I always meant to complete a speaker position detector using either a
Soundfield (or Bramha or Tetramic). By feeding each speaker in turn with a
PSRB and cross-correlating that with the outputs of the mic both the
position and the distance of each speaker could be acquired. If you don.t
have a B-format mic then four simple omni mics at the corners of a
tetrhedron (or related shape) could use relative timings to do the same
thing. Heck. if you are really cheapskate you could do it with one mic oved
to four positions, say on the arm of a boom stand.
Dave
On 13 July 2014 17:09, Marc Lavallée <[email protected]> wrote:
> Fri, 11 Jul 2014 21:12:42 -0400, Ronald C.F. Antony wrote :
> > There's a laser distance measuring device from Bosch with built in
> > incline measuring aka electronic level. Not too expensive and useful
> > for many other things as well.
> >
> >
> http://www.amazon.com/gp/product/B005AZZNXE/ref=as_li_tl?ie=UTF8&camp=1789&creative=390957&creativeASIN=B005AZZNXE&linkCode=as2&tag=cubiculumsyst-20&linkId=BE6RN3HLUWWGVJYK
> >
> > There's also a bundle with an aluminum bar that turns it into a
> > level, but for this purpose the device tripod-mounted would be good
> > enough, as long as the tripod head has markings for the horizontal
> > angles.
> >
> > Ronald
>
> The Bosch tool is nice and affordable, but maybe difficult to use.
>
> One (overkill) option would be to use a theodolite:
> https://en.wikipedia.org/wiki/Theodolite
> Some theodolites have an integrated laser pointer,
> like the Topcon DT-209L, but it is very expensive.
>
> A much cheaper (DIY) solution: a (large enough) dodecahedron made of
> cardboard could work as a guide for a laser pointer, by cutting holes
> at the vertices and faces. Inserting the laser pointer in two opposite
> holes would project a dot in the desired direction. The appropriate
> laser pointer should be in the form of a long tube; a small laser
> pointer could be inserted at the end of a tube. Even if not precise, the
> resulting laser pointer could be rotated to project a circle, its
> centre being precisely the right spot.
>
> In order to register (and remember) the centre position in the room, I
> suggest to first find its projection on the floor and mark it. Then
> find its projection on the ceiling using a plumb-line, and attach a
> hook to later suspend the plumb-line at the same position. Do the same
> on two side walls, using a level meter, and attach hooks for a string.
> The intersection of the vertical and the horizontal strings is a
> reference.
>
> On 11 Jul 2014, at 20:43, I wrote:
> > First you need the angular positions of the loudspeakers from
> > the listening spot. It shouldn't be too difficult to calculate for
> > your layout, knowing the properties of the dodecahedron (and some
> > trigonometry).
>
> Angular coordinates are usually required to design Ambisonics decoders.
> Here's how to find them from the cartesian coordinates:
> https://en.wikipedia.org/wiki/Dodecahedron#Cartesian_coordinates
> It's a matter of converting the cartesian ratios to polar coordinates
> (azimuth and elevation) in degrees, using the arctangent (inverse
> trigonometric) operation.
>
> According to:
> https://en.wikipedia.org/wiki/Ambisonic_decoding
> "The coordinate system used in Ambisonics follows the right hand rule
> convention with positive X pointing forwards, positive Y pointing to
> the left and positive Z pointing upwards. Horizontal angles run
> anticlockwise from due front and vertical angles are positive above the
> horizontal, negative below."
> So in the picture of the dodecahedron vertices (on Wikipedia), z becomes
> y and z becomes y.
>
> The golden ratio is phi, After conversion to polar coordinates, the
> angles for the green, blue and pink vertices are: a=31.7 and b= 58.3.
> The angle for the orange vertices (the cube) is (obviously) c=45.
>
> Here's the conversion from cartesian to polar coordinates:
>
> For the green vertices (0, +-phi +-1/phi):
> (90,a), (90,-a), (-90,a), (-90,-a)
>
> For the blue vertices (+-1/phi, 0, +-phi):
> (0,b), (0,b+2*a), (0,-b), (0,-b-2*a)
>
> For the pink vertices (+-phi, +-1/phi, 0):
> (a,0), (-a,0), (180-a,0), (a-180,0)
>
> For the orange vertices (+-1, +-1, +-1):
> (+c,+c), (+c,-c), (3*c,+c), (3*c,-c),
> (-c,+c), (-c,-c), (-3*c,+c), (-3*c,-c)
>
> I hope there's no mistakes; it's easy to double-check...
>
> Steve wrote:
> > 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).
>
> Calculating the positions of the described layout, rotating it to get a
> pentagonal face on the floor, is left as an exercice. ;-)
>
> But I would not recommend it, because it would mean more speakers on the
> floor, with more possible obstructions from the listening chair and
> listener's body. Worst: it would make the calculation and placement more
> difficult and counter-intuitive.
>
> I would simply rotate the hexagon by 90 degrees (horizontally), in order
> to get only two speakers on the floor and two on the ceiling (the blue
> vertices), on the right and left sides of the listener. The speakers on
> the side walls would on the pink vertices. The front and rear speakers
> would be on the green vertices (leaving room for a possible
> television/computer screen). The other speakers are on the cube formed
> by the orange vertices, and could be installed further apart. Think of
> the vertices as elastic sub-layouts; each of them is a valid
> Ambisonics layout. That said, I don't know how much elasticity is
> allowed by decoders.
>
> > Does anyone know of a simpler and maybe more accurate method?
>
> Another solution is to create an irregular layout with three
> horizontal layouts (middle, upper, lower), with more resolution for the
> middle layout.
>
> > Thanks
> >
> > Steve
>
> Good luck!
> --
> Marc
> _______________________________________________
> Sursound mailing list
> [email protected]
> https://mail.music.vt.edu/mailman/listinfo/sursound - unsubscribe here,
> edit account or options, view archives and so on.
>
--
As of 1st October 2012, I have retired from the University.
These are my own views and may or may not be shared by the University
Dave Malham
Honorary Fellow, Department of Music
The University of York
York YO10 5DD
UK
'Ambisonics - Component Imaging for Audio'
-------------- next part --------------
An HTML attachment was scrubbed...
URL:
<https://mail.music.vt.edu/mailman/private/sursound/attachments/20140713/e892d105/attachment.html>
_______________________________________________
Sursound mailing list
[email protected]
https://mail.music.vt.edu/mailman/listinfo/sursound - unsubscribe here, edit
account or options, view archives and so on.
