Subwoofer bracing not important? What?

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transmission loss

(sorry for my english)
Hi paulspencer,
right idea , the "double symmetrical closed box".
I have done one some years ago:) and the result is CATASTROPHIC:
Below 100 Hz the transmission loss is near to zero!
After, I "redirected" all my energy in the mechanics field.
Big work for next 50 years!!
best regards from Venezia
 
I've been doing some hunting after inertial's comments. This turned up:

An externally hosted image should be here but it was not working when we last tested it.


Figure 1. Transmission losses of typical single-leaf walls, A: 16 mm plywood, 10 kg/m², STC 21; B: 13 mm wallboard, 10 kg/m², STC 28; C: 1.3 mm steel, 10 kg/m², STC 30; D: 100 mm concrete, 235 kg/m², STC 52.

([url]http://irc.nrc-cnrc.gc.ca/cbd/cbd239e.html)[/URL]

Notice how the 16mm plywood and the 1.3mm steel is about the same? Not what I would have expected. It has got me thinking about the possibility of using a sandwich laminated panel of sheet metal and 3mm MDF for future planned curved subwoofer walls ...

As you can see in the chart, the TL drops below 10dB and this is a bit or a worry! Concrete 100mm thick is approx 30dB.

As far as TL goes, according to one site "The best barriers are heavy, high mass, limp, highly damped materials with a high weight to stiffness ratio such as sheet lead or mineral loaded polymeric vinyl." ([url]http://www.domesticsoundproofing.co.uk/tloss.htm)[/URL]

"At low frequencies the stiffness of the material is the main controlling factor. Just above this point, various resonances cause major variation in sound transmission. "

This demonstrates why bracing is needed in subwoofers - stiffness control.

Here's a chart which is interesting:

An externally hosted image should be here but it was not working when we last tested it.


Figure 2. Effect of air space on ideal double walls with 0.5 mm steel on each face, sound absorbing material in the cavity and no rigid mechanical connections between the faces. A has an airspace of 100 mm, a resonance dip at 135 Hz, and an STC of 29; B has an airspace of 5 mm, a resonance dip at 630 Hz, and an STC of 24. Curve C represents mass law predictions for a single 1 mm steel sheet and has an STC of 28.

This is effectively "constrained layer damping" and what is interesting is that it only works in the mid to high frequencies! It appears to have no benefit for bass! This is ignoring the benefit of CLD as vibration control. Vibration control from the driver and transmission loss are two different things.
 
(sorry for my english)
Hi Paul,
Well done!
It is time to serious things! :)
These are normal diagrams , BUT I see "IDEAL", "IDEAL", "IDEAL".
I assicure all the interesting people, in the real behaviour all it is
MUCH WORSE!
Who is researching in this field ? Maybe High-end audio designer?

By
 
These are normal diagrams , BUT I see "IDEAL", "IDEAL", "IDEAL".

They appear as based on measured data to me. Perhaps they have been smoothed a lot, but you get an idea.

They don't really show bass, but if you extrapolate the curves, you get a fair idea. Better performance might be achieved, however, due to bracing, as this is data intended to be used for noise reduction in rooms, where the materials span further.

The attached file gives an idea of how transmission loss through a material is different through different parts of the audio bandwidth. In a subwoofer, the mains concerns would appear to be mass and stiffness. You can see how with a fullrange speaker, things are different. In the resonance controlled region, a speaker box made from solid materials with a high Q and low daming (hardwood rather than softer timber or MDF, or metal or plastic perhaps) would suffer in the lower midrange.

In the midrange, damping in the speaker box has more effect. I'd say the midrange is perhaps the area most likely to suffer from box coloration, but in this range it is easier also to have a more inert box.

It all becomes more complicated where you have to also consider both transmission loss and the vibration of the box via the mechanical coupling of the driver to the box. This is where I see value in attaching the driver to an inner box, then placing another box around it with a damping layer in between - constrained layer damping. If you built a push pull arrangement with the drivers on opposite sides of the box coupled to a constrained layer vibration damping design with matrix bracing on the inner box, then you would have a box that would perform very well. Use a sealed box with a pair of XBL2 drivers and you would have something very accurate.
 

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(sorry for my english)
Hi Paul,
" Mass law" presume much ideal semplifications.
Anyway, the last diagram show TL caused by aerial waves .
Structural transmission is much more efficient ( and complicated)
( how you said) .
For max stiffness :
a) shape
b) size
c) material
These determinate the numbers of vibration modes , etc,etc.
What is most stiffness shape? ( easy :) )
About thick for closed or bass-reflex: this is never enough!!
About two opposite wf .....
I think this: Imagine your hand between two hammers ( one from left
and one from right) Could be the result = zero? I think not. :)

regards
 
I think this: Imagine your hand between two hammers ( one from left
and one from right) Could be the result = zero? I think not.

Not a good illustration! You had will be crushed on both sides, but this is different!

In high school physics the concept of wave cancellation was demonstrated by having waves move towards each other from opposite ends. At the moment when the waves pass, they cancel each other out if their polarity is opposite. I believe this is what happens with mounting drivers opposite each other on a box - one on front, one on back.
 
(sorry for my english)

Oh,incredible!
Imagine two identical earthquake one at north and one at south,
are the city in the middle intact?
What type of waves are you speaking ?
Put your 2 wf at the opposite side of a tube. It vibrate!
Structural transmission are not a my opinion but if you can show
the opposite you are welcome.
By
 
paulspencer said:
In high school physics the concept of wave cancellation was demonstrated by having waves move towards each other from opposite ends. At the moment when the waves pass, they cancel each other out if their polarity is opposite. I believe this is what happens with mounting drivers opposite each other on a box - one on front, one on back.

Suppose you have two woofers on opposite sides of the box Deliver a pulse to each, pushing the speakers "out" for an instant, this will push the baffle "in" for an instant on each side, This inward wave will pass around the box, and if there is symmetry and no damping, the waves will add when they meet. This is what "inertial" is talking about.

If you take a box and mount a driver on one side, another box and mount the woofers antipodally and put it on a track that constrains motion in the direction of the cones' axis, then measure the vibration of the cart, it will be less with a push-push system with antipodal woofers. THis sort of vibration is what is reduced in this manner. This is reaction vibration, not box vibration, per se...

Things are more complicated than that to judge overall vibration. Using two woofers means a bigger box with bigger panels which have a lower resonant frequency. Bigger panel area means more overall force on the panels for a given pressure change in the box and more panel vibration given the same contruction. The bigger you make the box, the more you have to brace it, regardless of how you mount the woofers.
 
I mentioned this earlier. Push push only cancels "whole box vibration" Panel vibration is an entirely different thing. When the entire box vibrates due to inertia forces from the moving cone of the speaker I would imagine it acts as a dipole. The entire box moves slightly one way then the other following the signal of the speaker to some degree. The vibration can easily be coupled to the floor (in down or upward firing subs) or a side firing design can be used which will couple alot less. Ports should have a similar effect. If you have one port on the box the air mass moving in the port should have a slight inertia to it that pushes and pulls on the box. About ported boxes needing less panel thickness, I dont think thats true at all. In my experience a small hole such as a screwhole that was misplaced, will whistle more in a vented box at resonant frequency than a sealed box. This would lead me to believe a higher pressure difference between the inside and outside of the box and more panel flexing. I may be wrong here but my idea why this occurs is because the port resonance creates higher peaks and dips in the pressure wave at or near resonance. Anyone else have ideas on this?
 
the tuning is where i think pressure increases in a vented box versus a sealed box. Thats why the excursion would decrease, think about it... If the pressure is higher in the box then when the cone moves in it is stiffened more, if the pressure is lower in the box then when the cone out it is stiffened more. Therefore the motion would be decreased. The port and air in the enclosure resonate and this is what causes the increase and decrease in pressure inside the box. I guess I still have no scientific proof to show for it.
 
If you consider the vent as a controlled tuned leak, then the pressure would be less overall. At least, intuition suggests a fully sealed box would have more pressure for the same cone excursion. However, the compound the confusion, vented boxes reduce cone motion near tuning. Perhaps someone can clear this up!

you are probably confusing resonace of the cone with resonance of the system (box tuning). At resonace the movement thru the port is at a maximum and the output is in phase. Therefore the box offers greater resistance to the cone movement. As the cone moves in the box pressure is greatly increasing due to port intake and out is the opposite.

Try this put a mass on a long rubber band and hold it with you your hand and mone your hand up and down till you find resonance not e that the mass movement (port volume) is at a max and the resistance to your hand movement is alo at a max (cone movement)

Or take a driver and attach a spring mass to it that matches the box and watch for a peak in the drive current.

for a spring mass f = 1/2pi*root(k/m) wher k is spring constant and m is mass and for a helmholtz k = p*a*a*v*v/V and m=pal where p is the air density, a is the port area, v is the velocity of sound in meters and V is the volume of your box in I believe liters and l is the length of the port witch must be adjusted for the open ends.

May be some one understands this better
 
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