db-- TODAY'S QUIZ

[Lonesomedave]

Omg you guys are crazy , I just hope I learn to play by the time I'm 60 ,no less go as deep as you guys are.

son, if you can play at all, you're miles ahead of me------

i could not play a stringed instrument to save my life.

what i am is the "sound engineer" for my 14 year old son. and from the looks of things, he might want to consider firing me and going pro!

/s/ Dave

 

[Lonesomedave]

kjung,

now wait a minute. i thought from the wikipedia excerpt that doubling the sound pressure resulted in a 3 db increase. -----

have i read that wrong?

/s/ Dave

 

[KJung]

kjung,

now wait a minute. i thought from the wikipedia excerpt that doubling the sound pressure resulted in a 3 db increase. -----

have i read that wrong?

/s/ Dave

Yes, I believe that is the source of most of your confusion. Doubling the sound pressure is represented by an increase of 10db on the log decibel scale, and all things being equal, would require either a 10 times increase in powe (assuming you have a speaker that can handle that massive power), or in your example, approximately one additional full rig added to your example (which would technically give you a 12db increase in total versus a single identical rig, or about a 120% increase in sound pressure level measured by db's.

Edit: Doubling the WATTAGE results in a 3db increase, which I think is what you are remembering and misinterpreting.

 

[Son of Bovril]

you doubled the power = 3db increase, but by adding the extra speakers you are also adding another 3db. therefore 6bd increase

in fact just adding another cab into the room, even without the extra amp, would give you +3db

 

[Lonesomedave]

you doubled the power = 3db increase, but by adding the extra speakers you are also adding another 3db. therefore 6bd increase

in fact just adding another cab into the room, even without the extra amp, would give you +3db

Yes, I believe that is the source of most of your confusion. Doubling the sound pressure is represented by an increase of 10db on the log decibel scale, and all things being equal, would require either a 10 times increase in powe (assuming you have a speaker that can handle that massive power), or in your example, approximately one additional full rig added to your example (which would technically give you a 12db increase in total versus a single identical rig, or about a 120% increase in sound pressure level measured by db's.

Edit: Doubling the WATTAGE results in a 3db increase, which I think is what you are remembering and misinterpreting.

thanks guys, i think that was indeed the source of my confusion.

what they said was "doubling the power" and i took it (wrongly, as it turns out) to mean "doubling of the sound pressure"

silly me

/s/ Dave

 

[KJung]

thanks guys, i think that was indeed the source of my confusion.

what they said was "doubling the power" and i took it (wrongly, as it turns out) to mean "doubling of the sound pressure"

silly me

/s/ Dave

Nothing silly about it. Complicated stuff to get you head around. I enjoyed the thread!

 

[seamonkey]

It varies. It's different at different frequencies and at what initial volume level you start at.
This is a good start
Equal-loudness contour - Wikipedia, the free encyclopedia

It also varies with what loudness your ears have been last listening at.

The alarm clock clicking on in the morning can be annoying loud, but I bet when you set it the night before it was a fairly quiet level.

 

[billfitzmaurice]

Doubling the sound pressure is represented by an increase of 10db

Doubling of sound pressure is 6dB. That's the increase you get by doubling the excursion of one driver, or by doubling the total system excursion using two drivers to achieve twice the displacement. Doubling of power is 3dB, but that doesn't get you twice the sound pressure because driver excursion isn't linear with respect to applied power but rather to applied voltage, and doubling of voltage results in quadrupling power. See Ohm's Law.
Doubling of perceived volume is 10dB not because speakers or amps operate on a logarithmic scale but because our ears do.

 

[KJung]

Doubling of sound pressure is 6dB. That's the increase you get by doubling the excursion of one driver, or by doubling the total system excursion using two drivers to achieve twice the displacement. Doubling of power is 3dB, but that doesn't get you twice the sound pressure because driver excursion isn't linear with respect to applied power but rather to applied voltage, and doubling of voltage results in quadrupling power. See Ohm's Law.
Doubling of perceived volume is 10dB not because speakers or amps operate on a logarithmic scale but because our ears do.

+1 in that I figured sound pressure level was not exactly the same thing as 'perceived volume'. Also, +1 that the log scale is based on how we perceive things. However, the relationship between the mechanical stuff and db's is also non-linear, from my understanding. That was the point I was making there.

 

[Balog]

Whoa...

So every 6db is twice the total sound pressure level, but our ears require 10db to perceive twice the "volume"?

 

[KJung]

Whoa...

So every 6db is twice the total sound pressure level, but our ears require 10db to perceive twice the "volume"?

That, I guess, is the non-linear relationship between 'real volume and perceived volume' (the human ear thing) that I think Bill is talking about. That is over my head though!

 

[billfitzmaurice]

Whoa...

So every 6db is twice the total sound pressure level, but our ears require 10db to perceive twice the "volume"?

Yes.

 

[SSCBA]

Lets say we have the two acoustic b200's measuring 106 db as we stated before..
If I wanted to reach 112db would I have to add two more b200's, correct?

Because I understood that kjung said that adding a full rig would double perceived volume reaching 12db more (what he firstly refered as sound pressure but Bill corrected him), but I think that adding a full rig (or two more b200's) two the previous two b200's will result in an addition of 6db, giving the four b200's (or two b200's + a full rig) a total of 112db. I am correct?

Sorry for the messy redaction but didn't find other way to ask and dont enjoy too much typing on a smartphone

 

[KJung]

Lets say we have the two acoustic b200's measuring 106 db as we stated before..
If I wanted to reach 112db would I have to add two more b200's, correct?

Because I understood that kjung said that adding a full rig would double perceived volume reaching 12db more (what he firstly refered as sound pressure but Bill corrected him), but I think that adding a full rig (or two more b200's) two the previous two b200's will result in an addition of 6db, giving the four b200's (or two b200's + a full rig) a total of 112db. I am correct?

Sorry for the messy redaction but didn't find other way to ask and dont enjoy too much typing on a smartphone

Interesting. You might be right. As you suggest, since we now have a new starting point.... 106db with the double rig, we would now have to double that power and double that speaker count to get an additional 6db from that higher base of two rigs.

I guess in my example, you would reach 109db with the addition of the third rig, which is pretty close to doubling the perceived volume of the first rig. Again, I'm posting in this thread as much to learn and make sure I have this right than being any sort of 'expert' (obviously!). I enjoy the topic.

 

[billfitzmaurice]

Lets say we have the two acoustic b200's measuring 106 db as we stated before..
If I wanted to reach 112db would I have to add two more b200's, correct?

Correct.

 

[Lonesomedave]

thanks loads guys; you have really straightened me out here and i think i understand it.

let me put it in laymen's terms and tell me if i am correct or not.

we start at 1 b200 and 100 decibels.

when i move a second b200 in the room, i HAVE doubled the sound pressure

it's just that, because of my initial confusion, i thought that doubling the sound pressure resulted in only a 3 db increase, not the 6 db increase that it actually does.

this doubling of the sound pressure does not result in a doubling of the perceived volume.

to do that, you need to move a third b200 in the room (which will yield 109 db-- not quite reaching the 10 db that is an actual doubling, but close enough)

/s/ Dave

 

[KJung]

thanks loads guys; you have really straightened me out here and i think i understand it.

let me put it in laymen's terms and tell me if i am correct or not.

we start at 1 b200 and 100 decibels.

when i move a second b200 in the room, i HAVE doubled the sound pressure

it's just that, because of my initial confusion, i thought that doubling the sound pressure resulted in only a 3 db increase, not the 6 db increase that it actually does.

this doubling of the sound pressure does not result in a doubling of the perceived volume.

to do that, you need to move a third b200 in the room (which will yield 109 db-- not quite reaching the 10 db that is an actual doubling, but close enough)

/s/ Dave

That is now my understanding. I assume BillF will validate or correct our understanding of this. Thanks in advance Bill.

 

[Robby Hoinsky]

Another little nugget - this all assumes that the sound sources are correlated (meaning the same inputs to both rigs). If you had two different inputs, say one was a bass and one a guitar, both at 100 dB, the result would be 103 dB.

Robby

 

[Lonesomedave]

robby;

OH, WOW...



please explain... i am back to the land of the ignorant again!

/s/ Dave

 

[KJung]

robby;

OH, WOW...



please explain... i am back to the land of the ignorant again!

/s/ Dave

What he is saying is that the assumption (which is one that we all made I believe) is that the input source is the same... that you are literally adding a second rig to the same input source (bass, CD player, whatever). So, in other words, your 'multiple rig' example should behave pretty much like adding a second cab to a solid state amp, which pretty nearly also doubles the wattage).

If you have each rig amplifying a different signal, things would obviously change.

At least that is what I think he is saying!