I can’t seem to find the measurements of the total tube length of each valve for a CC tuba. I know it has been posted several times in the past but my initial search was unsuccessful.
I would greatly appreciate a reference link to any of the previous postings on the subject, and I would also appreciate it if someone would send me their own measurements.
Thanks in advance.
Mark Howle
CC Tuba valve tubing measurements....I know it’s in here s
-
markaustinhowle
- bugler
- Posts: 93
- Joined: Mon Aug 21, 2006 9:23 pm
-
iiipopes
- Utility Infielder
- Posts: 8580
- Joined: Tue Sep 06, 2005 1:10 am
OK, this is not absolute, because there are such variables as the diameter of the valves, the flare of the bugle, etc., but as a starting point, a CC tuba open bugle, if laid out straight mouthpiece to bell, is approximately 196 inches long, and the proportion for each half step is to multiply the base length by 2^1/12, or @1.0594631, for each half step, then subtract the length of the open bugle. So, here's the math:
196 X 2^1/12, or 1.0594631 = 207.65, 2nd valve length @ 11.65 inches
196 X 2^2/12, or 1.122462 = 220.00, 1st valve length @ 24 inches
196 X 2^3/12, or 1.1892071 = 233.08, 3d valve length @ 37.08 inches
196 X 2^5/12, or 1.3348399 = 261.63, 4th valve length @ 65.62 inches
Thumb valve, if used to get T4 low F in tune, the long whole step is (196 X 2^7/12) - (196 X 2^5/12) = (196 X 1.4983071) - (196 X 1.3348399) @= (293.66 - 261.63) = 32.33 inches.
Remember, these are theoretical lengths, not actual lengths, although they are close, to within an inch of actual.
196 X 2^1/12, or 1.0594631 = 207.65, 2nd valve length @ 11.65 inches
196 X 2^2/12, or 1.122462 = 220.00, 1st valve length @ 24 inches
196 X 2^3/12, or 1.1892071 = 233.08, 3d valve length @ 37.08 inches
196 X 2^5/12, or 1.3348399 = 261.63, 4th valve length @ 65.62 inches
Thumb valve, if used to get T4 low F in tune, the long whole step is (196 X 2^7/12) - (196 X 2^5/12) = (196 X 1.4983071) - (196 X 1.3348399) @= (293.66 - 261.63) = 32.33 inches.
Remember, these are theoretical lengths, not actual lengths, although they are close, to within an inch of actual.
Jupiter JTU1110
"Real" Conn 36K
"Real" Conn 36K
-
Chuck(G)
- 6 valves
- Posts: 5679
- Joined: Fri Mar 19, 2004 12:48 am
- Location: Not out of the woods yet.
- Contact:
What iiipopes said, but taking a different tack...
These are the numbers that I use when I'm building a tuba and it works for me. First, my numbers:
Valve 1 = 64.3 cm or 25.3"
Valve 2 = 31.2 cm or 12.3"
Valve 3 = 99.3 cm or 39.1"
Valve 4 = 175.7 cm or 69.2"
Valve 5 = 68 cm or 26.8"
This is for A440 at 20 C, so you'll need to shorten things a bit for tuning. I use a rather simple way to compute slide lengths. I first compute the wavelength of the second open partial (524.9 cm), then compute the wavelengths of the note one step, a half step, one and a half steps, etc. lower and subtract from them the open wavelength. I'm using 65.4 Hz as C, by the way.
Note that this is the entire length through the center of the tube, including the crook and the knuckles. Also, these will be a tiny bit bit long because of the effects of the conical bore and the bell, but I'd rather have my slides a bit too long and trim them back to be in tune rather than having them too short!
Another way is to compute lengths by the "rule of 17". This is based on the property that 18/17 is a fairly close approximation of the the 12th root of 2. There are two problems with this however. The first is that it's just an approximation and a bit on the short side; the second is that the rule must be applied iteratively, so the error accumulates.
Thus, to compute the length of the tube sounding B natural, one would take our wavelength for C (524.9 cm) and multiply it by 18/17; to get the length for Bb, we'd take that and multiply again by 18/17 and so on. As a "back of the envelope" method, it's not bad, but I wouldn't use it to compute where I was going to set my hacksaw.
These are the numbers that I use when I'm building a tuba and it works for me. First, my numbers:
Valve 1 = 64.3 cm or 25.3"
Valve 2 = 31.2 cm or 12.3"
Valve 3 = 99.3 cm or 39.1"
Valve 4 = 175.7 cm or 69.2"
Valve 5 = 68 cm or 26.8"
This is for A440 at 20 C, so you'll need to shorten things a bit for tuning. I use a rather simple way to compute slide lengths. I first compute the wavelength of the second open partial (524.9 cm), then compute the wavelengths of the note one step, a half step, one and a half steps, etc. lower and subtract from them the open wavelength. I'm using 65.4 Hz as C, by the way.
Note that this is the entire length through the center of the tube, including the crook and the knuckles. Also, these will be a tiny bit bit long because of the effects of the conical bore and the bell, but I'd rather have my slides a bit too long and trim them back to be in tune rather than having them too short!
Another way is to compute lengths by the "rule of 17". This is based on the property that 18/17 is a fairly close approximation of the the 12th root of 2. There are two problems with this however. The first is that it's just an approximation and a bit on the short side; the second is that the rule must be applied iteratively, so the error accumulates.
Thus, to compute the length of the tube sounding B natural, one would take our wavelength for C (524.9 cm) and multiply it by 18/17; to get the length for Bb, we'd take that and multiply again by 18/17 and so on. As a "back of the envelope" method, it's not bad, but I wouldn't use it to compute where I was going to set my hacksaw.