catapult said:
This would be true with monopole bass (spherical wave), but the setup I am describing is planar (plane wave), so no attenuation. In practice you'd get some attenuation because the surfaces of the room, the items in it, as well as the air itself will absorb a little and, but pretty minimal at this frequency.
John
454Casull said:
If you want some feedback on this you'll have to give me some more details on why you think this is the case. I am pretty confident that you will get a plane wave.
It's a bit more complicated than just slope, but basically yeah...
Here is an example of a low-pass linear-phase filter that doesn't ring:
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(Dang, can someone tell me how to set spaces in this thing?)
Subtract that impulse response from a unit impulse response and you've got yourself a high-pass linear-phase filter than also doesn't ring. Put the two together and you've got yourself a linear-phase crossover. If you did an FFT of these filters, though, you'd see that it has a pretty gentle roll-off in the crossover region.
John
Bill F. said:
That sounds pretty cool too. I love extreme solutions. I remember seeing somewhere on the net a guy who built stereo bass horns into his room with bricks. I wish I still had that link.
The logic behind my idea is to really make a small room sound like a large room, to do that you need to maximize the left/right energy of the bass. I really think this would be the ultimate.
John
Bill F. said:
What matters here is really the shape of the wave front. You want to cancel out the wave when it hits the boundaries of the room so that there are no reflections. That's only going to be practical when you are setting up plane waves in opposite directions in a rectangular room. If you use two monopoles on either side of the room, they won't cancel the reflections. I hope that is clear. It is a bit hard to describe.
John
My vision of ultimate speaker system is related to simplest is the best pricipe.
So, to make that kind of system I will use a little more exotic drivers like Manger for poin source or Bohlender-Graebener for line source.
Those are alomost a full range drivers so I do not need to use a crossover points higher then 200-300Hz. I will make the best enclosures for those units , and they will be connected directly to my (or someon else) power amp without passive components.
The bass section will be reproduced by high speed active woofer system made by four 8" drivers in large volume sealed box.
Next step is design of high resolution active crossover with adjustable parameters. This unit will take low frequecies away from "full range" before power amp that drives them. For subwoofer is the same thing but in reverse . On active crossover unit you can fine adjust levels , phase, Q factors for bass section , and it will be made with step rotary swiches for precise control , becouse with pots you never know exactly position.
OK , so this is it , more or less
So, to make that kind of system I will use a little more exotic drivers like Manger for poin source or Bohlender-Graebener for line source.
Those are alomost a full range drivers so I do not need to use a crossover points higher then 200-300Hz. I will make the best enclosures for those units , and they will be connected directly to my (or someon else) power amp without passive components.
The bass section will be reproduced by high speed active woofer system made by four 8" drivers in large volume sealed box.
Next step is design of high resolution active crossover with adjustable parameters. This unit will take low frequecies away from "full range" before power amp that drives them. For subwoofer is the same thing but in reverse . On active crossover unit you can fine adjust levels , phase, Q factors for bass section , and it will be made with step rotary swiches for precise control , becouse with pots you never know exactly position.
OK , so this is it , more or less
Bill F. said:
The pressurization happens because of the reflections off of the opposing walls that reinforce the source signal--they reinforce because the walls are less than 1/2 wave away. However, if you don't have the reflections, you won't get any pressurization.
Another way to think of this is the following:
In free space, the pressure of a plane wave is proportional to the partical velocity of that wave:
free space SPL=v (v=velocity)
Put that plane wave into an acoustically small box (like your room) and add the refections, though, and SPL is proportional to the displacement of the wave. In effect, the room acts as an integrater:
room SPL=integral(v)=d (d=displacement)
The integral of velocity equals displacement.
What I am proposing is to subtract out a time-delayed version of the original signal from the room:
my room SPL=d(t) - d(t+delta),
where delta is the time delay. If delta is small relative to a wave length, then:
my room SPL=d(t)-d(t+delta)~=derivative(d)=v
and we're back to the free space SPL.
I hope that helps,
John
454Casull said:
But that sound will be reflected back. Essentially you are using the room as a transmission line. Perhaps the easiest way to think of this is with a mirror image trick:
http://www.silcom.com/~aludwig/Physics/Main/Image_analysis.html
http://www.steve-m.us/rir.html
In a mirror image model, the wall of woofers becomes an infinite plane of woofers.
John
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