To be clear, a mathematical definition of "paper" has been used: An object of thickness 𝑇₀ and sufficient mass density to make such a folding possible.
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𝑇₀: thickness of paper
𝑇(𝑛): thickness after 𝑛 folds
𝑇(102): the size of the OBSERVABLE UNIVERSE
That is, after folding paper 102 times its thickness will exceed the size of the observable universe.pic.twitter.com/ARd7j3N46A
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Pedants: It is mathematically well defined, but physically impossible ℓ(𝑛): minimum length of paper needed for n-folds 𝑤(𝑛): minimum width of paper needed for n-folds
𝑤(𝑛) > 𝑇(𝑛) for 𝑛 >3 ⇒ papers width would itself have to be larger than the observable universepic.twitter.com/SHimzugJ9D
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The real question is: how large should this sheet of paper be for 102 foldings to be possible?
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*idealized paper has been assumed*
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I did round down from my estimate, the actual answer was 102 point somthing.. guess I should have done one more fold just for good measure
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call in the myhtbusters
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Rougly speaking: Thickness sheet of paper: 0.01mm T=0.01 2^102=0.01×5.1×10^30mm= 5.1×10^22km While radus "observable Universe" is 46 billion ly (and not 13.7 billion ly) = about 4.6x10^22 km~5.1x10^22km (1ly=roughly 10^4 billion km=10^13 km) That is quite fair ;-)
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But 0.01 mm is too thin I suppose a sheet of paper has a (larger) thickness of about 0.1mm in that case it differs a factor of 10 and one has to fold it about 3 times less since 2^3=8~10;-) Yielding: T=T_0*2^99
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