Why Bi-wiring Makes No Sense.

[2*pi*f*t]

Bi-wiring revisited....

Cool...

I will not hold you to your hat offer.

Give me a few minutes to draw up a supporting diagram..for now, here's a graph to consider, it shows the dissipation differences between a mono and a biwire configuration..

Ok..now with that to see, a little splainin...

The currents and the resistance are normalized, so the peak woof and tweet currents are 1, and the resistance of the wire is also normalized to 1.

The woof current is the blue line, the diss is (Asquared) I²R, so as you can see, it peaks at 1, and is always positive (real dissipation cannot be negative). It is your basic sine squared waveform, a direct result of the woof current flowing through a wire by itself. Note that for an ideal load, this is also exactly the same dissipation time profile scaled differently.

The tweet current is the magenta line, called Bsquared. again, I squared R...

With biwiring, the total wire dissipation is of course, the sum of the two, A squared plus B squared, and the load dissipation at the speakers is an exact scale of the wire dissipation.

You have been after me about not understanding the bi-wire thingy, so I moseyed on over
here to see all thess big doin's. But I'm having a little trouble here, John. I might could see
your arguement if it were, say:

But, we don't have two sources as your graph indicates, we have the lower (revised)
one with only a single voltage source (amp.) As in

Now, I learned that cos(a) + cos(b) = 2*cos(a+b)*cos(a-b)
So, for say:
v1(t) = 1*cos(2*pi*100*t) (Woofer Signal = cos(a))
and
v2(t) = 1*cos(2*pi*5000*t) (Tweeter Signal = cos(b))
where f1 = 100Hz and f2 = 5,000Hz, respectively. And V1=V2. Our output signal is
vS= cos(5100*pi*t)&cos(4900*pi*t) and looks like this:

But, I'm sure I'm in error, so I wait for your correction.

 

[jneutron]

You have been after me about not understanding the bi-wire thingy, so I moseyed on over here to see all thess big doin's..

Welcome to the discussion..

Your statement which I did not feel was correct was this:

2.) The true ampacity of wire is a function ONLY of its impedance. This IS a current limiting device due to Ohm’s law i = v/Z. And its impedance is SOLEY a function of length. So the shorter the wire, the more current it will carry. A 6 ft length of #14 copper at 120 vac 60Hz will carry 120/(6*0.002525) A or 7,920 A continuously. Note that impedance is frequency dependent. This is the basis for bi-wiring speakers..

Again, when you said impedance in that, you actually meant endpoint resistance.. And the biggest issue I had was you stating a #14 can carry continuously...7920 amperes...it cannot..ever. Your statement that impedance is frequency dependent and that being the basis for biwiring...I've read some interesting white papers from vendors which claim such, but they are not founded in reality.

But I'm having a little trouble here, John. I might could see
your arguement if it were, say:.........

First, to text..My graphs do not indicate two sources. They detail the power that is dissipated within the wires. That is all.

The pics...

1. You depict "my biwire model" with two sources...I have never done so.

2. You depict "my mono model" with two sources in parallel..I did not do that, nor is it even possible to do. And, you depicted the load as a single entity, which I did not do.

3. You do not indicate any frequency dependent elements to control which load receives which signal. Without that branching, there is no discussion as the wire dissipation will temporally match that of the loads. I am not sure if you were implying the crossover components in the designations Rw and Rt.

As for your voltage discussion, the graph looks just fine.. While it depicts a more realistic picture of a high and low frequency mix, I chose a 3.5 to 1 frequency ratio simply for ease of discussion.

But you do not discuss the wire dissipation, which is the crux of my discussion.

The dissipation math is what I provided...your summation drawing and voltage waveform is exactly what I consider as the system stimulus. So you are not disagreeing with what I have stated.

But, I'm sure I'm in error, so I wait for your correction.

Where you have been in error technically, I have stated such. That is all.

Go back and study the drawing I provided (well, ok, it's not as good as yours). The difference between bi and mono is very clear..then we can further discuss if you wish.

Cheers, John

 

[FLZapped]

Who cares??

The resistive loss within the wires has absolutely no concern for the voltage potential of either wire. It only deals with the current.

The instantaneous power that is going through the wire, of course relies on both the amp voltage and the current within the wire.

Note that I have chosen this scenario very carefully. I use one amp to guarantee that both the biwire and the monowire case receive the exact same voltage stimulus at the exact same time. That way, the losses that occur within the monowire case can be clearly defined with respect to that of the biwire currents.

Note also that the capacitor and inductive values are also of no concern to the problem..Once one assumes that the loads and the crossover values are identical, the problem is easy....if biwiring makes no difference, then how is it the wire carrying the current to the load dissipates power differently depending on mono/bi setup.


Cheers, John

Answer the question, John. What is the voltage output of the amp, John. With the loads normalized to 1 ohm and 1 amp of current supposedly going through the two halves of the circuit in opposite directions, you can't have both +1 and -1 volts driving it, now can you?

-Bruce

 

[FLZapped]

Welcome to the discussion..

First, to text..My graphs do not indicate two sources. They detail the power that is dissipated within the wires. That is all.

The pics...

1. You depict "my biwire model" with two sources...I have never done so.

That's what you're alluding to.

2. You depict "my mono model" with two sources in parallel..I did not do that, nor is it even possible to do. And, you depicted the load as a single entity, which I did not do.

But that is what you're alluding to.


There is something else going on here that you missed, and I did too until I took another look at it - which is why I'm harping about the ouput voltage of the amplifier. When you answer that question, it should be clear what is going on.

-Bruce

 

[jneutron]

Answer the question, John. What is the voltage output of the amp, John. With the loads normalized to 1 ohm and 1 amp of current supposedly going through the two halves of the circuit in opposite directions, you can't have both +1 and -1 volts driving it, now can you?

-Bruce

No, you cannot have two voltages at the same time. That is not what I've said.

I am talking current.

Edit:

Bruce...try this with two sine generators and an input mixer, an amp, and a two way speaker.., two clamp on current meters.

1. inject hf signal until the tweeter current is at one ampere sine. (yah, assuming the clamp meter is accurate at hf..)

2. inject lf until the current at the woofer is 1 ampere sine.

Both drivers are seeing exactly one ampere of sine current at their respective frequencies.

This is the problem setup I have been discussing..

Please explain why one amplifier cannot do this?

Cheers, John

 

[jneutron]

There is something else going on here that you missed, and I did too until I took another look at it - which is why I'm harping about the ouput voltage of the amplifier. When you answer that question, it should be clear what is going on.

-Bruce

Well, tell us. Stop beating around the bush.


The crossover elements charge and discharge...(I consider an inductor with current within it to also be "charged". You note I've not discussed the reactive elements with the phase shifts..that doesn't enter..The wire doesn't care what the phase shifts are.

Don't worry about the voltages Bruce, worry about the current. The loads with their selection elements worry about the voltages..

But the feed wire worries about the current. And so does the resistive dissipation of the feed wires.

If it'll help you, reconsider the problem without either the amp or the loads. Just the wires.

Draw a node at the center of the page. From the left, one wire. From the right, two wires.

Now, put the hf sine current on one of the right side wires, the lf sine current on the other wire on the right. On the left, the current within the wire is the summation of the two currents on the right. (the node sum has to be zero net current..) (note there is no voltage discussion here).

Now, on the left wire, calculate the power loss per unit length...repeat the same for the two wires on the right.

Note the extra term on the left wire. That is the "2AB" part that doesn't happen on the right side..

I can draw it another day, if you wish...I won't have time for a coupla days due to bein in "class". They're tryin ta larn me stuff..don't they know I'm an old dog??

Cheers, John




Cheers, John

 

[adwilk]

Please forgive me, i'm just really confused after reading the last couple of pages...

Are we arguing that there is no improvement in SQ by biwiring or are we arguing that there are no audible differences?

 

[2*pi*f*t]

Please forgive me, i'm just really confused after reading the last couple of pages...

Are we arguing that there is no improvement in SQ by biwiring or are we arguing that there are no audible differences?

Actually neither. John drew some curves of current flow through a wire or wires and some of us can follow what exactly he's done.

 

[highfihoney]

Please forgive me, i'm just really confused after reading the last couple of pages...

Are we arguing that there is no improvement in SQ by biwiring or are we arguing that there are no audible differences?

Im right along with ya being confused but i have been watching the thread,kinda like a train wreck ya cant quit thinking about.

No offense to the guys who are working on this problem,i think its interesting to read.

I think jneutron is trying to show that there is a difference between mono wire & bi wire in certian cases.

 

[2*pi*f*t]

Im right along with ya being confused but i have been watching the thread,kinda like a train wreck ya cant quit thinking about.

No offense to the guys who are working on this problem,i think its interesting to read.

I think jneutron is trying to show that there is a difference between mono wire & bi wire in certian cases.

No doubt, and we, too, are trying to digest what he's saying. But, some of us are just having trouble with the form of the argument; i.e. how it was framed or presented exactly. There is certain method as to the way these kinda problems are analyzed.

 

[jneutron]

No doubt, and we, too, are trying to digest what he's saying. But, some of us are just having trouble with the form of the argument; i.e. how it was framed or presented exactly. There is certain method as to the way these kinda problems are analyzed.

I agree, many are having difficulty with the form of the argument.

But remember, a construct..a model as it were, must be true regardless of the approach tried.. If a model is accurate, then it doesn't matter what the methodology is that is used to test it. In point of fact, if the model can withstand tests which take different approaches, that just reinforces it.

I question the model which states that monowire and biwire are identical. I have provided instances in time where that model falls apart..specifically, when the currents are opposite and cancel within the monowire, and when they add up.

The difference is a physical entity, that of dissipative loss, that has a form which is unexpected, and very likely, transparent to test methods we take for granted.

Cheers, John

 

[FLZapped]

Well, tell us. Stop beating around the bush.


The crossover elements charge and discharge...


Cheers, John

Well, you've finally answered it correctly, which means that indeed these things are NOT uncorrelated, it also means that the amplifier is in an off state (hi Z) and what you are seeing is energy transfer between the two halves of the crossover, which takes place anyway, you've just moved the path ourside the speaker box with the bi-wiring - which explains the difference between the single wire and bi-wire dissipation. (which I missed at first) So this shouldn't be unexpected at all.

-Bruce

 

[jneutron]

Well, you've finally answered it correctly, which means that indeed these things are NOT uncorrelated, it also means that the amplifier is in an off state (hi Z) and what you are seeing is energy transfer between the two halves of the crossover, which takes place anyway, you've just moved the path ourside the speaker box with the bi-wiring - which explains the difference between the single wire and bi-wire dissipation. (which I missed at first) So this shouldn't be unexpected at all.

-Bruce

I've known that this is what you've been thinking.. I needed for you to say it..

That was specifically why I drew the biwire and monowire as fed from one amp. To remove the amp from the equation.

And, recall that I did not specify the crossover components nor the actual frequencies..they do not matter. In other words, the capacitance can be chosed arbitrarily small, so that it's charging current cannot continue to support the low frequency current...and the same applies to the inductor. You are thinking of a resonant topology.

Go back to the diagram, and explain the zero mono feedwire current with the understanding that the crossover CANNOT support cross currents...if it did, the tweet would get the low frequency current and the woof the high..

That is also why I chose a simple first order crossover bruce..to keep it clear that there is no cross currents...

Your "energy transfer between the two halves of the crossover REQUIRES that the woofer and tweeter see the same current, which clearly they do not..

It was a well though out try....fortunately for my premise, it is incorrect.

Do not stop trying..this is of course peer review, which I am thankful for.

Cheers, John

 

[FLZapped]

I've known that this is what you've been thinking.. I needed for you to say it..

That was specifically why I drew the biwire and monowire as fed from one amp. To remove the amp from the equation.

But the way you've gone about trying to make your point, you've not taken it out. You've lead the readers to assume that it is an active component for that instantaneous moment and even tried to argue with me that it was as well. Others have rightly pointed that error out too.

And, recall that I did not specify the crossover components nor the actual frequencies..they do not matter. In other words, the capacitance can be chosed arbitrarily small, so that it's charging current cannot continue to support the low frequency current...and the same applies to the inductor. You are thinking of a resonant topology.

Nope. Bad assumption. Who said anything about continue, you described an instantaneous case. However, the whole thing IS a resonant circuit.

Go back to the diagram, and explain the zero mono feedwire current with the understanding that the crossover CANNOT support cross currents...if it did, the tweet would get the low frequency current and the woof the high..

Current is not frequency dependant and I already did explain the zero mono-wire configuration, go back and re-read what I said. Furthermore, you're talking an instantaneous moment, therefore, there is no frequency component. Sorry, but you cannot take the position that the crossover/speakers circuitry cannot support cross currents. As long as there are reactive components and you have real speakers that can be behave both as a motor and a generator there will indeed be cases where energy is transferred from one half of the circuit to the other.

That is also why I chose a simple first order crossover bruce..to keep it clear that there is no cross currents...

Doesn't matter. It could be a 24 element crossover - as long as there is one.

Your "energy transfer between the two halves of the crossover REQUIRES that the woofer and tweeter see the same current, which clearly they do not..

Sorry, but we're not talking about what they are seeing, the amplifier is in a high impedance state (or "off") and you're instantaneous model is showing that there is an amp of current flowing from one half of the circuit to the other. And for the record, I never said they were seeing the same current from the amplifier. Never. I was in error when I said the proper answer was no current in either leg, but I figured out that error remembering that there are reactive components in the circuit.

It was a well though out try....fortunately for my premise, it is incorrect.

Do not stop trying..this is of course peer review, which I am thankful for.

Cheers, John

Sorry John, this is one time you're off kilter somewhere. What you've drawn out is indeed energy transfer from one half a reactive circuit to the other - for that instantaneous moment. Take out the crossover components and replace the speakers with resistors and your scenario cannot exist.

It's time for you to figure it out, I'm done.

-Bruce

 

[Sheep]

Would you people give it a rest! Some of us are trying to sleep!

SheepStar

 

[stratman]

Would you people give it a rest! Some of us are trying to sleep!

SheepStar

As Nelson would say: ha,ha! Page 17, post 165, has all the answers, and from a cavamen to boot.

 

[jneutron]

It's time for you to figure it out, I'm done.

-Bruce

That would be unfortunate, and not what I wish.


Bruce, think about it. You have two errors..

1. Your position is that the crossover supports the current? To do so, the current must pass through both elements and loads, as they are in series with each other for "cross conduction".. So the tweet must support the lf, the woof must support the hf. Clearly, that does not happen.
2. The amp is a pure voltage source..zero series resistance. It does not care what the load is, it will do exactly what it is told.

Go through the wires Bruce, look at the dissipation for case 2.

Or prove to me that it is not possible for the load currents to be equal and opposite at any point in time.

You are confusing the model, as you are only considering resistive branches. The reactance comes into play.

Did you try the two signal setup I told you to do? The one which shows that it is possible to have both loads see the 1 amp peak but opposite polarity, with the cable current equal to zero? It's an easy setup, it just requires 3 current probes..

It will show you your error.

Re-read my posts, you will see.

Cheers, John

 

[mtrycrafts]

Would you people give it a rest! Some of us are trying to sleep!

SheepStar

You mean this wakes you up? Rewire the alarm mode in the computer so it won't.
It invigorates me! I am still learning something

 

[gene]

Jim Lesurf was kind enough to write an article dispelling some of the fallacies posted in this thread. I hope this article puts this issue to bed and we can move on to real topics in audio that affects system performance.

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/audio/biwire2/page1.html

 

[jneutron]

Jim Lesurf was kind enough to write an article dispelling some of the fallacies posted in this thread. I hope this article puts this issue to bed and we can move on to real topics in audio that affects system performance.

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/audio/biwire2/page1.html

Other than a simple math error and not understanding the conceptual argument at all, I see nothing incorrect.

It is good to see jim involved. It is unfortunate he does not understand the argument...so incorrectly dismisses it.

Does the fine print of this forum allow the unrestricted use of another's IP, or allow others to put into "e-print", the words of another without the permission of the author?


PS...I've e-mailed jim his math error, and invited him to discuss the argument so he doesn't do such silly rebuttals without understanding the argument..
Cheers, John

ps...as I told him, I enjoy his site, it is for the most part, very well done. It saddens me to see one of his stature (in my eyes) present a strawman argument to rebuke..

pps. I am dissapointed in you, Gene. The fact that you do not believe my argument is inconsequential..I expect disagreement. The fact that Jim missed the salient point of my argument, is also inconsequential.

You are guilty of a heinous act. You have hired a "gun" to try to trash me without allowing jim to contact me for a proper, scientifically based and academically correct discussion. Rather, a stab in the back, as it were.

I believed you to be above that.