Line array size vs near field/far field transition distance

[wonderfulaudio]

Why does the near field/far field transition distance of an array depend on the height squared and not on the height?
d=1.5fh2. Its not intuitive.

Thanks,
WonderfulAudio

 

[Jim Griffin]

You can do your own research if you wish. Start with my NFLAWP paper which includes the approximate equation d=1.5 x f x h^2 in Table II on page 7. https://audioroundtable.com/misc/nflawp.pdf

I include some references of note (my references #1 and 2). Reference 1 is to the Urban et al work of L'Acoustics work reported at the AES Convention (New York) in 2001. Go to the 1992, 2001, and 2003 "Wavefront Sculpture Technology" papers at:
About us - R&D fields - L-ACOUSTICS

Click on the links within the L'Acoustic link to get pdf links for their papers. My approximate equation was developed from their work.

Follow the Ureda (reference 3 in the NFLAWP) for JBL/Harmon's work on line arrays. That paper is: http://www.jblpro.com/pub/technote/convention_paper_5304.pdf

Later in 2004 work by researchers in the Netherlands follow similar lines here: https://hectorsaavedra.files.wordpress.com/2012/03/linearraytheory.pdf

Have fun reading through the references.

 

[weltersys]

Why does the near field/far field transition distance of an array depend on the height squared and not on the height?
d=1.5fh2. Its not intuitive.

I don't find formulas "intuitive" until plugging in enough examples, then they become "obvious" ;^).

The approximate equation as given by Jim in post #2:
d=1.5 x f x h^2 requires “f” to be in kHz, not Hz, an important distinction!

"f", if substituted for the wavelength of the frequency in question (“λ”) would be the denominator for 1.5xh(length) squared, not another multiplier.

The near field in a line array exists for any frequency and distance where moving farther away will result in more drivers summing coherently (close to “in phase”), instead of incoherently. Once all the drivers sum coherently, the far field is attained.

From:
Line Array Limitations
By Dale Shirk

"The near field distance can be defined by the following relationship:
D=1.57 L squared/λ

where

D is the distance to the far field transition
L is the physical length of the line source
λ = the wavelength of the frequency in question (all lengths in identical units).

Beyond this distance the listener is in the far field and there is 6 dB drop in level per doubling of distance. The transition distance can be quite long at short wavelengths, that is, high frequencies, but it is shorter at low frequencies. For each octave lower in frequency, the transition distance is cut in half."

 

[Jim Griffin]

Yes sir. You have to have the correct units each definition to compare. Apples to oranges won't work.

My equation is good for frequency in kHz and height in meters. That pesky speed of sound in the medium (I use air and it is 343 meters/sec) can trip you.

My previous reply timed out before I added my reference to the L'Acoustics 2001 paper

About us - R&D fields - L-ACOUSTICS

See page 2 of the 2001 paper for their equation. I assume that the square root multiplier on the right trends to 1--especially as height and frequency increases. That assumption yields my simplified equation.