Wayne and other MathGuys: a little help with volume and port length please? [message #38636] |
Fri, 11 October 2002 18:21 ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
BillEpstein
Messages: 886 Registered: May 2009
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Illuminati (2nd Degree) |
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The almost 2Pi in my post below consists of a bottom cubisty deal 10X15X3 and an upper trapezoidal thingy shape that tapers back from the front so the top is 6 1/2X15, 22 1/2" up from the top of the base. Overall height of the entire speaker Q.E.D. = 25 1/2. Facil, N'est pas? So is that somewhere close to the normal 2Pi volume? I have plastic ports,1 7/8" I.D. X 4" long max. Enough? Hey, I know enough math that I was long 300 OEX puts on October 19, 1987, so there!
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Composite solid geometry [message #38637 is a reply to message #38636] |
Fri, 11 October 2002 19:03 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
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Wayne Parham
Messages: 18758 Registered: January 2001
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Illuminati (33rd Degree) |
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The way I would calculate volume of your cabinet is in two parts: First calculate a box with dimensions of height x width x depth of your cabinet at its smallest depth. Then calculate the volume of the wedge shape seperately. The dimensions of the wedge would be found as the difference in the cabinet depth at the bottom and depth at the top times the height x width divided by two. So do two things: Box = H x W x D (at the thinnest section) Wedge = (H x W x D) / 2 (calculates the "slanted section") Add 'em together and you have the total volume. This is called "composite solid geometry" and is simply a matter of dealing with complex shapes by modeling them as a composite group of primitive shapes.
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Re: hows 1.3 cu/ft sound [message #38640 is a reply to message #38636] |
Fri, 11 October 2002 19:27 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
bmar
Messages: 346 Registered: May 2009
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Grand Master |
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Till this is what I calculate for your inside volume for this box. I'm guessing the MDF is 3/4"?
![](http://www.woodhorn.com/till2pi.jpg) The cross sectional area is 167 square inches * 13.5" inside width = 2,254 cubic inches 2,252 / 1728= 1.30 cubic foot inside volume your on your own with the port! bill
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Perfect! [message #38644 is a reply to message #38642] |
Fri, 11 October 2002 20:08 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) |
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Wayne Parham
Messages: 18758 Registered: January 2001
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Illuminati (33rd Degree) |
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That's perfect - The cabinet is designed for 1.4 cubic feet internal volume, so the port should be as specified. Did you get a pair of those 0.5mH coils to add in-line with the midwoofers? Silver label Alpha 10's really need the coils and they really sound good with 'em. The change is pretty subtle, but I like the results.
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