How to: NET Enclosure Volume/ Slot Port - Realm of Excursion



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Old 05-13-2008   #1 (permalink)
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Default How to: NET Enclosure Volume/ Slot Port

I have come to realize lately that a lot of people dont know how how to find the net volume of an enclosure in cubic feet. i thought this might help some people that are having a hard time.

I am using the example that Aroc gave me.

So now we have
14.5H X 32.5W X 12.5D=5890.625 cubic inches
5890.625 divided by 1728 = 3.4 cubic feet


now you take all the dimensions you have and minus 1.5"(if your using 3/4" MDF on each side of the enclosure)

so are new dimensions are:
13"x31"x11"=4433
divided by 1728=2.565 cubic feet.
now, if you were going to add a port to this, i would shoot for 32" of port area(width of the port x the height). a good rule of thumb is to have 12-16" of port area per cubic foot(this box will have a little more but its just an example).

to achieve this, you will have a slot port that is 2.5" wide and 13" tall(13" is the internal height of your box) this gives you 32.5", pretty close eh?

now lets just say the port is 25" long. to find out how much space this port takes up, you take the port width and add the thickness of your wood to it(that is the outside of the port). so now we have 3.25". times that by the height(13") and then by the depth(25"). lets put it into a formula.

3.25x 13 x 25=1056.25
divided by 1728=.611 cubic feet.

subtract this number by the internal volume of your box

2.565-.611=1.953 cubic feet internally.

now you have to subtract sub displacement. lets say we have a pretty beefy 10" woofer(hdc3 10" for example) and it has a displacement of oh maybe .2 cubes?

1.953-.2=1.753 cubic feet after all displacements.

i hope this helps you guys out. this box went from 3.4 to 1.753 cubes pretty quick didnt it? with some braces this example box would work great for a decently sized woofer like a hdc310. dont forget to subtract the displacement from braces if you using them.



oh and i thought i would add how to find the displacement of round ports:

use this formula, Pi x R^2 x D/1728 (3.14 x Radius x Radius x Depth of Port/1728)

i will use a 6" port that is 25" long as my example

3.14 x 3 x 3=28.26"

28.26" x 25"=706.5

706.5/1728= .408 Cubic Feet

just as you did with the slot port, minus this from the internal volume
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Old 05-13-2008   #2 (permalink)
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I actually just was looking this up yesterday. Good writeup.
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Old 05-13-2008   #3 (permalink)
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I actually just was looking this up yesterday. Good writeup.
thanks i appreciate it.
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Old 05-13-2008   #4 (permalink)
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EDIT: round port displacement added

Sticky maybe?
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Old 05-13-2008   #5 (permalink)
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good descirption. this should help alot of people out there.

btw were is kaukauna? is that by greenbay?
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The easy way that I calculate up a NET enclosure is like this.

1) Figure out what net volume you want
2) Figure out the port size and length needed to tune it to what you want
3) Figure out the port displacement
4) Add 1&3 together
5) Design a box that has that much gross airspace

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Old 05-14-2008   #7 (permalink)
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Quote:
Originally Posted by tommyk90 View Post
The easy way that I calculate up a NET enclosure is like this.

1) Figure out what net volume you want
2) Figure out the port size and length needed to tune it to what you want
3) Figure out the port displacement
4) Add 1&3 together
5) Design a box that has that much gross airspace

dont forget sub displacement! but yeah this is the way i do it also.
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Old 05-14-2008   #8 (permalink)
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Quote:
Originally Posted by tymoto21 View Post
good descirption. this should help alot of people out there.

btw were is kaukauna? is that by greenbay?
in between appleton and greenbay
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Old 05-14-2008   #9 (permalink)
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Quote:
Originally Posted by kun3racer View Post
dont forget sub displacement! but yeah this is the way i do it also.
Yeah, forgot about that.

Usually sub displacement doesn't affect the volume/tuning too much unless it's a BIG sub or lots of them.
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Old 05-14-2008   #10 (permalink)
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Originally Posted by tommyk90 View Post
Yeah, forgot about that.

Usually sub displacement doesn't affect the volume/tuning too much unless it's a BIG sub or lots of them.
Agreed...or its a small box. .1cuft into a 1cuft box is noticeable.

When I design boxes I usually make the a bit too big, then I use the space up with bracing.



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Old 05-14-2008   #11 (permalink)
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well hey i posted this because i seriously can find some people who come here who cant do this, and i tried putting it into the simplest terms i could.
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Old 05-14-2008   #12 (permalink)
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Originally Posted by tommyk90 View Post
Yeah, forgot about that.

Usually sub displacement doesn't affect the volume/tuning too much unless it's a BIG sub or lots of them.

well like my setup im running 4 d6 12's with a common chamber so since each sub is .1 ft3, they add up to .4 which does affect the tuning.
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Old 05-15-2008   #13 (permalink)
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What about the woofer xman?? Does that affect the enclosure too? Also i didn't read the whole thing carefully but how to figure out the tuning frequency of ported box??
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Old 05-15-2008   #14 (permalink)
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What about the woofer xman?? Does that affect the enclosure too? Also i didn't read the whole thing carefully but how to figure out the tuning frequency of ported box??
thats not in there but i can get you the forumula.
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Old 05-15-2008   #15 (permalink)
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Fb = 0.159*SQRT[(Av*(1.84*10^8)]/(Vb*1728*[Lv+(.823*SQRT(Av))])

LV is the length of the port

AV is your port area

VB is your boxes internal volume(every displacement accounted for)

Fb will be your tuning frequency once you work the formula through.
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Old 05-15-2008   #16 (permalink)
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Quote:
Originally Posted by ndstrctbl View Post
Fb = 0.159*SQRT[(Av*(1.84*10^8)]/(Vb*1728*[Lv+(.823*SQRT(Av))])

LV is the length of the port

AV is your port area

VB is your boxes internal volume(every displacement accounted for)

Fb will be your tuning frequency once you work the formula through.

Alright, i write that down and i will see if come up with anything that is right. Also, i want to know if the woofer xmas is important while tuning a box?? Anyone know??
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xmas is important while tuning a box?? Anyone know??
I usually wait till Christmas to tune my boxes
Otherwise it will adversely affect its response.

haha

xmax :P

And xmax has little to do with port velocity. Motor strength has more to do with it.

Think about a pro audio sub...they have less than 10mm xmax...but they need large ports to keep the port velocity down.

Look at it this way: The driver moves the LEAST when the port is at tuning and is moving the MOST!



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Old 05-24-2008   #18 (permalink)
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gust bought 2 AQhdc315 need to know what size port i need i want to tune it to 40 hz
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Old 05-24-2008   #19 (permalink)
 
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One thing I noticed is that you figured the port displacement by taking the entire length of the port. Don't forget, the 1st 3/4" is already accounted for by outside of the box, so you can subtract that from the length. Also, if the port turns, you have to take that into account, as it will not take up the entire length as you figured it out, you would end up tuned lower than expected.

I always do it the way TommyK suggested also.
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Quote:
Originally Posted by matt caudill View Post
gust bought 2 AQhdc315 need to know what size port i need i want to tune it to 40 hz

Port depends on the enclosure size, not the woofer.
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Old 05-31-2008   #21 (permalink)
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Quote:
Originally Posted by Fallen View Post
I usually wait till Christmas to tune my boxes
Otherwise it will adversely affect its response.

haha

xmax :P

And xmax has little to do with port velocity. Motor strength has more to do with it.

Think about a pro audio sub...they have less than 10mm xmax...but they need large ports to keep the port velocity down.

Look at it this way: The driver moves the LEAST when the port is at tuning and is moving the MOST!

uhmm, You saying that it has nothing to do with the volume enclosures? Let say an Re audio xmax is 22 so how big do you thing the port should be? Sorry I'm kinda new to with enclosures, but I'm here to learn.
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Old 05-31-2008   #22 (permalink)
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about the round port so is that example port 6 inches tall by 6 inches wide and you figure it 3.14x3"x3". or is it really just a 3" pipe
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Old 05-31-2008   #23 (permalink)
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Quote:
Originally Posted by ranmansnaple View Post
about the round port so is that example port 6 inches tall by 6 inches wide and you figure it 3.14x3"x3". or is it really just a 3" pipe
the formula to find the aera of a round port is (Pi times radius times radius). the diameter is 6", and half of that is the radius which is 3". so it looks like this:

3.14x3x3.

then to figure out the displacement you times it by the depth and divide by 1728.

so, no it is not a 3" pipe, it is infact a 6" pipe.
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Old 06-06-2008   #24 (permalink)
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Explaination with diagram will be very appreciated, i'm still confused about where to start to measure the start of the port
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Old 07-20-2008   #25 (permalink)
 
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Nice write up! Thank you.
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