© 2000 by Joe Monzo
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.
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'.
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).
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.
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 |
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 |
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
Rearranging Robson's transliteration and translation:
40 GAM KI.LA.BI |
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 |
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
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.
Robson, Eleanor. 1999.
_Mesopotamian Mathematics, 2100 - 1600 BC
Technical Constants in Bureaucracy and Education_.
Oxford editions of cuneiform texts, vol. XIV.
Clarendon Press, Oxford.
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
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.
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.
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.
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 12-edo; see my Simpler sexagesimal approximation to 12-edo.
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
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