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View Full Version : Trying to understand a particular box design



Goldtaz1
04-07-2005, 12:32 PM
Guys, I need some help with a box design I am looking at. See the attachment below because I made a drawing of what I am going for. When you see the picture, please assume that the port height is the same as the internal height of the enclosure. Here are my questions:

1) Would this design be considered as having two ports? If the enclosure has two ports then I assume I would be correct by doing the port length formula for two ports?

2) Would the woofers see both chambers for airspace since I am not putting in a divider?

3) If the width of the mouth opening is say 3 inches, do I stay off of the back wall 3 inches with both bends or do I stay from the backwall 1.5 inches?

req
04-07-2005, 01:03 PM
put a divider down the middle and make it wider so that the middle section is double the *internal* width of the bottom 'shafts' i suppose. each sub would be in its own seperate box. but you can NOT do a port like that where they both have two tails that share the same middle shaft.

________
|___|||___|

see the divider? then put the back two beside them or whatever. but you cant share the same middle one off to the others

jellyfish420
04-07-2005, 03:04 PM
i would disagree with req.
you can do it the way you said. if you wanted the mouth to be 3" then you would figre the port for 1/2 the box. using 1.5" port. pretend that its 2 seperate boxes w/ 2 1.5" ports 'joined' together. i just did a box like this not to long ago;)

(internal pics)
http://forums.caraudio.com/vb/showthread.php?t=76593

(finished pics)
http://forums.caraudio.com/vb/showthread.php?t=78936

Goldtaz1
04-07-2005, 03:22 PM
Thanks a bunch jellyfish, that definitely sheds light on my project. However, I just want to make sure I understand you correctly, here goes. If I am looking for 2 cubes PER woofer and say 42 square inches of port where the mouth is 3 inches wide and a tuning of let's say 33 Hz then in order to find port length I would perform my calculation using 1) 2 cubes net 2) 21 square inches of port and 3) the 33 Hz tuning frequency??? If this is correct then I knew I was making it more difficult then it really is?

Also, on your second link the pictures would not pull up, so would you mind sending those via PM or to Goldtaz1@charter.net?

Insomniac119
04-07-2005, 04:23 PM
Also, on your second link the pictures would not pull up, so would you mind sending those via PM or to Goldtaz1@charter.net?

they worked for me.

Goldtaz1
04-07-2005, 05:20 PM
they worked for me.


Still can't get em to work.

BoomCrew
04-07-2005, 05:48 PM
worked for me at 6.03 on april 7th.

jellyfish420
04-07-2005, 05:53 PM
Thanks a bunch jellyfish, that definitely sheds light on my project. However, I just want to make sure I understand you correctly, here goes. If I am looking for 2 cubes PER woofer and say 42 square inches of port where the mouth is 3 inches wide and a tuning of let's say 33 Hz then in order to find port length I would perform my calculation using 1) 2 cubes net 2) 21 square inches of port and 3) the 33 Hz tuning frequency??? If this is correct then I knew I was making it more difficult then it really is?

Also, on your second link the pictures would not pull up, so would you mind sending those via PM or to Goldtaz1@charter.net?
yes you are correct sir.
the pics don't work? huh, but the first ones do? they're both on the same server...
i'll see if i can find them.

for instance if you calculation comes to 30" long port and your box is 17" deep, you would take 3" down the middle, then you split it, and have 1.5" running each way 12.25"...make sence?

jellyfish420
04-07-2005, 06:13 PM
Can anyone say, "Cancellation" ???
you were the one that helped me figure mine out when i did it, and there was no mention of cancelation...:crap:

Goldtaz1
04-08-2005, 08:33 AM
Can anyone say, "Cancellation" ???


This box design is still about as clear as mud to me. Jmac please clarify this issue, I don't think that either Jellyfish nor myself wants to build a completely worthless box?? Additionally, we don't want to go around advising other folks incorrectly, so please help us out on this one.

Jellyfish, I am quite positive that we are on the same sheet of music and I just wanted to let you know I appreciate your help.

Goldtaz1
04-08-2005, 11:01 AM
Need some response on this.

jellyfish420
04-08-2005, 03:03 PM
Need some response on this.
me too. i would like to hear about this.

i have already built my box. it sounded great! louder than the 'regular' ported that i just built for the 2-10's

Goldtaz1
04-08-2005, 03:06 PM
me too. i would like to hear about this.

i have already built my box. it sounded great! louder than the 'regular' ported that i just built for the 2-10's


Do you know Jmac? Do you suppose we could go to him directly for some assistance? Anyone else familiar with this design please feel free to chime in.

bcarpenterfhl
04-08-2005, 04:32 PM
yea. i thought there could be some cancellation issues, but was never vertain. so fo the boxes i design i give them two seperate ports just incase. (like the one in my sig).

Goldtaz1
04-08-2005, 04:39 PM
http://img221.exs.cx/img221/2871/portedbox7rw.png
Anyone see a problem here ?

The waves will be completely out-of-phase when the meet up in the center of the port and cancellation will occur ...

You should have a divider in the middle of the port ...

But I have seen many boxes built this way and they absolutely slammed. I am talking about boxes that were built by the professionals at my local car audio shop. One guy in particular has won numerous national awards for his builds. If you require his name then I can certainly provide that.

I guess the main reason I am questioning you on this is because Jellyfish stated that you helped him design his enclosure.

bcarpenterfhl
04-08-2005, 04:41 PM
just because he can build boxes, doesnt mean crap about his design/math. i work at a car audio shop, and the people there build absolutely amazing boxes, but they no ***** when it comes to the proper calculations and physics of a box.

tRiGgEr
04-08-2005, 04:44 PM
You wont have to worry about it for a daily bumper. If your going to be building a comp/show box worry about it.

Goldtaz1
04-08-2005, 04:52 PM
just because he can build boxes, doesnt mean crap about his design/math. i work at a car audio shop, and the people there build absolutely amazing boxes, but they no ***** when it comes to the proper calculations and physics of a box.


I was not saying that because he works at a car audio shop he automatically knows how to do calculations/math. I just find it highly improbable that he would not know that the waves would AUTOMATICALLY die in the enclosure upon the crash effect illustrated by Jmac due to his lengthy experience in the car audio industry. When Jmac posed the question about seeing a problem with his diagram, I did see the problem, however I would just think that the waves would continue to move through the rest of the port even though they are crashing into one another.

bcarpenterfhl
04-08-2005, 04:58 PM
A lot of people won't make the divider visible with this type of design, also ...

how does one conceal the divider?

Goldtaz1
04-08-2005, 05:09 PM
Make the divider shorter and paint the port black ...


How much shorter could you make the divider?? Just to the point that it doesn't allow the waves to collide??

jellyfish420
04-08-2005, 06:12 PM
Basically ...

Once the waves are both going in the same direction, they won't cancel each other out ...
so the divider would only have to be 1.5" long?

DBfan187
04-08-2005, 06:19 PM
Well **** me sideways! I've built several boxes this way and the sound awesome, but I've never did the divider thingy.:censored:

Goldtaz1
04-09-2005, 06:24 PM
so the divider would only have to be 1.5" long?


I will assume this comment is correct?? You could maybe add another inch just to be on the safe side of things.

SPL140.2
04-09-2005, 06:40 PM
Thats why I like to build boxes with the port on the corner of the enclosure, as mentioned above the sound waves collide with each other before heading out of the port, (loss of energy) :crazy: , I dont know I have found it to be louder though.

Goldtaz1
04-11-2005, 11:03 AM
Thats why I like to build boxes with the port on the corner of the enclosure, as mentioned above the sound waves collide with each other before heading out of the port, (loss of energy) :crazy: , I dont know I have found it to be louder though.


What have you found to be louder? The port in the corner or doing it the way I am looking to build it?

chubby
04-11-2005, 08:40 PM
least obtrusive

Goldtaz1
04-12-2005, 08:30 AM
least obtrusive

Cool, but I still fail to understand why it is necessary to put that in there.

req
04-12-2005, 08:36 AM
or you could make a flat fiberglass piece thats like 1\8" thick and use the design jellyfish did, but the BACK port width has to be HALF the center port width.

duh.

Exaran
04-12-2005, 04:08 PM
Wait a minute guys, If the subs are wired to the same phase, why would the waves be outta phase in the port?!

Say both speakers pull back at the same time, this compresses the air in the box momentarily, and sends a high pressure wave down the ports.... both high pressure waves would hit in the center port at the same time, and COMBINE to a wave with a higher amplitude or power. You dont have cancelation unless a high pressure wave and a low pressure wave hit each other...

Raven
04-12-2005, 06:45 PM
Well, that's the argument. JMac says the waves will cancel each other out. Others say the wave will combine. Right now, noone on this board has the technology to figure it out.

You'd basically have to trace the air pressure wave from the sub to the intersection and out the port. Even then, you're going to find that different frequencies will have different amounts of cancellation.

My theory: It'll work, but now you're looking at a second tuning frequency based on the width of that port. I have no idea how long a 30hz wave is opposed to a 60hz wave, so you're on your own there.

That's a box tuned to 30hz grand with a 53hz bend for that extra kick. ;)

Goldtaz1
04-12-2005, 11:27 PM
Well, that's the argument. JMac says the waves will cancel each other out. Others say the wave will combine. Right now, noone on this board has the technology to figure it out.

You'd basically have to trace the air pressure wave from the sub to the intersection and out the port. Even then, you're going to find that different frequencies will have different amounts of cancellation.

My theory: It'll work, but now you're looking at a second tuning frequency based on the width of that port. I have no idea how long a 30hz wave is opposed to a 60hz wave, so you're on your own there.

That's a box tuned to 30hz grand with a 53hz bend for that extra kick. ;)


I'm not sure I follow your last comment. My understanding of a ported enclosure is that you tune to ONE tuning frequency and that bends in a port length are done to achieve the port length at a given tuning frequency. I am going to go to an engineering buddy of mine and see if he can provide some scientific evidence of cancellation or a combination of waves.

pooksta
04-13-2005, 09:00 PM
Hopefully, first of all, the subs are getting the same signal (mono). This would be the first way to avoid cancellation. Adding the wave guide (the little part where the port splits) may help, but probably not noticeably. If the subs are putting out the same signal, then their output should sum where the ports come together. I'd recommend running a test if you have an SPL meter, build the box with a couple of variations and see what works best.

Raven
04-13-2005, 10:21 PM
^ Agreed.

As for the "dual tuning", well, ignore it. I'm thinking pretty 2 dimensional. First things first, this is just air pressure. Up wave is high, down wave is low.

Theory A:
If two high waves meet, they could possibly reflect each other, and immediately go back to occupy the low pressure space immediately behind it. The top wave will strike, cancel itself out, AND affect the backwave. Uber cancellation occurs.

Theory B:
You make the entire box a high pressure zone (back wave) and the air has to go somewhere. So whats it do? Since both sides of the enclosure have an equally high amount of pressure, the air goes out the port, where there is less air pressure and more empty space to occupy. The front wave occurs, and the exact opposite happens, you make a low pressure zone, so air rushes in to fill the space, and divides evenly between the two equally depressurized chambers. Cancellation is neglegible.

Overthought this yet? Here's Chubby's take:
During the back wave, the excess pressure races for it's only exit (again). Rather than getting stuck in a corner and creating turbulence, it bounces off the corner block at a nice angle. Now in the port, the second bend it hit the same way. It' only at the end that both waves meet, and they're now both going in the same direction, so they naturally combine.

Yay, I'm bored.
END Self proclaimed expert mode; Wheeeee.

Goldtaz1
04-14-2005, 08:44 AM
Thanks a bunch for your help Raven. Hopefully I can take what I've learned here and do some testing and just see what works and what doesn't work.