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17th November 2007, 07:41 AM  #1 
diyAudio Member
Join Date: Sep 2007
Location: Syracuse, NY

Confusion about port length with multiple ports
I've been following along with Ray Alden's book in designing my latest project and I'm confused about the use of multiple ports.
Using two ports of diameter D1 and D2 is the equivalent of using one port of the diameter: SQRT(D1 + D2). So two 2" diameter ports is equvalent to using one of 2.828". We're good so far, but he leaves one thing unclear... When calculating the length of each port, do I calculate each port's length using 2.828" as the diameter (which comes out to 12.1"  much too long for my cabinet)? Or, do I calculate each port's length using their ACTUAL diameter (which comes out to 5.63"  perfect for my cabinet)? I suspect it's the latter, but definitely want to be sure before cutting any holes! Thanks for the advice. 
17th November 2007, 09:35 AM  #2 
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

You should do like this (pardon me for translating to the metric system):
Using one port, a box volume of for example 50 litres, a vent radius 3 cm and a tuning frequency of 40 Hz. L=(c/wh)²*S/V=(343/(2*pi*40))²*(0.03²*pi)/0.05=0.105 m Now this is the effective length of the tube, by removing two end corrections, the actual required length becomes L'=L2*0.85*r=0.1052*0.85*0.03=0.054 m ie 5.4 cm. Using two ports the crosssectional area is doubled: L=(c/wh)²*S/V=(343/(2*pi*40))²*2*(0.03²*pi)/0.05=0.211 m The end corrections should be subtracted from each of the tubes: L'=L2*0.85*r=0.2112*0.85*0.03=0.160 m ie 16 cm. It is worth noting that this length actually is different from that of a single tube of twice the area: L=<same as the two tube example>=0.211 m But now the port radius is a factor sqrt(2) larger, so the end correction becomes greater: L'=L2*0.85*r=0.2112*0.85*sqrt(2)*0.03=0.139 m ie 13.9 cm. The difference between the two latter examples is based on that the two vents are well separated, ie not mounted very close to one another. End corrections (0.85*r) assumes that the port openings are baffled, the 0.85 drops to 0.6 if the port end is free. The volume V of the box will also appear larger if there is any damping material inside it. All in all this highlights the importance of measuring the resulting port frequency once the system is built. There are many small corrections here and there that are quite hard to have full control of. Measuring the port frequency and adjusting the port length ensures that the design ends up as desired. 
17th November 2007, 10:00 AM  #3 
Did it Himself
diyAudio Member

I presume the end corrections are due to the abrupt step in the portbaffle interface? So would 0.85r be used in all cases, i.e. square and slot ports as well?
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17th November 2007, 09:53 PM  #4  
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

Quote:
Square and slot ports have a different factor (which I don't have here), but as I said, there are so many little things that affect the actual length that is needed so the calculations should only be seen as approximations. The final length has to be determined by measurement. 

17th November 2007, 11:16 PM  #5 
Did it Himself
diyAudio Member

Thanks for the clear explanations.
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