The Amount of Slide Adjustment Needed for a Given Pitch Change


Imagine moving a trombone slide out about halfway from the first to second position, causing the pitch to go down by half a semitone, or by 50 "cents". (A "cent" is one hundredth of a semitone in the same sense that a "semitone" is one twelfth of an octave.) The overall length of the instrument would be increased from Lo to Lo plus roughly half of L2. ("Lo" is the overall effective length of the instrument in first position or with no valves depressed, and "L2" is the length of tubing added when you go from first to second position, or the length of the second-valve tubing.) Half of L2 is 0.0595 × Lo/2, or Lo(S - 1)/2. To lower the pitch by just one cent, you would move the slide out roughly one hundredth of L2, or Lo(S - 1)/100 from first position. (Remember that "S" represents the semitone ratio, as explained in section 3.)

But a one-cent adjustment starting from a different position would have to be bigger, because the positions are farther apart. The “Lo” in that formula would have to be replaced with a longer value. For example, the length of a trombone in 6th position is about 1.333Lo because sixth position brings it down a perfect fourth. To make a one-cent adjustment from 6th position we then must move the slide a distance of approximately (1.333 Lo)(S-1)/100. As mentioned earlier, it is more convenient to express length adjustments in terms of L2 rather than Lo, and L2 is just (S-1) Lo. Using that trick, we conclude that a one-cent adjustment from 6th position is approximately (1.333/100) L2. Starting from a different position would require us to replace the “1.333” in that formula with a different ratio. In general, if “N” represents the number of semitones down from first position, then we must replace the 1.333 with S to the Nth power. So now we have a general slide-adjustment formula:

One-cent adjustment on the Nth semitone = approximately (L2/100) times SN.


Example: On a BBb tuba the length of the second valve tubing is about 6 inches, measured from piston to pull-ring. To lower the open instrument by one cent we must pull the tuning slide this far: (0.06 in.)(1.0595) = 0.0636 inches. (That's roughly one sixteenth of an inch, or 0.025 centimeters.) But when the 4th valve is down, we need 1.333 times as much pull, or 0.0848 inches to make a one-cent adjustment. In either case, pulling the slide ten times farther will result in approximately ten times as much pitch change. Similarly, a 13-cent adjustment of the first valve tubing length will be (13)(1.0595)(1.1224)(0.06 in.) = 0.93 inches for a BBb tuba. For a CC tuba it would be about 12% less, and for an F tuba it would be 2/3 as much.

If your 4th valve tubing is set for a perfect 4th and your 2nd valve tubing is set for a semitone on the open horn then using the second and fourth valves together will be pretty sharp, although better than 123. To correct that note you need to lengthen the combined tubing by 33% of the second valve tubing length. Since the second valve tubing length on a BBb tuba is about six inches from piston to pull-ring, you need to pull a slide about two inches when using 2&4.

Next: Tuning the Third Valve

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