Relationship between Fcb, F(3) and Qtc in a closed box sub

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I'm trying to work out the relationship between Closed box resonance frequency (Fcb) and F(3) - as well as why Fcb is significant - and why F(3) goes down when Qtc goes up when the box volume is descreased???

I understand the Fcb is the resonant frequency of the driver when put into a box - in other words the the peak Qtc of the system and frequency where the driver exhibits its largest impedance.

1. Does it make sense to have a F3 lower than Fcb?.

2. As box size decreases - F3 goes down, whereas Fcb and Qtc go up. Unless I've got my fromulas wrong - I thought according to the text books that higher Q actually causes a steeper -dB slope - hence why would F3 be lower with a higher Qtc???

3. I still don't really understand what Fcb has to do with sound output (or is it more of an issue with impedance compensation)

diyAudio Moderator Emeritus
Joined 2001

There are several programs here that predict response in both closed and ported boxes:

There is also WinISD which I don't use but many others do, and Unibox has drawn good reviews on this forum.

Here is a brief overview of the sealed box situation. Most unenclosed subwoofer speakers have have Qts of .4 or below. Let us use an example of a speaker with a Qts of .4 and an Fs of 23 Hz.

Every speaker has a Vas. That is the volume of the enclosure where the springiness of the air is equal to the springiness of the speaker's own suspension-when you press a speaker's cone down, it bounces back. Fs, Qts and Vas are your key numbers-sometimes Qts is broken up into it's key components, Qes and Qms, by some programs. But we can use Qts for our calculations.

As I am sure you know, SPL is the dB that a speaker plays at throughout most of it's range. A typical number would be "88 dB @1Meter/1Watt". Meaning that at a distance of 1 Meter, 1 Watt of power will drive the speaker to a volume of 88 dB throughout most of the speaker's frequency range. This number is called the "midpoint" of the speaker. We often consider the lowest frequency the speaker can adequately play-the low frequency "cutoff"-as the place where it is (-3 dB) from the "midpoint". In this case, that would be 85 dB.

Okay, so our key numbers-our "parameters"- for our imaginary 12 inch subwoofer are as follows:

Vas = 4 cu. ft
Qts = .4
Fs = 23 Hz
SPL = 88 dB @ 1Meter/1Watt

Suppose we want to put this speaker in a 3 cubic foot box. Then Vb = 3 cu. ft.

Let's see what happens.

When you put an unenclosed speaker in a sealed box, you raise both the Fs and the Qts. The new numbers, Fc and Qtc, (What Fs and Qts are converted to when placed in a closed box) are raised according to the following formula:

Fc = {square root of: [(Vas/Vb) + 1]} X Fs
Qtc = {square root of: [(Vas/Vb) + 1]} X Qts

In our 3 cubic foot box, this yields the following numbers:

Fc = {square root of: [(4/3) + 1]} X 23 means Fc equals 35 Hz
Qtc = {square root of: [(4/3) + 1]} X .4 means Qtc equals .61, or .6.

Plug these numbers into any speaker software, and you will see that at Fc, the speaker is about 4.4 dB down from the midpoint. That means that F3 will be somewhat above Fc, which is 35 Hz.

Your Qtc is going to determine how far below, or above, the "midpoint" your speaker is when it hits Fc.

Let's take this same speaker and put it into a 2 cubic foot box. That means our Vb equals 2 cu. ft.

Using the same formulas as above:

Fc = {square root of: [(Vas/Vb) + 1]} X Fs
Qtc = {square root of: [(Vas/Vb) + 1]} X Qts

We get:

Fc = {square root of: [(4/2) + 1]} X 23 means Fc = 39.8 Hz
Qtc = {square root of: [(4/2) + 1]} X .4 means Qtc = .69

Taking a look at our graph in any program, we find that at Fc, our speaker is down 3.2 dB from our midpoint. So F3 is right around Fc.-when our Qtc is around .69.

I am trying not to fill you up with too many formulas. I just want to give you some idea what to expect when you start plugging numbers into a speaker program, to save you a lot of confusion when you see the results.

To know just how far Fc will be below, (or above), the midpoint, we have the following formula:
20 Log Qtc.


If our Qtc is .5, we will be -6 dB from the midpoint at Fc.
If our Qtc is .7, we will be -3.1 dB from the midpoint at Fc.
If our Qtc is 1.0, we will be at the midpoint exactly.
If our Qtc is 1.25, we will be +1.93 dB from the midpoint.

To briefly summarize:

For any given speaker, the smaller the box, the higher the Fc and Qtc.

The higher the Qtc, the higher the Fc is compared to the midpoint.

One more thing. The lower the Qtc, the better the transient response-less "hangover". For many years, in a sealed box, builders used to go for a Qtc between .7 and 1.0. Above 1.0, and the bass becomes sloppy. Below .7, and the response droops. However, some builders have begun to build their subwoofers with a Qtc of .5. It has the droopiest response, but they make up for it by running a big amplifier. It does have the advantage of the best transient response.

Sorry for the length of the post. I just wanted to give you a general guideline.

Good luck!:)
To AudioFreak

Thanks - thats what Ray Alden also says in his book. "-12dB fall off for closed boxes regardless". However - what is misleading is in Weem's book - he has a variable dB falloff - depending on Qtc which got me confused (well variable according to the graphs which look a bit freehand so it could be my poor interpretation)

To AudioFreakand jteef

Thanks for the formulas and statement on F3 being lowest at Qtc=0.707. Ray Alden's F3 formula is stuffed - it causes a -ve sqrt() for any Qtc < 0.5 or Qtc > 1.0

jteef - I used the formula you provided and it makes much more sense.

To kelticwizard
Thanks for the detail in your post - it makes more sense than some of the books I've been reading! and also gives me more figures I can sanity check with

It really pisses me off when I pay $70 (New Zealand here, overseas books are expensive!) - only to find the formulas are wrong! It makes me wonder what other formulas are broken...

I know people could say "just use a speaker design program" - but I like to know what the consequences of changes to parameters will be to the design.

The one book I didn't buy was Dickasons... might get it... I didn't know about it when I bought the other books - Dickason's seems to be a bit of a "bible"

Thanks once again people for your help.
Just one more question :)....

What doesn't make sense to me is how transient response improves with a lower Qtc....

Say Qtc = 0.707
alpha = 2.02
Vb (ft3) = 4.90

Now reduce Qtc...
Qtc = 0.5
alpha = 0.51
Vb (ft3) = 19.42

I'm using the dayton titanic 12" driver in my formulas:
Qts = 0.407
Qes = 0.428
Fs = 16.25
Vas (ft3) = 9.89

You'd think with a higher apha - the box air volume is assisting the cone to return to "rest" = therefore smaller box = higher alpha - therefore a Qtc = 0.5 would therefore give *worse* transient response than a Qtc=0.707

Have I either mis-understood what transient response is? are my equations flawed?

Thanks again for your help,
At first I was thinking what you were. Then, I thought how the smaller air volume actually resists cone movement. That could tend to hold it back and damp its movement, actually slowing its initial movement.

But the blade is double edged. I think you are right, the smaller box tends to let the cone return to center faster. In my listening, I find I prefer to sacrifice the former to have more of the latter. the larger box muddies the bass reproduction because it tends to resonate more (internal reverberation). It can also help compensate for a speaker whose Qts is too high, avoid woofers above .40 or so. I should also mention that with a purposefully undersized box, it is important not to go too far because the active equalization needed to flatten the response peak and raise the extra low end roll-off can end up being very hard or impossible to achieve. Basically, it takes more amplifier power, and makes bottoming out the voice coil more likely too. Plus, cone stiffness is more important. I use a 7 band equilizer set up for just bass frequencies, say 28, 45, 60, 80, 100, 125, 150 to get flat response. Also, that is why I fill my enclosure completely with polyester fill, to lower the peak and raise the low end response, plus to fight all internal resonance. Besides, the cabinets are very hard and stiff, and need to be muffled. Push the insulation back away from the woofer cones. I don't use fiberglass or anything like that for fear that the sonic energy will break the strands and release particles. I have seen this problem, though it may have been caused by using bad quality fiberglass or whatever it was.

I would not have been able to get it right without the software's graphing ability. It was able to similtaneously show the how the cone excursion, power, and SPL all tied together. I don't recommend anyone forgoing using the software. I realize that what has worked for me may not for someone else since my situation is so fine-tuned to a specific goal (disclaimer). Sorry for having such a tall soapbox. :(
Actually maybe muddiness is rather a kind of distortion, a lack of definition, which may be able to extend close to the top of the bass range, while boom may also be able to extend down to the bottom. But those areas may not be where they are usually associated. I can see what you mean in that if the speaker were mounted on an infinite baffle, the boom and muddiness should be at a minumum.
diyAudio Moderator Emeritus
Joined 2001

Thanks for the kind words. I am glad to see we are on the same page about speaker software-the individual should have a basic grounding in what principles are involved, the software should only serve to bring things into fine focus.

I have never read Dickason's book. I learned the basic principles from David Weems' first book, which I think is just excellent. Since then, I have read two compliations on loudspeakers from the Journal of the Audio Engineering Society, from whcih I have learned much. I would say that unless I read Weems' book first, "Designing, Testing and Building Loudspeakers", I would not have gotten much out of the Audio Engineering Society compilations.

Weems is not decieving you. In a closed box of .51 Qtc, you will be -6dB at Fc, (20.4 Hz with the Titanic), and about -9 dB half an octave lower. That is only 3 dB difference per half octave. You will be about -14 dB from midpoint at 10 Hz, which is a rolloff of about 8 dB for the octave below Fc. After that, it does indeed rolloff at 12 dB per octave. However, that is hardly important because by the time you get 14 dB below the midpoint, you really can't get any usable bass out the speaker anyway. You are pretty well out of the passband by that time.

If your speaker has a sensitivity of 90 dB @ 1Meter/1Watt, then it is playing at 76 dB at 1 Watt if it is -14 dB from midpoint. Pumping 1,000 Watts through it will yield an SPL of 106 dB. If you are down at 20 Hz or thereabouts, normal room gain might give you another 6 dB or so. But your ability to perceive sound drops off quickly in those low ranges too. Believe me, 106 dB is not that loud at 20 Hz. And it is only achievable with a 1,000 Watt amplifier.

Weems was dealing with the characteristics of the rolloff at the frequencies where a speaker is actually going to be played. In those frequencies, the rolloff is indeed variable with Qtc.
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