2 Woofers = 2 x Volume = What Port?

[sdclc126]

Hi,

If this has been answered before (maybe in Subwoofers) I apologize, but I couldn't think of good keywords to search with...

I know that if you use two woofers, either midwoofers or for "dedicated" bass, you should double cabinet volume to achieve the same tuning as for a single woofer. But what then happens with port size calculation? Does that also double? Or?

Thanks.

 

[richie00boy]

Everything doubles. You simply use two of the ports that you specced for the single size box. Or one port of the same cross sectional area.

 

[sdclc126]

Thanks Richie,

It's nice that it's that simple and straightforward - makes for a short thread though.

 

[Dave Bullet]

I don't think it is simply 2 ports at the same length as a single port (unless the spreadsheet I am using is wrong). I don't have the formula handy. To maintain the cross sectional area by adding another port - don't you have to almost double the length of both ports for the same tuning frequency?

 

[richie00boy]

No it's simply a double up on everything, port length remains as per a single woofer box. Think of it as two single boxes joined siamese style.

Your last sentence does not make sense by the way, but I know what you are trying to say.

 

[Jay_WJ]

To maintain the cross sectional area by adding another port - don't you have to almost double the length of both ports for the same tuning frequency?

Dave, what you said is correct only when the box volume stays the same. If the box volume is doubled, you can achieve the same port tuning frequency by adding a port of the same length.

 

[Svante]

Yes, two drivers, double volume, two ports of the same length is correct. However, the port needs to be slightly shorter if a single port is used due to that the end correction becomes larger for a larger diameter port. Actually, if the two ports are placed close to one another, there will be an effect in this direction too since the ports share co-oscillating air.

 

[sdclc126]

It's nice that it's that simple and straightforward - makes for a short thread though.

Maybe I spoke too soon.

 

[Dave Bullet]

Ooops - yeah that was a bit dumb of me - cheers guys.

 

[augerpro]

Yes, two drivers, double volume, two ports of the same length is correct. However, the port needs to be slightly shorter if a single port is used due to that the end correction becomes larger for a larger diameter port. Actually, if the two ports are placed close to one another, there will be an effect in this direction too since the ports share co-oscillating air.

Could you elaborate on end correction please?

 

[sdclc126]

Could you elaborate on end correction please?

I second that.

 

[MJL21193]

My understanding is that end correction is accounting for the air mass at the end of the port (outside air), and how it moves with the air in the port. Not a good description, I know. Think of the air mass as part of the port, extending it out a little bit.
Having an effect on this would be the shape of the port exit, flared or flush. Also, whether this air mass is normally still, or being disturbed (like by another port).

 

[marchel]

correct me if I'm wrong,

I think , simply doubling the volume and doubling the number of the same port is wrong, I think you'll gonna have to recalculate the port dimension for the new box volume and a new port size altogether.

If you're calculating for a speaker system with 2 identical woofer, I think it's easier if you input 2 x the value of the Vas of a single driver, and make the piston area twice, so that the software will compute for the correct port dimention automatically.

Or if you'll be using flared ports to reduce the port diameter, There is a method of calculating it , somewhere in this link CLICK HERE

 

[Svante]

Could you elaborate on end correction please?

Sure. The air inside the tube oscillates, but also the air outside both ends of the tube contribute to the effective mass of air. There is a simple formula for calculating the effective mass of the air outside a baffled tube, and it can be expressed as if the tube was a bit longer than the actual length.

For each side of the tube, one should add 0.85*r to the length (r is the tube radius).

Obviously, having two (well separated) vents, and equal amount is added to both vents, and the amount is the same as for the single vent in the small box. However, if a larger single vent is used in the large box, the end correction becomes larger (due to a larger r), and so the tube length has to be shortened.

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An externally hosted image should be here but it was not working when we last tested it.

correct me if I'm wrong,

I think , simply doubling the volume and doubling the number of the same port is wrong, I think you'll gonna have to recalculate the port dimension for the new box volume and a new port size altogether.

One can also think like this: Take two subwoofers and stack them on top of each other. Then drill a hole through their walls, so that the air can flow freely between the two box cavities. Due to symmetry, no air will flow between the boxes. This means that the wall between the two boxes has no effect. This in turn means that a box with double box size and two drivers should have two vents, identical to the single vent in the single box.

 

[vasyachkin]

very simple:

if you have the right box for one woofer then:

1 - build a second identical one

2 - glue them together

3 - remove the part that separates the two volumes

4 - you're done.

5 - if you want to replace the two ports with one follow the same logic ...

really, its all common sense ... the key thing to realize is that when you have two speakers or ports together and you remove whatever separates them - nothing changes ... from that you can deduce that speaker volume has to double, port cross section area has to double and port length has to stay the same ...

 

[marchel]

You're right , and I was wrong.

I've beeen through port calculation, And indeed, If you 2x the volume , 2x the driver, you should 2x the number of ports of the same legnth, to achieve the same tuning.

What I discovered is , If you 2x the volume and use a single port that you would use for the 1x box. Common sense says that you should cut the lenght of the port in half, To be able to achieve the same tuning freq. Due to the 2x increase in box size, But it's wrong. In my calculation, the 2x box would require shorter than half the length of the original port lenght for a 1x box. I'm not sure though , If I'm making sense to you.

 

[Svante]

In my calculation, the 2x box would require shorter than half the length of the original port lenght for a 1x box. I'm not sure though , If I'm making sense to you.

Yes, that is also due to the end corrections.

 

[sdclc126]

Let's do a real-life example:

I'm considering making dual/stereo sub/bass units for my speakers. The baffles are 7.5" wide, and I want to keep that dimension and mount the woofers on the front.

I am considering the Peerless 830946 -

http://www.madisound.com/catalog/product_info.php?cPath=45_228_256&products_id=8234

- using two of them in each cab wired in series to give 8 Ohms. The Madsisound page recommends a vented box of 0.4 cubic feet with a 2" diameter vent by 7" long, for an F3 of 44Hz.

So - for two units I want a cab volume of .8 cubic feet - I've got plenty of room for that. So - do I use two ports each 2 x 7, or one port 4 x 7 - will they both achieve the same thing? Or, taking into consideration end correction, what then?

 

[sreten]

So - do I use two ports each 2 x 7, or one port 4 x 7 - will they both achieve the same thing?

Hi,

Talk about labouring a point and then getting it all wrong,
Root2x2 x 7 is nearly equivalent to two ports each 2 x 7.
As stated above it will tune slightly lower than two spaced ports.

/sreten.

 

[sdclc126]

Hi,

Talk about labouring a point and then getting it all wrong,
/sreten.

That's why I asked for an example.