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From: don@umd5.UUCP
Newsgroups: net.audio
Subject: Re: The physics of good sound a Bose speakers
Message-ID: <461@umd5.UUCP>
Date: Thu, 4-Apr-85 22:07:00 EST
Article-I.D.: umd5.461
Posted: Thu Apr  4 22:07:00 1985
Date-Received: Sun, 7-Apr-85 03:19:56 EST
References: <2407@drutx.UUCP>
Organization: U of Md, CSC, College Park, Md
Lines: 58

> Subject: How/why do Bose speakers work?
> "It's all done with mirrors!  And microprocessors use micromirrors."
>
--   [ after reading what follows, I tend to believe it !]
> 
> Bose uses nine drivers to achieve an effective piston area equivalent to that
> of about a standard 10" acoustic suspension driver.
>
--   [ nice speaker graphics in his posting ]
> 
> Cone pistons also possess a physical limitation of reproducing wavelengths
> shorter than their diameters.  (That's why good sounding tweeters have small
> diameter pistons).  As mentioned in another article, as the wavelength of the
> signal increases and approaches the diameter of the piston, the dispersion
> pattern of the driver changes from a hemispherical pattern (well dispersed) to
> a cardioid pattern (poorly dispersed).
> 
> -Phil Rastocny

[]
OK, now just hold on here a cotton-picking minute!

   The only formula in Phil's posting is F=ma, and I buy his statement that
for a small mass the transient response of the speaker is better than for
a speaker with a large mass.
   This wavelength stuff he talks about has got me confused ...
A wavelength shorter than the speaker's effective 10 inch diameter is the
speaker's physical limitation you say?
   OK, Let us calculate the wavelength of a 20kHz wave for starters --
   (300,000 km per second) divided by (frequency in kHz) = 15,000 meters
This is approximately 9.32 MILES. Somehow I don't think this figures into
Phil's explanation very well, but let's go with it still --
   The limiting frequency at a 10 inch wavelength is:
   (300,000 km/s) divided by (0.254 m) = 1,180,000 kHz  (about 1.18 GHz !!)
Something is still wrong here! If Phil's explanation was correct, Bose 
speakers would either have to have a 9.32 mile diameter, or they would
still be good tranducers at 1.18 GHz (even your dog couldn't hear that!!).
Well, maybe I haven't punched holes in the explanation big enough to sail
the Nimitz through just yet -- How about waveguides?
If I model the 10 inch speaker as a circular waveguide, the cutoff wavelength
is 3.41 times the radius of the waveguide. Now to calculate:
   (300,000 km/s) divided by (3.41 times 0.127 m) = 693 MHz
Closer to 20 kHz, but still 30,000 times 20 kHz ... Now I don't profess to
know exactly what this dispersion mechanism that Phil is talking about really
is, but I seriously doubt the explanation that he posted.
   Any takers ?

-- 
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"Space, the final frontier .." Final, hell! It's the frontier of frontiers !!
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