Dukk

JL has just a little more real world experience than either of us.

The statement is there, prove it is wrong.

I offer again my basic example of an inductor in series with a woofer. The inductor definitely influences the woofer. With two woofers in series, they definitely influence eachother. The degree of audibility is the only argument.

StickyFingaz

iTrader: (4)

This thread is awsome....

Anytime four or more senior members of this forum can debate something, I am hooked. Always trying to learn something new...

Thanks for emailing JL and posting that response Zoomer, and all the hard work you have put into your resposes. Same to everyone who has posted.

Sassmaster

iTrader: (2)

wow... I've read this whole thread, and I have heard nothing that even comes close to a legitimate answer electrically, except that the inductance of the coils would affect each other.

of course, take the AC wave from the amplifier, run it through 2 coils in series and you end up with a much worse power factor, which would put the voltage and current even more out of phase relative to each other causing, for the same wattage output, a severely increased volt-ampere requirement, which would strain the amplifier more, causing more heat, etc. for the same wattage, but since the impedance would be increased even more greatly by the power factor effect having the coils in series, you would end up with less volt-ampere output, which, in turn would lead to less watts out of the speakers.

that's basic AC electrical theory.

zoomer

iTrader: (3)

Sorry Dukk, but I must respectfully dissagree. Science is based on prooving that your statement is correct by showing how you came to your conclusion. You publish the theory and measurements to prove your statement so that others can verify it and agree with your work, or point out where you may have made an error. This is known as "peer review".
Otherwise you get snake oil.
JL has confirmed that they have no clue as to why they made the statement. Had they published some theory or measurements or any kind of data we could then review it and agree, or point out where they may have made a mistake. They have done neither.

SQ Civic

iTrader: (3)

That's not accurate either.. just because the guy that answered your email said HE had nothing to tell you on why you shouldn't do it, or why JL made the statement, that doesn't really mean anything. It means nothing really. You just happened to get a guy looking at your comments to JL that didn't have on his "answers for a given question" sheet, the question you asked him. I would bet if you talked to the engineers at JL they would have a completely differencet response for you, weather it's right or wrong is irrelevent, but they would still have an answer...

if you think the 'smart' people at JL are answering emails like how do I make this w7 blow windows, or I want some stickers for my ride, or even, why can't I wire sub in series, I would bet your mistaken...

When it comes down to it, the question you asked most likely wasn't a quesiton they get often, or at all for that matter so they didn't have a prepared answer for you; and most likely didn't want to take the time to find out for you, so just passed off the response you got. Take it with a grain of salt..

Mark

zoomer

iTrader: (3)

Not sure who posted something about power factor..but

I learned that Basic AC electrical theory says:

The average power in an AC circuit expressed in terms of the rms voltage and current is
Paverage= VI cos A
where A is the phase angle between voltage and current.

cosA is also the Power Factor = R/Z
were
R is the resistance and
Z is the impedance Z = sqrt(R^2 + XL^2)
where XL=2*pi*freq*L

If we put 2 woofers in series, both the R and the X double leaving the power factor the SAME.

Using numbers:

For a typical subwoofer, such as an Alpine Type R 1241D with dual voice coils in parallel, we have
R=1.8 ohms and
L = 1.9mH and XL=1.2ohms at 100hz

Therefore th power factor at 100Hz is R/Z
R=1.8
Z=sqrt(1.8^2+1.2^2)
and PowerFactor=R/Z=1.8/2.1= .833

This does not change if they are in series or parallel. (you do the math with 2 of these woofers in series. R=3.6 and L=3.8)

Now, what does this actually mean for a power amp?

Lets take an amp that is rated at 500 watts RMS into 2 ohms. The test signal is a sine wave.
The spec is given with a 2 ohm resisitive load.
Using our power equations, P=V^2/R we get
VRMS= Sqrt(500*2) = 31.6 volts RMS
And either using P=I^2 or Ohms law I=V/R
We get a current of 15.8 amps RMS
And of course we can verify our math
By doint P=VI =31.6*15.8=500 watts rms

The 31.6 volts RMS corresponds to a peak voltage of
31.6* sqrt2 = 44.6 volts peak.

And by the same calculations the peak current will be
22.3 amps
This is all driving a resistor with a constant sinusoid. Do not confuse this with the Peak Power that an amp can deliver during brief musical peaks. That is another topic!

Now what happens if we hook up our trusty Alpine sub. The voice coil has 1.8ohms and 1.9mH giving it a power factor of .833

The max power delivered to the sub is not 500 watts, but

500*PowerFactor= 500*.833=390 watts rms.

And the peak current is now a bit LESS because of the reactive inductance in series with VC resistance = Z=2.1 ohms.
Peak current is therefore 44.6/2.1=20.7 amps

If indeed the power supply is limited to +/- 45 volts, there is no way to get the full
500watts RMS out of this amp to your sub.
In order to actually output 500 watts to your sub, the amp has to be a conservatively rated 500 watt amp. It must be able to deliver 500/.833=600 watts into a 2 ohm resistor so that it can deliver 500watts into your sub. This will also having a higher power supply rail and also higher current of course. This conservative rating, among other things, is what will differentiate good quality amps from cheap ones. Another issue will be what is the power dissipated as heat in the amp, and how it is influenced by the power factor of the sub…or any inductive load…And that will be the subject for another day!