Exploder

08-18-2007, 05:36 PM

were can i find info on 2 dual slot ports?

any help would be great

any help would be great

View Full Version : dual ports info

Exploder

08-18-2007, 05:36 PM

were can i find info on 2 dual slot ports?

any help would be great

any help would be great

skadude016

08-18-2007, 05:38 PM

what do you wanna know about em?

dyohn

08-18-2007, 05:38 PM

What kind of info do you need? Dual slot ports function no differently than dual round ports. You still calculate for the enclosure volume and target tuning frequency in the same way.

Exploder

08-18-2007, 06:25 PM

I guess i should be looking for port square area?

what do you guys do on avg.?

37.5 square (would that be all right for 2 12's in 3.6" box)?

what do you guys do on avg.?

37.5 square (would that be all right for 2 12's in 3.6" box)?

skadude016

08-18-2007, 06:49 PM

3.6 inch box?

guessing you mean 3.6 cubes? If so that is a little on the small side of port area. I would atleast bump it up to 45-50 sq inches of port :)

guessing you mean 3.6 cubes? If so that is a little on the small side of port area. I would atleast bump it up to 45-50 sq inches of port :)

spoonraker

08-19-2007, 04:46 AM

*edited for retardedness*

jellyfish420

08-19-2007, 02:10 PM

Just do the calculations the same way you would for a box with a single port, only the specs of the single port come from two ports added together. Port area, length, and displacement must be added together from each port to get the total.

completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.

completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.

Ninteens

08-19-2007, 06:02 PM

completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.

Thank you!:graduate:

Thank you!:graduate:

wesl56

08-19-2007, 06:28 PM

completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.

only if the box is going to be divided... if its all one open area, the ports need to be calculated together, right?

eaisest way would be to use a port calculator that allows to figure dual ports

only if the box is going to be divided... if its all one open area, the ports need to be calculated together, right?

eaisest way would be to use a port calculator that allows to figure dual ports

jellyfish420

08-19-2007, 10:27 PM

Thank you!:graduate:

one thing that i just cannot stand, is people spreading mis-information. i'm pretty smart...but if i'm not for positive, i wont guess. i'll let you know.

one thing that i just cannot stand, is people spreading mis-information. i'm pretty smart...but if i'm not for positive, i wont guess. i'll let you know.

spoonraker

08-20-2007, 10:07 AM

He never said the box had two chambers did he? Or does that not matter? If it does then yeah...I guess I was misinformed along with a lot of people on here because that's where I read it at.

The way I was told is take the NET volume of the entire box. Take the total port area (both ports) and then figure your port length. Divide it in half and each of your ports should be that long...this is for a single chamber box btw, a dual chamber box is obvious....just design two boxes.

The way I was told is take the NET volume of the entire box. Take the total port area (both ports) and then figure your port length. Divide it in half and each of your ports should be that long...this is for a single chamber box btw, a dual chamber box is obvious....just design two boxes.

Trixter

08-20-2007, 01:44 PM

The way I was told is take the NET volume of the entire box. Take the total port area (both ports) and then figure your port length. Divide it in half and each of your ports should be that long...this is for a single chamber box btw, a dual chamber box is obvious....just design two boxes.

lets see if it works.....i don't believe it will....

for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

if you divide the box by the numbers of ports first......

Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)

Lv = (111773200 / 3118500) - 4.04

Lv = 35.84 - 4.04

Lv = 31.0

now lets add the port area to get our 48sq" and use the total box volume...

Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)

Lv = (223546400 / 6220800) - 5.72

Lv = 35.94 - 5.72

Lv = 30.22

now you say to divide this by 2 and you get...

Lv = 15.11

as you can see the results are very different from each other.

YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.

lets see if it works.....i don't believe it will....

for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

if you divide the box by the numbers of ports first......

Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)

Lv = (111773200 / 3118500) - 4.04

Lv = 35.84 - 4.04

Lv = 31.0

now lets add the port area to get our 48sq" and use the total box volume...

Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)

Lv = (223546400 / 6220800) - 5.72

Lv = 35.94 - 5.72

Lv = 30.22

now you say to divide this by 2 and you get...

Lv = 15.11

as you can see the results are very different from each other.

YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.

jellyfish420

08-20-2007, 02:07 PM

lets see if it works.....i don't believe it will....

for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

if you divide the box by the numbers of ports first......

Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)

Lv = (111773200 / 3118500) - 4.04

Lv = 35.84 - 4.04

Lv = 31.0

now lets add the port area to get our 48sq" and use the total box volume...

Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)

Lv = (223546400 / 6220800) - 5.72

Lv = 35.94 - 5.72

Lv = 30.22

now you say to divide this by 2 and you get...

Lv = 15.11

as you can see the results are very different from each other.

YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.

very well put. i didn't feel like going that in depth. but now maybe people can comprehend...

if anyone feels like reading heres a good one that tells all about it...

http://mobile.jlaudio.com/support_pages.php?page_id=165

for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

if you divide the box by the numbers of ports first......

Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)

Lv = (111773200 / 3118500) - 4.04

Lv = 35.84 - 4.04

Lv = 31.0

now lets add the port area to get our 48sq" and use the total box volume...

Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)

Lv = (223546400 / 6220800) - 5.72

Lv = 35.94 - 5.72

Lv = 30.22

now you say to divide this by 2 and you get...

Lv = 15.11

as you can see the results are very different from each other.

YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.

very well put. i didn't feel like going that in depth. but now maybe people can comprehend...

if anyone feels like reading heres a good one that tells all about it...

http://mobile.jlaudio.com/support_pages.php?page_id=165

spoonraker

08-20-2007, 02:08 PM

Wait...maybe I'm missing something. The two results were only different by seventy-eight hundeths of an inch....is that a big difference?

spoonraker

08-20-2007, 02:21 PM

Well after reading that JL tutorial and the examples they provided there is no doubt that you are correct, which I never doubted. I guess the arbitrary numbers we chose didn't end up being a good example :)

Notice that Method 1 produces the same port length as did our single 4" diamter port as it should (after all, we have the same total port cross-sectional area which this school of thought proclaims is correct!). But the first method is incorrect because it neglects the frictional losses encountered by using many smaller ports--there is a higher port wall surface area to cross-sectional area ratio which raises the total amount of frictional losses in the ports and thus shifts the tuning!

That's the part that explains why this occurs for anybody wondering. I learned something today.

Notice that Method 1 produces the same port length as did our single 4" diamter port as it should (after all, we have the same total port cross-sectional area which this school of thought proclaims is correct!). But the first method is incorrect because it neglects the frictional losses encountered by using many smaller ports--there is a higher port wall surface area to cross-sectional area ratio which raises the total amount of frictional losses in the ports and thus shifts the tuning!

That's the part that explains why this occurs for anybody wondering. I learned something today.