How many decimals does NASA use to make it’s calculations? This was a question posed to NASA engineers. The full answer is amazing.
We can bring this down to home with our planet Earth. It is 7,926 miles in diameter at the equator. The circumference then is 24,900 miles. That’s how far you would travel if you circumnavigated the globe (and didn’t worry about hills, valleys, obstacles like buildings, rest stops, waves on the ocean, etc.). How far off would your odometer be if you used the limited version of pi above [3.14]? It would be off by the size of a molecule. There are many different kinds of molecules, of course, so they span a wide range of sizes, but I hope this gives you an idea. Another way to view this is that your error by not using more digits of pi would be 10,000 times thinner than a hair!
I found the finale particularly mind blowing:
Let’s go to the largest size there is: the visible universe. The radius of the universe is about 46 billion light years. Now let me ask 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)? The answer is that you would need 39 or 40 decimal places. If you think about how fantastically vast the universe is — truly far beyond what we can conceive, and certainly far, far, far beyond what you can see with your eyes even on the darkest, most beautiful, star-filled night — and think about how incredibly tiny a single atom is, you can see that we would not need to use many digits of pi to cover the entire range.
25 Responses
We can bring this down to home with our planet Earth. It is 7,926 miles in diameter at the equator. The circumference then is 24,900 miles
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golu dolls
Chris, you should be embarrassed by this. You wrongly added put in the 2 digits representation, instead of 15, when you wrote “pi [3.14] above”.
The answer is 15 digits: >>” For JPL’s highest accuracy calculations, which are for interplanetary navigation, we use 3.141592653589793. ”
The wrongness of this note is really your wrong understanding of what they wrote “in pi above” without specifying 15 digits.
Please revise your bad edit.
I really like most of your posts (like the evl paper), this error makes you look bad.
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According to my Matlab, using 3.14 you’d be off by 12.62 miles.
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Your editor’s note – “[3.14]” – is wrong. For the Earth’s diameter example, you need fifteen decimals places: 3.141592653589793.
Your [3.14] suggests that NASA’s version of pi uses two digits, but the article is discussing NASA’s version of pi which uses 15 digits.
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