"chung" wrote in message
rvers.com...

See above. The objection is, of course, your saying that the drop is
proportional to the resistance of the conductor. And since you have not
admitted that mistake, we can go on .

Look, the only physical property that a wire has with relation to current
and voltage, is it's resistance. Voltage is not a property of wire. Current
is not a property of wire.

To say the voltage drop on a wire is proportional to it's resistance is just
plain not wrong.

It's a simple case of a analysis with a dependant variable and an dependent
variable.

You tell me the current, I'll apply the proportion of the resistance and
calculate the voltage drop.

This is exactly like an analogy of the length of a building's shadow. It's a
function of it's height and the sun's angle. Nothing I do to the building
can impact the sun angle. The only parameter of the building I can adjust is
it's height. Therefore, the length of a building's shadow is indeed
proportional to it's height. When the sun angle changes, so does the shadow
length, but sun angle is not a property of the building. It is perfectly
valid syntax to state that the length of a building's shadow is proportional
to it's height. In much the same way it's perfectly valid to state that the
voltage drop on a conductor is proportional to it's resistance -- that being
the only physical property the wire possesses that can change the voltage
drop.

-afh3

I'll try not to mention, over and over, as has been your style so far, that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far, that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far, that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far, that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far, that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

"Bruce" wrote in message
...
"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

I suppose that has some merit when considering the back-emf generation of
the woofer as it fails to precisely track the applied signal. Has anyone
done any modeling or measurements to see what level of output impedance as a
shunt would be neccesary to create even the slightest benefit? It seems to
me that even if there were appreciable back-emf in the woofer driver
circuit, the high-pass filter section of the tweeter driver cross-over
network would block it anyway in the single-wire configuration.

I'm thinking that a green pen on the CD would probably make more audible
difference ;-)

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

I suppose that has some merit when considering the back-emf generation of
the woofer as it fails to precisely track the applied signal. Has anyone
done any modeling or measurements to see what level of output impedance as a
shunt would be neccesary to create even the slightest benefit? It seems to
me that even if there were appreciable back-emf in the woofer driver
circuit, the high-pass filter section of the tweeter driver cross-over
network would block it anyway in the single-wire configuration.

I'm thinking that a green pen on the CD would probably make more audible
difference ;-)

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

I suppose that has some merit when considering the back-emf generation of
the woofer as it fails to precisely track the applied signal. Has anyone
done any modeling or measurements to see what level of output impedance as a
shunt would be neccesary to create even the slightest benefit? It seems to
me that even if there were appreciable back-emf in the woofer driver
circuit, the high-pass filter section of the tweeter driver cross-over
network would block it anyway in the single-wire configuration.

I'm thinking that a green pen on the CD would probably make more audible
difference ;-)

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:fNp1c.35731$PR3.677741@attbi_s03:


"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:5mo1c.450966$I06.5065611@attbi_s01
"Arny Krueger" wrote in message
...
"afh3" wrote in message
news:IOn1c.450824$I06.5064681@attbi_s01

Is it not true that the voltage drop across the conductors is
only proportional to the resistance of the conductor itself, and
not impacted by the load presented at the output end?

Nope, its a voltage divider and the impedance of he wire and the
load are relevant.

The statement I made was an assertion that the voltage drop across
the conductor is proportional to the resistance of the conductor.
The voltage at the end of one of the conductor pairs will be
exactly the same as the voltage at end of the other conductor
pair, minus the difference in the voltage drop across each pair of
conductors REGARDLESS of what passive or reactive component is
connected to each end. Read Kirchoff's laws if this is unclear.

However, the voltage drop will depend on whatever passive or reactive
component is connected to each
end.


No it won't. Put whatever you want at either end of the conductor
pair, and Kirchoff's Laws GUARANTEE that the voltage at both ends will
be identical -- minus the difference of the voltage drop in each
conductor pair.

This is patently not isolation.

My ASCII art probably sucks but I'll try anyway.

---------(+)--------
Z1 Z2
---------(-)--------

The generator(power supply) impedance isn't infinity. Actually, it's
nearly a short circuit at the + and - point....so....

Z1 and Z2 are the loads presented by the two separate speaker driver
circuits in a biwired arrangement. No values for Z1 or Z2 will ever
result in the voltages across Z1 or Z2 varying by more than the
difference of the voltage drop of the conductors to the (in this case)
left of the power-supply (in the middle) and those on the right.

(+)-------------------
Z1 Z2
(-)-------------------


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

I suppose that has some merit when considering the back-emf generation of
the woofer as it fails to precisely track the applied signal. Has anyone
done any modeling or measurements to see what level of output impedance as a
shunt would be neccesary to create even the slightest benefit? It seems to
me that even if there were appreciable back-emf in the woofer driver
circuit, the high-pass filter section of the tweeter driver cross-over
network would block it anyway in the single-wire configuration.

I'm thinking that a green pen on the CD would probably make more audible
difference ;-)

-afh3

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far,
that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................

Cute. Nice dodge.

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far,
that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................

Cute. Nice dodge.

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far,
that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................

Cute. Nice dodge.

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 20:32:40 GMT, "afh3" wrote:

I'll try not to mention, over and over, as has been your style so far,
that
in your last sentence above you claimed I have "not" admitted a mistake.
Turns out, I didn't have to.

Since the EE is clear, maybe we should all agree that English isn't
your first language, and move on..................

Cute. Nice dodge.

"Bruce" wrote in message


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

Yes, I've noticed that most bi-wiring proponents are weak on the concept of
networks with passive versus active components.

"Bruce" wrote in message


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

Yes, I've noticed that most bi-wiring proponents are weak on the concept of
networks with passive versus active components.

"Bruce" wrote in message


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

Yes, I've noticed that most bi-wiring proponents are weak on the concept of
networks with passive versus active components.

"Bruce" wrote in message


The bi-wiring proponents look at z1 and z2 as also being generators,
therefore, with the amplifier impedance in between them as in the top
pictoral, you get more isolation from Z1 to Z2 than in the lower one,
because you're used the amplifier and wiring scheme to form a Tee
network.

Yes, I've noticed that most bi-wiring proponents are weak on the concept of
networks with passive versus active components.

"afh3" wrote in message
news:wlR1c.45144$PR3.910451@attbi_s03

I suppose that has some merit when considering the back-emf
generation of the woofer as it fails to precisely track the applied
signal.


The so-called "back emf" is well-modelled as an increase in the impedance of
the driver.

"afh3" wrote in message
news:wlR1c.45144$PR3.910451@attbi_s03

I suppose that has some merit when considering the back-emf
generation of the woofer as it fails to precisely track the applied
signal.


The so-called "back emf" is well-modelled as an increase in the impedance of
the driver.

"afh3" wrote in message
news:wlR1c.45144$PR3.910451@attbi_s03

I suppose that has some merit when considering the back-emf
generation of the woofer as it fails to precisely track the applied
signal.


The so-called "back emf" is well-modelled as an increase in the impedance of
the driver.

"afh3" wrote in message
news:wlR1c.45144$PR3.910451@attbi_s03

I suppose that has some merit when considering the back-emf
generation of the woofer as it fails to precisely track the applied
signal.


The so-called "back emf" is well-modelled as an increase in the impedance of
the driver.

"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be neccesary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifer impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

I have seen calculations that show some theoretical improvement, but the
calculated stock cofiguration was something like 40dB already at the
crossover frequency, so the extra few dB come out to give you what(?)...we
don't really know as far as audiblity is concerned, personally, I agree
that it wouldn't be audible in the vast majority of cases. In those cases
where it is, I doubt any audiophile would own that sort of speaker system.
(Although they still believe they are making some wonderful improvement to
their gold-plated rumpty-thunp speaker system, since they know better than
the manufacturer's engineers)

"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be neccesary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifer impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

I have seen calculations that show some theoretical improvement, but the
calculated stock cofiguration was something like 40dB already at the
crossover frequency, so the extra few dB come out to give you what(?)...we
don't really know as far as audiblity is concerned, personally, I agree
that it wouldn't be audible in the vast majority of cases. In those cases
where it is, I doubt any audiophile would own that sort of speaker system.
(Although they still believe they are making some wonderful improvement to
their gold-plated rumpty-thunp speaker system, since they know better than
the manufacturer's engineers)

"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be neccesary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifer impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

I have seen calculations that show some theoretical improvement, but the
calculated stock cofiguration was something like 40dB already at the
crossover frequency, so the extra few dB come out to give you what(?)...we
don't really know as far as audiblity is concerned, personally, I agree
that it wouldn't be audible in the vast majority of cases. In those cases
where it is, I doubt any audiophile would own that sort of speaker system.
(Although they still believe they are making some wonderful improvement to
their gold-plated rumpty-thunp speaker system, since they know better than
the manufacturer's engineers)

"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be neccesary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifer impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

I have seen calculations that show some theoretical improvement, but the
calculated stock cofiguration was something like 40dB already at the
crossover frequency, so the extra few dB come out to give you what(?)...we
don't really know as far as audiblity is concerned, personally, I agree
that it wouldn't be audible in the vast majority of cases. In those cases
where it is, I doubt any audiophile would own that sort of speaker system.
(Although they still believe they are making some wonderful improvement to
their gold-plated rumpty-thunp speaker system, since they know better than
the manufacturer's engineers)

"Bruce" wrote in message
...
"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be necessary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big
time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifier impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

Yeah, this is why I was asking about measurements. Adding the amp's variable
output impedance over frequency to the mix of the impulse response of the
cross-over networks, it's starts to get to be a bit much. I wasn't even
going to suggest modeling, but I thought maybe somebody had been crazy
enough to try it -- like I was once with a related question.

In the early 80's, right after EE school, when I thought I pretty much knew
everything about this stuff, I brought an RLC bridge, HP sweep generator,
one of the first digital o'scopes made, and some other sundry equipment home
from work and completely modeled every part of my speakers -- drivers,
cross-over networks (even the individual specific components) and then threw
them into a SPICE model that I ran on their TI thermal printing terminal
(with acoustic coupler!) that I brought home and connected surreptitiously
to the (company that I worked for's) big mainframe (a way-high-end MULTICS
machine that cost about as much money as God had at that time.) Since the
lab was a classified area, it was easier to get all this equipment *home*,
than it would have been to get a pocket full of capacitors and coils *in*.
(I could just imagine the face on the guard if I had tried to explain why I
was bringing my woofers into the lab, heh.)

My goal was to try to develop cross-over networks for *my* specific drivers
that would minimize the phase shift at the cross-over point, and (yup, I
wanted the world) present a nearly constant 8 ohm non-reactive load across
the entire 20-20k frequency range. I had originally started the project
doing the entire calculation by hand -- and still have the 20+ pages of
complex-number-space math where I tried to solve about 20 simultaneous
equations with about as many variables. I went to visit one of my old profs
too see if he could give me any tips for simplifying it, and he seemed
pretty impressed that I was even trying it -- but I'm real sure I heard him
laughing as I walked down the hall when I left.

So, back the TI Silent 700. Took me quite a while to get it all punched in,
and then, oh yeah, she cranked for a while. Even thought the department I
was working for at the time didn't pay for processing time (we were doing
MMW RF patch antenna R&D at the time for a DOD contract -- Ron Reagan and
all, you know) the job still submitted the stats at the end of it's nearly
full-day run -- about $20,000 if I recall correctly. Seems it used a bit o'
CPU.

Anyway, I remember finding out (I still have the output around here
somewhere, although the thermal paper had faded to the point of
near-unreadability last time I dug it up) that all I needed were some
capacitors with negative values in the uF range and some coils that couldn't
be physically represented too. Disappointed, I then let the design program
create it's own network, rather than vary the values in the existing one.
This time, it ran for a day and a half, and I recall coming up with about a
16th order filter -- the specific type of which I no longer remember -- that
only partially met my criterial (and spending another $25k worth of computer
time in the process.)

So, I humbly reassembled my speakers and decided they sounded just fine as
they were.

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be necessary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big
time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifier impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

Yeah, this is why I was asking about measurements. Adding the amp's variable
output impedance over frequency to the mix of the impulse response of the
cross-over networks, it's starts to get to be a bit much. I wasn't even
going to suggest modeling, but I thought maybe somebody had been crazy
enough to try it -- like I was once with a related question.

In the early 80's, right after EE school, when I thought I pretty much knew
everything about this stuff, I brought an RLC bridge, HP sweep generator,
one of the first digital o'scopes made, and some other sundry equipment home
from work and completely modeled every part of my speakers -- drivers,
cross-over networks (even the individual specific components) and then threw
them into a SPICE model that I ran on their TI thermal printing terminal
(with acoustic coupler!) that I brought home and connected surreptitiously
to the (company that I worked for's) big mainframe (a way-high-end MULTICS
machine that cost about as much money as God had at that time.) Since the
lab was a classified area, it was easier to get all this equipment *home*,
than it would have been to get a pocket full of capacitors and coils *in*.
(I could just imagine the face on the guard if I had tried to explain why I
was bringing my woofers into the lab, heh.)

My goal was to try to develop cross-over networks for *my* specific drivers
that would minimize the phase shift at the cross-over point, and (yup, I
wanted the world) present a nearly constant 8 ohm non-reactive load across
the entire 20-20k frequency range. I had originally started the project
doing the entire calculation by hand -- and still have the 20+ pages of
complex-number-space math where I tried to solve about 20 simultaneous
equations with about as many variables. I went to visit one of my old profs
too see if he could give me any tips for simplifying it, and he seemed
pretty impressed that I was even trying it -- but I'm real sure I heard him
laughing as I walked down the hall when I left.

So, back the TI Silent 700. Took me quite a while to get it all punched in,
and then, oh yeah, she cranked for a while. Even thought the department I
was working for at the time didn't pay for processing time (we were doing
MMW RF patch antenna R&D at the time for a DOD contract -- Ron Reagan and
all, you know) the job still submitted the stats at the end of it's nearly
full-day run -- about $20,000 if I recall correctly. Seems it used a bit o'
CPU.

Anyway, I remember finding out (I still have the output around here
somewhere, although the thermal paper had faded to the point of
near-unreadability last time I dug it up) that all I needed were some
capacitors with negative values in the uF range and some coils that couldn't
be physically represented too. Disappointed, I then let the design program
create it's own network, rather than vary the values in the existing one.
This time, it ran for a day and a half, and I recall coming up with about a
16th order filter -- the specific type of which I no longer remember -- that
only partially met my criterial (and spending another $25k worth of computer
time in the process.)

So, I humbly reassembled my speakers and decided they sounded just fine as
they were.

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be necessary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big
time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifier impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

Yeah, this is why I was asking about measurements. Adding the amp's variable
output impedance over frequency to the mix of the impulse response of the
cross-over networks, it's starts to get to be a bit much. I wasn't even
going to suggest modeling, but I thought maybe somebody had been crazy
enough to try it -- like I was once with a related question.

In the early 80's, right after EE school, when I thought I pretty much knew
everything about this stuff, I brought an RLC bridge, HP sweep generator,
one of the first digital o'scopes made, and some other sundry equipment home
from work and completely modeled every part of my speakers -- drivers,
cross-over networks (even the individual specific components) and then threw
them into a SPICE model that I ran on their TI thermal printing terminal
(with acoustic coupler!) that I brought home and connected surreptitiously
to the (company that I worked for's) big mainframe (a way-high-end MULTICS
machine that cost about as much money as God had at that time.) Since the
lab was a classified area, it was easier to get all this equipment *home*,
than it would have been to get a pocket full of capacitors and coils *in*.
(I could just imagine the face on the guard if I had tried to explain why I
was bringing my woofers into the lab, heh.)

My goal was to try to develop cross-over networks for *my* specific drivers
that would minimize the phase shift at the cross-over point, and (yup, I
wanted the world) present a nearly constant 8 ohm non-reactive load across
the entire 20-20k frequency range. I had originally started the project
doing the entire calculation by hand -- and still have the 20+ pages of
complex-number-space math where I tried to solve about 20 simultaneous
equations with about as many variables. I went to visit one of my old profs
too see if he could give me any tips for simplifying it, and he seemed
pretty impressed that I was even trying it -- but I'm real sure I heard him
laughing as I walked down the hall when I left.

So, back the TI Silent 700. Took me quite a while to get it all punched in,
and then, oh yeah, she cranked for a while. Even thought the department I
was working for at the time didn't pay for processing time (we were doing
MMW RF patch antenna R&D at the time for a DOD contract -- Ron Reagan and
all, you know) the job still submitted the stats at the end of it's nearly
full-day run -- about $20,000 if I recall correctly. Seems it used a bit o'
CPU.

Anyway, I remember finding out (I still have the output around here
somewhere, although the thermal paper had faded to the point of
near-unreadability last time I dug it up) that all I needed were some
capacitors with negative values in the uF range and some coils that couldn't
be physically represented too. Disappointed, I then let the design program
create it's own network, rather than vary the values in the existing one.
This time, it ran for a day and a half, and I recall coming up with about a
16th order filter -- the specific type of which I no longer remember -- that
only partially met my criterial (and spending another $25k worth of computer
time in the process.)

So, I humbly reassembled my speakers and decided they sounded just fine as
they were.

-afh3

"Bruce" wrote in message
...
"afh3" wrote in
news:wlR1c.45144$PR3.910451@attbi_s03:


I suppose that has some merit when considering the back-emf generation
of the woofer as it fails to precisely track the applied signal. Has
anyone done any modeling or measurements to see what level of output
impedance as a shunt would be necessary to create even the slightest
benefit? It seems to me that even if there were appreciable back-emf
in the woofer driver circuit, the high-pass filter section of the
tweeter driver cross-over network would block it anyway in the
single-wire configuration.

I'm thinking that a green pen on the CD would probably make more
audible difference ;-)

-afh3




This answer won't be very comprehensive. The flu has bitten me big
time....

So far, I've only seen static calculations at the crossover frequencies,
which is where I believe the effect will be maximum.

The real problem is that there are so many variables. The individual
speaker impedances, the amplifier impedance, the effects of the crossover
sections, the impedances of the wires used. The solution is set-up
specific.

Yeah, this is why I was asking about measurements. Adding the amp's variable
output impedance over frequency to the mix of the impulse response of the
cross-over networks, it's starts to get to be a bit much. I wasn't even
going to suggest modeling, but I thought maybe somebody had been crazy
enough to try it -- like I was once with a related question.

In the early 80's, right after EE school, when I thought I pretty much knew
everything about this stuff, I brought an RLC bridge, HP sweep generator,
one of the first digital o'scopes made, and some other sundry equipment home
from work and completely modeled every part of my speakers -- drivers,
cross-over networks (even the individual specific components) and then threw
them into a SPICE model that I ran on their TI thermal printing terminal
(with acoustic coupler!) that I brought home and connected surreptitiously
to the (company that I worked for's) big mainframe (a way-high-end MULTICS
machine that cost about as much money as God had at that time.) Since the
lab was a classified area, it was easier to get all this equipment *home*,
than it would have been to get a pocket full of capacitors and coils *in*.
(I could just imagine the face on the guard if I had tried to explain why I
was bringing my woofers into the lab, heh.)

My goal was to try to develop cross-over networks for *my* specific drivers
that would minimize the phase shift at the cross-over point, and (yup, I
wanted the world) present a nearly constant 8 ohm non-reactive load across
the entire 20-20k frequency range. I had originally started the project
doing the entire calculation by hand -- and still have the 20+ pages of
complex-number-space math where I tried to solve about 20 simultaneous
equations with about as many variables. I went to visit one of my old profs
too see if he could give me any tips for simplifying it, and he seemed
pretty impressed that I was even trying it -- but I'm real sure I heard him
laughing as I walked down the hall when I left.

So, back the TI Silent 700. Took me quite a while to get it all punched in,
and then, oh yeah, she cranked for a while. Even thought the department I
was working for at the time didn't pay for processing time (we were doing
MMW RF patch antenna R&D at the time for a DOD contract -- Ron Reagan and
all, you know) the job still submitted the stats at the end of it's nearly
full-day run -- about $20,000 if I recall correctly. Seems it used a bit o'
CPU.

Anyway, I remember finding out (I still have the output around here
somewhere, although the thermal paper had faded to the point of
near-unreadability last time I dug it up) that all I needed were some
capacitors with negative values in the uF range and some coils that couldn't
be physically represented too. Disappointed, I then let the design program
create it's own network, rather than vary the values in the existing one.
This time, it ran for a day and a half, and I recall coming up with about a
16th order filter -- the specific type of which I no longer remember -- that
only partially met my criterial (and spending another $25k worth of computer
time in the process.)

So, I humbly reassembled my speakers and decided they sounded just fine as
they were.

-afh3

Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Oh don't I hear you. In college I once saw a kid drop down to his knees and
start sobbing as the person in front of him using the card reader (standard
issue of course for computer science class back then) turned around more
quickly than he was ready for and caused the several-hundred-card deck in
his hands to fly out if his grasp and flutter to the floor like so much
confetti -- apparently just hours before the project was due.

I remember trying to help him pick them up, and he just kept sobbing "No,
no, go away, I can do it, I can do it."

God I felt sorry for him.

"Bruce" wrote in message
...
Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Oh don't I hear you. In college I once saw a kid drop down to his knees and
start sobbing as the person in front of him using the card reader (standard
issue of course for computer science class back then) turned around more
quickly than he was ready for and caused the several-hundred-card deck in
his hands to fly out if his grasp and flutter to the floor like so much
confetti -- apparently just hours before the project was due.

I remember trying to help him pick them up, and he just kept sobbing "No,
no, go away, I can do it, I can do it."

God I felt sorry for him.

"Bruce" wrote in message
...
Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.

Oh don't I hear you. In college I once saw a kid drop down to his knees and
start sobbing as the person in front of him using the card reader (standard
issue of course for computer science class back then) turned around more
quickly than he was ready for and caused the several-hundred-card deck in
his hands to fly out if his grasp and flutter to the floor like so much
confetti -- apparently just hours before the project was due.

I remember trying to help him pick them up, and he just kept sobbing "No,
no, go away, I can do it, I can do it."

God I felt sorry for him.

"Bruce" wrote in message
...
Wow, what a great trip down memory lane. Brings back memories of the room
full of ADM-3s running a whopping 110 baud on the college campus I went to
that could access the mainframe there, although the main input was still
punch cards. That was in the late 70s. We sure have come a long way.