> Thank for all this guidance. > > If the min is (N+1)². > Surely for 3rd order that is > (3+1)² = 16 speakers? > > This is what has been confusing me as I see reports of third order over 8 > speakers which seems to go against the rules. > > Cheers Garth
Eight 'is' a cube, the classic first order periphony. (I say 'is' as you can arrange eight in an infinite number of ways ;-)> (It is also a classic pantophonic ring ... which may explain the "reports"?) You have to face two other problems: 1) The number of symmetric solids with (n+1)^2 faces or vertices is fairly limited. You can depart from them, but it gets trickier. 2) Once you pass first order, you have to start (and I hyperbolise) digging holes in the floor, placing sound-transparent floors, placing overhead loudspeakers ... You can go for domes (hemispheres), but that rather leaves the original question behind. One compromise is to have two (or three) rings of loudspeakers. A cube is one such (and a 'perfect' solution for first order). Rings are practical to install. One argument in their favour is that vertical information is not 'as great' (a symphony orchestra has a wide soundstage, but not a very high one). The counter-argument is that, for a symphony orchestra, as the vertical differences are smaller, they need higher precision to catch the nuances ... A good point to discuss in a bar ... but for most people (I suspect) the mechanical problems of placing speakers are decisive. Not sure these musings take you much further ... Michael _______________________________________________ Sursound mailing list Sursound@music.vt.edu https://mail.music.vt.edu/mailman/listinfo/sursound
