[John Phillips]

On 2007-06-30, William Sommerwerck wrote:
OK, so you get substantially all unwanted acoustic energy
out of the vinyl and into the platter by matching the acoustic
impedance. But where does it then go?

It's absorbed by the plastic itself. I'm not claiming _all_ of it is
absorbed (or for that matter, that "all" the energy leaves the LP). Rather,
the platter provides a much bigger chunk of substance to dissipate the
energy than the LP by itself.

If you do need to remove acoustic energy from the vinyl I suspect that
plastic(s) alone in the platter may not have a great acoustic loss factor
[0] so, as you said, you do need to make sure the platter is acoustically
lossy with some proper engineering.

By the way, about 10 years ago a well-known Canadian manufacturer produced a
platterless LP player -- the disk was supported only at the center! The
designer (whose name I will not repeat) claimed that air made a better
impedance match to the LP than a metal or plastic turntable platter! Not
surprisingly, this product didn't last long. If nothing else, it failed to
provide a stable azimuth for the pickup.

Yes indeed. The acoustic impedance of air is very low indeed [1].
It's about 10 million times less than vinyl. It's a VERY poor match so
the reflection coefficient is very close to 100%. At least in a vinyl -
aluminium interface the acoustic impedance change is just 3:1 or so and
the reflection coefficient is about 25% (i.e. 75% of the acoustic energy
is transmitted).

[0] A quick Google search didn't reveal that much detail so I am assuming
this and I may be wrong. I was also looking for the loss factor of
vinyl itself but with no luck so far.

[1] For longitudinal waves, acoustic impedance is the square root of the
product of the material's density and its Young's modulus.

--
John Phillips