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Old 4th May 2004, 11:18 PM   #1
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Default Theoretical SPL vs distance

Ok, if I remember correctly:

My A7 boxes are 97 dB with 1 watt/ 1 meter
My A4 boxes are the same.
Each running at one watt (2 watts total) = 100 dB
Around the 2048 watt mark I should reach the magical 130 dB plateau.

Now, how do I calculate the what the SPL drop is from being further from the speakers. I wish to know what the dB level is at 10 meters. This is in open air, no room gain.

Is there a formula or does it have to be measured because of differing dispersion patterns and other factors?

I think I just answered my own question.

Cal
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Old 4th May 2004, 11:28 PM   #2
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Not that I have any idea, but WinISD has a calculator for that.

At 2048W and 10 meters it says you should have 110dB... seems kinda low to me.
And 80dB at 2W and 10M
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Old 4th May 2004, 11:30 PM   #3
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It's a 6db decrease each time distance is doubled.
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Old 4th May 2004, 11:33 PM   #4
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Your loss is 20 log distance
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Old 4th May 2004, 11:36 PM   #5
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Line arrays have a 3db decrease per doubling of distance up to a certain point because the sound is focused within the vertical plain of the array while horizontal dispersion is the same as with a point source speaker.
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Old 5th May 2004, 12:08 AM   #6
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It depends on what BW you use to measure efficiency, but an A7/A4 in the majority of their BW are 104/107dB/W/m with Altec drivers, so are the woofers you're using that much lower eff.?

Anyway, sounds falls in free air at 6dB/doubling of distance, so using 97dB and 10m:

dB = 97 - (20*log10 (1m/10m)) = 77dB

at 2048W:

77+(10*log10 (2048) = 110.11dB/10m

Really, due to your tall stacks, it's best to measure fairly far away where the soundfields have 'jelled', then calc backwards to find the real 1m eff.. The old standard, and one that is still used by some in the prosound market is 0.001W/30ft, though 10m is within 0.78dB, so will work well also.

Note that due to atmospheric conditions and the decreasing acoustic pressure with increasing frequency, it falls quicker up in the HF. WRT off axis power response, you'll have to measure it.

GM
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Old 5th May 2004, 12:49 AM   #7
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Thanks for your responses. I have a pretty good handle on it now. I didn't know it was calculable.

GM,

You've always been a big help with the Altec stuff, maybe you can help again.
If you remember, I am using this woofer:

15"

And this tweeter on the 511B on the A7 boxes:
1" comp

The tweeter is heavily resisted (resistored?) to make the two match.
I am simply guessing at the 97dB.
Might it be more?
I have no way of measuring anymore. The only SPL meter I owned was the Radio Shack one about 25 years ago.

Thanks for your help,
Cal
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Old 5th May 2004, 03:44 AM   #8
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These aren't a good match to either cabs, but can be FR shaped with digital EQ. In the A4s, they would have ~99.81dB/2.83V/m in half space if wired in parallel, plus the ~7dB horn gain beginning around 100Hz. Since these are outdoors, subtract 6dB, or ~93.8dB ramping up to ~100.8dB/2.83V/m.

The A7 will be down to ~88dB/W/m outdoors, ramping up to ~94dB/W/m, not including whatever the ground plane adds, which won't be a lot IIRC, but I lost my outdoor measurements in a house fire in the early '70s.

I still use a 1st generation RS SLM, though it requires a correction factor.

GM
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Old 5th May 2004, 02:24 PM   #9
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Default Re: Theoretical SPL vs distance

Quote:
Originally posted by Cal Weldon
Ok, if I remember correctly:

My A7 boxes are 97 dB with 1 watt/ 1 meter
My A4 boxes are the same.
Each running at one watt (2 watts total) = 100 dB
Around the 2048 watt mark I should reach the magical 130 dB plateau.

Now, how do I calculate the what the SPL drop is from being further from the speakers. I wish to know what the dB level is at 10 meters. This is in open air, no room gain.

Is there a formula or does it have to be measured because of differing dispersion patterns and other factors?

I think I just answered my own question.

Cal
This is not what you wera asking about, but playing the two speaker simultaneously doubles sound pressure, ie + 6 dB. So running both at 1 watt each you end up with 103 dB, not 100 dB.

Yes, the efficiency is doubled, at least for low frequencies where there is no cancellation in any direction.

Adding amplifier from 2 to 2048 Watts, will increase the level by 10*log(2048/2)=30 dB, so you are now at 133 dB.
If the speakers are point sources (not line sources), the level drops by 6 dB for each doubling of the distance, or Lf=Lc+20*log (dc/df) where Lf/Lc are far/close levels, df/dc are far/close distances.

If df=10 m and dc=1 m level will drop by 20 dB, to 113 dB.

All under the conditions low frequency, free field, and point sources.
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Old 5th May 2004, 02:41 PM   #10
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At these power levels one has to take thermal power compression into consideration !

But they would still be loud enough to annoy the whole quarter.

Regards

Charles
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