1. ## sound1

if concorde registers at 128dB and another aeroplane registers at 94 dBwhat is their combined loudness?...

2. Originally Posted by jaswinder
if concorde registers at 128dB and another aeroplane registers at 94 dBwhat is their combined loudness?...
The long way first:
$\displaystyle 128 = 10 log_{10} \left ( \frac{I_1}{I_0} \right )$
(where $\displaystyle I_0$ is the reference level, the value of which at the moment escapes me, but we don't need a value for it anyway.)

$\displaystyle \frac{I_1}{I_0} = 10^{12.8}$

$\displaystyle I_1 = I_0 \cdot 10^{12.8}$

For the second sound
$\displaystyle 94 = 10 log_{10} \left ( \frac{I_2}{I_0} \right )$

etc.
$\displaystyle I_2 = I_0 \cdot 10^{9.4}$

So
$\displaystyle I_1 + I_2 = I_0 (10^{9.4} + 10^{12.8})$

So the "loudness" will be:
$\displaystyle 10 log_{10} \left ( \frac{I_0(10^{9.4} + 10^{12.8})}{I_0} \right )$

$\displaystyle = 10 log_{10} (10^{9.4} + 10^{12.8}) \approx 128.002 \, dB$

I have corrected the above calculation (thanks again to CaptainBlack). The result is still practically the same as my original result in that it would be VERY difficult to hear the 94 dB sound as it barely contributes compared to the 128 dB sound.

-Dan

3. Originally Posted by topsquark
The long way first:
$\displaystyle 128 = log_{10} \left ( \frac{I_1}{I_0} \right )$
(where $\displaystyle I_0$ is the reference level, the value of which at the moment escapes me, but we don't need a value for it anyway.)
Except:

$\displaystyle 128 = 10 \ \log_{10} \left ( \frac{I_1}{I_0} \right )$

We have a 10 here as $\displaystyle I$ is a power like unit, if we were
using the pressure ampltude we would have a 20 here.

RonL

4. Originally Posted by topsquark
The short way:
Loudness is on a logarithmic scale so a 2 dB sound is 10 times as loud as a 1 dB sound. The 128 dB sound is 128 - 94 = 34 orders of magnitude louder than the 94 dB sound. You'll never hear the 94 dB sound in the background. It's like listening for your cellphone to ring while sitting in front of the speaker at a rock concert.

-Dan
10 dB is 10 times the power density, thus 128-94=34dB difference which
is a factor of 10^3.4, or between 1000 and 10000 times the power density,
or 10^1.7 times the pressure amplitude.

RonL

5. Originally Posted by CaptainBlack
Except:

$\displaystyle 128 = 10 \ \log_{10} \left ( \frac{I_1}{I_0} \right )$

We have a 10 here as $\displaystyle I$ is a power like unit, if we were
using the pressure ampltude we would have a 20 here.

RonL
Originally Posted by CaptainBlack
10 dB is 10 times the power density, thus 128-94=34dB difference which
is a factor of 10^3.4, or between 1000 and 10000 times the power density,
or 10^1.7 times the pressure amplitude.

RonL
(Ahem!) This is what I get for not double checking my memory before I start a problem! Thanks for the correction.

-Dan

6. Originally Posted by topsquark
(Ahem!) This is what I get for not double checking my memory before I start a problem! Thanks for the correction.

-Dan
Your welcome, its one of the few things that I am an expert on (dB's of
power and/or amplitude like units/ratios).

RonL

7. ## ok.....

ok i sort of got that..bt disreguarding the fact if u can or cannot hear certain things.....what actualy is the combined loudness if u did add both sounds together....

the sounds being...

concorde registering 120dB

and

an aeroplane registering 94dB?

8. Originally Posted by jaswinder
ok i sort of got that..bt disreguarding the fact if u can or cannot hear certain things.....what actualy is the combined loudness if u did add both sounds together....

the sounds being...

concorde registering 120dB

and

an aeroplane registering 94dB?
I have gone back and corrected my original answer to your question. You'll see that even with CaptainBlack's correct to the equation the result is essentially the same as I stated before: you won't hear the 94 dB sound.

-Dan