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12th July 2006, 06:42 AM  #1 
diyAudio Member
Join Date: Jun 2004
Location: Indiana

Port length  multiple ports
I looked in the help files for both of the box design programs that I have and neither one could clarify something for me. When using multiple ports to reduce mach number the specified length more than doubles when the number of ports is doubled. What is not specified is whether the length listed is the length of each port or the sum of all port lengths. So which is it? And if it is the sum does the total length need to be divided evenly between the ports?
mike 
12th July 2006, 07:08 AM  #2 
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Join Date: Feb 2004
Location: Chinook Country.Alberta

multiple ports and tuning....
dt=sqrt(d1^2+d2^2+...+dn^2).. that's the effective diameter .
calculate the effective diameter then calculate the total area and put her into Helmholtz equation.area
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stew ☮ "A sane man in an insane world appears insane." 
12th July 2006, 09:43 AM  #3 
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Join Date: Feb 2004
Location: Stockholm

The simplified version is this: The (individual) port length is proportional to the sum of all crosssectional areas.
In real life there is a bit of a difference due to the cooscillating air outside and inside the tube (which makes the tube appear a bit longer). That makes an exact calculation of the port length a bit more difficult. But roughly: add a second port  double their lengths. 
12th July 2006, 10:50 AM  #4 
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Join Date: Feb 2004
Location: Chinook Country.Alberta

Svante..an over simplified view..
as an example , use 2 ports of 2" diameter..
dt=sqrt(2^2 + 2^2) =sqrt(8)=2*(sqrt(2))=2*1.414=2.828 >r=1.414 Area,S, is pi * r^2=2*pi=6.2832 so the area doubles, therefore the length needs to be expanded. The Helmholtz equation is as follows f=(c/2*pi)*sqrt(S/VL) so if f is constant, V is constant (the enclosure), c is a constant as is 2*pi f is proportional to sqrt(S/L), but 2 ports of 2" diameter does not = the same as 1 port of 4", it =2.824 . (which is 2*sqrt(2)), so the length must be multiplied by sqrt(2) or 1.414 now take it to 3" port.. L would have to be multiplied by sqrt(3) and so on...
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stew ☮ "A sane man in an insane world appears insane." 
12th July 2006, 04:08 PM  #5 
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Location: Indiana

Thanks
OK. I think I have it. For a design with 1 4" port of 11.53" length the Mach # came out to .17. To reduce this I add one more 4" port. This gives a port length of 25.98 for each of the two ports and reduces Mach to .09. Does that sound about right.
mike 
13th July 2006, 10:44 PM  #6  
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Re: Svante..an over simplified view..
Quote:
There is no need to involve equivalent diameters, unless the software requires it (if it does, the effective diameter is multiplied by sqrt(2) as you say). 

13th July 2006, 11:10 PM  #7 
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Location: Chinook Country.Alberta

doubling S, etc...
in this case yes, using 2 ports requires a doubling of the port length.
but what if you want a single 5" or 6" (unusual but not out of the question). Or a rectangular port? The answer provided is the most generalized one, else if others read into the explaination, they may not see the important relation and could make assumptions. As the most generalized solution, it can be used in any case, not just simply putting in a second port, perhaps maybe 5 ports or some other geometry. Why not an oval? It's like algerbra or calculus. Solve the equation in most general terms first, then find a specific solution based on provided information, boundary conditions, etc.
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stew ☮ "A sane man in an insane world appears insane." 
13th July 2006, 11:54 PM  #8  
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Re: doubling S, etc...
Quote:


14th July 2006, 01:39 AM  #9 
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Join Date: Feb 2004
Location: Chinook Country.Alberta

not my formula
it is the Helmholtz equation.
I agree for the same tuning S/L is to be kept constant. How long should a 4" X 5" port be(rectangular)? Or a 7" diameter port. you get the idea, I hope. Understanding the whole equation and understanding where the tuning comes from is important as a whole. What if you change the Volume of the enclosure..? Limiting the possibilites to only doubling the number of ports to attain the correct tuning and mach number ..is , well ...limiting. In the example I am using ..a 4" X 5" rectangular port, the length needs to be increased by a factor of 1.592.
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stew ☮ "A sane man in an insane world appears insane." 
14th July 2006, 10:15 AM  #10  
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Re: not my formula
Quote:


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