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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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 Red Star Posts: 136 Registered: 6/28/08
Re: Theory of errors: center of gravity
Posted: Mar 14, 2013 2:40 PM

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.
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> 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.
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> In effect, you are averaging presumed independent random variables, and averaging reduces variance, which I think is what David was getting at.
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> Paul

Date Subject Author
3/7/13 Red Star
3/7/13 David Jones
3/7/13 Paul
3/10/13 Red Star
3/10/13 Paul
3/10/13 Paul
3/14/13 Red Star
3/7/13 Steve Oakley