
Vented Box Larger Than Optimal??
Hey, so I was reading through Ray Alden's Speaker Building 201 and trying to play around with some of the new Dayton Designer series woofers. And I'm running into something I don't quite understand here.
I know that Vas is the volume of air with equivalent compliance of the driver's suspension. And the equation in this book for the optimal box volume is:
Vb=15*Vas*(Qts^2.87)
I don't know how common this formula is among everyone here but I've been playing around with it for a few drivers and it seems to make sense. The last one I was looking at was the 61/2" (295428). It has a Vas of .61 cu.ft. and a Qts of .30
What I'm trying to grasp here is that if I get an optimal Vb of .289 cu.ft. and an F3 of 59Hz, what happens if you increase the volume even more?? I know that if you decrease it, the F3 will increase and there will be a peak, but I've not read about what a larger box will do. I guess it wouldn't have enough air spring? What impact would that have? The equations I'm using are pretty simple and obviously don't take any other factors into account as far as acoustics.
To find the F3 I'm taking the square root of (Vas/Vb) and multiplying by Fs (40.9Hz). So according to that equation, any driver will have an F3 the same as the Fs if Vb=Vas. Keep in mind, I'm very new at this and trying to find the best way to get the right cabinet sizes and port sizes. I'm starting to understand more about crossovers, but haven't begun trying to design them yet. I was just thrown off when I saw that it said "oh this driver can play at full rated volume down to 40Hz" but then the recommended boxes show an F3 no less than 59Hz.
Thanks in advance for all your wisdom

Re: Vented Box Larger Than Optimal??
The area under the speaker curve is like a bag of water. You can pull it up with a high Q but then it has a bump. Or you can let it relax and stretch the furthest but then it gets kind of low all over. Q = 0.7 represents the historically flattest spot to grab the bag and pull.
Bass reflex is like having two places to grab the bag. The bag still has the same amount of water though. If you grab them too close together or put one too high then the bag has a big bump and sounds all boomy. But if you pick just the right spot then it can actually be flatter longer (before falling off really fast).
Yeah, it's not mathematically correct but it will give you an intuitive understanding of the tradeoffs involved.

Re: Vented Box Larger Than Optimal??
Formulas are almost as obsolete as slide rules. Download WinISD Alpha Pro freeware, learn to use it. You can see what happens to response, impedance, excursion and more in real time when you adjust box dimensions and tuning.

Re: Vented Box Larger Than Optimal??
If you try to tune it too low, you'll run out of excursion faster, limiting your SPL. Basically you're stretching the bag of water out, and it gets shorter.

Re: Vented Box Larger Than Optimal??
If you compare a B4 vented alignment to a B6 you might notice that they are very similar except for the B6 having 1/2 the box volume. The B6 requires a 2nd order high pass filter with 6 dB of peaking at Fb because you loose 6 dB by making the box half size. The B6 is back to 3dB at Fb with the boost filter.
Increasing the box size makes it a more efficient radiator and provides more output at Fb. Those equations are optimal from some mathematical standpoint.
You can always tune lower and it will behave more like a second order system but you will get less output at the higher frequency where it was previously tuned. Vented systems have their power handling and output drop off fast below vent tuning and they are often tuned too high. 40 Hz is a good tuning frequency for smaller systems since it is roughly bottom E on a regular bass guitar. 28 Hz for slightly larger systems since it is the lowest note on a grand piano, and even lower for larger systems 20 Hz and below. Home theater sound effects have content down to 12 Hz perhaps even lower.
Experiment with a simulator that shows displacement limited output to see the effects on both FR and power handling.

Re: Vented Box Larger Than Optimal??
OK, good advise, thank you. I downloaded WinISD a while ago and looked at it real quick and didn't get it (couldn't find anywhere to enter anything), but I'm gonna get on that and find out how to use it.
Yeah I've heard about Q=.707 being the best spot for transient response, I'm really trying to get a feel for how this works underneath before I get too dependent on programs doing it for me but I suppose there's really nothing wrong with it as long as I have a good understanding of HOW it all works.

Re: Vented Box Larger Than Optimal??
, I'm really trying to get a feel for how this works underneath before I get too dependent on programs doing it for me
Think of it as a performance envelope with a range.
The Loudspeaker Design Cookbook describes the effect of performance for variations of Qts and Tuning Ratio (H)
"If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

Re: Vented Box Larger Than Optimal??
Originally Posted by Sydney
Think of it as a performance envelope with a range.
The Loudspeaker Design Cookbook describes the effect of performance for variations of Qts and Tuning Ratio (H)
That's the next book I was planning on getting

Re: Vented Box Larger Than Optimal??
Originally Posted by monoiz15
Yeah I've heard about Q=.707 being the best spot for transient response,
Q isn't so much about transient response as it is about frequency response. High Q can result in a midbass peak, with boomy sound the result. Low Q can result in weak midbass, which can sound thin. Q of 0.7 is generally considered the best compromise.

Re: Vented Box Larger Than Optimal??
Originally Posted by monoiz15
OK, good advise, thank you. I downloaded WinISD a while ago and looked at it real quick and didn't get it (couldn't find anywhere to enter anything), but I'm gonna get on that and find out how to use it.
Yeah I've heard about Q=.707 being the best spot for transient response, I'm really trying to get a feel for how this works underneath before I get too dependent on programs doing it for me but I suppose there's really nothing wrong with it as long as I have a good understanding of HOW it all works.
If you have MS Excel you might want to try Unibox which is also free.
Yes it is certainly good to understand what is really going on in the system.
Q, or Qtc only applies to 2nd order (closed box) systems. 4th order systems (vented) can be described by two cascaded second order sections but this is not how it is usually done with speaker design.
You ought to consider what a speaker does in a highly resonant room before worrying too much about transient response.
An interesting characteristic of 2nd order systems is that the relative amplitude response at Fb is 20 log(Qtc). Qtc = .5 is critically damped and 6dB, .707 is maximally flat 3dB, and anything higher has some peaking, Qtc = 1.0 is 0dB at Fb with peaking in the response above Fb. Consider that a system with a Qtc of .5 requires four times as much power for the same output as a system with a Qtc of 1.0.
Small has defined a volumebandwidthefficiency figure of merit K and this peaks around a Qtc of 1.2 IIRC.

Re: Vented Box Larger Than Optimal??
Originally Posted by monoiz15
Hey, so I was reading through Ray Alden's Speaker Building 201 and trying to play around with some of the new Dayton Designer series woofers. And I'm running into something I don't quite understand here.
      This is the best book for you right now.
You don't need any other until you've read it through six times and understand 60% to 80% of it.
I know that Vas is the volume of air with equivalent compliance of the driver's suspension. And the equation in this book for the optimal box volume is:
Vb=15*Vas*(Qts^2.87)
I don't know how common this formula is among everyone here but I've been playing around with it for a few drivers and it seems to make sense. The last one I was looking at was the 61/2" (295428). It has a Vas of .61 cu.ft. and a Qts of .30
      So . . . Vb=15*.61*(.30^2.87)
      Vb = 9.15 * .0316 = .289
We'll just call it 0.29 cf.
What I'm trying to grasp here is that if I get an optimal Vb of .289 cu.ft. and an F3 of 59Hz, what happens if you increase the volume even more?? I know that if you decrease it, the F3 will increase and there will be a peak, but I've not read about what a larger box will do. I guess it wouldn't have enough air spring? What impact would that have? The equations I'm using are pretty simple and obviously don't take any other factors into account as far as acoustics.
      "enough" air spring? The more air you've got (the bigger the box) the "softer" the spring gets. A smaller box has a stiffer spring.
To find the F3 I'm taking the square root of (Vas/Vb) and multiplying by Fs (40.9Hz). So according to that equation, any driver will have an F3 the same as the Fs if Vb=Vas. Keep in mind, I'm very new at this and trying to find the best way to get the right cabinet sizes and port sizes. I'm starting to understand more about crossovers, but haven't begun trying to design them yet. I was just thrown off when I saw that it said "oh this driver can play at full rated volume down to 40Hz" but then the recommended boxes show an F3 no less than 59Hz.
      you're saying F3 = Fs * (Vas/Vb)^.5
      so F3 = 40.9 * (.61/.29)^.5
      or F3 = 40.9 * 1.45 = 59.3 Hz
I'm with ya. I don't see that formula in Ray's book, but
      On pg. 33 I see: F3 = (.26 * Fs) / Qts^1.4
      or F3 = ( .26 * 40.9 ) / .30^1.4
      or F3 = 10.6 / .185 = 57.3 Hz
That's basically the same answer.
Thanks in advance for all your wisdom
Your last observation however; "any driver will have an F3 the same as the Fs if Vb=Vas" isn't entirely correct. While Vb (your box volume) CAN be whatever size you want, IN THIS INSTANCE the equation is only true if the Vb in the equation WORKED OUT TO BE EQUAL to Vas in your previous equation.
Vb=15*Vas*(Qts^2.87)
We can ask, "under what circumstances would Vb be equal to Vas", or "does there exist a value for Qts that would make Vb equal to Vas?"
solving: Vb/Vas = 15*Qts^2.87
or (if Vb = Vas) 1 = 15*Qts^2.87
or 1/15 = Qts^2.87
or 0.067 = Qts^2.87
solving for Qts we get 0.39
testing: if Qts = 0.39 then 0.39^2.87 = 0.067
So, your statement that F3 will equal Fs only holds true for a driver whose value for Qts = 0.39. For higher Qts values, F3 will be lower than Fs, and for Qts values < 0.39, F3 will be higher than Fs.
Given 2 drivers, BOTH with an Fs of 45 Hz, but one with a Qts of .50 and the other with a Qts of .25, the higher Q driver will probably have an F3 in the mid 30's, whereas the low Q driver will most likely end up with an F3 of 60 Hz or higher. That's where the statement that "low Q kills bass" came from.
That makes it SEEM like higher Qts drivers are the way to go, but in the previous example, if BOTH drivers also had the same Vas (let's say, 2.00 cf), the low Q driver would need about a 0.2 cf box (I'm guessing here, but probably pretty close  the box would be VERY small  AND therefore hard to tune with a vent), whereas the higher Q driver would require a large box, more like 3.2 cf. Not that big a deal for a single corner sub that can hit in the 30's, but getting fairly large for a pair of floorstanders in the livingroom.
Your second entry starts talking about "system" Q's of 0.707 being maximally flat. That's only for sealed boxes though. Don't mix the two together.
Ray will show you how to use WinISD in a later chapter. Don't worry.
Chris

Re: Vented Box Larger Than Optimal??
the best transient response for a sealed system is slightly less than .7.
A Qtc of .7 produces the most extended "flat" passband to cutoff, but not the best step response.

Re: Vented Box Larger Than Optimal??
Originally Posted by moron#99
The area under the speaker curve is like a bag of water. You can pull it up with a high Q but then it has a bump. Or you can let it relax and stretch the furthest but then it gets kind of low all over. Q = 0.7 represents the historically flattest spot to grab the bag and pull.
Bass reflex is like having two places to grab the bag. The bag still has the same amount of water though. If you grab them too close together or put one too high then the bag has a big bump and sounds all boomy. But if you pick just the right spot then it can actually be flatter longer (before falling off really fast).
Yeah, it's not mathematically correct but it will give you an intuitive understanding of the tradeoffs involved.
Nice way of explaining the principles!

Re: Vented Box Larger Than Optimal??
Originally Posted by Pete Schumacher ®
the best transient response for a sealed system is slightly less than .7.
A Qtc of .7 produces the most extended "flat" passband to cutoff, but not the best step response.
If you want to know some information about the time domain (e.g. transient response) I suggest you download and use Jeff Bagby's Woofer Box and Circuit Designer to model the vented enclosure and then look at the step response. You can compare it for various types of enclosures and then make up your own mind.
http://audio.claub.net/software/jbabgy/WBCD.html
This is an excellent program, and you can also include the response of an active filter(s). This makes it easy to model a B6 or other filterassisted alignments.
Charlie

Re: Vented Box Larger Than Optimal??
Wow, this has been really helpful. I did kind of notice that Qtc was more for sealed systems. What are B4 and B6??
What I originally said about any driver having an F3 equaling Fs if you make the Vb the Vas was really my way of asking what other factors are involved in that relationship, and you cleared that up for me very well. The formula of F3=SQRT(Vas/Vb)*Fs is on the next page, 34 where he talks about designing a smaller box. I didn't understand why the maximally flat response was in such a small box, but I now see that it's because that driver has a relatively low Qts = .3 which means the suspension has more control of the driver so you'd need a box with a "stiffer" spring so smaller. If the box were bigger, it would have too loose a spring when you get to lower frequencies and the response fades off. Right? I guess it's matching the compliance of the suspension with the stiffness of the air spring to give whatever results you desire that the driver can offer. And for the most part I suppose you would normally aim to not stray too far from that maximally flat area but you can stretch it a little bit to fit your needs if its not too far off.
OT: I got a Behringer iNUKE 1000 and an ART CleanBox Pro for my D8s and put some speakon connectors on them, they are LOVIN it. Also, I'm waiting on my roommate to decide what kind of wood she wants for the Noahs 8s and once I'm ready I'll be posting a final details thread to confirm everything before I start.

Re: Vented Box Larger Than Optimal??
WinISD and WinISD Pro suffer from one big problem today: the help sections no longer work on Win7, so I can see them being harder to use. Regardless, if you're smart enough to be manhandling the equations, you're smart enough to use Unibox. ;)
While Qtc doesn't matter much for vented, Qts DOES matter for vented boxes, just as much as it does for sealed ones and may have a larger impact on the overall response. A driver will ALWAYS ring at Fs, and since the air volume is so large (compared to a sealed box), the system is highly underdamped so the driver is free to flop around to its heart's content. This is why it's almost impossible to get flat response out of a driver with a high Q in a vented box.
nothing can stop me now

Re: Vented Box Larger Than Optimal??
Originally Posted by Dirk
WinISD and WinISD Pro suffer from one big problem today: the help sections no longer work on Win7, so I can see them being harder to use. Regardless, if you're smart enough to be manhandling the equations, you're smart enough to use Unibox. ;)
I've tried WinISD Pro 0.5a7 under Windows 7 64bit, and when I attempt to close the program, it gives me an access violation, a read of address 0. Then I have to shut it down from Task Manager. And as you say, the help does not work when you try to launch it from within the program. A workaround for this is to create a shortcut to the WinISD.chm help file and just invoke the help directly from Windows. That works fine.
There is also a newer WinISD 0.7 alpha version that's compatible with Windows 7 but not linked from the linearteam home page. There is only a link to it in their forum for some reason. It works fine under Windows 7, but has no help file yet. I'd recommend installing both these programs in different directories, and using the direct link to the WinISD Pro 0.5a7 WinISD.chm file as mentioned above for help only, while using the WinISD 0.7 version for simulation only. The old help file won't exactly match up with the UI of the new version, but it's a lot better than having no help.
The new 0.7 version is working quite well for me.

Re: Vented Box Larger Than Optimal??
Originally Posted by monoiz15
Wow, this has been really helpful. I did kind of notice that Qtc was more for sealed systems. What are B4 and B6??
Qtc is only for closed box systems in fact it is Q t=total, c=closed box.
Qts is Q t=total, s=system for the driver as a system in free air.
Qes and Qms are the two sources of damping that contribute to the total (Qts)
B4 = 4th order (vented) Butterworth which is monotonically flat.
B6 = 6th order (vented + 2nd order peaked high pass line level filter) Butterworth.
The legendary B&W801 Matrix SII is a low tuned B6 design. They mention that
it is a Bessel alignment without the electrical filter and is 9 dB down at 19 Hz
which is about right when you leave out the boost circuit. The peaking electrical
highpass provides 6 dB of boost at 19 Hz making it 3 dB down as a Butterworth
alignment should be:
http://www.stereophile.com/floorloud...506/index.html
The classical B4 and B6 alignments are covered in Thiele and Small's work:
See alignments #5 and 15 here where the types are given as B4 and B6:
http://www.readresearch.co.uk/thiele..._article_1.pdf
More T&S papers:
http://www.readresearch.co.uk/articles.php
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