If we have a sealed box subwoofer in a very small room, 8’x8’x8’, playing 40Hz wavelength 28’ or even lower, will a TubeTrap that absorbs 40 Hz or lower be useful?
Essentially, if a standing wave cannot be created, will TubeTraps do anything
Short answer is yes. If there are pressure fluctuations near the TubeTrap within its frequency range of operation it will absorb acoustic energy.
Longer answer includes the answer to another question: Is it useful to use a TubeTrap in such a situation?
Let’s go over the situation in slow motion. When the listening room is less than say 1/5th wavelength the concept of distributed acoustics tends to fade, transitioning in the “pressure pot” or “breathing mode” realm of acoustics. Envision a large hollow dumb bell, two spheres connected together with a large tube. One sphere might be 8’ in diameter and the other might be 4’ in diameter. The 8’ sphere is a model for a small room and the 4’ diameter is a model for a sealed box subwoofer cabinet. The tube between them is essentially the 15” woofer.
As the woofer shuttles air back and forth between the two sphere volumes. Here is the “standing wave” effect. At one instant we have the speaker cone pushing far forward, creating a positive pressure in the big sphere and pulling a vacuum in the smaller sphere.
At this moment we have no speaker movement, it’s fully extended, and we have no kinetic energy in the system but lots of potential, pressure energy. A quarter cycle later the piston if flying past the neutral point of the driver and the pressure in both spheres has been returned to Po, ambient pressure. When the pressures are zero, atmospheric, we have no PE and lots of moving air, which is KE, kinetic energy.
Another quarter cycle later the speaker has continued on, pulling further back until it reaches its maximum extension in reverse position, where the pressures in the two spheres are also reversed from before. Clearly, we have a resonance which alternates between potential and kinetic energy. The fraction energy absorption or leakage each cycle compared to the total energy of the system is the Q or damping factor of the resonant system. We know from experience that a Q of 10 seems to be reasonable enough in the realm of room acoustics.
There is a formula that relates room Q to Room RT60. I presented many room Q relationships in my second AES paper in 1986, Listening Room and Low frequency Acoustics. All my AES papers are available on the ASC web site. I didn’t cover this breathing mode of the room in the paper, however the formula doesn’t change.
Q = f xT60 / 2.2. At 30 Hz and with a room Q = 10 the RT60 = 2.2 Q/f = 2.2 x 10 / 30 = 2/3rds second. Pretty fast RT for 30 Hz.
So, without any energy absorbing system at work the speaker will simply oscillate between its inertial condition and the collective stiffness of the air spring of the two air cavities. And finally, we have the answer as to why adding a bass trap to this air vibration mode is useful, which is because it helps to dampen the spring in the spring-mass system.
This discussion has excluded any consideration of the walls, floor and ceiling as if the room was a solid concrete sealed room. Most room surfaces are not solid but vented through doorways and have flexible surfaces, like sheetrock and stud walls or windows. In general, TubeTrap or other bass traps do not significantly load with friction or dampen flexing wall, floor, and ceiling movement. This is why ASC developed the WallDamp system, which extracts energy through shear stress damping from each movement of constructed wall and ceiling assemblies, back in1987 and continues to provide WallDamp through this day.
How many TubeTraps will it take to get a Q of 10 or 15 in a small volume space?
Basically, it’s about the volume ratio between the volume of the TubeTraps and the volume of the room. Let’s say we are going for Q of 10. This means 1/10th or 10% of the energy must be removed each cycle. There are two pressure changes each cycle, one positive pressure and the other is negative pressure. The TubeTrap extracts energy out of each pressure change. That means the same TubeTrap absorbs energy out of the room twice per cycle. That means we only need 1/20th or 5% of the energy in the room volume to be removed each half cycle.
If the volume of the room is 500 cubic feet we need 1/20th of that volume performing 100% energy absorption each half cycle, which is 25 cubic feet of air volume. A 16×4 IsoThermal TubeTrap has 5 cubic feet of volume. It also should produce 50% efficiency at 30 Hz and so we need 50 cubic feet of this size trap in the room or ten of these units. We put two in each vertical corner and we are up to 8 traps already. That is a difference between Q = 10 and Q = 12.5, which is close enough.
A seasoned acoustic person might be thinking about the pressure boost, increased sound-absorbing power due to TubeTraps in corners. In the breathing mode, all pressure is same everywhere in the room, and so there is no pressure buildup in the corner.
Thanks for asking…..Art Noxon