# Why are physical simulations more accurate with floating point numbers "closer to one"?

I've often heard it said that any sort of "dimensional" (involving length, time, mass, charge, etc.) calculation should be put in a dimensionless form for two reasons

1. Getting a value ~ 1e14 when all of your inputs are near 1 is rather obvious (and unlikely) making it easier to pinpoint algorithmic errors.
2. Conventional IEEE floating point numbers are denser near zero so dimensionless calculations are more accurate.

I have a pretty intuitive grasp of the first point (if all my lengths ≈ 1 and my masses ≈ 1 and my times ≈ 1 then I can expect all of my forces will reasonably ≈ 1) but I'm having a more difficult time digesting the second. I totally understand why the epsilon between floats depends on your location on the real line, but I don't see how this fact can be exploited for better accuracy. For instance, if you're using the SI speed of light (c = 3e8) and add to it a small epsilon (eps = 1e-8), then c + eps == c and you've "lost" the epsilon to the floating-point void. I don't see how using a dimensionless speed of light (c = 1) diminishes this sort of issue. Why are "dimensionless" simulations said to be more accurate?

• Ok, this is forming a better picture but I'm still not quite sure I get the whole thing. Using your intervals example, if there were more numbers in [0.5, 1] than in [4,8], everything would make perfect sense but I'm given to understand this is not the case; each interval has the same quantity of numbers distributed over smaller intervals near zero (thus "denser"). If I use $c = 3e8$, then, presumably, I care about fluctuations on the order of epsilon relative to $3e8$ -- I don't see where the additional accuracy comes from. Nov 13 '17 at 15:35