Box size constant, shorter port...

[Bent]

No, I'm not re-porting my Heretic,
But I don't think a post somewhere else is correct so I wanted confirmation.

If a speaker box (enclosure) volume reamains constant, and port diameter remains constant, and I shorten the port length - will my tuning frequency go up or down? (I've always thought it would go down, but I haven't run the numbers).

[JonMarsh]

Shorter port means the tuning freqeuncy goes up. Less airmass in the port volume.

Think of blowing across the mouth of a bottle- large bottles, low notes. Small bottles, higher notes.

~Jon

[Bent]

Oh well, at least I still no the difference between a hot and a neutral.
(some day I'll get my head around this concept)

[Davey]

Most all of the bottles are the same size now, but awhile back I tended to prefer the low tuning frequency of Corona and XXequis vice the higher frequencies of Bud, Miller, etc. Of course, with enough testing the port length that you're looking to "target" doesn't really matter.

Cheers,

Davey.

[cjd]

Most all of the bottles are the same size now, but awhile back I tended to prefer the low tuning frequency of Corona and XXequis vice the higher frequencies of Bud, Miller, etc. Of course, with enough testing the port length that you're looking to "target" doesn't really matter.

Cheers,

Davey.

You really should hear the 5L wine bottle I have. Well, once we empty it (done two others though). Banfi Col di Sasso. Excellent stuff. Italian wine from around Tuscany.

C

[joecarrow]

Port diameter?

What about increasing port diameter? It seems like that should increase air mass in the port and decrease the tuning frequency, but WinISD clearly shows the opposite effect when you change the port diameter.

By the way- speaking of old Klipsch speakers, the shop two blocks from me just got a pair of Klipschorns. I had no idea of their size! :E

[Brian Bunge]

For the same length port, a larger diameter will give a higher tuning frequency. In other words, the larger the diameter of the port, the longer it must be for a given tuning frequency.

[cjd]

By the way- speaking of old Klipsch speakers, the shop two blocks from me just got a pair of Klipschorns. I had no idea of their size! :E

Very interesting sound if you get to listen to 'em. Depends what vintage they are too, for that matter.

C

[joecarrow]

Shorter port means the tuning freqeuncy goes up. Less airmass in the port volume.

Think of blowing across the mouth of a bottle- large bottles, low notes. Small bottles, higher notes.

~Jon

For the same length port, a larger diameter will give a higher tuning frequency. In other words, the larger the diameter of the port, the longer it must be for a given tuning frequency.

I'm confused- why does a larger diameter port for the same length and box volume give a higher tuning frequency? For the sake of this discussion, the port volume doesn't count toward box volume- the port is external to the box.

Larger diameter seems as though it should provide a larger mass of air in the port and decrease tuning frequency- what's conteracting that? Is it decreased resistance to motion of the air in the port?

Thanks in advance!

[Dennis H]

Yeah, think of box volume as the spring and air resistance through the port as the mass. Bigger port area gives less resistance so you have to balance it with longer port length.

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/cavity.html

[Feyz]

From Mr. Leach's Intro. to Electroacoustics book

The "mass" that moves in the port is an "acoustic mass" under the scope of ported box frequency response behaviour. The regular mass m, the mechanical one, is defined by the relationship of force and accelaration (or velocity) which is:

F = m x a = m x dv/dt, where a is acceleration, v is velocity and dv/dt is rate of change of velocity with respect to time

Acoustic mass, call it MA, on the other hand is defined by the relationship between pressure change and "volume velocity", call it U, and given by:

P = MA x dU/dt, dU/dt being volume velocity's derivative with respect to time.

Volume velocity U is the rate of change of displaced volume with respect to time. For a port, volume velocity U is equal to the volume of air that goes in to (or leaves out of, which is equal) the port, which is given by the equation:

U = pi x r^2 x v, where pi is 3.14..., ^2 means to the power 2, v is velocity of air in the port. IOW, U = S x v, where S is cross section area of the port.

Pressure change between the open ends of the port is equal to net force applied to the mechanical air mass inside the port divided by cross section area:

P = [F / (pi x r^2)]
= (m x dv/dt) / (pi x r^2)

Since v = U/ (pi x r^2), dv/dt = (1/(pi x r^2)) x dU/dt


Then P = [m/(pi x r^2)^2] x dU/dt

This means MA, the acoustic mass is equal to m/((pi x r^2)^2)
, or m/(S^2)

Since m = density x volume = density x length x pi x r^2, where length is length of port,

MA = density x length / (pi x r^2)

Which says that acoustic mass increases with increased length of port and decreases with radius of port.

Something like that anyways....

[joecarrow]

Thanks, guys! It's all clear now.