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25th February 2011, 09:28 AM  #1 
diyAudio Member
Join Date: Oct 2008

Tangential room modes, calculating frequency, propagation etc.
By my understanding, the travelled path of a diamond between the side walls and the front/rear walls (mode 1,1,0?) will be longer than any single axial dimension and yet, the frequency of a tangential mode appears to be higher.
Does this have something to do with the propagation pattern...that it can't just be seen as a diagonal mode? I'm using this info to set up subs so I'd like to understand where the nodes would be and at what frequency. 
25th February 2011, 09:50 AM  #2 
R.I.P.
Join Date: Nov 2003
Location: Brighton UK

Hi,
As I understand they are lower not higher, take a 3x4x5 room  3 main modes are related to 3, 4, 5 lengths. 3 double modes (pythagoras) related to 5, root 34 and root 41. Lowest triple mode is all three axis, root 50, very near 7. So if the three main modes are 60, 45 and 36 Hz, Secondary modes are 45 (again, not good), 31 and 28 Hz. The triple lowest mode is 25.5Hz. rgds, sreten. These are of course the fundamentals of each mode, adding in the harmonics does considerably complicate the picture. Last edited by sreten; 25th February 2011 at 09:53 AM. 
25th February 2011, 10:07 AM  #3 
diyAudio Member
Join Date: Oct 2008

I've taken a screen shot of an online calculator that suggests otherwise. I became suspicious of a formula I noticed on another site but many calculators seem to be giving the same result.
(The tangential modes are the middle height bars on the graph.) 
25th February 2011, 11:07 AM  #4 
diyAudio Member
Join Date: Oct 2008

It is not the overall path length that matters, but the distance between the surfaces along the path (the arms of the diamond).

25th February 2011, 11:21 AM  #5 
diyAudio Member
Join Date: Oct 2008

Hmmm, now I recall something. Is this because a pressure node must be found at a boundary in a standing wave, which therefore can only happen when there are at least four wavelengths in the round trip?
EDIT: this still doesn't match the chart? Last edited by AllenB; 25th February 2011 at 11:25 AM. 
25th February 2011, 11:28 AM  #6 
diyAudio Member
Join Date: Nov 2009
Location: The Mountain, Framingham

Thats the way the math works out. The Tangential and oblique modes are higher. In the case of the tangential modes the path length isn't from corner to corner, a longer dimension. It is more like bouncing from mid wall point to mid wall point, a shorter overall path. Hence frequences are higher.
The math has a component: square root sum of the inverse of the squares of the 3 dimensions, so going from first length mode to first length and width tangential mode is a higher number or necessarily a higher frequency. Longer dimensions may lower frequency but adding consideration of width or height will raise frequency. David S. 
25th February 2011, 11:42 AM  #7 
diyAudio Member
Join Date: Oct 2008

OK, so if i have L=7.2m and W=3.6m and,
F = c/2 * sqrt( 1/L^2 + 1/W^2 ) then F = 53.4 So one wavelength would be 6.44m, but one arm of the diamond is 4.02m. It isn't clear where the nodes would end up. 
25th February 2011, 12:11 PM  #8 
diyAudio Member
Join Date: Nov 2009
Location: The Mountain, Framingham

Yes, your math is correct. It isn't clear to me why the apparent "path" gets shorter when width is added. In the limit as the room gets narrower the tangential mode is the same as length. Boncing around the room mid points is the same path as the longest diagonal.
Huh? 
25th February 2011, 01:32 PM  #9 
R.I.P.
Join Date: Nov 2003
Location: Brighton UK

Hmmm......
So if its centre to centre distances I'm missing a 1/2 scale factor. The double and triple modes would be double what I said ..... rgds, sreten. 
25th February 2011, 01:48 PM  #10 
diyAudio Member
Join Date: Oct 2008

I found this look at figure 5
In my example, the wavelength of 6.44m doesn't fit either the room diagonal of 8.04m or one side of the diamond at 4.02m 
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