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First venture into ported box
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<blockquote data-quote="Moble Enclosurs" data-source="post: 7538793" data-attributes="member: 634917"><p>Well, your'e right about the 3.5" port cubic volume. BUT.....and this is very important for everyone to understand. You cannot retain a ports physical dimensions based on volume and retain tuning!!! Here is why:</p><p></p><p>Example: Take a 8-9" round port, or a 10x6 square port...same volume. Now, if you give is a 23.25" length, you get a certain tuning, say with a chamber volume of about 2 cubic ft for a specific driver ( I am using one that I am working on right now as an example). You get about a 44Hz tuning with a 4th order Im working on for this driver. Now, if we get volume of the port and keep the chamber volume, we get 1464.75in^3. now, if we want to use a smaller port for example, say 10" and retain volume of the port...naturally it should be shorter right? Well, we know making it shorter, we are decreasing the area to maintain tuning. This is well known. But if we keep volume, making it shorter and also making it smaller, well, physics will say, you will have a pretty large area compared to the longer port if you are to keep volume. So, lets say the ideal of 44Hz for a 10" port in this example is about 5.88" round, but with retaining volume, you get about 146.48in^2 area, which is a 13.65" round port.</p><p></p><p>So, that is not going to give you the same tuning AT ALL. In fact it will be around (for my example box/driver combo) 80hz tuning! Almost a full octave above the original tuning! not going to work that way.</p><p></p><p>best to figure for minimum port area needed, then tune it based on length required afterwards. Not by retaining port volume. Not a good idea.</p><p></p><p>Note: My math is a little off only because I averaged and rounded fast to get this typed. But the idea is solid that it increased tuning by retaining port volume if making the port shorter, and vice versa due to physics.</p></blockquote><p></p>
[QUOTE="Moble Enclosurs, post: 7538793, member: 634917"] Well, your'e right about the 3.5" port cubic volume. BUT.....and this is very important for everyone to understand. You cannot retain a ports physical dimensions based on volume and retain tuning!!! Here is why: Example: Take a 8-9" round port, or a 10x6 square port...same volume. Now, if you give is a 23.25" length, you get a certain tuning, say with a chamber volume of about 2 cubic ft for a specific driver ( I am using one that I am working on right now as an example). You get about a 44Hz tuning with a 4th order Im working on for this driver. Now, if we get volume of the port and keep the chamber volume, we get 1464.75in^3. now, if we want to use a smaller port for example, say 10" and retain volume of the port...naturally it should be shorter right? Well, we know making it shorter, we are decreasing the area to maintain tuning. This is well known. But if we keep volume, making it shorter and also making it smaller, well, physics will say, you will have a pretty large area compared to the longer port if you are to keep volume. So, lets say the ideal of 44Hz for a 10" port in this example is about 5.88" round, but with retaining volume, you get about 146.48in^2 area, which is a 13.65" round port. So, that is not going to give you the same tuning AT ALL. In fact it will be around (for my example box/driver combo) 80hz tuning! Almost a full octave above the original tuning! not going to work that way. best to figure for minimum port area needed, then tune it based on length required afterwards. Not by retaining port volume. Not a good idea. Note: My math is a little off only because I averaged and rounded fast to get this typed. But the idea is solid that it increased tuning by retaining port volume if making the port shorter, and vice versa due to physics. [/QUOTE]
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