Relation between Vas and Enclosure volume

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I have a big question.
How big can be the volume of the box for a given Vas of the driver? Nobody seems to treat this subject on the forum. Does it matters if the box is sealed or bass-reflex?
Are there any clear formulas or just an rule of thumb?
 
My problem is with a ported box.
The Vas of the driver is 4,4 l and Sd=65 cm^2.The box volume is 27 l
In any simulation that I made it (WinISD, MJK MathCAD sheets), The Fb of the box should roll off around 55 Hz...But my FR plots show that actually it rolls off at 85 Hz (very close to the Fs of the driver). So I'm questioning myself if the problem is not the Volume Box of the driver, 6 times bigger than Vas. Why in any simulation that I made this problem is not appearing?
 
Let's try to straighten out some terminology and a few realities:

Fb is a box tuning frequency. It is a single number, hence can't "roll off." You can determine it by looking at the impedance curve.

The box tuning frequency does NOT necessarily correspond to the box/driver's rolloff frequency except in the special case of a discrete alignment.

With a tiny Vas like 4 liters, actual boxes will not have the effective volume you think they have, especially once you mount a driver and a port into the box. Presumably, you'll want to add some damping materials on the walls so that driver reflections don't come back through the cone (a real problem in small boxes), and that will affect both the effective volume and the box Q.

Driver parameters are not generally the same as they are on spec sheets. Vas in particular can vary wildly, even with small changes in temperature.

Trying to get a driver with such a tiny Vas to produce bass much below 100Hz is going to be an exercise in frustration.
 
SY,
You are right about last thing:
<Trying to get a driver with such a tiny Vas to produce bass much below 100Hz is going to be an exercise in frustration.
My common sense tells me the same think. I know the box is just too big and probably is impossible to reach 55 Hz with this small driver. BUT I want an answer from the mathematical and physically point of view. I want to know why in every simulation that I made it the port response looks very good, when actually should be almost negligible???
I want to know if is there any rule for Ported Box regarding VAS and volume of the enclosure. For example: “You cannot put a driver in a ported box with VAS smaller than half of the enclosure volume”
You see, this questions really frustrates me for the last week J
 
SY's Third Law

"An analogy is not an equality."

Because simulations assume the correctness of the assumptions. Let's take an example: shape the cabinet so that it's really, really thin and really, really tall. The simulation will show no difference from a box with normal proportions. Measure something like that in reality and you'll find that the model is not predictive.

In this case, the cabinet is effectively infinite in size. The driver can't effectively drive the slug of air in the port- it's like trying to push a golf ball uphill using a soggy piece of spaghetti. The assumptions in the Thiel-Small model (note that last word!) are not met.
 
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Joined 2001
Well, I cannot tell you precisely how big an enclosure has to be compared to Vas of the speaker before the rule book gets thrown out.

But I can tell you why nobody has bothered to investigate much, if that puts your mind at ease.

If you take a look at the Thiele-Small charts, you will see that there really are not any useful alignments which require much over Vas times 2. After that, distortion takes over.

The famous audio writer David Weems tried this out. He took a speaker with the Qts of 0.6, put it in a box over 2 times Vas, and pronounced the distortion too high to be useful. Thiele-Small were being generous when they put their last useful alignment at Qts=0.57.

You see, from a practical standpoint, nobody even wants to build a reflex box with a volume much over Vas times 2, in fact, most people don't even want to go that big in the first place.

The larger the box volume compared to Vas, the higher the distortion and the worse the transient response. So somewhere above Vas times 2, the Thiele-Small parameters mean progressively less and less until they don't matter at all. Few have bothered to investigate, or at least publish their findings, because if you want to build a good reflex, you aren't going anywhere near those enclosure sizes anyway.

Perhaps you can add blocks of wood and bricks to your enclosure to reduce volume and let us know where the Thiele-Small parameters kick in. It might be fun, and you already have built the test enclosure. :)

Below is a graph for the step responses for various Thiele-Small alignments. The sloppier the response, the worse the distortion.

I took this chart from Small's article on vented boxes. He has charts that deal with K and B in his article, so I took it from there. All the colored printing is my addition. It was unclear to me what the Keibs and SC4 alignments were, so I added no additional info. However, four of the charts I added the Vas÷Vb ratio, and the F3÷Fs ratio.

a=(Vas÷Vb)

Note the response gets smoother and less distorted the smaller Vb is compared to Vas, and the higher F3 is compared to Fs. Of course, this is the precise opposite to the way you tried to build your box.
 

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Joined 2002
It seems to me (being at work and not having any of my resources at hand) that as the box gets bigger (compared to VAS) you start to "grow" into an Infinite Baffle design, which will be akin to the open baffles that been all over the forum for the past months. The drivers response will start to resemble that of the driver in free air (fc=fs).

I'm not quite sure what distortion your speaking of KW, but I would think that the level of distortion will depend a lot on the quality of the driver. If it were strictly an alignment issue then Bozak and Linkwitz were mistaken in some of their efforts.
 
You also need to quote the Fs and Qt of the driver to allow analysis.

One of the major issue of commercial speakers is volume effiency.
What can you get out of a box of a given size.

Normally the Box size is much less than Vas and this dominates,
the real issue being the Qt at the open air resonant frequency.

Using extreme alignments the following is generally true :
going beyond Box volume = Vas is generally not worth it,
the driver suspension begins to dominate.
Once box volume = 2 Vas then there hardly any point going any
larger, a 2 Vas box will give uncontrollable reflex alignments.

E.G. for a Tb driver Vas=3 litre, Fs = 81 Hz and Qt =0.49,
reflex alignments above 6 litres are impossible without
the reflex peaking.

:) sreten.
 

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Ok,
The thinks are pretty clear to me. Yestarday I put some bricks inside the box, and the graphs start to look good when I reach Vb=2.5 * Vas. Of course the roll off start at 85 Hz in the simulations, just like in the FR measured graph. I was searching for a couple of hours on the Internet to find some subwoofers with the Vas smaller the ½ Vb, but I didn’t find any. Most of them have the Vas LARGER than Vb.
So, in conclusion, now I’m sure that is impossible to reach 55 Hz with such a small driver;
And second, EVERY TIME I START BUILDING A ENCLOSURE I HAVE TO CHECK IF VB IS MAXIMUM TWICE THE VAS OF THE DRIVER.

Thank you very much for your patience and for you very helpful replies.
 
I researched a little bit more on the Internet and on Bob Speaker Stuff (http://www.geocities.com/rbrines1/) I found a project called Peerless Pipe. Well, I calculate the volume of the TL enclosure, and guess what: Vb=70 l, Vas of the Peerless Driver=15 l…. so Vb is 4 times bigger than Vas. This project is very well known in TL diy world, and is based on MKJ MathCAD sheets. Also the measured FR plots looks very good, just like in the simulations. So, where is the theory that says nobody wanted to build something bigger than twice of the Vas?
I think in most of the cases of TL enclosure, Vb is bigger than Vas…
Maybe the actual problem is the fs of the driver…maybe it’s impossible to achieve an fb much more lower than fs.
I’m in trouble again…J
 
But the Peerless pipe is a quarter-wave transmission line, which is a very different beast to a reflex enclosure.

I use Winspeakerz to analyse sealed-box and Reflex speakers, but a transmission line is difficult to analyse. TL speakers seem to sound better than the simulations would suggest!
 
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Joined 2002
tda said:
I researched a little bit more on the Internet and on Bob Speaker Stuff (http://www.geocities.com/rbrines1/) I found a project called Peerless Pipe. Well, I calculate the volume of the TL enclosure, and guess what: Vb=70 l, Vas of the Peerless Driver=15 l…. so Vb is 4 times bigger than Vas. This project is very well known in TL diy world, and is based on MKJ MathCAD sheets. Also the measured FR plots looks very good, just like in the simulations. So, where is the theory that says nobody wanted to build something bigger than twice of the Vas?
I think in most of the cases of TL enclosure, Vb is bigger than Vas…
The VAS to box volume ratio limitation holds for closed and ported (and PR) designs. As I pointed out in my earlier posts, this ratio limitation doesn't hold for infinite baffle and open baffle designs. It also doesn't hold for TL designs, although it will effect the FR as the box gets smaller. TL designs depend more on the wavelength of the tuning frequency than on the volume of the air in the box.

You should provide some more information about your driver.
 
Yes,
I know all this things very well, my problem started when I tried to simulate on Ported Box MathCad sheet an enclosure for my small driver. The result where disastrous so I'm beginning to ask myself if I know how to use this MathCAD sheet. I used Ported Box sheet because MJK told me that is the most advanced and complete sheet. So, the simulations was for a quarter wave system (I used MJK sheets) but the mesured results indicates a classic ported box system where the Vb/Vas problem ocurred.
 
Hello tda,

I have been thinking about the question of driver Vas and enclosure Vb. The driver Vas is a constant and known from measurements. If we look at the method used to evaluate Vas, the formaula Vas = Vb/(rho x c^2), then we see that an inherent assumption is made that the pressure is constant everywhere in the enclosure. If standing waves exist then the expression for enclosure impedance becomes very complicated and exhibits many peaks and nulls associated with the resonances. I don't think that there is any magic ratio of Vb/Vas that will tell you that your simulation is inaccurate. But if your internal dimensions are big enough to support standing waves, then I think that the simple expression for Vas is no longer applicable and the simulations will be inaccurate.

The Peerless TL referenced above has a large Vb compared to the Vas, and one very long dimension. The MathCad worksheet takes the long dimension into account and predicts and accounts for the standing waves in the system response. The response is accurately predicted and correlates well with the computer model. The MathCad worksheets can accurately handle one long dimension. But if several long dimensions exist in the design, the MathCad worksheet will not be accurate. The MathCad worksheets take you one step further then then the lumped parameter models like WINISD. A true 3D representation of the enclosure is the next level of calculation.

If you really want to see if your enclosure is behaving like the computer model predicts, then I recommend you remove all fiber stuffing and measure the impedance magnitude and phase, as a functions of frequency, for the driver in a baffle and the driver mounted in the enclosure. These curves will tell you how accurate the computer models are and show all system resonances. If there are unpredicted resonances they should be obvious. The sorce of the resoance can be determined and then corrective actions can be tried in the enclosure.

Hope that helps,
 
Hello MJK,
If I understood you right, the MathCAD sheets are good as long as the So and Sl are small (close to the Sd)...so the enclosure is very tall and thin. If the enclosure starts to look like a cube for example, the results are not predictable. Am I right? My test enclosure is almost a cube, so all 3 dimensions are almost equal...Could be this the reason of my problem?

And another general question: You think that finally all the enclosure (ported, TL, closed, horns, etc) is just special cases and all of them should be theoretically simulated with just one set of formulas, no matters the enclosure shape, volume, etc?
 
Hi tda

If we look at a TL that is basically a square enclosure with the bottom end open then the length of the TL is equal to the width and depth of the cross-section. Along the length you get a quarter wavelength standing wave at frequency fo. From side to side and from front to back you will have half wavelength standing waves of frequency 2 x fo. These frequencies are probably too close for the enclosure to act as a TL and there will be destructive interference between the modes, this seems to be what your test data is telling us. If I have to guess about the minimum length to width or height ratio, I would probably use a factor of 5 as a minimum. If the classic TL is five times longer then the other two dimensions, then the difference between the frequencies of the fundamental standing waves would be a factor of 10. This is just a guess and should not be taken too seriously, but I hope it makes the point.


So what can be done to "save" your enclosure. Have you tried dividing the cross-sectional area internally with pieces of cardboard. If, looking down from the top, you put two pieces of cardboard inside the enclosure forming a plus sign then you would effectively divide the TL into four parallel TL's each with 1/4 of the cross-sectional area of the original enclosure. You would need to allow communication at the top and bottom of the line so that the boundary conditions at each end of the four TL's would be the same. The next step might be to divide the areas again forming more parallel TL's. If this were done then I think the system performance would start to converge to the simulation's predictions. Sounds like an easy thing to try.

Hope that helps,
 
Hmm, this is all very interesting.

My Emerald Towers use the same driver as the Peerless Pipe (850122 - 7" CSX). Now, I didn't measure my the T/S on my drivers, but they are of a recent batch, so can only assume they probably have the newer specs, with a low Vas and high Fs. I did some quick and dirty Fs measurements, and they came out to the higher version, at 50Hz. Assuming Vas also jives with the new version, my enclosure is almost three times Vas. It's vented at the bottom. Here's where it gets interesting. My speakers are more or less pipe shaped, with the drivers up at the top, and the length of the pipe being just over five times the greatest width. I wasn't intentionally making a TL design...

What concerns me is all this talk of distortion. I certainly feel I should say this is not what I have experienced. The bass out of my speakers is among the most accurate I've ever heard, and has no problems maintaining control.

I don't have the answers, just wanted to throw out that such a large Vb/Vas ratio isn't always a recipe for disaster. :)


Aaron Gilbert
 
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