Aero Vs Slot Ports

[tommyk90 5,000+ posts]

with little bit of bracing it would be fine... //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

bracing=displacement. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

Pretty much negates any added airspace.

 

[Haunz 10+ year member]

^ I don't think 4 3/4" X 3/4" strips of MDF is going to displace much volume //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[tommyk90 5,000+ posts]

^ I don't think 4 3/4" X 3/4" strips of MDF is going to displace much volume //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

I wouldn't even bother.

Just make it out of 3/4" like it should be.

Do some simple math. Say he does a simple slot port that only requires one piece of wood.

Say that port is 3" tall, 20" wide, and 12" deep (60 sq. in. of port). Port displacement for a 3/4" piece of wood is .49 cubes.

The same port made out of 1/4" ply displaces .42 cubes.

A SINGLE .75" x 4.75" x 11.25" (the length of the port) piece of bracing displaces .02 cubes. And one of those isn't sufficient bracing for such a large port. You would have use at least two. So that brings up the bracing displacement up to .04 cubes. Add that to the 1/4" ply port displacement and it brings it up to .46 cubes.

You would want to put forth all that effort of measuring and cutting bracing (and buying different wood) for .03 cubes?

Gimmie a break. //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[Haunz 10+ year member]

^^ Math > you ?

I don't know what kind of backwords example you are trying to use... but If you are looking at overall port displacment including the wood.... the total volume taken up by a port with

20"x12"x3" internal dimensions made with a single 3/4" wall that measures 20"x12" is .5208 cubes.... the total volume with the same port with a single 1/4" port wall with four 3/4"x3/4"x12" braces is .4670 cubes......

The point is moot.... no significant difference would be made....

However, the point is that a slot port with equal port area and tuning will displace less volume then a regular round or aero port... even if you use a 3/4" wall....

if you want to debate something lets look at the difference between a 3/4" wall slot port and a typical pvc port if the exit velocity is kept the same....

 

[tommyk90 5,000+ posts]

^^ Math > you ?
I don't know what kind of backwords example you are trying to use... but If you are looking at overall port displacment including the wood.... the total volume taken up by a port with

20"x12"x3" internal dimensions made with a single 3/4" wall that measures 20"x12" is .5208 cubes.... the total volume with the same port with a single 1/4" port wall with four 3/4"x3/4"x12" braces is .4670 cubes......

The point is moot.... no significant difference would be made....

However, the point is that a slot port with equal port area and tuning will displace less volume then a regular round or aero port... even if you use a 3/4" wall....

if you want to debate something lets look at the difference between a 3/4" wall slot port and a typical pvc port if the exit velocity is kept the same....

Sorry, I should have made myself clear. The OVERALL port length is 12", so the port wall itself is only 11.25" long since you have to add on the the depth of the enclosure wall.

But the argument is still there. Really no benefit to wasting your time with 1/4" hard ply to gain less than .1 cube.

And how do you figure a slot port will displace less air? Take two 4" aeroports for example. 25 square inches of port area.

Say we do a 2 cube net enclosure. Port length will have to be 31".

(31x(3.14 x 2^2))/1728=.23 cubes. Multiply that by 2 for two ports and you get .46 cubes.

A similar sized slot port will have dimensions of 2.5 x 10 x 26 to acheive the same tuning frequency. (3.25 x 10 x 25.25)/1728=.475 cubes.

These numbers were made using psp's website (the maker of aeroports) and my box program which is dead on with every box program out there and uses mathematical formulas.

Granted this is just an example, but your statements aren't true in all cases. And on top of that, you can usually get away with less port area using aeros because of the flared ends and increased efficiency, thus making the difference between them even larger, but once again an insignificant amount.

 

[arizona 10+ year member]

Thanks for all the replies guys.... Don't know if it matters but I am hoping to get 7-7.5 net for the enclosure. I have been busy with work and school so I haven't had time to figure out the new dimensions on what I want to do in the trunk. So I guess after I get that all final I will just calculate if its worth the extra money etc.

 

[Haunz 10+ year member]

But the argument is still there. Really no benefit to wasting your time with 1/4" hard ply to gain less than .1 cube.

It was a suggestion since he is hard up for space.... and every bit of extra space certainly helps...

how do you figure a slot port will displace less air?

Because the end correction "k" is significantly larger for a slot port sharring three walls of an enclosure then it is for a regular port; so the port does not need to be as long...

Take two 4" aeroports for example. 25 square inches of port area.
Say we do a 2 cube net enclosure. Port length will have to be 31".

(31x(3.14 x 2^2))/1728=.23 cubes. Multiply that by 2 for two ports and you get .46 cubes.

A similar sized slot port will have dimensions of 2.5 x 10 x 26 to acheive the same tuning frequency. (3.25 x 10 x 25.25)/1728=.475 cubes.

These numbers were made using psp's website (the maker of aeroports) and my box program which is dead on with every box program out there and uses mathematical formulas.

Granted this is just an example, but your statements aren't true in all cases. And on top of that, you can usually get away with less port area using aeros because of the flared ends and increased efficiency, thus making the difference between them even larger, but once again an insignificant amount.

It's convienient that you factor in the air displaced by the MDF but not the pvc pipe... that stuff is 3/16" thick and as far as I am concearned the 3/4" mdf slot port will still displace less volume... and what may I ask did you use as an end correction value for the slot port ? I doubt you even know...

since you like to do examples go ahead and compare a box with a 2.5"x10" slot port made with 1/4" stock braced with metal brakets and k = 2.2... to a box with two 4" dia regular ports with k=.73 (factoring in 3/16" PVC this time)

 

[Mr Cabinetry 10+ year member]

I didn't want to jump in to this discussion because the last few post arguing over port size's and displacement is pretty much a matter of fighting over what amounts to nothing more than a minor difference.

But, to be on point, if the enclosure volume is fixed in terms of net volume only and you calculate aero vs slot:

aero ports are longer, BUT, displace less volume

Slot Ports are shorter, BUT, displace more volume

This is Based on 3.0 cu.ft. tuned @ 35hz - 2 4" dia aero port vs 1 slot port of the same port area.

The aero's ports - 16.5" vs Slot Port - 15.75"

And as far as end correction factor goes, you don't subtract from the length, you add to the length, so in actuality the slot port would be longer in you want to be totally anal about the whole " MATH " business.

One other thing, when you calculate for the length of a slot port, you only account for the part of the port that's displacing the volume within the enclosure, the part of the port beyond the interior is not part of the displacement, not the total length of the port.

You could also split the same hairs over what the actual displacement being accounted for for an aero port as well, though I didn't see anyone bring that part up.

For you learning enjoyment:

http://www.subwoofer-builder.com/port-flares.htm

Scroll down to Effective Length

 

[tommyk90 5,000+ posts]

It's convienient that you factor in the air displaced by the MDF but not the pvc pipe... that stuff is 3/16" thick and as far as I am concearned the 3/4" mdf slot port will still displace less volume... and what may I ask did you use as an end correction value for the slot port ? I doubt you even know...

3/16"? What the hell kind of PVC are you using?

The PVC used in aeroports is less than 1/8" thick, I just went out and measured it. Overall diameter comes out to ~4 1/8".

A slight miscalculation on my part, but a gross overestimate on yours.

I didn't want to jump in to this discussion because the last few post arguing over port size's and displacement is pretty much a matter of fighting over what amounts to nothing more than a minor difference.
But, to be on point, if the enclosure volume is fixed in terms of net volume only and you calculate aero vs slot:

aero ports are longer, BUT, displace less volume

Slot Ports are shorter, BUT, displace more volume

This is Based on 3.0 cu.ft. tuned @ 35hz - 2 4" dia aero port vs 1 slot port of the same port area.

The aero's ports - 16.5" vs Slot Port - 15.75"

And as far as end correction factor goes, you don't subtract from the length, you add to the length, so in actuality the slot port would be longer in you want to be totally anal about the whole " MATH " business.

One other thing, when you calculate for the length of a slot port, you only account for the part of the port that's displacing the volume within the enclosure, the part of the port beyond the interior is not part of the displacement, not the total length of the port.

You could also split the same hairs over what the actual displacement being accounted for for an aero port as well, though I didn't see anyone bring that part up.

For you learning enjoyment:

http://www.subwoofer-builder.com/port-flares.htm

Scroll down to Effective Length

Thanks for the info. I will say, though, that some of the things you stated are the ways I calculate my enclosures (the port displacement inside the box only for example, thus why I said a 12" long port will only have a 11.25" length inside the box).

I don't want to say that you're backing me up, but people would rather hear it from you than me. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[Haunz 10+ year member]

I didn't want to jump in to this discussion because the last few post arguing over port size's and displacement is pretty much a matter of fighting over what amounts to nothing more than a minor difference.
But, to be on point, if the enclosure volume is fixed in terms of net volume only and you calculate aero vs slot:

aero ports are longer, BUT, displace less volume

Slot Ports are shorter, BUT, displace more volume

This is Based on 3.0 cu.ft. tuned @ 35hz - 2 4" dia aero port vs 1 slot port of the same port area.

The aero's ports - 16.5" vs Slot Port - 15.75"

And as far as end correction factor goes, you don't subtract from the length, you add to the length, so in actuality the slot port would be longer in you want to be totally anal about the whole " MATH " business. [/i]

huh ?

it's not rocket science... 16.5 x a given area will always displace more volume then 15.75 x that same area... no more bs 'examples' are needed...

Lv = (23562.5*Dv^2*Np/(Fb^2*Vb))-(k*Dv)

Dv = port diameter (cm)

Fb = tuning frequency (Hz)

Vb = net volume (litres)

Lv = length of each port (cm)

Np = number of ports

k = end correction

(for a slot port Dv = 2*((W*H)/pi)^0.5)

Clearly, for the same tuning frequency the length of the port DECREASES with an INCREASING end correction factor.... (ie slot port)

 

[tommyk90 5,000+ posts]

huh ?
it's not rocket science... 16.5 x a given area will always displace more volume then 15.75 x that same area... no more bs 'examples' are needed...

How do you come to that conclusion?

If the walls are thinner on the 16.5" length, then it will displace similar (or possibly less) than a thicker wall on a 15.75" length port.

Even with the addition of 1/8" to my calculations, that still brings up the aeroport displacement up to .48 cubes, the same displacement as the slot port.

WHY THE FUCK ARE WE STILL ARGUING ABOUT THIS?!?! It's completely irrelevant.

 

[Haunz 10+ year member]

3/16"? What the hell kind of PVC are you using?
The PVC used in aeroports is less than 1/8" thick, I just went out and measured it. Overall diameter comes out to ~4 1/8".

A slight miscalculation on my part, but a gross overestimate on yours.

alright, you got me... 3/16" is standard for ABS platic which is commonly used for ports and is what I was thinking of... you would have 1/8" thick PVC, which is a whopping 1/16" difference in thickness..

At this point it would seem we are into semanticst, but I will stand by my assertion that a slot port will displace less volume then a regular port, if not at least brace the box...

 

[tommyk90 5,000+ posts]

alright, you got me... 3/16" is standard for ABS platic which is commonly used for ports and is what I was thinking of... you would have 1/8" thick PVC, which is a whopping 1/16" difference in thickness..

At this point it would seem we are into semanticst, but I will stand by my assertion that a slot port will displace less volume then a regular port, if not at least brace the box...

It's actually about 1/16" in thickness, so now we're talking a 1/8" difference. //content.invisioncic.com/y282845/emoticons/tongue.gif.6130eb82179565f6db8d26d6001dcd24.gif

 

[Haunz 10+ year member]

How do you come to that conclusion?
If the walls are thinner on the 16.5" length, then it will displace similar (or possibly less) than a thicker wall on a 15.75" length port.

Even with the addition of 1/8" to my calculations, that still brings up the aeroport displacement up to .48 cubes, the same displacement as the slot port.

WHY THE FUCK ARE WE STILL ARGUING ABOUT THIS?!?! It's completely irrelevant.

the next box I build is going to use a 1/32" sheet of carbon fiber laminate for the slot port... FTMFW...

humor aside, End Correction Factor is serious business

 

[tommyk90 5,000+ posts]

the next box I build is going to use a 1/32" sheet of carbon fiber laminate for the slot port... FTMFW...


humor aside, End Correction Factor is serious business

Whenever I calculated end correction for an enclosure (using winISD, which I usually hate) it never came out to more than a 1 hz difference in tuning. Completely unnoticeable IMO.

Thus, I never do it.

You have to remember, I'm in the SPL competition world, so tuning on the fly is a must for me. I really like aeroports. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif