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Loudest SPL (and how easily it *could* be broken) 
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#1
Apr610, 10:10 PM

P: 26

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 colocated, it gains 6dB. I'm not sure how colocation works for 3+ speakers, so I'm going to pretend colocating 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? 


#2
Apr610, 11:28 PM

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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/f...8/m/9511927169 


#3
Apr710, 01:22 AM

P: 66

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! 


#4
Apr710, 01:28 AM

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



#5
Apr710, 01:38 AM

P: 66




#6
Apr710, 01:49 AM

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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.phyastr.gsu.edu...elelin.html#c3 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.phyastr.gsu.edu...elelin.html#c1 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 xy 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, dE_{z} = dE (z / r) = dE sin(θ). [tex]E = 2 k \sigma \int_{\infty}^{+\infty} \frac{{sin}(\theta)dx}{r}[/tex] [tex]{sin}(\theta) = z / r[/tex] [tex]E = 2 k \sigma \int_{\infty}^{+\infty} \frac{z {dx}}{r^2}[/tex] [tex]r^2 = z^2 + x^2[/tex] [tex]E = 2 k \sigma z \int_{\infty}^{+\infty} \frac{dx}{z^2 + x^2}[/tex] [tex]E = 2 k \sigma z \left [ \frac{1}{z} tan^{1}\left (\frac{x}{z}\right )\right ]_{\infty}^{+\infty} [/tex] [tex]E = 2 k \sigma z \left ( \frac{1}{z} \right) \left ( \frac{+ \pi}{2}  \frac{ \pi}{2} \right ) [/tex] [tex]E = 2 \pi k \sigma [/tex] 


#7
Apr710, 02:24 AM

P: 66

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. 


#8
Apr710, 03:01 AM

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P: 7,049

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. 


#9
Apr710, 08:47 AM

P: 2,292

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! 


#10
Apr710, 07:40 PM

P: 26

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 100dB@ 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 100200 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 170190dB 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 272292dB if you convert it to 1 meter. 


#11
Apr710, 08:26 PM

P: 66

HOWEVER. YOUR UNDERSTANDING STILL WRONG TOO! :p 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! 


#12
Apr710, 08:46 PM

P: 26




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