Cubic Feet Help

[nicker10 10+ year member]

I was using this

http://www.reaudio.com/speaker_box/LPort_Box_Calc.html

1728

as a guide to help figure out cubic feet..but it doesnt make sense to me..I entered in this

Box width: 33.5

Box height: 17

Box top Depth: 14.5

Box bottom Depth: 14.5

Wood Thickness: .75

Square Port Width: 3

Square port Length 2: 14

And it says cubic feet would be 2.9 but I thought you would multiply

width*height*Depth and for that I got 8257.75

Then since 1 cubic foot would be 1728 (12*12*12) I divided 8257.75/1728 and ended up with 4.77..

The only thing I can think of is when it says box volume(ft3) it isnt talking about cubic feet..but it must be right?

Click to read more...

 

[plugitin 5,000+ posts]

go to google...and type in rockford fosgate box wizard...im too lazy to find it

 

[Pioneer~Saturn 5,000+ posts]

The cubic feet it's telling you is the net airspace left in the main chamber...meaning it took away the space taken up by the port...unless im missing something..?

 

[nicker10 10+ year member]

is that called displacement? not too sure on any of my terms so i really have no clue..lol

 

[JBerger 10+ year member]

And when you do your own math, your not taking into consideration that you want internal measurements not external. Plus, the RE one is also calculating port displacement.

 

[nicker10 10+ year member]

okay so when people say such and such box would sound good at 3.5 cubes that would mean the internal measurements and after displacement?

 

[JBerger 10+ year member]

okay so when people say such and such box would sound good at 3.5 cubes that would mean the internal measurements and after displacement?

yup

 

[nicker10 10+ year member]

Alright, thanks for the help

 

[Highs on bass]

here is how this works. You have 3/4 mdf. thats .75 inches thick. If you have 2 walls on each side of the box both sides top, bottom, front back. thats 1.5 inches you have to take from internal airspace. so if its 33.5 inches wide inside its only 31 inches wide. 17 deep thats 15.5 and top and bottom depth thats 14.5 is now 12. so you really should mulitply 31x15.5x12 which is 5580 or 3.229 now if you take and measure the size of your port. now if the port exits on the 17 inch side the port takes up 162.75 inches. if its on the 14.5 side the inches it takes up is 126. so if the port exits on the 17 inch side your internal air space before the sub disp. will be 5580 - 162.75 = 5417.25 (3.13) or the 14.5 side which is 5580 - 126 = 5454 (3.15)

so your airspace before sub displacement will either be 3.13 cubes or 3.15 cubes. get it?

 

[nicker10 10+ year member]

"now if the port exits on the 17 inch side the port takes up 162.75 inches. if its on the 14.5 side the inches it takes up is 126"

That seems to be the only part im confused on..wouldnt it have to be 14.5 since the top and bottom depths are the same?

 

[Highs on bass]

"now if the port exits on the 17 inch side the port takes up 162.75 inches. if its on the 14.5 side the inches it takes up is 126"
That seems to be the only part im confused on..wouldnt it have to be 14.5 since the top and bottom depths are the same?

no the port only takes up the space inside the top and bottom. you build a box and put the port inside correct? so the space between the top and bottom plate are not 14.5 if both sides are 3/4 of wood or .75 inches thick that 1.5 inches of wood that is already there. so at 14.5 inches of height subtracting the 1.5 inches of wood that is the top and bottom plate the port will be 12 iches by 14 inches by .75 Understand?

 

[nicker10 10+ year member]

hmm..im still lost..i understand how you would subtract 1.5 from the 14.5 to make up for the wood..but i dont really get the 12 by 14 by .75..square port length 2 would be the 14? and the wood would be the .75? what is the 12?

 

[Highs on bass]

you have height by length by thickness. the port you said was 14 inches long so it goes into the box 14 inches... and it is 12 inches tall. so there is 12 inches between the top plate and the bottom. making the port 12 inches tall by 14 deep. understand now?

 

[nicker10 10+ year member]

so the port would not take up the whole height inside of the box? If the height was 17 you would subtract 1.5 and get 15.5..so wouldnt the port be 15.5 inches tall? Sorry im quite new to this..

 

[nicker10 10+ year member]

here is how this works. You have 3/4 mdf. thats .75 inches thick. If you have 2 walls on each side of the box both sides top, bottom, front back. thats 1.5 inches you have to take from internal airspace. so if its 33.5 inches wide inside its only 31 inches wide. 17 deep thats 15.5 and top and bottom depth thats 14.5 is now 12. so you really should mulitply 31x15.5x12 which is 5580 or 3.229 now if you take and measure the size of your port. now if the port exits on the 17 inch side the port takes up 162.75 inches. if its on the 14.5 side the inches it takes up is 126. so if the port exits on the 17 inch side your internal air space before the sub disp. will be 5580 - 162.75 = 5417.25 (3.13) or the 14.5 side which is 5580 - 126 = 5454 (3.15) so your airspace before sub displacement will either be 3.13 cubes or 3.15 cubes. get it?

I re read this and think I know why I am confused.. the 17 deep should be height correct? and for the top and bottom depth it should be 14.5-1.5 which is 13 not 12..maybe thats why I was getting so lost..