I got a flute being delivered this week… old German flute. Very excited about it.
However, for safety I had it delivered to work. I walk to work- it’s about a mile from home. I’d probably be taking it home wednesday evening- expecting a low of 15 for that day. I could arrange for someone to pick me up, but would it be safe to carry it home (in a pistol case) in the cold?
If it were me I would put
the pistol case in my coat in some secure way, so that body heat would keep it
warm. Terry’s experiment notwithstanding, I would not walk
home with the flute exposed to that low a temperature. Better
safe than sorry, especially with a flute you’ve never seen.
If the flute gets cold, let it warm up gradually.
Might the flute be quite cold upon delivery? The probability of its having sat in a cold delivery vehicle for some hours is high. I would follow the advice already given and add that perhaps it would be wise to leave the flute in its case inside the warm house/apartment for a day to allow it to warm slowly. I know this would be hard to do with a new (to you) flute, but I’d surely hate to have it crack on you due to a sudden change in temperature.
A flute with a metal sleeve, in the barrel and/or in the head joint, likely won’t crack at sub-freezing temperatures, and that’s because metal “shrinks” at a faster rate than wood. However, the opposite is also true, that metal “expands” faster than wood, which is why it’s ever so important to ever so gradually warm a flute, before any attempt to play it.
If nothing else, warm a flute in your hands, completely, from tip to toe, before playing it.
Yes the rate of thermal contraction/expansion would be higher for metal than wood, given metal’s higher thermal conductivity, but the coefficient of thermal expansion for wood is higher. Still the real question is just how much dimensional expansion/contraction is there?
Let’s assume we bring a room temperature flute to the freezing cold, i.e., 68degF → 14degF (or 20degC → -10degC). I measured one of my flute liners and it is 0.7846 inches (19.93mm) OD and appears to be made of stainless steel. Following the example here:
dl = (8x10^-6 in/in degF) x .7846 in x (14 - 68 degF) = -0.00034 in
In other words, the liner does not effectively shrink at all. The coefficient of linear thermal expansion for oak across the grain is about 3 times of steel. But again, you multiple -.00034 in by 3 and you get almost negligible change. So there you go, Terry’s limited experiment pretty much validates the math. But I doubt either will bust the myth.