How to get the right volume in a trapezoid box?
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 12th March 2013, 09:22 PM #1 cross reference   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 Cal Weldon   Speakerholic diyAudio Moderator     Join Date: Jan 2004 Location: Near Vancouver Are these the inside dimensions? If not, give them, including the height. __________________ planet10 needs your help: Let's help Ruth and Dave
 12th March 2013, 09:38 PM #3 xrk971   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 picowallspeaker   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 Cal Weldon   Speakerholic diyAudio Moderator     Join Date: Jan 2004 Location: Near Vancouver 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" __________________ planet10 needs your help: Let's help Ruth and Dave
 12th March 2013, 09:52 PM #6 picowallspeaker   diyAudio Member     Join Date: Apr 2007 There's a fourth dimension...
xrk971
diyAudio Member

Join Date: Oct 2012
Location: Metro DC area
Quote:
 Originally Posted by Cal Weldon 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"
Why are you averaging the dimensions of each face? If you take the extreme example of a trapezoid that is straight like a regular rectangular box say 6x9x12, we would get doing it your way (6+9)/2=7.5 and 7.5x7.5x12= 675. Whereas the true vol is 54x12= 648.

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.

xrk971
diyAudio Member

Join Date: Oct 2012
Location: Metro DC area
Quote:
 Originally Posted by picowallspeaker Just discompose the trapezoid in regular forms like parallelepipedes, triangular base prisms, cubes.
gives same answer as mine but way more complicated.

 12th March 2013, 10:10 PM #9 picowallspeaker   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 ....
Cal Weldon
Speakerholic
diyAudio Moderator

Join Date: Jan 2004
Location: Near Vancouver
Quote:
 Originally Posted by xrk971 Why are you averaging the dimensions of each face?
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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