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

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Steve Oakley

Posts: 10
Registered: 12/17/04
Re: Theory of errors: center of gravity
Posted: Mar 7, 2013 5:05 PM
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> 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 well-known relation:
>
> X = (m1*x1 + m2*x2)/(m1+m2) = 0.7467 m
>
> In order to find the associated uncertainty, I used
> the relationship for non-linear 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 well-known 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



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