Question about db spl

[itsallgoodman]

Hi folks, I'm new here and looking for some help. I'll cut to the chase.

If 1 watt @ 1 metre gives 96dbspl, how do I work out how many watts I need to deliver 90 dbspl at 18 metres?

I am going to allow the room reflections to provide me with extra headroom so I'm not counting them in the calculations.

Anybody know how to do this?

 

[JMFahey]

IF your speaker is a point source, radiating equally in all directions, you lose 6dB pr each distance doubling.

Your reference distance is 1 meter, so as a mental calculation, we'll use 16m simply because it's on an even doubling scale (2/4/8/16) and close enough.

4 doublings so 24dB loss.

Since you want 90dB there, you need 114dB at the source.

Or 114-96=18dB above 1 Watt driving the speakers, some 63W RMS

I mentally calculated at 16m, for a closer result we'll have to fire the calculator or ask:

For an online calculator:
Sound level distance damping decibel dB damping calculation calculator change distance versus sound level apps reduction drop dissipation SPL sound transmission loss free field loss sound and distance - decrease drop fall sound over distance versus d

which gives us a more precise (since we use 18m) 115dB at the source or 80W

For a dB to power ratio calculator:
dB conversion

 

[itsallgoodman]

Thank you very much. That's awesome.

 

[Gnobuddy]

I don't think you have to worry about radiating uniformly in all directions unless the speaker is also suspended 18m above the ground.

Then again, radiation into a half-sphere also follows the same law (surface area quadruples when distance from source doubles, so -6dB for each distance doubling.)

But we know that only very low frequencies are so widely dispersed from an actual speaker (as opposed to an ideal point source). I think that means in practice that the SPL in front of the speakers will be considerably higher than the point-source estimate. (And the SPL behind the speakers, far off to the sides, etc, may be lower than the point-source estimate.)

I have had the same question, and never found an easy way to get a reasonably accurate answer.

-Gnobuddy

 

[AndrewT]

20*log(18) gives the dB required for 18m compared to 1m.
i.e. +25.1dB if the radiation is not directed as mentioned by Gnobuddy.

Take off the 6dB for your listening level and you find that Fahey's mental arithmetic is pretty close.

 

[sreten]

Hi,

In the room the sound level beyond the critical distance,
(= direct spl = reflected spl), the apparent level will be
perceived as the same. Below the CD the level will
increase as you get nearer the speakers.

The CD depends on the speakers dispersion and
how lively that particular room reflections are.

One can say the shorter the CD the louder the
perceived volume beyond the CD will seem.
I.e. speakers seem a lot louder in lively acoustics
than they do in dull acoustics, so 90dB @ 18m
is a moving target, unless you are outside.

rgds, sreten.

 

[djk]

"critical distance"

Most rooms this is less than 20', so the level will not drop beyond this distance.

20' is about -16dB from 1M distance.

Speech become difficult to understand at a little less than 3x the critical distance, due to loss of consonants.

 

[dotneck335]

IF your speaker is a point source, radiating equally in all directions, you lose 6dB pr each distance doubling. Your reference distance is 1 meter, so as a mental calculation, we'll use 16m simply because it's on an even doubling scale (2/4/8/16) and close enough.
4 doublings so 24dB loss.Since you want 90dB there, you need 114dB at the source. Or 114-96=18dB above 1 Watt driving the speakers, some 63W RMS

Precisely. Plus 14 db for headroom=1,582 watts.