Calculate Volume for Triangular Prism..I think

Bose(o) · 11th June 2002, 08:36 PM

OK....I'm trying to calculate the volume of my speaker boxes that I'm building for the heck of it. The box when standing it's tall height of 100", and you looking at it from the side is a right-angle triangle. Dimensions:H=100" W=12" L=20" My guess to the formula was a simple...l*w*h/2 And I get 196.68 L....Is this correct? With the regular l*w*h equation I get, 393.36L. Here's my thinking...Take the volume of the rectangular prism that is created by two speakers of right-angles and divide that volume by two. It seems right to me.

UrSv · 11th June 2002, 09:59 PM

Your calculations are correct if the shorter sides are 12" and 20" in the right angle triangle. Just remember that those are external dimensions from what you say and given 1" material you would come closer to 144 l which is quite different from 197 l.

/UrSv

Bose(o) · 11th June 2002, 10:33 PM

Oh yeah, I always forget about that part! With stuffing here and there, I'm guesstimating about 139L.

Keld · 11th June 2002, 11:15 PM

When you caculate the volume substract bracing, port(if any) and driver space. NOT the stuffing. Stuffing is actually making the box appear bigger to the driver, dont ask me how

Keld

Bose(o) · 12th June 2002, 12:55 AM

1. This sub will be used for some Party action in my friend's backyard. He wants some *** kickin' bass... Any ideas on how to go about this? More drivers help, right? Watts does too, but we'll base it on a100W RMS amp right now because if you can't make a low-fi driver sound like mid-fi than you're not a good engineer.

11th June 2002, 08:36 PM	#1
Bose(o) diyAudio Member Join Date: Dec 2001 Location: Ottawa, Ontario, Canada	Calculate Volume for Triangular Prism..I think OK....I'm trying to calculate the volume of my speaker boxes that I'm building for the heck of it. The box when standing it's tall height of 100", and you looking at it from the side is a right-angle triangle. Dimensions:H=100" W=12" L=20" My guess to the formula was a simple...lwh/2 And I get 196.68 L....Is this correct? With the regular lwh equation I get, 393.36L. Here's my thinking...Take the volume of the rectangular prism that is created by two speakers of right-angles and divide that volume by two. It seems right to me.

11th June 2002, 11:15 PM	#4
Keld diyAudio Member Join Date: Jan 2002 Location: Borås, Sweden, Tellus	When you caculate the volume substract bracing, port(if any) and driver space. NOT the stuffing. Stuffing is actually making the box appear bigger to the driver, dont ask me how Keld __________________ "What is done by what is called myself is, I feel, done by something greater than myself in me." James Clerk Maxwell. 1879

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11th June 2002, 09:59 PM	#2
UrSv diyAudio Member Join Date: Jan 2002 Location: Sweden	Your calculations are correct if the shorter sides are 12" and 20" in the right angle triangle. Just remember that those are external dimensions from what you say and given 1" material you would come closer to 144 l which is quite different from 197 l. /UrSv

11th June 2002, 10:33 PM	#3
Bose(o) diyAudio Member Join Date: Dec 2001 Location: Ottawa, Ontario, Canada	Oh yeah, I always forget about that part! With stuffing here and there, I'm guesstimating about 139L.

12th June 2002, 12:55 AM	#5
Bose(o) diyAudio Member Join Date: Dec 2001 Location: Ottawa, Ontario, Canada	1. This sub will be used for some Party action in my friend's backyard. He wants some *** kickin' bass... Any ideas on how to go about this? More drivers help, right? Watts does too, but we'll base it on a100W RMS amp right now because if you can't make a low-fi driver sound like mid-fi than you're not a good engineer.

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