# 2 small ports=1 larger??



## the_dealer

I've been finding mixed information on round ports. I'm going to use these numbers as examples. Are two 2" round ports equal to one 4" round port? would the length for each of the smaller ports be half the length of the larger?

For example if the length of the 4" port is 10", would the length for the two 2" ports be 5" long each?

some say they have the same port space, others say they don't. What's the truth? I'm trying to get some box building skills, and there are some scenarios where using 2 smaller ports that are shorter in length (then one larger)would be a benefit


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## minbari

the_dealer said:


> I've been finding mixed information on round ports. I'm going to use these numbers as examples. Are two 2" round ports equal to one 4" round port? would the length for each of the smaller ports be half the length of the larger?
> 
> For example if the length of the 4" port is 10", would the length for the two 2" ports be 5" long each?


for the first part, this is simple math. Pi x R^2 is the formula for area. (2) 2" ports will not equal a single 4".

3.14 x 4^2 = 50.26
3.14 x 2^2 = 12.56

so (2) 2" would be 25.13. you actually need (4) 2" ports to equal (1) 4"

for the second part, all the ports have to be the same length. so if you need a single 4" at 15" long, then (4) 2" will be 15" long to tune to the same freq. all four have to be 15" long, you cant divide the 15" between them.


> some say they have the same port space, others say they don't. What's the truth? I'm trying to get some box building skills, and there are some scenarios where using 2 smaller ports that are shorter in length (then one larger)would be a benefit


this has more to do with port velocity than tuning. I could tune a 2 cuft box with (1) 15" sub that has 30mm of xmax with a 1" port. but it will sound terrible with more than 50 watts, because the port will stop being effective once more air is being force through it than the port is capable of.

get WinISD and do some experimenting, you will see pretty fast how it is


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## amalmer71

Never thought I'd say this to a guy, but play with this. It'll give you the answer you're looking for. It's an online port calculator from Linear Team (the same creators of WinISD). 

LinearTeam


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## the_dealer

That shows the vent length of 2 small ports being half as long as 1 bigger port?


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## the_dealer

For example, a box 1.5cft tuned at 35hz. 1 4" port will need to be 15.684" long. 2 2" ports would be 7.842" long(ea). But no matter what it doesn't get the port mach below .16

even with a 6" port, on 350 watts, mach is .65

Something must be up with that calculator


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## emilime75

A better way to think about converting from a single to multiple ports or from circular to slot ports is to match up the volume of air contained by the actual port. So, as an example, if a driver/box combo requires a 4x7 inch port, said port contains 87.96 cubic inches. So, if you want to do multiple ports, or a slot port, you have to calculate to make sure they equal 87.96 cubic inches. So, like minbari said above, a 2x7 inch port contains 21.99 cubic inches or air, x4 that equals 87.96, the same as the single 4x7.

Multiple ports of smaller diameter will not reduce velocity or the possibility of chuffing compared to a single large port. I find the biggest hurdle with modern subwoofers is the small box requirement and the length of port needed for low tunes. Often, they simply don't fit inside the box. That's where slot ports help, or you can always run the port outside the box, if your install allows it.


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## minbari

the_dealer said:


> That shows the vent length of 2 small ports being half as long as 1 bigger port?


yes, you have 1/2 port area, it will be half as long. BUT you will have higher port velocity. you have to look at the whole picture. If the 2 smaller ports keep port velocity low enough, then there is not much gain to a larger port. if port velocity is high, then you need a larger port, no way around this.


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## minbari

the_dealer said:


> For example, a box 1.5cft tuned at 35hz. 1 4" port will need to be 15.684" long. 2 2" ports would be 7.842" long(ea). But no matter what it doesn't get the port mach below .16
> 
> even with a 6" port, on 350 watts, mach is .65
> 
> Something must be up with that calculator


did you use the acoustic power calculator? that is NOT amplifier power in that variable.

it is alot easier to use the downloaded version IMHO


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## amalmer71

minbari said:


> did you use the acoustic power calculator? that is NOT amplifier power in that variable.


Affirmative.

T/S parameters need to be entered into this calculator in order to obtain the acoustic power.

LinearTeam


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## undone1

hmmm,I remember somewhere it was said the the multiple smaller ports actually need to be longer to be tuned the same to make up for additional friction of the walls...something like that can't remember exactly...


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## ATOMICTECH62

That's true.You cant just make 2 ports with the same sqin as 1 the same length.
You have to recalculate it.


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## minbari

another reason to use a modeling software like WinISD. it will do those kind of calculations for you.


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## undone1

minbari said:


> for the first part, this is simple math. Pi x R^2 is the formula for area. (2) 2" ports will not equal a single 4".
> 
> 3.14 x 4^2 = 50.26
> 3.14 x 2^2 = 12.56
> 
> so (2) 2" would be 25.13. you actually need (4) 2" ports to equal (1) 4"
> 
> for the second part, all the ports have to be the same length. so if you need a single 4" at 15" long, then (4) 2" will be 15" long to tune to the same freq. all four have to be 15" long, you cant divide the 15" between them.
> 
> 
> 
> get WinISD and do some experimenting, you will see pretty fast how it is


Maybe I am misunderstanding,are you sure the calculation for the 4'' is 50.26...? 4 x 4 x .7854 = 12.56 area....?

Also the length on the (4) 2'' need to be longer correct...?


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## minbari

undone1 said:


> Maybe I am misunderstanding,are you sure the calculation for the 4'' is 50.26...? 4 x 4 x .7854 = 12.56 area....?
> 
> Also the length on the (4) 2'' need to be longer correct...?


Yes pi = ~3.14. 4 squared is 16. Pi x 16 is ~50.26 sqin of area. NOT Length.

Yes because of some loading aspects of smaller tubes, 4 2" will be slightly longer overall

sent from my phone using digital farts


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## undone1

minbari said:


> Yes pi = ~3.14. 4 squared is 16. Pi x 16 is ~50.26 sqin of area. NOT Length.
> 
> Yes because of some loading aspects of smaller tubes, 4 2" will be slightly longer overall
> 
> sent from my phone using digital farts


4'' port is not 50.26 sq in of port...


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## minbari

undone1 said:


> 4'' port is not 50.26 sq in of port...


Whoops you are right. I did pi d squared.

That makes. 

4" - 12.56 sqin
2" - 3.14 sqin

Math was wrong, but the idea is still the same. (4) 2" = (1) 4"


sent from my phone using digital farts


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## undone1

minbari said:


> Whoops you are right. I did pi d squared.
> 
> That makes.
> 
> 4" - 12.56 sqin
> 2" - 3.14 sqin
> 
> Math was wrong, but the idea is still the same. (4) 2" = (1) 4"
> 
> 
> sent from my phone using digital farts



(slowly wipes sweat from forehead)..man that had me second guessing all of my current boxes...:laugh:


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## minbari

Ha ha. Anytime. 

sent from my phone using digital farts


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## subwoofery

undone1 said:


> (slowly wipes sweat from forehead)..man that had me second guessing all of my current boxes...:laugh:


Yep, _Minbari_'s right, that doesn't solve your problem - you still need 4 x 2" round ports to equal a single 4" round port... 
Then the length is about the same as the single one. 

Kelvin


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