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Date: August 07, 2002 at 12:36:23
From: Henrik, [12.22.223.3]
Subject: Re: sub volume


My assumption is that you are talking about the volume displaced by the driver and port. The port volume is 3.14159*(r^2)*L, where r is the outer radius of the port and L is the length.
For the driver, if you want to be technical, you would get the volume by submerging the driver in water (protected… needless to say). The amount of water displaced in liters, or gallons represent the volume of the driver. However, for most of us, submerging our drivers in water, protected or not, does not sound to enticing… so here is an alternative approach:
Integrate from r2 to r1 Pie*r^2 *h1
= 1/3*3.14(r2^3 – r1^3)*h1, where r2 is the radius of the basket where it meets the mounting ring and r1 is the radius of the basket at the magnet and h1 is the height of the basket (ie the vertical distance between the mounting ring and the magnet structure).
Add to this the volume of the magnet structure Pie*r1^2*h2 where h2 is the height of the magnet structure.
Hence:
Ball park volume of the driver = 3.14(1/3* (r2^3 – r1^3)*h1 + r1^2*h2)
As to your question: is it necessary? Yes. Depending on the driver and port, the combination can significantly affect the predicted responses.


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