db-- TODAY'S QUIZ

[Lonesomedave]

OK-- so, posting on another thread this weekend made me realize how little i actually know about things like sound pressure, loudness, db rating etc.

and i got to thinking -- which is rare and dangerous i know-- but it led me to this question, i am gonna propose it and then give my best guess as to the answer.

anybody who either knows or thinks he knows the answer is welcome to give it.

SO... suppose you have a closed room and two identical rigs, say acoustic B200's just for laughs. and suppose you set them up and measure the sound output from each, and it is for the sake of this experiment 100db.

now suppose you move one of the rigs into the closed room and again measure the sound and it is still 100db.

now, without touching the first rig, you move the second rig into the room.

what would your db measure register?

here is my answer. please feel free to tell me if i am right or wrong, tell me why, or give your own answer. hell, feel free to do any damn thing----

i think in the case set forth above your db meter would go up 3db. you have doubled the sound pressure, and Munjibunga and others showed me that doubling the sound pressure equals a 3 db increase.... not twice as loud, but simply one notch or 3 db on the loudness scale.

am i right or wrong? chime in

/s/ Dave

[KJung]

First, 3db increase is not 'double the volume'. I believe double the volume would be represented by a 10db increase on that non-linear scale.

I think what you are remembering is that 'double the wattage' (all other things being equal, and having a cabinet that can actually take advantage of doubling the wattage) will give you roughly a 3db increase in volume (which is significant).

In your example, two idendical cabs with two identical heads will give you pretty much the same increase in volume as adding a second identical cab to a rig with a solid state head (i.e., you are basically doubling the wattage and doubling the number of identical drivers).

Roughly doubling the wattage (pretty similar between using two identical heads and going from 8ohm to 4ohms with a solid state head), and doubling the drivers will give you a roughly 5 or 6db increase in volume (which is HUGE, but still not technically 'double the volume').

[projectMalamute]

6 dB

3 from doubling the wattage, 3 from doubling the speakers.

[Lonesomedave]

you have not convinced me.

i still say, in my example, moving a second identical rig in the room, doubles the sound pressure- not the loudness, but the sound pressure.

when you double the sound pressure you go up 3db.

now, where am i wrong?

/s/ Dave

[KJung]

Originally Posted by Lonesomedave

you have not convinced me.

i still say, in my example, moving a second identical rig in the room, doubles the sound pressure- not the loudness, but the sound pressure.

when you double the sound pressure you go up 3db.

now, where am i wrong?

/s/ Dave

Based on my layman's understanding of this, db is the measurement unit of sound pressure. Since the relationship of db's to mechanical increases in power and number of speakers is non-linear, a doubling of power+speakers will not give you a doubling of sound pressure, and hence you will get less of an increase in the non-linear db scale than the 10db that I believe represents true doubling of volume.

The increase in your example will be roughtly 6db, which is a lot, but less than doubling given the non-linear relationship between mechanical measures like wattage and sound pressure, as measured in db's and perceived as 'loudness' by human beings.

IMO there... I assume Bill or Greenboy or whoever will drop by eventually to verify.

[Lonesomedave]

kjung.

so you are saying, in my example, doubling the sound source (ie moving the second identical rig into the room) MORE than doubles the sound pressure?

how can that be?

/s/ Dave

[billfitzmaurice]

Originally Posted by Lonesomedave

what would your db measure register?

106dB. You get 3dB additional from the doubled power, 3dB from the doubled radiation efficiency realized through mutual coupling of two cabs. There are some other details having to do with dispersion and the high frequency -3dB Ka limit dropping by a factor of 0.7, but with a basic sound meter and a pink noise source you'll get a 106dB reading.

[projectMalamute]

Originally Posted by Lonesomedave

you have not convinced me.

i still say, in my example, moving a second identical rig in the room, doubles the sound pressure- not the loudness, but the sound pressure.

when you double the sound pressure you go up 3db.

now, where am i wrong?

/s/ Dave

Why should we be trying to convince you?

You've got a correct answer. If you don't understand it go educate yourself. They make books for this purpose.

[Lonesomedave]

projectmalamute;

gee, i thought that's what i was attempting to do by starting this thread.

if i wanted to spend a lot of time going elsewhere i would not have asked the question on TB in the first place.

thank you so much for your condescending attitude, very helpful.

/s/ Dave

[KJung]

Originally Posted by Lonesomedave

kjung.

so you are saying, in my example, doubling the sound source (ie moving the second identical rig into the room) MORE than doubles the sound pressure?

how can that be?

/s/ Dave

Again, I could be wrong, but 3db is NOT a doubling of sound pressure. 10db is a doubling of sound pressure.

If doubling the wattage/drivers resulted in doubling of the sound pressure as measured by db's, then db's would be based on a linear scale. DB's are based on a log scale, since the relationship you are talking about is not linear.

So, the 6db increase you would achieve is very large, but less than the 10db increase you would need to actually double the sound pressure.

Again, my layman's understanding. I hope BillFitz or someone pops in here to verify (or to clean up my mess!)


Edit: I see BillF posted above, and I believe verified my posts.

[Lonesomedave]

kjung; well i am woefully ignorant, and i have read bill's post above.

here is what wikipedia has to say about sound pressue:

"A change in power ratio by a factor of 10 is a 10 dB change. A change in power ratio by a factor of two is approximately a 3 dB change"

now i took the phrase "power factor of two" to mean doubling or halving the sound pressure

am i correct in this?

/s/ Dave

[KJung]

Channeling Jack Nicholson in 'A Few Good Men'....

'Are we clear'

[JHAz]

THe answers above are correct. 3 dB for doubling power, 3 dB for doubling radiating area, 6 dB total increase (aauming placement of the rigs so you don't get significant cancellations, etc.

That's how it works.

IDK what the air pressure is that equates to 100 versus 106 dB. I actually don't need to know the physical air pressure involved, so I guess the bottom line is I don't cae about it for my practical purposes . . . What I do know is 3 dB is not subjectively twice as loud. People differ, but 3 dB is something like one notch louder in most of the stuff I've read. Twice as loud is generally considered to be 10 dB, although again folks differ . . . .

A fella with a radio shack meter (or a home theater receiver denominated in (hopefully accurate) dB on its volume control could experiment with this stuff and get his own sense of it . . .

[KJung]

Originally Posted by Lonesomedave

kjung; well i am woefully ignorant, and i have read bill's post above.

here is what wikipedia has to say about sound pressue:

"A change in power ratio by a factor of 10 is a 10 dB change. A change in power ratio by a factor of two is approximately a 3 dB change"

now i took the phrase "power factor of two" to mean doubling or halving the sound pressure

am i correct in this?

/s/ Dave

You are correct there... a DOUBLING of power (power 'ratio' increase of 2), all other things equal, gives you a 3db change. Increasing power by a factor of 10 (10 times... i.e., going from 100 to 1000 watts) will double the sound pressure, which is 10db.

This is all theoretical though. Doubling a 100 watt amp to 200 watt can easily give you that magic 3db increase with most speaker cabinets (i.e., a noticeable increase in volume). Doubling a 500 watt amp to 1000 watts is often a waste, since most single cabs can't really mechanically make use of all that juice.

However, again, what you have done in your example is both increase the power by 2 (double), which gives you roughly 3db increase (again, less than a 'third more sound pressure), and since you also doubled the drivers, you get another 3 db increase, resulting in a sound pressure level that is about 66% higher than a single amp.

Again, you can see the non-linearity in the translation of increasing wattage and cone area with volume as measured by db's. It takes a lot more than double the watts and drivers to actually doubling the sound pressure level.

[Lonesomedave]

well, i guess i am less ignorant then when i started, but i still have questions.

if you are right (and i assume you are right), how could simply doubling the rig, and not changing anything else, MORE THAN DOUBLE THE SOUND PRESSURE?

if 6 db is right for the increase, then simply doubling the rig did that. and it just blows my mind.

i realized going in that db was a non-linear measurement, but if anything, before this thread i would have assumed that doubling the rig produced a slightly less than doubling of the sound pressure.

now you tell me that not only is this wrong, but in fact, doubling the rig much more than doubles the sound pressure.


oh, well...

/s/ Dave

[Lonesomedave]

Originally Posted by KJung

..... and since you also doubled the drivers, you get another 3 db increase, resulting in a sound pressure level that is about 66% higher than a single amp.

Again, you can see the non-linearity in the translation of increasing wattage and cone area with volume as measured by db's. It takes a lot more than double the watts and drivers to actually doubling the sound pressure level.

starting to come clearer. thanks.

/s/ Dave

[Mike 257]

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.

[KJung]

Originally Posted by Lonesomedave

starting to come clearer. thanks.

/s/ Dave

Hey, I learned all this stuff from BillF and other AE's an EE's on this site. That is what the site is for. I enjoy trying to explain it, since that is a good way of making sure I understand it (at least at layman's level).

This is why all those questions about 'is a 4ohm cab louder than an 8 ohm cab' or 'how much wattage do I need' are impossible to answer without knowing more info about speaker SPL, mechanical capabilities of speakers (which have little to do with those max wattage ratings on speaker cabs), and also at what level are you starting from when considering increasing wattage (i.e, going from 'not enough' or 'low' to 'just enough', versus going from massive to stupid massive)

[KJung]

Originally Posted by Mike 257

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.

I do admit that understanding this stuff would mean more to me if it would result in me not rushing coming out of the bridge of a tune

[KJung]

Originally Posted by Lonesomedave

well, i guess i am less ignorant then when i started, but i still have questions.

if you are right (and i assume you are right), how could simply doubling the rig, and not changing anything else, MORE THAN DOUBLE THE SOUND PRESSURE?

if 6 db is right for the increase, then simply doubling the rig did that. and it just blows my mind.

i realized going in that db was a non-linear measurement, but if anything, before this thread i would have assumed that doubling the rig produced a slightly less than doubling of the sound pressure.

now you tell me that not only is this wrong, but in fact, doubling the rig much more than doubles the sound pressure.


oh, well...

/s/ Dave

Again, just to make sure we are being clear 6db is LESS, not more than doubling the sound pressure, which would be a measurement of a 10db increase on the sound pressure scale.