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BFM Jack Lite

Just keep this in mind - if you find yourself boosting the bass to fill in the frequencies below what the horn is handling, you're largely giving up the advantages of having a horn loaded cab in the first place.
I run a Jack 12, and I don't have to boost the bass. Using my head I've compared it side by side with lots of 112s, I prefer it to all of them, including the fearful 12/6 and Baer ML112.
The Jack will behave like any old direct radiator cab in the bass frequencies,
It doesn't, because the ports load into the horn, so even in the lows its more efficient than a regular ported cab. You see that comparing the SPL charts to regular ported cabs. Its 2.5dB more efficient at 50 hz than the Simplexx 12, and the Simplexx 12 is more efficient than most 112s. The Jack 15 is 4dB more efficient than the Simplexx 15 at 45 hz, and the Simplexx 15 is more efficient than most 115s.
 
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Same 3015's or PR400's
I built the boxes to spec, no mods.

Right, you will gain efficiency in the mids, but it is not a 1:1 increase across the response.

It makes no difference, efficiency is relative to voltage, not power. When you double the cab count with constant voltage efficiency goes up by 6dB.

You double cabs you get 3db.
 
Really?
So two of them would be much louder than a 810? My 100 watts would love that.

The graph posted is 1 Jack 15 vs an 810. Your 100w amp would rule everything with 2 Jack 15's! I know mine does! :thumbsup:

Horn loaded cabs work better in groups, 2 jack 15s would have a much smoother response than a single cab. I can't imagine a gig I couldn't handle with an Ampeg V4B and 2 Jack 15's.
 
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It makes no difference, efficiency is relative to voltage, not power. When you double the cab count with constant voltage efficiency goes up by 6dB.
Much louder.

Please help me out on this as it has been making my brain hurt. On a tube amp, the wattage is the same into 4 or 8 ohms. Using the formula for voltage of watts=(volts^2)/ohms, wouldn't the voltage be higher going into a higher impedance load?

My math:
100W=(V^2)/4
(V^2)=400
20V at 4 ohms

100W=(V^2)/8
(V^2)=800
28.28V at 8 ohms

Am I missing something?