Predicting Pitch from a picture
Apr 10th, 2006

I've been working on a chart to use to try to predict the pitch of an instrument from a picture.

Sometimes a photo will have accompanying information such as bell size, overall length, etc. as a reference point. On other occasions, there is something else in the picture that will give a clue. For example, sometimes instruments are photographed sitting on a deck made of 2x4s. (The 4" side is closer to 3.5").

One of the problems with deal with photographs is that perspective distorts what the true measurement is. For example, one picture I checked recently had a length and a width measurement. But the picture had the instrument slanted. When I tried to measure the length of the 3rd valve slide based on length, I got a result that was in-between likely options. But when I used the width (which was less subject to perspective distortion in this example.) I got a measure that was within the chart.

I expect that the chart will be more accurate the larger the instrument. However, it is usually the horns in the alto to contrabass range that are confusing. Also, it seems preferable to first try for measuring the 3rd valve slide as it is longer and therefore more tolerant of measurement error. However, sometimes that is not possible as in the case of many sousaphones, for example that have their tuning slides wrapped up in various turns and twists. In which case, it may be necessary to use the 1st or 2nd valve tubing. Additionally, this is too rough a measure to be able to distinguish between high pitch and low pitch. In fact, it is probably too rough a measure to distinguish between adjacent pitches (Bb vs C, F vs Eb, etc.)

The logic of the table is: A straight line measure is going to be more accurate than trying to account for lengths of tubing in bows and curves to enter the valves, etc. There is no advantage to measuring both sides of the tube. One will be sufficient. So, the chart is based one side of the straight section of valve slide tubing. Additionally, the table is based on only a few examples, and could/should be expanded to be more comprehensive. So, here goes:

Horn Bell Dim Length 1st 2nd 3rd Pitch
Marchand alto 8.5 22 2 1.25 5.25 Eb HP
York Alto 7.75 27 2.5 1.25 6.5 Eb LP
Jaubert Tenor 9.25 22 2.5 1 7.5 Bb HP
L&H Baritone 11 27 3 1.5 7 Bb LP
L&H Tuba 11.5 28.5 5.75 1.5 12 Eb

Reply #1 - Apr 11th, 2006

As an example: A picture of an instrument was described as an Eb Alto horn. But the measurement given was length, which was 36 in. (Now, it should be a clue that a 36in Alto horn would be one big honking alto horn).

So, if the horn is 36 in long, by measuring the straight section of the 3rd valve slide, we get 12 in. Egads, its an Eb Tuba!


Reply #3 - Apr 11th, 2006

OK, here's how I helped a guy on brass-forum.co.uk figure out the pitch of an unknown horn. But you must know your algebra and your semitone constant for it to work, as well as some practicality in the observations of the playing details of the horn (I know, this sounds a little bit like Sherlock Holmes, because it is!):

Measure the third valve slide, if possible, and if not, any slide where you can get a reasonably close measurement as to its actual length. The slide must be fully pushed in.

A note one semitone lower than another requires @ 1.0594631 times the length of the open tubing, as this number is the 12th root of 2, the multiplier needed to get 12 equal semitones in equal temperament out of an octave, which is, of course, double the pitch, and therefore double the length. The length of the entire horn is then determined from the function of multiplication of the proportions derived from the known valve slide lengths, which are geometric and not linear in nature.

If you know the length of the second valve slide, one semitone, you use 1.0594631. If you know the length of the first valve slide, two semitones, then you use 1.0594631^2. If you have the length of the third valve slide, three semitones, you use 1.0594631^3. (1.0594631^12, which = @2, is the octave)

So, if you have the total length of the third valve slide, as y, then x, the length of the horn is as the following equation:

x + y = x(1.0594631^3)

If you have the total length of the first valve slide, the equation is:

x + y = x(1.0594631^2)

And if you have the length of the second slide, the equation is:

x + y = x(1.0594631)

For academic sake, if by chance you have the length of the 4th valve slide, 5 semitones lower, then the equation is:

x + y = x(1.0594631^5)

The long whole step of a thumb valve won't work, as it is made to be used in conjunction with the 4th valve, so it can't be used to determine the length of the open horn.

Solve for x, and you have the pitch of the horn:
BBb = @18 feet, or @216 inches
CC = @16 feet, or @192 inches
Eb = @13.5 feet, or @162 inches
F = @12 feet, or 144 inches


Of course, your result will always be a little off, as it is very hard to measure the exact curves of the valve slides into the valve casings, but you will be close enough that it will be apparent which it is.

But here's the kicker that I didn't post on TubeNet: if your horn is high pitch instead of low pitch, then the length of the horn will be @440/452, the respective pitches of high and low pitch expressed as another proportion, or @ .97345 times low pitch length. So, for high pitch horns, the lengths are:

BBb = @17.5 feet, or @210 inches
CC = @15.6 feet, or @187 inches
Eb = @13.1 feet, or @158 inches
F = @11.7 feet, or @140 inches

I hope this helps!

Let's see on your example below, if one leg of the 3rd valve slide is 12 inches, then the whole slide is @24 inches. Applying the formula, then:

x + 24 = x(1.0594631^3)
x + 24 @= 1.19X
24 = @.19x
24/.19 = x
x = @126 inches, which is 10.5 feet, which is more like an F tuba of some sort. But Besson only made a very few F tubas, and they were comp, not non-comp as in the picture, and everybody else's F tubas are either front or rotary valves. So, you must add an approximation of the lateral length of the curve in valve slide connecting the two legs and the return crook into the valve, to get a true total length of the 3rd valve slide. From the picture proportions the lateral additions may be estimated at @2.5 inches for each one, @5 inches total, bringing the total length of the third valve slide to @29 inches. Doing the algebra, this gives us @152 inches, which is indeed close to the theoretical length of an Eb high pitched tuba. It's longer than an F tuba. Additionally, the picture shows the valve slides pulled considerably, which is presumably for current playing, and which also lends credence to the horn originally being in high pitch. So, my vote is for the picture of a high pitch Eb tuba.

Reply #4 - Apr 12th, 2006

Thanks for the formulas. I knew they were out there and have used some similar, but it is good to have the high science posted.

But my method is intended for those more routine questions, such as looking at a horn on eBay.

And I think that I may have not explained one part of it well enough. I'm eliminating the curved sections of tubing from the calculations. They are too hard to estimate from a photograph. So, in my example, 12 inches is NOT 1/2 of the 3rd valve tubing. It is just one of the straight sections.

To use the scientific formulas, you need to account for all the tubing that diverts the air from the main run; in the valve, the tubing connecting the tuning slide to the valve, the bow, etc. But no measurement is any more accurate than its least accurate component.
So, a simpler method is what I was proposing.


Reply #6 - Apr 12th, 2006
Remember that the other main point, which I admit I did not dwell on as much as the math, is to observe the horn carefully, expand your knowledge about the models you do or should know about, what models are similar, and therefore be able to extrapolate a reasonable conclusion or guess from what you can tell from the picture, all as set forth in my last paragraph.

For example, if you've never "seen" a Reynolds Contempora, you can tell a lot about it from extrapolation from both King and Olds, because Reynolds was hired from York to work for H. N. White, where the patterns were developed, then after his first retirement went to work for Olds, so the cosmetics are more Olds like.

This just takes time to learn, so you can get a better idea about what's going on.

Reply #7 - Apr 12th, 2006,
Another TubeNetter recommends using the 1st slide, instead to remove possible regional valve length problems.

Reply #8 - Apr 12th, 2006
I assume the Tubenetter is thinking of those that use something other than 1 1/2 steps for the 3rd valve. (Course I have a horn that has the 1/2 step on the 1st valve!)
I suppose the prudent thing would be to check one and verify with the other.
Quote:
Remember that the other main point, which I admit I did not dwell on as much as the math, is to observe the horn carefully