Is the most efficient speaker the best speaker?

06-26-11: Weseixas
Hello Al,
Per meter = doubling of measured reference distance..

So therefore in your example, when you said ...

06-26-11: Weseixas
@ 15 feet the 100db/w/m horn would have an out put of 76 db/w ... To achive 94 db@5M the SET.horn combination in my eg would require 64 watts from it's 20 watter

... the 76db/w figure should be 86db/w (as I indicated in my previous post), and the power requirement would be only 6.4 watts, not 64 watts.

Best regards,
-- Al

Gentlemen, I believe that the losses that are being referred to as being "per meter" should be "per doubling of distance."

I'm not particularly knowledgeable about horns, but I suspect that the 6db number is correct, *on that basis*.

If so, the 100db/W/1m would correspond to 94db/W/2m, and 88db/W/4m, and 86.06db/W/5m, according to my calculations.

Best regards,
-- Al

06-26-11: Weseixas
Hello AL,
No your calculations are off the distance increase is 4M making the reduction 24 DB for a total of 76..

If 6db is lost going from 1 meter to 2 meters (and I have no particular knowledge of whether or not that is true for any particular horn design, but I'm making that assumption, as I indicated in my initial post above), then it seems to me that another 6db would be lost going from 2 meters to 4 meters (not 2 meters to 3 meters). And then another 1.94db, according to my calculations, going from 4 meters to 5 meters, for a total loss of 6 + 6 + approx. 2 = 14db.

I can't see how the loss can be 6db per meter for each of the additional 4 meters. That makes no sense to me. The loss is due mainly to the wave "spreading out," not to the attenuation of the air.

Best regards,
-- Al