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Old 6th January 2003, 02:29 AM   #21
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Quote:
Originally posted by SY
Rodd, NB that in most systems, Rap is assumed to be close to zero so that the port masses are in parallel. It's a good assumption if you don't put stuffing in the ports.
Sy,

Yes, I agree and I almost used the simplified version of the Small equivalent circuit, but want to keep Rap in there for the lossy port discussion.

My problem is in BassBox Pro, // ports are treated approximately as a single port with a total port area equal to the sum of the individual ports. This definitely goes against my theory and circuit theory. BB-Pro (and I assume the other box design programs) summing the port air masses of all the // ports and treating it a single large port is analogous to adding parallel inductances in a circuit and treating it a single large inductor. Parallel inductances do not add in this fashion (assuming no coupling occurs). Parallel inductors will sit there as a pure resistance until ac tries to pass, and will resonate at its resonant frequency.

What am I missing? I canít believe the Small and Thiele missed this so I must be missing something.

Rodd Yamashita
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Old 6th January 2003, 03:51 AM   #22
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I cannot speak in circuit theory terms, so I will just try to visualize.

Suppose a speaker is tuned to 42 Hz. Under 42 Hz, the response rolls off at a fast 24 dB/octave. The air in the port "sees" an open box below 42 Hz.

Now if I add an additional port to tune the box half an octave lower-30 Hz-won't all the 30 Hz frequencies "see" an open box and go out out the port tuned to 42 Hz?

In short, I would suggest that box with two ports that would tune a box normally to 42 Hz and 30 Hz respectively essentially add up to a box tuned to 42 Hz.

Just a guess on my part.
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Old 6th January 2003, 04:23 AM   #23
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Rodd, the equivalent port inductance is proportional to length/area. So, if we take a simple case of two duct tubes of equal length, doubling the area is the same as halving the inductance, i.e., paralleling the inductors. So, what your program is doing makes sense to me.

keltic, as soon as you add the second port, the first one is no longer tuned to the same frequency it originally was. The new box frequency will be something different, which you can calculate by considering the two equivalent inductances of the port air masses in parallel. It will still look, assuming that both ports are simple lossless tubes (i.e., not stuffed), like a plain vanilla reflex.
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Old 6th January 2003, 05:22 AM   #24
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Quote:
Originally posted by SY
Rodd, the equivalent port inductance is proportional to length/area. So, if we take a simple case of two duct tubes of equal length, doubling the area is the same as halving the inductance, i.e., paralleling the inductors. So, what your program is doing makes sense to me.

keltic, as soon as you add the second port, the first one is no longer tuned to the same frequency it originally was. The new box frequency will be something different, which you can calculate by considering the two equivalent inductances of the port air masses in parallel. It will still look, assuming that both ports are simple lossless tubes (i.e., not stuffed), like a plain vanilla reflex.
Than, according to what youíre saying, // ports of the same dimension will sum as // inductors to a single tuning frequency. Furthermore, // ports of different dimensions will also sum to a single inductor with a single tuning frequency. So contrary to the theory I presented earlier in this thread, // ports work together as a single port (or single inductance). Sorry. All of this is assuming that Rap is negligible.

I guess this make sense. If I were to construct a oscillator circuit with // inductors and a // capacitor (Cab), I would expect there to be a single oscillating frequency. Itís still hard to comprehend two ports of different dimensions resonating at a single frequency. There has to be some mechanical or acoustical mechanism (that I donít see at this point) to couple the ports together.

The idea of a wider band width (lower Q) should still be realizable through the use of port stuffing or similar techniques (the lossy port discussion), but dual resonances seem to be out of the picture at this point.

How about a port and a PR in the same box?

Rodd Yamashita
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Old 6th January 2003, 05:28 AM   #25
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Quote:
Originally posted by roddyama

How about a port and a PR in the same box?

Rodd Yamashita
I have heard of speakers with a "Slot loaded Passive Radiator". Always wondered what that meant.
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Old 6th January 2003, 05:30 AM   #26
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Port and a PR? Now you've got the L in parallel with an LCR series circuit (equivalent of the PR). There's just no end to this Helmholz resonator stuff, is there?

If it helps to grasp the analogies, one can think of the two ports in composite as representing summed masses on a single spring (the box volume compliance). There's only going to be the single resonant frequency despite the two masses. The ports are coupled because they're being driven by the same air volume and in phase.

Yes, the analogy isn't exact, but if it were, it wouldn't be an analogy, it would be an identity.
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Old 6th January 2003, 04:41 PM   #27
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Quote:
Originally posted by kelticwizard
Suppose a speaker is tuned to 42 Hz. Under 42 Hz, the response rolls off at a fast 24 dB/octave. The air in the port "sees" an open box below 42 Hz.

Now if I add an additional port to tune the box half an octave lower-30 Hz-won't all the 30 Hz frequencies "see" an open box and go out out the port tuned to 42 Hz?

In short, I would suggest that box with two ports that would tune a box normally to 42 Hz and 30 Hz respectively essentially add up to a box tuned to 42 Hz.
Kelticwizard,

You may have made a good point. It would be like having 2 inductors of different values (with no DC resistance) in //. If you apply a very low frequency, both inductors are seen as a short circuit. As you increase the frequency and reach the frequency at which the larger of the // inductors would resonate, it cannot because the smaller inductor is shorting out the // capacitance (Cab). If you than continue to increase the frequency, you will eventually reach the resonance of the smaller inductor, at which point it will resonate with the // capacitor and the larger inductor will be and impedance.

Now mind you, this is a static analysis. The resonant frequencies of the port are not brick wall values. It may be very possible that Sy is still correct, and the value of impedance created by the larger inductor, and the phase relationships between the components may sum to single resonant frequency that is different than that of the smaller inductor by itself. It would have to be analyzed in the frequency domain to know for sure, but there may be some thing I can look at still (it has been a long time since Iíve had to deal with this stuff).

Rodd Yamashita
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Old 6th January 2003, 05:06 PM   #28
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Rodd, you've mentioned "inductors resonating" a few times. What do you mean by that? I'm familiar with LC resonance, but I don't know what an "L" resonance is.
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Old 6th January 2003, 05:15 PM   #29
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Quote:
Originally posted by SY
Rodd, you've mentioned "inductors resonating" a few times. What do you mean by that? I'm familiar with LC resonance, but I don't know what an "L" resonance is.
The inductor in // with a capacitor (in this case, Cab compliance of the box) will form an oscillator. Check the equivilent circuit on the first page of this thread.

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Old 6th January 2003, 05:24 PM   #30
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Ah, that makes things clearer- it's a semantic, not a technical issue. Tnx!
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