1. ## 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.

2. 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.

3. Bacterius,

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

4. Originally Posted by 1LISTEN
Bacterius,

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

5. Bacterius,

Thank you very much!

P.s: sorry i edited, my mistake.

6. 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 !

7. Uhm ... how to tell you ... that's a 211-digit-long number. Here is the approached value :
26500000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00
I think it is better to keep it in scientific format.

$\mathfrak{I} \ \mathfrak{told} \ \mathfrak{you} \ \mathfrak{that} \ \mathfrak{was} \ \mathfrak{a} \ \mathfrak{lot} \ \mathfrak{of} \ \mathfrak{metres}$ ...

8. Thanks for your kind help.

9. 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

10. 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!