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9 message(s)
started Friday September 13, 2002
last updated 21 years ago
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Posted Friday September 13, 2002 - 21 years ago (#2966) |
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can anyone explain to me the correlative relationship between watts and decibals i.e. how many watts/amps does it take to produce some specific amount of decibals, does increasing by one db double the actual volume? |
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1724 post(s). Member since 22 years ago. |
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Posted Friday September 13, 2002 - 21 years ago (#2968) |
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actually, yes and no. the signal strength going across a wire is measured in voltage. the power of the signal is measured in amps. the resistance is always constant, thus wattage is the relationship between voltage ...
[dbwatts.html @ home.pacbell.net in /base10]
here is a decent article ... very mathematical in nature. i will try to find something written in english. |
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"underwater, darkness is as bright as daylights comes." |
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1724 post(s). Member since 22 years ago. |
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Posted Friday September 13, 2002 - 21 years ago (#2974) |
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Electrical stuff: [ohms_course.htm @ www.contractorcafe.com in /]
P = Power (watts or volt-amps) I = Intensity (current in amps) E = Electromotive Force (Voltage) R = Resistance (Ohms)
P = I x E
which means: watts = amps * volts
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So, if your raw signal is volts, then the decibel is ***relative voltage*** meaning it's measured in dbV (relative decibel volate). You will see this on you Vu meter as -3, -9, -12 etc. In the voltage context, wattage does not relate to a decibel at all.
However, when it gets pumped through a wire to the speakers, it's get measured differently. In that context, wattage is almost directly proportionate to decibel level (because they are both inverse squares). Wattage is a measurement of Power and Decibel are measurements of intensity, which in physics can be called "the same thing." One is electrical energy and the acoustical energy.
I wish I could find some good articles online. I'll keep looking. |
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"underwater, darkness is as bright as daylights comes." |
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1724 post(s). Member since 22 years ago. |
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Posted Friday September 13, 2002 - 21 years ago (#2975) |
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correction to previous mail:
So, if your raw signal is volts, then the decibel is relative voltage meaning it's measured in dbV (relative decibel volate). You will see this on you Vu meter as -3, -9, -12 etc. In the voltage context, wattage does not relate to ___wattage___ at all. |
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"underwater, darkness is as bright as daylights comes." |
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1724 post(s). Member since 22 years ago. |
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Posted Friday September 13, 2002 - 21 years ago (#2976) |
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ack! nevermind! i retract the correction! |
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"underwater, darkness is as bright as daylights comes." |
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Posted Saturday September 14, 2002 - 21 years ago (#2993) |
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thanks for the tech specs, i know, this one is a bit squirelly i think. |
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4331 post(s). Member since 22 years ago. |
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Posted Saturday September 14, 2002 - 21 years ago (#2996) |
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these are interesting correlations - i would love to hear more on the subject, just in layman's terms as i am an uneducated idiot... |
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158 post(s). Member since 22 years ago. |
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Posted Wednesday February 19, 2003 - 21 years ago (#5564) |
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short story:
in general, decibels are a kind of power ratio: Watts divided by Watts
sound decibels are a measure of sound pressure (ie: sound power): how hard does the air push between waves
a 10 db increase is needed for the average listener to hear something "twice as loud"
the electrical watts of your system compared to the sound decibel output depends on how good your system is: you always lose some coverting one energy to another energy: ie. a crappy system does not convert electricity into sound very well
long story (math): [dbwatts.html @ home.pacbell.net in /base10] |
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Posted Monday February 24, 2003 - 21 years ago (#5655) |
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thank you thank you for this great link!:D |
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