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<blockquote data-quote="TheOtherSide" data-source="post: 3005344" data-attributes="member: 570955">Alright, to answer a few questions.&nbsp; Yes I am good enough with power tools to make what I am asking about.&nbsp; Yes, I am very good at math so if your answer is an equation, I can handle it.&nbsp; No, I am not crazy.Here&#039;s the scoop.&nbsp; I have a Silverado Ext Cab.&nbsp; I am going to be putting a box under the seat with a 1.5&quot; lift.&nbsp; I am looking to get about 2.25 net tuned to 28Hz.&nbsp; I may have to settle with 2.0 net though.&nbsp; By net, I mean gross volume minus port volume.&nbsp; My problem lies here though.&nbsp; The box can&#039;t be a typical rectangle. Sure, it&#039;d make things easier and I wouldn&#039;t have to ask but I need all the space I can get to reach 2.0+ ft^3.Now the question:When figuring port length on a port that is just straight in a rectangular box, i.e. the 20 sq in of port area is equal at the output of the box and the internal input to the port, is it all based on volume of air moved through the port?&nbsp; If this is the case and the port dimensions change (both height and width), can you just alter the dimensions to make the available port area the same at the input (internal port) and output (the part you see on the outer surface of the port)?&nbsp; Also, can I just calculate necessary port volume and make the port long enough, wide enough to fill that given the 20 in^3 of port area?See this drawing for an idea of what I am talking about:<img src="http://www.everythingbacon.com/images/box3.bmp" alt="" class="fr-fic fr-dii fr-draggable " style="" />Assume port lenth P(L) is 28 inches.Port Area on the outside is given as P(Hi) {height} and P(Wi) {width}.I need P(Wf) and P(Hf).&nbsp; The final port height is going to be variable though due to incident angle of the bottom of the box.I am using 3/4&quot; MDF.{By the way, I just made these #&#039;s up too. But they aren&#039;t totally random.&nbsp; I want 20 in^2 of port and the external dimensions are close but not exact.}</blockquote>

[QUOTE="TheOtherSide, post: 3005344, member: 570955"] Alright, to answer a few questions. Yes I am good enough with power tools to make what I am asking about. Yes, I am very good at math so if your answer is an equation, I can handle it. No, I am not crazy. Here's the scoop. I have a Silverado Ext Cab. I am going to be putting a box under the seat with a 1.5" lift. I am looking to get about 2.25 net tuned to 28Hz. I may have to settle with 2.0 net though. By net, I mean gross volume minus port volume. My problem lies here though. The box can't be a typical rectangle. Sure, it'd make things easier and I wouldn't have to ask but I need all the space I can get to reach 2.0+ ft^3. Now the question: When figuring port length on a port that is just straight in a rectangular box, i.e. the 20 sq in of port area is equal at the output of the box and the internal input to the port, is it all based on volume of air moved through the port? If this is the case and the port dimensions change (both height and width), can you just alter the dimensions to make the available port area the same at the input (internal port) and output (the part you see on the outer surface of the port)? Also, can I just calculate necessary port volume and make the port long enough, wide enough to fill that given the 20 in^3 of port area? See this drawing for an idea of what I am talking about: [IMG]http://www.everythingbacon.com/images/box3.bmp[/IMG] Assume port lenth P(L) is 28 inches. Port Area on the outside is given as P(Hi) {height} and P(Wi) {width}. [COLOR=DarkRed]I need P(Wf) and P(Hf). The final port height is going to be variable though due to incident angle of the bottom of the box.[/COLOR] I am using 3/4" MDF. {By the way, I just made these #'s up too. But they aren't totally random. I want 20 in^2 of port and the external dimensions are close but not exact.} [/QUOTE]

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