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Ports Are Counter Intuitive/Don't Make Sense

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  • Ports Are Counter Intuitive/Don't Make Sense

    Howdy. I'm working on porting the 1.70 ft3 volume in which I've housed a PE Reference 10 inch sub. Thanks to Chris R.'s replies here, I understand that he recommends a 4 inch port over a 3 inch diameter one.

    Here's the confusing part. It seems that when you increase port diameter, you need to increase the port length... by a lot. Insane amount. That just doesn't make sense.

    Let's say that for a 25 hz tuning for that 1.70 ft3 volume, a 3 inch diameter port has to be 16 inches long. So, the total volume of that port's = 113.1 square inches.

    Then, if I increase the diameter to 4 inches, keeping the length the same at 16 inches, the total volume = 201 square inches. That's 88 additional square inches. So, if it's more cubic displacement, then the port should be shorter, right? For the same cubic displacement, the 4 incher should be about 8 inches long.

    But, it turns out that if you go from a 3 inch diameter port for 25 hz to a 4 incher, you need to increase the length to about 30 inches! Wow. That's nuts. Why?
    Why do you need to almost double the length if you're increasing diameter? It seems that the opposite should be true. Baffled.

  • #2
    Formulas explain why.... You can see here the diameter of the port if increased, will increase the length:

    Port Length
    The port length required to tune a volume of air to a specific frequency can be calculated by using the following equation: Lv = (23562.5*Dv^2*Np/(Fb^2*Vb))-(k*Dv), where:
    Dv = port diameter (cm)
    Fb = tuning frequency (Hz)
    Vb = net volume (litres)
    Lv = length of each port (cm)
    Np = number of ports
    k = end correction (normally 0.732)
    The value for k, the end correction, can be fine-tuned by using the following values to derive the appropriate end correction figure for each end of the port, then adding them together
    Both ends flanged: k = 0.425 + 0.425 = 0.850
    One end flanged, one end free: k = 0.425 + 0.307 = 0.732
    Both ends free: k = 0.307 + 0.307 = 0.614
    Normally, k=0.732 is assumed.

    But for people like me, here's how I think of it.....

    Forget port length for a minute - and think of the resistance a hole in the box makes.

    A small hole - provides more resistance. There's less surface area for air to escape.

    Increase the size of the hole - now the resistance reduces and the box becomes leaky.

    If we put a tube behind the hole - we increase resistance again - due to the loaded air mass in the tube. If we had an infinite length tube on a very small hole (port) we'd approach a sealed enclosure as we have a very big air mass acting as a spring.

    Now... if we increase the size of the hole - we reduce resistance.... so to make up for this - we must increase the resistive air mass behind the hole. We do this by increaseing the length of the tube to increase the mass to offset the increased diameter.

    I'll let others critique my logic.

    Comment


    • #3
      Originally posted by supermike View Post
      Why do you need to almost double the length if you're increasing diameter? It seems that the opposite should be true.
      Box Fb goes down as box volume goes up. Box Fb goes up as port area goes up. It's as simple as that.
      www.billfitzmaurice.com
      www.billfitzmaurice.info/forum

      Comment


      • #4
        You're not comparing the right parameters, or you're comparing them the wrong way.
        Your math is ok, but...
        you're tuning a box with a "chunk" of air, and the things that matter are the mass of the air (in the port - which is equivalent to your cu.in.), and the square area of the port.

        You COULD run the actual numbers, but - (relatively) all you really need to know is that the 4" port has NEARly twice the area (4^2/3^2, which is 16/9 or about 1.78 times) of the 3" port. For that larger area, the mass (or cu.in.) has to increase by 1.78 SQUARED (so, you already got 1.78x from the bigger dia. - you get the other 1.78 by making the slug of air (which is the port) 1.78 times LONGER. Off the top of my head, I get 29" (you got 30!). It all makes sense to me.

        For more detail, Google an equation for finding how to tune a box to a given freq. (I'm SURE it's out there).
        (OOPs - there it is (18min ago), it came to YOU)

        Comment


        • #5
          OK, guys. Thanks. That makes it clearer. Bottom line, ports are really wily, sneaky things.

          Comment


          • #6
            Originally posted by supermike View Post
            OK, guys. Thanks. That makes it clearer. Bottom line, ports are really wily, sneaky things.
            And you choose any old port in a storm. Obviously moisture must affect them.
            Francis

            Comment


            • #7
              I find that tuning below 30Hz you usually run into a catch 22 where either the port is too small to avoid air velocity issues, or too long to fit the cabinet and/or avoid pipe resonance issues. 30Hz is enough for most people though, below that there are alternatives to a ported design, either way you end up with a physically big system to get that sort of low end performance.
              "I just use off the shelf textbook filters designed for a resistor of 8 ohms with
              exactly a Fc 3K for both drivers, anybody can do it." -Xmax

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