It's common at schools to use $3.14$ as an appropriate approximation of $\pi$. However, here it's stated that for some purposes, $\pi$ should be approximated to $32$ decimal places. I need an example of such a purpose, accessible to a middle school student.


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    Not $32$, but more than $2$: astronomy. – André Nicolas Jun 18 '14 at 13:47
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    When you but a very big cylindric tank with oil, digits of $\pi$ does matter because **you** pay for this oil :) – Norbert Jun 18 '14 at 13:47
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    You will get $\sin(314)\approx -0.16$ instead of $\sin(100\pi)=0$. – gammatester Jun 18 '14 at 13:51
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    The reason they use 32 decimal places because the value of fine structure constant both from experiment and from theoretical calculation are now known to about 13 decimal places (with some uncertainty in last two digits) This is reaching the limit of accuracy of ordinary floating point numbers used by computer. To avoid any round-off errors for these hard and expensive calculations, they move up to the next level of accuracy offered by computer. This translates to about 32 decimal places of accuracy. – achille hui Jun 18 '14 at 14:06
  • @AndréNicolas Can the need to "more than 2" decimal places be explained in simple words for a middle school student? – Behzad Jun 18 '14 at 14:15
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    @Behzad circumference of our sun or the orbit a satellite takes around earth since it has to be very precise because if the satellite will be too low, it can burn from friction, too far can make harm his purpose (GPS for example). – GinKin Jun 18 '14 at 14:34
  • @GinKin: I don't think either is a good example. The error in the circumference of the sun (or any circle) is less than 3 parts in a thousand-who needs better? Satellites burning up again will not depend on a few parts in a thousand. GPS is so high that it won't burn up, though the time would be much too far off to be acceptable. – Ross Millikan Jun 23 '14 at 05:08
  • @RossMillikan now I remember someone told me that in finding distances to planets they sometimes use $x$ instead of $\sin x$ and the error is just about $1/1000$ so it's good enough for a lot of the calculations. I guess it should be good enough for satellites too. – GinKin Jun 23 '14 at 08:18

5 Answers5


If you want to calculate the volume of an $n$ dimensional sphere with radius $r$ you must include a power of $\pi^{n/2}$.

In general, we have $$V_n(r) = \dfrac{n\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}r^{n-1}$$

If we consider the ratio $R_n$ of the actual volume to the approximate volume we find that $$R_n = \frac{V_n(r)}{V_n(r)_{\text{approx}}} = \left(\dfrac{\pi}{\pi_{\text{approx}}}\right)^{n/2}$$

If we set $n = 2\cdot 40^{20}$ we find that $R_n \approx 1.10621$ rather than the expected value of $R_n = 1$.

This example is completely contrived but it demonstrates that the error will become apparent when large powers are involved. This is just a simple application of the use of significant digits.

There are many more examples using periodic functions with a period involving $\pi$. Consider finding


Essentially, you must find $10^{100}\! \pmod{\pi}$ which would require knowing $100$ digits of $\pi$ (to calculate the answer with some precision). You can find more information here. Again, this involves large powers requiring an excessive number of digits.

The easiest example of the need for an exact value most likely comes from numbers of an astronomical (or even extra-astronomical) scale. I don't know how convincing this argument would be to students and you can consider cross posting to MathEducators.

If you want an explanation for younger mathematicians you should explain how large powers of any number can an estimate unreliable and then present an applications of large powers of numbers or even just large powers of $\pi$. You can show how estimates like $\frac{1}{3} \approx .333$ are unreliable and then move up to more complicated examples.

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The BBP formula, a spigot algorithm for $\pi$, required knowing the value of $\pi$ to high precision in order to use the integer-relation algorithm which generated the formula. Perhaps the circularity (knowing $\pi$ to find a formula for $\pi$) makes this less interesting, but in general integer relation algorithms are one of the best reasons for caring about extremely high precision in constants.

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You could tell a middle school student:

32 digits of pi is enough to calculate the circumference of the visible universe to within an error invisible under a light microscope.

Be prepared to acknowledge that 32 digits of pi is therefore more than enough for "practical" purposes, while two digits of pi is often (but not always) satisfactory precision.

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Dynamical systems. E.g. the earth, sun plus a communications satelite system: this is a classical example of a 3-body problem where the mass of the satelite doesn't affect the earth-sun system, who follow an elliptical trajectory (2-body problem), but the earth and sun do affect the satellite. The trajectory of the satelite, however, doesn't have an analytical expression combination of known functions, and numerical methods and power expansion is needed to predict it. The number $\pi$ appears in some expressions of the series, and it is necessary to have a lot of decimals as the error may increase exponentially and the satellite crash.

Of course in real satelites the trajectory is analyzed and may be corrected. Nonetheless, high precision is required.

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The approximation required for any number depends on the purpose, as you can see. It is a good approximation to take $\pi$ being approximately equal to 3.14.

So, generally in mechanical engineering we can assume this approximation to be true because we take into account the factor safety which serves the purpose causing no damages. However, suppose you want to be build a spaceship you require things to be accurate. So, this approximation won't help!

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