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 Multi-Way Conventional loudspeakers with crossovers

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 24th July 2013, 03:38 AM #11 dumptruck   diyAudio Member   Join Date: May 2010 Location: MN Sorry, I missed the part about having converted your cabinet to a circle and was using a 14" square. So for the cylindrical cabinet with 7" radius: end of inner port = pi*1.5ē = 7.07 inē quickly transitions to: 3*pi*1 = 9.42 inē transitions to exit of: 14*pi*1 = 43.98 inē You could think of it like a flat horn where the top and bottom stay 1" apart but the width expands from 7 to 44 inches, quickly. Last edited by dumptruck; 24th July 2013 at 03:42 AM.
 24th July 2013, 06:35 AM #12 BlueWizard   diyAudio Member   Join Date: Jun 2007 Off topic, but that is actually an interesting point about the external port. At the port it is the circumference of a 3" circle times the height of the port, whether 1", 2", or 3". The with each concentric ring 1" farther out, the opening area expands by PIxDiaxH (PI times the Diameter (to give circumferance) time the height (1", 2", 3", whatever). That is not a linear expansion, though I confess I don't know what kind of expansion it is. Let's see if we can figure out what it is. The area of the actual port is PI X Rē = 3.14 x 1.5ē = 7.1 inē. If we assume a 2" external port height, then the area of the first concentric circle around the opening of the port is (PI x 3") x 2" = 18.85 inē. (Circumference X Height) The next concentric circle spaced at 1" increments would be PI x 4" x 2" = 25.13 inē So, let's examine the flare rate at 1" increments from the internal port to the outside of the external port - Int.port = 7.1 inē 3" ring = 18.85 inē 4" ring = 25.13 inē 5" ring = 31.42 inē 6" ring = 37.70 inē 7" ring = 43.98 inē Now the rate of change or Delta - P-3" = 11.75 3"-4" = 6.28 4"-5" = 6.29 5"-6" = 6.28 6"-7" = 6.28 So, I guess I was wrong, it is a linear flare, it is essence a straight CONE expansion. So, to a 3" opening, we have added a Cone Flared Port starting at 18.85 inē and expanding to 43.95 inē. That is the same as flaring from 4.89" to 7.5" in diameter. Keep in mind in this example we are using a 2" external port height. The underlying question is, is an external flare of these dimensions sufficient to add anything to the port. Or is it so large as to be un-loaded. I chose to make the bottom of the speaker round so we would have consistent numbers and a uniform radius to work with. But I suspect a rectangular base would act very much the same. But once again, while we have information, we don't seem to have knowledge. I'm still open to any discussion on the subject. Steve/bluewizard Last edited by BlueWizard; 24th July 2013 at 06:58 AM.
 24th July 2013, 07:43 AM #13 4Torr   diyAudio Member   Join Date: Sep 2012 Location: Southern California How about locating the port near the edge of the cabinet so as to reduce the "tunnel" effect of being centered on the bottom panel ? Maybe even an angled port ? Last edited by 4Torr; 24th July 2013 at 07:47 AM.
jwmbro
diyAudio Member

Join Date: Nov 2007
Location: München, Bavaria, Germany
Quote:
 Originally Posted by BlueWizard From an even different perspective, what if both tubes were internal? Say we had a 4.58"x3" exiting the box, but internal to the box, the 3" tube had an additional 10.6" diameter tube 7" long attached to it. Is that the same thing Will the Internal+Internal system act the same as an Internal+External system? Is the second 10.6"Wx7"L tube so large that it doesn't matter? Can we simply average to two together and come up with a fixed consistent diameter tube with the same total length that acts the same way? And what is the equivalent model, do we average the diameters in proportion to their lengths and come up with an average proportional diameter?
Hi Steve,

no, you can't. The reason being, two internal ports are two parallel ports, which can be considered the same as one port with the total combined surface area (assuming both are the same length). Same length, but larger area --> tuning gets higher.

Whereas an internal and an "external" port as you call it, are two ports in series. ie the area stays the same (for a simplified model) but the port gets longer, so the tuning gets deeper.

Quote:
 Originally Posted by BlueWizard So, I guess I was wrong, it is a linear flare, it is essence a straight CONE expansion. So, to a 3" opening, we have added a Cone Flared Port starting at 18.85 inē and expanding to 43.95 inē. That is the same as flaring from 4.89" to 7.5" in diameter. Keep in mind in this example we are using a 2" external port height.
Also not quite correct. The area of a cone increases quadratically with distance. So this is not a conical expansion, but rather a parabolic expansion, where the area increases linearly with distance. Why? Consider the diameter of a cone. If you go twice as far out along a cone, the diameter becomes twice as large. Since the area is (d/2)^2*pi, that means the area gets quadruple as large.

If you want the area of a (regular, round or square) horn to get twice as large every time you move twice as far away, you need the diameter to only get 1.41 times as large, which is a parabola shape rather than a cone.

But, addressing your original question of how this affects the total system response, this is actually quite easy to simulate precisely in HornResp.

Here we can see the simulation results for a basic vented box, 10 liters internal volume, with a 3" internal diameter round port which is 15cm long. The port is perfectly straight with sharp unrounded edges on either side, and produces a 81.5Hz tuning.

Using additional flare segments, we can add on the "horn expansion" of the boundary loading. Here's a simulation of a 14" diameter round cabinet, placed exactly 1" from a flat surface. The port exits the box precisely in the middle o the cabinet bottom.

In order to simulate this, we can first add a transition from the round port to the parabolic expanding flat "horn". It really doesn't make a difference what expansion is used here, I simply measured from the middle of the end of the cylindrical port to the edge of the port, halfway between cabinet and wall to reach an effective segment length of 4 cm. Then, from this point to the edge of the cabinet, length is the remaining 5.5" inches of radius to the edge of the cabinet, and the areas are simply the respective circumferences times the 1" distance to the wall. The whole thing put together looks like this:

As you can see, it has lowered the port tuning by about 15Hz. Despite the flare of the port, it still basically behaves like a vented box, no real horn effects to speak of.

Hope that helps clear things up.
__________________
Regards, James

BlueWizard
diyAudio Member

Join Date: Jun 2007
Quote:
 Originally Posted by 4Torr How about locating the port near the edge of the cabinet so as to reduce the "tunnel" effect of being centered on the bottom panel ? Maybe even an angled port ?
All possible, but I suspect the location of the drivers will determine the location of the port. Since this is a hypothetical design, I think we can put the ports just about anywhere.

But in a real cabinet, again, it will depend on the proximity of drivers and other internal components.

Still, that does throw an additional consideration in to things, especially if we are trying to find an equivalent model to the bottom port in order to understand it better.

Thanks for the feedback.

Steve/bluewizard

 24th July 2013, 11:24 AM #16 mayhem13   diyAudio Member   Join Date: Sep 2008 Hey Blue.....I've used the Polk Powerport spreadsheets to do downfiring ports before on subwoofers where the practical approach did not allow for the long ports required for the desired low tuning. The Polk design drastically reduces port length and turbulence. I'm on my iPad so no access to the spreadsheets but google it and I'm sure they're still out there.

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