how bad is this?

[Trixter 10+ year member]

(10.25 X 14.5 X .75) / 1728 = 0.0645

i don't know where you got the 10.25 + .75 from??? and divided it by 12???

volume = (H X W X L) / 1728

hope this clears thing up...if not, lmk

 

[thai_creeper 10+ year member]

(10.25 X 14.5 X .75) / 1728 = 0.0645
i don't know where you got the 10.25 + .75 from??? and divided it by 12???

volume = (H X W X L) / 1728

hope this clears thing up...if not, lmk

thanks for all the replies trixter, if u look at the plans http://img23.imageshack.us/my.php?image=295cubes34hz3625sqin2dd.png in the first picture, i have my port .75" to the right from where its supposed to be, so to make the port width right i'll have to cut out a piece that's of length: 10.25(the original length of the longer port) plus .75"(covering the width of the smaller port piece) to fully cover the side to the left of the (upside down L) port, see wat im sayin? if not ill try to explain differently..why is it divided by 1728? for some reason i thought if i multiplied the 3 things it would give me the volume in inches and id divide by 12 to get the volume in cubic feet

 

[Trixter 10+ year member]

the piece that your adding will not be added as displacement inside the box. the port volume and the materials that it is made out of are subtracted from the box volume. the is the port's displacement. the volume of this piece does not matter and does not need to be figured or accounted for.

if you have 1cu' that 12" x 12" x 12". 12x12x12=1728.

when you measure you get the cu" and then divide by 1728 to get the cu'. your theory would work if you divided your numbers by 12 first and then multipled them.

whatever the length is from the front to the back is what it needs to be. to me it looks like 10.25 + .75 (for the rear port wall section) and another .75 (to make it flush with the front of the box.

 

[thai_creeper 10+ year member]

the piece that your adding will not be added as displacement inside the box. the port volume and the materials that it is made out of are subtracted from the box volume. the is the port's displacement. the volume of this piece does not matter and does not need to be figured or accounted for.
if you have 1cu' that 12" x 12" x 12". 12x12x12=1728.

when you measure you get the cu" and then divide by 1728 to get the cu'. your theory would work if you divided your numbers by 12 first and then multipled them.

whatever the length is from the front to the back is what it needs to be. to me it looks like 10.25 + .75 (for the rear port wall section) and another .75 (to make it flush with the front of the box.

with that last paragraph i think you might have it confused, if i add a piece 10.25+.75 it will touch the front baffle, right now the front over hangs the port by .75". i didnt know that the port's volume isn't accounted for. if you do everything exactly by the plan though but you have one port wall "doubled" in a sense, wouldnt that extra wall need to be accounted for if it wasn't in the original plan? i dont know much about ported boxes, just thought to get the best sound you'd have to get your box as exact to the plan as you can.

well for the most part, if ur opinion still is that it's not worth the trouble adding an extra square baffle to even out the volume difference then i'll take your word for it, if it would make my box sound significantly better i wouldn't mind doing it.

 

[Trixter 10+ year member]

ok...i didn't know that the front had an over hang...then yeah 10.25 + .75.

the only part that will be considered extra displacement is the piece that is currently in the box, the messed up piece. it would be the same as if you built the box correctly and simply added another piece inside the box. .06cu' will not make a difference. add the piece to the inner port wall and you will be all good.

just as fyi....

the port's volume is accounted for in terms of displacement, just like the sub. if your sub takes up .5cu' and your port .4cu' and the sub calls for 3cu' box then you need to make a box that is 3.9cu' total.

to figure out port displacement...

-the length of the port minus the front baffle (front already accounted for because you measure the inside of the box. usually the front baffle is an added .75" of port length which will not be counted.)

-the inner heigth

-the port width + the thickness of the inner port wall(s) (usually .75)

multiply these together and divide by 1728 to get the port's displacement.

 

[thai_creeper 10+ year member]

cool, thanks again for all your help trixter //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[Trixter 10+ year member]

no prob, man...need anything give a shout.