figuring out that third dimension...but bracing_displacement is gettin on my nerves!

[sumone 10+ year member]

I'm trying to design a ported box, but I'm unsure on whether I'm thinking about this the right way, that is, the way you figure out the bracing_displacement.

(internal) dimensions 1 & 2 = box height & box width = 15 x 32 in

Vb = net volume: 3.0 (1.5 per sub) ft^3

Fb = tuning: 32 hz

Av = port area: 1.75 in wide x 15 in high = 26.25 in^2

plugging all this into

Lv = Av*1.84*10^8/[Vb*1728*(Fb/0.159)^2] - 0.823*sqrt(Av)

I get a port length of 18.786 inches.

The port will run along the right-most wall from top to bottom all the way to the back of the box, and then when it hits the back, just hit a left and basically the port is done with in just one change of direction. Just an L shape basically.

What I'm unsure about is is the area of the bracing for the port just

(port_length - port_width) * mdf_thickness????

Click to read more...

 

[jellyfish420 10+ year member]

how deep is your box?im gonna assume 10". after you figure out 18" deep port then you have to subtract the port displacement ~.4ft^3 which leaves you 2.4ft^3. there for altering your tuning freq.

 

[DBfan187 5,000+ posts]

BBP6 says: total VB = 3.35^3

*external*

height: 16.5"

width: 33.5"

lenght/depth: 13.56" *internal: 12.6"*//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

32Hz = 20.57"

 

[jellyfish420 10+ year member]

i got 29.5x18.25x11.25*internal*

with a 15x1.75 port 19" long @32Hz

3.5ft^3 internal. . . i added a little for sub displacement.

 

[DBfan187 5,000+ posts]

I fugged up a lil'!

What's the displacement?

 

[sumone 10+ year member]

how deep is your box

I figure that out AFTER I figure out bracing_displacement. it would make it easier but the thing is, that third dimension (deepness/depth) is a dependent variable .

4) Draw out a diagram of your enclosure on a sheet of paper and figure out the displacement of the port.5) Add up all displacements (port, driver, brace)

6) Add displacements to desired net volume and multiply by 1728

7) Divide by the two desired interal dimensions to get the third internal dimension

BBP6 says: total VB = 3.35^3
*external*

height: 16.5"

width: 33.5"

lenght/depth: 13.56" *internal: 12.6"*

32Hz = 20.57"

1. what's total Vb? can't be the flat out exterior l x w x h cause I get 4.338 ft ^ 3

2. did I screw up in my math figuring out the port length? I just double checked and I still get 18.786

 

[sumone 10+ year member]

ok, all measurements are *internal*

1. width = 32in

2. height = 15in

3. depth = UNKNOWN...will figure out after finding out bracing displacement

4. (Vb) net volume = 3.0 ft^3 (2 * 1.5 ft ^ 3)

5. (Fb) tuning = 32hz

6. (Av) port area = 1.5 in x 15 in = 26.25 in^2

7. sub displacement = 0.1 ft^3 (2 * 0.05 ft^3)

I get:

Lv = 18.786in

port_displacement = port_length * port_area = 18.786*26.25 = 493.1325in^3

sub_displacement = 172.8 in^3

net_volume = 5184in^3

So, I would do:

gross volume = net_volume + sub_displacement + port_displacement + bracing_displacement

therefore dimension 3, the depth would be (gross_volume / 32) / 15

 

[DBfan187 5,000+ posts]

1. what's total Vb? can't be the flat out exterior l x w x h cause I get 4.338 ft ^ 32. did I screw up in my math figuring out the port length? I just double checked and I still get 18.786

I said:

I fugged up a lil'!
What's the displacement?

So I came up with

and these

 

[sumone 10+ year member]

now you done confused the shit outta me. am I just totally wrong now??????

 

[supa_c 5,000+ posts]

you are such a nice young man

 

[EFFENDI 10+ year member]

you are such a nice young man

in a carol channing voice:

"ooh, i agree, he's such a nice, helpful sweetheart of a boy..."

//content.invisioncic.com/y282845/emoticons/wow.gif.23d729408e9177caa2a0ed6a2ba6588e.gif

 

[supa_c 5,000+ posts]

wanna see his pic.

he isnt a charmer

 

[DBfan187 5,000+ posts]

I want to correct you guys but I also want money, so ...
1) The port walls are part of the port displacement, count it as a solid object.

2) Mike, how did you get aim on the driver mounting ? I only have front, flush, and rear, but I can't invert.

3) Mike, I sure hope you don't design enclosures like that ... Your port displacement is WAAAAAAAAY off ...

I wasn't fininshed with it yet My version is one off the net. Then I d/l the updates.

 

[LoneRanger 10+ year member]

Okay, I'm in an agitated mood, so I'm going to try to cheer myself up by doing something extremely generous ...Lv = 26.25*1.84*10^8/[3*1728*(32/0.159)^2] - 0.823*sqrt(26.25)

Lv = 18.786"

Port Correction = 1.75/2 = 0.875"

Physical Lv = 17.911" (round up to 18")

Since we want an L-shaped port, we'll just use any old number that's smaller than 18" for depth, so plug in 10". The second port wall = Lv - External Depth ... Second port wall = 8"

Port wall 2 displacement = 8*(1.75+0.75)*15 = 300 ci

Port wall 1 displacement = 15*2.5*8.5 = 318.75 ci

Total port displacement = 618.75 ci (.358073 ft^3)

This will be the displacement no matter the depth, as long as it remains an L-shapped port (basically, as long as the external depth is less than the port length within the enclosure)...

3 ft^3 + 0.358073 ft^3 + 0.1 ft^3 = 3.458073 ft^3 (5975.55 ci)

5975.55/(15*32) = 12.45" (round up to 12.5")

So ...

External Height = 16.5"

External Width = 33.5"

External Depth = 14"

Port wall 1 = 15" x 10.75"

Port wall 2 = 15" x 4"

And, just to prove the length ...

0.75 (port passing through front wall) + 10.75 (wall 1) + 0.875 (half way between port wall 1 and back wall) + 0.875 (half way between port wall 1 and side wall) + 0.75 (thickness of port wall 1) + 4 (wall 2) + 0.875 (end correction) = 18.875"

And just to prove the displacement ...

4*2.5*15 = 150 ci

12.5*2.5*15 = 468.75 ci

Total = 618.75 ci (EXACTLY THE SAME) ... //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

End Result

 

Vb = 3.01415 ft^3 (5208.45 ci)

Fb = 31.86319 Hz

mmmmm....sexy math skills...i'd hump you if you weren't a trashcan-headed, beady eyed canadian //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[sumone 10+ year member]

Okay, I'm in an agitated mood, so I'm going to try to cheer myself up by doing something extremely generous ...

tell the person who agitated you thanks! //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

Ok, so I went through each thing line by line making sure I understood it, except for one thing I didn't get:

Port wall 2 displacement = 8*(1.75+0.75)*15 = 300 ci

Port wall 1 displacement = 15*2.5*8.5 = 318.75 ci

Total port displacement = 618.75 ci (.358073 ft^3)

I'm confused where the 8.5 came from. Is it not 7.5 (10 - 0.75 - 1.75)?

Would that then calculate to a :

- third internal dimension of 12.371 in,

- wall 1 of 12.371 - 0.875 - 0.875 = 10.621 in

- and a wall 2 of 5.004 in?

Also how exact does stuff have to be? Cause I notice you round up and I'm thinking stuff has to be on point (like no less than 3 significant digits)

I think I made a mental note but I forgot it as soon as I started writing this: when you think of your walls (say for this situation). would wall 2 go on top of wall 1? or would wall 2 go on the left side of wall 1?

like

-

|

or like

¯|

to make the right angles?

thanks a lot though!!!