jostenmeat (08-21-2008)
A sealed box will absorb differently than a box with holes. With approximately 5.5" of depth, no fuzz, and no holes, the result will be narrowband absorption centered at a (more or less) single frequency. With holes only, the same result, but at a different frequency. (Or frequencies if different sized holes are used.) With holes and fuzz, it will be more broadband, with lower overall absorption. This is generally the preferred approach (more forgiving for a wider range of possible LF problems), though it will be room dependent.
Best regards,
Jeff D. Szymanski
jostenmeat (08-21-2008)
jostenmeat should be listened to
Got it, thanks for the lucid explanation. I can't believe my riser may also get upgraded, lol. Nothing is immune from upgraditis!?!![]()
Could you please direct me towards Helmholtz formulae, explanations for laymen, and even really nice would be a plug 'n play calculator! (I was never very good at math).
Thanks a lot, Savant.
-jostenmeat
Hmmm... That's actually a tougher question than it looks. I recently wrote a chapter on acoustical treatments for a textbook that will be released in October (more details when it comes outCould you please direct me towards Helmholtz formulae, explanations for laymen, and even really nice would be a plug 'n play calculator! (I was never very good at math).). In my research, I found that there are many inconsistencies with the published formulae on Helmholtz resonators. It's not that the equations are wrong (far from it, in fact); they are simply derived and hence expressed differently, or they use the author's preferred symbology (for which there is no standard), or there are even multiple approximations for opening effects. In other words, it's not as easy as the familiar speed of sound / frequency / wavelength relationship: c = f*λ.
But that's the dodging answer.
The best resource, IMO, is Cox & D'Antonio. However, that book is (a) expensive, (b) very detailed, and (c) probably not what you're looking for if you're mathematically-challenged.
The other references I have are even more complex than Cox & D'Antonio.
So, in order to spare you a formula and subsequent explanation, I will post a spreadsheet in this post later. (It's on a different computer.)
Best regards,
Jeff D. Szymanski
jostenmeat (08-21-2008)
jostenmeat should be listened to
That is a lot, over half the cost of the riser. FWIW, I have a couple of friends with solid math background. One I see rarely, the other I see pretty often, and he has a masters degree in stats. Maybe he could decipher what you throw at me.
Thanks.![]()
Great article! I'm tempted to tackle this project now.
jostenmeat should be listened to
meat,
Apparently, I cannot edit the previous message anymore? (Don't know if there's a time limit on that...)
Anyway, I've obviously taken a while to get back to this and for good reason: I'm hopefully going to have quite a useful spreadsheet tool that anyone can use to calculate the resonant properties of a riser or riser cavity.
Since that is, naturally, taking more time than I'd anticipated, I will at least post the basic formula for calculating the resonant frequency, fr, of a Helmholtz resonator, which is what is created by punching a hole in the end of a joist cavity in a riser:
fr = (c/(2*pi))*[sqrt(S/(V*L'))] ................... Eq. 1
where:
c = speed of sound in air
S = area of opening
V = volume of cavity
L' = effective length of opening
There are two tricky parts for Eq. 1:
1. Mind your units. E.g., if you calculate V in ft³, use c in ft/s, S in ft², and so on.
2. The effective length of the opening is calculated by:
L' = L + (x*d) ................... Eq. 2
where:
L = actual length of the opening
x = mass correction term (unitless)
d = diameter of the opening
I have seen values of x between 0.70 and 0.85. I like x = 0.80, which should get the resulting fr in the ballpark.
For the example case of a riser joist cavity, where a hole is being cut or drilled into a "2x" piece of lumber, L would equal 1.5 inches and d would be the diameter of the hole. (Think of the hole in the joist as a cylinder with diameter d and "height" L; I'm calling the "height" the length here, just to confuse.)
Calculating fr, it will become apparent that it can change significantly for small changes in opening or cavity properties, especially since S and L' are interrelated. I hope to have (much later) a spreadsheet with some macros that will optimize the hole diameter, even if the plan is for more than one hole to open up the cavity. If you (or your math friend) come up with anything in the meantime, let me know.
Best regards,
Jeff D. Szymanski
jostenmeat (09-04-2008)
jostenmeat should be listened to
Savant, so I printed out your post, and checked it out last night. Seems* straighforward, on paper. The variable being port size. I want to list some things that I am mulling over, and trust me, I'm a mega-noob. Never built speakers either.
- the idea of placing a port of any particular length. A new variable?
- recommend any specific material for a port?
- the actual frequency that I would like to target.
- or frequencies, if I was to use multiple ports of varying size (and the accompanying complex math that two different ports would create).
- the effect of significant fill in relation to your mathematical explanation
- the idea that this math works when the riser is thought of as completely inert
just blabbering. In regards to targeted frequency, am I to have better luck with lower bass, as opposed to high bass? For instance, is tuning for 110hz going to be as effective as tuning for 15hz?
Can't wait to see the spreadsheet.oh, I'm patient.
![]()
No. By "port," I assume you're talking about the opening. Consider a joist cavity as a rectangular box. Drilling a 2" diameter hole anywhere on the outside of the box creates an opening, or "port." If the hole is drilled in a joist, the length is the thickness of the joist that's been drilled through. If the hole is drilled in the top of the box, the length is the thickness of the materal that's been drilled through.
Now, if a tube of some sort is fitted into the hole that's been drilled, the length of the tube is the new length of the opening, or "port." This length equals L in Eq. 2 above. The only tricky part would be accounting for a new diameter if the tube has a non-negligible wall thickness.
If the "port" is simply a drilled hole, there is no other material necessary. If some sort of tube is inserted in the hole, it could be made of anything, really. Piece of PVC, old TP roll, whatever.- recommend any specific material for a port?
If the "port" is a rectangular opening, it could be fitted with a vent cover, like something that would be used to cover an a/c opening in a wall or ceiling. I've done this before. The math is all the same (but with a different way to calculate L'), with some allowance necessary for the restriction introduced by slots in the vent cover. A little trickier, but it can be done.
This is a good question. If you have a side-to-side axial mode, that might not be a good frequency to target, especially of the "ports" will be across the face of a riser. The best application, IMO, would be for floor-ceiling or front-back axial modes. For the latter, ports on the back of the riser would be best...which may mean they need to be in the top of the riser if the riser is up against the back wall.- the actual frequency that I would like to target.
I've also done this. I.e., different joist cavities to address different frequencies. Careful planning of where to put which "ports" is needed in this case.- or frequencies, if I was to use multiple ports of varying size (and the accompanying complex math that two different ports would create).
Also an excellent point. Fill for a Helmholtz resonator generally broadens its frequency range while reducing the overall absorption. The center frequency of absorption will still be fr. Full (100%) fill is generally discouraged as is fill right behind the "port." I've had best results with approximately 50-75% loose fill that doesn't come within a few inches of the "port."- the effect of significant fill in relation to your mathematical explanation
If by "inert" you mean "rigid and sealed," then yes.- the idea that this math works when the riser is thought of as completely inert
In theory, you will get the same effect regardless of frequency. Running through the math, you will find that tuning a joist cavity to something like 110 Hz is going to be significantly more difficult than something like 15 Hz, simply due to the volume you're dealing with. Low bass will generally be easier with something on the order of a riser joist cavity. However, planning ahead for segmented joist cavities, each with their own "ports," could help get something into the higher bass range. But the real question at that point becomes whether you cannot address the higher bass with some good, thick wall panels or "traps."In regards to targeted frequency, am I to have better luck with lower bass, as opposed to high bass? For instance, is tuning for 110hz going to be as effective as tuning for 15hz?![]()
Best regards,
Jeff D. Szymanski
jostenmeat (09-05-2008)
jostenmeat should be listened to
Yes, regarding port length, I realized how redundant that was to this discussion . . . after posting. Thanks for addressing that nonetheless!
The reason I think of a different "port" than the actual joist is only for any possible flexibility in obtaining the desired Fr. I was indeed planning on the back side of the riser, and I have the option to make a hole through the thinner plywood that completes the seal, or through the ply + 2x6s that run horizontally at the top and bottom borders. I figured a port of desired length in the ply would be easy . . .
Regarding the different target frequencies among different joist cavities: I only have one cavity. It is simply one huge box. I suppose I could make more cavities. It might be difficult to produce an excellent seal. Maybe not. But, since I know nothing about this stuff, are we (or namely you) going to address multiple ports for the same cavity? Is this too wishful?
Thanks for the all of the various tips, everything from x=.80 to the idea of vent covers.
I am using traps, and have more on the way. I might even obtain round three of treatments, which would be for the ceiling, and not so much bass response.
Oh, yet another question comes to mind. For a very roughly estimated box of 70 cubic ft, how imperative is it to get the volume perfectly right, before the calculations? Of course I'd try to get the # as accurate as possible, but if I was off by, say, one cubic foot, how fruitless will my efforts become?
Could the target frequency change be very small, like a couple of hertz? I suppose with the spreadsheet to come, that I will find my answer then.
Thanks. You're the best.
-jostenmeat