Decibel DC/AP?

[RRTheEndMan]

why do you keep posting already known things? Also, where's the problem with that formula?

 

[CNBA3]

sorry, it is just I am trying to find that formula on the internet but nothing shows up, does your formula have a name that I can search by?

 

[RRTheEndMan]

I just inverted what I found on wikipedia

 

[CNBA3]

is that from the sound pressure level section?

B= 10log^10(P/P0), that is what is used for this equation.

 

[RRTheEndMan]

Wattage=10^((SPrL-92.0182013165)/20)

 

[CNBA3]

can't find it on the wikipedia

 

[RRTheEndMan]

https://en.wikipedia.org/wiki/Sound_power#Relationship_with_sound_pressure_level

 

[CNBA3]

still can't find it, be most specific with an exact name and/or image of the equation because what your wrote doesn't show up.

 

[CNBA3]

are referring to Lw= Lp+10log^10(2¤Ç ├ù 3.826898430053^2/A0)dB?

 

[RRTheEndMan]

$ L_W=L_P+10log(2pir^2/1) $Since standard distance is 1 m $ L_W=L_P+7,98179869 $ Solving:
since $ L_w=10logWE12 $ $ W=10^{L_W/10}/10^{12}=10^{L_W-120/10}=10^{L_P-112,018201/10} $.
I messed up the formulae for the previous values

 

[CNBA3]

ok, so I did this 1000+10log(0.1062/1) and I got 1030.26124517 for Lw, I guess there is something wrong with that.

 

[RRTheEndMan]

why 0,1062?

 

[CNBA3]

why 0,1062?

that is the surface area of the hemisphere in m^2 in radius

 

[RRTheEndMan]

pretty sure that if you are given no radius you assume 1 meter distance

 

[CNBA3]

even with that, I get 1000 lw for the same formula

 

[RRTheEndMan]

I don't understand.
where's the problem?

 

[CNBA3]

I applied the the 1 meter into the same formula by it doesn't show anything different.