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2nd February 2006, 07:23 AM  #1 
diyAudio Member
Join Date: Feb 2006
Location: Burleson TX

Line Array Isosceles Trapezoid Question
Hello All,
I have never been mistaken for a math wizbut maybe I can get some help here. I plan on constructing a 4 driver per side line array of The Smaller Advent woofers. I want to build the cabinet in the shape of a standing Isosceles Trapezoid with the woofers on one side measuring 10" for the driver width. My question is how does one caluculate the length, width, depth of the other 3 sides to equate to 4 cubic feet of internal volume insing 3/4" mdf? Better yet just what is the damn pannel sizes needed....... Thanks for your hopeful help here, now I must go lie down and tend to my head ache...... DC
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youi can't do that...why it's heresey....or so they said to Paul Klipsch! 
2nd February 2006, 08:57 AM  #2 
diyAudio Member
Join Date: Nov 2003
Location: Brighton UK

Hi,
are we talking about a truncated very tall pyramid, or do two faces and two sides remain parrallel. /sreten. 
2nd February 2006, 09:18 AM  #3 
diyAudio Moderator Emeritus
Join Date: Apr 2002
Location: Chatham, England

I hope you're good with compound mitres!
Seriously, a scale drawing would help.
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2nd February 2006, 07:50 PM  #4 
diyAudio Member
Join Date: Feb 2006
Location: Burleson TX

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From my understanding of the mathmatical diagrams
the driver face will be the 10" I need for the driver width (length of baffle unknown but 48" would be fine for my concerns) left and right sides should be an equal width and same length as baffle face above Cabinet back will be wider and same length as all above Here is the link; http://mathworld.wolfram.com/IsoscelesTrapezoid.html The original Smaller Advent had a net internal volume of close to 1 cu./ft. My 4 drivers should have a 4 cubic feet net. I like the "footing" & narrow baffle desiogn this offers. The (original) green fried egg tweeters are going to be piano hindge open mounted next to the front baffle for an aimability feature. Yes I can cut compound anglesonce I know them! Thanks again all DC
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youi can't do that...why it's heresey....or so they said to Paul Klipsch! 
3rd February 2006, 06:17 AM  #5 
diyAudio Member
Join Date: Jun 2005
Location: Austin Texas

I'm not a math wiz either but...
On the link that was posted....if you look at the diagram on the right side you'll see that they put two dotted lines that formed a box inside, flanked by two triangles. If you can imagine rotaing one of those triangles around and fitting it on top of the other, then it forms a box as well. Now you have two boxes...just calculate the area of those two boxes. In other words, if the inner box is 10 across by 12 deep you have 120 inches of area. and if the two triangles form another box that is 12 by 12 then that is another 144 inches of area. So, in this example the total area of your "footprint" (as you called it) would be 264 inches. If you are working in inches, then find the total number of cubic inches of volume you need in your cabinet (1cu ft = 1728 cu inches). Once you have the total number of cubic inches you need in the cabinet (don't forget to add enough to account for the driver's volumes and bracing), then divide that number by the total area from the calculation above, that will give you the internal heigth of the cabinet. There's a starting point. Adjust your dimensions as needed. Good Luck. The Ribbon Project 
3rd February 2006, 06:25 AM  #6 
diyAudio Member
Join Date: Jun 2005
Location: Austin Texas

By the way....regarding your box volume you mentioned, I really suggest you download one of the free woofer box modeling programs, and use the T/S parameters of your drivers to culculate box volume.
1 speaker = 1 cu ft, so 4 speakers = 4 cu ft (is inaccurate). Best Regards, The Ribbon Project 
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