Building an Enclosure Around a D.A.S D-401 2395 Clone

[ivica]

Hi Ian,

Re your sept 8 post # 86 with a tricky question about excursion when doubling cone area. I see nobody replied to you 4 days later... For myself I didn't reply since I had already seen part of the answer, and may be told I cheated.

However, If you look at my post # 82 dated sept 4 (4 days before yours) you can see I quoted from Dickason's Cookbook, 5th ed., p. 30-1. On those same Dickason pages I read 4-5 days before your question, a partial or incomplete answer to your question appears. So no real contest. Read below for more.

However again, I'm NOT satisfied with his explanation of driver coupling effect on sensitivity, doesn't seem to make sense, looks like a contradiction or typo made its way in his text. This is why in my sept 4 # 82 post I quoted John Eargle instead for the coupling effect on sensitivity issue, and Dickason on the voltage/impedance issue.

That speaker experts seem to disagree on science (!) is surprising, but does exist. Look at Bullock's book on p. 62 bottom right column where he says + 6 db for two woofers in parallel, and on p. 66 left column from the middle and on where he repeats that, as opposed to Eargle's clear explanation of + 3db for coupling and another possible + 3 db for power handling I quoted in post # 82 (Handbook of Sound System Design, p. 114, and Loudspeaker Handbook, p. 79). Then have a look at Dickason's, 5th ed., p.30 for a confusing explanation of 3 vs 6 db in 1, 2, and 4 woofer formats ...

I tend to give more credibility to Eargle since his manuscript was reviewed before publication by W. J. J. Hoge, another well-known Speaker Engineer, who would normally have picked-up a gross error... JBL's own Sound System Design Reference Manual says the same as Eargle (Authors: Augspurger and Eargle).

Same power input to identical single and double-woofer (in parallel and closely mounted) boxes, gives a 3 db sensitivity advantage to the double-woofer box and cone excursion would be half that of a single woofer box. Dickason stops here, the rest is mine.

This makes sense because a larger cone area can move more air or "take a bigger "bite" at it" and need not go as far (excursion) to reproduce the same sound. If I remember correctly the good old days, the E-V 30" woofer and the Hartley 21" woofer didn't have nor needed a lot of Xmax considering their huge cone area that moved lots of air more efficiently.

But since you mention specifically "for the same output spl" then that would imply a half reduction in input power to the double-woofer box (to get - 3 db) for spl to be equal to that of single woofer box. Logically, lower input also means less excursion and distortion, therefore the answer would be "quarter", if its already half the cone travel at + 3 db as mentioned by Dickason for double-woofer compared to single. Regards,

Richard

Hi RMC,

Please read
http://trace.tennessee.edu/cgi/viewc...t=utk_gradthes

I think there are a lot of answers. especially section : "Mutual coupling.."
"....2.3 Mutual Coupling
The concept of mutual coupling between loudspeakers is familiar to anyone who has mounted two loudspeakers close together. The power output of the two loudspeakers is approximately four times (+6dB) that of a single loudspeaker. Also, if you double the area of the diaphragm of
a loudspeaker drive-unit, given the same diaphragm velocity, the power output will again increase by +6dB. Reference to equation (2) shows that introducing a second loudspeaker close to a first will approximately double the pressure on each of the diaphragms, thereby doubling the power output of both loudspeakers.
What is perhaps less obvious however, is how introducing a distant second loudspeaker can double the power output of a loudspeaker. For the 3m separation and 0.15m radius of the pair of loudspeakers in the above examples, the magnitude of the pressure on loudspeaker A due to the operation of loudspeaker B is approximately one twentieth of the pressure on A due to its own velocity. How can an increase in pressure of 5% cause a doubling of power output? The answer lies in the phase of the two pressures. At low frequencies, the pressure on the surface of A due to its own velocity is almost in phase quadrature with the velocity - the radiation impedance is almost totally reactive - whereas that from B arrives almost in-phase with the velocity due to the propagation distance involved. Equation (2) tells us that it is only the in-phase part of the pressure that is responsible for power output. As the distance d is decreased, the magnitude of the pressure due to the second source increases but its phase approaches that of the pressure due to the velocity of the first source - the power increase remaining at +6dB but extending higher in frequency - until the "two close loudspeakers" situation exists. As can be seen from equation (3), the frequency up to which the mutual coupling occurs is determined by the distance between the two sources; as the propagation distance approaches half a wavelength the phase of the pressure from the second source is no longer in phase with the velocity. The distance over which mutual coupling occurs is known as the extent of the hydrodynamic near field of the loudspeakers.
...."

or read here:
https://books.google.com/books?id=Ts...upling&f=false


regards
Ivica