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req
11-10-2004, 05:59 PM
here we go-

http://www.asterixband.com/andy/pictures/27hz.gif

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? :confused:

BASS OUTLAW
11-10-2004, 06:35 PM
i think u can...
but dont quote me, wait for other replies

swimfreak26
11-10-2004, 07:11 PM
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.
http://img126.exs.cx/img126/7263/27hz.gif

jellyfish420
11-10-2004, 07:47 PM
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.
http://img126.exs.cx/img126/7263/27hz.gif
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
11-10-2004, 07:50 PM
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
11-10-2004, 08:14 PM
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.
http://img126.exs.cx/img126/7263/27hz.gif

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
11-10-2004, 09:49 PM
k guys heres the box.

http://www.asterixband.com/andy/pictures/box.gif

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
11-10-2004, 10:08 PM
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
11-11-2004, 06:37 AM
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
11-11-2004, 07:47 AM
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
11-11-2004, 07:58 AM
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 $$$ :banghead: )

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

ill draw another plan up tonight.

LBX2G
11-11-2004, 08:30 AM
that seems fine to me. I think if u search the forum u will find the info on making other ports like this.

req
11-11-2004, 09:05 AM
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?





http://www.jlaudio.com/tutorials/ports/images/duct3-D.gif

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
11-11-2004, 09:46 AM
It'll work, just seems more complicated to me.

OLD_SCHOOL
11-11-2004, 09:51 AM
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.

req
11-11-2004, 11:26 AM
old school, you said that the corners also have to be angled to make up for the excess area there... so would i make 45 angles that are 5" away from the wall of mdf across from the corner?

because the distance from the edge of that piece of MDF there, to the corner of the port is more than 5" right? so id put the angle to fix that, correct?

like this;

http://www.asterixband.com/andy/pictures/27hz2.gif

would that fix the problem?

req
11-11-2004, 01:08 PM
bump?

any help?

req
11-11-2004, 02:40 PM
bump!

i gotta build this box!

req
11-11-2004, 03:10 PM
cmon, 183 views and NOONE knows if this works?

LBX2G
11-11-2004, 05:50 PM
at first i knew what u trying ot do but now i am lost. Are u talking about bracing.

IgnoreMe
11-11-2004, 06:09 PM
just measure it is all i can say

req
11-11-2004, 06:48 PM
ill get home and draw up a image of what im talkaing about. its hard to explain. ill show ya when i get to my house, im at class right now (public computers that connect to the internet. i dont have autocad at work =P

OLD_SCHOOL
11-11-2004, 11:54 PM
i don't understand what your trying to figure out. or what your trying to measure. if your trying to find the displacement of the port and you have not found the correct port length it is going to be incorrect. the length of your physical port in this type application will not = LV, you must take into consideration corners or turns, wood thickness and end correction. before you can figure out the displacement of the port you must have the correct final port wall length.

req
11-12-2004, 12:16 AM
ok, does this one work.

you should be able to figure it out with no problem. the angles in the corners of the port are the major change (aside from smaller port area and a few other minor things)

http://www.asterixband.com/andy/pictures/30.7hz.gif

jellyfish420
11-12-2004, 01:58 AM
thats a pretty drawing. what did you use to do that with? did it calculate the lengths and areas for you??
the angles are not nessicary(sp?) but they DO help with better air flow which in turn will help it to be louder. but, that is nice looking box.

ngsm13
11-12-2004, 02:07 AM
WOW, 2.42 cube net for a 10" sub????

In all honesty, i'd go with 2 cubes @most net for a 10" sub...the box above should be ok. But if it was me:
2 cubes net @30hz, 40 sq. in. of port

you'll lose a little upper end response with such a large net volume, as well as a low tuning. Low tuning is not bad, but when coupled with a large box for a 10" sub...it happens. peace

NG

req
11-12-2004, 06:35 AM
ok, thats good news to hear.

i used AutoCAD 2002 for the drawings, and it dimensions (in real time) the lines for me. so i have to do little measuring (with rulers and calculations) and bypass all the "port wall volume" and just measure the internal volume... then the sizes are all there for me. it takes a little getting used to, but its a very user-friendly program. and i bet the new versions (autocad 2005 maybe?) are even better.

ill try and design a box with 2cubes@30hz & 40sqin of port area. that shouldnt be too bad. thanks ngsm!

the whole point of this thread was to see if i could measure the bends the way i am (dashed yellow line)

and the reason for the corners is because that area is more than 4" wide, so i had to eather make a rounded edge (hard to do with wood, for me) or angle it so the volume isnt completly different. ill show you later tonight.

req
11-12-2004, 08:29 AM
alright, here is waht im talking about. since from the edge of the first port wall to the corner is more than 4 inches, the port is not 4" wide continiously all the way through. and this will deviate the port volume\port length formula. so to fix this - i am going to add diagonials in the corners, to keep a 4 inch width of the port *roughly* all the way through (it may be a few fractions of an inch wider in some spots, but i dont have the tools to make a curved piece of wood)

http://www.asterixband.com/andy/pictures/portvol.gif

here is a diagram where the bottom represents the WRONG way to do port bends, because the area is too large because the corners have a longer width than the straight parts of the port... the angles in the top image keep the distances equal throughout the length, and hopefully the port volume is correct now. i will go home and measure the volume of these shapes inside the port, and compare it to the port volume that winISD requires. and hopefully they will match. i have another idea if they dont match\dont work.

is that what you were taking about OLD_SCHOOL? :thumbsup:

ChuteBoxe515
11-12-2004, 09:05 AM
you got it this time...that first drawing is what you want to go with...

req
11-12-2004, 09:12 AM
yea. ill go home and make the box smaller (in regards to ngsm's smaller box\smaller port area suggestion) and then calculate everything from there.

im gettin pretty good at this :)

now hopefully i can build the **** thing with all those 45's :p:

req
11-12-2004, 12:34 PM
k.

this is what i came up with.

2ft^[email protected] (so i can round the port length to an inch)
12x4x65 - 48in^2 port area - 1.8ft^3 of port volume

will this box hit hard as a motherfcuker?

is that better ngsm?

LBX2G
11-12-2004, 01:28 PM
ok i get what u are trying to do. I am not trying to confuse u but, if u take the width of the port internaly. and just adjust the wood lenght horizontaly, so the port width stays the same.

OLD_SCHOOL
11-12-2004, 01:33 PM
why do yall insist on doing boxs backwards ?

your port displacement will be incorrect if your final port wall is not the correct length

figure the right displacment first then add it to the VB + braces and sub

OLD_SCHOOL
11-12-2004, 01:35 PM
here is georges box which is box just like your trying to do a small box with a very low tuning


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

Internal Volume = 16 H x 20.5 D x 33 W

Port walls 1-3 = 16.5 x 16
Port wall 4 = 8.72 x 16

External measurements are 17.5 H x 22 D x 34.5 W

Cut sheet
Top/ bottom 34.5 x 22 (2)
Sides 16*22 (2)
Back 33x 16
Front 29 x 16
Ports 1-3 16.5 x 16 (3)
Port 4 8.75 (1)

req
11-12-2004, 01:40 PM
ok, here is a compairison.

when a runner in a race goes around a track, the outside runner is going to be running a longer arc than the inner runner. however they are both going the same distance.

if you took the same track, but put 90 degree corners on the inside and outside track, the outside track would be significantly longer than the inside.

look at this picture;
http://www.asterixband.com/andy/pictures/corners.gif

here we go - line C is 4"... the distance of line B is 4", the distance of line A is 5.3"

the port calculator thinks that the port im using is a straight line. and 4" wide ALL the way down. since it bends, and the distance widens to 5.3" there, it changes the volume by quite a bit. so i have to insert some angles (a 4" radius arc would work much better - but i cant make arcs like that with the tools i have) to keep the width as close to 4" in the bends as possible.

does that make sense?

OLD_SCHOOL
11-12-2004, 01:51 PM
this is why you use the internal depth to figure port dispalcement because the port extends not to the end of the port wall but all the way to the back of the box.

also if your not doing an SPL comp box which by your tuning your not don't worry about 45 those corners you won't hear a difference and you usually only pick up less than a DB by doing it. when spl guys compete every 1/10th of a db counts thats why they do it. you have plenty enough port area to keep down port noise given the power your using.


msg me on aim or msn

req
11-12-2004, 02:19 PM
this is why you use the internal depth to figure port dispalcement because the port extends not to the end of the port wall but all the way to the back of the box.

also if your not doing an SPL comp box which by your tuning your not don't worry about 45 those corners you won't hear a difference and you usually only pick up less than a DB by doing it. when spl guys compete every 1/10th of a db counts thats why they do it. you have plenty enough port area to keep down port noise given the power your using.


msg me on aim or msn

but wont the different in port volume change the tuning frequency? im not concerned with gaining a DB from the corners, im concerned with the difference in volume from what the winISD tells me, and what ill actually have.

the VOLUME of the port is what effects tuning, it can be tall and skinny or short and fat, but as long as the same net volume for the box (2.0ft^3) and port volume are used, it shuold still be tuned to the same frequency, right? this is why im worried about the corners.

ill be home around 5:30 EST, ill message you then. leave me a message so i dont forget - LreqL is my AIM tag.

OLD_SCHOOL
11-12-2004, 03:19 PM
but wont the different in port volume change the tuning frequency? im not concerned with gaining a DB from the corners, im concerned with the difference in volume from what the winISD tells me, and what ill actually have.

the VOLUME of the port is what effects tuning, it can be tall and skinny or short and fat, but as long as the same net volume for the box (2.0ft^3) and port volume are used, it shuold still be tuned to the same frequency, right? this is why im worried about the corners.

ill be home around 5:30 EST, ill message you then. leave me a message so i dont forget - LreqL is my AIM tag.

no it won't this is why your last port wall length being correct is important it acounts for length in the corners whether your port dispalces 1.5 cubes or 3 cubes isn't the key it is being sure it is the crrect length. with that said though volume will dictate length. a port with 40 sq inches will have a diffferant length than a port with 200 sq inches at the same tuning. so if you have a port and change it's AV but keep the length the tuning will change.

this is why when you build an L shaped port the depth of the box doesn't matter once you have the correct length of the second port wall the dispalcement won't change. this is also why you need to have a correct port length and displacement before you go trying to figure out how to build your box

i figure out how much box volume i need VB then figure out the port dispalcement add it to VB then add the braces and sub for a total box volume

a port works the same way a straw does by changing the length of a straw at a constant volume you can change it's pitch when blown. however the same accounts for volume if length is constant

this is why tubas are so much larger than flutes.

req
11-12-2004, 03:23 PM
ill contact you when i get home if thats ok.. i have a few questions to ask if its not too much of a problem.

ngsm13
11-12-2004, 04:14 PM
k.

this is what i came up with.

2ft^[email protected] (so i can round the port length to an inch)
12x4x65 - 48in^2 port area - 1.8ft^3 of port volume

will this box hit hard as a motherfcuker?

is that better ngsm?

looks great! it'll do well, sound great, and be loud. peace

NG

req
11-12-2004, 05:55 PM
you there old_school?

req
11-12-2004, 06:55 PM
hey. here we go again. after a little while in autocad, i came up with a FANTASTIC design. tell me what ya think.

http://www.asterixband.com/andy/pictures/lastbox.gif

now i MIGHT have a good idea on how to make dem curves. if it dosnt work, ill just 45 them, and say that its close enuf. i am almost posative that this box will work. NGSM, any feedback? old_school? jmac? anyone?

req
11-12-2004, 10:08 PM
Ttt

ngsm13
11-13-2004, 12:26 AM
looks great man, it'll sound good and wang! peace

NG

req
11-13-2004, 12:27 AM
k i got a new one. and its correct. that last one was too wide - i didnt realize i made the PORT LENGTH on the width 30 inches instead of the entire width (my max is 32) and i got all messd up.

so i changed it a bit, and it works now. i suppose ill build it.

OLD_SCHOOL
11-13-2004, 11:20 AM
dude your making this 9000 times harder than what it has to be sorry my puter isp went down and just came back up today

req
11-13-2004, 02:02 PM
k old_school, i *think* this is right, after the talk we had.

http://www.asterixband.com/andy/pictures/newbox.gif

i couldnt get the box volume to 2'^3+driver displacement with the port at 30hz, so i had to raise the tuning a little to keep the sencond port wall short enough. so i got the box to 2.14'^3, and the port is 34"-1\2 width (like you said)

but im not sure if that second wall is the correct length. i dont think you actually told me what to do to compensate for the turn. you just told me that the second length had to be shorter to compensate for the turn. lemme know - thanks mang!

and see how i did it with cad, i draw the box relative to the size, and i ignore the wood and i only worry about the internal dimensions. makes it much eaiser.

OLD_SCHOOL
11-13-2004, 02:09 PM
i think you need to let me draw a box for you..... :wacky:

req
11-13-2004, 02:35 PM
****it.

whats wrong this time? did you like the woofer drawing tho :wave:

im an artist, not a ****ing mathmetition. :toast: its hard for me.