db, gain, and loudness. Simple math
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 7th November 2005, 05:18 PM #1 Pbassred   diyAudio Member     Join Date: Nov 2003 Location: N London db, gain, and loudness. Simple math My head is tangled from too many posts. If I increase speaker efficiency by 3db, I actually double the output. but don't I need 10 times the output to double the apparent volume? er......... how many db would I need to increase by to double the apparent volume? I percieve that I have lost the (simple) plot. __________________ Jeeees! I only asked https://www.facebook.com/dave.potter.5815
 7th November 2005, 05:26 PM #2 Cal Weldon   Speakerholic diyAudio Moderator     Join Date: Jan 2004 Location: Near Vancouver Best I can describe it is: double the power input = 3dB increase 10X the power input = twice the apparent volume or 10dB increase dB is logrithemic so every 10 dB increase is twice the volume. ie: 100 is twice as loud as 90, 90 is twice as loud as 80 etcetera. Cal __________________ planet10 needs your help: Let's help Ruth and Dave
 7th November 2005, 05:44 PM #3 Hornlover   diyAudio Member   Join Date: Sep 2004 Location: Central California The ear is not linear, so it takes an average of 10 db to percieve a 2:1 difference. An increase of 3db is a doubling in power, but it is only an increase in amplitude of .414. For a doubling of amplitude, it takes an increase of 6db.
Pbassred
diyAudio Member

Join Date: Nov 2003
Location: N London
I already knew that the ear was logrithmic
Quote:
 I need 10 times the output to double the apparent volume
I already knew that 3db equates to a doubling of power output
Quote:
 If I increase speaker efficiency by 3db, I actually double the output
But that's not the answer to the question.
Quote:
 For a doubling of amplitude, it takes an increase of 6db
I could live with this answer if I could get my head around the numbers, but I just can't make the numbers prove it. There is no way that I can make 2, 3, of 10 crunch into .414 using logs or natural logs. Its buggin me.
__________________

 7th November 2005, 08:24 PM #5 simon5   diyAudio Member   Join Date: Nov 2004 Location: Québec, Québec The formula for the power in watts is : x dB = 10 log y x = number of dB increase y = times the power you get Look at what the guy said, he said an increase of .414 maybe he should have said power of 1.414 times versus the original power imput. You want it to use two times more power. 10 log 2 = 3.0102999566398119521373889472449 dB increase The formula for the amplitude in voltage is : z dB = 20 log w z = number of dB increase w = times the voltage amplitude you get You want it to use twice the amplitude. 20 log 2 = 6.0205999132796239042747778944899 dB increase. Understand a bit better now? Keep the questions coming! __________________ DIYaudio for President !
 7th November 2005, 08:29 PM #6 simon5   diyAudio Member   Join Date: Nov 2004 Location: Québec, Québec So for fun, since alot of people agree that 10 dB is twice as loud, you could invent a formula... u dB = 33.219280948873623478703194294948 log v u = number of dB increase v = times the loudness you get Want it twice as loud so : 33.219280948873623478703194294948 log 2 = 10.000000000000000000000000000014 dB increase Wonderful, isn't it? __________________ DIYaudio for President !
 7th November 2005, 11:00 PM #7 Pbassred   diyAudio Member     Join Date: Nov 2003 Location: N London So anyway, working in 3 sig fig if we can; Again, I get the power equation: 10 X the Log of 2/1 and the Voltage equation.; 20 x the log of the ratio 2/1 - which works for SPL I also get the 10db required for the perception of double volume. ( said 10 times. I was wrong). I still don't see 1.414 ( I already tried it). I can see that; 33.2 log 2 = 10db is derived from the voltage/SPL equation, but i can't see what use it is. (un herring rouge?) How much do I need to raise the speaker power by to increase the SPL by 10db? Or is it as simple as saying that to increase SPL by 10db I must raise the power by 10db? So If I needed to raise the power by 10db; 20 log 3.2 = 10db ? __________________ Jeeees! I only asked https://www.facebook.com/dave.potter.5815
 7th November 2005, 11:18 PM #8 simon5   diyAudio Member   Join Date: Nov 2004 Location: Québec, Québec The guy was talking about voltage amplitude... he wanted to tell you what 3 dB of increase would do to the voltage amplitude. If we take that formula... 20 log 1.414 = 3 dB Now, to get 10 dB more SPL, you need to increase the power by 10 dB, yes it's as simple as that. 10 dB = 10 log y y = 10^(10/10) = 10^1 = 10 That means to get 10 dB more SPL, you need 10 times the power. If with 1W you get 85 dB, to get 95 dB you'll need 10W, and to get 105 dB you'll need 100W, and to get 115 dB you'll need 1000W and etc... __________________ DIYaudio for President !
simon5
diyAudio Member

Join Date: Nov 2004
Location: Québec, Québec
Quote:
 Originally posted by Pbassred So If I needed to raise the power by 10db; 20 log 3.2 = 10db ?
You used the voltage equation there...

Yes, you need to increase the voltage by a factor of 3.2 and the current by the same factor... why? Because power is equal to current x voltage so 3.2 x 3.2 = 10.24.

More simple to use the power equation directly and forget about the voltage amplitude equation. You'll probably not need it.
__________________
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 7th November 2005, 11:36 PM #10 Pbassred   diyAudio Member     Join Date: Nov 2003 Location: N London Life is simple. So, more efficient speakers costs less than bigger amplifiers (bank for bucks). I'm my case, its a matter of jumping from 800W to 8000W, or find drivers that are 10db more efficient. So, ok, neither is going to happen but given my cheap chinese drivers, Eminence at 98db is a worthwhile swap. Life is simple __________________ Jeeees! I only asked https://www.facebook.com/dave.potter.5815

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