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
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
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.
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.
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?
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?
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 !!
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 !!
An externally hosted image should be here but it was not working when we last tested it.
Re: math is fun
I hope you have an accurate ruler!!! A hundred-millionth of an inch!?!?
break out the electron-microscope...
I thought I was going a little overkill with my calculation to four decimal places!!!!Equilibrium said:
I hope you have an accurate ruler!!! A hundred-millionth of an inch!?!?
break out the electron-microscope...- Status
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