System Q for BR in hornresp

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Hi

This may be a stupid simple question, but I cannot google the answer.

Optimal Q for a BR-box is said to be around 0.7 (depending on the source).
But can hornresp calculate the the Q of a simulated BR box?
If HR cannot calculate it, any good guidelines?

Kind regards TroelsM
 
THere is no such thing as "System Q" for a bass reflex box as it is a 4th order system. System Q is a 2nd order concept.

Driver Qts for a vented box can be anywhere from around 0.15 to 0.6 or so, but anywhere above 0.4 and the boxes start getting really large, really fast for anything like a flat response.
 
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Q

@ TroelsM

Optimal Q for a BR-box is said to be around 0.7

I know what you mean, & i don't have a problem with that definition ! See next.

I must have misunderstood something. I was pretty sure that the alignment of a BR-box would determine the Q and thus make the system possibly overdamped or underdamped.

The alignment of a BR-box DOES indeed dictate whether it's overdamped or underdamped.
 
There is no such thing as "optimal" q for anything, it depends on what your goals are.

For a closed box a qtc of 0.7 is a max flat alignment, it will have the flattest possible response to the lowest possible frequency. This may or may not be a good thing depending on how flat you want the response and how low the response extends and what type of environment you are in (a room or outside). A qtc of 0.5 is sometimes referred to as critically damped and in room it may be flatter than 0.7 qtc. For a sealed box you can't get a qtc lower than the driver's qts, so in infinite baffle the qtc = qts and in any box smaller than infinite baffle qtc will be higher than qts.

There is no qtc for ported boxes but the frequency response curve shape can be compared to qtc values of sealed boxes and you could give the curve shape a q value if you wanted to but it's not something most people do. There is a "max flat" ported alignment which will do the same thing as the 0.7 qtc sealed alignment - give the flattest possible response to the lowest possible frequency. This ported "max flat" alignment will usually extend to a fairly low frequency and may (probably will) sound pretty boomy in room. Max flat is probably the most popular diy alignment for ported boxes and it's very likely the reason ported boxes get bad reviews - they sound boomy as hell in room because the room gain boosts the low end and it sounds "slow", "boomy", and all the other audiophile terms for bloated bass.
 
Hi,

Q is a property of second order resonant system, 0.7
is maximally flat, known as a Butterworth alignment.

Vented boxes are effectively 4th order, so that is two Q's in series.
Maximally flat is still known as Butterworth, but is generally a very
bad choice for vented boxes for the reasons outlined above.

Of interest for a vented box is the equivalent Q of a sealed version.
Simply put you want about 0.4 to 0.6 for vented and the higher the
Q the lower you set Fport compared to Fresonance of sealed.

If Qbox is higher than about 0.6 you'll have a hard time venting.

IMO the best vented alignments are overdamped and emulate
a 1st order roll-off above Fport. Fport is necessarily tuned low.

rgds, sreten.
 
A maximally flat vented box has a driver Qts of about 0.4, and has Vb~=Vas and Fb=Fs. For Qts below 0.4, Fb is usually above Fs and vice versa.

For a vented box, a good build rule of thumb (Small-Margolis, 1981) is:
Vb=Vas*20*Qts^3.3
Fb=0.42*Fs*Qts^-0.96
F3~=0.26*Fs*Qts^-1.4

It could be said that any box with this box size and a tuning below the recommended above is "overdamped" (bass response will droop), and if tuning is above that given above, it is "underdamped" (response will peak)

In vented boxes you can also adjust volume, and port output at Fb goes up by 6dB for each doubling of volume (or down by 6dB if you cut in half - all else equal).

A vented box will sound fine as long as there is no peaking, IMO. Sreten's comment about first order above Fb is just going to confuse you.

Vented boxes essentially always have a less damped response than sealed, but once Fb is low enough, you will not have any of hearing the ringing. In a room, the room's own resonances will drown all of this out.
 
Vented boxes are effectively 4th order, so that is two Q's in series.

This is an oversimplification, and essentially useless as an aid in understanding a VB.

If Qbox is higher than about 0.6 you'll have a hard time venting.

What are you trying to say here? I would interpret Qbox as box losses (combined leakage, absorption and port) and if Qbox is 0.6, there isn't much point at all in venting...
 
This is an oversimplification, and essentially useless as an aid in understanding a VB.

What are you trying to say here? I would interpret Qbox as box losses (combined leakage, absorption and port) and if Qbox is 0.6, there isn't much point at all in venting...

Hi,

1) Its not an oversimplication, all 4th order is cascaded 2nd order.

2) So you completely agree, unless you want a boombox alignment.

rgds, sreten.
 
Hi,
1) Its not an oversimplication, all 4th order is cascaded 2nd order.

So you won't have any trouble deriving two "real, unique" Q values for "any arbitrary" BR alignment, will you? ;) And their combination will ultimately solve the question of what is over and underdamped in a vented box entirely, right? That's sarcasm, in case you didn't catch that...

2) So you completely agree, unless you want a boombox alignment.

You misunderstand entirely. What is Qbox? I interpreted it as 1/Qb=1/Qa+1/Ql+1/Qp, or any one of those on its own.

If Ql or Qp or Qb are 0.6, the box is essentially an extremely leaky sealed box, or a really lossy port vented box, which is the same as a leaky sealed.
 
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But can hornresp calculate the the Q of a simulated BR box?

Hornresp calculates the total quality factor Qtc for direct radiator closed-box systems only.
 

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Optimal vented box alignment uses a driver with free air Q=.383 not including amplifier low damping or other losses. This gives most flat response and strongest coupling over the sympathetic resonator bandwidth in the properly designed vented system. Of course, whatever comes out of the vent is 1/2 cycle behind what the main driver is doing. And of course, any small vent is going to quickly reach saturation meaning power compression limiting output. As example, a 2" vent tuned to 35Hz will have a max SPL output of about 85dB at 35Hz. There were many papers written on why ports and vents do not work due to power compression. Passive radiator or large vent is a better choice. The general rule is the port should be at least 1/3 the area of the active driver(s).
 
Thanks a lot for the informative answers. Whish I had more time and brainpower to read up on all the theory again.

Reason for the original question is that I'm looking at stuffing a 10" SP252E into a "smallish" cabinet. Maybe 30-40Liter. It will be active and I can do simple EQ'ing within the electronics cross-over. So I'm looking at different options: Closed, BR, small TH and other configs in Hornresp.
The smaller BR cabinets in Hornresp give a little extra low-end, but I'm concerned about the alignment being too "boomy/onenote/ect..".
I could just close the box and add a little low-end with EQ, but a smallish closed box may end up being overdamped and thus not very fun for pop/rock/edm.

40-50Hz lowpass (-3dB) is OK.

Driver-details for those who are curious:

Impedance (Z) : 4 Ohm
Frequency range : f3-6,000 Hz
Resonant frequency (fs) : 42 Hz
Music power (PMAX) : 150 WMAX
Power rating (PRMS) : 75 WRMS
Sensitivity : 94 dB
Suspension compl. (Cms) : 0.33 mm/N
Moving mass (Mms) : 43 g
Mech. Q factor (Qms) : 3.13
Electr. Q factor (Qes) : 0.60
Total Q factor (Qts) : 0.50
Equivalent volume (Vas) : 57 l
DC resistance (Re) : 3.75 Ohm
Force factor (BxL) : 8.5 Tm
Voice coil induct. (Le) : 1.1 mH
Voice coil diameter : 35.5 mm
Voice coil former : aluminium
Linear excursion (XMAX) : ±4 mm
Eff. cone area (Sd) : 353 cm2
Magnet weight : 960 g
 
So you won't have any trouble deriving two "real, unique" Q values for "any arbitrary" BR alignment, will you? ;) And their combination will ultimately solve the question of what is over and underdamped in a vented box entirely, right? That's sarcasm, in case you didn't catch that...



You misunderstand entirely. What is Qbox? I interpreted it as 1/Qb=1/Qa+1/Ql+1/Qp, or any one of those on its own.

If Ql or Qp or Qb are 0.6, the box is essentially an extremely leaky sealed box, or a really lossy port vented box, which is the same as a leaky sealed.

Hi,

All vented alignments can be modelled actively by two cascaded
2nd order filters because in theory that is what a vented box is.
(In the same way all sealed boxes are a 2nd order filter.)

Over and under damped can be derived from the step response.

Qbox is simply driver Q in the box if it was sealed. Venting is
generally not a good idea if Qbox > 0.6 for hifi, but that hasn't
stopped boombox vented bass aligments in the slightest.

rgds, sreten.
 
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Hi,

All vented alignments can be modelled actively by two cascaded
2nd order filters because in theory that is what a vented box is.
(In the same way all sealed boxes are a 2nd order filter.)

I know what a cascade is. In general cascaded filter sections are not coupled. The bass reflex is a coupled system
Pull out those two Q's you talked about and show us how useful they are.

Qbox is simply driver Q in the box if it was sealed.

I've never seen a reputable paper use this terminology, or this concept.
Show us how it's useful!
 
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I know what a cascade is.
Pull out those two Q's you talked about and show us how useful they are.

I've never seen a reputable paper use this terminology, or this concept.
Show us how it's useful!

Hi,

I'm not interested in pugilism. The shape of a vented boxes roll-off
is best explored in a speaker simulator, and the best I can say is
the curve of the roll-off is directly related to the transient response.

(The more gradual the curve, the better the transients. Flat with
a sharp roll-off rings at the roll-off. Nice is a gradually accelerating
roll-off towards the bottom end, with no sharp corners or knees.)

Its all standard filter theory, but I agree two Q's don't help most,
because for vented you also have two F's, related to to each Q.
It should be obvious what the two F's are.

I can't believe you don't understand vented boxes in terms of what
the driver would do in the same sealed box, its pretty fundamental.
FWIW if Qbox < 0.6 then venting generally is a very sensible option.

rgds, sreten.
 
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