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12th March 2013, 09:22 PM  #1 
diyAudio Member
Join Date: Jun 2011
Location: Canton Michigan

How to get the right volume in a trapezoid box?
Hello all ,
I have been working on a trapezoid design for 1 year now for my speakers and well it's not a new box desing to anyone it's my own design. Is there a easy way to figure out the volume for the inside of a trapezoid box? Here's a example. If the front and back of the box is 9 inch's on the bottom and 6 Inch's on top. Now for the the side's of the box. The bottom is 7 inche's on the bottom and 4 inch's on the top. So again how can I find the cubic feet inside this box? Well again thank you for all you help it's just I am new to this trapezoid box designing. Thanks Jm 
12th March 2013, 09:28 PM  #2 
Speakerholic
diyAudio Moderator

Are these the inside dimensions? If not, give them, including the height.

12th March 2013, 09:38 PM  #3 
diyAudio Member
Join Date: Oct 2012
Location: Metro DC area

Depends of course on how tall the box is which you did not specify. I think the basic formula is very simple: take the average area of the top and bottom faces and multiply by the height. So it sounds like the base is 6in x 9in = 54in^2 and top is 4in x 7in = 28in^2. Average area is (54in^2+28in^2)/2=41in^2. Let's assume the height is 12in. So 41in^2 x 12in = 492 in^3.
Last edited by xrk971; 12th March 2013 at 09:40 PM. 
12th March 2013, 09:43 PM  #4 
diyAudio Member
Join Date: Apr 2007

Just discompose the trapezoid in regular forms like parallelepipedes, triangular base prisms, cubes.

12th March 2013, 09:48 PM  #5 
Speakerholic
diyAudio Moderator

How the heck can we get so close but be different there X?
(6+9)/2=7.5 (4+7)/2=5.5 7.5x5.5x12=495 in^3 495/1728=0.286 ft^3 Based on ID of 12" 
12th March 2013, 09:52 PM  #6 
diyAudio Member
Join Date: Apr 2007

There's a fourth dimension...

12th March 2013, 10:04 PM  #7  
diyAudio Member
Join Date: Oct 2012
Location: Metro DC area

Quote:
I just think about it from a limits point of view as the area has to be the average between the two extremes of a rectangular box 6x9x12 and 4x7x12. 

12th March 2013, 10:06 PM  #8 
diyAudio Member
Join Date: Oct 2012
Location: Metro DC area


12th March 2013, 10:10 PM  #9 
diyAudio Member
Join Date: Apr 2007

Oh man! I never calculate anything! Just make a pipe well..er..about the longest in minimum space, make the section about speaker's diameter ....
Put some Helmoltz resonators here and there, tar, bitumen, fluffy something .... 
12th March 2013, 10:33 PM  #10 
Speakerholic
diyAudio Moderator

OK, let's do it a little different:
(6+9)/2=7.5X12=90 in^3 = area of one trapezoid (4+7)/2=5.5X12=66 in^3 = area of the other trapezoid 90x66X144=855,360/1,728= 495in^3 or 0.286 ft^3 
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