According to JPL’s Chief Engineer for Mission Operations and Science, Marc Rayman-
Let's go to the largest size there is: the known universe. The radius of the universe is about 46 billion light years. Now let me ask (and answer!) a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom, the simplest atom? It turns out that 37 decimal places (38 digits, including the number 3 to the left of the decimal point) would be quite sufficient.
The approximation is only around 4.28*10^-101 off from pi.
https://www.wolframalpha.com/input?i=integral+from+-a+to+a+of+(integral+from+-a+to+a+of+e^-(x^2%2By^2)dy)dx
https://www.wolframalpha.com/input?i=(1-erf(e^e)^2)*pi
And because it always bears repeating;
According to JPL’s Chief Engineer for Mission Operations and Science, Marc Rayman-
Let's go to the largest size there is: the known universe. The radius of the universe is about 46 billion light years. Now let me ask (and answer!) a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom, the simplest atom? It turns out that 37 decimal places (38 digits, including the number 3 to the left of the decimal point) would be quite sufficient.So the number 3 should be close enough for home use. Good to know. Thanks!
My maths exam asked me to consider pi=5.
“I will… consider it.”
As an engineer, I approve this message!
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Technically you need another 20 digits if you want to get down to a Planck length. (57 digits in total)