, Sumerian had ceased to be spoken (i.e., was
a 'dead' language) but was still widely used in its literary
form. (This situation was very similar to the medieval use
of Latin as a technical lingua franca long after
its spoken forms had transformed into the modern Romance languages.)
In the further details of the citations, the capital
roman numerals refer to the column in which that text
is inscribed on the tablet, and the arabic numbers are
the lines. The lower-case roman numerals in parentheses
give the numbering of the problems.
In the Roman-alphabet transliteration, numbers are represented
without any indication of absolute value (i.e., without
'sexagesimal points'). In Robson's translation and commentary
a semicolon marks the 'sexagesimal point' and a space is left between
sexagesimal places; I will show that in some places in her
translation the 'sexagesimal point' is misplaced, and will
always give decimal (= base-10) equivalents in my commentary.
Text between <> indicates inadvertent omissions by the scribe,
and text between [] indicates destroyed sections. These can
both usually be restored by means of the calculation.
'Problem text' refers to a calculation presented as a
problem with its solution.
'Logogram' refers to symbols derived from Sumerian words, which
represent key concepts and are a kind of shorthand. Robson says:
> [Robson 1999, p 8]
> ... at the same time they exhibit the esoteric nature of the text:
> because some cuneiform signs take on values and meaning specific to
> the genre, only the educated and initiated can comprehend them.
These logograms attest to the staying power of the Sumerian concepts
and writing, 500 or possibly even 1500 years after Sumerian had ceased
to be spoken, and some 2000 to 3000 years after their earliest recording.
Even tho by this time everyone in Mesopotamia spoke Akkadian, which
was Semitic, the non-Semitic Sumerian was still heavily used in writing.
(Sumerian is apparently unrelated to any other language on earth,
and Sumerologists are still not sure where they originated; it is
clear from their own writings that they migrated to Mesopotamia
from elsewhere.)
Unfortunately I cannot reproduce some of the diacritical marks
which appear over or under the letters in Robson's Roman-alphabet
transliteration. The two particular logograms which concern us
here are IM.LA, which should have an acute accent over the 'A',
and GAM, which appears in Robson exactly like that but in larger
letters - Robson gives no explicit explanation of that larger type,
but presumably it was used because of some doubt about its meaning
also. The pseudo-Sumerian logograms are reproduced, as in Robson,
in all capitals.
I reproduce in lower-case letters the syllabic Akkadian parts
of the texts, which appear in italics in Robson; these have
numerous accents and other diacritical marks which I cannot reproduce.
To start off, let's take a look at one of the tablets, BM 85194:
The math problems under consideration here appear in the third
column (on the right side of the tablet), second to fifth from
the top, with the text continued onto the side of the tablet.
On p 63-64, Robson presents the coefficient 18 (base-60)
[= 3/10 base-10] and tabulates the coefficient lists in which
it appears, but admits that the name is puzzling:
> [Robson, p 64]
> BM 85210 (xviii) and (xix) also concern IM.LA, but do not
> use this coefficient (see chapter 8.3). It is tempting to
> conclude that the phrase 18 IM.LA means '18 00 00, weighed
> clay', from the logograms _IM/tidum_ 'clay, earth', and _LA/saqalum_
> 'to weigh' - especially considering its proximity to the two
> other brick weight coefficients in lists D and F, and the
> fact that it is once written KI.LA, which may be read _suqultum_
> 'weight', in BM 85194 (xi): IV 10. However, it is difficult
> to reconcile this interpretation with the evidence of the
> problem texts themselves, which are discussed further in
> chapter 8.3. It *may* be, then, that IM.LA is not a logogram
> for _imlum_, and that we are dealing with two separate coefficients,
> albeit with the same numerical value: the latter is a brick
> density coefficient, which occurs only in proximity to others
> in the series, while the former describes some as yet undetermined
> structure.
From p 131, Chapter 8.3: Obscure coefficients in Problem Texts
8.3.2 Mathematical and philological commentary
> [Robson, p 131]
>
> The mysterious IM.LA is the subject of six problems, yet it is
> still impossible to determine exactly what the word refers to.
> BM 85194 (x)-(xiii), two sets of essentially identical problems,
> use the coefficient attested in the lists [18/60 = 3/10].
> The first two read:
>> (x) 1 IM.LA GAM EN.NAM ZA.E 4 u 3 NUMUN GAR.RA
>> III 6 4 a-na 3 i-si 12 ta-mar IGI 18 IM.LA
>> 3 20 ta-mar 3 20 a-na 12 i-si
>> 40 ta-mar 40 GAM
>> ki-a-am ne-pe-su
>>
>> The IM.LA is 1. What is the GAM? You: put down 4 and 3,
>> the ratios(*15*). Multiply 4 by 3. You will see 12. Take
>> the reciprocal of 18, the IM.LA. You will see 0;03 20.
>> Multiply 0;03 20 by 12. You will see 0;40. The GAM
>> is 0;40. This is the method.
>>
>>
>> (xi) 40 GAM KI.LA.BI 4 a-na 3 i-si 12 ta-mar
>> III 11 IGI 12 DU.A 5 ta-mar 18 IM..BI
>> a-na 5 i-si 1 30 ta-mar 1 30 a-na 40 GAM i-si 1 ta-mar
>> ki-a-am ne-pe-sum
>>
>> The GAM is 0;40. What is its KI.LA? You: multiply 4 by 3.
>> You will see 12. Take the reciprocal of 12. You will see 0;05.
>> Multiply 18, its _clay_, by 0;05. You will see 1;30. Multiply
>> 1;30 by 0;40, the GAM. You will see 1. This is the method.
> In the second pair of problems the IM.LA is 0;30 and the GAM 0;20,
> but apart from minor textual differences, the problems set and their
> solutions are the same as these ones. BM 85210 (xviii)-(xix)
> present the same subject in an even terser fashion, using a variant
> coefficient not attested in the lists:
>
>> (xviii) sum- 1 DAGAL IM.LA [GAM EN.NAM] ZA.[E IGI] DAL.BI
>> IV 5 DU.A 40 ta-mar 40 [...] 4 su.si
>> ne-pe-sum
>>
>
>
> --------------------
> (*15*) See 7.1.3 for a discussion of the logogram NUMUN.
>> [Robson, p 132]
>> If the width of the IM.LA is 0;01, [what is the GAM]? You: take
>> [the reciprocal] of its diameter. You will see 0;00 40, [...] or
>> 4 fingers. The method.
>>
>>
>> (xix) sum-ma [4 SU].SI GAM IM.[LA EN].NAM ZA.E 1 30 sa DAL
>> a-na 40 i-si 1 ta-mar 1 IM.LA
>> ne-pe-sum
>>
>> If the GAM is [4] fingers, what is the IM.LA? You: multiply
>> 1;30, of the diameter, by 0;00 40. You will see 0;01. The
>> IM.LA is 0;01. The method.
>
> There are several obstacles to a satisfactory interpretation,
> not least of which is the ambiguous terminology for the two
> parameters. The sign transliterated here as GAM can also be
> read GUR 'circle, diameter' as well as BUR 'depth'; it is used
> in both senses elsewhere on these tablets.
Robson goes on in this paragraph to present some of the interpretations
of IM.LA suggested by various scholars; these need not detain us here.
Robson's discussion of NUMUN is also instructive:
> [Robson, p 114]
> The logoram NUMUN 'ratio' is only found in the Sippar
> mathematical texts. It is occasionally used to denote
> geometrical coefficients(*15*), as well as the metrological
> conversions listed here [i.e., in this chapter]. It may
> also mean 'ratio' in other mathematical contexts: in
> VAT 6597(*16*) it is not simply a synonym of IGI.GUB.
> The Akkadian equivalent might be _zerum_ 'seed', which
> also has the logographic writing NUMUN, or there may
> have been a separate technical term.
>
> ------------------------
> (*15*) e.g. in BM 85210 (viii): see chapters 3.2, 3.4.
> (*16*) e.g. in VAT 6597 NUMUN refers to the share each
> 'brother' receives in the division of silver.
Indeed, I would say that NUMUN here has a meaning that is
much closer to 'seed', while at the same time it still also
expresses a
'ratio'.
MY COMMENTARY
Starting out, the GAM is the whole string, and
The IM.LA is the first ratio to be calculated.
Subsequently, the GAM is the last ratio calculated,
and the IM.LA is the next.
By continuing the process arbitrarily far, and using
the reciprocal where necessary in order to keep all
tones within one
'octave', one obtains the usual
Pythagorean series of
'5ths' and
'4ths', as described in
my paper on Indian Tuning.
One giveaway for me was that at _BM 85194 (x): III 5_
and _BM 85210 (xviii): IV 4_, which both start their
respective problems, the IM.LA is given the value
of 1, whereas subsequently at _BM 85194 (x): III 6_
the IM.LA is 18 (base-60) [= 3/10]. Anytime I see a
value of '1' for some variable at the beginning of
an ancient calculation and a smaller value for it later
on, I begin suspecting that the subject is division of
a string on a monochord. Seeing '4 and 3, the ratios'
further supported that idea.
I also got quite excited about seeing the '12' in
_BM 85194 (x)_ and _(xi)_, thinking that it may have
referred in some way to the recognition of the
Pythagorean Comma and its role as a limiter in scale
construction, and thus to the carrying-out of the calculation
to 12 notes. But after a deeper look it seems to be functioning
not as 12 (base-10) but as 12/60 = 1/5 (base-10).
ANALYSIS IN MODERN MATH
Let us examine the problems in modern mathematical
notation with base-10 numbers. I have deliberately
disregarded the editorial 'sexagesimal point' placing
by Robson and strictly followed the base-60 numbers as
written.
BM 85194 (x)
Rearranging the transliteration and translation:
1 IM.LA
GAM EN.NAM
ZA.E
4 u 3 NUMUN GAR.RA
4 a-na 3 i-si 12 ta-mar
IGI 18 IM.LA
3 20 ta-mar
3 20 a-na 12 i-si 40 ta-mar
40 GAM
ki-a-am ne-pe-su
The IM.LA is 1.
What is the GAM?
You:
put down 4 and 3, the ratios/seeds.
Multiply 4 by 3. You will see 12.
Take the reciprocal of 18, the IM.LA.
You will see 3,20.
Multiply 3,20 by 12. You will see 40.
The GAM is 40.
This is the method.
| |
Stripping out the Akkadian, leaving the Sumerian:
1 IM.LA
GAM EN.NAM
ZA.E
4 u 3 NUMUN GAR.RA
IGI 18 IM.LA
40 GAM
The IM.LA is 1.
What is the GAM?
You:
put down 4 and 3, the ratios/seeds.
Take the reciprocal of 18, the IM.LA.
The GAM is 40.
| |
IM.LA = 1
SEEDS = 3 and 4
GAM: (4*3)/60 = 12/60 = 1/5
1/(IM.LA) = 1/(18/60) = 10/3
(10/3) * (1/5) = 2/3
GAM = 2/3 = 40/60
BM 85194 (xi)
Rearranging Robson's transliteration and translation:
40 GAM
KI.LA.BI
4 a-na 3 i-si 12 ta-mar
IGI 12 DU.A 5 ta-mar
18 IM..BI
a-na 5 i-si 1 30 ta-mar
1 30 a-na 40 GAM i-si 1 ta-mar
ki-a-am ne-pe-sum
The GAM is 40.
What is its KI.LA? You:
multiply 4 by 3. You will see 12.
Take the reciprocal of 12. You will see 0;05.
Multiply 18, its _clay_,
by 5. You will see 1,30.
Multiply 1,30 by 40, the GAM. You will see 1.
This is the method.
| |
Stripping out the Akkadian, leaving the Sumerian:
40 GAM
KI.LA.BI
IGI 12 DU.A
18 IM..BI
40 GAM
The GAM is 40.
What is its KI.LA? You:
Take the reciprocal of 12.
Multiply by 18, the IM.LA
by 40, the GAM.
| |
GAM = 2/3
KI.LA: 1/((4*3)/60) = 1/(12/60) = 5
5 * (18/60) = 3/2
(3/2) * (40/60) [= 2/3, the GAM] = 1
KI.LA = 1
TEXTS NOT TRANSLATED BY ROBSON:
BM 85194 (xii)
--------------
IM.LA = 1
SEEDS = 4 and 3
GAM: (4*3)/60 = 12/60 = 1/5
IM.LA = 30/60 = 1/2
1/(IM.LA) = 2
2 * (1/5) = 2/5
GAM = 2/5
BM 85194 (xiii)
-------------
GAM = 20/60 = 1/3
KI.LA: (4*3)/60 = 12/60 = 1/5
1 / (1/5) = 5
(1/2) * 5 = 5/2
(5/2) * (1/3) [i.e, GAM] = 5/6
KI.LA = 5/6
===
But even here, GAM = 2/3 IM.LA.
1/3 = 2/3 * 1/2
1/3 (GAM) / 1/2 (IM.LA) = 2/3
===
============================================
BM 85210 (xviii)
----------------
sum- 1 DAGAL IM.LA If the width of the IM.LA is 1
[GAM EN.NAM] [what is the GAM]?
ZA.[E You:
IGI] DAL.BI DU.A take [the reciprocal] of its diameter.
40 ta-mar You will see 40
40 [...] 4 su.si 40 [...] or 4 fingers.
ne-pe-sum The method.
IM.LA = 1
GAM =
BM 85210 (xix)
--------------
sum-ma [4 SU].SI GAM If the GAM is [4] fingers
IM.[LA EN].NAM what is the IM.LA?
ZA.E You:
1 30 sa DAL a-na 40 i-si multiply 1,30 of the diameter by 40.
1 ta-mar You will see 1.
1 IM.LA The IM.LA is 1.
ne-pe-sum The method.
GAM = 4 fingers = 40
IM.LA = (diameter * 3/2) * 40 = 1
------------------------------------------------
Robson summarizes all these formulae as follows:
GAM = (2/3)*IM.LA
IM.LA = (3/2)*GAM
Thus it appears that GAM refers to the higher exponent
of 3 in any given calculation, while IM.LA is the
lower exponent.
>
> There are several obstacles to a satisfactory interpretation,
> not least of which is the ambiguous terminology for the two
> parameters. The sign transliterated here as GAM can also be
> read GUR 'circle, diameter' as well as BUR 'depth'; it is used
> in both senses elsewhere on these tablets.
The clincher was the description of the GAM at
_BM 85210 (xviii): IV 5_ (and presumably also at
_BM 85210 (xix): IV 7_) as 4 SU.SI [= '4 fingers'].
It turns out that 'finger' is a translation of a
Sumerian-Babylonian unit of length-measure, the width
(not length) of a finger [~1.&1/3-cm = ~2/3- or ~5/8-inch],
but I think that its name may provide a clue to a deeper
interpretation. Assuming that the generating process is
carried out symmetrically from the central 2/1, we may
obtain a Pythagorean heptatonic [= 7-tone] system as follows:
string-length 2^x 3^y ratio Semitones
~0.844 | 5 -3 | 32/27 2.94 C
~0.563 | 4 -2 | 16/9 9.96 G
0.75 | 2 -1 | 4/3 4.98 D
0.5 | 1 0 | 2/1 12.00 A == 1/1 0.000
~0.667 |- 1 1 | 3/2 7.02 E
~0.889 |- 3 2 | 9/8 2.04 B
~0.593 |- 4 3 | 27/16 9.06 F#
Placing these in ascending order of pitch gives
the following scale in the dorian minor mode:
string-length 2^x 3^y ratio Semitones
1.0 | 0 0 | 1/1 0.00 A
~0.889 |- 3 2 | 9/8 2.04 B
~0.844 | 5 -3 | 32/27 2.94 C
0.75 | 2 -1 | 4/3 4.98 D
~0.667 |- 1 1 | 3/2 7.02 E
~0.593 |- 4 3 | 27/16 9.06 F#
~0.563 | 4 -2 | 16/9 9.96 G
0.5 | 1 0 | 2/1 12.00 A
It is certainly possible that the scale was determined
asymmetrically; for example, substituting 3^-4 [= 128/81
= 7.92 Semitones] for 3^3 [= 27/16 = 9.06 Semitones], which
would give the standard Greek Pythagorean aeolian minor mode.
Given what appears to me to be a Sumerian love of codification
and organization, however, I feel that the symmetrical dorian
arrangement is more likely. (If the reader will take a look
at any collection of Sumerian art, he will be immediately
struck by the constant application of symmetry.)
In either case, it can clearly be seen that if 1/1 is the open
string on a fretted instrument, 3/2 is the fret played by the
4th finger; or alternatively, if this tuning scheme is meant
for a lyre or harp with the 1/1 [i.e., lowest in pitch] string
held closest to the player and played with the thumb, the 3/2
would be the string played with the 4th finger:
A B C D E
1/1 9/8 32/27 4/3 3/2
Thumb 1st 2nd 3rd 4th
If one assumes that the 'octave' was used as the
basic division, and that its value is 1, and if
use is made of both formulae as well as their
reciprocals, one obtains exactly the same 12-tone
Pythagorean scale that I posit for
ancient Indian tuning
and for Aristoxenus,
which is exactly the
same as the 12-tone Pythagorean scale noted by
the theorists in the 1300s in medieval Europe:
string-length 2^x 3^y ratio Semitones
~0.712 | 10 -6 | 1024/729 5.88
~0.949 | 8 -5 | 256/243 0.90
~0.633 | 7 -4 | 128/81 7.92
~0.844 | 5 -3 | 32/27 2.94
~0.563 | 4 -2 | 16/9 9.96
0.75 | 2 -1 | 4/3 4.98
0.5 | 1 0 | 2/1 12.00 == 1/1 0.000
~0.667 |- 1 1 | 3/2 7.02
~0.889 |- 3 2 | 9/8 2.04
~0.593 |- 4 3 | 27/16 9.06
~0.790 |- 6 4 | 81/64 4.08
~0.527 |- 7 5 | 243/128 11.10
~0.702 |- 9 6 | 729/512 6.12
Placing the tones in ascending order of pitch gives the following scale:
string-length 2^x 3^y ratio Semitones
1.0 | 0 0 | 1/1 0.00
~0.949 | 8 -5 | 256/243 0.90
~0.889 |- 3 2 | 9/8 2.04
~0.844 | 5 -3 | 32/27 2.94
~0.790 |- 6 4 | 81/64 4.08
0.75 | 2 -1 | 4/3 4.98
/ ~0.712 | 10 -6 | 1024/729 5.88 \ 2 different
\ ~0.702 |- 9 6 | 729/512 6.12 / 'tritones'
~0.667 |- 1 1 | 3/2 7.02
~0.633 | 7 -4 | 128/81 7.92
~0.593 |- 4 3 | 27/16 9.06
~0.563 | 4 -2 | 16/9 9.96
~0.527 |- 7 5 | 243/128 11.10
0.5 | 1 0 | 2/1 12.00
That should look extremely familiar to everyone on the
Tuning List.
Most likely, only one of the two
'tritones' (which are only
a
Pythagorean comma
apart) would be selected. Alternatively,
perhaps the Babylonians carried the calculation out farther,
and made use of
schismatic substitutions
(i.e., the '3==5
bridge'
in my theory) to acheive
pseudo-5-limit, as I describe in
my
paper on Indian Tuning.
While I noted in my last post about this that, if Ernest McClain's
conjectures are correct, the Babylonians certainly included the
prime-number
5 in their musical system, this system is
strictly 3-limit. Considering the vast span of time associated
with the writing of these cuneiform tablets, I think there's
no reason to believe that the 3-limit system necessarily was
prior to the 5-limit, altho the simplicity of this formula
seems to argue in favor of that opinion.
It's very interesting to me that the coefficient that describes
the relationship between the repeating 'octave' of 360 degrees and 1200
cents
is exactly the one used here: 360/1200 = 3/10 = 18/60.
Could this be an indication that the Sumerians had discovered
temperament?
(See the footnote to my post in Yahoo Tuning Groups
Message 11107
(Sat Jul 8, 2000 9:28 pm) for more on this; it's reproduced in
expanded form below.)
RELATED:
Robson, p 52:
Coefficient that may be needed for sound-hole on a lyre.
REFERENCE
Robson, Eleanor. 1999.
_Mesopotamian Mathematics, 2100 - 1600 BC
Technical Constants in Bureaucracy and Education_.
Oxford editions of cuneiform texts, vol. XIV.
Clarendon Press, Oxford.
Part 2:
How a Sumerian could approximate 12-tone equal-temperament
In this section I propose to explore the method by which
it might be possible for a Sumerian to calculate string-lengths
which divide a string into a tuning which closely approximates
12-tone equal-temperament.
The Sumerians used the sexagesimal (= base-60) place-notation
numbering system in their written calculations and math tables.
I will therefore use this system to show how a Sumerian could
obtain successively closer approximations to a desired
irrational value.
Since our perception of pitch is related logarithmically to
the mathematics of rational division and multiplication, what
we perceive as equal division of
frequency, is mathematically
calculated by taking roots of frequencies or string-lengths.
Another preliminary fact that must be noted is that measurement
of string-lengths is inversely proportional to the perception
of the resulting frequencies. Thus, the mid-point of a length
of string would be 1/2 the length of the total string, and
the frequency would be double that of the total string, thus
giving a
ratio of 2:1.
We will assume that our hypothetical
Sumerian is measuring the frequencies on a monochord string.
For convenience, the length of the entire string will be 60 units, which we'll
also label as 0
cents.
The "8ve" (1200 cents) occurs at the
midpoint of 30 units.
As I demonstrate the basic calculations to determine the
so-called
"Pythagorean" scale, I will also review basic
sexagesimal (= "base-60") math.
For the purpose of aiding in understanding for a modern
musically-literate reader, let's call our string by the
modern letter-name "F". (All ancient people used very different names.)
Of course, the "8ve" will also be "F". I'll supply
crude ASCII diagrams which don't show the lengths
at the proper proportions, but at least give a general
idea of the string divisions, which are given in both
base-60 and regular decimal fractions.
∞ + (bridge)
|
|
|
|
|
|
|
|
30 + F 1/2
|
|
|
|
|
|
|
|
|
|
|
60 + F 1/1
Calculating the "Pythagorean" scale
To descend a
"perfect 4th"
(ratio 4:3, ~498 cents) from the "8ve", we have
to multiply 1/2 by 4/3, or 30/60 by 80/60. This
80/60 = 1 + 20/60, so is written sexagesimally as 1;20.
(1/2) * (4/3) = 2/3 = 40/60. So the problem and its answer
is written sexagesimally as:
1;20
* 30
-----
40
In sexagesimal math, multiplying by 30 is the same as
dividing in half (just as multiplying by .5 is the same
for us in decimal). So the above can be translated into:
(60 + 20)
* .5
----------
(30 + 10) = 40
Thus, the "C" a "perfect 4th" below our "8ve" of "F",
and thus simultaneously a
"perfect 5th"
(ratio 3:2, ~702 cents) above
the lower "F", occurs at 40 units from the bridge.
∞ + (bridge)
|
|
|
|
|
|
|
|
30 + F 1/2
|
|
|
|
40 + C 2/3
|
|
|
|
|
|
60 + F 1/1
We repeat this procedure to find the "perfect 4th" below "C",
(ratio 9:8, ~204 cents) which would be
13 carry row
1;20
* 40
------
53,20 answer
In sexagesimal math, whenever answers in any "cell"
exceed 60, one divides by 60, writes
down the remainder in the "cell", and carries the quotient over into
the next column to the left (noted here as "carry rows").
Here, 20 * 40 = 800. Since 60 * 13 = 780, which is the closest
mupltiple of 60 to 800, and 800 - 780 = 20, we write down "20"
and carry 13. 40 * 1 = 40, then we add the 13 + 40 to get 53.
So the note "G" is 53,20 (which we call 53 & 1/3) units
from the bridge.
∞ + (bridge)
|
|
|
|
|
|
|
|
30 + F 1/2
|
|
|
|
40 + C 2/3
|
|
|
|
53,20 + G 8/9
|
60 + F 1/1
Now, to find the "D" a "perfect 5th" *above* "G", we
*divide* the measurement for "G" by the ratio 3:2, which
is the same as multiplying by 2/3, which in sexagesimal
is again a multiplication by 40:
35 13 carry row
53 20
* 40
---------
35 33 20 answer
Again, 20 * 40 = 800, 60 * 13 = 780, 800 - 780 = 20, so
write down remainder of 20, and carry 13.
53 * 40 = 2120, add 13 = 2133. Since 60 * 35 = 2100,
and 2133 - 2100 = 33, write down
remainder of 33 and carry 35 into next column. There
is no multiplicand for this column, so 35 is carried
down by itself into the answer.
Thus, "D" occurs at 35,33,20 units from the bridge.
∞ + (bridge)
|
|
|
|
|
|
|
|
30 + F 1/2
|
|
35,33,20 + D 16/27
|
40 + C 2/3
|
|
|
|
53,20 + G 8/9
|
60 + F 1/1
This procedure can be easily carried out by hand, until
one reaches the thirteenth pitch (our "E#") and finds that
it is very close to the "8ve":
------------ string-length --------------
note fraction sexagesimal decimal
+ ∞
|
.
.
.
+ E# 262144/531441 29,35,46,21,35... 0.493270184
+ F 1/2 30 0.5
+ E 128/243 31,36,17,46,40 0.526748971
+ D# 32768/59049 33,17,44,39,17... 0.554928957
+ D 16/27 35,33,20 0.592592593
+ C# 4096/6561 37,27,27,44,11... 0.624295077
+ C 2/3 40 0.666666667
+ B 512/729 42,8,23,42,13,20 0.702331962
+ A# 131072/177147 44,23,39,32,22... 0.739905276
+ A 64/81 47,24,26,40 0.790123457
+ G# 16384/19683 49,56,36,58,55... 0.832393436
+ G 8/9 53,20 0.888888889
+ F# 2048/2187 56,11,11,36,17... 0.936442615
+ F 1/1 60 1
Thus it would be immediately apparent (without finishing the
calculation, as soon as one arrives at 29,35) that the
ratio
for "E#" would be slightly less than 30/60, and this would
constitute numeric proof that the
"8ve" is not equal to 6
"whole-tones"
(of ratio 9:8). This ratio of 531441/524288
is known to us today as the
"Pythagorean comma",
but obviously the recognition of it occurred long before Pythagoras's lifetime.
It might be useful at this point to check some of these
numbers, to show explicitly the conversion between sexagesimal
and decimal math.
We may choose as our example the ratio of the Pythagorean
"major 3rd",
which we know as ratio 81:64. What I will show
here is the method for converting from our decimal notation
into sexagesimal notation, or converting the denominator from
units of 81 to units of 60.
For our example we may choose the measurement for the
Pythagorean "major 3rd" with ratio 81:64.
81 is 60 + 21, or sexagesimal 1;21. The reciprocal of 1;21
is 44,26,40, well attested in many Babylonian math texts.
64 is 60 + 4, written 1;4.
2 1 2 multiplication carry row
44,26,40
* 1; 4
===========
1 1 addition carry row
-----------
2 57 46 40 results of multiplying 44,26,40 by 4
44 26 40 - results of multiplying 44,26,40 by 1, shifted one place
-----------
47,24,26,40 final answer
This result is fairly close to 47,30, which is the
exact midpoint between 47 and 48 units.
The Babylonian tablets attest that the Sumerians had
worked out a sophisticated _thetic_ system of modal
classification. It would be easy to understand this
system in the _dynamic_ sense if the discrepancy at
the end of the cycle (the "Pythagorean comma") was
divided up and distributed among all the other intervals,
which would result in
12-tone equal-temperament.
The Sumerians were capable of finding good approximations
for both square- and cube-roots, so in determining a measurement
for a
tempered scale, the starting point would be to find a
string-length for the tempered
"major 3rd", since that would be the
cube-root of 1/2, or to a Sumerian, the cube-root of 30.
Calculation of tempered "major-3rd": cube-root of 2
So we begin by knowing that three successive Pythagorean
"major 3rds" exceed an "8ve", but that three successive JI
(i.e., 5-limit) "major 3rds" produce an interval which is
smaller than an "8ve". Thus, we are looking for a tempered
"major 3rd", and that tempered version lies between the
two familiar rational intervals.
So we'll restate the above paragraph in sexagesimal mathematical
terms, and also relate it to our familiar measurement of
cents.
The JI "3rd" is the 5:4 ratio (~386.3 cents), which is
4/5 or 48/60 of the string-length as measured
from the bridge. The Pythagorean "3rd" is the 81:64 ratio
(~407.82 cents), which is almost midway between 47/60 and 48/60.
We want to find the fraction of 60 which gives the 400-cent "3rd".
As before, I will illustrate by creating a schematic diagram of the
string-lengths, not exactly to scale.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ A 64/81 47,24,26,40 0.790123457 407.8200035
|
|
|
+ A 4/5 48 0.8 386.3137139
So first we find the cube of 4/5, or in modern notation,
(4/5)^3.
48
* 48
-------
38,24
38,24
* 48
--------
30,43,12
Thus, the measurement of (4/5)^3 is slightly farther from the bridge
than the string's midpoint of 30, making the interval composed of
three 5:4's somewhat smaller than the "8ve". In our modern
rational notation we call it 125:64, ~1158.94 cents.
In our calculations above, we have already found that (64/81)^3
is measured at ~29,35,46,21,35... units from the bridge, making
that interval somewhat larger than the "8ve". In our notation
it's 531441:262144, ~1223.46 cents.
So we begin by taking the mean value of these two "major 3rds".
Start by adding together the two terms:
1
47 24 26 40
+ 48
----------------
1 35 24 26 40
Then divide each of these numbers by 2:
0.5 17.5 12 13 20
We need to remove the decimal components, so we remove
.5 from 17 and add 30 to the 12 in the column to the right of it:
0.5 17 42 13 20
Then we also remove .5 from the first column and add 30
to the 17 in the column next to the right of it:
47 42 13 20
So our first answer for the tempered "major 3rd" is
47,42,13,20 units from the bridge.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ A 64/81 47,24,26,40 0.790123457 407.8200035
|
|
+ A 47,42,13,20 0.795061728 397.0334636
+ A 4/5 48 0.8 386.3137139
Next we check on the accuracy of this result by cubing it.
I will assume from this point on that the reader is capable
of performing the intermediate steps in the arithmetic.
In practice, it is unnecessary to carry the accuracy out
to more than three places, because differences beyond that
would be inaudible -- within the reference "8ve" between
30 and 60 parts (i.e., from 1/2 to the whole string),
the second sexagesimal place represents a maximum variation
of ~57 cents, the third place of only ~1 cent.
47 42 13
* 47 42 13
------------
~37 55 39
* 47 42 13
------------
~30 9 17
So we see that stacking three of these tempered "major 3rds"
results in an interval measured at 30,9,17 units from the bridge,
or ~1191.09 cents. This is considerably closer than either of
the untempered "3rds", but can still
be improved. So we find another mean, this time between the
result we just calculated, which is still a bit too small, and the
Pythagorean "major 3rd", which is too large.
1
47 24 26
+ 47 42 13
------------
1 35 6 39
/ 2
------------
47 33 19 30
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ A 64/81 47,24,26,40 0.790123457 407.8200035
+ A 47,33,19,30 0.792590278 402.423389
|
+ A 47,42,13,20 0.795061728 397.0334636
+ A 4/5 48 0.8 386.3137139
Now we check the accuracy of this subdivision by cubing it:
47 33 19
* 47 33 19
-----------
~37 41 30
* 47 33 19
-----------
~29 52 26
This result, measured 29,52,26 units from the bridge, is
~1207.3 cents. Now this is a bit too large, but still
the closest result so far.
We'll take one more mean and that should give a sufficiently
accurate result. So now we find the mean between the two
tempered "3rds" which we've calculated:
47 42 13
+ 47 33 19
-------------
1 35 15 32
/ 2
-------------
47 37 46
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ A 64/81 47,24,26,40 0.790123457 407.8200035
+ A 47,33,19,30 0.792590278 402.423389
+ A 47,37,46 0.793824074 399.7305356
+ A 47,42,13,20 0.795061728 397.0334636
+ A 4/5 48 0.8 386.3137139
Cube this to check accuracy:
47 37 46
* 47 37 46
-----------
~37 48 34
* 47 37 46
-----------
~30 0 51
Here we have finally found an interval which is audibly
indistinguishable from the "8ve". It would be measured
at 30,0,51 units from the bridge, which for all practical
purposes is the same as 30. If it could be produced
accurately, it would measure ~1199.183 cents.
So we will set down the value for our tempered "major 3rd"
as 47,37,46 units. This is ~399.73 cents.
Calculation of tempered "whole-tone": square-root of cube-root of 2
Our next task is to determine the value of the tempered
"whole-tone",
which we obtain by finding successively
closer approximations to the square-root of our tempered
"major 3rd".
We do this by again taking the mean of two values which
we already know to be close to our target. In this case,
our starting ratios will be the Pythagorean "whole-tone"
of ratio 9:8 (~203.91 cents, 53,20 units from the bridge)
and the other "whole-tone" which commonly occurs in 5-limit
JI, of ratio 10:9 (~182.4 cents, 54 units from the bridge).
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ G 8/9 53,20 0.888888889 203.9100017
|
|
|
|
+ G 9/10 54 0.9 182.4037121
53 20
+ 54
--------
1 47 20
/ 2
--------
53 40
In this case the mean is actually very easy to compute
mentally: 54 is the same as 53,60, and so the difference
between the two terms is 0,40, whose mean is 20. Add 0,20
to 53,20 to get 53,40.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ G 8/9 53,20 0.888888889 203.9100017
|
|
|
+ G 53,40 0.894444444 193.1234619
+ G 9/10 54 0.9 182.4037121
Now square 53,40 to see how closely it matches our tempered
"major 3rd":
53 40
* 53 40
------------
48 0 6 40
This is not too close to our goal, but note in passing that it
*is* almost exactly the same as the JI "major 3rd" of ratio 5:4,
which is 48 units from the bridge. Thus, 53,40 units gives
an extremely accurate
"meantone", ~193.124 cents. Compare
with the ~193.157 cents for the true meantone of (5/4)^(1/2).
(Could meantone thus also be 5000 years old? To my mind, it
is entirely possible that the Sumerians may have grappled
with the calculation of this tuning also ... perhaps i'll
write another webpage about that someday.)
So we continue by finding another mean, this time between
the Pythagorean "major 3rd" and the "meantone" just calculated.
This one is a trivial mental calculation, as one can easily
see that the mean between 53,20 and 53,40 is 53,30.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ G 8/9 53,20 0.888888889 203.9100017
|
|
+ G 53,30 0.891666667 198.508331
+ G 53,40 0.894444444 193.1234619
+ G 9/10 54 0.9 182.4037121
Square it to check the accuracy:
53 30
* 53 30
--------
47 42 15
This measurement, ~397.017 cents, is indeed quite close to
our tempered "major 3rd" of 47,37,46 units.
Let us try to improve on this by taking the mean of this
new "whole-tone" and the Pythagorean "whole-tone". Again
the calculation is trivial: the mean of 53,20 and 53,30
is 53,25.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ G 8/9 53,20 0.888888889 203.9100017
+ G 53,25 0.890277778 201.2070597
|
+ G 53,30 0.891666667 198.508331
+ G 53,40 0.894444444 193.1234619
+ G 9/10 54 0.9 182.4037121
Square it to check the accuracy:
53 25
* 53 25
----------
~47 33 20
This is ~402.42 cents.
Let us do one more calculation to get closer.
This time we will take the mean between the last two
calculated "whole-tones". The exact midpoint of
53,25 and 53,30 is 53,27,30.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ G 8/9 53,20 0.888888889 203.9100017
+ G 53,25 0.890277778 201.2070597
+ G 53,27,30 0.890972222 199.8571695
+ G 53,30 0.891666667 198.508331
+ G 53,40 0.894444444 193.1234619
+ G 9/10 54 0.9 182.4037121
Square it to check accuracy:
53 27 30
* 53 27 30
----------
~47 37 47
That's almost exactly our tempered "major 3rd". Thus, we
can set down our tempered "whole-tone" at 53,27,30 units,
~199.857 cents.
Calculation of tempered "semitone": square-root of square-root of cube-root of 2
The final remaining task is to find the tempered
"semitone"
by taking the square-root of this tempered "whole-tone".
Here we can begin with the two familiar Pythagorean semitones,
the _limma_ of ratio 256:243 (~90.225 cents) and the _apotome_
of ratio 2187:2048 (~113.7 cents). The measurement of the
former will be ~56,57,11 units from the bridge, and of the
latter, ~56,11,12 units.
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
|
|
|
|
|
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
First we find the mean:
56 57 11
+ 56 11 12
------------
1 53 8 23
/ 2
------------
56 34 11 30
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
+ F#/Gb 56,34,11 0.942828704 101.9188967
|
|
|
|
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
Then we square it to see how close the result comes to
our tempered "whole-tone" of 53,27,30 units:
56 34 11
* 56 34 11
----------
~53 20 8
This is good, but we can do better.
Since our first calculated "semitone" is a bit larger
than the target, we take the mean of this and the _limma_:
56 34 11
+ 56 57 11
------------
1 53 31 22
/ 2
------------
56 45 41
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
+ F#/Gb 56,34,11 0.942828704 101.9188967
|
|
|
+ F#/Gb 56,45,41 0.946023148 96.06313167
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
Square this to check accuracy:
56 45 41
* 56 45 41
----------
~53 41 51
This is not as close as our last answer, so we now take
the mean between these two results:
56 45 41
+ 56 34 11
----------
1 53 19 52
/ 2
----------
56 39 56
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
+ F#/Gb 56,34,11 0.942828704 101.9188967
|
|
+ F#/Gb 56,39,56 0.944425926 98.98853833
+ F#/Gb 56,45,41 0.946023148 96.06313167
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
Square it to check accuracy:
56 39 56
* 56 39 56
----------
~53 31
We'll try to get closer by taking the mean between
our two most accurate values:
56 39 56
+ 56 34 11
----------
1 53 13 7
/ 2
----------
~56 36 33
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
+ F#/Gb 56,34,11 0.942828704 101.9188967
+ F#/Gb 56,36,33 0.943486111 100.7121775
|
+ F#/Gb 56,39,56 0.944425926 98.98853833
+ F#/Gb 56,45,41 0.946023148 96.06313167
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
Square to check accuracy:
56 36 33
* 56 36 33
----------
~53 24 36
Now find the mean between these last two:
56 39 56
+ 56 36 33
-----------
1 53 15 29
/ 2
-----------
~56 37 44
------------ string-length --------------
note fraction sexagesimal decimal ~cents
+ ∞
.
.
.
+ F#/Gb 2048/2187 56,11,12 0.936444444 113.6816247
+ F#/Gb 56,34,11 0.942828704 101.9188967
+ F#/Gb 56,36,33 0.943486111 100.7121775
+ F#/Gb 56,37,44 0.943814815 100.1091332
+ F#/Gb 56,39,56 0.944425926 98.98853833
+ F#/Gb 56,45,41 0.946023148 96.06313167
+ F#/Gb 243/256 56,57,11 0.949217593 90.22710661
Square to check accuracy:
56 37 44
* 56 37 44
----------
~53 26 50
Here at last we have come within 1/60 of our target value.
At ~100.11 cents, this "semitone" of 56,37,44 units
is certainly audibly indistinguishable from 2^(1/12).
In fact, the actual 3-place sexagesimal value for the
12-tET semitone 2^(1/12) is 56,37,57.
Our hypothetical Sumerian could then easily check his
work by multiplying this value by itself 12 times, to
see how closely his final result comes to the target of
30 units.
60 F
56,37,44 F#/Gb
53,26,50 G
50,26,39 G#/Ab
47,36,36 A
44,56,06 A#/Bb
42,24,37 B
40,01,39 C
37,46,43 C#/Db
35,39,22 D
33,39,10 D#/Eb
31,45,43 E
29,58,39 F
This final result is ~1201.3 cents, again, audibly
indistinguishable from the actual "8ve" under all
ordinary circumstances.
So by finding successively closer approximations, by
using a method of finding a mean of two known approximations and squaring
or cubing that mean and comparing it to the squares or
cubes of the known approximations,
that's how a Sumerian could have, by hand, calculated the
string-length measurements which produce a very accurate
approximation of 12-tET, 5000 years ago.
In fact, even using a less accurate method with only two
sexagesimal places, the resulting tuning contains no more than
a couple of cents error from true 12edo; see my
Simpler
sexagesimal approximation to 12edo.
See my letter to
Jacky Ligon and my webpage
about the nefilim
for some of my more unusual speculations about the Sumerians.
Also see my posts the the Yahoo Tuning Group in
message 10930 (Mon Jun 26, 2000 11:12 pm),
the footnote to
message 11107 (Sat Jul 8, 2000 9:28 pm),
and
message 11624 (Sat Aug 19, 2000 4:48 pm)
for more on Sumerian and Babylonian music.
-monz
