What if I increased cabinet size to 12cft?
Quote:
Originally Posted by
Ian Mackenzie
What if I increased cabinet size to 12cft?
If You pick a larger cabinet size Vb then the following formulas apply.
f3 = fs * sqrt( Vas/Vb)
fb = fs (Vas/Vb)**0.32
You will have a "ripple" (deviation from flat response):
R = 20 * log10(2.6 * Qts * (Vas/Vb)**0.35)
There is a missunderstanding about fb and the box resonant frequency, fb *IS* the box resonant frequency, this is when the system of air mass in the ports together with the air "spring" in the box does resonate.
If You build a passive xover then the Qt of Your speaker will change because of the DC resistance of the series coil in the xover. This has been explained several times here, and it is explained in the Thiele paper. See the technical references thread.
Ruediger
There are more alignments than just one...
Quote:
Originally Posted by
kartsmart
I see going larger than 10 cu ft but what happens when going smaller to the 8 cu ft box ? like what jbl has sold. what would be the trade off for 30 HZ to 340 HZ
... see the Thiele paper. Many well-defined filters are possible, Butterworth of order 4 is just one possible choice, and even that only when the driver fits.
You may as well apply the formula given in my last response. The response will have a positive ripple, the response will be peaked somewhat. The corner frequency will be higher. In too small a box a smaller driver may be better than a larger one.
Ruediger