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Old 21st February 2004, 09:31 PM   #1
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Post Qb and Vb in acoustic suspension systems

The volume Vb of an acoustic suspension system in terms of its damping factor Qb is typically given by:

Vb = Vas / [(Qb^2/Qts^2) - 1]

Division by zero results in an undefined quantity, so [(Qb^2/Qts^2) - 1] cannot be zero. Thus (Qb^2/Qts^2) cannot be 1. In turn, Qb^2 cannot equal Qts^2, so Qb cannot equal Qts. Also, if Qb < Qts, Vb is negative, which cannot be so in reality.

For practical use we must restrict the above equation to values of Qb in which Qb > Qts. So we never see a driver with Qts equal to or greater than 0.707. In a sealed box, such a driver will never be able to attain a "flat" response!

For drivers with a very high Qts, say 0.68, we can never achieve a Qb of less than that value in a "real" enclosure. So how do we get slightly lean bass response (the result of a Qb less than 0.707)? Never!

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Old 21st February 2004, 09:47 PM   #2
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Your'e right Qb cannot be lower than Qt of the speaker. When you want Qb == Qt you'll need an infinite box volume. So use a very, very big box or better an open baffle.

The only other option is to use horn loading or a LW transform.

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Old 21st February 2004, 10:14 PM   #3
Svante is offline Svante  Sweden
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Default Re: Qb and Vb in acoustic suspension systems

Quote:
Originally posted by coolkhoa
So how do we get slightly lean bass response (the result of a Qb less than 0.707)? Never!

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Lower Qts by a negative series resistance or by stuffing in the box (in the case of a closed box).
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Old 21st February 2004, 10:57 PM   #4
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Quote:
Originally posted by Pjotr
Your'e right Qb cannot be lower than Qt of the speaker. When you want Qb == Qt you'll need an infinite box volume. So use a very, very big box or better an open baffle.

The only other option is to use horn loading or a LW transform.


In practice, that's all we can do. The volume provided by an open baffle is theoretically "infinite," but take into consideration the dimensions of the surrounding space (i.e. room) and it is no longer. Of course, one can always take his open-baffle speaker outside, where the volume of air is truly infinite.
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Old 21st February 2004, 11:19 PM   #5
sreten is offline sreten  United Kingdom
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Default Re: Qb and Vb in acoustic suspension systems

Quote:
Originally posted by coolkhoa
The volume Vb of an acoustic suspension system in terms of its damping factor Qb is typically given by:

Vb = Vas / [(Qb^2/Qts^2) - 1]

Division by zero results in an undefined quantity, so [(Qb^2/Qts^2) - 1] cannot be zero. Thus (Qb^2/Qts^2) cannot be 1. In turn, Qb^2 cannot equal Qts^2, so Qb cannot equal Qts. Also, if Qb < Qts, Vb is negative, which cannot be so in reality.

For practical use we must restrict the above equation to values of Qb in which Qb > Qts. So we never see a driver with Qts equal to or greater than 0.707. In a sealed box, such a driver will never be able to attain a "flat" response!

For drivers with a very high Qts, say 0.68, we can never achieve a Qb of less than that value in a "real" enclosure. So how do we get slightly lean bass response (the result of a Qb less than 0.707)? Never!

Comments?
Qb is always greater than Qts.

Your other suppositions are entirely incorrect,
because for a start your equation is wrong :

Qb/Qts = root (Vas/Vb+1)

(Qb/Qts)^2 = Vas/Vb +1
Vas/Vb = (Qb/Qts)^2 - 1
Vb= Vas/ [(Qb/Qts)^2- 1]

the first equation tells you :
for Vb = Vas Qb = 1.414Qts
for Vb = 1/3 Vas Qb = 2Qts.

the other rearrangements tell you the same thing in reverse.

There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
Though for bass units the range is usually 0.20
to impractically high values above 0.7 up to 1.
(unless open baffle or car free air mounting)

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Old 22nd February 2004, 02:12 AM   #6
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Not sure about this discussion. What do you mean by Qb? Do you mean a speaker with Qts put into a closed box to yield Qtc?

If that is the case, I just thought I would post the formula for how we arrive at Qtc. Hope this helps.
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Old 22nd February 2004, 02:21 AM   #7
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Default Re: Re: Qb and Vb in acoustic suspension systems

Quote:
Originally posted by sreten



There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
I wondered about that too.

I am guessing that when KBK said that, he was assuming that it was understood that we were not interested in speaker/box combos that did not yield absolutely flat response. That "acoustic suspension" system implied flat response.

My understanding is the "classic acoustic suspension " system is as follows:

Qts = 0.38
Vb = 1/3 Vas

Since that time, I believe "acoustic suspension" system has broadened a little to mean a sealed box where the box is appreciably smaller than Vas.
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Old 22nd February 2004, 08:08 AM   #8
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Default Re: Re: Qb and Vb in acoustic suspension systems

Quote:
Originally posted by sreten


Vb= Vas/ [(Qb/Qts)^2- 1]

the first equation tells you :
for Vb = Vas Qb = 1.414Qts
for Vb = 1/3 Vas Qb = 2Qts.
Errh... How did you get from

Vb= Vas/ [(Qb/Qts)^2- 1]
to
Vb = Vas Qb = 1.414Qts ??

Quote:
Originally posted by sreten

There are no restrictions on the value of Qts,
which in practise can vary between 0.1 and 2,
Though for bass units the range is usually 0.20
to impractically high values above 0.7 up to 1.
(unless open baffle or car free air mounting)

sreten.
I think coolkhoa assumed butterworth response and closed box. In that case, since Qb>Qts Qts cannot be larger than 0.7, unless Qts is altered with negative resistance or stuffing in the box.
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Old 22nd February 2004, 08:24 AM   #9
sreten is offline sreten  United Kingdom
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Default Re: Re: Re: Qb and Vb in acoustic suspension systems

Quote:
Originally posted by Svante


Errh... How did you get from

Vb= Vas/ [(Qb/Qts)^2- 1]
to
Vb = Vas Qb = 1.414Qts ??



I think coolkhoa assumed butterworth response and closed box. In that case, since Qb>Qts Qts cannot be larger than 0.7, unless Qts is altered with negative resistance or stuffing in the box.
For a Butterworth alignment Qb=0.707, end of story.
It is a maximally flat alignment with no peaking.

Other Q's give other alignments.

The first equation :

Qb/Qts=root(Vas/Vb-1)

If Vb=Vas, then Qb = root2 x Qts.

This is the equation repeated by KW
with the helful addition of :

Fb/Fs= root(Vas/Vb-1)

For sealed boxes alignments range from Qb = 0.5 to Qb =1.1,
though most are between 0.6 and 0.9, with the 0.7 Butterworth
being the most common for builders.

For a sealed Butterworth alignment any driver with Qts
below 0.7 can be used though Vb gets silly with Qts
near 0.7 and Fb gets silly with very low Qts values.

A Qts of 0.35 indicates for Qb = 0.7, Vb = Vas/3.
A Qts of 0.50 indicates for Qb = 0.7, Vb = Vas.


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Old 22nd February 2004, 09:36 AM   #10
sreten is offline sreten  United Kingdom
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coolkhoa,

I now see what you where describing, apologies.....

simply put :

if the design target is an inbox Qb of 0.7,
a sealed Butterworth alignment,
Qts must be less than 0.7.

A driver with Qts > 0.7 cannot be used by definition.

Drivers with Qts near 0.7 will give silly box volumes.

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