DLHgn Posted September 28, 2017 Report Share Posted September 28, 2017 You could keep them 40" and it would be fine. The difference would be minimal at worse. To find the volume of the curved rectangle that would be the port, you could just take the port area and multiply by the center line length (in this case 40") but why do you need to know the volume of the port? Link to comment Share on other sites More sharing options...
06RTCharger Posted September 29, 2017 Author Report Share Posted September 29, 2017 1 hour ago, DLHgn said: You could keep them 40" and it would be fine. The difference would be minimal at worse. To find the volume of the curved rectangle that would be the port, you could just take the port area and multiply by the center line length (in this case 40") but why do you need to know the volume of the port? Oh ok, i was hoping four 10in long aeros would be equal so ill scratch that idea. I wouldve needed it to calculate the displacement the internal portion of the port causes. Link to comment Share on other sites More sharing options...
06RTCharger Posted September 29, 2017 Author Report Share Posted September 29, 2017 The measurements i got, left me with a max box internal volume of 14,731.5 cubic inches or 8.5 cubic feet. Thats without port and bracing displacment. Link to comment Share on other sites More sharing options...
06RTCharger Posted September 29, 2017 Author Report Share Posted September 29, 2017 I could possibly squeeze out a couple more cubes but the box would no longer be a "box", probably look a lil like an everlasting gob stopper lol. Link to comment Share on other sites More sharing options...
DLHgn Posted September 29, 2017 Report Share Posted September 29, 2017 With that large of a port, you'll be left with like 6 cubes. Then bracing and driver displacement will probably leave you with 5.5 cubes. With the same port area and length, you would be tuned to about 29Hz. This would give you better power handling and lower port velocities but you would lose quite a bit of low end output. It falls below the larger box at about 30Hz. However, if you reduce the port area to 91in2 but retain the 40in length, you would end up to a net volume of around 6ft3. You would be tuned to ~26Hz and your port velocity would come up right below 24 m/s. Cone excursion would be at about 18mm @ 3000w. Unfortunately you would lose output below 30Hz but this might be your best be honestly. Link to comment Share on other sites More sharing options...
DLHgn Posted September 29, 2017 Report Share Posted September 29, 2017 Also to note, my above volumes were assuming the port was completely internal. Link to comment Share on other sites More sharing options...
06RTCharger Posted September 29, 2017 Author Report Share Posted September 29, 2017 Is it one program ur using to get these numbers and see the results? Or is it multiple programs ur using. What would be a good "entry level" program you think i could play with numbers and see the predicted output graphs. Link to comment Share on other sites More sharing options...
DLHgn Posted September 29, 2017 Report Share Posted September 29, 2017 Just now, 06RTCharger said: Is it one program ur using to get these numbers and see the results? Or is it multiple programs ur using. What would be a good "entry level" program you think i could play with numbers and see the predicted output graphs. I would suggest using WinIsd. It is pretty straight forward and has proven to be accurate with sealed and ported enclosures. I'm not too sure how accurate it is with other designs. Link to comment Share on other sites More sharing options...
06RTCharger Posted September 29, 2017 Author Report Share Posted September 29, 2017 Thanks Ill try it out. Link to comment Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.