increasing power?

[mustangowner 10+ year member]

how much will it help on output? i have the setup in my sig as of now. later i am gonna install a Power Acoustik A1800DB which will increase my power from 250 watts rms to each sub...to 500 watts rms to each sub.

I used the db equation posted on here a week or two ago and according to that it only gained 2 db using the same equipment just more power. Is this correct or did i mess up somewhere? any suggestions.

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[hoss 5,000+ posts]

there is no equation that will tell you how much you will gain... there never has been one, there never will be one, so whatever math you did was useless.

there are too many variables that will effect the output increase or decrease that comes along with increasing power. and yes, often times when you go to a "more powerful" amp, it is possible to reduce the output of your stereo.

and before you go listing how it is installed, what kind of vehicle it is in, dont waste your time... no one will be able to tell you. the only way you will be able to find out is on a term-lab //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[squeak9798 5,000+ posts]

*Theoretically* doubling your power nets an increase of 3db.

However, that's only "in a perfect world". You are going to have issues like power/Bl compression, that are going to affect how much output is actually realized with the higher power level.

 

[squeak9798 5,000+ posts]

PS: What "db equation" are you referencing?

 

[mustangowner 10+ year member]

there is no equation that will tell you how much you will gain... there never has been one, there never will be one, so whatever math you did was useless.
there are too many variables that will effect the output increase or decrease that comes along with increasing power. and yes, often times when you go to a "more powerful" amp, it is possible to reduce the output of your stereo.

and before you go listing how it is installed, what kind of vehicle it is in, dont waste your time... no one will be able to tell you. the only way you will be able to find out is on a term-lab //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

first off chill out, here is the formula i got. i believe it was posted on this forum. i may have found it on another forum (not sure). but it was said it was tested and it came within +/- 3 dB.

This formula gives a rough estimate +/- 3 dbs.

-This formula is assuming that all things are perfect. It is a Logical "Guessimate" that is always within 3 db of actual!

-That voltage is near 14 volts/plenty of

battery/alternator

-Using at least 1/0 Ot guage wire

-Some kind of Line Driver is being used

1st Things first

Start off with sensitivity of the "raw" driver

Then total power of the amplifier (peak AC

volts measured with a volt meter)

For example: Take a Concept CC-D1200 amp, connected to 2 SS-12( a load of 1 ohm) You've measured the AC volts out into this speaker load as 42 volts, which calculates to 1764 "peak" watts. A/C volts squared (times itself) divided by the impedance equals power. So...42 x 42 =1764 divided by the impedance, which is 1 ohm in this instance, so its still is 1764!!! We now know the real power (cause you measured it into a real load...The Concept SS-12 woofer)

2nd Thing

Calculate the toal efficiency of the woofers

Add each woofer as +3, doubling of cone area

equals output

The concept SS-12" subs are 87 db 1W/1M (typical of all small box woofers"

Lets do the dB Log:

1 woofer = 87 dB

2 woofers = 90 dB

4 woofers = 93 dB

3rd Thing

Do the "dB log" starting with the total subwoofer(s) efficiently (which is 93 in this instance)

93 dB = 1 watt

96 dB = 2 watt

99 dB = 4 watt

102 dB = 8 watt

105 dB = 16 watt

108 dB = 32 watt

111 dB = 64 watt

114 dB = 128 watt

117 dB = 256 watt

120 dB = 512 watt

123 dB = 1024 watt

(for the example, we have 1764 watts, (comtemplate... 125 dB)

126 dB = 2048 watt

4th Thing

Add up the special "side deals" Like is the box sealed (no extra dB), hatchback or coupe? Etc.

Use the following to get "close to the REAL DEAL in SPL.

-Transfer Function (sedan) = 12 dB

-Transfer Function (hatchback with woofers aiming towards the rear glass) = 16 dB

-Sealed Box = no extra dB

-Vented box = 3 dB (UNLESS you have an odd order alignment)

-Vented box = 6-9 dB (If an odd order alignment - Big A$$ box)

-BandPass Box = 6-10 dB (depending on how small the enclosure is)

-Mic Placement = 6 dB for DB drag dash/window mic placement

Example: let's total it all up

124 dB (basic "dB log") + 12 dB (standard cab car with a direct facing woofer box) = 137

Woofer box is vented in something other than optimum or flat + 6 dB = 143 dB

+6 dB (DB drag dash/window mic placement)

So...this car will play

= 149 db (total) (+/- 3 dB)

 

[squeak9798 5,000+ posts]

Promptly forget that formula ever existed.

It's just horseshit. Has zero bearing on the real world.

 

[hoss 5,000+ posts]

first off chill out

i was calm, you asked for an answer, and i had the answer you were looking for. dont shoot the messenger //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

and squeak here backed me up, dont go and jump on him either because its not the answer you want //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[mustangowner 10+ year member]

ok thanks for your input guys.

 

[mustangowner 10+ year member]

there is no equation that will tell you how much you will gain... there never has been one, there never will be one, so whatever math you did was useless.

i took this the wrong way then. sounded like u were jumping on me for asking a dumb question. i didnt jump on sqeak or you. the equation was close with the increased output it said i would see 2 db increase and someone said "theorectically" i would net 3 db.

 

[squeak9798 5,000+ posts]

the equation was close with the increased output it said i would see 2 db increase and someone said "theorectically" i would net 3 db.

The problem with that "equation" is that it tries to pinpoint an actual sound pressure level; 149db (for example), which makes it totally useless.

If you want to use any equation, this is the one to use:

10*log(power1/power2) = db

10*log(2/1) = 3.0103db

10*log(500/250) = 3.0103db

So, *in theory* you should gain 3.0103db by doubling your power. But, like I said....this doesn't account for things like power/Bl compression, etc etc, so it's only theoretical.

 

[Nikuk 5,000+ posts]

i took this the wrong way then. sounded like u were jumping on me for asking a dumb question. i didnt jump on sqeak or you. the equation was close with the increased output it said i would see 2 db increase and someone said "theorectically" i would net 3 db.

The difference betweem 2db & 3db increase is not "close"

 

[gHOstFaCe 10+ year member]

It was very noticable when I gained ~2db just by opening the hatch leading through my back seat to my trunk. Then I went from 300w to 1200w rms, all I can say is wow. Huge difference.

 

[Nikuk 5,000+ posts]

while that is true, You have to understand that going from 130-135 dB reference is NOWHERE near as easy as going from 148-150 dB.

The scale is a logorythmic one, meaning it's harder to gain as You get louder.

About as simple as it can be made

I'm looking for James to post 1200 charactors of formulae //content.invisioncic.com/y282845/emoticons/naughty.gif.94359f346c0f1259df8038d60b41863e.gif