Conical Port for Bass Reflex-How To Calculate?

[kelticwizard]

Suppose I have a two cubic foot box, (56L). If I use a 2" diameter port 5 inches long, any of a number of programs can tell what the box will be tuned to.

But suppose instead of a piece of straight pipe, I use a conical port-one that is 2" diameter on the side which faces the outside of the box, but a 4" diameter on the interior side. No flare, just a conical expansion, 5 inches long, from 2" to 4".

Does anyone know how to calculate which frequency the 56L box will now be tuned to?

I can probably simulate one using Martin J King's software, but I prefer an equation, if possible.

[DcibeL]

I would also be interested in a simple formula to calculate this, because Solen does sell parabolic port tubes, which I think would be nice to use (however it is uncertain from Solen's specs on whether the diamater of the port is measured at the inside or the outside).

You could calculate the tuning frequency in a rather lengthy task of dividing the port into sections, for example 1/2". Take the diameter of every 1/2" section and calculate the port for that section. Combine the result of all the sections to get an approximated result. I think this should work, though I haven't done the calculations myself to prove it.

[Nanook]

not sure of the need for this. Basically you are creating a small hjorn or a small transmission line, depending on whether the largest diameter is inside the volume or not.

However, to be effective, either would have to longer than 5"...

you may be able to approximate the length (very roughly by looking at the length of a Helmholtz resonator with , say 5" Length, and a diametre of 3.5 inches. That will give you some idea.

did a quick calculation: f(h) would 52 Hz.

[Patrick Bateman]

Originally posted by kelticwizard
Suppose I have a two cubic foot box, (56L). If I use a 2" diameter port 5 inches long, any of a number of programs can tell what the box will be tuned to.

But suppose instead of a piece of straight pipe, I use a conical port-one that is 2" diameter on the side which faces the outside of the box, but a 4" diameter on the interior side. No flare, just a conical expansion, 5 inches long, from 2" to 4".

Does anyone know how to calculate which frequency the 56L box will now be tuned to?

I can probably simulate one using Martin J King's software, but I prefer an equation, if possible.

Yes, you can model it with Martin's software.
I did that a couple of years back. You can see what I came up with
here.

By the way, I would *never* recommend tuning any kind of vented box using a piece of software. You *must* measure the impedance peak to tune the box properly. No matter how good your model is, it will ALWAYS be off by a few hz, due to boundary reinforcement, the location of the port mouth and exit, etc. Go get yourself a multimeter - the software model is only a starting point.

[Svante]

You can calculate the acoustic mass* of the port by using the integral

Ma=I rho0/S(x) dx

where I is the integration sign, rho0=1.2 kg/m³, S(x) is the cross-sectional area of the vent as a function of the position along the tube. Integrate between the tube ends.

The acoustic mass of the co-oscillating air has to be added.

Thereafter, the Helmholtz angular frequency can be calculated as

wh=1/sqrt(Ma*Ca)

Ca=Vb/(rho0*c²)

*Acoustic mass is NOT measured in kg, but kg/m^4

...but I dunno if that was simple, though.

[Ron E]

Originally posted by Svante
The acoustic mass of the co-oscillating air has to be added.

Ah, the beloved end corrections.

Intuitively, making a conical port is somewhat like making a lossy acoustic diode. Flow in one direction is easier than in the other, meaning that the damping on the cone due to box resonance will be unequal over a full cycle, increasing distortion in the region where the port acts....

[sreten]

Originally posted by kelticwizard

......a conical port-one that is 2" diameter on the side which faces
the outside of the box, but a 4" diameter on the interior side ......

Hi,

As described a fairly pointless exercise, what is the purpose ?

/sreten.

[Patrick Bateman]

Originally posted by sreten


Hi,

As described a fairly pointless exercise, what is the purpose ?

/sreten.

I just re-read the post, and realized it's completely backwards. The wide end is on the INSIDE?

The use of a conical port IS fairly pointless, but if the wider end is on the INSIDE of the box, it would actually be *less* efficient than a traditional port.

The only advantage I could envision would be to stagger the port resonances - and there are better ways to do THAT.

[Svante]

Originally posted by Ron E


Ah, the beloved end corrections.

Intuitively, making a conical port is somewhat like making a lossy acoustic diode. Flow in one direction is easier than in the other, meaning that the damping on the cone due to box resonance will be unequal over a full cycle, increasing distortion in the region where the port acts....

Well, yes, at high levels (at low levels, there will be no "recitfication effects). There will certainly be a jet of air on the outside, but not on the inside.

Maybe this is the whole purpose of the device, maybe it is not a loudspeaker at all .

[kelticwizard]

Originally posted by Svante
You can calculate the acoustic mass* of the port by using the integral

Ma=I rho0/S(x) dx

where I is the integration sign, rho0=1.2 kg/m³, S(x) is the cross-sectional area of the vent as a function of the position along the tube. Integrate between the tube ends.......


...but I dunno if that was simple, though.

Integral. That, um, means calculus, right?

For most of the people on this forum, yours would certainly be exactly the right answer.

For myself, however, that means I'm in deep trouble already.