I really need expert advice on this one

[mboyce84]

Single Sub vs. Dual Subs Setup?
I really need some help on this one. I already have on of the Audiobahn flame subs that has 1100 Watts RMS, in a ported box, plus a JBL 1200.1. I was planning on having two of these subs and two amps, one per sub. But, with all that stress on my battery, alternator, and pocket trying to upgrade, Im not so sure anymore. How much less of SPL would I have by just opting for just a single setup vs the dual

[lakersfan1]

Theoretically, 6 dB.

[zx3chris]

add a second sub, not a second amp.. you may blow that sub with the jbl, that amp is underrated and those audiobahns are waaay overrated. and dont worry about stress from your amps -- you can run several 1200.1's before u need to upgrade the alt.

[non_affiliated]

Originally posted by mboyce84:
Single Sub vs. Dual Subs Setup?
I really need some help on this one. I already have on of the Audiobahn flame subs that has 1100 Watts RMS, in a ported box, plus a JBL 1200.1. I was planning on having two of these subs and two amps, one per sub. But, with all that stress on my battery, alternator, and pocket trying to upgrade, Im not so sure anymore. How much less of SPL would I have by just opting for just a single setup vs the dualNot factoring in all the variables, and just going off of the basic principles. As you stated, One sub, and one 1200W amp will end up being roughly 6 db's less than a system with two subs and two 1200W amps.

Normally if you double the power you'll gain 3 db, and if you double the power you'll normally gain 3 db. That doesn't always hold exactly true for the fact that linear excusion has to be accounted for and power compression, and a few other variables.

[zx3chris]

yeah, but you're system will be infinty db quieter when the sub blows up :D ... audiobahn cannot handle what they are rated for, atleast this is the case in everytime i've used their subs.

[PBCrunch]

Get two subs. Lots of people get one-woofer setups, try to play it like they have two subs, and end up breaking ****. Play it safe and put one amp on two subs. If you are worried about the electrical end of things, get a second battery.

Also, if you pick the right woofers you dont need 1000 watts to have a loud system.

[kkant]

Shouldn't it be a 3 db difference? Ignoring the amps for a second...since the same power is applied to each sub,you have the same excursion with double the cone area, so double the Vd which means +3db difference between single and dual setups.

[Audiolover]

Double the cone area= +3db
Double the power= +3db
Double both= +6db

You've got to look for the final impedance to get the "real" power going to the final configuration.

Audiolover

[kkant]

Double the cone area with half the power to each sub = +0db (power compression issues aside)

Double the power = +3db

Total = +3db

[johnl]

Originally posted by kkant:
so double the Vd which means +3db difference between single and dual setups.Double the VD also makes it hurt more when you pee

[808BUFF]

"Double the VD also makes it hurt more when you pee"

You sure about that? tongue.gif

[Audiolover]

Originally posted by kkant:
Double the cone area with half the power to each sub = +0db (power compression issues aside)

Double the power = +3db

Total = +3dbI suppose when you say "half the power" you are reffering to the power distributed to both speakers. Doubling cone area with the same total power give's you 3db, guarateed.

Audiolover

[kkant]

Don't think so. Doubling cone area also doubles moving mass. Thus efficiency doesn't increase. Therefore, since you have the same amount of power going in to the pair of subs, you have the same amount of power coming out.

So, adding a sub and keeping total input power the same means +0db.

But...adding an extra amp doubles the total input power. That doubles output. +3db.

[lakersfan1]

Originally posted by kkant:
Don't think so. Doubling cone area also doubles moving mass. Thus efficiency doesn't increase. Therefore, since you have the same amount of power going in to the pair of subs, you have the same amount of power coming out.

So, adding a sub and keeping total input power the same means +0db.

But...adding an extra amp doubles the total input power. That doubles output. +3db.Wrong. You're not doubling the mass of any driver. Just adding another woofer with it's own motor structure to do it's own job.

Plus one sub same amp, same total power = 3dB
Two amps and two subs = 6 dB
End of story.

[Audiolover]

Originally posted by lakersfan1:
You're not doubling the mass of any driver. Just adding another woofer with it's own motor structure to do it's own job. That was a very very correct statement.

Audiolover

[ October 19, 2002, 07:30 PM: Message edited by: Audiolover ]

[kkant]

Originally posted by lakersfan1:
Two amps and two subs = 6 dB
End of story.6 db SPL = quadruple acoustic power. You're telling me if one sub/amp combo radiates x acoustical watts, then two sub/amp combos will radiate 4x watts. How is this physically possible. Where is the extra energy created. What I see here is two pressure sine waves which add together constructively to form a new sine wave that has twice the pressure amplitude of the original. Twice the peak pressure means 3db. Right???

[BigVinny]

KKant seems to be correct. Atleast it is what I always thought.

Double the cone area would require the same setup twice. Theoretically, if you put two subs on the one amp, the power is halved between the two, making them excurse half as much as the one. This is 0db gain. (Theoretically!!)

[kkant]

OK, I think I've figured out why we are wrong, Vinny. dB is always a measure of power, whether electrical or acoustical. dB is not a measure of pressure or voltage.

But electrical power is proportional to voltage squared, and acoustical power is proportional to pressure squared. The amount of excursion is proportional to voltage (we are talking in the linear range), and the amount of acoustic power output is proportional to electrical power input (again, linear).

So. If we put half power to a speaker, the voltage is actually 0.707 of the original voltage. Because 0.707 squared is 0.5. Since the voltage is 0.707, the excursion/pressure is also 0.707. The acoustic power output is pressure squared, and 0.707 squared is 0.5...so output power is also half its original value, which means 3 db down. So far there are no problems...we are 3db down on input power, and 3db down on output.

But when we put two speakers together, how do they add up? Do the pressures add up, or does the power add up? Ideally, it is the pressures that add up. Low frequencies in a car are close to an ideal situation, since it is closer to uniform pressure changes than travelling waves. So we add the two presures together...but 2x the pressure = 4x the acoustic power, since power = pressure squared. That's 6db up from what each speaker was putting out individually.

Applying that to the stuff 2 paragraphs above, we have one pressure of 0.707 added to another pressure of 0.707, total pressure is 1.414. Pressure squared is power, so acoustic power is 2. That's 3db up from what only one speaker can do with the same *input* power.

So, two speakers are twice as *efficient* as one speaker of the same type. That's how the extra output power is created from the same input.

Also, twice the power = +3db (of power), but twice the pressure = +6db (of power). So "dB SPL" is kind of a misnomer...you are not really figuring out decibels of pressure, you are calculating decibels of power by measuring pressure and squaring (which is the same as multiplying by two on the log dB scale).

[ October 20, 2002, 07:20 AM: Message edited by: kkant ]

[lakersfan1]

That's correct. Sound is a logarithimic scale. So a doubling of the pressure wave is 10 dB.

[kkant]

Originally posted by lakersfan1:
That's correct. Sound is a logarithimic scale. So a doubling of the pressure wave is 10 dB.No. A "bel" is log(power ratio). A decibel is a tenth of a bel, so it is 10*log(power ratio). 1 bel (equivalent to 10 decibels) corresponds to a power ratio of 10:1. So 10 db means 10 times the acoustic power. 3 db means twice the acoustic power.

By convention, decibels are used ONLY to measure power (whether electrical or acoustic). Other quantities like voltage and pressure must be converted into power before calculating db. Now pressure is the square root of power. That means that in order to calculate decibels of power from a pressure ratio, we have to take the normal db formula 10*log(ratio) and multiply it by 2. The new formula is 20*log(pressure ratio). Only one thing left...sometimes we have only a single pressure, how do we get a ratio? We need a reference pressure to compare to. The reference pressure that is used by everyone is 20 micropascals. This number was chosen because it is about the threshold of human hearing. So, when any particular SPL figure is given without comparison to something else...the comparison is implicitly against 20 micropascals.

So, that means that a doubling of pressure causes a quadrupling of power, so it is +6db. +10db is about 3.16 times the pressure, and 10 times the power (3.16 squared).

[lakersfan1]

OK. I guess this needs to be broken into two segments. One is perceived sound by a human being. This is commonly referred to as dB (decibel).

The other segment is actual pressure. This can probably be measured in many different units, but a common one seems to be microbar.

The equation is
decibels = 20 log (Sound Pressure Level/0.0002)

The 0.0002 is the minimum threshold of hearing and is the reference point.

So, if you take the eqation and use 100 microbars, you'd get 114 dB. If you double the amount of pressure and make it 200 microbars, you'd get 120 dB. A 6 dB increase.

I think what I meant to say above, is 10 dB is a PERCEIVED doubling of sound. Not a doubling of the pressure wave. That was wrong. A doubling of the pressure wave is 6dB.

[kkant]

Originally posted by lakersfan1:
OK. I guess this needs to be broken into two segments. One is perceived sound by a human being. This is commonly referred to as dB (decibel).No I wouldn't say that. There is only one dB, and it is decibels of power. You can use dB's to derive pressure ratios since there is a straightforward formula relating power to pressure. You can sort of derive perceived loudness from db, because there is a fuzzy formula that applies to some frequency ranges and loudness ranges but probably breaks down outside of that.

Originally posted by lakersfan1:
I think what I meant to say above, is 10 dB is a PERCEIVED doubling of sound. Not a doubling of the pressure wave. That was wrong. A doubling of the pressure wave is 6dB.Agreed. +10 db can generally be taken as a perceived doubling of volume. That means it takes about 3 times the pressure and ten times the acoustic power to double volume.

BTW, most of this info comes from The Audio Dictionary written by a Mr. Glenn White. With a little analysis from me. smile.gif This is a great reference book.

[lakersfan1]

well, considering decibels were invented to have a scale that refers to how we hear (read: percieve) sound, I think it's safe to say decibels relate only to percieved sound.

"In the early 1900's, Bell Laboratories did the most comprehensive study of human hearing ever. They decided to come up with a standard measurement of how loud we hear things. The initial idea was that they would increment the measurements by making each sound twice as loud as the previous. Each increment would be called a "Bel" out of respect for Alexander Graham Bell. The plan was to divide the Bel into 10 equal parts (the decimal system) and call each part a "deciBel". A base reference was needed so that the measurement system could be used accurately. Bringing in extremely sensitive barometric pressure sensors, they measured the changes in the air pressure of the room. Using a 1kHz tone, they found that the quietest sound their subject could hear was at 0.0002 Micro-Bars pressure. This is the point that Bell Labs referenced to 0 dB-SPL (Sound Pressure Level) which is commonly referred to as the "Threshold of Hearing." The "Threshold of Pain" is at approximately 200 Micro-Bars pressure. One of the more interesting findings during the study was that human hearing is not linear, meaning that twice the barometric pressure does not create a sound that we perceive as being twice as loud.
The equation for converting Sound Pressure Level to decibels is
decibels = 20 log (Sound Pressure Level/0.0002)

Knowing that the Threshold of Hearing is at 0.0002 Micro-Bars pressure and that the Threshold of Pain is at 200 Micro-Bars pressure, this is why using logarithms is a good way to measure sound."

The whole purpose was to relate it only to our perception of loudenss, otherwise, we could have just stuck to measuring the atmospheric pressure.

[ October 20, 2002, 02:07 PM: Message edited by: lakersfan1 ]

[Haunz]

G/D.... I am glad you cleared that up kkant..

Since you seem like one of the real audio guru's on the site your earlier posts scared the hell out of me...

I was like WTF is he smoking graemlins/freak.gif

[kkant]

Originally posted by lakersfan1:
well, considering decibels were invented to have a scale that refers to how we hear (read: percieve) sound, I think it's safe to say decibels relate only to percieved sound.No, I don't agree. This is like saying the foot was originally referenced to the size of a human foot. We no longer relate the foot to feet, for the obvious reason that eveyone's feet are a different size. Now a foot is related only to an unchangeable uniform quantity. Similarly, although we can use dB to relate to loudness, the real definition of dB is in power.

Originally posted by lakersfan1:
The whole purpose was to relate it only to our perception of loudenss, otherwise, we could have just stuck to measuring the atmospheric pressure.My source disagrees with this.

Glenn White:
The bel had its origin in the Bell Telephone Labs, where workers needed a convenient way to express power losses in telephone lines as power ratios.That is why to this day *all* db numbers, whether used to describe power or pressure or loudness or voltage, are always referenced to power and power alone.

Think of it this way. A bel is log base 10. That means that +1 bel should result in ten times the quantity. If perceived loudness was the reference, then +1 bel (or +10 db) would mean 10 times the perceived loudness. If pressure was the reference, then same thing...+10 db would mean 10x the pressure. Both these statements are false. However, notice that power alone is multipled by 10 when you go up 1 bel.

[lakersfan1]

I think it's kind of hard to dispute when the denominator of 0.0002 is the minimum threshold of human hearing. If it was referenced to dog hearing, then the denominator would be different, and dB while it could still be on the same logarithmic scale, would be not the same as it is currently.

I'm thinking perhaps dB can be referenced to different baselines when used in different applications such as measuring S/N ratios and such. But as for the commonly used dB for SPL, it is clearly referenced to the human threshold of hearing, at least as we commonly use it.

[kkant]

No problem Jason. I think at this point we both know how it is. smile.gif Let me just say this though...pressures are referenced to a certain pressure to calculate the ratio...but dB's are *defined* by power. You can reference to anything you want...e.g. a 0dBFS sine wave, or -115 db S/N...but you can't change the *scale* of dB's, and the scale is defined by power ratios.

Originally posted by Haunz:
I was like WTF is he smoking graemlins/freak.gif Yeah man...I guess we all make mistakes and everyone has something to learn and all the rest of that rot. :D Some of this db stuff has always confused me to some extent, so I'm glad I got it cleared up.

[Sarsos]

Ok, I hate to throw a monkey in the wrench and make you guys rehash this but here it goes.

This is a summary of what the age old physics book written by Halliday and Resnick says:

Because the ear is so sensitive, we introduce a logarithmic scale of INTESITY called

sound level = 10 log (I/I0)

Where I0 is a reference INTENSITY referenced to the threshhold of human hearing.

Since it is I/I0 the actual measurement is in reality unitless (no power no nothing). However sound level is measured in decibells. Unless I am interperting this completely wrong that means that the unit decibells just tells us what we are referencing things to and in reality everything has no untis of import (i.e. Newtons mass and so on).

So the unit deciballs is a normalised measure of intensity, not power. Furthermore it is normalized to the human ear so that we have a rough measurement of how humans hear things. If you double the intensity, not the power (pressure is a measurement of intensity) and you get a +3 dB increase in "SL" which I think you guys relate to SPL but am not certain

For those of you who wish to know but don't preassure or intensity is generally related to power by I=P/(m*m) (m = meters).

This info is strait out of a very respected physics book with only a little interperting on my part. Most of what you have said is right, but I think that what I have read out of the physics book is saying something slightly different.Please feel free to comment

Sarsos

[ October 21, 2002, 03:52 AM: Message edited by: Sarsos ]

[kkant]

Sarsos, I believe you are misinterpreting H&R.

Originally posted by Sarsos:
sound level = 10 log (I/I0)SPL = 20*log(p/p0)

I don't know what intensity is...sounds like H&R is using intensity to mean power density or something...not pressure.

Furthermore, dB is unitless...but the scale of dB is defined by power, not pressure. The ratio is determined by the reference value (pressure, voltage, whatever), but after you take the log of the ratio, you have to scale it so that you get it back to the log power ratio scale. This is why when you take a pressure ratio, you multiply the log by 20 instead of the normal 10.

You see, a dB is more than just a log base 10. It is a log base 10 of power ratios. Why invent the term "Bel" after all...it is just as easy to say "deciLog" as in "I hit 140 dL last weekend". The reason for the new term was to specify that the Bel unit applied to power ratios, and could only be used with other units after an appropriate scaling factor. Why do this at all...because that way we can say 130 db SPL is the same as 130 db power, if referenced to the same base pressure (or base power). But 130 dL SPL is NOT the same as 130 dL power, even if both are referenced to the same base pressure or power.

Originally posted by Sarsos:
If you double the intensity, not the power (pressure is a measurement of intensity) and you get a +3 dB increase in "SL" which I think you guys relate to SPL but am not certainIncorrect. Doubling power gets you 3db. Doubling pressure gets you 6db (this is why two woofers with two amps is +6db rather than +3db). The reason for this is that power is proportional to pressure *squared*. Again, I don't know what exactly H&R is referring to when they say "intensity".

To illustrate this, let us derive the relationship between power and pressure. I don't know how to do the most general case, but I think I can do a specific derivation. Say we have a sealed room, and on one of the walls there is a small piston in a cylindrical track. There is an air compressor with a pressure regulator attached to the room. When we turn up the compressor, the piston is forced down the cylindrical track from the increased pressure. As the piston moves down the track, the regulator keeps the compressor going so the room pressure stays constant at the higher pressure. The track has the property that the frictional back-force it exerts on the piston is proportional to the speed of the piston, so we get a terminal piston velocity (just like the terminal velocity you get when skydiving).

OK, that's the system. Now we want to calculate how much power the pressure is applying to the piston. We know power = F*V, or force times velocity. F is force from the pressure, and we know F = P*A where P is room pressure and A is piston area. The terminal velocity occurs when the force from the piston equals the force from the friction. The frictional force is K*V, where K is the friction coefficient and again V is velocity. So we have F = k*V or P*A = K*V. Solving for V, V = P*A/K. That is the terminal velocity. Now we know both F and V...power = F*V = (P*A)*(P*A/K).

So finally we have power = P^2 * (A^2/K), and since A and K are just constants, we come to the conclusion that power is proportional to the square of velocity.

[Audiolover]

Kkant... I applaud the effort of briging this up. It's all interesting and such, but I have to remember what I said... witch is correct... 2 woofers with double total power is +6db... 2 woofers with the same original power (going to one, previously) is 3db... double the power is +3db... tongue.gif
The 18th post of this thread (by kkant) explained the reason pretty well, for those who doesn't what to read all this (cool) stuff... graemlins/thumb.gif

Audiolover

[ October 21, 2002, 09:04 AM: Message edited by: Audiolover ]

[kkant]

Yup, you got it right Audiolover. smile.gif

[Audiolover]

Originally posted by kkant:
Yup, you got it right Audiolover. smile.gif Nononono... YOU got it right. I was right since the beggining :D

Audiolover

[kkant]

LOL :D Yeah man, that's what I meant. smile.gif

[lakersfan1]

Hey kkant, are you the reason that they even bother teaching summation in calculus class?? Just wondering. :D

[kkant]

Yeah, it's my fault. Remind me to tell you sometime what it's like to be constantly accompanied by an team of armed guards, fending off attacks from legions of enraged H.S. seniors. :D

[non_affiliated]

Well all of a sudden I see this post reversing. Point being, under normal situations. If you double the cone area you will see a nominal 3 db increase in SPL . Now if you double the power too you will see a 6 DB incrase SPL.

You guys throwing up all these wonderful equations and functions is great, but for the average joe blow, you'll lose him.

I watch a guy with 4 18" subs running two amps at around 1000W. 250 to each sub. he hit like 148 db. he then removed 2 18" subs. Halved the cone area, yet bridged his amps to the to subs for 2000W. Can you guess what happened to his score? 148 DB on the button. Halved the cone area -3 DB, doubled the power +3 db, which menat he had a gain of 0 DB. Now if he were to double his power yat keep the same about of cone area he probably would have seen roughly a +3 db gain in his score. If he would have doubled the cone area he would of probably gained another 3 db.

What I am trying to say i guess is it looks all nice on paper. Yet when you do real world experimenting the results can be way different from the calculations. It seems whe really have a massive issue withe db's in reapect to where they are used. Correctme if I am wrong but I always kinda understood that db was a unit of measure, but with out a reference point it was kind of a usless number. Cause of various scalse db is used on. I mena from what I know a in RF antenna systems there are dbi, and dbd. Which are to totally different measures. dbi is referenced to a isotropic antenna which is a theroretical antten that doesn't exsit, but only on paper. Then there is dbd, witch is a referenced Di-pole antenna, a real world working antenna. Now if you have 2 anttenas with the exact gain, yet one is referenced to db, and one dbd. those numbers may look the same yet the actaul gain of the antenna referenced to dbd will be greater as it is referenced to an actual real world antenna that already has the ineefficencies added to the reference , rather than an isotropic antenna that is only a antenna that only exsists on paper. Maybe I am babling aou my a$$, which I probably am.

[kkant]

Originally posted by non_affiliated:
I watch a guy with 4 18" subs running two amps at around 1000W. 250 to each sub. he hit like 148 db. he then removed 2 18" subs. Halved the cone area, yet bridged his amps to the to subs for 2000W. Can you guess what happened to his score? 148 DB on the button. Halved the cone area -3 DB, doubled the power +3 db, which menat he had a gain of 0 DB. Now if he were to double his power yat keep the same about of cone area he probably would have seen roughly a +3 db gain in his score. If he would have doubled the cone area he would of probably gained another 3 db.This is consistent with the math and formulas above.

Originally posted by non_affiliated:
What I am trying to say i guess is it looks all nice on paper.It applies to the real world too.

The rest of your post is also consistent with what we have said above. A dB value is always referenced to something, because it is proportional to the log of a ratio, and the ratio needs a reference for the denominator.

[non_affiliated]

Originally posted by kkant:
</font><blockquote>quote:</font><hr />Originally posted by non_affiliated:
I watch a guy with 4 18" subs running two amps at around 1000W. 250 to each sub. he hit like 148 db. he then removed 2 18" subs. Halved the cone area, yet bridged his amps to the to subs for 2000W. Can you guess what happened to his score? 148 DB on the button. Halved the cone area -3 DB, doubled the power +3 db, which menat he had a gain of 0 DB. Now if he were to double his power yat keep the same about of cone area he probably would have seen roughly a +3 db gain in his score. If he would have doubled the cone area he would of probably gained another 3 db.This is consistent with the math and formulas above.

Originally posted by non_affiliated:
What I am trying to say i guess is it looks all nice on paper.It applies to the real world too.

The rest of your post is also consistent with what we have said above. A dB value is always referenced to something, because it is proportional to the log of a ratio, and the ratio needs a reference for the denominator.</font>[/QUOTE]Ok well stated earlier where was the confusion that adding two subs would net a nominal 3 db increase in SPL? Then doubling the power would net another 3 db increase which would make a over all increase of 6 db.

the orginal question that was as was:
Single Sub vs. Dual Subs Setup?
I really need some help on this one. I already have on of the Audiobahn flame subs that has 1100 Watts RMS, in a ported box, plus a JBL 1200.1. I was planning on having two of these subs and two amps, one per sub. But, with all that stress on my battery, alternator, and pocket trying to upgrade, Im not so sure anymore. How much less of SPL would I have by just opting for just a single setup vs the dual Last sentence contains the question. He was given the answer. Not the History of the decibel. You said you didn't see how he could get a 6 db increase with the dual setup and two amps, over the single amp and sub. The point of this whole post was lost with all the techni-babble. Some people could give a poo, unlike some of us that breath this poo day in day out. I am no expert but, nor a Guru, yet I know some of use look like a know it all *** when well sit here and nitpick the darn varables. SOme people like measuremnts that are dragged out 12 decimal places, some like the numbers rounded to the nearest whole number.

Do you think the original poster of the is thread was looking for a 12th decimal place answer. I doubt it.

I think sometimes we all get so caught up in little itsy-bitsy discrepencys and make it a point to pshow flaw that we lose the whole perspective on the post. It does get old.

[ October 21, 2002, 03:37 PM: Message edited by: non_affiliated ]

[kkant]

His question was answered by several people in several places, including myself. I thank you for adding the answer again. As for the rest, I learned something new and that was valuable to me. So did other people. If you don't like it, or don't want to understand it, then don't read it.

[Lex Luther]

I like the discussion! graemlins/thumb.gif

But in the case of the "Flame" subs, its take one POS add another POS and still have a POS.