How do i calculate triangular ports? - diyAudio
 How do i calculate triangular ports?
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 15th June 2012, 09:45 PM #1 diyAudio Member   Join Date: Apr 2007 Location: Kgs. Lyngby How do i calculate triangular ports? Hi I'm building some PA speakers and i wanted to make the refleks ports triangular so i can place the on the front around the speaker unit. But i can't find any calculator or formulas to calculate the size of the ports (that provide the same answer...) Can you guys help me? Thanks a bunch!
 15th June 2012, 10:15 PM #2 diyAudio Member     Join Date: Nov 2010 Location: Glasgow, UK As a first approximation I would probably calculate based on a square/rectangular port of the same area, as a triangle would have turbulence closer to a square than a round port, as they both have sharp corners. A triangle has half the area of a square port of the same width and height, so I would just enter the width of the triangle base as the rectangle width and half the height of the triangle as the rectangle height. Bear in mind that calculation of ports is only a rough approximation anyway - even if you use a smooth round tube or a rectangle it will never be exactly what is calculated because there are other variables at play that are only estimated in the calculation, such as proximity of the internal end of the port to other cabinet walls, the size of the baffle the port is located on and how close it is to the edge of the baffle etc. (Most programs assume the port is flush mounted in an infinite baffle) Always allow room for an extra 30% length beyond that calculated and expect to trim the length down based on measurement... __________________ - Simon Last edited by DBMandrake; 15th June 2012 at 10:19 PM.
 15th June 2012, 10:19 PM #3 diyAudio Member     Join Date: Jun 2007 Location: Norlane; Geelong: Victoria: Australia Just a wild guess here but the true value of a triangular port would probably be closer to the diameter of the circular tube that would fit inside the triangle __________________ QUOTE" The more I know, the more I know, I know (insert maniacal laugh >here<) NOTHING"
 16th June 2012, 12:31 AM #4 diyAudio Member RIP   Join Date: Nov 2003 Location: Brighton UK Hi, Most sims include the necessary end corrections and rectangular ports. So simply halving the shortest dimension should work, i.e triangle area is half base time height, make the base the longest dimension and plug in a port base x half height into a simulator, should be near. rgds, sreten.
 16th June 2012, 12:33 AM #5 diyAudio Member   Join Date: Dec 2009 Location: Md I too would guess closer to the round. You'll have to prototype and measure, then report back. I can see the attraction of putting ports in the corners.
 16th June 2012, 12:44 AM #6 diyAudio Member RIP   Join Date: Nov 2003 Location: Brighton UK Hi, It depends on the end corrections built into simulators. You can guarantee the end corrections are wrong for slot ports formed with a cabinet edge or triangular ports across corners. rgds, sreten.
diyAudio Member RIP

Join Date: Nov 2003
Location: Brighton UK
Quote:
 Originally Posted by Moondog55 Just a wild guess here but the true value of a triangular port would probably be closer to the diameter of the circular tube that would fit inside the triangle.
Hi,

Clearly totally wrong for a say 2" high 5" wide triangle.

rgds, sreten.

diyAudio Member

Join Date: Jun 2007
Location: Norlane; Geelong: Victoria: Australia
Quote:
 Originally Posted by sreten Hi, Clearly totally wrong for a say 2" high 5" wide triangle. rgds, sreten.
OK point taken, although most of the designs I have seen are of equilateral triangles, as these fit the corners of most boxes more easily
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QUOTE" The more I know, the more I know, I know (insert maniacal laugh >here<) NOTHING"

diyAudio Member RIP

Join Date: Nov 2003
Location: Brighton UK
Quote:
 Originally Posted by Moondog55 OK point taken, although most of the designs I have seen are of equilateral triangles, as these fit the corners of most boxes more easily
Hi,

They are not equilateral (60 degrees each vertex), they are half squares.
Equilateral does have the least difference between its area and a circle
that fits inside, but the same for half a square is clearly much worse.

rgds, sreten.

diyAudio Member

Join Date: Jun 2007
Location: Norlane; Geelong: Victoria: Australia
Quote:
 Originally Posted by sreten Hi, They are not equilateral (60 degrees each vertex), they are half squares. Equilateral does have the least difference between its area and a circle that fits inside, but the same for half a square is clearly much worse. rgds, sreten.
This is why I failed geometry LOL
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QUOTE" The more I know, the more I know, I know (insert maniacal laugh >here<) NOTHING"

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