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I am consistently obsessed with transmitting as much audible sound as possible. A lot of progress has been made with your help.

What I would like to ask is to increase the frequency moderately in order to transmit the audible sound further, and to increase the near field distance by arranging the loudspeaker in a rectangular shape. Given a frequency, loudspeaker dimensions and arrangement, how do we calculate the boundary between the near field and the circle field?

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  • $\begingroup$ Not sure what you mean with the "circle field", would you care to clarify a bit? Now, regarding the speaker arrangement, maybe you could go through the paper "Sound Field Calculation for Rectangular Sources" by Ocheltree and Frizzell (brl.uiuc.edu/Publications/1989/Ocheltree-UFFC-242-1989.pdf) where they give an analytical solution. I haven't go through the paper to be able to say how much of a help could be, but I believe it could act as a first order approximation for your speaker arrangement well below the spatial aliasing frequency. $\endgroup$
    – ZaellixA
    Feb 8 at 18:46

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The standard rule of thumb is that the transition between near field and far field is less than or equal to about one wavelength at the lowest frequency you wish to radiate. This means for maximum radiation, the dimension l of a square array of loudspeakers must be made about equal to the cutoff frequency f of the system. For f = 1000Hz, l is about one foot. For 100Hz, l becomes 10 feet and 50Hz yields l = 20 feet, etc.

You can see that to bathe the listener in near-field sound requires an enormous speaker array. The band Grateful Dead actually assembled and toured with just such a system in the early 1970's, for just that reason. The band members stood in front of a speaker array 20 feet tall which stretched across the entire stage.

It was loud and impressive, but ultimately impractical.

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  • $\begingroup$ Thank for your kind answer. But I am so unhappy about having doubt for your answer. If my comprehension is not incorrect, you mentioned that the higher frequency, the shorter the near-field, right? I have already known that the higher the frequency, the longer the near-field. What do you think about this? $\endgroup$
    – Tiny
    Feb 9 at 0:45
  • $\begingroup$ Hello Tiny, do you have any access to the standard texts on acoustic engineering? issues like this are simple to research if you have the best resources. I wish you could find those texts to study from (Beranek & Boyd is one), let me try to remember the others- Niels $\endgroup$ Feb 9 at 5:08
  • $\begingroup$ Thank you. Could you please tell me the best resources you mentioned? I can't understand what "Beranek &Boyd" is. $\endgroup$
    – Tiny
    Feb 9 at 8:00
  • $\begingroup$ Those are names of authors. Look for books written by Beranek or Everest, they are the standard resources in the field. $\endgroup$ Feb 9 at 18:29
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Borrowing from the RF world, a rule of thumb for defining the far field distance, $d_{ff}$, is: $$d_{ff} \geq \frac{2D^2}{\lambda}$$ where $D$ is the maximum linear distance of the radiating array and $\lambda$ is the wavelength of the radiated signal.

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    $\begingroup$ This is one of the many proposed near field - far field boundary distances in the literature. I just wanna add a minor detail which wasn't clear to me when I first started working in the field. The maximum linear dimension may not be one of the most intuitive ones. For example, for a rectangular array, like the one in the original question, this dimension may be the diagonal. It wasn't very clear to me at the beginning and I just wanted to add this tiny bit of information. +1 from me :). $\endgroup$
    – ZaellixA
    Feb 10 at 11:37

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