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Ports? One or two?

Posted: Tue Jun 17, 2008 8:54 am
by KHPower
I have a question about ports for a box. Some manufactures recommend two ports some just 1. And if you use Bassbox or other software programs you cab obtain the same tuning with just one port even when some specs call for two.

What is the advantage to having two ports? Wouldnt it be easier to just build one? In some cases , using just one port , its usually smaller and shorter (if I remember correctly) .

One time I was talking to a tech guy and we were talking about using two 4'' diameter ports in a box for a certain sub and I asked if I could use just one 8'' diameter port and I think he said yes , but that was years ago.
Whats everyones thoughts?

Posted: Tue Jun 17, 2008 9:22 am
by Bfowler
i personally don't think their is any advantage to building 2. imo it just adds more potential for error into the equation.

i think when the manufactors are reccomending 2 ports, it is to ensure thier is enough port area. if you build on large port you should be fine.

Posted: Tue Jun 17, 2008 9:26 am
by stipud
As long as the area of the port is the same, one port is just as good as two.

Posted: Tue Jun 17, 2008 9:32 am
by KHPower
So in my case , my box requires (2) 4 x 11.25 I could just build 1 port that is 8"" x 11.25?

The only other advantage I would think to building two ports would be spreading the bass around your vehicle.

Posted: Tue Jun 17, 2008 12:01 pm
by KHPower
When you say area of the port is the same , what exactly do you mean? If it Says 2 ports 4 x 11.25 do you mean One 4'' x 11.25'' would do?

Please explain

I entered in a 8'' by 8'' port by 11.25 inches long and it raise's the tuning up 10 hz :?

Posted: Tue Jun 17, 2008 12:34 pm
by Bfowler
those apear to be circle ports, so you would have to use some different math.

you can't just double the diamiter

for a 4" port:

(pie)R^2 would be (in this case R = 2) 12.56 square inches. x 2 (for 2 ports) ~25 square inches.

so (pie)R^2 of a single port would have to = about the same area. (25 square inches^)

3.14R^2=25

divide by Pie

R^2=7.9617

square root 7.9617
R=2.8213

so a single port of 5.64" would have the same overall area as 2 4" ports.

Posted: Tue Jun 17, 2008 3:00 pm
by KHPower
Bfowler wrote:those apear to be circle ports, so you would have to use some different math.

you can't just doule the diamiter

for a 4" port:

(pie)R^2 would be (in this case R = 2) 12.56 square inches. x 2 (for 2 ports) ~25 square inches.

so (pie)R^2 of a single port would have to = about the same area.

so a single port of 5.64" would have the same overall area as 2 4" ports.


Thanks for the info but I am lost on how you found the 5.64'' , I kept up until whats outlined in BOLD came into the equation. Sorry :cry:

Posted: Tue Jun 17, 2008 3:08 pm
by KHPower
Also when I change from circle ports to square there is not that big of a difference in tuning for example: Circle port tuning 38.83-Square port 38.84

I always thought there was not a difference in circle and square ports but the louder noise that comes from circle ports. I know a slot port is different but I thought that circle and square ports that use the same size in inches were the same , just the square ports have better results noise wise :?:

Thanks.. I really need the help :wink:

Posted: Tue Jun 17, 2008 3:54 pm
by joyride
The purpose of the dual port is to eliminate potential noise caused by air being forced through a single port.

Posted: Tue Jun 17, 2008 4:30 pm
by stipud
joyride wrote:The purpose of the dual port is to eliminate potential noise caused by air being forced through a single port.
That would only be the case if you ran a single smaller diameter port. One port with equal displacement to two will obviously not have the same noise.

Posted: Tue Jun 17, 2008 4:32 pm
by stipud
KHPower wrote:Also when I change from circle ports to square there is not that big of a difference in tuning for example: Circle port tuning 38.83-Square port 38.84

I always thought there was not a difference in circle and square ports but the louder noise that comes from circle ports. I know a slot port is different but I thought that circle and square ports that use the same size in inches were the same , just the square ports have better results noise wise :?:

Thanks.. I really need the help :wink:
Circle, square, slot... it's generally all the same thing as long as they have the same area.

So you basically need 25 square inches of displacement (equal to two 4" circular ports. A 5x5" square would be the same, so would a 2x12.5" one, or a single 5.65" circular one.

Posted: Tue Jun 17, 2008 5:19 pm
by KHPower
I think I am getting it :wink: a little , I need to get some more examples of the whole procedure. Your info is great and I appreciate it.

Is there anywhere online that I can see more examples? I have found a lot of places with regards to building boxes but nothing too in depth on this subject.

Thanks everyone 8)

Posted: Tue Jun 17, 2008 6:23 pm
by stipud

Posted: Wed Jun 18, 2008 8:04 am
by Jopop
You guys are overcomplicating math :lol:

So your ports are 4". There are two. You only want one.
Use this formula.

D x √n

Where D is the port diameter and n is the number of ports.

In your case:
4" x √2 = 5.66".



You can even type this into google: 4 * (sqrt 2)

Posted: Wed Jun 18, 2008 8:05 am
by Bfowler
KHPower wrote:
Bfowler wrote:those apear to be circle ports, so you would have to use some different math.

you can't just doule the diamiter

for a 4" port:

(pie)R^2 would be (in this case R = 2) 12.56 square inches. x 2 (for 2 ports) ~25 square inches.

so (pie)R^2 of a single port would have to = about the same area.

so a single port of 5.64" would have the same overall area as 2 4" ports.


Thanks for the info but I am lost on how you found the 5.64'' , I kept up until whats outlined in BOLD came into the equation. Sorry :cry:
i added the math in there. what i did was find the area of two 4" circles which ended up being ~25 sqaure inches.

so then i used that total to set up a new equation for only 1 port

pieR^2=25
3.14R^2=25. then solved for R

Posted: Wed Jun 18, 2008 9:44 am
by KHPower
3.14
4 ×
2 ×
--------------------------
25.12 =

sqrt(25.12)
--------------------------
5.011985634456667 =

Did I do the math correct? When I have the square area of 25.12 do I then square that? to get 5.011985634456667 then wouldnt that be 5.02?

Posted: Wed Jun 18, 2008 10:06 am
by Bfowler
not quite...i'm not sure what Jopop did there...but it seems to have worked


ok...lets start over so i can get this straight too..first your two 4" ports


Pi = 3.14
R = 2

Pi((R)^2)= X
Pi(4) = 12.56 (times 2 ports) =25.12 ok...so far everyone is on board with this part...

now we change the variable and solve for R

pi = 3.14
X = 25.12

Pi((R)^2) = 25.12 - divide the whole equation by pi

(R^2) = 8 - sqrt 8

R = 2.8284 - times by 2 to give us a port diameter of 5.6568542 inches

Posted: Wed Jun 18, 2008 10:09 am
by Bfowler
ok...i just figured out what he did...much eaiser....its not magic...its math!!!

well done jopop

Posted: Wed Jun 18, 2008 10:35 am
by KHPower
I finally got it!(Again I think) Thanks guys

I'll try a eqution to figur (2) 3 " ports:

for (2) 3'' ports:
R=1.5x3.14 x 2=4.71
4.71x 2 =9.42x2 =18.84
18.84 / 3.14=6
6 SQRT=2.449489742783178
2.45 x 2= 4.9

So (2) 3 '' ports converted to 1 port would be 4.9 "

Am I correct?

Posted: Wed Jun 18, 2008 10:49 am
by Bfowler
KHPower wrote:I finally got it!(Again I think) Thanks guys

I'll try a eqution to figur (2) 3 " ports:

for (2) 3'' ports:
R=1.5x3.14 x 2=4.71
4.71x 2 =9.42x2 =18.84
18.84 / 3.14=6
6 SQRT=2.449489742783178
2.45 x 2= 4.9

So (2) 3 '' ports converted to 1 port would be 4.9 "

Am I correct?

close, it looks like you multiplied by 2 above instead of squaring it.

for (2) 3'' ports:
R=1.5^2x3.14 =7.065
7.065 x 2 =14.13

14.13 / 3.14= 4.5
4.5 SQRT= 2.1213203
2.1213 x 2= 4.2426 diameter

or jopops way

3 x sqrt of 2 (1.414) = 4.242 also

Posted: Thu Jun 19, 2008 7:09 am
by KHPower
Ok , I have entered in the 5.65 into the Bassbox software I am using. I am building a 3.75 cf box after displacement. Heres what I got in tuning.

Original design: (2) 4 inch square ports that are 11.25 " long to achive the tuning of 38.68 hz

I then typed in (1) 5.65 circle port that is 11.25''=33.96 hz :shock:

I changed from a circle port to a square port (1) square port 5.65 '' H and W= 37.51


Now the fomulas that we have been doing is for circle ports and I would like to make (1) big square port , what do I do? as you can see using 5.65 for the square port didnt change the tuning too bad and I would actually like it tuned at the 37.51 hz or lower

thoughts?

Posted: Thu Jun 19, 2008 7:48 am
by Bfowler
thats....odd...i have no explanation for that.

if you are going to build a square port, you can dial it in exactly. you can keep caging it until you get exactly the frequency you want

Posted: Thu Jun 19, 2008 8:18 am
by KHPower
Bfowler wrote:thats....odd...i have no explanation for that.

if you are going to build a square port, you can dial it in exactly. you can keep caging it until you get exactly the frequency you want
Do You mean by also keeping the port at least 5.65 " for 1 port correct? and O would just increase the length?

Does anyone else hae the Bassbox software? I could send someone(s) the file to double check everything

Posted: Thu Jun 19, 2008 8:58 am
by Bfowler
KHPower wrote:
Bfowler wrote:thats....odd...i have no explanation for that.

if you are going to build a square port, you can dial it in exactly. you can keep caging it until you get exactly the frequency you want
Do You mean by also keeping the port at least 5.65 " for 1 port correct? and O would just increase the length?

Does anyone else hae the Bassbox software? I could send someone(s) the file to double check everything
i just meant you can increase or decrease the width as you see fit until bass box gives you exactly the frequency you want.

where as with round ports, you are limited to the size pipe you can find.

after thinking about it...i suspect why they said to go with 2 4" round ports is because 4" is an easy size to find...a 5.65" round port isnt exactly available at home depot

Posted: Thu Jun 19, 2008 9:18 am
by KHPower
Gotcha!

I think I will go with a port that is 5.65" H x 5.65" W by 11.25" in length :wink: I can also increase the length to 13.87" to achieve a 35 hz frequency.

Quick Question: The rule of thumb is to keep the port end inside the box at least the diameter of the port away from any walls and in my case that would be 5.65 inches away from the rear wall. The deptgh of my box is 16.81 inches internally. Does this rule of thumb also pertain to square ports? Will I be alright if I come close to the rear wall? If I use the 11.25 length minus the .75 thicknes of the front baffle I should have around a 1 inch to spare if I stick by this "rule of thumb for ports"