# Loudest SPL (and how easily it *could* be broken)

I read from several sources that the loudest sound ever made is over 200dB either from earthquake or *insert powerful bomb name here* explosion. Unforunately I don't know at what distance these volumes are measured.

I thought for every speaker you add, you gain 3dB. If they're co-located, it gains 6dB.

I'm not sure how co-location works for 3+ speakers, so I'm going to pretend co-locating speakers only result in 3dB increase per speaker.

You can easily find speakers that can play over 100dB at 1 meter. If you get 50 speakers playing 100dB at the same time, right beside each other, you'd get 250dB at one meter, right? You'd also get that with very little energy. Assuming each speaker has a sensitivity of 88dB with 1 watt, it only takes 16 watts to reach 100dB. 16x50=800 watts.

I realize I'm talking about 1 meter, but you only need two extra speakers for every doubling of distance. So even a mile away would only need 20ish more speakers for the same SPL.
Am I right? If I'm not, could you point out what is wrong?

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rcgldr
Homework Helper
At around 194db, the pressure varies from 0 to +2 atm. Since pressure can't go negative, SPL above 194db is no longer a sound wave but a shock wave. Volcano expostions have exceeded 300db.

A list of spls:

http://community.discovery.com/eve/forums/a/tpc/f/7501919888/m/9511927169 [Broken]

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I read from several sources that the loudest sound ever made is over 200dB either from earthquake or *insert powerful bomb name here* explosion. Unforunately I don't know at what distance these volumes are measured.

I thought for every speaker you add, you gain 3dB. If they're co-located, it gains 6dB.

I'm not sure how co-location works for 3+ speakers, so I'm going to pretend co-locating speakers only result in 3dB increase per speaker.

You can easily find speakers that can play over 100dB at 1 meter. If you get 50 speakers playing 100dB at the same time, right beside each other, you'd get 250dB at one meter, right? You'd also get that with very little energy. Assuming each speaker has a sensitivity of 88dB with 1 watt, it only takes 16 watts to reach 100dB. 16x50=800 watts.

I realize I'm talking about 1 meter, but you only need two extra speakers for every doubling of distance. So even a mile away would only need 20ish more speakers for the same SPL.
Am I right? If I'm not, could you point out what is wrong?
Well there are a two big problems.

The biggest being is that the dB scale is logarithmic: 106dB is twice as loud as 100dB, 112dB is twice as loud as 100dB. So let's say you can get the full +6dB for your second speaker. To get from 106 to 112 you don't need 3 speakers, you need 4! And at 118 you need 8
124 needs 16, 130 needs 32, and 136 needs 64. You're out of speakers already!

The next problem is you can't just line them up. They all have to be the same distance from the point you've measuring. If you wanted to get a spot with 130dB, it needs to be 1 meter away from all 32 of your 100db at 1 meter speakers at the same time!

rcgldr
Homework Helper
speakers ... you can't just line them up. They all have to be the same distance from the point you've measuring.
If you arranged the speakers into a plane, you'd reduce the effect of distance. If the size of the plane is relatively large compared to distance from the plane, then distance would not be much of a factor. The issue with this is that you'd need magic speakers that weren't directional.

If you arranged the speakers into a plane, you'd reduce the effect of distance. If the size of the plane is relatively large compared to distance from the plane, then distance would not be much of a factor. The issue with this is that you'd need magic speakers that weren't directional.
Why wouldn't it? If a speaker is 2 meters away, it's going to have 12dB less than a speaker 1 meter away (presuming they're point sources). I don't see how arranging them into a plane would help.

rcgldr
Homework Helper
Why wouldn't it? If a speaker is 2 meters away, it's going to have 12dB less than a speaker 1 meter away (presuming they're point sources). I don't see how arranging them into a plane would help.
If you have a bunch of point sources arranged into a plane, and they are close enough together, and if the distance from the plane is large compared to the distance between points, but small compared to the size of the plane, then the intensity versus distance from the plane within that range of distances is approximately a constant.

Link to the math for intensity of a field relative to a solid disk. As the radius of the disk approaches infinity, then the intensity approaches a constant indenpendent of distance:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c3 [Broken]

From an old post I wrote here:

I couldn't find the rectangle based derivation, so I made a summary of that approach:

For an infinite line charge, E = 2 k λ / r, where λ is charge per unit length and r is distance from the line. Here is the derivation:
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c1 [Broken]

For an infinite plane case, infinitely long strips (rectangles) approximate a line as their width -> 0. The area of a strip of lengh L is L dx, and the charge dq = σ L dx. The charge per unit length dλ = dq/L = σ dx. Assume the plane exists on the x-y plane, then the magnitude of the field at any point in space from a strip is dE = 2 k σ / r, where r is the distance from a strip to that point in space. In the case of the entire plane, the x components cancel because of symmetry, with only a net force in the z direction, and for each strip of the plane, dEz = dE (z / r) = dE sin(θ).

$$E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{{sin}(\theta)dx}{r}$$

$${sin}(\theta) = z / r$$

$$E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{z {dx}}{r^2}$$

$$r^2 = z^2 + x^2$$

$$E = 2 k \sigma z \int_{-\infty}^{+\infty} \frac{dx}{z^2 + x^2}$$

$$E = 2 k \sigma z \left [ \frac{1}{z} tan^{-1}\left (\frac{x}{z}\right )\right ]_{-\infty}^{+\infty}$$

$$E = 2 k \sigma z \left ( \frac{1}{z} \right) \left ( \frac{+ \pi}{2} - \frac{- \pi}{2} \right )$$

$$E = 2 \pi k \sigma$$

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If you have a bunch of point sources arranged into a plane, and they are close enough together, the intensity versus distance from the plane is approximately a constant as long as distance from the plane is small compared to the size of the plane.
That doesn't really have anything to do with my statement

If you want 32 100dB @ 1 meter point source speakers to produce 130dB at a spot, they have to all simultaneously be 1 meter away from that spot. If you put them in a plane, you aren't going to get 130dB because they all won't be 1 meter away, there might be several spots where moving toward the speakers and away doesn't change the db levels much, but that's irrelevant, because you aren't going to get 130dB in any spot.

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rcgldr
Homework Helper
If you want 32 100dB @ 1 meter point source speakers to produce 130dB at a spot, they have to all simultaneously be 1 meter away from that spot.
So arrange them into a sphere that surrounds the target point, although that would be a standing wave.

The point I was trying to make is that with sufficient density and number of speakers arranged in a plane, you can reduced the effect of distance from that plane. For example, in a stadium filled with screaming people, distance from the crowd doesn't make that much difference with the intensity of the sound you hear. The sound you hear at 10 feet to 30 feet above the crowd would be about the same.

At around 194db, the pressure varies from 0 to +2 atm. Since pressure can't go negative, SPL above 194db is no longer a sound wave but a shock wave. Volcano expostions have exceeded 300db.

A list of spls:

http://community.discovery.com/eve/forums/a/tpc/f/7501919888/m/9511927169 [Broken]

"310 (N) Krakatoa volcanic eruption, 1883. Heard 3,100 miles away. Barometers fluctuated up to 100 miles away, indicating sound levels of at least 170-190dB (P) at 100 miles. Instantaneous fog at 100 miles. Debris thrown to a height of 34 miles. Debris feel, continously, for 10 days. Ejected 4 cubic miles of the Earth."

Instantaneous fog at 100 miles. Wow!
And being heard 3,100 miles away. Dang, that's like someone in New York City hearing your stereo from beyond Los Angeles. Yikes!

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Interesting insights.

i'd like to answer a few questions.

@Perfection and Jeff Reid about the comment that if you line the speakers, you won't get [email protected] 1m because you lined them up and they're far apart.

Well, you can use small speakers like the size of a CD case. Quality small speakers can reach 100dB @ 1m as well. You can stack it and you can have 100-200 speakers stacked in a way that no speaker is more than 2 meters away from the mic.

@Perfection. I think your understanding of the dB scale is wrong. Every 3dB increase is twice as loud. For every speaker you add, you get 3dB more if you put them side by side. So to go from 100dB to 200dB you theoretically only need 34 speakers together playing 100dB at the same side. But of course you lose a little SPL because the speakers on the outer rings will be farther that the "listening" point, so just add a few speakers to it to compensate.

@Pallidin. If you can measure 170-190dB from 100 miles away, then our speaker example is nothing, I guess.

EDIT: actually, according to the inverse square law, you lose only 6dB for every doubling of distance. So it is only 272-292dB if you convert it to 1 meter.

@Perfection. I think your understanding of the dB scale is wrong.
It was! My only defense was it was late and I was tired. 3dB is twice as loud.

HOWEVER. YOUR UNDERSTANDING STILL WRONG TOO! :p

Every 3dB increase is twice as loud.
Correct

For every speaker you add, you get 3dB more if you put them side by side.
You agree that any two speakers of X dB when placed side by side out X+3 dB, Correct?

So let's look into our 100dB speakers, you could imagine that speaker being composed of 2 97dB speakers could you not?

And we could call our two 100dB speakers side by side a 103dB speaker right?

Now how do we get 106dB? Well, by our rules. We need two 103dB speakers. And since each 103dB speaker is composed of two 100dB speakers, we need 4 100dB speakers to make our 106dB!

And we need two 106dB speakers to get 109db, that's 8 100dB speakers.

And heaven help us if we want 199db! Because that requires 33 doublings or 8,589,934,592 speakers!

HOWEVER. YOUR UNDERSTANDING STILL WRONG TOO! :p

Correct

You agree that any two speakers of X dB when placed side by side out X+3 dB, Correct?

So let's look into our 100dB speakers, you could imagine that speaker being composed of 2 97dB speakers could you not?

And we could call our two 100dB speakers side by side a 103dB speaker right?

Now how do we get 106dB? Well, by our rules. We need two 103dB speakers. And since each 103dB speaker is composed of two 100dB speakers, we need 4 100dB speakers to make our 106dB!

And we need two 106dB speakers to get 109db, that's 8 100dB speakers.

And heaven help us if we want 199db! Because that requires 33 doublings or 8,589,934,592 speakers!

You're right! I didn't think of that! I knew I had to be wrong somewhere.