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Old 11th October 2002, 06:07 PM   #1
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Default A problem for a geometry wizard

I need the help of someone who is better at geometry than I am. Here's the problem...

A speaker cabinet design has internal dimensions that are 13"x13"x45"(wdh). Not liking square cabinets, I want to transform this square cross-section cabinet into a circular cross-section cylinder with a flat front baffle. Like a straw sliced end to end. However, I need to maintain the original internal cross-sectional area (13"x13"=169"sq) and the flat front baffle width (13"). Ignore the cabinet height.

So the problem is, given a chord of length 13" (the front baffle) that bisects a circle, calculate the radius where the larger piece of the circle bisected by the chord (the internal speaker cross section) has an area of 169"sq. Logically we can tell that the angle of the chord must be somewhere between 90 and 180.

Any takers? The speaker is the Hammer Dynamics Super 12. Thanks!!

Xylenz
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Old 11th October 2002, 06:17 PM   #2
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Default One more thing...

The internal baffle width and the external baffle width will be identical (13") so dont worry about compensating for the thickness of the cylinder. All i need is the internal radius. Thanks.
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Old 11th October 2002, 08:11 PM   #3
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Your inside radius would be 7.974"
assuming the inside of your baffle is 13"
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Old 11th October 2002, 08:11 PM   #4
Schaef is offline Schaef  United States
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Default Here's my guess

If I'm reading your question correctly, you want the area of a parabola with a base width of 13" and a total area of 169" sq.

Assuming that, then a quick search on: (And bookmark this folks, its a great site for problems like this!) http://mathworld.wolfram.com/ reveals that the area of a parabola of a base width of 2a and height h is given as:

area = 4/3 * a * h

or an answer for you would be a height of 16.9". Does that seem correct?
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Old 11th October 2002, 08:36 PM   #5
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I got 7.9708" radius...
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Old 11th October 2002, 10:06 PM   #6
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Talking Thanks much!

Im not sure about the parabola thing. Its just a circle. But thanks anyway.
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Old 12th October 2002, 12:31 AM   #7
Andy G is offline Andy G  Australia
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Default my results, with working

internal radius = 8.6cm
angle subtended by baffle =98.2 degrees

the following is worked in radians !
if you or any wants a copy of the spreadsheet, email me, its a pretty simple one !!

Click the image to open in full size.
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Old 12th October 2002, 01:42 AM   #8
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Default math is fun

if radius = 7.97077469 inches then the area = 169.00000178 sq. inches
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Old 12th October 2002, 03:17 AM   #9
Andy G is offline Andy G  Australia
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Default whoops !! error in spreadsheet

I left out a sinØ, doh !!

yep, close as!

r = 7.97
Ø=109.26º
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Old 12th October 2002, 03:55 AM   #10
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Default Re: math is fun

Quote:
Originally posted by Equilibrium
if radius = 7.97077469 inches then the area = 169.00000178 sq. inches
I thought I was going a little overkill with my calculation to four decimal places!!!!
I hope you have an accurate ruler!!! A hundred-millionth of an inch!?!?
break out the electron-microscope...
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