[Request] How accurate is this?(i.redd.it)

Unfortunately the post is not accurate!

To add equal signal levels you add 10*log( the number of sources ) [dB] to your single source power.

This gives us 70 [dB] + 10 [dB] * log(2,000,000,000 [cats])

=163 dB from all the cats. That's real loud but nowhere near 130 (billion?!) dB.

Given a single jet takeoff of 120 dB, the 163 dB cat serenade would be equivalent to about 20,000 jets taking off at once. This is by the same formula, and no where close to the posting. I'd still call it catastrophically loud.

Interestingly, this is also very close to the noise of a gunshot.

https://www.engineeringtoolbox.com/adding-decibel-d_63.html

https://ehs.tamu.edu/media/1279860/NoiseThermometer(English).pdf.pdf)

I'm pretty sure 130 billion dB would end the entire Universe.

Yeah, at that point you are creating black holes from cats meowing

Why arent we?

if you're being serious it's because you need a absolute fuckton of energy to create a wave powerful enough to compresses air under the schwarzschild radius

Because cats are constructs of pure entropy. You'd accelerate the heat death of the universe to "instantaneously now" upon any attempt.

Because of math. You should stay here for a while.

a

The inevitable end of our universe to be fair

I don't mind having my gravestone say "Death by Meow".

You only need like 250 db for that really

Good explanation for black cats! (A.K.A - Void cats)

If you tape a freshly buttered toast to the back of a cat it will create a black hole. Since cats always land on their feet and buttered toast always lands on the buttered side, the object of those two taped together starts spinning increasingly fast while approaching but never reaching the ground.

A little bit of googling shows it would only take several thousand dB

"With energy as great as 1100 dB, it would create enough gravity to cause a black hole to form, and an incredibly large one at that. Decibels are a logarithmic unit. That means 20 decibels isn't 2 times more powerful than 10 decibels, it's 10 times more powerful."

*The whole article was written from a reddit discussion though.

Idk. Source seems sketchy af

Yeah, some random reddit user with a name like /u/hazar815 is NOT the kind of reliable testimony I'm looking for. Your vigilance and caution against trusting this kind of ill reputed source is much appreciated, /u/hazar815!

About how many notifications a week do you get because of that article? and did you give them permission or did they just rip off your math?

A sound can't be louder than ~194 dB in Earth's atmosphere.

(A sound is a pressure wave in a medium, and the volume of that sound is caused by the difference in pressure between the peaks and valleys of that wave. At 194dB in Earth's atmosphere at sea level, the valleys are a perfect vacuum. If you add more energy, you get a shock wave, a wall of high air pressure travelling at the speed or of sound, but you don't get "sound" in any meaningful sense.)

This is very interesting!!

290 something dB is as loud as it gets, because that's the speed of sound in the atmosphere. Shockwaves have a max intensity limited by the speed of sound through the material their propagating in.

0 dB is 20μPa. Each 20 dB is a tenfold increase in pressure. 1 atmosphere is roughly 100kPa. Taking the log of the ratios and multiplying by 20 gives 194dB. Any higher than that would result in negative pressure (or clipping).

Each 20 db is a tenfold increase? I thought it was each 10 db?

Only 194 actually

What happens to the excess energy if you have a spherical source of some kind outputting 300DB in earth atmospheric conditions?

Explosive shockwaves that will start crashing into things and transferring energy in a rather destructive manner

Is that because a 10 dB increment is 10 times the unincreased amount? (Sorry, I don't remember much from logarithms and logarithmic scales)

Let’s do the math.

It would have ended way way way before that

Also, that's supposing all the cats are in exactly the same place

Exactly, distributed evenly, this sounds like <1 cat meowing just about everywhere on earth

As opposed to the 2 or 3 "meowing" outside my window some nights.

Exactly. At any given point in time, there are countless billions of things making noise all over the world. If noise levels worked like this for things that were far apart, we'd be deafened the moment we were born.

I love how the mother has the strength to withstand the excessive energy in your scenario

She's also been deaf since birth, obviously

Not necessarily.

It assumes that all cats are at the same distance from the listener as the single cat that the 70dB figure comes from.

Let's say that each cat needs an area of 200cm^{2} to poke it's head in to meow. This translates to 40M m^{2} for 2B cats. Thus, to arrange 2B cats in such a way that they're all the same distance, you'd need a sphere around 1.7km in radius (surface area of the sphere needs to be 40M m^{2).} I don't think the average cat can manage 70dB from a distance of 1.7km. If you squish them all togethar and squeeze their heads into an area of 100cm^{2} (which should be doable without causing pain/injury for most of them) the required radius is still over a km at 1.2km.

and by exactly, we mean EXACTLY, cats in full superposition, not just herded into a big amphitheater.

That's like probably twice as hard as herding them into any room.

Right? 20,000 could possibly have taken off at the same time around the world. Over 100k take off each day. Under the right circumstances it could have happened. Since it is dispersed all over the globe, we haven't noticed some massive blast.

What would the “average” noise level be? Assuming all cats were equally spaced across the globe, how would one catculate how loud it would be? Would it just be 163 dB divided by the area of earth? ~197million mi^{2} would give ~8.3e-7 dB if it’s a direct division, but I am guessing there’s logarithmic shenanigans at play

dB is a measure of pressure level. Two equidistant sources can both double the effective pressure level at some measurement point or even cancel eachother out completely, depending on the interference pattern (this is how noise cancelling headphones work). If we'd take the extreme case where all cats make the exact same meow in-phase, we tesselate the cats as triangles and we measure at the centers of those triangles (where the constructive interference is highest), you'd find the entire exercise is moot: there are too few cats in the world to get a dense tesselation, and the centers of these triangles are in the orders of hundreds of meters away from your three closest cats. This means the theoretical optimal interference point (3 times local pressure) doesn't even come close to just standing next to one cat who's meowing at your face.

To clarify, Decibels are a unit of proportion, a ratio between two quantities.

dBA is the measure of pressure level. dBV is a measure of voltage, and dBm is milliwatts.

You're completely right otherwise though

To get even more specific, dBA is A-weighted sound pressure level, which is the measure if sound pressure corrected to the spectrum of human hearing.

This formula also assumes that all the sources are concentrated at the same distance to the observer, but cats have volume and we can include this.

An average cat weights around 5 kg, and is essentially water with stuff. We approximate the density of a cat with that of the water, and a cat is around 5 litres = 0.005 m^{3} in volume. 2 billion cats have a combined volume of 10^{13}m^{3}. This is the volume of a cube 21.5 km long.

We distribute the cats in this enormous cube and front one of the faces at a distance of 1 m. The intensity that we receive is not 2bn times the original, because the sound intensity dissipates with the distance (particularly as an inverse square law, double the distance, quarter the intensity). The actual intensity received is 91673 time bigger than that of a cat alone. This number is the form factor. If all the cats were in a single point the form factor would be the number of cats

The formula for the power in dB correcting for distance is then

Power in dB = 10*log(Intensity)+10*log(form factor)

The signal corrected for the distance 70[dB] +10*log(91673) = 70 + 49.6 = 129dB. It is indeed loud. But only around 8 to 10 jets taking off at once (and at the same distance)

*This number is the form factor, and comes from calculating the distance to 2 billion points in the cube. One can use numerical methods or integrating the inverse square of the distance to the point of observations over the volume of the cube

Makes sense. I didn’t see your post at first so I derived things slightly differently and got a similar answer.

Let’s say the cats are spaced in concentric rings on a big open field (ignoring questions of herding). The number of cats in ring r is equal to 2 * pi * r. The number of cats in total is pi * r^{2} . 2 billion cats comes out to 14,200 rings.

Conveniently, the sound intensity of each cat decays according to 1 / r. So the loudness contribution of each ring is 2 * pi * r / r = 2 * pi. (Correct me if I’m wrong, but I think I’m using the right values of intensity vs. pressure). Therefore all rings contribute the same amount to the loudness.

That also means that the sound intensity of spatially distributed sources on a plane increases according to sqrt(n). Or to make the math convenient, cut the decibel increase in half.

So instead of increasing from 70 to 163 dB, we should only expect the volume to increase to 117 dB.

Did you integrate this numerically, or maybe there a table where one can find common fits for form factors?

I integrated numerically. The integrand is the inverse of the square of the distance from the observer to any point of the cube.

Integrate [(x^{2} +(y+d) ^{2} + z^{2)} ^{-1} , {x,-L/2,L},{y,0,L},,{z,0,L}]

Corresponds to an observer at a distance d of a cube of side L. The observer sits on the floor in the center of one of the sides of the cube

Is there a Formular that considers distance for the sound level at the point of the observer?

formular

Mr. Krabs, what are you doing on Reddit

It depends a bit on what the source looks like, and the frequency of the sound. From a point source, like a single cat, in free space, with a high frequency meow that behaves more like rays it would drop by 6dB per doubling of distance. The 65dB figure for a cat would be at a standard 1 meter. The area of the sound energy drops by the square or the distance: https://www.sweetwater.com/insync/media/2016/02/Line-Array-Inverse-Square-Law.jpg

Now if some poor soul tries, and manages, to get all those cats to sit in a row we get what's called a line array effect, where it only drops by 3dB per doubling instead, since the area is proportional to distance. This is utilised at conserts and such using those large stacks of speakers at the sides of the stage to be able to cover a larger area in front of the stage without having to build islands with additional speakers in the middle of the sea of people.

https://www.sweetwater.com/insync/media/2016/02/Line-Array-Line-Source.jpg

But to get the effect the distance between the cats depend on the frequency of the meow and a bunch of other stuff.

People always seem to forget dB is a log scale

I found a near-exact analog: a rocket launch, on the launchpad! 165 dB or firing an M1 Garand, 1m away 168 dB

http://www1.lasalle.edu/~reese/decibels.htm

Doesnt that assume all the cats are placed together. Which in real life they of course are not.

okay...are the cats in a single location, and there's no destructive interference in the sound waves?

🤔 I mean, it seems like it’s entirely based on the position the cats are in? If they were in a tightly contained circle, it would be much different than if they were equally spaced across the world… 🌎

That is correct! Based on the current spacing of cats I imagine I would personally hear about 3. Quite short of 2,000,000,000,000. :)

I'm no scientist but I'm pretty sure the fact that the cats are evenly distributed throughout the world would have a diminishing effect

It'd also be interesting to look at how far sound can propagate and how much volume decays over distance traveled, and therefore approximately how wide a radius to consider for the volume at any given point

Edit: Taking a look at: https://physics.stackexchange.com/questions/415409/how-far-can-a-shout-travel, we can determine the sound absorption coefficient to be approximately 0.4 dB/100m (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7000907/ puts a cats meow at approximately 700Hz, and we can assume 20°C air at 1 atm, giving 0.4 dB/100m for most humidities). Using the same 0.3m distance for the initial meow volume in OP's post, we get:

L = 70dB - 20log10(r/0.3m) - 0.4r/100m

Again we'll solve for L = -9dB. the approximate threshold for human hearing, which gives a maximum distance an average cat's meow can be heard by an average human, r, of ??? - I got stuck here, I realised I didn't know how to solve that equation!

Also all the calculations are condensing the cats in one point and not accounting for the distribution of the cats around the globe...

catastrophically

Isn't this also true only if the sound signal arises from a single point in space? Given that 2 billion cats are gonna occupy about 2 billion sq.ft. (about one cat per 1 sq.ft.) or about 185 million sq.m., the sound is going to a lot more muted. Imagining all cats are arranged in a cube, the cube would measure 1260 ft. in all three directions.

Cats' meow is typically about 45db, not 65db. The loudest cat meow (Guinness record) is 67.8 dbs. As per the inverse square law, 45db would die down in about 181m or 590ft. We can even consider all those cats in a 590ft. cube to be a point source (which might give the highest value). That's about 205 million cats.

So maybe it works out to = 45db + 10db*(log250,000,000) = 128db.

Even if that is loud, there are single animals that produce that loud sounds. A single Moluccan Cockatoo makes 129 db. A lion's roar is 114 db. African elephant- 117 db. A howler monkey- a freaking 140db.

Also, I have no idea if that working is correct. And that sound will be a lot lesser IRL, because it is not made by a point source. The meow would become a whisper level 30db at a distance of 8 meters or almost inaudible at around 30m (about 100 ft.) I am guessing the SPL might be around 80db IRL. I might be completely wrong.

8 meters is the the same distance as 11.59 replica Bilbo from The Lord of the Rings' Sting Swords.

8 meters is 8.75 yards

Yeah, you only need to reach 1,100 db to be in the right range to create black holes. Billions is just a stupid amount of energy. I have absolutely nothing to confirm it as i can't be bothered learning how the formula works, and the online calculator just wouldn't accept 1 billion (let alone 130 billion) but I'd be very comfortable saying that large stars don't come close to putting off that much energy over their lifetime.

It would be even lower if you can't gather all cats into a small area.

The marginal increase per cat goes basically to zero, so it may be less of a difference than you may expect!

Wait a second.... 120 dB = one jet, then how does 163 dB = 20000 of them??

Just very confused

It's a logarithmic scale. Each additional cat, (or jet) adds less power to the sound.

The same way 70 dB cat * 2,000,000,000 cats is not more than 3x the dB value than one cat!

Thanks for the explanation! But a thing I still dont get... How exactly does (log) work? If you have 70 + (10*log) wouldnt it be greater than 20.000.000.000.000 dB?

I am so sorry for spamming questions I am just so confused about simple arithmetics :((

It is the power of one cat plus 10 times the log of 2 trillion.

Log (2,000,000,000,000) = 11.3

Power = 70 [dB] + 10 [db] * (11.3) = 163 [dB]

The logarithm is a function, log (2 trillion) is equivalent to: 10^(x) = 2 trillion, or ten to the power of x is 2 trillion, where x is log(2 trillion).

Ten times the cats increases x by one. So if instead of 2,000,000,000 cats we had 20,000,000,000 it would be 70 + 10 * 12.3.

Does this make sense?

:)

Also, this would assume they are all arranged in a sphere around the microphone. If they did it in place, we wouldn't even notice unless you're near a dozen cats.

Keep in mind that 10 DB more means that the sound is double

thank you and this is in acoustic pressure, and it even neglect the volume of the cats. It could be even lower than that. a cat is about 11 litres (0.011 cubic meter. 2B cat is a ball of 36m of diameter. (4/3 pi r³)

we cannit take a measurement closer than 18m from the center of the serenade, thus the sound pressure is way lower.

https://www.evilmadscientist.com/2007/computing-the-volume-of-a-cat/

This would be sound energy, not pressure. dB measures energy!

A 36 meter diameter seems very small. I imagine the actual measurement would be very close to 163ish dB, given the outside cats, although far away, contribute so little to the total energy!

dB means it's on a log scale.

you can have sound power in dB, the base unit is watt.

you can have sound energy in dB, base unit is Joules

You can have sound pressure in dB, base unit is Pascal.

the subscript wil change, usually, we use "level" to let know the reader that the unit will be in dB. Sound pressure level (SPL) and sound powet level (SWL) are both in dB. whereas Sound power is in Watt and Sound pressure is in Pa.

I'm going back to my test but I'll use my software here to check the math and use sound power everywhere before midnigh EST.

Goodness, I did not know that. Thank you for the info. I'm curious what you find out!

It’s important to note that 2 billion cats meowing cannot be considered a point source and even if co-located but not stacked vertically, would consume considerable land area, so large that if you were standing in the middle of it, there would be a certain radius that would be too far away for the pressure wave to propagate to you.

If you really want to get technical, since not all cats meow at the same pitch, there would be some destructive interference with the sound waves so it would be somewhat quieter than 163 dB. Also, it would depend on how the cats are arranged around the listener.

Minor correction to the number of jets though.

A jet takes off at about 140 dB. Let the number of jets required to get to 163 dB be x

140 + 10 * log(x) = 163

10 * log(x) = 23

log(x) = 2.3

x = 10^2.3

x = 199.52 ≈ 200 jets

The second link I posted even has it as 150 dB! This would mean 20 jets! The distance from the jet and the type of engine is not specified! :O

I think this is a case of using the wrong formula to get a technically correct numerical answer. A decibel is often not used as a unit in itself but rather as a way to express scaling a unit logarithmically, the way "kilo" or "micro" alone are not units but instead convey scale.

There are multiple commonly used decibel scales for sound that often get confused, each of which scales differently.

Pressure (Pa); "SPL", what loudness is most often measured using Power (W) Intensity (W/m^{2)}

Sounds are virtually always measured in SPL (sound pressure level). This post uses sound power level, which is a location-independent expression of how much acoustic energy over time is present. The units are incorrect (which technically renders the answer meaningless; you need a reference distance to convert power to pressure or intensity).

However, identical sound sources are assumed, which is also not quite true in this scenario. Identical sound sources at identical distances scale as 6dB/doubling using intensity/pressure vs. 3dB/doubling using power (since power doesn't account for phase dependence, and identical sources create interference). But because cats should be assumed to be not identical and not perfectly spaced, they are incoherent sound sources, which do scale as 3dB/doubling with pressure (this is where you could convert to power as an intermediate step for simple addition (or notice that integrating over the phase difference gives a factor of sqrt(2)), and convert back to pressure). In the end, both errors made above perfectly cancel each other out and only the units are incorrect.

This also assumes that all the cats are concentrated into a single point. Although technically you can use the power formula and correctly determine the total sound power independent of the locations, this has essentially no effect on perceived loudness (imagine that all the cats were placed infinitely far away; intuitively you'd expect to get a loudness of 0, yet the power result is not any different because it has no positional dependence).

And that's not minding the numbers in the prompt. A cat's meow being 65dB loud should be assumed as an SPL value, yet a reference distance is not given, rendering it completely meaningless. The best I could find was an oft-quoted but never cited 45dB, with no distance given. The numbers as a whole look to be almost entirely made up; the number of cats on earth should be significantly lower than 2 billion.

Would this be enough to make people deaf? Might have to call out the experiment.

I once saw a subwoofer playing at 177db. They but a paper in there and the vibrations shreaded it to pieces

I think it was the start of a mythbuster or something

So a football stadium full of people all yelling together is louder? Thats fascinating and I can calulate that myself with a single line? Brilliant.

Right on!!

CAT astrophic

r/beatmetoit

A few questions because I have no idea as to what's going on here....

Why are you adding 10dB?

Is the log only of the 10dB? Why?

Is this assuming that all cats are in the same room?... I can't really see the cats in Australia really adding much to the Japanese cats, much less the Canadian ones.

This does assume all the cats are at equal listening distance.

The derivation of the equation I used can be found on the first link under my post. :)