How to get the right volume in a trapezoid box? - Page 2 - diyAudio
 How to get the right volume in a trapezoid box?
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Speakerholic
diyAudio Moderator

Join Date: Jan 2004
Location: Near Vancouver
Quote:
 Originally Posted by xrk971 Why are you averaging the dimensions of each face?
Because the area of a trapezoid is
h(a+b)/2
where:
h=height
a=shorter of the parallel dimensions
b=longer of the parallels
__________________
Let's help Ruth and Dave

 12th March 2013, 09:56 PM #12 diyAudio Member   Join Date: Apr 2006 Location: MA Math
 12th March 2013, 09:58 PM #13 Speakerholic diyAudio Moderator     Join Date: Jan 2004 Location: Near Vancouver Sorry, I'm supposed to have forgotten all this by my age. __________________ planet10 needs your help: Let's help Ruth and Dave
 12th March 2013, 10:08 PM #14 diyAudio Member     Join Date: Mar 2010 Location: Sweden Shouldn't integrals work for this? Or am I wrong? __________________ My audio and DIY blog: http://phimusic.blogspot.se/
 12th March 2013, 10:12 PM #15 diyAudio Member   Join Date: Aug 2002 Location: victoria BC for those with simple constant CSA just draw them out in CAD, and use the measure area function, snapping to intersection points, then times height __________________ on hiatus
 12th March 2013, 10:13 PM #16 Speakerholic diyAudio Moderator     Join Date: Jan 2004 Location: Near Vancouver Ghosts of departed quantities. __________________ planet10 needs your help: Let's help Ruth and Dave
 12th March 2013, 10:17 PM #17 frugal-phile(tm) diyAudio Moderator     Join Date: Oct 2001 Location: Victoria, BC, NA, Sol III Blog Entries: 5 Cal & X fighting over the use of the Associative property. dave __________________ community sites t-linespeakers.org, frugal-horn.com, frugal-phile.com ........ commercial site planet10-HiFi p10-hifi forum here at diyA
Speakerholic
diyAudio Moderator

Join Date: Jan 2004
Location: Near Vancouver
Quote:
 Originally Posted by planet10 Cal & X fighting over the use of the Associative property.
Associative property, really?
Quote:
 Originally Posted by xrk971 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.
Is that what you get from the dimensions in post 1?
__________________
Let's help Ruth and Dave

 12th March 2013, 10:25 PM #19 diyAudio Member     Join Date: Aug 2012 Location: Los Angeles, California Has everyone actually forgotten there basic geometry? Cal, glad someones brain still works here and you go that right. Why would you have to use a cad program to figure this out? That is like asking what the area of a circle is! Now if we start talking about the sq. rt of -1 I imagine that many would not know the term i, but a simple trapezoid you could look that up on Google in two seconds. How do you do ac circuit work if you don't remember some basic math?
diyAudio Member

Join Date: Oct 2012
Location: Metro DC area
Quote:
 Originally Posted by Cal Weldon Associative property, really? Is that what you get from the dimensions in post 1?
I think we are on same thinking. Without a picture I interpreted trapezoid in the vertical dim with top wall and bottom wall being rectangular csa. I think you see this as a trapezoid csa on top and bottom wall of varying area - trapezoid in all dimensions. Not clear what op meant based on words. That's why we have diagrams. We are both right. Depends on interpretation. Heck op did not even give height. My formula of average areas on top and bottom multiplied by height applies to any cross section, round, trap, triangle, etc. As long as same type of cross section top and bottom.

Last edited by xrk971; 12th March 2013 at 10:44 PM.

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