Room mode visualization attempt relative to bass trapping

[harmonic]

Hi All:

I have a question about room modes, and how they behave in general relative to bass traps. I'll give my room data, but I don't think it matters much for my real concern further below: drywall for the walls and ceiling, carpet floor, 4.3 m long, 3.9 m wide, 2.4 m high; and starting at 1.2 meters high along the parallel long walls, each wall turns in about 45-50 degrees from vertical, following this angle up to the ceiling (bye-bye easy mode calculations).

In any case, I'm trying to visualize general, generic room mode behavior in 3D in my head relative to what a pair of opposing, parallel bass traps will do to the modes.

Consider a generic rectangular room. The Hunecke modes calculator for such a room shows a nice visual of room modes at different frequencies that made me think of fog. In this analogy, if a room mode (standing wave) could be thought of as stationary fog, then if you walk around in this foggy room, the fog is cyclically dense throughout the whole room, with some areas being dark fog (peaks of the standing wave - overly loud at the mode frequency) and some areas of bright fog (nulls of the standing wave - overly silent at the mode frequency), neither of those two fog types being what we want instead, "clear air" (i.e. true/accurate level for that frequency). (this analogy also being applicable, with different peak/null spots, for each of the front-back mode, the side-side mode, the up-down mode). If this analogy doesn't work so far, please point out my misunderstanding.

Next, the listening position. While I'm sitting stationary at my listening position in the rectangular room, I don't want any room modes doing their nasty business in my ears. So maybe I'd start by following generic advice and pick a close to ideal sitting position given the room dimensions, then put some bass traps in the corners (pressure being highest there, etc.). However, my question is, if I put traps on the two walls opposite the sides of my head, on the two walls in front of and behind my head, and on the ceiling and floor above and below my head (ok, in reality I'd skip below my chair), would or would not each of these pairs of traps be "clearing fog" in the parallel space between them, creating something like cavities of non-mode air for my head to sit in? If not, where has my visualization/understanding gone wrong? For example, if my visual is off, maybe it's because standing wave "fog" has dispersant like properties? so the traps would reduce some fog, but the fog just immediately fills in the space again between the traps, albeit after the traps have weakened the overall strength of the fog in terms of total density for the room as a whole?

Thanks.

[Soundman2020]

Well, it's a little more complex than that, unfortunately. To start with, you are only talking about axial modes in your example, but in reality there are also tangential and oblique modes. Axial modes are ones that involve only two parallel walls, tangential modes involve any four of the six "walls" of the room (where "wall" also means ceiling and floor in this context), and oblique modes involve all six "walls".

Also, that diagram only shows a static state at a single point in time, and only for pressure, whereas in reality standing waves exhibit changes over time, as the energy on the wave moves around the room. And in addition, air pressure changes are only half of the energy involved in a sound wave: the other half is velocity. A sound wave has two components to it: pressure and velocity. Where pressure is high, velocity is low and vice versa. Think of a tennis ball for example: Throw it at a hard wall, and it bounces back. Draw a graph of velocity, and you will see that the velocity drops to zero as the ball hits the wall, but then increases again as it bounces back. You can't really talk about pressure from a single tennis ball, bit picture millions of them hitting the wall and bouncing around: The pressure where they hit the wall is high, but further away is lower. That's not a terribly good analogy, but it helps to get your head around the concept that a sound wave has both pressure and velocity components that are interchanging all the time as the wave propgates around the room, and interacts with boundaries.

neither of those two fog types being what we want instead, "clear air" (i.e. true/accurate level for that frequency).

You are talking about the instantaneous static picture of the room, again, but like I said before, it is changing all the time. The "dark" areas of your fog will fade to gray, to white, to gray and back to black again, while other areas will do the opposite. A standing wave doesn't really "stand" in the room, in the sense of it being static, with no pressure or velocity changes: If it did, then you wouldn't hear it! (Think about it...). The point about a standing wave is not that it stands immobile, creating a "frozen" pattern of high and low pressure areas that don't change (which is what you often see in text books), but rather that the peaks and troughs always fall in the same position in the room. So even though the "dark band" for a certain mode might occur at (for example) 50% of the height of the room, that does not mean that the pressure at that point remains static. Rather, it means that the pressure at that point will change in prefect synchronicity with the frequency of the wave (duh!), and that the peak value in the room for that frequency will be at that location. So if for example, we were talking about a 100 Hz. mode, then the pressure will still increase and decrease 100 times per second at that point (while the air molecule velocity does the same, but 180° out of phase). So the wave only "stands" in the sense that the light and dark areas mark the maximum and minimum pressure points, but the pressure is still changing. Sometimes I think that "standing wave" isn't a good term, since it gives the wrong impression about what the wave is actually doing.

then put some bass traps in the corners (pressure being highest there, etc.).

Depending on the type of bass trap you are talking about, that might or might not work! Some traps (such as membrane traps) work on the pressure component of the wave, so they do well right up against the walls. Other traps (such as absorbers) work on the velocity component, and they don't do so well against walls....

However, my question is, if I put traps on the two walls opposite the sides of my head, on the two walls in front of and behind my head, and on the ceiling and floor above and below my head (ok, in reality I'd skip below my chair), would or would not each of these pairs of traps be "clearing fog" in the parallel space between them,

No, because at best you'd only be treating axial modes whose peak pressure (or velocity, depending on your trap design) happened to fall at those exact points on the walls.

In any event, you cannot "clear the fog"! (If you did, you would have no sound in your room!) You cannot realistically get rid of a room mode: it will still be there, simply because it is related to the room dimensions, not the treatment. If you change the shape of the room, you don't' destroy modes: you just move them around, so now they fall at different frequencies.

Bass "trapping" is also a misnomer, IMHO: Traps don't really "trap" sound at all: they just convert energy from one form to another. Installing a tuned trap that hits, for example, 100 Hz. does NOT meant that it will kill any room modes at 100 Hz. It just means that it will reduce reverb times a bit at 100 Hz, which is good. It doesn't kill the mode: It just removes some energy at the same frequency as the mode, hence helping to knock a bit off the peak for that frequency, and smooth out the room response curve a bit.

The goal of room ratios and treatment isn't really to "clear the fog", nor is it to eliminate the modes: The best you can do is to try to reduce the intensity of the peaks and nulls, and to spread the modes around evenly, so that you don't have large frequency gaps between them, and neither do you have several modes that overlap the same frequency, or are really close to one specific frequency. You want a nice smooth spread of modes across the lower end of the spectrum. In fact, you want as many modes as you can get in the low end! The problem with small rooms is not that they have too many modes in them, but rather that they don't have enough! A small room cannot support modes corresponding to frequencies whose wavelength is longer than the longest path around the room, so it cannot support modes at low frequencies. This means that there are only a handful of modes available in small rooms, so the best you can do is to choose ratios that spread those precious few modes out as evenly as possible.

Too many modes is not a problem: they overlap and merge into each other. Too few modes IS a problem, which is one of the reasons why small rooms don't sound good: modes are few and far between. The ideal is that, as you go up the scale, each octave should have at least twice as many modes in it as the octave below. That is what a Bonello curve shows.

In your room, with your dimensions, that isn't happening. You have no modes at all below 40 Hz, and two large dips on your Bonello curve, where higher octaves have FEWER modes than the next lower octave. So your room does nto give you good, smooth, evenly spaced modes, but rather "clumped" modes. Your dimensions are not good.

The Schroeder frequency for your room is 128 Hz, which basically means that only above that frequency doe things start smoothing out, and modes are not too much of a concern any more. But below 128 Hz you have modal issues. You have a total of 18 modes below that frequency. In some places you have "holes" in your modal spread, such as for example at 107 Hz: the next lowest mode is 97, the next highest is 114. So you have big empty gaps in between. But then you also have the case at around 120 Hz, of three modes practically on top of each other: 119.2 Hz. is your 2,2,0 Tangential mode, 120.1 Hz. is 3,0,0 axial mode, and 120.6 Hz. is your 1,2,1 oblique, all within about 1 Hz of each other. So any note that excites one of those modes is likely to excite all three of them, while a note at 101 Hz is unlikely to excite any modes at all.

I'm not sure if that helped much: maybe I just confused you even more (I sure as hell confused ME even more! ), but the point is that sound waves and acoustics aren't very intuitive, and things are very dynamic, not static, which is what text books and charts often show: static images from one point in time.

So, getting back to your basic question:

In any case, I'm trying to visualize general, generic room mode behavior in 3D in my head relative to what a pair of opposing, parallel bass traps will do to the modes.

.. the answer is "basically nothing"! The bass traps don't do anything to the mode itself: they just remove energy from the room at that modal frequency, so the mode is less likely to be excited, and even if it is excited, it won't "ring" for as long as it would have without the treatment. But the mode is still there. It didn't go away just because you "treated" it.

Anyway, that's a very long-winded way of saying that the best way you can deal with modes in a small room is to get a bigger room that has more of them in it!

- Stuart -

[harmonic]

Thanks for the in depth reply!

I wasn't totally, er, clear, with my use of "clearing the fog", but I think you kind of addressed the underlying issue anyway. By wanting to "clear the fog" (for at least a pocket of the room where my head goes) I really meant wanting to get non-resonant wave behavior at the subject frequency (not absence of the wave) in that spot, but what I gathered from your explanation is that in a small room that will necessarily create a narrow band axial mode at frequency X, in practical terms (i.e. with bass traps < 1/4 wavelength thick) one simply cannot have that frequency in that room without substantial contribution from the mode too, i.e. if the speaker puts out frequency X, one cannot practically keep the original/source wave and ditch the narrow band mode build up. At best, using regular bass traps, is a slight reduction in resonant peak & null amplitudes for the resonance.

I'm further curious about the sound pressure vs. velocity part. You used tennis balls as a analogy (qualifying it as an imperfect analogy), with velocity dropping to zero at the wall and the pressure correspondingly going up. For a tennis ball itself though, the velocity would drop to zero at the wall because it is has elasticity, right? I would have expected sound waves to be more like millions of pebbles, losing essentially no velocity, only changing direction, when colliding with a surface. But alas, pressure is higher in corners, so I'm still missing something.

[Soundman2020]

I would have expected sound waves to be more like millions of pebbles, losing essentially no velocity, only changing direction,

The velocity still drops to zero at the point where the pebble hits the wall. It cannot possibly be any other way! It is moving forwards at 340 m/s, and then a moment later it is moving backwards at 300 m/s. so at some point it MUST have passed through 0 m/s. Try this: Get in your car, and drive forwards down the street at 10 km/h. Now drive backwards down the same street at 10 km/h, but WITHOUT STOPPING.... You cannot do that. It is impossible. You HAVE to stop. So do the air molecules that are transmitting the energy of the sound waves. Yet with sound waves it is different, since air is compressible. So the molecules start to decelerate before they hit the wall, as the pressure goes up. Then they come to a stop at the wall, zero velocity, where pressure is maximum, and they accelerate again as they leave the wall, velocity going up while the pressure goes down again.

But of course, to make things even more complex, this only applies to a wave that is absolutely normal to the wall. The vast majority of waves are not: they strike at all angles between 0 and 90°, so the effect changes depending on the angle of incidence. A wave striking at a glancing angle doesn't stop, and it's speed only changes slightly, but neither does the associated pressure increase much.

And then to make it REALLY complex: real world sounds are not perfect individual sine waves, but a huge mess of interacting waves, with all kinds of shapes, frequencies and wavelengths, all muddled together, and going in different directions at different angles. And add in the fact that wavefronts are not flat, but curved, and that directionality depends on frequency, and that temperature and humidity also play a role, plus you have diffraction, refraction going on as waves interact with other objects in the room, with the degree of diffraction and refraction once again being governed by the size of the objects and the relative dimensions of the wavelengths involved, .... So it really isn't simple at all. !

For a tennis ball itself though, the velocity would drop to zero at the wall because it is has elasticity, right?

Air molecules act pretty elastically! If they didn't, then sound wouldn't reflect off hard surfaces. Collisions between air molecules are actually almost perfectly elastic. Once again: if they weren't then the relationship between temperature and pressure would be really strange. The equations in physics that describe how gases behave would be drastically different from what they are.

Think of air not as pebbles bouncing around, but as perfect "superballs" bouncing around. Remember "superballs"? Little rubber spheres, about the size or marbles, that bounce like crazy? Well imagine that superballs could be made to bounce even more perfectly. Imagine putting thousands of them in a very large glass box, and shaking the box like mad to get them all going. Now observe. That's pretty much how air is, on the microscopic scale.

(i.e. with bass traps < 1/4 wavelength thick) one simply cannot have that frequency in that room without substantial contribution from the mode too,

Well, there is still the issue of "exciting" the mode, which is rather nebulous in itself. If you play a pure tone corresponding to a mode at a level of 30 dB (C), then most likely the mode will not add anything to the level, and the decay will be fast once the tone stops. But play they same tone at a level of 120 dB (C) and the mode certainly will add something, storing energy at that frequency, then releasing it slowly even after the tone has stopped. The level at which each mode is triggered or excited is different. So playing the same piece of music loudly or softly could also sound different in the room, depending on which modes are doing what and at what level. I don't really understand this part of modal behavior ("excitation") very well, but I do know that it is an issue. Check it yourself: Set up your bass guitar and amp in the room, and play a note that is close to one of your axial modes. Play it very softly, and nothing happens: the room sounds fine. Play it very loud, and the room resonates along with you, because you have "triggered" the mode. But what actually "triggers" it, and why that occurs at some levels and not others, is a bit of a mystery to me.

So it isn't just a matter of all the other stuff I mentioned, but also a matter of level.

I wish acoustics were a bit easier than this!


- Stuart -

[seamus]

<snip>It is moving forwards at 340 m/s, and then a moment later it is moving backwards at 300 m/s. <snip>

Stuart, how did you arrive at this conclusion?
Is there a formula for calculating velocity loss as it pertains to angle of incidence?

Thanks,
Seamus