
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
12th March 2013, 08:22 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, 08:28 PM  #2 
Speakerholic
diyAudio Moderator

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

12th March 2013, 08:38 PM  #3 
diyAudio Member
Join Date: Oct 2012
Location: Virginia

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 08:40 PM. 
12th March 2013, 08:43 PM  #4 
diyAudio Member
Join Date: Apr 2007

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

12th March 2013, 08:48 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, 08:52 PM  #6 
diyAudio Member
Join Date: Apr 2007

There's a fourth dimension...

12th March 2013, 09:04 PM  #7  
diyAudio Member
Join Date: Oct 2012
Location: Virginia

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, 09:06 PM  #8 
diyAudio Member
Join Date: Oct 2012
Location: Virginia


12th March 2013, 09:10 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, 09:33 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 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
box volume for sub  thefox  Subwoofers  5  29th December 2011 03:25 PM 
Parts Express knockdown trapezoid cabinets  cyclecamper  PA Systems  11  13th December 2011 10:39 PM 
Line Array Isosceles Trapezoid Question  dc270  MultiWay  5  3rd February 2006 06:25 AM 
Box volume  eitanwaks  MultiWay  7  19th April 2005 02:45 AM 
New To Site?  Need Help? 