Can you measure Sd?

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I tried to search on this but the search engine says my search term contains too few characters.

I bought an MCM aluminum woofer before they changed their website. As a result, most of the T/S params are now missing. Because I want to build a transmission line enclosure, how do I get the Sd? Borrow the value from an 8" woofer whose value is known?
 
Assuming a circular driver, sd can be approximated by an area calculation.

Measure the driver from the outer edge of the surround of one side, to the inner edge of the surround on the other side (diameter). (eg. the piston diameter is approximately half surrond to half surround)

then sd is given by
sd = ((0.5 * diameter) ^2) * pi

(eg. circle area = pi * radius ^ 2)

Dave.
 
Not to be a stick in the mud but... when they refer to the area, wouldn't you want to approximate with a cone area, rather than a circle? Becuase the woofer is closer to cone-shaped than circular?

Or a paraboloid if it's like many car speakers are now?

Or hyperbolic paraboloid if it's a hyperbolic paraboloid speaker?

Or toroid if it's closer to a toroid shape?

Or, will they be close enough together that it doesn't really matter that much?

Or, is this a parameter that can be fudged in one direction and be just fine?

Or, is this a parameter that, if you got it wrong, you can just change the stuffing to fix it?

*devil's advocate*
 
diyAudio Moderator Emeritus
Joined 2001
Basically, I think you will find that an 8" woofer has around 32 sq inches. A 10" has 56 sq inches, a 12" has 86 sq inches, and a 15" has 132 sq inches. These are approximations-it is very rare indeed that you need any exact figures, and besides, measuring methods vary.

PS: 32 sq in equals 200 sq cm.
 
frugal-phile™
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ultrachrome said:
Because I want to build a transmission line enclosure, how do I get the Sd? Borrow the value from an 8" woofer whose value is known?

The entire "classical" rule-of-thumb set for building TLs is based on Sd. It turns out, now that we have accurate models, that Sd doesn't matter*. What matters is the Vas (just like every other kind of enclosure).

* it is convenient to use Sd to define the geometry in Martin King's worksheets but the Sd does not have to be accurate.

dave
 
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