So if i do Scary's 2nd diagram, aka: parallel wiring, there will be a 4 Ohm final load I picked up a QSC RMX 1450 pro audio amp a couple days ago. 6 months old for $250 cash. it'll do 900watts in the bridged mode to an 8 Ohm Driver. i have that Adire Audio Tempest in that ported box as well. BTW, is mr marv still here? He did all the mdf - cabinet work. I had thought i'd use the QSC for the Maelstrom and the 250watt parts express plate amp for the Tempest. For years now i had just been using the Tempest and the P.E. plate amp. So i was hunting for an amp for the Maelstrom so i could play both at the same time. i decided on the QSC 1450 to do 900w to the Maelstrom setting it to the bridge mode. Now i see the Maelstrom has two eight ohm VCs. So four Ohm is the Obvious wiring configuration i guess i really shouldn't complain about that either btw So the QSC does 450w to each of the two channels to a 4 ohm load. So i should use the QSC to push both adire subs. And that prolly makes more sense than bridging a 1,400watt pro amp for one sub and using a 250w $99 plate amp for the other. And that 15 in the ported mr marv box is great too
The Tempest i can't remember anything about either. Well it must be 4 Ohms final load wiring if i'm using the P.E. plate amp. I established this equipment in 2003.
FYI for the wiring to try out the Maelstrom with the PE plate amp i did the positive wire from the amp to one of the positive terminals on one side Then the negative wire from the amp to the negative terminal On The Other side. Then one single piece of wire connecting the remaining two unused terminals, ie the remaining negative terminal on one side connected to the remaining positive terminal On The Other side. I supposed the PE plate amp was seeing a 16 Ohm final load
What if by chance the Tempest is an 8 Ohm final load. Can the QSC take a 4 Ohm load on one channel and an 8 Ohm load on the other