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Old 6th January 2003, 10:07 AM   #11
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This is what I meant when I was talking about more elegant foldings. The length ratio is also 1:3 (i.e. driver moved in by 1/4 of the total line-lenght) as with the straight wave cannon.
It needs less space than a straight wave cannon but has less bends than the labyrinth proposed in the B@£& patent.

I have never tried it myself though I was interested in this woofer principle some years ago. This was the folding i wanted to use back then.
Although I think a straight cannon would work better if you have the space for it.

Regards

Charles
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Old 7th January 2003, 03:42 PM   #12
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Quote:
Originally posted by kelticwizard

I think the long tube is between 1/4 and 1/2 the wavelength of the F3 frequency aimed for, and the short tube behind is 1/3 the length of the long tube. It might not be clear from ther illustration, but both ends are open to the air.
You know, I have read that patent over, and I am still not exactly sure if the length of the two tubes added together amount to one wavelength of the lowest frequency to be reproduced well or one quarter the wavelength of that frequency.

The illustration shown uses a 4 1/2 inch driver similar to the one used in the Bose 802, and the back tube is about 4 feet or so.

If you add the length of the front tube, that comes to a total length of 5 1/2 feet. Which is about the wavelength of 200 Hz, but one quarter the wavelength of 50 Hz.

The drawing is just there to illlustrate a principle, but one must ask why anyone would go through all that to make a 4 1/2" speaker that goes down to only 200 Hz!

I shall reread the patent and try to come up with an answer as to length of the tube vs. frequency.

The patent also says that a good cross section of the tubes would be 2/3 to 1/2 the cross section of the speaker cone.
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Old 7th January 2003, 04:14 PM   #13
Wizard of Kelts
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Is this the unit you were referring to? It is 12 feet long.

Twelve feet is one quarter the wavelength of 24 Hz, and the Cannon is rated down to 30 Hz or 25 Hz. However, I do not know if the tube is straight, or if there is a folded or even a spiral arrangement inside. The senstitivity is listed at 90 dB, with a peak of 115 dB.

The width is 17". There is a bracket on it-I don't know if they refer to the width including the bracket or just the tube. If they include the brack et in the specs, that would make the outside dimensions of the tube 14", I suppose.

Even with a figure of 14" outside diameter for the tube, that would still make the outside dimensions over 13 cubic feet. There are ways to produce 25 or 30 Hz at 90 dB sensitivity in enclosures much smaller than that.

For instance, an Adire Shiva could be put into a box about a third that size and produce 25 Hz at the rate of 88 dB/watt.

Anyway, I will reread the patent, unless someone else comes up with an answer as to the ratio of tube length to lowest wavelength reproduced.
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Old 7th January 2003, 05:23 PM   #14
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Quote:
Originally posted by DrewP
As far as I can recall the Bose cannon can be approximated by placing a long stroke driver (Peerless 8 ohm XLS 10 inch perhaps would be good) 1/4 of the way down a 6m long pipe.

I'll need to crunch the graphs again to look at the predicted response but from memory it was fairly flat (peaks and troughs less than +/-2dB ) from around 18Hz up to about 90.
Drew's configuration would appear to be consistent with a 12.5 foot pipe with an F3 of 27 Hz, which is what the Bose Cannon is. Drew's pipe is about 50% longer than the Cannon, but his F3 has a wavelength that is about 50% longer as well.

So it looks like there is no special routing inside the tubes-apparently they are just straight tubes. It looks like a good bet to just scale your F3 and the length of your cannon to Drew's and Bose's guidelines.

By the way, the promotional flyer that I got that picture from is located here, along with Bose's specs:
http://www.ccidaho.com/product/techn...dfs/awcs2t.pdf
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Old 7th January 2003, 08:39 PM   #15
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Default I made a quick test

I bought a 10 " speaker (pentivent PV-1030) with Fs of 19
2 sonotube and put the driver in the 1/4 of the tube (12 foot long)

the result is poor .It have a resonance at one frequency (deep base!!) but not very loud. (not as loud of my energy es-18xl) The sound is terrible.... I don't know why My cross-over is set to 80 Hz 12 db/ octave

I don't have time this week to make other test.....

Any idea is welcome

René
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Old 7th January 2003, 09:38 PM   #16
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Remember the graph of response in the Bose patent? I'm thinking that a 12 foot pipe will place the lower of the 2 peaks at just below 50Hz (possibly 45). If that's the case then your 80hz xo is likely missing the upper of the 2 peaks.

for deep bass I think you need longer than 19foot of pipe.

Could be that current setup corresponds with your room modes as well which wouldn't help matters any.

Drew
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Old 9th January 2003, 07:31 PM   #17
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Hey, does it matter what the T/S parameters of the driver are, as long as the tubes on each side add up to a quarter-wavelength of the lowest frequenciy desired?
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Old 9th January 2003, 11:03 PM   #18
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Bam:

I think the patent mentioned something about resonant frequency and wavelength. I will check.

The only thing that they mentioned was that the 4 1/2" speaker in the patent illustration was similar to the one used in the Bose 802. However, I was unable to locate Thiele-Small parameters on the 802 speaker.

PS: What an appropiate nickname in a thread about the Acoustic Cannon .
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Old 9th January 2003, 11:35 PM   #19
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Here we go about the resonant frequency. This is from the patent:

"The free air resonant frequency of the loudspeaker driver may be chosen to be that at which the length of the longer of the tubes is a half wavelength and thereby lessen response irregularities that might be produced by resonances between reactive components of the loudspeaker driver and the tube. Preferably, the loudspeaker driver is overdamped to avoid undesired resonances between the loudspeaker and the tube."

Okay, if we do the math, (and if I'm doing the math, are we ever in trouble), that means the resonant frequency should have a wavelength 1 1/2 times the length of the tube, front and back together.

So a 12 foot tube should have an Fs wavelength of 18 feet, which means that Fs = 62.5. Uh-oh.

Overdamped means a low Qts, I believe. Since .38 seems ot be a dividing line of sorts when discussing Qts, I guess that means a speaker with a Qts of .30 or below. this is speculation on my part.

Maybe we shuld look among speakers designed for bass guitar?

Perhaps that line from the patent, "The free air resonant frequency of the loudspeaker driver may be chosen to be that at which the length of the longer of the tubes is a half wavelength" might mean that it should be at least twice the wavelength of the longer tube, in which case the speaker with the Fs = 19 Hz might be fine.
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Old 10th January 2003, 12:03 AM   #20
Wizard of Kelts
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The diameter of the tube also affects frequency response. I mentioned the ratio of the area of the tube to tarea of the cone as being perhaps 2/3 to 1/2. This is from the patent:

"For a given ratio of (B1).sup.2 /R.sub.e the ratio of tube to cone areas (ATCR) typically controls the size of the system response peaks at the frequencies where the tube length is an odd multiple of a quarter wavelength for a single tube. For some typical speakers and an ATCR of 1 these peaks are relatively large. For ATCR of 0.5, the system response is relatively smooth. For ATCR less than one half, system response decreases because the tube provides increased load on the loudspeaker cone."

A 12" louspeaker typically has a cone area of 86 sq. inches. An 8" tube has an area of 50 sq inches. That is 58% of the 12 incher's cone area. Perhaps putting an 8" tube sown each 12" tube, held in place by 2 12 inch round brackets, would smooth the response.
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