Slashdot Mirror
How Many Digits of Pi Does NASA Use? (kottke.org)
An anonymous reader quotes an article on Kottke.org: Mathematicians have calculated pi out to more than 13 trillion decimal places, a calculation that took 208 days. NASA's Marc Rayman explains that in order to send out probes and slingshot them accurately throughout the solar system, NASA needs to use only 15 decimal places. Rayman explains, "The most distant spacecraft from Earth is Voyager 1. It is about 12.5 billion miles away. Let's say we have a circle with a radius of exactly that size (or 25 billion miles in diameter) and we want to calculate the circumference, which is pi times the radius times 2. Using pi rounded to the 15th decimal, as I gave above, that comes out to a little more than 78 billion miles. We don't need to be concerned here with exactly what the value is (you can multiply it out if you like) but rather what the error in the value is by not using more digits of pi. In other words, by cutting pi off at the 15th decimal point, we would calculate a circumference for that circle that is very slightly off. It turns out that our calculated circumference of the 25 billion mile diameter circle would be wrong by 1.5 inches. Think about that. We have a circle more than 78 billion miles around, and our calculation of that distance would be off by perhaps less than the length of your little finger."
And similarly, 39 digits of pi will let you calculate a sphere the size of the observable universe with an error the width of a single hydrogen atom.
Liberty in your lifetime
The NASA computers will be using IEEE 754 floating point format, which in 64 bits (double precision) yields about 16 decimal digits of precision. So: what came first, NASA deciding that 1.5 inches the needed accuracy in the solar system or their computers being that accurate being deemed an acceptable accuracy ?
How many digits would it take of PI to bring the error down to the Planck length? Anything smaller in any measure of distance would be meaningless.
E Proelio Veritas.
Duh... What?!? Reference, please??? What hardware constant built into the FPU?
Pi is built in to x86s. You can load it into a register with the FLDPI instruction. It is built in to many other processors as well. Using it will give you the maximum precision, and will be faster to execute since it uses less bandwidth and cache.
If you plug that number into an iterative algorithm that uses any ill-conditioned functions, the 1.5-inch error can grow exponentially at every step. Ensuring that computations have acceptable error margins is an extremely complex and tricky subject.
How many digits to use depends on the application. For a satellite trajectory the 15th decimal is OK, but if you want to make a sharp mirror the precision in the calculation have to be higher.
Err, no. The radius of a proton is around 10^-15 meters. Atoms are 10.000 times larger than that. Visible light is around 5 * 10^-7 m, depending on color. Polishing a mirror at 10^-15 accuracy would be physically impossible and pointless.
78 billion miles is nothing when it comes to the universe.
You're saying that it's still in the same ballpark as a trip down the road to the chemist, then?
"Slashdot - News and Chat Sites Deviant". (Click "homepage" link above for details).
The Rydberg constant is known to 14 decimal places, but I don't think it's very relevant for NASA calculations. https://en.wikipedia.org/wiki/...
How so? For high-precision applications, you'd typically want a mirror with a deviation from its calculated surface that is better than lambda/20. For UV light at 200 nm and a mirror size of 1 m, that would be about 8 significant digits to describe the surface. Then you have 7 more digits to deal with intermediate results during the calculation.
Note: For X-ray mirrors, the wavelength is much shorter, but because X-ray mirrors are grazing-incidence mirrors, the surface tolerance is more like lambda/1 - - which also boils down to about 8 significant digits. Anyway, to describe the surface with an accuracy down to the size of an atom, you still only need 10 significant digits.
Avantslash: low-bandwidth mobile slashdot.
Posting because I've seen too many programmers misunderstand this and screw things up when using floats. This does not mean you can do whatever you want with floats and your accuracy will be at least 1.5 inches within a 25 billion mile radius. It means an individual double float number has a precision of 17 significant figures, nothing more. When you start doing math with that number, the precision can quickly go out the window.
For example, if you add a large float to a small float, say 3.14159265359x10^8 + 2.7182818284x10^-3, the latter number doesn't actually matter because it's largest digit is smaller than the uncertainty (the last significant figure) in the first number. In other words, floating point's accuracy decreases the further you get from zero. A solar system modeled with ints will have the same resolution everywhere, which is how people want to think significant figures work. But a solar system modeled with double floats will have very fine resolution close to the origin, lousy resolution out near the edges.
To do this sort of math accurately using float (e.g. calculating docking coordinates for two spacecraft orbiting Jupiter), you are better off first doing a coordinate transform to center your zero closer to where everything is happening, do your math, then transform the results back to your original coordinates (or keep them in your new coordinate system if you plan to do more math there later). Obviously that's a pointless exercise with this simplified math problem, but it can make a big difference in accuracy with more complex math.
For a real-world example, we once networked two flight simulators together and tried to make them fly in formation. The position of the second plane appeared to jump all over the place when viewed from the first plane, and vice versa. What was happening was our coordinate system was fixed to the Earth. Planes fly on the order of 1000 kph. At that speed, a transient network lag error of only 10 milliseconds results in a position error of several meters (dp = v * dt). It didn't matter that we knew the velocities to 7 significant figures or even if we'd known them to 17 significant figures, because the error in the timestamp overwhelmed that accuracy. The fix was to define a new coordinate system centered on the first plane. The velocity of the second plane was then only a few cm/sec relative to the first plane (since they're flying in formation), and the jumping disappeared since a 10 ms error only resulted in only a few millimeters of error.
eng1: How many digits of pi did you use?
eng2: 16.
eng1: So how come the spacecraft isn't on track? It's off by several tens of meters already, and we just launched.
eng2: Meters?
I've fallen off your lawn, and I can't get up.