Fb Hz lower than F3?????

[sumaudioguy]

Parts-Express.comeerless 830669 12" Paper Cone SLS Subwoofer | tymphany09 Peerless SLS 830669 12" Subwoofer bass
Parts-Express.comayton ST305-8 12" Series II Woofer | woofer st305-8 12" woofer midbass dayton series II series 2 dayton audio dayton loudspeaker
Parts-Express.comioneer A30FU20-52F 12" Woofer | woofer 12" woofer midbass paper cone pioneer
Parts-Express.comayton DC300-8 12" Classic Woofer | 12" woofer dc300-8 12" woofers 12 inch woofer 12" speaker 12 inch speaker 12 speaker midbass dayton 12 " woofer dayton classic 12" woofer dayton audio dayton loudspeaker daytonref083108
Parts-Express.comayton ST305-8 12" Series II Woofer | woofer st305-8 12" woofer midbass dayton series II series 2 dayton audio dayton loudspeaker

Liked the pioneer especially and the paper cone peerless. Hope any of these help! Good luck.

 

[Ron E]

In any ported system designed for maximum flatness the lowest cutoff frequency is achieved with a woofer Qts of .383 or very slightly less if a passive crossover is used. Q other than this cutoff will always be above box frequency.

When mentioning 0.383 you are probably talking about the butterworth alignment (15?) from AN Thiele's papers from 1961. Well, Thiele mentions enclosure losses in his paper, then promptly neglects them for brevity. An actual B4 alignment has Q=0.383 for zero enclosure losses (not practical) and has Qts~0.4048 for Ql=7. For the true situation where you have port, enclosure and leakage losses, there is no such thing as a B4 alignment, although it can be approximated.

 

[sumaudioguy]

Actually enclosure losses as postulated have a lot more to do with the erroneous measurements by using the impedance curve to measure a transducers mechanical system. Losses and the R Small "fudge factor" all result from the that bad measurement method over many tries. If cutoff frequency versus driver Q is plotted the result shows Q=.383 to be the lowest cutoff and tightest coupling between the active and passive resonators. The problem with higher or lower Q driver is the coupling is "looser" allowing for the coupled resonator to resonate more freely and also store more energy in the coupled resonator which means less predictable response in real rooms. Ported systems are bad enough that way all ready and to have looser coupling simply aggravates the problem. Having correct data for a transducer I was unable to even find box losses as described in so many papers in the laboratory. All that was based on bad transducer data from using the impedance curve to model the mechanical system. On the other hand making a box that was twice as long in one direction and another and then a box 2.5 times as long we found the bigger box only appeared 2.2 times as large. From a physical standpoint a cube is the best for a box but a cube obviously has other problems. Never could validate any box loss theory though. Could validate bad transducer data though many hundreds of times. The Small method of measuring a drivers mechanical system is pretty much useless.

I have never been a fan of resonant peaked tuning of any kind because the peaking also means high output impedance which leads directly to unpredictable response in a real room. Far better to have very low Q sealed system and electronically equalize that to flat. This results in much flatter response in a real room with a single woofer especially.