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Topic:
Theory of errors: center of gravity
Replies:
7
Last Post:
Mar 14, 2013 2:40 PM




Re: Theory of errors: center of gravity
Posted:
Mar 7, 2013 5:05 PM


> Let's suppose we have two point masses: > > m1 = 4 kg, m2 = 3.5 > > Their associated standard deviation are: > > dm1 = 0.5 kg, dm2 = 0.2 kg > > Their distance from the origin of the reference frame > are (1D problem): > > x1 = 0.7 m, x2 = 0.8 m > > Their associated standard deviation is: > > dx1 = 0.01 m, dx2 = 0.01 m > > I have to find the center of gravity (CoG) of this > system and its uncertainty. The nominal value can be > found with the wellknown relation: > > X = (m1*x1 + m2*x2)/(m1+m2) = 0.7467 m > > In order to find the associated uncertainty, I used > the relationship for nonlinear combinations (cf. > http://en.wikipedia.org/wiki/Propagation_of_uncertaint > y), evaluating the four partial derivatives of the > previous relationship. > > Hence dX = 0.0079 m > > In my opinion is not an intuitive result: in fact the > uncertainty of the center of gravity is less than the > uncertainty on the position of each mass. > Are you able to justify this result? > > Regards >
You are calculating an average. Intuitively, it should be more accurate than the individual components.
Your estimate of dX is excellent. A simulation of 10 million variates yielded a dX of 0.007903 (I used normal distributions.)
You may think your nominal value is wellknown and correct, but you need to be very careful. The simulation yields X = 0.746837
The two are close enough to make you think you got it right. And for applied work you did. But X is a nonlinear function of random variables, and expectation is a linear operator. You have stumbled upon something deeper than you thought. Look up Jensen's Inequality to see why your answer is too small.
Steve Oakley



