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Copyright © 2000, 2001 by Richard I. Schwartz

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CHAPTER 1

Early History Part 6

Acoustical Problems of Three Valves

The following discussion is offered to illustrate intonation problems of three-valve cornets. The mathematical and acoustical principles discussed here can also be applied to other brass instruments. Measurements below are given for explanatory reasons and should be considered only approximate. Actual bore lengths of nineteenth-century cornets varied greatly, A=440 (and therefore, the length of the cornet bore) was not universally stabilized until after 1920, and bores varied greatly from instrument to instrument in terms of their cylindrical/conical proportions, a fact which alters the exactitude of precise mathematical figures.

To lower an open pipe ½ step, it must be lengthened approximately 1/17 (.059) of its actual length (Beck 1999; Davenport 1999; Peterson 1957, 13-15; Rossing 1990, 226-228; White and White 1980, 255-257). The obvious value of 1/12 (8.3333333333 %) does not work! See below for an explanation of the validity of the use of the fraction 1/17 (5.8823529411 %). If a given pipe is approximately 56 inches, as the cornet is in its open valve position (and tuning slide/mechanism adjusted for the nineteenth-century cornet to be anything near A=440), it must therefore be lengthened approximately 3.3 inches to be lowered ½ step. This 3.3 inches is the theoretical length of the tubing of the second valve slide. The total length of tubing at this point is approximately 59.3 inches. To lower the pitch another ½ step, we must determine to which length this ½ step corresponds, i.e. is the next ½ step to be determined in relation to the open tubing length of "c1" or to the 59.3 inch measurement of "b." This double standard becomes even worse as the instrument progresses through its seven valve positions. We achieve two different sets of possible tubing lengths as is listed below:

Valve position

[Valve combination]

Length determined from first position in inches:

(1/12 of 56 inches) or

3.294117647016" added to

each measurement to get next measurement

Length determined from each previous position in inches: 1/12 or 5.8823529411% of each measurement added to itself to get next measurement

First [0]

56

56

Second [2]

59.3

59.3

Third [1]

62.6

62.8

Fourth [1,2]

65.9

66.5

Fifth [2,3]

69.2

70.4

Sixth [1,3]

72.5

74.6

Seventh [1,2,3]

75.8

79.0

 

After each new number was calculated in the above chart, the resulting value was rounded-off to the nearest 1/10th for easier visual access. The above values only tell half of the story, since the lengths of the first and second valve slides are already fixed as soon as (or truthfully, even before) they are opened. As stated above, the second valve tubing is fixed at 3.3 inches, so the first valve tubing should be constructed to lengthen the open pipe by approximately 6.6 inches, thereby fixing the length of the first valve slide to approximately 6.6 inches.

If the lengths of 3.3 and 6.6 inches are used respectively for the first and second valve slides, the following chart is the result:

Valve position

[Valve combination]

Adjusted length of tubing in inches

First [0]

56

Second [2]

56 + 3.3 = 59.3

Third [1]

56 + 6.6 = 62.6

Fourth [1,2]

56 + 6.6 + 3.3 = 65.9

Fifth [2,3]

56 + 3.3 + 11 = 70.3 (see below for 11)

Sixth [1,3]

56 + 6.6 + 11 = 73.6

Seventh [1,2,3]

56 + 6.6 + 3.3 + 11 = 76.9

The length of the third valve slide creates an interesting dilemma. It is true that the third slide can be long enough to accommodate a "truer" pitch for Ab, since this valve is not used as frequently as the other two. Perhaps the best length for this slide (a major third lower than C) is approximately 14 inches, which is fine for Ab, but much too flat for G#. This third slide, however, can be constructed to approximately 11 inches to accommodate either Ab or G# (Peterson 1957, 13-15). There are compromises to be made at all steps of the process! Any fixed length for the third valve slide affects the lengths of all of the lower valve positions and compounds the gradual sharpening of their pitches. The lengths of tubing in the above chart perhaps represent "true" valve lengths, but can never resolve the inherent intonation problems of a 3-valve system.

Another acoustical anomaly is that the fifth harmonic of any given harmonic series is manageably flat and the sixth one, sharp (Peterson 1957, 13-15). The pitch e2 (in open position, played as the fifth harmonic of C) is manageably flat in the C major, but would be noticeably sharp using valve combination 1-2 (as the sixth harmonic of A). In the D major scale, d2 (with first valve, played as the fifth harmonic of Bb) is manageably flat, while e2 (in open position, played as the fifth harmonic of C) sounds flatter than it actually is, because the next pitch in the scale, f 2 , is normally played with the first valve (as the sharp sounding sixth harmonic of Bb). As can be seen here, the intonation situation on a three valve system has many variables and is quite complex. It requires a good ear and years of training to be able to adjust quickly and naturally to different compositional environments.

The fraction 1/17 and not 1/12 is used in The Cornet: Practical advise on the production, valve limitations, mouthpieces, mutes, and other valuable information to the cornetist (Peterson 1957, 13-15). James Davenport, Chair of the Department of Physics, Virginia State University, Petersburg, Virginia; and James Beck, Professor of Chemistry, Virginia State University, Petersburg, Virginia confirm Peterson’s use of the fraction 1/17 (Beck 1999; Davenport 1999). These two Professors cite the references The Science of Sound (Rossing 1990, 226-228) and Physics and Music: The Science of Sound (White and White 1980, 255-257), but also state that 5.89% is actually more accurate than 1/17 or 5.8823529411%.

Why lengthen the pipe 1/17th and not 1/12th of its length? All of the following three subjective experiments and an objective mathematical discussion provide an answer:

The first experiment was simple. A note was played on a 65 inch-long bugle (the measurement was carefully taken in the middle of the entire length of the pipe). The tuning slide on the instrument was extended to a point where ½ step lower was aurally perceived. It was measured at a total length of 68.75 inches, or 65 + (5.78% of 65) inches. Not exactly 5.88%, but extremely close! For the second experiment, the tube was extended to the calculated length of 65 + (5.88% of 65) inches (or 68.82 inches), and was played. The pitch was perceivably the same as in the first experiment. For the third experiment, the tube was extended to 1/12 longer than its 65 inch length (or 70.4 inches) and the pitch was perceivably much too low!

The objective mathematical discussion provides a scientific explanation and is as follows: Theoretically, the length of any given tube is inversely related to its frequency. In other words, the longer the tube is, the smaller the number of vibrations per minute (and thus, a lower pitch is created). Another basic principle applies as well, i.e., if any given tube is twice as long as another, it will produce a pitch which is one octave lower. If 1/12th of the length of any given tube is gradually added to itself to lower its pitch one half step, the resulting length of the tube for the lower octave is much too long!

By logic, the process of adding any fraction (in this case 1/12 or 8.3333333333%) must be extended for each new length of tubing that is created:

Length in inches

Pitch relationship to the given length

Written pitch

56

The given length

c1 [also known as middle C]

60.6

Minor second lower

b

65.7

Major second lower

b flat

71.2

Minor third lower

a

77

Major third lower

a flat

83.6

Perfect fourth lower

g

90.5

Augmented fourth lower

g flat

98.1

Perfect fifth lower

f [theoretical length]

106.2

Minor sixth lower

e [theoretical length]

115.1

Major sixth lower

e flat [theoretical length]

124.7

Minor seventh lower

d [theoretical length]

135.1

Major seventh lower

d flat [theoretical length]

146.3

The octave lower?

c [theoretical length]

146.3" is much too large and if 1/12 of any given measurement is added to the measurement itself, the system does not work!

 

If 1/17th (or 5.8823529411%) of any given measurement is added to the measurement itself, however, the system works, as shown below:

Length in inches

Pitch relationship to the given length

Written pitch

56

The given length

c1 [also known as middle C]

59.3

Minor second lower

b

62.8

Major second lower

b flat

66.5

Minor third lower

a

70.4

Major third lower

a flat

74.5

Perfect fourth lower

g

78.9

Augmented fourth lower

g flat

83.6

Perfect fifth lower

f [theoretical length]

88.5

Minor sixth lower

e [theoretical length]

93.7

Major sixth lower

e flat [theoretical length]

99.2

Minor seventh lower

d [theoretical length]

105.0

Major seventh lower

d flat [theoretical length]

111.2

The octave lower!

c [theoretical length]

111.2" divided by 2 is equal to 55.6" (a number which is much closer to 56" than the final number of 146.3" in the previous chart!).

THE CONCLUSION IS AS FOLLOWS: adding approximately 1/17th to the length of any given tube to itself will lower it ½ step! It is true that the most correct number is probably somewhere between 5.8823529411% and 5.89% , but any mathematical discussion at this point becomes much too excessive!

To remove any possible doubts about these conclusions or to alleviate any possible anxiety that may have arisen from this section, the author of this document must explain that any fraction that was chosen, in fact, needed to lower a tube ½ step! The intuitive fraction of 1/12, however, is based on the principle that any tube must logically be divided into 12 equal parts, one for each ½ step. Its use also assumes that lowering any tube ½ step is a linear one. This process is not linear, however; but rather, exponential (Beck 1999). Using 1/12 can not work, because the tube is already one octave higher by the time the tube becomes half of its original length. The fraction 1/17, however, does work, since its use never implied that the pipe was to be divided into 17 equal ½ steps. 1/17 was only a value added to a given length to achieve a new length.

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