Hi!
Looking at power and air pressure - we
don't have to care about them as long as we don't want to
include thermal effects
Isn't the heat proportional to the mean power ? Then you just do
[*~] with itself and then some kind of [rpole~] to account for the
accumulation thereof. After that I don't really know what to do
with that.
one could feed the output of this [rpole~] into a [*~] to the input
signal (or at any place later, but then it has to be cared for the
delay of the doppler vd~ too, so better do it first). The exact
parameters can be found by experiments, but as written in an earlier
mail the signal reduction can be up to 7dB (I would use slightly
more). I don't have time right now, but maybe I'll experiment
later...
or nonlinearities of the air I think.
I don't know them at all. I've never heard of anyone taking them
into account.
I think there aren't many too. The only case I can imagine and I
heard of is for huge negative signals, where you can't go beyond 0
pascal of air pressure and the signal is clipped physically - but I
doubt this ever happens in small speakers. Also I'm not sure if this
only happens in compression speakers.
The pressure directly in front of the cone
is related to the acceleration I think, but I'm not sure about
that.
It has to : the speaker makes a sound by pushing and pulling on
the air, and that changes the pressure.
yes, but it could also be seen in that way: the cone's travel
compresses air in an enclosed volume. when the cone is at its travel
peak (the acceleration is 0) the air is compressed at maximum. And
I'm not sure if this is the case for a speaker in "free field" (not
sure if that's the exact term) too. I know that there's something
called radiation resistance - but I know nothing useful about it
yet.
Can anybody confirm that? I think that's
not trivial to answer anyways, because already 10cm farther from
the speaker the pressure and air velocity are different.
It has to... the reason why you hear the sound and why sound has a
speed, is because neighbouring pressure differences cause pressure
differences to propagate. It's a second-order differential
equation, as the position of air particles is proportional to
their acceleration. The Laplacian of the wave function along x,y,z
is proportional to the 2nd derivative of the wave function along
t... I'd write it like :
D[D[f,x],x] + D[D[f,y],y] + D[D[f,z],z] = D[D[f,t],t] / v²
Where v² is the square of the speed of sound.
With a slight coordinate change using imaginary numbers, you can
see it as a Laplacian along x,y,z,t instead, in 4-dimensional
spacetime, and the Laplacian is equal to zero. But that's only if
the air is considered frictionless :)
sorry, I don't understand that equation I think (I'm not
understanding difference equations very well in general) and also I
have no idea of Laplacians until now (I'll read about it and maybe
understand later)
Ah, that means that the mechanical amplitude (travel) of the wave
is much smaller for treble than bass, is that right ?
exactly.
cheers
Martin