Re: Brand and Source for OFC speaker cable ?

[124]

Walt wrote:

> James wrote:
>
> > My amp is 50 watts per channel, so pretty modest. But, my "run" to my
> > speakers is 35 feet.
> >
> > I am not a "nut" for monster cable or magic cable that claims to work
> > wonders, but I do want very good quality, and use wire that will perform
> > well at what I believe it a bit of an extended distance.
>
> Any 12 gauge wire will be fine. You can probably go as far as 16
> without ever hearing any difference.

He can probably go as far as 18 without ever
hearing too much of a difference.

--124

[George M. Middius]

4 of 12 said:

> > Any 12 gauge wire will be fine. You can probably go as far as 16
> > without ever hearing any difference.

> He can probably go as far as 18 without ever
> hearing too much of a difference.

Some speaker manufacturers recommend thicker wire. Mine, for one. What
does the Hive have to say about manufacturers recommending 12-ga wire?

[Arny Krueger]

"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
in message ...
>
>
> 4 of 12 said:
>
>> > Any 12 gauge wire will be fine. You can probably go as far as 16
>> > without ever hearing any difference.
>
>> He can probably go as far as 18 without ever
>> hearing too much of a difference.
>
> Some speaker manufacturers recommend thicker wire. Mine, for one. What
> does the Hive have to say about manufacturers recommending 12-ga wire?

Just goes to show that your mind is really shot, Middius. You've commented
here on 12 gauge speaker wire in the past - not that long ago. What did you
say then?

"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
in message ...
>
>
> 4 of 12 said:
>
>> > Any 12 gauge wire will be fine. You can probably go as far as 16
>> > without ever hearing any difference.
>
>> He can probably go as far as 18 without ever
>> hearing too much of a difference.
>
> Some speaker manufacturers recommend thicker wire. Mine, for one. What
> does the Hive have to say about manufacturers recommending 12-ga wire?

The ability to hear an instantaneous difference in sound output as a
function of wire size is not the only reason to use heavier wire. There is
also the value of the amplifier power lost through the wire. Frequently
that value is greater than the increased price of heavier wire. Not only is
there a power loss, there is also an even greater loss in maximum power.

e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
..0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
power. Worse yet, it's a reduction of 5.4% in the maximum output of your
amplifier. If you choose #18 wire, multiply everything by about 4 !

If you paid big money for your amplifier, these are big numbers--well worth
considering.

Norm Strong

[ScottW]

> wrote:
> "George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
> in message ...
> >
> >
> > 4 of 12 said:
> >
> >> > Any 12 gauge wire will be fine. You can probably go as far as 16
> >> > without ever hearing any difference.
> >
> >> He can probably go as far as 18 without ever
> >> hearing too much of a difference.
> >
> > Some speaker manufacturers recommend thicker wire. Mine, for one. What
> > does the Hive have to say about manufacturers recommending 12-ga wire?
>
> The ability to hear an instantaneous difference in sound output as a
> function of wire size is not the only reason to use heavier wire. There is
> also the value of the amplifier power lost through the wire. Frequently
> that value is greater than the increased price of heavier wire. Not only is
> there a power loss, there is also an even greater loss in maximum power.
>
> e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
> .0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
> power. Worse yet, it's a reduction of 5.4% in the maximum output of your
> amplifier.

How do you figure this? Why is the power loss double in max power?
I don't think it is.

ScottW

[Pooh Bear]

ScottW wrote:

> > wrote:
> which is a loss of about 2.8% of your amplifier's
> > power. Worse yet, it's a reduction of 5.4% in the maximum output of your
> > amplifier.
>
> How do you figure this? Why is the power loss double in max power?
> I don't think it is.

For a sine wave peak power = 2x 'rms' power.

Graham

[Pooh Bear]

ScottW wrote:

> Pooh Bear wrote:
> > ScottW wrote:
> >
> > > > wrote:
> > > which is a loss of about 2.8% of your amplifier's
> > > > power. Worse yet, it's a reduction of 5.4% in the maximum output of your
> > > > amplifier.
> > >
> > > How do you figure this? Why is the power loss double in max power?
> > > I don't think it is.
> >
> > For a sine wave peak power = 2x 'rms' power.
>
> Ok.... I don't see how this affects power loss. If the cable is
> eating 2.8% of the power delivered by the amp... it will do so no
> matter the power level.

You're right. I expect that's where Norm got the idea from though.

Graham

[Chung]

ScottW wrote:
> > wrote:
>> "George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
>> in message ...
>> >
>> >
>> > 4 of 12 said:
>> >
>> >> > Any 12 gauge wire will be fine. You can probably go as far as 16
>> >> > without ever hearing any difference.
>> >
>> >> He can probably go as far as 18 without ever
>> >> hearing too much of a difference.
>> >
>> > Some speaker manufacturers recommend thicker wire. Mine, for one. What
>> > does the Hive have to say about manufacturers recommending 12-ga wire?
>>
>> The ability to hear an instantaneous difference in sound output as a
>> function of wire size is not the only reason to use heavier wire. There is
>> also the value of the amplifier power lost through the wire. Frequently
>> that value is greater than the increased price of heavier wire. Not only is
>> there a power loss, there is also an even greater loss in maximum power.
>>
>> e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
>> .0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
>> power. Worse yet, it's a reduction of 5.4% in the maximum output of your
>> amplifier.
>
> How do you figure this? Why is the power loss double in max power?
> I don't think it is.
>
> ScottW
>

I'll answer for Norm. Assume the max output voltage of the amp is V
(volts rms), and that the amp is an ideal voltage source with zero
output impedance.

Without the loss in the cable, the power available to drive the 4 ohm
speaker is V*V/4. Now conider the resistance of the cable being 0.112
ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
max power into that speaker is the square of that divide by 4, or

V*V*4/(4.112**2)

The ratio of that to the max available in the case of no cable
resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
in max. available power.

The power consumed in the cable resistance is (using I**2*R)

(V/4.112)**2*.112=V**2*(6.624E-3)

Compared that to the max power available to the 4 ohm speaker without
cable resistance of V**2/4, that cable power consumption is

(6.624E-3)*4=2.65%

So the cable power consumption is 2.65% of what's available to drive
that 4 ohm speaker from the amp.

You get a double hit because (1) there is consumption in the cable
resistance, and (2) the voltage available to the speaker is now reduced.

[ScottW]

Chung wrote:
> ScottW wrote:
> > > wrote:
> >> "George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
> >> in message ...
> >> >
> >> >
> >> > 4 of 12 said:
> >> >
> >> >> > Any 12 gauge wire will be fine. You can probably go as far as 16
> >> >> > without ever hearing any difference.
> >> >
> >> >> He can probably go as far as 18 without ever
> >> >> hearing too much of a difference.
> >> >
> >> > Some speaker manufacturers recommend thicker wire. Mine, for one. What
> >> > does the Hive have to say about manufacturers recommending 12-ga wire?
> >>
> >> The ability to hear an instantaneous difference in sound output as a
> >> function of wire size is not the only reason to use heavier wire. There is
> >> also the value of the amplifier power lost through the wire. Frequently
> >> that value is greater than the increased price of heavier wire. Not only is
> >> there a power loss, there is also an even greater loss in maximum power.
> >>
> >> e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
> >> .0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
> >> power. Worse yet, it's a reduction of 5.4% in the maximum output of your
> >> amplifier.
> >
> > How do you figure this? Why is the power loss double in max power?
> > I don't think it is.
> >
> > ScottW
> >
>
> I'll answer for Norm. Assume the max output voltage of the amp is V
> (volts rms), and that the amp is an ideal voltage source with zero
> output impedance.
>
> Without the loss in the cable, the power available to drive the 4 ohm
> speaker is V*V/4. Now conider the resistance of the cable being 0.112
> ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
> max power into that speaker is the square of that divide by 4, or
>
> V*V*4/(4.112**2)
>
> The ratio of that to the max available in the case of no cable
> resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
> in max. available power.
>
> The power consumed in the cable resistance is (using I**2*R)
>
> (V/4.112)**2*.112=V**2*(6.624E-3)
>
> Compared that to the max power available to the 4 ohm speaker without
> cable resistance of V**2/4, that cable power consumption is
>
> (6.624E-3)*4=2.65%
>
> So the cable power consumption is 2.65% of what's available to drive
> that 4 ohm speaker from the amp.
>
> You get a double hit because (1) there is consumption in the cable
> resistance, and (2) the voltage available to the speaker is now reduced.

Ok... but you always get that hit... not just at peak levels. His
original post implied there would be a non-linearity introduced which
would be a horrific problem... but that isn't true.

The power delivered to the speaker is always reduced by the factor
(Rsp/Rsp+Rcab)**2 which in the example comes .946 or 5.37%

My mistake was accepting the 2.8% number. That number is wrong... and
in his example under no circumstances will the power loss be limited to
2.8% of what the amp delivers.

ScottW

[chung]

ScottW wrote:
> Chung wrote:
>
>>ScottW wrote:
>>
> wrote:
>>>
>>>>"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
>>>>in message ...
>>>>
>>>>>
>>>>>4 of 12 said:
>>>>>
>>>>>
>>>>>>>Any 12 gauge wire will be fine. You can probably go as far as 16
>>>>>>>without ever hearing any difference.
>>>>>
>>>>>>He can probably go as far as 18 without ever
>>>>>>hearing too much of a difference.
>>>>>
>>>>>Some speaker manufacturers recommend thicker wire. Mine, for one. What
>>>>>does the Hive have to say about manufacturers recommending 12-ga wire?
>>>>
>>>>The ability to hear an instantaneous difference in sound output as a
>>>>function of wire size is not the only reason to use heavier wire. There is
>>>>also the value of the amplifier power lost through the wire. Frequently
>>>>that value is greater than the increased price of heavier wire. Not only is
>>>>there a power loss, there is also an even greater loss in maximum power.
>>>>
>>>>e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
>>>>.0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
>>>>power. Worse yet, it's a reduction of 5.4% in the maximum output of your
>>>>amplifier.
>>>
>>> How do you figure this? Why is the power loss double in max power?
>>> I don't think it is.
>>>
>>> ScottW
>>>
>>
>>I'll answer for Norm. Assume the max output voltage of the amp is V
>>(volts rms), and that the amp is an ideal voltage source with zero
>>output impedance.
>>
>>Without the loss in the cable, the power available to drive the 4 ohm
>>speaker is V*V/4. Now conider the resistance of the cable being 0.112
>>ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
>>max power into that speaker is the square of that divide by 4, or
>>
>>V*V*4/(4.112**2)
>>
>>The ratio of that to the max available in the case of no cable
>>resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
>>in max. available power.
>>
>>The power consumed in the cable resistance is (using I**2*R)
>>
>>(V/4.112)**2*.112=V**2*(6.624E-3)
>>
>>Compared that to the max power available to the 4 ohm speaker without
>>cable resistance of V**2/4, that cable power consumption is
>>
>>(6.624E-3)*4=2.65%
>>
>>So the cable power consumption is 2.65% of what's available to drive
>>that 4 ohm speaker from the amp.
>>
>>You get a double hit because (1) there is consumption in the cable
>>resistance, and (2) the voltage available to the speaker is now reduced.
>
>
> Ok... but you always get that hit... not just at peak levels.

The power hit is the reduction in max. available power. If you are not
trying to get the max power out of the amplifier, then you can still get
the amp to deliver the required power to the speaker.

For instance, if you only need V/2 across the speaker terminals, the amp
can still do that even though the cable has a fairly large resistance.
It's only when you try to get as high a voltage as possible into the
speaker that there is a reduction in the power available.


> His
> original post implied there would be a non-linearity introduced which
> would be a horrific problem... but that isn't true.

I didn't get that impression at all, from his post. Everything is
linear, and we simply are using Ohm's Law.

> The power delivered to the speaker is always reduced by the factor
> (Rsp/Rsp+Rcab)**2 which in the example comes .946 or 5.37%
>
> My mistake was accepting the 2.8% number. That number is wrong... and
> in his example under no circumstances will the power loss be limited to
> 2.8% of what the amp delivers.

He said that 2.8% of the amp's power is lost in the cable resistance. As
I showed, 2.65% of the max power available without cable loss is
dissipated in the cable resistance. If you compare the power dissipated
in the cable as a fraction of the power delivered to the speaker, then
that number is .112/4 = 2.8%. (The same current flows through cable
resistance and speaker.) He was referrring to the loss in the cable as a
fraction of the power delivered to the speaker.

>
> ScottW
>

[ScottW]

chung wrote:
> ScottW wrote:
> > Chung wrote:
> >
> >>ScottW wrote:
> >>
> > wrote:
> >>>
> >>>>"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
> >>>>in message ...
> >>>>
> >>>>>
> >>>>>4 of 12 said:
> >>>>>
> >>>>>
> >>>>>>>Any 12 gauge wire will be fine. You can probably go as far as 16
> >>>>>>>without ever hearing any difference.
> >>>>>
> >>>>>>He can probably go as far as 18 without ever
> >>>>>>hearing too much of a difference.
> >>>>>
> >>>>>Some speaker manufacturers recommend thicker wire. Mine, for one. What
> >>>>>does the Hive have to say about manufacturers recommending 12-ga wire?
> >>>>
> >>>>The ability to hear an instantaneous difference in sound output as a
> >>>>function of wire size is not the only reason to use heavier wire. There is
> >>>>also the value of the amplifier power lost through the wire. Frequently
> >>>>that value is greater than the increased price of heavier wire. Not only is
> >>>>there a power loss, there is also an even greater loss in maximum power.
> >>>>
> >>>>e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
> >>>>.0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
> >>>>power. Worse yet, it's a reduction of 5.4% in the maximum output of your
> >>>>amplifier.
> >>>
> >>> How do you figure this? Why is the power loss double in max power?
> >>> I don't think it is.
> >>>
> >>> ScottW
> >>>
> >>
> >>I'll answer for Norm. Assume the max output voltage of the amp is V
> >>(volts rms), and that the amp is an ideal voltage source with zero
> >>output impedance.
> >>
> >>Without the loss in the cable, the power available to drive the 4 ohm
> >>speaker is V*V/4. Now conider the resistance of the cable being 0.112
> >>ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
> >>max power into that speaker is the square of that divide by 4, or
> >>
> >>V*V*4/(4.112**2)
> >>
> >>The ratio of that to the max available in the case of no cable
> >>resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
> >>in max. available power.
> >>
> >>The power consumed in the cable resistance is (using I**2*R)
> >>
> >>(V/4.112)**2*.112=V**2*(6.624E-3)
> >>
> >>Compared that to the max power available to the 4 ohm speaker without
> >>cable resistance of V**2/4, that cable power consumption is
> >>
> >>(6.624E-3)*4=2.65%
> >>
> >>So the cable power consumption is 2.65% of what's available to drive
> >>that 4 ohm speaker from the amp.
> >>
> >>You get a double hit because (1) there is consumption in the cable
> >>resistance, and (2) the voltage available to the speaker is now reduced.
> >
> >
> > Ok... but you always get that hit... not just at peak levels.
>
> The power hit is the reduction in max. available power. If you are not
> trying to get the max power out of the amplifier, then you can still get
> the amp to deliver the required power to the speaker.

Ok... like anyone cares about that little semantic.
>
> For instance, if you only need V/2 across the speaker terminals, the amp
> can still do that even though the cable has a fairly large resistance.
> It's only when you try to get as high a voltage as possible into the
> speaker that there is a reduction in the power available.

Reduction in power available... no. Reduction in power delivered...
always.
>
>
> > His
> > original post implied there would be a non-linearity introduced which
> > would be a horrific problem... but that isn't true.
>
> I didn't get that impression at all, from his post. Everything is
> linear, and we simply are using Ohm's Law.

Well I did and that would be a big problem... this problem is really
insignificant.
Its always been there and always will be.

>
> > The power delivered to the speaker is always reduced by the factor
> > (Rsp/Rsp+Rcab)**2 which in the example comes .946 or 5.37%
> >
> > My mistake was accepting the 2.8% number. That number is wrong... and
> > in his example under no circumstances will the power loss be limited to
> > 2.8% of what the amp delivers.
>
> He said that 2.8% of the amp's power is lost in the cable resistance. As
> I showed, 2.65% of the max power available without cable loss is
> dissipated in the cable resistance.

Now read what you said... suddenly you can have no loss but still
have resistance.
Sorry... not possible. The loss is due to the resistance.... no
resistance...no loss.
No loss...no resistance. You don't get to separate one from the
other.

>If you compare the power dissipated
> in the cable as a fraction of the power delivered to the speaker, then
> that number is .112/4 = 2.8%. (The same current flows through cable
> resistance and speaker.)

The same current always flows through the cable resistance and the
speaker... they are in series you know, and being in series.. the
current through both is dependent on the sum of the resistance of
both... never just one.

>He was referrring to the loss in the cable as a
> fraction of the power delivered to the speaker.

Yes and its a situation that never happens. If the cable has
resistance .... and all do... it will both have a voltage drop and
reduce the current delivered by the amp. You simply cannot have one
without the other. Never happen. You will ALWAYS have the cable
reduce the power delivered to the speaker by 5.37% and never by 2.8%.

ScottW

[chung]

ScottW wrote:
> chung wrote:
>
>>ScottW wrote:
>>
>>>Chung wrote:
>>>
>>>
>>>>ScottW wrote:
>>>>
>>>>
> wrote:
>>>>>
>>>>>
>>>>>>"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
>>>>>>in message ...
>>>>>>
>>>>>>
>>>>>>>4 of 12 said:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>>>Any 12 gauge wire will be fine. You can probably go as far as 16
>>>>>>>>>without ever hearing any difference.
>>>>>>>
>>>>>>>>He can probably go as far as 18 without ever
>>>>>>>>hearing too much of a difference.
>>>>>>>
>>>>>>>Some speaker manufacturers recommend thicker wire. Mine, for one. What
>>>>>>>does the Hive have to say about manufacturers recommending 12-ga wire?
>>>>>>
>>>>>>The ability to hear an instantaneous difference in sound output as a
>>>>>>function of wire size is not the only reason to use heavier wire. There is
>>>>>>also the value of the amplifier power lost through the wire. Frequently
>>>>>>that value is greater than the increased price of heavier wire. Not only is
>>>>>>there a power loss, there is also an even greater loss in maximum power.
>>>>>>
>>>>>>e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
>>>>>>.0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
>>>>>>power. Worse yet, it's a reduction of 5.4% in the maximum output of your
>>>>>>amplifier.
>>>>>
>>>>> How do you figure this? Why is the power loss double in max power?
>>>>>I don't think it is.
>>>>>
>>>>> ScottW
>>>>>
>>>>
>>>>I'll answer for Norm. Assume the max output voltage of the amp is V
>>>>(volts rms), and that the amp is an ideal voltage source with zero
>>>>output impedance.
>>>>
>>>>Without the loss in the cable, the power available to drive the 4 ohm
>>>>speaker is V*V/4. Now conider the resistance of the cable being 0.112
>>>>ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
>>>>max power into that speaker is the square of that divide by 4, or
>>>>
>>>>V*V*4/(4.112**2)
>>>>
>>>>The ratio of that to the max available in the case of no cable
>>>>resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
>>>>in max. available power.
>>>>
>>>>The power consumed in the cable resistance is (using I**2*R)
>>>>
>>>>(V/4.112)**2*.112=V**2*(6.624E-3)
>>>>
>>>>Compared that to the max power available to the 4 ohm speaker without
>>>>cable resistance of V**2/4, that cable power consumption is
>>>>
>>>>(6.624E-3)*4=2.65%
>>>>
>>>>So the cable power consumption is 2.65% of what's available to drive
>>>>that 4 ohm speaker from the amp.
>>>>
>>>>You get a double hit because (1) there is consumption in the cable
>>>>resistance, and (2) the voltage available to the speaker is now reduced.
>>>
>>>
>>> Ok... but you always get that hit... not just at peak levels.
>>
>>The power hit is the reduction in max. available power. If you are not
>>trying to get the max power out of the amplifier, then you can still get
>>the amp to deliver the required power to the speaker.
>
>
> Ok... like anyone cares about that little semantic.
>
>>For instance, if you only need V/2 across the speaker terminals, the amp
>>can still do that even though the cable has a fairly large resistance.
>>It's only when you try to get as high a voltage as possible into the
>>speaker that there is a reduction in the power available.
>
>
> Reduction in power available... no. Reduction in power delivered...
> always.
>
>>
>>> His
>>>original post implied there would be a non-linearity introduced which
>>>would be a horrific problem... but that isn't true.
>>
>>I didn't get that impression at all, from his post. Everything is
>>linear, and we simply are using Ohm's Law.
>
>
> Well I did and that would be a big problem... this problem is really
> insignificant.
> Its always been there and always will be.
>
>
>>>The power delivered to the speaker is always reduced by the factor
>>>(Rsp/Rsp+Rcab)**2 which in the example comes .946 or 5.37%
>>>
>>> My mistake was accepting the 2.8% number. That number is wrong... and
>>>in his example under no circumstances will the power loss be limited to
>>>2.8% of what the amp delivers.
>>
>>He said that 2.8% of the amp's power is lost in the cable resistance. As
>>I showed, 2.65% of the max power available without cable loss is
>>dissipated in the cable resistance.
>
>
> Now read what you said... suddenly you can have no loss but still
> have resistance.
> Sorry... not possible. The loss is due to the resistance.... no
> resistance...no loss.
> No loss...no resistance. You don't get to separate one from the
> other.

Seems like I am not getting through.

I said "2.65% of the max. power available without cable loss is
dissipated in the cable resistance".

That means 2.65% of the max. power available *that was calculated based
on no cable loss* was dissipated in the cable resistance.

The math clearly shows that. Max power available with no cable= V**2/4

2.65% of that is dissipated if you now connected the speaker using cable
with a 0.112 ohm resistance.

Is it clear what I meant now?



>
>
>>If you compare the power dissipated
>>in the cable as a fraction of the power delivered to the speaker, then
>>that number is .112/4 = 2.8%. (The same current flows through cable
>>resistance and speaker.)
>
>
> The same current always flows through the cable resistance and the
> speaker... they are in series you know, and being in series.. the
> current through both is dependent on the sum of the resistance of
> both... never just one.

But you can calculate the ratio of power dissipated in the cable as a
fraction of the power that is dissipated in the speaker, when the
speaker is connected using the cable with known resistance. And that
ratio is 2.8%.

>
>
>>He was referrring to the loss in the cable as a
>>fraction of the power delivered to the speaker.
>
>
> Yes and its a situation that never happens. If the cable has
> resistance .... and all do... it will both have a voltage drop and
> reduce the current delivered by the amp. You simply cannot have one
> without the other. Never happen. You will ALWAYS have the cable
> reduce the power delivered to the speaker by 5.37% and never by 2.8%.
>

Please reread what I said. Clearly one can calculate the ratio of power
lost in the cable as a fraction of power dissipated in the speaker.

I think there is really no point in me beating this dead horse. I'm
finished here. BTW, Norm is an experienced engineer, and he knows what
he is talking about. His only oversight is saying that the cable loss is
2.8% of the power available, when he could/should have said 2.8% of the
power delivered to the speaker. But it is very close.


> ScottW
>

[chung]

ScottW wrote:
> chung wrote:
>
>>ScottW wrote:
>>
>>>Chung wrote:
>>>
>>>
>>>>ScottW wrote:
>>>>
>>>>
> wrote:
>>>>>
>>>>>
>>>>>>"George M. Middius" <cmndr [underscore] george [at] comcast [dot] net> wrote
>>>>>>in message ...
>>>>>>
>>>>>>
>>>>>>>4 of 12 said:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>>>Any 12 gauge wire will be fine. You can probably go as far as 16
>>>>>>>>>without ever hearing any difference.
>>>>>>>
>>>>>>>>He can probably go as far as 18 without ever
>>>>>>>>hearing too much of a difference.
>>>>>>>
>>>>>>>Some speaker manufacturers recommend thicker wire. Mine, for one. What
>>>>>>>does the Hive have to say about manufacturers recommending 12-ga wire?
>>>>>>
>>>>>>The ability to hear an instantaneous difference in sound output as a
>>>>>>function of wire size is not the only reason to use heavier wire. There is
>>>>>>also the value of the amplifier power lost through the wire. Frequently
>>>>>>that value is greater than the increased price of heavier wire. Not only is
>>>>>>there a power loss, there is also an even greater loss in maximum power.
>>>>>>
>>>>>>e.g. Let's consider a 35' run of #12 wire to a 4 ohm speaker. This is 35 x
>>>>>>.0032 ohms = .112 ohms, which is a loss of about 2.8% of your amplifier's
>>>>>>power. Worse yet, it's a reduction of 5.4% in the maximum output of your
>>>>>>amplifier.
>>>>>
>>>>> How do you figure this? Why is the power loss double in max power?
>>>>>I don't think it is.
>>>>>
>>>>> ScottW
>>>>>
>>>>
>>>>I'll answer for Norm. Assume the max output voltage of the amp is V
>>>>(volts rms), and that the amp is an ideal voltage source with zero
>>>>output impedance.
>>>>
>>>>Without the loss in the cable, the power available to drive the 4 ohm
>>>>speaker is V*V/4. Now conider the resistance of the cable being 0.112
>>>>ohm. The voltage available across the speaker is now V*(4/(4+.112)). The
>>>>max power into that speaker is the square of that divide by 4, or
>>>>
>>>>V*V*4/(4.112**2)
>>>>
>>>>The ratio of that to the max available in the case of no cable
>>>>resistance is (4/(4.112**2))/(1/4)=0.9463. So there is a loss of 5.37%
>>>>in max. available power.
>>>>
>>>>The power consumed in the cable resistance is (using I**2*R)
>>>>
>>>>(V/4.112)**2*.112=V**2*(6.624E-3)
>>>>
>>>>Compared that to the max power available to the 4 ohm speaker without
>>>>cable resistance of V**2/4, that cable power consumption is
>>>>
>>>>(6.624E-3)*4=2.65%
>>>>
>>>>So the cable power consumption is 2.65% of what's available to drive
>>>>that 4 ohm speaker from the amp.
>>>>
>>>>You get a double hit because (1) there is consumption in the cable
>>>>resistance, and (2) the voltage available to the speaker is now reduced.
>>>
>>>
>>> Ok... but you always get that hit... not just at peak levels.
>>
>>The power hit is the reduction in max. available power. If you are not
>>trying to get the max power out of the amplifier, then you can still get
>>the amp to deliver the required power to the speaker.
>
>
> Ok... like anyone cares about that little semantic.
>
>>For instance, if you only need V/2 across the speaker terminals, the amp
>>can still do that even though the cable has a fairly large resistance.
>>It's only when you try to get as high a voltage as possible into the
>>speaker that there is a reduction in the power available.
>
>
> Reduction in power available... no. Reduction in power delivered...
> always.
>
>>
>>> His
>>>original post implied there would be a non-linearity introduced which
>>>would be a horrific problem... but that isn't true.
>>
>>I didn't get that impression at all, from his post. Everything is
>>linear, and we simply are using Ohm's Law.
>
>
> Well I did and that would be a big problem... this problem is really
> insignificant.
> Its always been there and always will be.
>
>
>>>The power delivered to the speaker is always reduced by the factor
>>>(Rsp/Rsp+Rcab)**2 which in the example comes .946 or 5.37%
>>>
>>> My mistake was accepting the 2.8% number. That number is wrong... and
>>>in his example under no circumstances will the power loss be limited to
>>>2.8% of what the amp delivers.
>>
>>He said that 2.8% of the amp's power is lost in the cable resistance. As
>>I showed, 2.65% of the max power available without cable loss is
>>dissipated in the cable resistance.
>
>
> Now read what you said... suddenly you can have no loss but still
> have resistance.
> Sorry... not possible. The loss is due to the resistance.... no
> resistance...no loss.
> No loss...no resistance. You don't get to separate one from the
> other.
>
>
>>If you compare the power dissipated
>>in the cable as a fraction of the power delivered to the speaker, then
>>that number is .112/4 = 2.8%. (The same current flows through cable
>>resistance and speaker.)
>
>
> The same current always flows through the cable resistance and the
> speaker... they are in series you know, and being in series.. the
> current through both is dependent on the sum of the resistance of
> both... never just one.
>
>
>>He was referrring to the loss in the cable as a
>>fraction of the power delivered to the speaker.
>
>
> Yes and its a situation that never happens. If the cable has
> resistance .... and all do... it will both have a voltage drop and
> reduce the current delivered by the amp. You simply cannot have one
> without the other. Never happen. You will ALWAYS have the cable
> reduce the power delivered to the speaker by 5.37% and never by 2.8%.
>
> ScottW
>

Before I really leave (:)), let me illustrate this with an example.
Perhaps then it will become clear.

Let's assume the amp can deliver 100W into a 4 ohm load.

Without any cable loss, the 4 ohm speaker can dissipate that full 100W.

If we use a cable with 0.112 ohm of resistance, then 2.65W is dissipated
in the cable resistance, and only 94.6W is available to the 4-ohm speaker.

The ratio of 2.65W to 94.6W is about 2.8%. Is it now clear?

Note now that the total power available from the amp is no longer 100W,
but only 97.25W.

[dizzy]

Chung wrote:

>You get a double hit because (1) there is consumption in the cable
>resistance, and (2) the voltage available to the speaker is now reduced.

But that's assuming that the amplifier's peak output voltage is
unchanged. In fact, it will be be able to put out a higher voltage
into 4.112 ohms than it will into 4.000 ohms.

[chung]

dizzy wrote:
> Chung wrote:
>
>
>>You get a double hit because (1) there is consumption in the cable
>>resistance, and (2) the voltage available to the speaker is now reduced.
>
>
> But that's assuming that the amplifier's peak output voltage is
> unchanged. In fact, it will be be able to put out a higher voltage
> into 4.112 ohms than it will into 4.000 ohms.
>

Please read the very first paragraph I wrote:

"I'll answer for Norm. Assume the max output voltage of the amp is V
(volts rms), and that the amp is an ideal voltage source with zero
output impedance. "

Solid state amps are very close to ideal voltage sources. The output
voltage changes evry little whether the load is 4 or 4.112. The output
inmpedance is close to 0.

[Pooh Bear]

chung wrote:

> Before I really leave (:)), let me illustrate this with an example.
> Perhaps then it will become clear.
>
> Let's assume the amp can deliver 100W into a 4 ohm load.
>
> Without any cable loss, the 4 ohm speaker can dissipate that full 100W.
>
> If we use a cable with 0.112 ohm of resistance, then 2.65W is dissipated
> in the cable resistance, and only 94.6W is available to the 4-ohm speaker.
>
> The ratio of 2.65W to 94.6W is about 2.8%. Is it now clear?
>
> Note now that the total power available from the amp is no longer 100W,
> but only 97.25W.

Because the total load impedance has gone up !

Correct.

Mind you 5% loss is only ~ -0.2dB.

Graham

[Pooh Bear]

dizzy wrote:

> Chung wrote:
>
> >You get a double hit because (1) there is consumption in the cable
> >resistance, and (2) the voltage available to the speaker is now reduced.
>
> But that's assuming that the amplifier's peak output voltage is
> unchanged. In fact, it will be be able to put out a higher voltage
> into 4.112 ohms than it will into 4.000 ohms.

By a *miniscule* amount. Not enough to compensate.

Graham

[EddieM]

> chung wrote
>> dizzy wrote:
>>> Chung wrote:
>>
>>
>>>You get a double hit because (1) there is consumption in the cable
>>>resistance, and (2) the voltage available to the speaker is now reduced.
>>
>>
>> But that's assuming that the amplifier's peak output voltage is
>> unchanged. In fact, it will be be able to put out a higher voltage
>> into 4.112 ohms than it will into 4.000 ohms.
>>
>
> Please read the very first paragraph I wrote:
>
> "I'll answer for Norm. Assume the max output voltage of the amp is V
> (volts rms), and that the amp is an ideal voltage source with zero
> output impedance. "
>
> Solid state amps are very close to ideal voltage sources. The output voltage
> changes evry little whether the load is 4 or 4.112. The output inmpedance is
> close to 0.


Assuming that everything said to this point was technically 100% allright,
My hat is tipped.

[dizzy]

chung wrote:

>dizzy wrote:
>> Chung wrote:
>>
>>
>>>You get a double hit because (1) there is consumption in the cable
>>>resistance, and (2) the voltage available to the speaker is now reduced.
>>
>>
>> But that's assuming that the amplifier's peak output voltage is
>> unchanged. In fact, it will be be able to put out a higher voltage
>> into 4.112 ohms than it will into 4.000 ohms.
>>
>
>Please read the very first paragraph I wrote:
>
>"I'll answer for Norm. Assume the max output voltage of the amp is V
>(volts rms), and that the amp is an ideal voltage source with zero
>output impedance. "

So nice of you to hand-wave that away, while nit-picking about 0.112
ohms of speaker-wire impedance...

>Solid state amps are very close to ideal voltage sources. The output
>voltage changes evry little whether the load is 4 or 4.112. The output
>inmpedance is close to 0.

Speaker wires are close to 0, too.

[dizzy]

dizzy wrote:

>chung wrote:
>>
>>Solid state amps are very close to ideal voltage sources. The output
>>voltage changes evry little whether the load is 4 or 4.112. The output
>>inmpedance is close to 0.
>
>Speaker wires are close to 0, too.

In any case "output impedance" isn't the point. The point is the drop
in power-supply voltage.

[Chung]

dizzy wrote:
> chung wrote:
>
>>dizzy wrote:
>>> Chung wrote:
>>>
>>>
>>>>You get a double hit because (1) there is consumption in the cable
>>>>resistance, and (2) the voltage available to the speaker is now reduced.
>>>
>>>
>>> But that's assuming that the amplifier's peak output voltage is
>>> unchanged. In fact, it will be be able to put out a higher voltage
>>> into 4.112 ohms than it will into 4.000 ohms.
>>>
>>
>>Please read the very first paragraph I wrote:
>>
>>"I'll answer for Norm. Assume the max output voltage of the amp is V
>>(volts rms), and that the amp is an ideal voltage source with zero
>>output impedance. "
>
> So nice of you to hand-wave that away, while nit-picking about 0.112
> ohms of speaker-wire impedance...

Nit-picking? I was merely trying to explain what seemed like a mystery
to some people. I did not say anything about the importance of the
wasted power or capability.

Handwaving? I was stating the assumptions under which the subsequent
calculations apply.

>
>>Solid state amps are very close to ideal voltage sources. The output
>>voltage changes evry little whether the load is 4 or 4.112. The output
>>inmpedance is close to 0.
>
> Speaker wires are close to 0, too.
>

In this discussion, we were talking about the consequences of 0.112 ohm
of cable resistance. If you want to start discussing the importance (or
lack thereof) of output impedance, you are welcome to contribute. Show
some calculations that lead to whatever point you were trying to make.

[Chung]

dizzy wrote:

> dizzy wrote:
>
>>chung wrote:
>>>
>>>Solid state amps are very close to ideal voltage sources. The output
>>>voltage changes evry little whether the load is 4 or 4.112. The output
>>>inmpedance is close to 0.
>>
>>Speaker wires are close to 0, too.
>
> In any case "output impedance" isn't the point. The point is the drop
> in power-supply voltage.
>

You don't understand. Output impedance is used to model the effects of
how the output voltage change with different loads. If the output
voltage rises because the load resistance is higher, then the output
impedance cannot be 0.

[Pooh Bear]

Chung wrote:

> dizzy wrote:
>
> > dizzy wrote:
> >
> >>chung wrote:
> >>>
> >>>Solid state amps are very close to ideal voltage sources. The output
> >>>voltage changes evry little whether the load is 4 or 4.112. The output
> >>>inmpedance is close to 0.
> >>
> >>Speaker wires are close to 0, too.
> >
> > In any case "output impedance" isn't the point. The point is the drop
> > in power-supply voltage.
> >
>
> You don't understand. Output impedance is used to model the effects of
> how the output voltage change with different loads. If the output
> voltage rises because the load resistance is higher, then the output
> impedance cannot be 0.

That alone can't model PSU loading ( voltage droop ) on dynamic signals
though.

Graham

[Chung]

Pooh Bear wrote:
>
> Chung wrote:
>
>> dizzy wrote:
>>
>> > dizzy wrote:
>> >
>> >>chung wrote:
>> >>>
>> >>>Solid state amps are very close to ideal voltage sources. The output
>> >>>voltage changes evry little whether the load is 4 or 4.112. The output
>> >>>inmpedance is close to 0.
>> >>
>> >>Speaker wires are close to 0, too.
>> >
>> > In any case "output impedance" isn't the point. The point is the drop
>> > in power-supply voltage.
>> >
>>
>> You don't understand. Output impedance is used to model the effects of
>> how the output voltage change with different loads. If the output
>> voltage rises because the load resistance is higher, then the output
>> impedance cannot be 0.
>
> That alone can't model PSU loading ( voltage droop ) on dynamic signals
> though.
>
> Graham
>
>

For power capability calculations, you can still model any voltage drop
due to loading with a finite output impedance. Dizzy seems to think that
since the cable resistance increases the load resistance the amplifier
sees, there is a corresponding increase in output voltage, therefore one
can recover some of the loss in available power.

The interested reader can simply model the power supply droop with
loading by setting the output impedance to a small finite value. As an
exercise, let the output impedance be 0.112 ohms, and calculate the loss
in power available to the speaker when a 0.112 ohm cable is used. Dizzy
will be surprised at the result.

[dizzy]

Chung wrote:

>In this discussion, we were talking about the consequences of 0.112 ohm
>of cable resistance. If you want to start discussing the importance (or
>lack thereof) of output impedance, you are welcome to contribute. Show
>some calculations that lead to whatever point you were trying to make.

I did some calculations, and the effect was less than what I expected.
I assumed an amp that could put out 100W (28.3V) into 8 ohms, and 150W
(24.5V) into four ohms. Interpolating the output voltage into 4.112
ohms only gives 0.1V greater output voltage.

[chung]

dizzy wrote:
> Chung wrote:
>
>
>>In this discussion, we were talking about the consequences of 0.112 ohm
>>of cable resistance. If you want to start discussing the importance (or
>>lack thereof) of output impedance, you are welcome to contribute. Show
>>some calculations that lead to whatever point you were trying to make.
>
>
> I did some calculations, and the effect was less than what I expected.
> I assumed an amp that could put out 100W (28.3V) into 8 ohms, and 150W
> (24.5V) into four ohms. Interpolating the output voltage into 4.112
> ohms only gives 0.1V greater output voltage.
>

Using your numbers, the amp's output voltage with the cable resistance
in series with the speaker is 24.6V. But the voltage into the actual
speaker impedance of 4 ohms is only 24.6*4/4.112=23.93V. In other words,
the power available to the speaker drops by about 1-(23.93/24.5)**2 = 4.6%.

BTW, the output impedance calculates out to be 1.47 ohm, using your
numbers. Clearly that is the impedance when the amp is close to running
out of current and distortion starts to rise sharply. The small signal
output impedance is much less than that, for solid state amps. And a
good SS amp will have close to the same output voltage driving 8 ohms or
4 ohms.