Free air (OB) bass roll off?
 

Sealed subs are 12db

Ported can vary but 4th order is 24db

What is an open baffle woofer roll off?

I'm looking to use a 7" thats FS 35hz w/Qts of .45 so it starts to roll off at 77hz but how steep? It s going to be used with multiple subs but Im trying to model the midbass. Is my math off?

Any help appreciated.

Dipole roll-off itself is 6 dB/oct.

Driver roll-off below Fs is an other 12 dB/oct.

In theory:
above Fs: 6 dB/oct, below Fs: 18 dB/oct

It depends on the qts of the driver. A very high q driver can roll off at 24 db below fs. A very low q driver will roll off at 12 db total below fs.
Your average driver with a q between .5 and .7 will roll off around 18 db total below fs.
I'm guessing your driver will be around 16 db?

Thank to both responses.

Why would a higher Qts roll off faster? Thats interesting.

Any input into how to gauge power handling free air?

OB loading tends to suit a high Qts driver. Many would choose Qts between 0.6 and 0.8 i.e. close to 0.71 for a sort of Butterworth roll off.

I don't know, but that "Butterworth" might be the clue to the roll off rate near that frequency.

Mmmm, sort of. The fundamental roll off rates are, as stated before, 6dB/octave above the driver resonance and 18db/octave below the driver resonance. This assumes that the OB rolloff (the 6dB/oct one) starts above the driver resonant frequency... which is typically the case. The frequency at which the 6dB/oct OB rolloff starts is determined by the effective baffle size, e.g. the pathlength from the front to the back of the driver radiating surface. Below the driver resonance, the additional 12dB/oct contributed by the driver causes an increase in steepness to an asymptotic 18dB/oct rolloff.

So then, what about driver Q? The driver response is "superimposed" on, or added to, the above fundamental OB rolloffs. You are essentially operating the driver in free air, and the vast majority of woofers have Qts < 0.5, with some exceptions. A response with such a low Q value has a drooping response, which is further diminishing output where you need it. This is why high Q drivers (as high as Q=2) are actually a good match to open baffle loading - the natural low end response near the free air resonance is actually flat or even bumped up a couple of dB (which acts to compensate for the 6dB rolloff a bit).

I think that we are saying more or less the same thing, but it's a little clearer (IMHO) if you break it up into the baffle response (6dB/oct transitioning to 18dB/oct) and the driver response resulting from its free air Q.

-Charlie

It has to do with the definition of "Q". Andrew is correct, a driver with a Q of .707 will roll off at -12dB/oct. For a Qts above that, the initial slope will be steeper, below it will be more gradual. The ultimate slope will always be -12dB, but that occurs below meaningful output (and only in theory, as you'll always have room gain). And Charlie's right as to why higher Qts drivers are preferred for OBs. The catch is, a high Qts pretty much guarantees a high Fc.

As to power handling, you have to investigate how the manufacturer measures it. If they measure in free air, you're all set. If they measure in a box that restricts cone motion, it can be a bit trickier. Myself, I'd take being measured in a box as a strong indication the driver was not suitable for OB use (though it some cases it may simply be the company's "standard practice").

Ack! Sorry, I'm afraid I explained nothing, except to those who already know the answer. :o

Here's a link to a simple definition of "Q":
Passive Crossovers: The Filter Q

(Wikipedia used to be good for such things, but the internet show-offs have elevated every technical page to a doctoral level discussion. :mad:)

And yes, that's a page about filters, but what happens to the low end of a driver is basically a mechanical filter that follows the same rules as electrical ones do. This is the basis of all Thiele's work (Small was left with the job of sorting out the differences between the electrical and mechanical).

It's a "second order" filter (also called "two pole") and that accounts for the -12dB slope. Every order means -6dB. The cancellation of an open baffle creates a first order filter, and so adding the two together, we get the -18dB of a third order filter.

The speaker you mentioned, with a Qts of .45, exhibits the gradual slope of the green line in the graph on that page. To that you need to add the roll-off determined by your baffle width, and add in a factor for room gain. Then you can calculate a lowpass filter with the inverse roll-off for your subwoofers. Simple, no? :headbash:

This is why measurement tools are so popular here. It's much easier to just build the thing, measure it in your room, and tweak the subs until it's flat.

I can measure after but I'm looking to start with a trapezoid front panel of 15" wide at the base tapering to 9"at the top and 42"tall for an ear listening height of 36. (Think apogee)

It will be an 80-100hz 4th order elect roll off. Im not looking for huge SPL, just satisfying audio with great depth illusion. Its using a Neo 3PDRW dipole for the top end and RS180 for the midbass with a 2K 4th order cross, all active.

Its a 21 foot"x 21 x 9.5 room. I own all these drivers so I'm sort of testing here. Its not a final draft.

Any suggetions for a better midbass for use with the Neo3PDRW? I have a pair of focal 6V4311 with an FS of 52 and Qts of .57 if I recall. Those cones can cross really high.

15" wide seems more like a mid range OB.

10' wide is a bass OB.
Or better, a true infinite baffle. i.e. inserted into the wall separating two rooms. The other "room" can be the basement or the loft.