Look My Design Over! For my friends Mustang. 2 12" JL's

[Try2makeitwork 10+ year member]

Each bridged channel will be operating at 4ohms when bridged to an 8ohm load. So just take what the two channels do at 4ohms, add it together, and that'll be your output at 8ohms bridged.

Ok cause I have always been cautions with mine and my friends stuff and ive never bridged a 2ch amp to mono and had it looking at a 4 ohm 'wired' (1sub,svc4ohm) load cause most class-d amps are 2 ohm stable;stereo and 4ohm stable;mono.

but when you have 2 subs wired mono to the amp it dosent divide. Cause ive always done that, for years and years

 

[RAM_Designs 5,000+ posts]

so when did 12*2 not = 12*2...

It does. What you're proposing is that (2)12"x2" is = to (1)12"x2". Does that actually make sense to you?

when you split a port you still keep the area constant...

Yep, you just don't understand what that requires in this case.

what you are proposing is like using a reducer on a round port and going from 4" to 2" or 4" to 6" and it changes the tuning!! how can you not possibly understand that?

Incorrect. What I'm(and the OP is) doing is keep the port area the same. What you're saying is like having a 6" pipe branch off into two 6" pipes, which can account for twice the amount of water flow.

the port area and length determine the tuning!!! whether you split it or not!

Indeed, which is why when you split a port, you split the width as well, in order to keep the same port area, in order to keep the same(or close) port length.

omg it is like talking to a wall...

What's so hard for you to understand that splitting a single 12"x2" into two 12"x2" ports does not give you the same port area as the single 12"x2" port that he started with?

This reminds me of the time you were making boxes, but using the gross volume to find the port length instead of the net volume the sub will see after port displacement...you were wrong then, you're wrong now.

 

[SicAudio 5,000+ posts]

ok then fine since you seem to remember that thread ... riddle me this...

¾: MDF

32” wide x 14” tall x 17.5” deep

Port = 12.5” tall x 3.5” wide

Length will be an L on either side : 12.5” deep and then 3.75” behind each sub

what is the volume and tuning of this enclosure..

because it was designed and used by a world champion who is also an expert in the field as well..... and that is who gave it to me...

 

[SicAudio 5,000+ posts]

What's so hard for you to understand that splitting a single 12"x2" into two 12"x2" ports does not give you the same port area as the single 12"x2" port that he started with?

because 12*2 = 24 .... whether it is split or not 12*2 is still 12*2...

I have gotten the info after that thread came about from several experts... in fact it is probably on that thread...

 

[RAM_Designs 5,000+ posts]

because 12*2 = 24 .... whether it is split or not 12*2 is still 12*2...

You are not taking into account that you're wanting to make it into two ports, so why are you not saying 2x2"x12"? With two 12"x2" ports you have 48in^2 of port area, which is twice as much as the 24in^2 you started with. Jesus Christ this is getting ridiculous.

 

[RAM_Designs 5,000+ posts]

ok then fine since you seem to remember that thread ... riddle me this...

¾: MDF

32” wide x 14” tall x 17.5” deep

Port = 12.5” tall x 3.5” wide

Length will be an L on either side : 12.5” deep and then 3.75” behind each sub

what is the volume and tuning of this enclosure..

because it was designed and used by a world champion who is also an expert in the field as well..... and that is who gave it to me...

You tell me what it's supposed to be, that's the issue here. You thought you were building boxes to certain specs and they were off due to you not taking into account port displacement when figuring out port length.

 

[SicAudio 5,000+ posts]

I know what it is //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

I have the box in a personal vehicle lol

if I am wrong as you suggest then I need evidence of that.. otherwise you telling me I am wrong means nothing at all right? no learning involved unless evidence ( math ) can substantiate it..

you see... after that debacle I made it a point to be sure from then on regarding it so I spent a lot of time getting it right..I pm'ed people that I know are proficient at it, I emailed the experts , I made phone calls, ect ect ect...

1 of them gave me that box design as a test to see if I finally understood the math dead on... I admit it took 3 tries but I did finally get it...

so know if I am wrong still he is wrong and I would like to know how... so if you would... humor me..

I will even tell you who did the design ...

 

[SicAudio 5,000+ posts]

You are not taking into account that you're wanting to make it into two ports, so why are you not saying 2x2"x12"? With two 12"x2" ports you have 48in^2 of port area, which is twice as much as the 24in^2 you started with. Jesus Christ this is getting ridiculous.

I see what you are saying yes... but from what I learned that would be exactly the same as starting the slot off at 2" wide and tapering it to 1" ( ignoring the other side of the L and doing it as a single sub enclosure..)

or when you wrap the port back and forth in the box and forget to account for the bends... it changes the port volume...

area * length = volume = tuning... so explain it with math to me if you would.. I can understand that easier than words apparently

 

[RAM_Designs 5,000+ posts]

so know if I am wrong still he is wrong and I would like to know how... so if you would... humor me..
I will even tell you who did the design ...

You need to be more specific about the box. It'd be better if you posted a picture. There's too many questions about the port.

I don't care who designed it; if it's done wrong, then it's done wrong.

 

[SicAudio 5,000+ posts]

3.51 b4 port

3.23 after port but w/o sub

 

[Bob Saget]

You need to be more specific about the box. It'd be better if you posted a picture. There's too many questions about the port.
I don't care who designed it; if it's done wrong, then it's done wrong.

Didn't you get the memo? SicAudio is never wrong when it comes to car audio. That and Maxxsonics is the greatest thing since sliced bread.

 

[RAM_Designs 5,000+ posts]

I see what you are saying yes... but from what I learned that would be exactly the same as starting the slot off at 2" wide and tapering it to 1" ( ignoring the other side of the L and doing it as a single sub enclosure..)
or when you wrap the port back and forth in the box and forget to account for the bends... it changes the port volume...

area * length = volume = tuning... so explain it with math to me if you would.. I can understand that easier than words apparently

It's just pretty simple, in order to keep the cross-sectional area the same, you need to halve the width when splitting the port into two separate ports. I'll try to find something...

In this case, the 13.5"x3" port, which has a port area of 40.5in^2, is divided into two 13.5"x1.5" ports. This gives each individual port 20.25in^2 of port area, or 40.5" total, which is the same as the original single port.

 

[Bob Saget]

It's just pretty simple. I'll try to find something...

^^^ specs?

 

[SicAudio 5,000+ posts]

looking for a pic now

 

[RAM_Designs 5,000+ posts]

3.51 b4 port3.23 after port but w/o sub

You did something wrong. A 12.5"x3.5" port opening is going to take up 12.5"x5" inside the box, times whatever length you need. You're saying the port is only taking up .28ft^3, which would mean it's only 7.75" long inside the box.

12.5"x5"x7.75" = 484.375in^3

484.375 / 1728 = .2803ft^3