Q: confused about vents
William T. Rankin
In article <34C5F1CE...@cybec.com.au>,
Adam Seychell <asey...@cybec.com.au> wrote:
>What is the relation ship between the dimensions of vent in a speaker
>enclosure to its tuning frequency ?
From <http://www.spiceisle.com/audiodiy/portcal.htm> :
[begin excerpt]
Port Length
The port length required to tune a volume of air to a specific frequency
can be calculated by using the following equation:
Lv = (23562.5*Dv^2*Np/(Fb^2*Vb))-(k*Dv)
where,
Dv = port diameter (cm)
Fb = tuning frequency (Hz)
Vb = net volume (litres)
Lv = length of each port (cm)
Np = number of ports
k = end correction (normally 0.732)
The value for k, the end correction, can be fine-tuned by using the
following values to derive the appropriate end correction figure for each
end of the port, then adding them together
Flanged End: 0.425
Free End: 0.307
e.g. if both ends were flanged,
k = 0.425 + 0.425 = 0.850
if one flanged, one free,
k = 0.425 + 0.307 = 0.732
if both ends were free,
k = 0.307 + 0.307 = 0.614
Normally, k=0.732 is assumed
In practice, it's best to use ports that are slightly longer than
predicted by the above equations, then adjust their length until the
correct tuning is achieved. It is much easier to shorten a port than to
lengthen it!
[end excerpt]
>The volume is not constant ! i.e. if you half the vent area, the length
>does not exactly double.
This is because of the effect of the 'k' term above, which takes into
account the effects of the air mass at the port ends.
>regards,
>
>Adam
>
Hope this helps. Check out the site above. It has lots of useful
information. You may also look into picking up a copy of Dickason's
"Loud Speaker Design Cookbook -- 5th Edition".
-bill
--
bill rankin ...................................... philosopher/coffee-drinker
wra...@ee.duke.edu ........................................ doctoral wannabe
duke university dept. of electrical engr ......... scientific computing group
Richard D Pierce
In article <34C5F1CE...@cybec.com.au>,
Adam Seychell <asey...@cybec.com.au> wrote:
>What is the relation ship between the dimensions of vent in a speaker
>enclosure to its tuning frequency ?
>frequency and box volume, the port must have a specific area and length.
>The volume is not constant ! i.e. if you half the vent area, the length
>acts like a piston of air having a cretin mass. This air mass will
>resonate with the box compliance (like stiffness of a spring). Therefor,
>I would think the mass of air in the vent should be directly
>proportional vent volume.
Yes, this makes intuitive good sense, but, unfortunately for intuition,
it's also wrong.
The effective acoustic mass of a port is proportional to its length
DIVIDED BY the square of the diameter. It is not propoertional to the
volume of the vent.
The reasons for this are complex, but firmly rooted in physics. I am in
the process of preparing an article as to why this is, so be patient.
>Is there a simple equation to determine the vent tuning freq?
>I wish to construct my own vent of different shapes, but don't know how
>to design.
The acoustic mass of a vent is:
l'
Map = p----
2
r
where Map is the acoustic mass in units of kg/m^4, p (rho, actually) is
the density of air 1.18 kg/m^3m, 'l is the end-corrected length of the
port in m, and r is the radius of the port in m.
The end correction is applied due to the fact that inertia effects of the
medium at the ends of the tube make it appear somewhat longer than it
actually is. For the end terminated in the baffle (a "flanged" end), the
it's necessary to add about 0.42 diameters, while for the end that is
hanging in the cabinet, it's about 0.3 diameters, for a total of 0.72
times the diameter, so:
l' = 0.72 l
where l is the physical length of the port.
Now, the effective acoustic compliance enclosure needs to be determined as
well. That value is:
Vb
Cab = ------
2
p c
where Cab is the acoustic compliance in units of cm^4 sec^2/kg, Vb is the
enclosure volume in cubic meters, p (rho) is as above, and c is the speed
of sound, at about 345 m/sec.
The box tuning frequency is:
1
Fb = ------------------
2 pi sqrt(Mas Cab)
--
+---- Dick Pierce ---------------------------------------------+
| Professional Audio Product Development |
| Transducer Design and Measurement |
+---- (781) 826-4953 (Voice and FAX) DPi...@world.std.com -----+
Adam Seychell
What is the relation ship between the dimensions of vent in a speaker
enclosure to its tuning frequency ?
I've tested with speaker design software to show that for a given tuning
frequency and box volume, the port must have a specific area and length.
The volume is not constant ! i.e. if you half the vent area, the length
does not exactly double. This confuses me since I thought that the vent
acts like a piston of air having a cretin mass. This air mass will
resonate with the box compliance (like stiffness of a spring). Therefor,
I would think the mass of air in the vent should be directly
proportional vent volume.
I wish to construct my own vent of different shapes, but don't know how
to design.
regards,
Adam
Adam Seychell
Richard D Pierce wrote:
> In article <34C5F1CE...@cybec.com.au>,
> Adam Seychell <asey...@cybec.com.au> wrote:
> >resonate with the box compliance (like stiffness of a spring). Therefor,
> >I would think the mass of air in the vent should be directly
> >proportional vent volume.
>
> Yes, this makes intuitive good sense, but, unfortunately for intuition,
> it's also wrong.
>
> The effective acoustic mass of a port is proportional to its length
> DIVIDED BY the square of the diameter. It is not propoertional to the
> volume of the vent.
> The reasons for this are complex, but firmly rooted in physics. I am in
> the process of preparing an article as to why this is, so be patient.
>
> >Is there a simple equation to determine the vent tuning freq?
> >I wish to construct my own vent of different shapes, but don't know how
> >to design.
>
Thank you Richard , this makes more sence now. However after thinking about
end correction stuff, your equation l' = 0.72 l should be l' = 0.72d + l
where l = physical length and d = vent diameter.
I tested again with computer program and yes, if you double the AREA, the
length must also double (its actually its a bit more than double, probably due
to end-correction).
Where will you be posting your article ? looking forward to seeing it.
Adam
Richard D Pierce
Sorry, made a typing mistake:
In article <En4zG...@world.std.com>,
Richard D Pierce <DPi...@world.std.com> wrote:
>The end correction is applied due to the fact that inertia effects of the
>medium at the ends of the tube make it appear somewhat longer than it
>actually is. For the end terminated in the baffle (a "flanged" end), the
>it's necessary to add about 0.42 diameters, while for the end that is
>hanging in the cabinet, it's about 0.3 diameters, for a total of 0.72
>times the diameter, so:
>
> l' = 0.72 l
>
>where l is the physical length of the port.
The correct equation is:
l' = l + 0.72 d
Sorry for the confusion.
Robert Bullock
William T. Rankin wrote:
>
> In article <34C5F1CE...@cybec.com.au>,
> Adam Seychell <asey...@cybec.com.au> wrote:
>
> >enclosure to its tuning frequency ?
>
As a historical note, this formula should probably be attributed to L.
L. Beranek because of his discussion of an open tube as an acoustic mass
that can be found in his book "Acoustics" (McGraw-Hill, 1954, pp
131-133). The general form that can be inferred from his discussion is
Lv = (c^2/(4*pi^2))*(Sv/(fb^2*Vb)) - (C1 + C2)*Sqrt(Sv),
where Sv is the total cross-sectional area of the vent system, c is the
speed of sound in air, C1 and C2 are the end corrections derived by
Barenek, and length units are meters. Taking Sv = Nv*pi*(Dv/2)^2, c =
344 m/s and converting to centimeters (and liters) gives the formula
quoted by Mr. Rankin.
Note that the constant 23562.5 in the Rankin formula can differ from
derivation to derivation, depending on the value assumed for c. This is
one reason that I made c a program assignable parameter in my BoxModel
program. Sincerely, R. Bullock.