Please Read and express your feelings and opinions....

Those who may be interested in some background underlying the exchanges above between Ralph (Atmasphere) and CJ1965 will find the following thread to be relevant, beginning with the post by Atmasphere on 4-17-2018 and continuing to the end of the thread:

https://forum.audiogon.com/discussions/a-question-on-speaker-driver-efficiency

I believe that the posts in that thread by Ralph, CJ1965, me, and others will provide readers with a good perspective on the various protagonists.

Regards,
-- Al
 

You see, sir, that's how science works - full disclosure of the details.

Hm. Something that article certainly lacked! Based on what you say here, you don't think a little disclosure would have been a good idea? That certainly **seems** what you are saying; if anything, this should have been an objection that you raised about this article rather than stand behind it. Why the contradiction? Are you needing to make others wrong so much that you are willing to contradict yourself?  This slight on your part is a bit of hand waving.

As for your other claims regarding what I've posted elsewhere, they don't even merit a response.

Hm. You made the claim that when you cut the voltage applied to a speaker in half, its a 3db reduction. Here's the math:

2.83 volts is 1 watt into 8 ohms. I hope we can agree on that...

So 1/2 of 2.83 volts is 1.415 volts. Using Ohm's Law, to calculate current we divide 1.415 with 8 ohms and we get 0.178975amps.

Power is equal to amps x current, now we know the current so:

0.178975amps x 1.415volts = 0.25 watts. That's a 6 db reduction, not 3 db.

Do you still think it does not merit a response, like- 'oh, I meant 'power' not voltage'? You were being pretty specific about voltage on that thread; perhaps you could clear that up.

Looks like we're seeing that cult of anti-intellectualism rearing it's ugly head here. I can't believe I'm reading somebody lecture a legit engineer on why he should ignore blatant failures of reason. It doesn't get more obvious than that. 

@atmasphere

You’re using the wrong formula - equating electrical power to sound power. The db formula for sound power is :

10 X log 10 (P/Po) db

where P is the measured sound power and Po is the reference level.

If you doubt me, look it up. I believe it’s spelled out pretty clearly in Wikipedia.

And sound power is a function of the force applied, (in this case the voltage) along with the area and displacement over which that force is applied. When you halve the force and double the area, you wind up with roughly the same sound power that you began with. But improved acoustical coupling with increased area improves net efficiency (electrical power in versus sound power out).