I've come across the following YouTube video on the Numberphile channel: https://www.youtube.com/watch?v=FpyrF_Ci2TQ

which mentioned that 38 decimal places of pi (i.e. 39 digits) is all you need to calculate the circumference of the observable universe to the width of a hydrogen atom.

I can admit that I know pi to 40 decimal places (yeah, I'm a math geek, but let's leave that for another forum)... But the question then begs, to what level of accuracy would 40 decimal places of pi give to the calculation of the circumference of the observable universe? Surely much finer than a hydrogen atom's width... does anyone have any idea?

  • $\begingroup$ Your main worry here would have to be that the circumference of the observable universe is not an observable physical quantity and even if it were, it would be constantly changing... and not by a little but basically by a quantity of the order of one... $\endgroup$
    – CuriousOne
    May 11, 2015 at 2:40
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    $\begingroup$ When I was a teenager, I memorized pi to 100 digits. You might like A History of Pi by Petr Beckman. It has a lot of information on calculating pi to as much accuracy as people could at various times in history. In case the first sentence puts off normal people, let me hasten to add that it is a classic. Much like Flatland, it is short, readable, and filled with out of date social commentary. $\endgroup$
    – mmesser314
    May 11, 2015 at 3:47
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    $\begingroup$ Have you tried computing the circumference of a sphere with different degrees of precision? If not, why not? $\endgroup$
    – Kyle Kanos
    May 11, 2015 at 4:04
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    $\begingroup$ @KyleKanos Because lately I've been focusing on other things, namely the pending birth of my child within the week and my Master's degree assignments... $\endgroup$ May 11, 2015 at 4:15
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    $\begingroup$ @Eliseod'Annunzio: Congrats on those, but this is something that could take about 30 seconds of programming to compute. $\endgroup$
    – Kyle Kanos
    May 11, 2015 at 11:24

1 Answer 1


The assumption has 2 problems with it. 1) the universe is expanding, so it's larger than what we observe and the rate of that expansion isn't exactly known. 2) we don't know how old the universe is exactly (They have a pretty darn good estimate, but it's still an estimate) and 3) due to gravity and time dilation, the observable universe isn't a perfect sphere. It's sightly lopsided and bubbled - so we can't measure it's size in perfect terms.

But lets say that right "now", the universe is exactly 13.8 billion years old and lets pretend it's a perfect sphere with 13.8 billion light years radius around us. (We'll address the expansion later).

Diameter of a hydrogen atom (source): http://en.wikipedia.org/wiki/Bohr_radius

5.2917721092(17)×10−11 M, which means, 1.88972 x 10^10 hydrogen atom diameter in 1 meter

and meters in a light-year: 9.4605284 × 10^15, and that times 13.8 x 10 ^9.

So, right "now" assuming it's the same now it was before, our observable universe is (pretending it's a perfect sphere), about 2.467 x 10^36th hydrogen atom radii in the radius. Now, if we're calculating the circumference to within the radius of a hydrogen atom, we have to multiply this by 2 Pi (cause that's what we're estimating, we know the Radius exactly), so, the circumference is about 1.55 x 10^37 or, lets use E37.

So, how many digits of Pi would you need - lets look at Pi.


Source: http://digitsofpi.com/Top-200-Digits-Of-Pi.htm

The accuracy of each digit depends on the number that follows after it, for example, if we don't know the next digit, it could be a handful of 9s, so if we have the first digit (3) and we don't know the decimals, our maximum error is .99999999(etc)/3 or 33.3% - curious thing about numbers like Pi, there will always be at some point in it's infinity of numbers, every combination of consecutive 9s, for example 99, 999, 9999, 99999, 6 - 9s, 7-9s, nine hundred and ninety nine nines, even a million billion trillion 9s, all of them, at some point in it's infinite string of numbers, in a row. There has to be, otherwise it wouldn't be truly random. But I digress.

But back to the problem, while the error varies with the numbers that follow, the error generally is on average, improved by a factor of 10 with each digit.

So, we need a percentage error lower than 1/1.55 E 37. And the answer to that is fairly simple. 38 digits (I'm counting the 3 as a digit, if you only count digits as past the decimal, then you just need 37) - now, the youtube guy says 39. Maybe he's taking into account expansion. Not sure why I have different numbers than him, but we're close.

To add 2 digits to 40, it's just 100 times smaller - give or take.

Now, Planck Lengths, the answer is quite different. There are nearly as many Plank Lengths in the diameter of 1 hydrogen atom as there are hydrogen atom diameters in 1 light year. (Well, not almost as many - but close enough) - that's a pretty good way to define how small a Planck Length is - it's really tiny.

There's about 3.27 x 10^24th Planck length's in the radius of 1 hydrogen atom. So, the margin for error 1/5.08 x 10^51. so, you'd need 62 digits of Pi to measure the circumference of the universe to within one Planck length, and if we calculate for expansion (about 46 billion light years radius not 13.8), you'd need another digit, so 63.

All this is just exponent multiplication (which is basically addition) - nothing fancy.

A penny on the national debt for example is 1/1.8E15, so my Excel spreadsheet with 16 digits can count the national debt to the penny. Add just a couple more decimals and you can calculate daily interest rates on the changing national debt to fractions of pennies and fun stuff like that.

Now, if your working with public key cryptography (whether encrypting or decrypting) which uses very very big prime numbers, then you need over 100 digits (and a very very fast computer if you're decrypting), so there is at least one practical use for having over 100 digits in some calculations. Mostly it's overkill though.

Mersenne number hunters need to work with millions and millions of digits, but that's just perfect number chasing. Not much practical use there.

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    $\begingroup$ On a side note: It has not been proven that pi has all the combinations of numbers nor it is a random number. It is believed to be so but that may not be the case. $\endgroup$
    – Gonenc
    May 11, 2015 at 10:10

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