# Encasing Taylor Swift

August 13, 2015

I’m not much of a reddit guy, but I just found a neat one: r/theydidthemath. Here’s the gist: people ask silly questions, and other people looking to do some mathematical recreation answer their silly questions.

Here’s one that tickled my fancy:

If you took the net worth of Taylor Swift, in $20 bills, and used those as building blocks, how big of an enclosure could you build around her and still have it remain soundproof? Great question. Let’s see if we can figure it out. A quick google search indicates that Taylor Swift’s net worth is$240 million USD. WolframAlpha, the source of all sorts of fantastic and weird mathematical facts, states the volume of a $20 USD bill is 1138mm3, which gives us an easy 2.4 ⋅ 108USD ÷ 20 × 1138mm3/ USD = 136.5m3 of building material to work with1. Good start. We will assume an enclosure around Taylor Swift is a hemisphere shell, and for good measure, we’ll pad the floor of the enclosure as well. The volume of a hemisphere is $$v = \frac{4 \pi \left(r_o^3 - r_i^3\right)}{6}$$ where ro and ri are the outer and inner radii of the sphere, respectively. Since sound attenuates over distance, it would be more helpful to describe ri as the attenuation distance (which is to say, the necessary thickness of our shell). We rewrite the hemisphere’s volume as: $$v = \frac{4 \pi \left(r^3 - (r-s)^3\right)}{6}$$ where s is now the shell thickness, and r is the outer radius. We now need to add the floor of the enclosure, and since hemispherical shells with a floor aren’t very popular, we’ll need to do the mathematics ourselves. We can model the floor as a bunch of infinitesimally thin of cylinders of varying radii:2 Let l be the total height of the stacked cylinders, and thus volume of our floor’s enclosure is: $$\int_0^s \pi \left((r-s) \sqrt{1 - \left(\frac{l}{r-s}\right)^2}\right)^2 dl = \frac{1}{3} \pi s \left(3r^2 - 6rs + 2s^2\right)$$ Add this to our hemisphere, and the total volume of our enclosure, after simplifying, is thus: $$v = \frac{1}{3} \pi s \left(2s - 3r\right)^2$$ Onwards. I don’t know much about sound attenuation, but according to this page, sound dissipates according to the law: A = A0eαx where A0 is the initial loudness, A is the resulting loudness, α is the attenuation coefficient of the material, and x is the distance the sound travels through the material. The initial question doesn’t state if we’re trying to sound-proof Taylor Swift or a Taylor Swift concert, so let’s try doing both. We assume Taylor will be shouting, so we’ll aim to sound-proof her up to A0 = 78dB A rock (we’re being generous, here) concert is approximately A0 = 115dB. We assume dampening the sound down to the loudness of a whisper is acceptable, so this gives us A = 30dB. Solving our attenuation for x (which is s, the necessary width of our shell to achieve sound dampening), we get: $$s = \frac{\ln{A_0}-\ln{A}}{\alpha}$$ Great! So given α, we should be able to compute s. Strangely, there doesn’t seem to be much in the literature for the sound attenuation coefficient for paper, so we’ll have to use an educated guess here. This page gives coefficients for a few common engineering materials, the closest to paper of which is probably cork. If you were going to actually build this enclosure around poor Taylor, you would probably want to experimentally find a value for this (and have some really good lawyers), but we’ll assume cork is close enough to paper to continue our analysis. Substituting in α = 0.15, the mean coefficient listed for cork, we get: $$s_t = \frac{\ln{78}-\ln{30}}{0.15} = 6.4 \text{m} \\\\ s_c = \frac{\ln{115}-\ln{30}}{0.15} = 9.0 \text{m} \\\\$$ where st is the necessary shell width to silence Taylor Swift, and sc, to silence her concert. Because doing math is hard, we’ll use WolframAlpha to do the remaining calculations for us: …which gives us radii smaller than the necessary shell width for the speaking case. Since this is physically impossible, we conclude that such an enclosure can’t actually be constructed. Shame. But wait! We totally fudged that α = 0.15 number for cork. But maybe paper attenuates more like snow? Maybe not, but hey, we’ve got nothing to work with otherwise. We recompute st, sc: $$s_t = \frac{\ln{78}-\ln{30}}{0.75} = 1.3 \text{m} \\\\ s_c = \frac{\ln{115}-\ln{30}}{0.75} = 1.8 \text{m} \\\\$$ and again run them through WolframAlpha. This time things look better, we get: $$r_t = 4.2 \text{m} \\\\ r_c = 4.0 \text{m}$$ Success! If you want to silence just Taylor Swift, your enclosure can be 4.2m in radius, and to silence her concert, you lose less than a meter in total radius. Which is super cool, and actually pretty surprising. So, if paper attenuates like snow, you can enclose Taylor Swift in her own money and have her play a concert that no one will ever hear. Marvelous. 1. The original version of this post accidentally calculated this in terms of$1 USD bills, so this number was off by a factor of 20. Thanks to James Barton for spotting this error.↩︎

2. The original version of this post approximated this floor as a single cylinder, which is a good approximation only for small s. Thanks to Marius van Voorden for spotting this error.↩︎