2 Box Subwoofer Array

Kip Conner

Junior
Mar 13, 2011
370
0
0
Athens, GA
I'm attempting to do a 2 box sub arrangement to do some low frequency steering and need some help with my math. I used this article as a guide:

http://www.dasaudio.com/wp-content/uploads/2011/12/Cardioid_Subwoofer_System_V11.pdf

Since I'm a dumb american I decided to do my measurement in feet and inches and choose the upper xover point to find my 1/4 wave length. So, the wavelength is equal to the speed of sound divided by the frequency. I converted the speed of sound from 1130 ft/s to 13560 inches/s. So 13,560 divided by the upper xover point (105Hz) is equal to 129.14 inches. Divide that 4 to get the 1/4 wave length gave me 32.28 inches. This was the distance I used to separate my two cabinets (from front of box to front of box)

I did what the author suggested and measured from the backs of the box. Thus meaning that once I set my reference delay time in smaart for the front box I had to apply delay to the rear box (or in this case the box closest to the mic) to re-align the phase responses of the two sub boxes and decrease the amplitude to match the output levels.

The final step was to invert the rear box (again, the box closest to the mic) and the cardioid pattern should have presented itself in the acoustic realm.

Nope. I checked my procedure and math and can't find my errors.

Is there a better way to do this than as the article suggests?
 
Re: 2 Box Subwoofer Array

I'm attempting to do a 2 box sub arrangement to do some low frequency steering and need some help with my math. I used this article as a guide:

http://www.dasaudio.com/wp-content/uploads/2011/12/Cardioid_Subwoofer_System_V11.pdf

Since I'm a dumb american I decided to do my measurement in feet and inches and choose the upper xover point to find my 1/4 wave length. So, the wavelength is equal to the speed of sound divided by the frequency. I converted the speed of sound from 1130 ft/s to 13560 inches/s. So 13,560 divided by the upper xover point (105Hz) is equal to 129.14 inches. Divide that 4 to get the 1/4 wave length gave me 32.28 inches. This was the distance I used to separate my two cabinets (from front of box to front of box)

I did what the author suggested and measured from the backs of the box. Thus meaning that once I set my reference delay time in smaart for the front box I had to apply delay to the rear box (or in this case the box closest to the mic) to re-align the phase responses of the two sub boxes and decrease the amplitude to match the output levels.

The final step was to invert the rear box (again, the box closest to the mic) and the cardioid pattern should have presented itself in the acoustic realm.

Nope. I checked my procedure and math and can't find my errors.

Is there a better way to do this than as the article suggests?

Hi Kip,

your math seems okay, as far as a dumb German can judge this

two thoughts:

try a frequency in the middle of the sub passband, maybe 70 Hz not 105

and by inverting I hope you refer to polarity ?

Uwe
 
Re: 2 Box Subwoofer Array

You can check your work by inverting polarity of the boxes, to purposefully have the cancellation in the audience rather than on stage (if tuning from FOH area), adjust the delay times until you get as close to full cancellation as possible. When you find the best delay time, invert polarities to put cancellation back towards stage. Agreed on choosing 60-70Hz center.

This gets iffy if the subs are near a boundary, such as backed against a wall.
 
Re: 2 Box Subwoofer Array

Look for Bennett's article. The three box array he described can effectively be done with 2 boxes.

Yep, you just have to do a little gain matching. This is easiest with Smaart, measure behind the array a reasonable distance away (maybe 10') at a midpoint between your two boxes. Align them with delay, match their acoustic output (you will likely need to turn the rear box down), then polarity invert the rear box. It should be immediately obvious.

http://www.bennettprescott.com/downloads/AdvancedSubwooferArraying.pdf