need help figuring this out. im tryin to learn ;)

[IgnoreMe 5,000+ posts]

Your port has a displacement of .88cuft using 3/4" MDF ((5.75" x 6.75" [for 30sqin internally + the MDF] x 39") / 1728), if that's what you were asking. Which brings your box volume down to 1.34cuft internally before sub displacement.

hey squeak thanks for the help. 1.34 is way to small for a mag so i started over. here is what i got for my third box, if i cant get it done with this third box i think im done haha. anyways, here the info. 16" high 12" deep, 27" wide for a total volume of 3 cubic feet. my port length comes out to 27". im using 30 square inches of port area. now would the port discplacement be 5.75" x 6.75" x 39" / 1728 for a total displacment of .88 cubic feet again? or do i have to change the 6.75 and 5.75 to something difference since the box is bigger?

 

[squeak9798 5,000+ posts]

Your length will change. Notice in the equation the variable "Vb", which is net internal box volume. As the internal volume changes, so will the port's length (given the cross area stays the same). If you want, you can keep the same cross area (30sqin in your example), but the length will change.

 

[squeak9798 5,000+ posts]

Also, I think you may be going about this the wrong way. Remember, in the equation it uses Net internal volume, which is the volume after all displacements (port, sub, bracing, etc). So, for your new box, you would not use 3cuft for the volume of the box in the equation to get port length becuase that is gross and not net.

You really need to figure out what internal volume you want (say, 2.5cuft net for example), and use that to figure out the dimensions for the port. After you have both those figured out, then determine the physical dimensions of the box that will achieve that desired Net box volume.

 

[OLD_SCHOOL 10+ year member]

Ok let me try and take a stab at this. You want about 2.4 ^3 for you sub and I read that you wanted to stay under you window in your explorer which gives you a maximum height. And since you said your box was 16 inches inside volume I will assume that 17.5 external height is ok. So now we have one max height and since you have a little bit of width in the back of the explorer and you want to use 20 inches wide we have two measurements.

So internally we have 16 H and 20 W now we want 2.4 cubic feet to get that we have to find the port displacement, brace displacement and sub displacement and add it to the NET volume of 2.4 cubic feet.

So you want to use 30 in^2 port area so now to figure that you say ok my height is 16 inches so what times 16 gives me 30 or close to it well 16 x 2 equals 32 and that’s close enough to the 30 that you wanted. Now here is the thing different port sizes I.E. width and such can yield different results depending on the application but all you would need to do to get a wider port would be to increase the amount of port area you wanted … for example if you wanted 50 in^2 port area you would say what times 16 (your internal height) = 50. Well 16x3 is 48 which is might close to 50 you wanted. Trick is make sure you use a program that can give you port velocity and the magic number that many agree on here is 17.2 m/s port velocity

Ok so now we have our port area 30in^2 lets plug it in to the formula and get the port length

Lv= 30*1.84*10^8/ [2.4*1728*(29/0.159)^2]-0.823*sqrt(30) and for port length I get 35.50351466

So then you use a port correction which is half your port width to get the physical port length, so LV – 1 = 34.50

So 34.5 is close enough well right off I can tell I am going to have to turn this port in a L fashion. Because I am pretty certain my width is going to be less than 35 inches so I have to figure out the length of the second port wall before I can figure its displacement. Since as long as any old number shorter than port length will do because the displacement for the second wall will always be the same no matter length as long as it stays an L, what you can do is just choose a number and 12 seems like a good one to me.

So to find the length of the second wall you take the physical lv and subtract it by the external depth so in this case we have 34.5 - (12+1.5)=21 inches long now we can find the displacement of that port wall

21 * (2+.75) * 16 = .5347 ft^3

Now for the fist wall it is just the internal depth – port width so 12-2 is 10 so the front wall is 10 inches long and will displace

10* (2 +.75) * 16 = .2546 ft^3

Ok now since we have the port displacements we can find the sub displacement I don’t know what your sub displaces but lets say it displaces .070 ft^3 so what we have to do is add up all the displacements (.5347 + .2546 + .070 +2.4) = 3.2 ft^3 gross volume so you have to build a box to that size so that your sub will see its 2.4 feet cubed

Now since we have a height of 16 inches and width of 20 you can take 3.2*1728/ (20*16) and get a depth = 17.28 inches deep inside volume.

So port wall one would be 17.25-2 = 15.25 inches long

And port wall 2 would be 34.5-18.75= 15.75 inches (edit wrong number)

You can play with all the numbers now because your second port wall displacement will always be the same

Don’t take this as gospel because I am just learning this stuff as well and I could be off somewhere but this is how I understand it so far if one of the more knowledgeable guys comes on and oks what I have here feel free to use it i just learned this stuff from hanging around asking questions that usually piss off the mods cause my questions are usually lame hehehe

 

[IgnoreMe 5,000+ posts]

Ok let me try and take a stab at this. You want about 2.4 ^3 for you sub and I read that you wanted to stay under you window in your explorer which gives you a maximum height. And since you said your box was 16 inches inside volume I will assume that 17.5 external height is ok. So now we have one max height and since you have a little bit of width in the back of the explorer and you want to use 20 inches wide we have two measurements.
So internally we have 16 H and 20 W now we want 2.4 cubic feet to get that we have to find the port displacement, brace displacement and sub displacement and add it to the NET volume of 2.4 cubic feet.

So you want to use 30 in^2 port area so now to figure that you say ok my height is 16 inches so what times 16 gives me 30 or close to it well 16 x 2 equals 32 and that’s close enough to the 30 that you wanted. Now here is the thing different port sizes I.E. width and such can yield different results depending on the application but all you would need to do to get a wider port would be to increase the amount of port area you wanted … for example if you wanted 50 in^2 port area you would say what times 16 (your internal height) = 50. Well 16x3 is 48 which is might close to 50 you wanted. Trick is make sure you use a program that can give you port velocity and the magic number that many agree on here is 17.2 m/s port velocity

Ok so now we have our port area 30in^2 lets plug it in to the formula and get the port length

Lv= 30*1.84*10^8/ [2.4*1728*(29/0.159)^2]-0.823*sqrt(30) and for port length I get 35.50351466

So then you use a port correction which is half your port width to get the physical port length, so LV – 1 = 34.50

So 34.5 is close enough well right off I can tell I am going to have to turn this port in a L fashion. Because I am pretty certain my width is going to be less than 35 inches so I have to figure out the length of the second port wall before I can figure its displacement. Since as long as any old number shorter than port length will do because the displacement for the second wall will always be the same no matter length as long as it stays an L, what you can do is just choose a number and 12 seems like a good one to me.

So to find the length of the second wall you take the physical lv and subtract it by the external depth so in this case we have 34.5 - (12+1.5)=21 inches long now we can find the displacement of that port wall

21 * (2+.75) * 16 = .5347 ft^3

Now for the fist wall it is just the internal depth – port width so 12-2 is 10 so the front wall is 10 inches long and will displace

10* (2 +.75) * 16 = .2546 ft^3

Ok now since we have the port displacements we can find the sub displacement I don’t know what your sub displaces but lets say it displaces .070 ft^3 so what we have to do is add up all the displacements (.5347 + .2546 + .070 +2.4) = 3.2 ft^3 gross volume so you have to build a box to that size so that your sub will see its 2.4 feet cubed

Now since we have a height of 16 inches and width of 20 you can take 3.2*1728/ (20*16) and get a depth = 17.28 inches deep inside volume.

So port wall one would be 17.25-2 = 15.25 inches long

And port wall 2 would be 34.5-18.75= 15.75 inches (edit wrong number)

You can play with all the numbers now because your second port wall displacement will always be the same

Don’t take this as gospel because I am just learning this stuff as well and I could be off somewhere but this is how I understand it so far if one of the more knowledgeable guys comes on and oks what I have here feel free to use it i just learned this stuff from hanging around asking questions that usually piss off the mods cause my questions are usually lame hehehe

what the ****?! seems so easy for some people to do these things. i was never good at mathematics back in school so im kinda like the one legged man tryin to run a marathon. anyways, i think squeak was right, i was going about it the wrong way. you sound like one of the pros OldSchool. i mean i dont know if your right but it **** well seems like it. but one question, in the 7th paragraph you subtracted 13.5" from 34.5". that gives you 21" i thought the 12"(13.5 with the addition of the mdf) was the size that my first port wall was gonna be but later on it changed to 15.25. how does that happen?

 

[OLD_SCHOOL 10+ year member]

the 13.5 is 12 inches + 1.5 inches of mdf that gives you your second wall length.. it changed size because we just used an arbitrary number of 12 inches because your displacement will always be the same as long as it is an l shaped port

see lets do it with a depth of 15 inches this time

so for port wall 1 it is just internal depth - port width 15-2=13inches

for port wall 2 its physical lv - external depth so 34.5 - (15+1.5)=18 inches long

now figure displacements

port wall 1 16 * (2+.75) * 13 =.3310ft^3

port wall 2 16 * (2 +.75) * 18 = .45833 ft^3

those two added togther = .789351 ft^3

same as the first two displacements we found using a 12 inch depth 5347 + .2546 = .7893 ft^3

so see we wernt using any special depth at this point we are just trying to figure out the displacement of the second port wall once you have that then you can figure your bx

so we are going to come up with the same gross box volume every time all that will need to be adjusted is the port wall length ALSO these figures don't include bracing you have to account for those as well in your final dispalcement

and yes squeak is right it is best to start off with a net volume and add your stuff than try to figure out a round figure and subtract

now what i want to learn is how to figure he length and displacement on ports that hve 3 walls

 

[IgnoreMe 5,000+ posts]

the 13.5 is 12 inches + 1.5 inches of mdf that gives you your second wall length.. it changed size because we just used an arbitrary number of 12 inches because your displacement will always be the same
see lets do it with a depth of 15 inches this time

so for port wall 1 it is just internal depth - port width 15-2=13inches

for port wall 2 its physical lv - external depth so 34.5 - (15+1.5)=18 inches long

now figure displacements

port wall 1 16 * (2+.75) * 13 =.3310ft^3

port wall 2 16 * (2 +.75) * 18 = .45833 ft^3

those two added togther = .789351 ft^3

same as the first two displacements we found using a 12 inch depth 5347 + .2546 = .7893 ft^3

so see we wernt using any special depth at this point we are just trying to figure out the displacement of the second port wall once you have that then you can figure your bx

so we are going to come up with the same gross box volume every time all that will need to be adjusted is the port wall length ALSO these figures don't include bracing you have to account for those as well in your final dispalcement

and yes squeak is right it is best to start off with a net volume and add your stuff than try to figure out a round figure and subtract

ok sweet. thanks A LOT man! its really really helpful. just to make sure though, when i go to lowes to get my wood and get them to cut it, i will want 2 top and bottom pieces that will be 20 x 17.25, a back piece that will be 16 x 18.5, 2 side pieces that will be 17.25 x 16, and a front peice that will be 16 x what?

 

[OLD_SCHOOL 10+ year member]

well that is where your first port wall will mount flush against the back side of the front wall (baffle) so the front bafffle will be whatever you decide for width minus 2 inches for your port if you decided to go that way. remeber though those numbers above where just rough numbers with no bracing and guessed at sub displacement.

also those measurements are made using the two side ontop of the top and bottom and face and back insde so all your measurements are off.

also you may wanna wait and see if someone like jmac or steve comes in and gives a thumbs up on my numbers

 

[IgnoreMe 5,000+ posts]

well that is where your first port wall will mount flush against the back side of the front wall (baffle) so the front bafffle will be whatever you decide for width minus 2 inches for your port if you decided to go that way. remeber though those numbers above where just rough numbers with no bracing and guessed at sub displacement.
also those measurements are made using the two side ontop of the top and bottom and face and back insde so all your measurements are off.

also you may wanna wait and see if someone like jmac or steve comes in and gives a thumbs up on my numbers

you were right for the sub displacment. SI says its .7 so thats clear. im not sure exactly how much bracing im gonna need either //content.invisioncic.com/y282845/emoticons/uhoh.gif.c07307dd22ee7e63e22fc8e9c614d1fd.gif .

 

[IgnoreMe 5,000+ posts]

anyone wanna check old schools numbers for me?