small enclosure speaker

[bser]

wondering what exactly to look for in a speaker that will perform well in as small of an enclosure as possible. I know Vas plays a role as well as Qts. But looking at thischart I can't find any corellations between Vas, Qts, and manufacture recomended box size. I'm looking specificly for drivers to put in fiberglass kickpanels in my car. What should I look for when choosing a small box driver?

[Stocker]

You look for something that models well in the software program you use to design the box. Manufacturers are free to specify their box sizes as they wish, for whatever frequency response they think is ok that week. Don't trust their box size numbers unless you *really* want to got the easiest route possible.

[dooper]

Closed-box theory is fairly simple.

Qtc is defined as the system Q at fc, which is the system resonance frequency. Qtc will usually range from 0.5 to 1.2.

Vb is defined as the net internal volume of the enclosure.

a (alpha) is equal to Vas/Vb

a = (Qtc/Qts) ^ 2 - 1

Using the Dayton RS125S-8 as an example, and using Qtc = 1.0 as a target:

a = (1/0.52) ^ 2 -1 = 2.698

Therefore Vb = Vas / a = 0.12 / 2.698 = 0.0445 cu.ft.

You can see that the 0.15 cu.ft. enclosure on the chart has a Qtc equal to somewhat less than 1.0. Make sense?

dooper

[bser]

makes great sense! thanks so much for the helpful response in layman terms.

[bser]

Coil Height (Hvc) : 11.3 mm
Compliance (Cms) : 1.22 mm/N
Cone Area (Sd) : 132 cm
D.C.Coil Resistance (Re) : 3.2 ohm
Electrical Q (Qes) : 0.50
Equivalent Suspension Stiffness (Vas) : 30 Liters
Force Factor (Bl) : 4.3 T-m
Free Air Resonance (Fs) : 43 Hz
Frequency Response : 45 - 15kHz
Gap Height (Hag) : 5 mm
Impedance (Nominal) : 4 ohm
Inductance (Le) : 0.80 mH
Linear Excursion (X Max) : 3.15 mm
Mechanical Excursion (Peak-to-Peak) : 16 mm
Mechanical Q (Qms) : 4.90
Moving Mass (Mms) : 11.2 g
Sensitivity : 90dB
Total Loudspeaker Q (Qts) : 0.45


From my calculations (i'm terrible at math) a speaker such as this one would not be suitable for a small enclosure in mind.

using your formula with a goal Qtc of 1 again (not actually sure what I want this to be, inputs appreciated)

1.06/Vb=(1.0/.45)^2-1

After doing the math I get an enclosure size of 3.23 ft^3!!!! That can't be right can it? What am I missing?

[dooper]

Hope this helps:

a = 3.94

Vb = 30 / 3.94 = 7.62 liters = 0.27 cu. ft.

Cheers,

dooper

[bser]

maybe you can resubmit the equation you used without any numbers in it so i can get an idea of where everything goes, sorry to be a pain, but math doesn't come easy to me.

[dooper]

No bother

a = (Qtc/Qts) ^ 2 - 1

a = (1 / 0.45) ^ 2 - 1 = 2.2222 ^ 2 - 1 = 4.94 - 1 = 3.94

Vb = Vas / a

Vb = 30 / 3.94 = 7.62 liters

1 liter = 0.03531467 cu. ft.

7.62 liters = 7.62 * 0.03531467 = 0.27 cu. ft.

Cheers,

dooper

[bser]

thanks so much for your patience. Consider yourself lucky if math comes easy to you I'm not a stupid person I've just got a mental block with numbers and math