| physics of port tuning [message #38132] |
Mon, 02 September 2002 00:45  |
dbeardsl
Messages: 127
Registered: May 2009
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Master |
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I noticed something, messing around with a box calculator. When tuning a box I've thought that the tuning was simply using the air in the port as a mass and the air in the box as a spring in a simple spring/mass system. But then, how much the spring (air pressure inside the box) affects the mass (air in the port) is directly proporional to the area presented to it (port area). And if I follow that reasoning, the volume (port air mass) is directly proportional to the area, making the tuning frequency directly proportional to the length of the port no matter what the area... Whatever the case... I'm wrong and I noticed something that I really don't understand. If you have a certain tuning, and you divide the port in half down its length so it is now 2 ports, the overall length must increase to keep the same tuning... Why is that? Anyone care to explain more of the physics involved?
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| re: [message #38144 is a reply to message #38134] |
Tue, 03 September 2002 01:01  |
dbeardsl
Messages: 127
Registered: May 2009
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Master |
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I use WinISD for calculations, I did it correctly, two ports with the same total area as another single port end up having different lengths. Why?
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| port calcs [message #38145 is a reply to message #38144] |
Tue, 03 September 2002 02:02 |
vladimir4
Messages: 22
Registered: May 2009
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Chancellor |
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The right approach to calculate multiple identical ports is:
a) on port : normal formulae for Vb
b) two ports: do calculations as for one port for Vb/2, (then for Vb use two such ports)
c) in general, for m ports : calculate one port for Vb/m and use m such ports in Vb
I hope it's clear :)
v.
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| Re: physics of port tuning [message #38146 is a reply to message #38132] |
Tue, 03 September 2002 05:28 |
Jostein
Messages: 4
Registered: May 2009
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Esquire |
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The physics involved: Port resonance fport=(1/(2*Pi))*squareroot(K/M) K is stiffness of airspring, M is mass of moving air in port. K=beta*Ap*Ap/Vb, there Ap is port area, and Vb is box volume and beta is a stiffness constant for air. M=p*Ap(Lp+16/3sqrroot(Ap/Pi*Pi*Pi)) there p is air density and Lp is length of port. If you use one port with area A and port length Lp or 2 ports with area A/2 and port length Lp, the tuning frequency should be the same
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| Re: re: [message #38147 is a reply to message #38144] |
Tue, 03 September 2002 08:52 |
Adam
Messages: 419
Registered: May 2009
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Illuminati (1st Degree) |
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There is an "end correction" factor involved with multiple ports. I'm not completely aware of the physics behind it, but I believe it is because there is increased air drag on the inner walls of the port when you use two smaller ports verses a single large port. Thus the length has to be increased a small amount to compensate. Usually no more then a couple of inches. Is this what you're talking about? Adam
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| Exact value [message #38160 is a reply to message #38147] |
Wed, 04 September 2002 01:00 |
vladimir4
Messages: 22
Registered: May 2009
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Chancellor |
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Hi,
sorry form my redundant input, but when you use the approach I described below you get exactly what winisd (or other speaker cad) calculates:
Example:
Eminence Delta-12LF in 3.853 ft3 box tuned to 43Hz:
one square port 5"x5" require 5.53" length
"two" ports 2.5"x5" (same area) require 6.45" length
BECAUSE: length of one port 2.5"x5" in 3.853/"two"=1.9265 ft3 for 43Hz is 6.45"
This is exact and valid approach.
V.
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