Don't worry Paaul. Thank for the validation. Therefore my evaluation is correct, but not intuitive.
> Sorry, I'm wrong -- I read dx2 as 0.02 rather than 0.01. With that corrected, I match your dX value. > > > > As to an explanation/rationale, suppose for the moment (per David Jones) that dm1 = dm2 = 0 (no uncertainty in the masses). Assuming independence of the variations in x1 and x2, and letting a = m1/(m1+m2), dX^2 = a^2 dx1^2 + (1-a)^2 dx2^2. Suppose now, as with your example, that dx1 = dx2; then dX = dx1 * sqrt(a^2 + (1-a)^2). Since a^2 + (1-a)^2 < 1 for 0 < a < 1, which is the case here, dX < dx1 = dx2. > > > > In effect, you are averaging presumed independent random variables, and averaging reduces variance, which I think is what David was getting at. > > > > Paul