# Decibel loss question

• December 28th 2009, 05:07 AM
1LISTEN
Decibel loss question
If it was possible to have a sound source of 4225 db, i know that at 1m it looses 11dB then every time you double the distance it looses 6db.

IE: 1m = 11, 2m = 17, 4m = 23, 8m = 29 etc etc

Question:
how far would it take in m for the dB level to go from 4225 to zero? working on the above formula?

Edited for 1m and revised for above example.
• December 28th 2009, 05:20 AM
Bacterius
That means that the decibel loss for a distance in metres $d$ is $6 \times \log_2(d) + 11$.

You want $d$ such as $6 \times \log_2(d) + 11 > 4225$ (when the decibel loss is greater than $4225$). Thus :

$6 \times \log_2(d) > 4214$

$\log_2(d) > \frac{4214}{6}$

Thus :

$d > 2^{702.3333...}$

Conclude : we would require many metres (around $2.65 \times 10^{211}$) to dissipate the sound. That's a little more than $6.62 \times 10^{203}$ times the circumference of the earth.
• December 28th 2009, 05:23 AM
1LISTEN
Bacterius,

Sorry i needed to edit i made a mistake, can you look at my above edited post please?
• December 28th 2009, 05:25 AM
Bacterius
Quote:

Originally Posted by 1LISTEN
Bacterius,

Sorry i needed to edit i made a mistake, can you look at my above edited post please?

Done.
• December 28th 2009, 05:28 AM
1LISTEN
Bacterius,

Thank you very much!

P.s: sorry i edited, my mistake.
• December 28th 2009, 05:33 AM
1LISTEN
Bacterius,

my calculator won't accept 2.65 x 10^211?? it comes up math error, can you tell me how many mile it is taking 1 mile = 1609.34 m please ?

Thank you !
• December 28th 2009, 05:37 AM
Bacterius
Uhm ... how to tell you ... that's a 211-digit-long number. Here is the approached value :
Quote:

26500000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00
I think it is better to keep it in scientific format.

$\mathfrak{I} \ \mathfrak{told} \ \mathfrak{you} \ \mathfrak{that} \ \mathfrak{was} \ \mathfrak{a} \ \mathfrak{lot} \ \mathfrak{of} \ \mathfrak{metres}$ ...
• December 28th 2009, 05:40 AM
1LISTEN
• December 28th 2009, 08:03 AM
CaptainBlack
Quote:

Originally Posted by 1LISTEN
If it was possible to have a sound source of 4225 db, i know that at 1m it looses 11dB then every time you double the distance it looses 6db.

IE: 1m = 11, 2m = 17, 4m = 23, 8m = 29 etc etc

Question:
how far would it take in m for the dB level to go from 4225 to zero? working on the above formula?

Edited for 1m and revised for above example.

Lets ignore that and take it a face value as a souce level of 4214 dB re whatever.

At range $R$ the sound intensity level will be:

$IL(R)=4214-10 \log_{10}(R^2)$

(this is what it means by an additional loss of 6 dB per doubling in range)

Now solving this gives $R=10^{210.7}\text{ m} = 10^{207.7}\text{ km}\approx 5 \times 10^{207}\text{ km}$.

This is absurd since 1 light year is only $\approx 9.5 \times 10^{12} \text{ km}$.

2. There is a maximum sound intensity that can be supported in an acoustic medium before the underlying assumptions become invalid and I am pretty sure that a source level of 4214 dB re the usual reference levels is way beyoun that.

Are you sure that you are not missing a decimal point in the given level?

CB
• December 28th 2009, 10:58 AM
1LISTEN
CB,

someone asked a question on a forum what would two levels of 65 dB be, i replied 68.01 dB .

$10 \log\left(10^{6.5} + 10^{6.5}\right) = 68.01\:dB$

someone else replied no! it's 65 x 65 dB = 4225

so having the answer off here i replied how far they would hear the sound, lol!