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.
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.
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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.
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.
ok....
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.
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.
kelticwizard said:
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.
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.
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.
Svante said:
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....
Re: Re: Conical Port for Bass Reflex-How To Calculate?
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.
sreten said:
Hi,
As described a fairly pointless exercise, what is the purpose ?
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 said: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.
Re: Re: Conical Port for Bass Reflex-How To Calculate?
Just an outgrowth of my looking into Transmission Lines. I was playing around with Martin J. King's software and comparing tapered lines with straight lines of equal length and volume.
I noticed that tapering the lines had the effect of lowering the resonant frequency-the notch in the cone's response kept going downward the greater the taper.
The thought occurred to me that the reason this was happening was that tapering the line basically was a way of mass loading the Transmission Line. When you close off a Transmission Line and add a port, you will get a lower resonance frequency than the port size and enclosure volume would indicate. Also, the midrange resonances in the line get reduced in volume, as also happens with tapering the line.
So I thought I would do a little calculating of how large a conical port would have to be to give the same effect as tapering the line gives. Hence the question.
sreten said:
Hi,
As described a fairly pointless exercise, what is the purpose ?
Just an outgrowth of my looking into Transmission Lines. I was playing around with Martin J. King's software and comparing tapered lines with straight lines of equal length and volume.
I noticed that tapering the lines had the effect of lowering the resonant frequency-the notch in the cone's response kept going downward the greater the taper.
The thought occurred to me that the reason this was happening was that tapering the line basically was a way of mass loading the Transmission Line. When you close off a Transmission Line and add a port, you will get a lower resonance frequency than the port size and enclosure volume would indicate. Also, the midrange resonances in the line get reduced in volume, as also happens with tapering the line.
So I thought I would do a little calculating of how large a conical port would have to be to give the same effect as tapering the line gives. Hence the question.
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