Bear

Question for Mr. Nick D if he's still around:

Nick, I need a basic crash course in Physics of sound. Why does longer metals (more valves) create sharper pitches. Thxs.

Tim
Bear

nieuwguyski

Not to step on Nick's toes here, but I don't think anybody will be able to answer your question as posed. What do you mean? Are you asking why a four-valve piccolo trumpet is higher than a three-valve Bb trumpet?

rowuk

I think you have it backwards. The pitch is determined by the total length of the instrument, pushing a valve down makes the horn longer-lowering the pitch. The second valve is 1/2 tone lower, the first a whole tone and the third 1 1/2. The more valves that you push the longer the horn gets-the lower the tone.
Each valve combination has its series of "Partials"(not overtones). These are the notes that you can play without changing the valves that you are pushing. The intonation of these partials is basically controlled by the shape of the bell for low octave and primarily mouthpiece/leadpipe/gap for the upper octaves. This is of course way simplified, should get you started though.

Derek Reaban

Bear,

I think your question implies that when you play a 1-3 or 1-2-3 combination, the pitch needs to be adjusted (lowered / lengthened) to be in tune with the rest of the ensemble (i.e. pushing down more valves leads the horn to be sharper than what you would expect to hear for that specific note). This is by design. The 1st valve slide and 3rd valve slide are “a little” too short (sorry I don’t have the exact numbers). This is to allow the player to adjust easily to the tonal center of a specific piece and the degree of the scale being played.

For instance, let’s say that you are playing the 2nd trumpet part from the chorale in Mahler Symphony No. 2. There is a printed low Ab that repeats for several bars, and since the part is for trumpet in F, you are playing a low Db (i.e. 1-2-3). That pitch acts as the minor third in one chord followed immediately by acting as a major third in the next chord (everyone changes around you).

We all know that we need to throw our slides out to be “in-tune” in most circumstances when we have a 1-2-3 combination, but in this case you have to strongly consider the position of the note in the chord. The minor third need to be 16 cents higher than equal temperament and the major third needs to be 14 cents lower than equal temperament. If the slide were “just the right length” this would lead to problems in this specific example.

The note could be lowered (slide thrown out / lengthened), but the note would not be able to be raised (the slide all the way in would not be short enough). This movement of 30 cents [Minor 3rd (+16) / Major 3rd (-14)] happens infrequently, but horns are designed to be able to “cover all bases” and allow the player that focuses on Just Intonation in all playing situations to be “in-tune” as well as locking in to the most resonant placement of the note on the horn.

Having a horn with adjustable slides in the "all the way in position" be a little too short allows for these special situations.

Hope this helps!

Fred

Congratulations, I think you have correctly understood the question of Mr. Bear. When reading your answer, I understood the question!

But the answer isn't correct!

Basically, the valve pipe should be longer when combined with other slides.

Here is an explanation I found:

"How long should the second valve pipe be? To make the calculation simple, let's imagine that we are designing valves to make a narrow bore, chromatic didjeridu (ie we are going to add valves to a simple cylindrical pipe), whose natural length is 100.0 cm. If the second valve lowers the pitch of the natural pipe by an equal tempered semitone, then it must decrease the frequency by 5.9% and thus increase the length of the instrument by 5.9 cm. So the difference in frequency between - - - and - 2 - is now 5.9%. However, if we are to use the fingering scheme above, we also want the second valve to lower the pitch from 5 to 6 semitones below the natural frequency. Suppose that the first and third valves do lower it by 5 equal tempered, in other words they decrease the pitch by 33.5% and so must together add a length of 33.5 cm. So fingering 1 - 3 has pipe of length 133.5 cm. If we add our 5.9 cm, we increase this to 139.4 cm. This is an increase of 5.9 cm/133.5 cm, which is only 4.4%, which is only three quarters of an equal tempered semitone."

-> 5.9 cm is 5.9% (a semitone) of the pipe when only the second valve is down.
-> 5.9 cm is 4.4% (less than a semitone) of the pipe when 1 2 3 valves are down.

Hope it's understandable.

Fred

R.A.S.

Fred
Now that makes sense!
Could you share with us the source of your quote?
(It would be interesting to read more.)

Fred

Oh sorry, I've forgotten to post the link. Here is:

http://www.phys.unsw.edu.au/~jw/brassacoustics.html#valves

Fred

Derek Reaban

Fred,

You wrote:
I read through the information that you quoted and it is exactly right. I’m not sure why you would say that my answer isn’t correct, though.

The text you referenced looked only at the length of the second valve slide and assumed that the 1st and 3rd valve slides were the correct length to lower the pitch a perfect fourth. Then it says that to lower the pitch an additional half step the 2nd valve slide is too short (only lowers the pitch by ¾ of a equally tempered semitone). Yup! I agree.

The text that you referenced simply says that the instrument is too short when all three valves are depressed, which is what the original poster was asking about. I provided a specific situation that helps fill in the “why”.

I’ll stand by my answer!

Vulgano Brother

Oh yeah? When I play a c with valves 2&3 the pitch goes down. Explain that one, please. (hee-hee)

Fred

O.k, I will try to explain that (But consider that's it's late here and I begin to be tired).
And consider that all my explanation are basic explanation; the physic of brass instruments is a little bit more complex. You need to study a little more to understand fully this matter.

Let's go to the explanation.
I've said that when you combine two or three valves the pitch is sharper than it has to be.
But in your particular case, there is another phenomenon that occur.
Consider your C with -23. That's a major third higher than the Ab, which is your foundamental with valves 23.

But is this a tempered major third. No!
It's a harmonic major third. The pipe will basically produce the harmonics (that's not exactly right because a trumpet is not cylindrical).
In the series of harmonics the major third is the 5th partial, so the frequency of your C with -23 will be 5/4 times the frequency of your Ab.
5/4=1.25
The tempered major third is 1.2599. (=2^(4/12): 4 semitones)
since 1.25<1.2599, the harmonic third (Which in our case is C 23) will be down the C in open position.

Is it clear?
(If not, please wait until tomorrow for more explanations)

Fred