Answer to Andrew Randle

[Roger Andersson]

You are wrong. The speakers do NOT rock in motion with the music if
you use the type of feet I am talking about. On the contrary the
speakers are totally still and there is NOT the slightest vibration or
movement. If you play at 120 db level and put your fingertips on the
top of the speaker you will not feel any vibration or movement. I have
tried this. I can get you in tuch with the highly skilld and
experienced engineers and scientists that have designed the feet. I
can prove this on the highest possible scientifical level. You do NOT
know what you are talking about. Also: About biwiring. What I
actually ment was that the highpass and lowpass part of the filter
MUST get the same signal. This is not the case if you use
biwiring. The highpass filter and the lowpass filter should never be
on separate boards and must share the same conectors. I hope THIS
will clear things!

Regards Roger Andersson. local editor of the soundtechnical society of
sweden.

PS! You can NOT impress me with any titels . I am a highly educated
person and I have studied physics at the highest level.

[Stewart Pinkerton]

Roger Andersson <rog...@algonet.se> writes:

>You are wrong. The speakers do NOT rock in motion with the music if
>you use the type of feet I am talking about. On the contrary the
>speakers are totally still and there is NOT the slightest vibration or
>movement.

This statement is totally and ludicrously wrong. Basic Newtonian
physics INSISTS that action and reaction are equal and opposite, so if
you have reciprocating motion of a bass cone of a certain mass at a
certain height above the cabinet base, it will inject a rocking couple
into the speaker cabinet. If the mounting is compliant (e.g. 'rubber'
feet), then the cabinet will most certainly rock, the amplitude being
in direct ratio to the relative masses of driver and cabinet.

>If you play at 120 db level and put your fingertips on the
>top of the speaker you will not feel any vibration or movement.

I'd be fascinated to know which particular domestic loudspeaker you
have been using which is able to generate this kind of acoustic
output. A tenth of this power level is more than the majority can
manage.

>I have
>tried this. I can get you in tuch with the highly skilld and
>experienced engineers and scientists that have designed the feet. I
>can prove this on the highest possible scientifical level.

I seriously doubt that you can PROVE it on any level.

>You do NOT know what you are talking about.

Hmmmm. On the evidence so far presented, that would seem to be more
applicable to your good self.

>Also: About biwiring. What I
>actually ment was that the highpass and lowpass part of the filter
>MUST get the same signal. This is not the case if you use
>biwiring. The highpass filter and the lowpass filter should never be
>on separate boards and must share the same conectors. I hope THIS
>will clear things!

This is of course extremely arguable. Some designers do not agree with
bi-wiring, but simply because there is no guarantee that the same
wiring will be used for bass and treble - if it is, there is no
problem (if also no improvement!). Other designers favour bi-wiring,
but I cannot see what difference it makes whether or not the various
sections of the crossover are on the same board.

>Regards Roger Andersson. local editor of the soundtechnical society of
>sweden.

>PS! You can NOT impress me with any titels . I am a highly educated
>person and I have studied physics at the highest level.

Well, I've never yet come across anyone who made that kind of claim,
who could actually back it up. REAL scientists stand behind their
arguments, not their qualifications. The "I have four doctorates to
your three masters degrees" argument has been tried before, but
doesn't actually advance the debate, even if true.
--
Stewart Pinkerton | If you can't measure what you're making,
A S P Consulting | how do you know when you've got it made?
(44) 1509 880112 |
"I canna change the laws o' physics" - the other Scotty

[epotter]

Mr. Pinkerton, how does bi-wiring work? Do you have a passive
crossover right at the amplifier, and direct connexions at the
drivers, or what? Probably there are several people puzzled by all
this.

[Stewart Pinkerton]

epo...@mail.utexas.edu (epotter) writes:

I'm tempted to give the short answer - bi-wiring doesn't work!

That might however be seen as a little two-faced since I do bi-wire my
own speakers! :-)

There are two approaches to bi-wiring. By far the most common is where
the passive crossover remains inside the speaker box but is physically
divided into bass and treble sections which are brought to separate
pairs of terminals. A much less common, more expensive but potentially
superior system is where there is, as you suggest, an external
crossover box which may be situated close to the amplifier, with
direct connections from the crossover to the speaker drive units via
separate terminal pairs on the speaker box. The principal advantage of
such a system is that the external crossover may easily be replaced by
an active crossover feeding a separate power amp for each driver, the
only real (and major!) improvement over simple single wiring, IMNVHO.

One very important point is that - despite popular mythology -
bi-wiring of the first kind does NOT reduce the cable impedance below
that of a single cable (assuming similar cables are used for bass and
treble). This should be obvious if you consider what the crossover is
there for. You'll typically get better results from a bi-wired speaker
by strapping the cables together at the speaker end, which will revert
to single wiring but will halve both the resistance and the inductance
of the cable.

For those of a curious nature, the reason I bi-wire my Apogee Duetta
Signatures is to use 22AWG solid-core for the treble ribbon, allowing
the cable resistance to replace the internal equalising resistor and
thereby enabling fine tuning of the frequency balance to suit my
listening room. No mystical magical tweakery, just simple physics!

[Lee Meador]

Stewart Pinkerton wrote:

<a bunch of stuff including this paragraph>

> There are two approaches to bi-wiring. By far the most common is where
> the passive crossover remains inside the speaker box but is physically
> divided into bass and treble sections which are brought to separate
> pairs of terminals. A much less common, more expensive but potentially
> superior system is where there is, as you suggest, an external
> crossover box which may be situated close to the amplifier, with
> direct connections from the crossover to the speaker drive units via
> separate terminal pairs on the speaker box. The principal advantage of
> such a system is that the external crossover may easily be replaced by
> an active crossover feeding a separate power amp for each driver, the
> only real (and major!) improvement over simple single wiring, IMNVHO.

Another thing to remember is that in any passive crossover that
includes inductors (you may call them chokes or coils) you may cause
problems if things are positioned so that the coils are close
together. The actual orientation of the coils also matters. What you
do by placing them in the wrong places is to create a transformer. The
"sound" that goes through one coil is transferred electomagnetically
to the other coil.

To see if this is a problem try this. (Assume a two-way speaker.)
Unhook the input to the treble portion of the crossover where it comes
in from the amplifier. Unhook the output from the bass portion of the
crossover where it runs to the woofer. Now play something. Don't turn
the volume up to much.

If you hear sound coming from the tweeter, it is being transferred
"illicitly" from the low-pass portion of the crossover which should
only be feeding the woofer. You may be able to get rid of the sound
(or reduce it) by moving portions of the crossover around. It depends.

You should try the same thing with the tweeter unhooked and no input
to the low pass section. If you have threeway speakers or a passively
attached sub-woofer anything could bleed into anything else. Use good
sense, however, since the electromagnetic effects are severly limited
by increased distance. (eg. 18 inches or 50 cm)

Finally, this can really mess up the sound. A tweeter playing low
notes is going to distort something terrible. If the left and right
speakers have differing levels of "bleed through" the imaging will
likely suffer.

-- Lee Meador

[Ronald M. Hoering]

Here is my thoughts on Spikes, feet, and nothing but naked speaker
cabinet.

As the contact area goes down the force per unit area goes up. ie:
presure.

Spikes reduce this contact area to a minimum and so increase the
presure greatly over and unshod speaker.

Newton's #2 is not violated here it is just and issue of degree here.

The mass of the floor coupled to the building is much greater than
that of the speaker. So while the forces and therefore momenta are
equal the effect is that our house does not move much due to the
speaker's impulse.

Mr. Wizard time: Take a tall thing with a flat base and set it on the
floor. Kind of unstable isn't it. Now take three small things like
pennys or equal size marbles or BBs and place them equally spaced
under the base of the tall thing. Not so unstable now. If this did not
work for you doublecheck your execution of the experiment that is at
fault.

The 'diode' part comes from two notions.

First, is that the speaker being a transducer its energy in the form
of heat, sound, and cabinet borne vibration. As above the momentum of
the speaker due to the force is equal to the momentum of the
floor/building/earth due to the reaction force. Depending on were you
decouple, (at the floor, the room, the building, etc), you will notice
that the mass of the floor/building/earth acts as a mechanical ground
for the speaker. (In fact I believe that is what defines a mech
ground, huge differences in mass at the interface.)

Second, if the floor/building/earth were to move (like in California)
or near a highway the frequency of the even is usually very low, on
the order of tenths of cycles per second. The effect is to have
amazingly long wavelengths which on the scale of the base of a speaker
is effectively flat. That is the speaker will rise and fall
perdendicular to the support surface.

This all tends to let 'effective' vibration leave the speaker and not
get back in.

Sorry to be so windy.

--
______________________________________________________________________________
Ronald M. Hoering Physicist "What is all this math stuff anyway?"
University of Chicago Mathematician "How do I know, its just cool!"
Wanabe "Geometrodynamicist" Philosopher "What is physics?"

[Pete Goudreau]

> On Fri, 4 Oct 1996, rmho...@midway.uchicago.edu (Ronald M. Hoering) wrote,

[ cut -- rgd ]

> Newton's #2 is not violated here it is just and issue of degree here.

Well yeah, that's called static equilibrium...

> The mass of the floor coupled to the building is much greater than
> that of the speaker. So while the forces and therefore momenta are
> equal the effect is that our house does not move much due to the
> speaker's impulse.

One would get the same effect by bolting or glueing the spkrs to the
floor as well...spikes have nothing to do with this...

> Mr. Wizard time: Take a tall thing with a flat base and set it on the
> floor. Kind of unstable isn't it. Now take three small things like

hmmmm, no it isn't unstable at all unless the base and floor are not
flat...again this has nothing to do with spikes...

> pennys or equal size marbles or BBs and place them equally spaced
> under the base of the tall thing. Not so unstable now. If this did not
> work for you doublecheck your execution of the experiment that is at
> fault.

This isn't a stability "thing", this has to do with defining the plane
of contact between the spkr base and the floor...again, nothing to do
with spikes, etc. as you will notice that you have made the problem
disappear by using spherical objects as opposed to conical
objects. The simple reason for that is that they provide point contact
and thus define a plane geometrically. Naturally, the base of the
spkr cannot as it doesn't have three raised points, it will generally
have just one high spot and thus will rock...nothing magical here I'm
afraid...

> The 'diode' part comes from two notions.
>
> First, is that the speaker being a transducer its energy in the form
> of heat, sound, and cabinet borne vibration. As above the momentum of
> the speaker due to the force is equal to the momentum of the
> floor/building/earth due to the reaction force. Depending on were you
> decouple, (at the floor, the room, the building, etc), you will notice
> that the mass of the floor/building/earth acts as a mechanical ground
> for the speaker. (In fact I believe that is what defines a mech
> ground, huge differences in mass at the interface.)

Yes that is the definition of a mechanical ground but only in the
sense that the mechanical analog of voltage at this node is zero. The
mechanical analog of voltage (an "across" variable) is velocity...the
mech analog of current (a "through" variable) is force thus making
mass analogous to capacitance and the inverse of the spring constant
(compliance) the analog of inductance. This of course has absolutely
nothing to do with spikes, the audiophile concept of mechanical
ground, vibration conduction through mechanical diodes, or anything
else along these lines for that matter....

> Second, if the floor/building/earth were to move (like in California)
> or near a highway the frequency of the even is usually very low, on
> the order of tenths of cycles per second. The effect is to have
> amazingly long wavelengths which on the scale of the base of a speaker
> is effectively flat. That is the speaker will rise and fall
> perdendicular to the support surface.

Well, that seems intuitively true but it is not as can easily be seen
by the swaying of skyscrapers in LA during a temblor...and it's lots
of fun, too....%-{} Unfortunately this has not a thing to do with
spiking spkrs to the floor, or equipment to shelves, or racks to
floor, etc., etc.

> This all tends to let 'effective' vibration leave the speaker and not
> get back in.

This statement of course has absolutely nothing to do with the
preceeding ones either logically or semantically. This statement is
the equivalent of saying that vibration is like a fluid and can be
conducted out of a vibrating structure to some other massive structure
where it can harmlessly dissipate. Utter and complete nonsense.

Vibration is defined as a time dependent displacement of a node in a
mechanical structure, the frequency of which is dependent upon the
compliance tensor of the material and the effective mass at the
measured node and as such is the analog of voltage and you will notice
that the ground of an electrical circuit does not drain voltage away
from some node separated from ground by an impedance.

Spikes are very useful in coupling a spkr to mechanical ground in
order to define a node's vibe spectra relative to a fixed physical
reference so that added vibrations from a rocking spkr structure are
not added to the fundamental resonant spectra of the cabinet.

To think of them as "mechanical diodes" is quite decidedly
wrongheaded. Their use under equipment on shelves is a different
matter and is a bit more complex to visualize but they do not act as
"diodes" in that application either. When will this laughably
ignorant "theory" in the high-end audio community ever die the death
it so richly deserves?

> Sorry to be so windy.

That's OK, I do the same thing on occasion, like now...;-)

Cheers,
Pete