Agent,

I've read that when calculating modes, etc, for a room with an acoustic
ceiling, you should measure the height to the actual hard surface above the
acoustic tiles.

Yes.

It makes sense that as you APPROACH full absorption, you would APPROACH
infinite effective height (i.e. zero reflection). But in practice, these
are only approachable limits, not actually attainable, right?

Right. You'd need (guessing) about 3 feet of dense rigid fiberglass to
absorb 100 percent at 100 Hz. And then there's 60 Hz and 40 Hz etc which
would require even more absorbing material. I've been in the large anechoic
chamber at IBM's acoustic lab, and they have *huge deep* fiberglass pyramids
all over. Even with all that the room is guaranteed anechoic only down to
100 Hz.

Wouldn't any low frequencies that are not fully absorbed be reflected in
the same timeframe as if the absorption was not there? Or does the
absorption affect phase as well as amplitude?

Yes, any waves that are not absorbed will get through to the rigid boundary
behind and be reflected back. As for phase, I honestly don't know. I do know
that waves passing through absorption are slowed down a little. So you'd
think that would affect the phase. As you add absorption to a room you
actually lower its resonant frequencies a little. This is why loudspeakers
have fiberglass inside them, to make the box seem acoustically larger than
it really is. So, yeah, phase may be affected, but this is more a theory
issue. In practice, just add as much absorption as you can. It will never be
100 percent in a typical size room no matter what you do.

--Ethan