U port quick question

[req]

here we go-



so thats what im trying to do.

can i measure up the middle of the port with a ruler?

those are 45 degree angles there. does this work?

i need 67.47" long of port area... my box is ~15 inches deep and ~25" wide- can i measure it like that instead of doing math?

[BASS OUTLAW]

i think u can...
but dont quote me, wait for other replies

[swimfreak26]

I don't see why you would need to, it seems like making 45's complicates the problem.

I would just measure linear lengths and call it good. In other words, from the front to the back of the box should be a known lengthand that will be put into the calculation of the overall length of the port. As long as you have the port wall as far away from the wall as the width of the port (which you have to do anyways) then you can just add lengths of the individual sections.

I don't know if what I'm saying is making sense, but I really do not see the need for the angles as it really makes things A LOT more complicated IMO. It should work however.

EDIT:
This is what I'm saying...if the two shown dimensions are 5" then the port length is equal to 1+2+3+4 on the picture. You may have to calculate a little bit for port extension (I know that's the wrong termanology, but I can't think of the right word now) but I'm not knowledgeable enough to tell you exactly how to do that.

[jellyfish420]

I don't see why you would need to, it seems like making 45's complicates the problem.

I would just measure linear lengths and call it good. In other words, from the front to the back of the box should be a known lengthand that will be put into the calculation of the overall length of the port. As long as you have the port wall as far away from the wall as the width of the port (which you have to do anyways) then you can just add lengths of the individual sections.

I don't know if what I'm saying is making sense, but I really do not see the need for the angles as it really makes things A LOT more complicated IMO. It should work however.

EDIT:
This is what I'm saying...if the two shown dimensions are 5" then the port length is equal to 1+2+3+4 on the picture. You may have to calculate a little bit for port extension (I know that's the wrong termanology, but I can't think of the right word now) but I'm not knowledgeable enough to tell you exactly how to do that.

thats how i do it. you said 15 front to back. so 15+15=30+15=45+15=60 then your final length -60=your final wall.

[jellyfish420]

but now looking at it again your port alone is gonna be wider then 25". 5+5+5+5+5+3.75=28.75. so your gonna have to make something smaller or your box bigger.

[req]

I don't see why you would need to, it seems like making 45's complicates the problem.

I would just measure linear lengths and call it good. In other words, from the front to the back of the box should be a known lengthand that will be put into the calculation of the overall length of the port. As long as you have the port wall as far away from the wall as the width of the port (which you have to do anyways) then you can just add lengths of the individual sections.

I don't know if what I'm saying is making sense, but I really do not see the need for the angles as it really makes things A LOT more complicated IMO. It should work however.

EDIT:
This is what I'm saying...if the two shown dimensions are 5" then the port length is equal to 1+2+3+4 on the picture. You may have to calculate a little bit for port extension (I know that's the wrong termanology, but I can't think of the right word now) but I'm not knowledgeable enough to tell you exactly how to do that.

that image is not making any sense to me at all.

the angles would take 5 seconds with a protractor and a penicl. the only reason for doing that is to find the center of the port. then its just connect the dots, and measure.

the JL site said to measure down the center- or do the math. so im gonna go play with autocad real quick.

but CAN i do it with the angles? im pretty sure that should work.

[req]

k guys heres the box.



thats my real measurements in autocad2002 (god i love this program) using WinISD for the port length\box size - and a 2002 RE SE 10" woofer config.

i can make this no problem. but will it be @ ~30hz? my fiberglass box is tuned a bit high, and has a smaller internal volume. will this box be louder and\or hit lower than a 1.2@37hz?

basically. will that work?

[jellyfish420]

should sound great. i did a box the other day for a 10. 2.25ft^3@30Hz. loud and low. it only had 30in^2 port though. **** 75in^2 port. thats pretty big!!!! i don't no if youd want to go quite that big.

[req]

why wouldnt i go that big? the port is ****ing bigger than the **** subs volume.

does anyone else know if this would work - or sound nice?

[swimfreak26]

From what I hear, you're never supposed to go larger than the subs surface area for port area. SPL enclosures will come near to this level, but daily installs never should. Hopefully someone with more box building experience will shed light on this however.

[req]

ok - i can make the box like this then. since the driver has 75~80in^2 of cone area, i divided that by 2 to get a rough number... and im gonna give it a port thats 11x4x63 for a 2.2ft^3@28hz.

this would give me 1.6ft^3 of port volume. and it would be tuned lower, and the box will be alot smaller.

i dont have autocad here to draw this up. but it would be similar to that drawing above, just 12.5 (wood thickness) tall instaead of 15"

the maximum i can go with the box is 32 wide, 15 tall, and lets say 20 deep. that gives me 5.5 cubes to play with. but when i go snowboarding, i dont know if ill be able to fit all that in my car =p

im seriously thinking of a chevy S10, ford ranger (4x4 - winter in NY = bad) 2000 and up. i can prolly get a 2004 for ~10,000 brand new. and i am SERIOUSLY thinking of doing a blowthrough if i get it. maybe with a SX 15 if i can get one (i wont have any $$$ )

but this is a test to see if the port U idea works by measuring the centerline.

ill draw another plan up tonight.

[LBX2G]

that seems fine to me. I think if u search the forum u will find the info on making other ports like this.

[req]

not really, the search function is complete rubbish. it never comes up with anything that im looking for.

i searched for a thread yesterday, i posted in it, people quoted me, and the word ROPE and NYLON was said. so i searched for posts by REQ and the word rope. and i got no results. WTF. i did the same search with NYLON and i still got nothin. so i searched for rope without REQ as a poster, and i got like a billion threads that i didnt want.

so think about searching for 'port' in there. such a ton of nonsense threads with 2 posts.

this is a simple question.

can i measure down the center as JL said?

Lv(physical) - The physical length of the port is measured down the dead center of the port from end to end. In the picture above, this would correspond to L1 + L2, but the physical length of the port isn't really what is important, it's the effective length of the port (with end correction) that is important.

this is all im doing. does it work?

[swimfreak26]

It'll work, just seems more complicated to me.

[OLD_SCHOOL]

Port size: 16 X 4
Port Area: 64 in ^ 2
Peak Port Velocity: 16.49 m/s @27.05 HZ @ 1500 watts

Lv = (64* 1.84^8)/(2.5 * 1728 (29/0.159)^2) – (.823 * sqrt (64)
Lv= 75.3591
Physical Lv = 73.35 round to 73.25

Port displacements:
Port walls 1-3 = Id- PW = PWL
20.5 – 4 = 16.5
20.5* ( 4+.75) * 16 = 1558 in^2
(1558 in^2 /1728) (3) = 2.70 Ft^3

Port wall 4 = LV - (16.5 (wall length port 1-3) + 4 (Corners) + .75 (MDF) (3)) –(.75( MDF) + 2 (end correction) = 8.75 in

Displacement = 8.75 * 16 * (4+.75) = .385 Ft^3

Total Port Displacement= 2.7 Ft^3
.385 Ft^3
--------------
3.09 Ft^3
Sub Displacement= .70 Ft^3
Net volume = 2.5 Ft^3
Total box volume = -----------------
6.285 Ft ^3
this is an example how to do it this is from georges 12 inch mag box i did for him a while back

you have to find how many port walls your gonna need that is just the port wall minus the internal depth then use the above formula to find the length needed for the last prot wall.