"Paul Draper" wrote in message
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"Porky" wrote in message
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"Jim Carr" wrote in message
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"Paul Draper" wrote in message
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That's odd. I don't find this hard to imagine, slowing it down in my
mind. The highest speed of the diaphragm forward will result in the
highest "sweeping up" of air molecules, resulting in the highest
local
density. In this mental image, then, the middle of the throw (for a
fundamental tone) would produce the peak of the oscillation in the
medium.

Thanks for the feedback. Take me through a simple sine wave. It starts
at
the zero line. It does a semicircle above the line an a semicircle
below
the
line. Describe for me the motion of the speaker in relation to that
and at
what point the wave is created.

Actually it's an ellipse, not a semicircle, and that does make a
difference.
The cone starts from zero at a fairly rapid rate of acceleration, and
the
rate slows gradually until it reaches the peak excursion where it
reverses
direction and gradually starts accelerating in the other direction, the
rate
of acceleration increases to a peak value as it passes through zero
again,
then the rate starts decreasing again until it again reverses direction
again at the "bottom" of the cycle, where the rate starts increasing
again
until it again reaches zero, then the whole thing repeats again. Due to
the
inertia and compressability of air, the air pressure at the cone's
surface
will vary in proportion to the cone's speed, and this pressure variation
becomes a sound wave.
Personally, I think the "virtual" source point of the sound wave
being
radiated by the speaker in the rest point of the cone at the center of
the
excursion limits, but this has been a subject of debate for a long time,
and
there are quite a few other theories. One other theory is that the
instantaneous virtual sound source point is the position of the cone at
any
given point in time, which is where the notion of Doppler distortion in
a
speaker originates. Note that if the virtual source is really the rest
point
of the cone, then there is no Doppler shift in a loud speaker because
the
virtual source is not moving with respect to the listener. If the latter
theory is correct, then Doppler shift is produced by a loudspeaker
because
the virtual source is moving with respect to the listener.


I will note that even though this is a contentious thread, I am making
a
sincere request to help me envision this. Maybe I'm thinking too hard
or
have some sort of mental block on this issue.

This is perhaps why a square wave is hard to push through a speaker,
because it demands instantaneous accelerations of the diaphragm.

I always though it was because most speakers are round! Nyuk!

Nope, you can push a square wave through a speaker, but you can only do
it
once because the sharp edges will tear up the cone. :-)
Seriously, some very good speakers will produce a pretty good
approximation
of a square wave at higher frequencies, but none will do it at lower
frequencies, and no speaker will reproduce a true square wave at any
frequency because a reproducing a true square wave would require that
the
cone travel instantaneously from one excursion limit to the other, and
as
long as the cone has mass, that ain't gonna happen!

Actually, this isn't quite right. The pressure fluctuations that
correspond to a sound wave don't match with the *position* profile of
the speaker, but with the *velocity* profile of the speaker. Thus the
requirement of a square wave (in pressure) is that the speaker
instantaneously accelerate to its maximum velocity, proceed from one
end of its throw to the other at constant (maximum) velocity, and then
at the other end of the throw instantaneously accelerate to maximum
negative velocity. The way to think about the motion under such a
circumstance is like the little ball in a Pong game. For just the
reason you mention, real speakers won't achieve that hard,
instantaneous bounce at the extrema, but this sure isn't the same as
the even more extreme motion you describe.


In order to accurately reproduce a square wave, the time required for the
cone to travel from its negative excursion to its positive excursion must
equal the rise time of the square wave, and since a perfect square wave has
a rise time of zero the travel would have to be instantaneous. By "excursion
limit" in my previous post, I did not mean the speaker's excursion limits, I
meant the amount of excursion necessary to reproduce the wave at the desired
volume.