driver displacement

[joelives]

Hi. First post, great forum.

Ive recently finished an initial readthrough of The Loudspeaker Design Cookbook by Vance Dickason and searched this forum, and still have a simple, but unanswered question. How do I figure the displacement of the drivers in a given enclosure? Is there a calulation or just a standardized figure for a particular size of driver?

Im using 2-6.5" drivers and a tweet in a MTM config, and doing box calc with WinISD.

Sorry if this is a stupid question. Any help would be appreciated, thanks,

Joe

[kelticwizard]

Displacement, (Vd), is Area of Cone times Excursion.

Approximate Speaker Cone Areas (close is good enough for this):

15" = 132 sq in
12" = 84 sq in
10" = 56 sq in
8" = 32 sq in
6.5" = 25 sq in

Excursion, if not listed, is:
(length of voice coil - magnetic gap) divided by 2. Magnetic gap is also called air gap.

Speaker A: A 1 inch long voice coil with a .3" magnetic gap has an excursion of (+ or -) .35".

Speaker B: A .75" voice coil with a .25" magnetic gap has an excursion of (+or-) .25".


The (+ or -) figure is the correct one when figuring excursion, (and subsequently Vd).

So if Speaker A is a 10" speaker, then it's Vd is:
56 sq in times .35" equals 19.6 cu inches for Vd, or driver displacement.

If Speaker B is a 12" speaker, then Vd is:
84 sq. in times .25" equals 21 cu. in Vd.

WinISD has a Metric Converter in it, just plug those numbers in and convert the units to Metric, if that is what you need.

On the cone area, there is not much variation between speakers of those sizes. Not enough to make a difference anyway. Excursion can vary quite a bit from one model of speaker to another, even if they are the same diameter.

Hope this helps.

[joelives]

Thanks kelticwizard.