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12th December 2013, 12:47 PM  #1 
diyAudio Member
Join Date: Dec 2012

How to calculate volume of waterdrop shape
Hi
Im thinking about creating a speaker cabinet as a teardrop shape as b&w nautilus with one full range speaker. The speaker need 16.5l in volume and on a Sphere thats around 32cm in diameter (17.1l). I was thinking that the teardrop will be around 50cm long and around 25 in diameter as widest so like a waterdrop it will be round and the front and then smaller and smaller, almost like a sharp end as the nautlius and the backend. Full range speaker is 6.5 inch. But do you know some way to calculate a approximatly around 16.5 liter? Maybe the volume isnt so sensitive when someone use this shape, that should be the best shape for a speaker cabinet? Best regards, Mattias Johannesson 
12th December 2013, 01:12 PM  #2 
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

The volume < sphere+cone
I wonder if that would be close enough?
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regards Andrew T. Sent from my desktop computer using a keyboard 
12th December 2013, 01:37 PM  #3 
diyAudio Member
Join Date: May 2011
Location: Silicon Valley

This is a common problem in integral calculus, where it goes by the name volume of a solid of revolution (link 1)
If the shape's outline is defined by a set of (x,y) points, you can perform the integration numerically (link 2), perhaps using Riemann sums (rectangles) or the trapezoidal method (triangle + rectangle). 
12th December 2013, 01:37 PM  #4 
diyAudio Member
Join Date: Apr 2011
Location: Leesburg, VA

I assume that volume is for best performance of the driver in a sealed box. If that's the case, the volume is shape insensitive for the most part. I'd approximate the teardrop volume as being a sphere and a cone. That will get you close, and you can try different amounts of stuffing in the cabinet to get the results you like.
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Francis 
12th December 2013, 01:45 PM  #5 
diyAudio Member
Join Date: Dec 2012

Thanks.
Its true a hemisphere and a cylinder cone would do it then. About box volume, do you know whats thé alpint of 16.5l is equialent in a waterdrop shape? maybe i could go for a Little smaller volume as 1213l in this shape or what do you Think thé volume should be when 16.5l is recommended? Best Mattias 
12th December 2013, 01:56 PM  #6  
diyAudio Member
Join Date: Apr 2011
Location: Leesburg, VA

Quote:
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Francis 

12th December 2013, 02:06 PM  #7 
diyAudio Member
Join Date: Dec 2012

I dont know if thé volume was for closed or not, if it is 16.5l ported then what do you Think thé volume should be in closed box?
Thé material Will be 1mm thick stainless Steel. 
12th December 2013, 02:12 PM  #8 
diyAudio Member
Join Date: Apr 2011
Location: Leesburg, VA

Hard to tell what volume the driver needs without knowing its T/S parameters. Where did the 16.5 liter number come from?
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Francis 
12th December 2013, 02:22 PM  #9 
diyAudio Member
Join Date: May 2011
Location: Silicon Valley

If you set the cone's aspect ratio (height/radius) equal to PI, the formulas become especially simple, see attachment. A radius "r" of 14.53 centimeters gives a volume of 16.5 liters.

12th December 2013, 02:46 PM  #10 
diyAudio Member
Join Date: Dec 2012

Thanks,
It looks like the size i have come forward to also. I have calculated on hemisphere and cylinder cone and the specs would be radius 15cm on hemisphere=total 7l, cone radius 15cm at base, length 40cm =9.4l. total 16.4 The speaker would be 55cm deep then with a diameter of 30cm, do you think it would look akward or cool? 30cm diameter at the front should be quite a good size for fitting a 6.5 inch range? Shouldnt it be only a acoustic benefit with a deep/long cone? 
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