Are you asking how to figure out the angles or how to arrange to
hold the speakers in position?
One pattern you could use with 30 instead of 32 speakers would be to put
speakers at the midpoints of the edges of a regular dodehedron. (there are
30 edges). This is a quite regular pattern though these are not the
vertices of a Platonic(regular) solid(none of the five has 30 vertices!).
Calculating the angles (eg Euler angles) of these
is not hard. Use vector methods. You can look in Coexeter's book
Regular Polytopes(free on line as a pdf) if you do not feel like doing it
yourself. I think it is all worked out in there. Coxeter will show you
lots of quite regular patterns!
For the dodec. vertices(and hence edge midpoints) look here
http://paulscottinfo.ipage.com/polyhedra/platonic/dodecahedron.html
(Scroll down)
Robert
On Fri, 11 Jul 2014, Steve Boardman wrote:
Message: 4
Date: Wed, 9 Jul 2014 19:48:27 -0400
From: Marc Lavall?e <[email protected]>
To: [email protected]
Subject: Re: [Sursound] Calculating speaker placement
Message-ID: <20140709194827.694b2639@telecino>
Content-Type: text/plain; charset=US-ASCII
Hi Steve.
You can use "golden rectangles" (of ratio 1/1.618) to calculate
placements of your speakers. You can refer to:
https://en.wikipedia.org/wiki/Icosahedron
https://en.wikipedia.org/wiki/Dodecahedron
https://en.wikipedia.org/wiki/Golden_rectangle
--
Marc
Hi Mark
Of course, but not sure how easy this may be in practice.
Would I use the first golden rectangle on the smallest plane, and intersect the
others with that. Then use each rectangle corner as a line from centre until it
hits reaches a wall and then mark the speaker position?
The problem I have is the room has a sloping ceiling, low at front and then
high at the back. I would prefer to extend the angles and attach speakers to
the boundaries rather than build a frame to hold them, as that would use up
space and become an obstruction. It is also easier to attach to walls and
ceiling.
I was thinking of having the face of a Dodecahedron on the floor. This way
there will be less obstruction in the room and I will only have to embed one
speaker in the floor (i'm using both the vertices and faces of dodecahedron).
Does anyone know of a simpler and maybe more accurate method?
Thanks
Steve
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