Author: rsholmes
Date: 2007-04-23 19:25
Since the tenon rings are in close contact with the wood, and both are in contact with the air in the bore, I would expect all three to stay at about the same temperature -- that is, the metal rings, the air in the bore, and the layer of wood closest to the bore. The ice cube analogy isn't completely apt because the metal isn't cold. Metal feels cold to the touch because it conducts your body heat away more efficiently than wood does. To the extent that the wood and metal and bore air are all the same temperature, the metal will not cool anything off.
But of course an instrument in a cool room is warm on the inside and cool on the outside, and the bore does cool off. In a metal instrument (especially a single-wall one), the metal will conduct the heat from the bore to the room more efficiently than wood would -- so all other things being equal the bore on the metal instrument cools down faster (and it's the temperature of the bore, I think, that's most important here).
In the case of tenon rings, first of all the amount of metal involved is much smaller, and second, the metal isn't exposed directly to the room air -- heat still has to go through a (thinner) wood layer. In principle, yes, the bore would cool more quickly than in an otherwise identical instrument without metal tenon rings. In practice I'm guessing the difference would be minuscule, and that most of the cooling would take place at the bell and the tone holes anyway. But that's only a guess. Certainly it'd still cool more slowly than, say, a sax.
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The Clarinet BBoard
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