Cents

59.1 Cents (The Division of the Octave into 1200 parts)

"This system (by Alexander J. Ellis 1890) has left intact the definition of any individual note as the result of a certain number of vibrations per second... It cares only for describing distances between two such notes".

"It is more impressive to be presented with the ratio 524288 : 51441; but who understands that this means the distance called the Pythagorean comma, which amounts to exactly 12 % of a tone"?

"The ingenious system of Cents... describes any distance by one simple number. A Cent is one-hundredth part of an equal-tempered (piano) semitone:
the distance between two notes a semitone apart comes to one hundred, and the octave, consequently, to twelve hundred Cents. The essential standard distances are":

Semitone 100 C.
Fifth 700 C.
Second 200 C.
Minor Sixth 800 C.
Minor Third 300 C.
Major Sixth 900 C.
Major Third 400 C.
Minor Seventh 1000 C.
Forth 500 C.
Major Seventh 1100 C.
Tritone 600 C.
Octave 1200 C.

"Single distances as well as complicated scales become simple and intuitively evident: a second of, say, 180 C. means a distance by 10 % smaller than an equal-tempered second; a distance of 220 C. is by 10 % larger than a second, and so on".

59.2 Calculating Cents

"This can be done by a simple logarithmic operation. (Other) method(s) can be substituted if no table of logarithms is available:

59.3 First Method without Tables or Logarithms

"Multiply the difference of the two vibration numbers by 3477 and
divide the product by their sum. In case the triple of the larger vibration number exceeds the quadruple of the smaller one, multiply the greater number by three and the smaller number by four before starting the operation indicated above, and ...finally add 498 (the perfect, not the equal-tempered forth) to the result.

If, on the contrary, the proportion of the two vibration numbers is greater than two to three,
multiply the greater number by two, and the smaller number by three, before starting the operation indicated above, and finally
add 702 (the perfect fifth) to the result".

The Rise of Music in the Ancient World
By Sachs
781.8 Sa 14.

59.4 The Calculation of Cents from Interval Ratios Book: Sensations of Tone by Helmholtz, pg. 447

Second Method of Calculating Cents with Tables, but without Logarithms.
Third Method of Calculating Cents by Five Place Logarithms.
Forth Method, by Seven Place Logarithms.

Sensations of Tone

59.5 Table. Number of Cents Corresponding to given Frequency Ratios Helmholtz

Sensations of Tone

59.6 Table. Table for Calculation of Cents Helmholtz

59.7.1

Sensations of Tone

59.8 Logarithms

"Units for Intervals. Before examining our musical scale it is desirable to have some accurate means of describing the musical interval. We already know then an interval is characterized by its frequency ratio, and for many purposes it is entirely satisfactory to describe an interval by giving this ratio. There are other purposes for which frequency ratios prove to be rather awkward. Question 47 may illustrate one of the difficulties that we meet when we attempt to measure intervals by frequency ratios.

Question 47. Five notes have frequencies of 400, 500, 600, 800 and 1600 cycles/sec. These notes are to be laid off on a horizontal line in such a way that equal distances shall represent equal intervals. The distance between 400 and 800 marks is to be 4 cm. How far from the 400 mark should the marks for the rest of the notes be placed? Notice that the interval from 400 to 800 is an octave, and that the interval from 800 to 1600 is also an octave.

A somewhat different consideration is the following. Suppose that the musical interval from c3 to f3 is a certain number of small unit intervals of some sort, and that the interval from f3 to a3 is another number of these same units. We should like to be able to get the interval from c3 to a3 by adding the two numbers. If we measure intervals by means of their frequency ratios we must multiply. We should like to add instead of multiplying.

We recall that if we add the logarithms of the two numbers we get the logarithm of their product. This suggests that we measure intervals by the logarithms of their frequency ratios instead of by the frequency ratios themselves. By making use of this idea it is possible to devise systems in which intervals is given by the sum of the smaller intervals that compose it.

But the logarithms themselves are not entirely satisfactory. Let us see why. We know that among all musical intervals the octave occupies a position of unique importance (Footnote 29, below), and what we should like is a small interval that fits into an octave some whole number of times, and that is also convenient in measuring other intervals.

Now when two notes are an octave apart the upper vibrates twice as fast and the lower. Moreover, the common logarithm of 2 is 0.30103. This number is awkward, and it is not a whole number.

But if we multiply all our logarithms by some suitably chosen factor, it turns out, not only can we add the numbers which represent our intervals, but also that the intervals can be measured in terms of a small interval of any size we choose. In order to determine the factor by which we multiply the logarithm suppose that we wish to divide and octave into 1000 parts. Let us call one of these parts a millioctave.

And suppose that we wish to know how many millioctaves there are in an interval which has a given frequency ratio. Let f stand for the given frequency ratio, and n for the corresponding number of millioctaves. Then we may write the proportion:

n /1000 = log f/log 2

n = 1000/log 2 x log f

That is, to find the number of millioctaves that correspond to the frequency ratio f we multiply the logarithm of f by the factor [1000/log 2] = 3321.93..... For many acoustic purposes the millioctave is a convenient unit, and it is often used in the current literature.

The factor just found is a little awkward. It would certainly be easier to multiply by some such factor as 1000 instead of 3321.93. If we use the factor 1000 an octave is divided into 301.03 ...equal parts.

One of these parts is sometimes called an savart, although the number of savarts in an octave is not a whole number, it is nevertheless true that for some purposes it is sufficiently accurate to regard an octave as containing 300 savarts, and for such purposes the savart is a convenient unit.

Another division of the octave is into 1200 parts. In this case the factor by which to multiply the logarithm is [1200/log 2] = 3986.314....
The 1200th part of an octave is called a cent... (See Cents).

Footnote 29 ...If the partial tones of the notes with which we deal are harmonic it is not difficult to see why notes an octave apart seem to have something in common.

Helmholtz [Sensations of Tone], states the reason as follows. "Let any melody be executed.....the hearer must have heard not only the primes of the compound tones, but also their upper Octaves, and, less strongly, the remaining upper partials. When, then, a higher voice (frequency) afterwards executes the same melody an Octave higher, we hear again a part of what we heard before, namely the even numbered partial tones of the former compound tones, and at the same time we hear nothing that we had not previously heard. Hence the repetition of a melody in a higher Octave is a real repetition of what has been previously heard, not of all of it, but of a part.

If we allow a lower voice (frequency) to be accompanied by a higher in the Octave above it, the only part music which the Greeks employed, we add nothing new, we merely reinforce the evenly numbered partials. In this sense, then, the compound tones of an Octave above are really repetitions of the tones of the lower Octave, or at least of part of their constituents. Hence, the first and chief division of our musical scale is that into a series of Octaves".

Undocumented Source

59.9 Logarithmic Cents Calculation

59.10.1

"Since pitch depends upon frequency and interval upon ratio of frequencies, we have the following important result. Let it be required to measure intervals so that the sum of the measures of component intervals shall be the measure of the resultant interval. Then the only method possible is that of taking for each interval a number proportional to the logarithm of the ratio of the frequencies of the notes composing that interval.

Thus, let the frequencies of three notes, beginning at the highest and proceeding in the order of pitch, be L, M and N. Also let the intervals be I sub. 1 between L and M, I sub. 2 between M and N, and I between L and N. Then, if each interval be measured by k times the logarithm of the ratio of frequency we have:

I sub. 1 = k log L/M = k(log L - log M)....................(1)

I sub. 2 = k log M/N = k(log M - log N)..................(2)

I = k log L/N = k(log L - log N)............................(3)

But by addition of (1) and (2)

I sub. 1 + I sub. 2 = k(log L - log N).........................(4)

so by (3) and (4)

I = I sub 1 + I sub. 2.............................................(5)

For k any convenient number could be chosen, as (5) shows that the relation desired is independent of it. But the late Mr. A. J. Ellis (the translator of Helmholtz's Sensations of Tone) has adopted as the unit for this logarithmic measure the cent, 1200 of which make the octave. The name cent is used because 100 cents make the semitone of those instruments in which 12 equal semitones are the intervals occurring in the octave. Hence the clue to reduction of any intervals to their logarithmic cents would be found in the following equation, where I is the interval in cents between notes of frequencies M and N":

I = k log M / N...................................................(6)

1200 = k log 2....................................................(7)

Whence by (6) / (7).

I = 1200 log M - log N / log 2...............................(8).

Textbook on Sound
By Barton (1919)
534 B2

.59.11 Cents Calculation

cents = log r x 1200 / log 2
............ = log r / log 2 x 1200
............ = 1200 / log 2 x log r
log r = cents x log 2 / 1200
............ = log 2 / 1200 x cents

59.12 The System of Cents

"Cents are limited music logarithms that supply information about the position and frequency of a note, especially in relation to equal temperament. Thus, 105 would tell us that the frequency of C sharp is only slightly higher than the C sharp in equal temperament, while 150 cents would indicate a quarter-tone midway between C sharp and D.

In the mathematics of music the range is only 1 to 2, because this relationship expresses an octave. Hence, the cents are not merely common logarithms but, rather, common logarithms to base 2. Imagine that we are working with common logarithm, 0.1703. If we divide 0.1703 by the log of 2, the quotient will be 0.565724355, which we can call "the common logarithm, 0.1703 to base 2". As proof it can be shown as the antilog of 0.1703 is 1.48 and, also, that 2 with an exponent of 0.565724355 is, likewise, equal to 1.48. (This can easily be checked by using your calculator).

To obtain the actual number in cents, one additional step is necessary. After you have a logarithm to base 2, multiply it by 1200, for the number of cents. This operation has the effect of "spacing" the 1 to 2 interval. Thus, 0.565724355 x 1200 = 679 cents, approximately.

In Table 1(in the book) the letter r, stands for any decimal ratio between 1 and 2. Choose any such ratio; then, work thru the formulae with your calculator".

The Mathematics of Music.

59.13 Table. 120,000,000 parts to the Octave Found in the Mathematics of Music, p. 5,
781.1 L6485m

59.14 Table. 12 Parts of Two Commas and a Schisma
Found on p.7 of the book.

59.15 Table. 11 Parts of the Synodic Comma
Found on p.7 of the book.

59.16 Method for defining the Number of Parts in the Pythagorean System

"The unit of divergence from equal temperament is 1 / 12 diatonic comma, 1.955000865 cents.
In equal temperament, the interval, C-D, has 200 cents.
Since, in the Pythagorean intonation, the interval C-D, has two units of divergence, multiply 1.955000865 by 2.

955000865 x 2 = 3.910001731 cents.
Add 3.910001731 to 200 cents.
910001731 +200 = 203.910001731 cents.

In the Pythagorean intonation, the interval C-B sharp, has 12 units of divergence.

955000865 x 12 = 23.460010380 cents.

Therefore, the dieses is the diatonic comma, with 23.460010384 (corrected).
Divide cents for the dieses into cents for the interval, C-D.

910001731 / 23.460010384 = 8. 691812083.

460010384 x 8 = 187.680083072.

91001731 - 87.680083072 = 16.229918659.
229918659 / 2 = 8.114959330
114959330 is much too small to function as a break comma, which should be only slightly less in size as the dieses. Consequently, 7 should be the number for the dieses in the interval, C-D. We must retrace our steps, then, proceed as follows:
460010384 x 7= 164.220072688
910001731 - 164.220072688 = 39.689929043
689929043 / 2 = 19.844964522

We have multiplied cents for the dieses by 7 and subtracted the product from cents for the interval, C-D; also, we have divided the difference by 2, for there are 2 break commas in the interval, C-D. (We think of these commas as coming at the 'cracks" between piano keys, including the black notes). 19.844964522 will be the number of cents for each break comma.

Multiply 9.844964522 by 12 and subtract the product from 1200 cents. (There are always 12 break commas in the octave).

844964522 x 12 = 238.139574264
1200 - 238.139574264 = 961.860425736

860425736 / 23.460010384 = 41

There are 41 dieses and 12 break commas; in all, 53 parts".

See File, "30 Unique Divisions"

(Note: We will use these 53 parts since it corresponds to the Mayan Tzolkin).

Mathematics of Music
by Link

Impossible Correspondence Index