xmax and deep bass

[Aaron Clinton 2,372]

I really hope a lot of people read this topic, Brad worded it much better then I could. Not to mention, you are not going to get info like this on most other car audio focused forums.

[abmoore 8]

That's the calculation for Vas.

Should be pretty easy to figure out from there what will cause an increase in Vas

(p is the density of air & c is the speed of sound, both of which are constants, only really need to worry about Sd & Cms)

Oh, and Fs;

So what happens to Fs and Vas when Cms and Sd increases? Or decreases?

OK.... so generally the lower the Fs, the larger the Vas? It seems the tighter the suspension, the higher the Fs, which would be a decreased Cms correct? So Vas would increase with more Sd obviously, and a higher Fs driver, that would lower Cms, decreases Vas. This all assumes that Mms is the same. Speaking of which, the lighter the cone of the driver, and the lower the Vd, the lower the Mms. The lower the Mms, the lower the Fs. So there is generally** some correspondance between Fs and the amount of air a driver can move, therefore less spl? That is assuming a low Mms, low Cms, which would lead to the low Fs. Although high Mms, to a certain point, will create more spl? Am I catching on at all? This does show some relationship between xmax and Fs and spl, but proves it is not always important, especially with low Mms (but with high Mms, high Vd, high xmax, low Fs, high power handling, comes a true low end beast). Sorry for all of that lol.

[emh9009 199]

whats more important? cone area or xmax for low end?

There's much more to it than that.

First, cone area and Xmax are multiplied to arrive at a parameter known as Vd, or volume displacement. This is the amount of linear air displacement the driver is capable of achieving. The higher the Vd, the higher linear output capabilities the driver has as a larger Vd means the driver is capable of displacing more air.

Second, don't confuse Xmax with excursion. They are different. Excursion is how far the cone is physically moving. Xmax is the maximum amount of linear excursion the driver is capable of achieving. Excursion is the amount of movement, Xmax is the limit of how much of that movement is "linear". Generally a driver operating at Xmax will have ~10% THD. This is typically the highest amount of acceptable performance. As you exceed Xmax it's generally considered that the distortion performance would be at an unacceptable level (among other things).

In sealed enclosures, Vd is one of the important factors along with alignment and the driver's parameters in determining output capabilities. All things equal, the driver with the higher Vd will have higher output. In reality, things are never equal. So you have to look at the response of the driver in a given alignment along with your environment and available power. Having a higher Vd isn't any good if you don't have the requisite power to reach Xmax or if the driver is in an alignment that begins to roll off at a high frequency. Having a higher Vd is only a capability of higher output; whether or not you will be able to realize that higher output is dependent upon several other factors.

In terms of low frequency performance, you're required to increase displacement by a factor of 4 in order to maintain a given SPL level one octave lower in frequency. This means that you need to displace 4x the air at 25hz as you do at 50hz to maintain the same SPL level at both frequencies. You can see how having a higher Xmax could be beneficial as displacement requirements increase substantially in lower frequencies. In the low frequencies Vd can be very important as it can very much limit how much linear output your system will be able to achieve. Luckily in car audio we have substantial cabin gain in the low frequencies as well, so we can maintain SPL levels at lower frequencies without having to quadruple our displacement. A 12db gain at 25hz compared to 50hz, for example, is equivalent to quadrupling your displacement. So if you have a 12db gain at 25hz compared to 50hz in your vehicle, you've already accomplished the feat.

Generally you want the most cone area you can reasonably place within your space, environment and enclosure requirements. Why? A driver with a larger cone area will displace a greater amount of air for a given excursion level. This means that it takes less excursion to reach a given SPL level with more cone area. This, in turn, means that the driver(s) would ideally be operating further within it's linear range at that SPL level which, ideally, reduces distortion. This ofcourse assumes the drivers are of similar distortion performance as distortion performance will vary based on driver designs. For example, one driver may still have less distortion at 15mm than another more poorly designed driver (from a distortion perspective) driver operating at 10mm.

The reason Xmax "doesn't matter" for SPL is because 1) the subwoofers are generally "burped" near tuning where excursion is minimized and 2) Xmax is a linear parameter....people competing in SPL don't really care about having higher distortion due to possibly exceeding Xmax. But that doesn't mean Xmax is not important in ported enclosures. Xmax is not a limitation on how far the driver can physically move , only on how much of that movement is linear. Also, going back to what we said about power and alignment; Xmax does not tell you how much excursion you will have in your alignment with your power. And that is what ultimately determines how much output you will have from the enclosure, any enclosure. Vd is still "important" in ported enclosure as it still helps determine which driver has the capability for higher linear output, but you have to look at your power and alignment to determine which will ultimately have higher excursion, and hence output, in use. In a ported enclosure generally enclosure size and tuning are going to have a significant impact on low frequency response and excursion with regards to frequency; but that doesn't make Xmax irrelevant for daily listening. About a half octave above tuning, the enclosure will behave similarly to a sealed enclosure meaning excursion will increase to a point that is similar to that of a sealed enclosure. Xmax may become important in this region where it's possible to reach or exceed the Xmax of the driver.

That post feels like it was one giant ramble.....hopefully it makes sense.

nice explanation, i learned a bit

[Impious 1,476]

That's the calculation for Vas.

Should be pretty easy to figure out from there what will cause an increase in Vas

(p is the density of air & c is the speed of sound, both of which are constants, only really need to worry about Sd & Cms)

Oh, and Fs;

So what happens to Fs and Vas when Cms and Sd increases? Or decreases?

OK.... so generally the lower the Fs, the larger the Vas? It seems the tighter the suspension, the higher the Fs, which would be a decreased Cms correct? So Vas would increase with more Sd obviously, and a higher Fs driver, that would lower Cms, decreases Vas. This all assumes that Mms is the same.

As Cms decreases (suspension gets tighter), the Fs would increase and Vas decrease. And as Cms increases (suspension gets looser), Fs would decrease and Vas would increase. As you said, this assumes Mms (and Sd) remain constant.

Speaking of which, the lighter the cone of the driver, and the lower the Vd, the lower the Mms. The lower the Mms, the lower the Fs. So there is generally** some correspondance between Fs and the amount of air a driver can move, therefore less spl? Although high Mms, to a certain point, will create more spl?

Not sure if it was simply a misstatement; But a lower Mms would increase Fs. It's mass on a spring. The more mass you have on the end of a spring, the slower it will resonant.

I wouldn't go so far as to say that a lighter cone = lower Vd. Generally yes a decrease in Sd will be accompanied by a decrease in Mms assuming the same cone material and thickness is used. And a smaller Sd would yield a smaller Vd. But Mms is more than just the mass of the cone. Mms is the mass of the moving assembly (cone, coil, former, etc) including acoustic load. A driver with a long coil overhung motor could have a higher Mms than a driver of the same materials but using a short-coil motor topology like XBL^2 even though they may have the same Xmax. This is a case of not drawing too many conclusions from the "all things equal" scenario.

That is assuming a low Mms, low Cms, which would lead to the low Fs.

Low Mms + Low Cms would = higher Fs (light mass on a tight spring = higher resonant frequency). Or, think of it this way...since both Mms and Cms are in the denominator, Fs will be inversely proportional to both Cms and Mms. If those go down, Fs will go up. If those go up, Fs will go down.

Am I catching on at all?

Started off good, think you may have trailed off a little there towards the latter part

[abmoore 8]

That's the calculation for Vas.

Should be pretty easy to figure out from there what will cause an increase in Vas

(p is the density of air & c is the speed of sound, both of which are constants, only really need to worry about Sd & Cms)

Oh, and Fs;

So what happens to Fs and Vas when Cms and Sd increases? Or decreases?

OK.... so generally the lower the Fs, the larger the Vas? It seems the tighter the suspension, the higher the Fs, which would be a decreased Cms correct? So Vas would increase with more Sd obviously, and a higher Fs driver, that would lower Cms, decreases Vas. This all assumes that Mms is the same.

As Cms decreases (suspension gets tighter), the Fs would increase and Vas decrease. And as Cms increases (suspension gets looser), Fs would decrease and Vas would increase. As you said, this assumes Mms (and Sd) remain constant.

Speaking of which, the lighter the cone of the driver, and the lower the Vd, the lower the Mms. The lower the Mms, the lower the Fs. So there is generally** some correspondance between Fs and the amount of air a driver can move, therefore less spl? Although high Mms, to a certain point, will create more spl?

Not sure if it was simply a misstatement; But a lower Mms would increase Fs. It's mass on a spring. The more mass you have on the end of a spring, the slower it will resonant.

I wouldn't go so far as to say that a lighter cone = lower Vd. Generally yes a decrease in Sd will be accompanied by a decrease in Mms assuming the same cone material and thickness is used. And a smaller Sd would yield a smaller Vd. But Mms is more than just the mass of the cone. Mms is the mass of the moving assembly (cone, coil, former, etc) including acoustic load. A driver with a long coil overhung motor could have a higher Mms than a driver of the same materials but using a short-coil motor topology like XBL^2 even though they may have the same Xmax. This is a case of not drawing too many conclusions from the "all things equal" scenario.

That is assuming a low Mms, low Cms, which would lead to the low Fs.

Low Mms + Low Cms would = higher Fs (light mass on a tight spring = higher resonant frequency). Or, think of it this way...since both Mms and Cms are in the denominator, Fs will be inversely proportional to both Cms and Mms. If those go down, Fs will go up. If those go up, Fs will go down.

Am I catching on at all?

Started off good, think you may have trailed off a little there towards the latter part

Haha I tripped myself up there, trying to remember that Cms is a reciprical of tightness of the suspension. I see where I messed up. The mass on the spring analogy helps too. Just remembered back to my physics classes, I think I remember Fs being used, never connected the dots for some reason. When I said the lighter cone, I meant as well as a lower vd, would that equate to a lower Mms. Which would generally be less spl at a lower Fs because there is less force behind the air? That sounds like where Bl^2/Re becomes relevant right? The higher the Telsa*?* on a driver with a low Fs, has a lot of low end output, because that is when Re is the highest, so basically more force moving that large mass on the end of a loose spring, while being able to do this under a heavy resistance. But I guess that all depends on where Bl^2/Re is measured, which I have no idea.

[j-roadtatts 1,129]

I really hope a lot of people read this topic, Brad worded it much better then I could. Not to mention, you are not going to get info like this on most other car audio focused forums.

The man has a way with words, and when he speaks I get out my note book and pencil because SCHOOL IS IN.

[abmoore 8]

I'm suprised no one brought up sensitivity. So I will haha. It also looks like when Vd increases due to Sd increasing, that sensitivity increases. So larger Vas and lower Fs both correspond with a higher sensitivity?

[KU40 173]

I'm suprised no one brought up sensitivity. So I will haha. It also looks like when Vd increases due to Sd increasing, that sensitivity increases. So larger Vas and lower Fs both correspond with a higher sensitivity?

I was going to bring up sensitivity last night in a reply but got distracted. Higher Vas increases sensitivity, lower Fs lowers sensitivity.

[Impious 1,476]

I'm suprised no one brought up sensitivity. So I will haha. It also looks like when Vd increases due to Sd increasing, that sensitivity increases. So larger Vas and lower Fs both correspond with a higher sensitivity?

Subwoofer Sensitivity - SSA Car Audio Forum

There is the formula for sensitivity.

What will happen when Fs decreases? Sensitivity will decrease, not increase. Same with Vas. Sensitivity will move in the same direction of those two terms. If one of those two go up, so does SPL. If one of those two go down, so does sensitivity.

Sensitivity is inversely correlated to Qes...as Qes increases, sensitivity will decrease.

It's all a basic consequence of Hoffman's Law. If you want low frequency extension in a small enclosure (low'ish Fs, small'ish Vas, higher Qes) sensitivity will necessarily be sacrificed.

[abmoore 8]

I'm suprised no one brought up sensitivity. So I will haha. It also looks like when Vd increases due to Sd increasing, that sensitivity increases. So larger Vas and lower Fs both correspond with a higher sensitivity?

Subwoofer Sensitivity - SSA Car Audio Forum

There is the formula for sensitivity.

What will happen when Fs decreases? Sensitivity will decrease, not increase. Same with Vas. Sensitivity will move in the same direction of those two terms. If one of those two go up, so does SPL. If one of those two go down, so does sensitivity.

Sensitivity is inversely correlated to Qes...as Qes increases, sensitivity will decrease.

It's all a basic consequence of Hoffman's Law. If you want low frequency extension in a small enclosure (low'ish Fs, small'ish Vas, higher Qes) sensitivity will necessarily be sacrificed.

Cool, I just worked out some more senarios. Thank you this has been very helpful, dare I say eye opening. This will definately lead to more educated decisions when deciding what drivers to use and when to use them.

[Impious 1,476]

When I said the lighter cone, I meant as well as a lower vd, would that equate to a lower Mms. Which would generally be less spl at a lower Fs because there is less force behind the air?

Eh? Not really sure I'm following the question.

That sounds like where Bl^2/Re becomes relevant right? The higher the Telsa*?* on a driver with a low Fs, has a lot of low end output, because that is when Re is the highest, so basically more force moving that large mass on the end of a loose spring, while being able to do this under a heavy resistance. But I guess that all depends on where Bl^2/Re is measured, which I have no idea.

Re is the DC Resistence (DCR) of the coil. This means the measurement isn't done at a "frequency" as it's measured with direct current rather than alternating current.

Bl^2/Re describes the motor force factor. Using Bl alone to compare "motor strength" can be misleading because it ignores the current through the coil which will affect how much actual "force" the motor has. Essentially dividing by Re compensates for this as a higher Re will have less current through the coil. As Re increases the ratio of Bl^2/Re will decrease. This makes comparing the "actual" force of the motor between different drivers more of an apples-to-apples comparison.

There really isn't a direct correlation between Fs and Bl^2/Re as they are defined by different parts of the subwoofer (Fs by the compliance and mass, Bl^2/Re by the motor and DCR of the coil). But the force of the motor at Fs will be at it's minimum (in free air) as impedance is at it's maximum which means there is the least amount of current through the coil.

[KU40 173]

I'm suprised no one brought up sensitivity. So I will haha. It also looks like when Vd increases due to Sd increasing, that sensitivity increases. So larger Vas and lower Fs both correspond with a higher sensitivity?

Subwoofer Sensitivity - SSA Car Audio Forum

There is the formula for sensitivity.

What will happen when Fs decreases? Sensitivity will decrease, not increase. Same with Vas. Sensitivity will move in the same direction of those two terms. If one of those two go up, so does SPL. If one of those two go down, so does sensitivity.

Sensitivity is inversely correlated to Qes...as Qes increases, sensitivity will decrease.

It's all a basic consequence of Hoffman's Law. If you want low frequency extension in a small enclosure (low'ish Fs, small'ish Vas, higher Qes) sensitivity will necessarily be sacrificed.

Whoops I wrote that if vas increases, sensitivity increases. Not sure where I got that, apparently not from a formula.