Re: ATT: Rick Denney


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Posted by fred boyd on October 16, 2003 at 11:45:00:

In Reply to: ATT: Rick Denney posted by tim on October 16, 2003 at 03:57:29:

OK, I'm not a physicist and I don't even play one on TV. I'm not even a very good measure counter. But, try this answer. In the back row, we sometimes don't have enough to do, and we've finished our rehearsal magazine. So , there's nothing left to do but ponder these important questions in life.

In the real world, you would adjust your slides so that it sounds in-tune regardless of the temperature or those pesky oboists (who think they can hear the freaking grass grow). "Whatever works" is the correct answer. You know that, I know that, we all know that, but let's commence to ponderin' and cypherin' (ala Jethrow Bodine) the theoretical.

I'll have to take your word on the various frequencies and wave lengths. It sounds like you've done your homework. Try this amateur logic:

If you increase a wave length of 262.4 cm up to 262.9cm because of temperature, thats an increase of 0.5 cm, or roughly 0.19054878048780487804878048780488 percent. Oh, what the heck, lets call it 0.2 percent...2 tenths of a percent, or a factor of 0.002. So, it stands to reason that you would want to increase the length of tubing in play by a factor of 0.002 to get the proper tubing length for the increased temperature and resulting wave length. So, now all you have to do is determine the length of tubing and multiply it.

I have no idea what the length of open tube of a CC tuba is. It doesn't matter if its a CC tuba, BB flat tuba, French horn, or ram's horn...the math is the same...length of tubing times the correction factor. My guess of the length of unwrapped tubing of an open CC tuba is around 15 feet, or around 500cm. Correct me if I'm wrong, but don't forget that we're dealing only with the open horn...not the valves. So, 0.002 times 500 cm equals one cm. Ergo, you'd have to add one cm to the tubing length to theoretically compensate for the temperature rise, using your figures and my illogic.

Don't forget that a tuning slide has two legs, so the actual pull would be .5 cm.

Yeah, sure, we usually play C in the staff on the fourth harmonic. No matter. The tuning slide affects the entire instrument, regardless of the harmonic. The math is the same if you calculate using a factor (0.2%).

What do you think? There must be one of those DMA-type or PHD-type tuba players out there. I think I've just missed another entrance because of devoting limited brain power to this instead of counting rests.

keep on honkin'

fred boyd




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