Anyone familiar with graphical analysis?

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The discussion centers on issues with error bars in graphical analysis, specifically when small uncertainties result in disproportionately large error bars on the graph. Participants suggest that the problem may stem from miscalculations or the axis range being too small. One user describes their method of calculating uncertainties related to an Atwood lab experiment but expresses confusion about their arithmetic, particularly regarding the treatment of mass uncertainties. Another contributor asks for clarification on the nature of the uncertainties and encourages sharing complete calculations for better assistance. Overall, the conversation highlights the importance of accurate calculations and proper representation of uncertainties in graphical data analysis.
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After having typed in the absolute uncertainties for my values, I get error bars that stretch across the graph, when my uncertainties are rather small compared to the initial value. What am I doing wrong?! Should I change the error bars to percentage or keep it at fixed value? (Those who use this graphing software will understand me...hopefully). Thank you.
 
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What are you plotting, i.e. what are the abscissa and ordinate, or rather what are the independent and dependent variables?

If the uncertainties are small with respect to the value, and the magnitude of the error bars are larger, then it would seem to be a miscalculation of the error bar magnitude. If the error bars stretch across the graph, but the magnitudes of the error bars are much less than the associated value, it could be that the range of the axis (axes) are too small.

See this page - http://www.ncsu.edu/labwrite/res/gt/gt-stat-home.html

This might be helpful - http://www.rit.edu/~uphysics/graphing/graphingpart1.html#errorbars
 
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Hmmm. I have re-calculated the uncertainties...and I obtain the same results. Since this is an Atwood lab, I divide the sum of the relative uncertainties of time and distance by the acceleration value. For example, for situation 2, where the difference of the masses is constant, the acceleration value is 2.55 +/- 1.08 (this is quite big)...is my arithmetic incorrect? Also, for the masses, they have an uncertainty of .03 each. Thus, when I find the sum of the masses, I double the uncertainty. Then, when graphing, I have 1/m+m2 representing the x-axis which I convert to kilograms. So, would it be correct to divide 0.06 by 150 kilograms (for example) and then multiply the obtained value by 150 so to get the absolute uncertainty of the sum of the masses when it's in the denominator (argh, I'm sure you lost me by now...Well, I hope you understand the main parts!) Thank you. I think my error bars are big because our time measurements are inconsistent, and there is a great deviation between the average time and the smallest time measurement! argh!
 
I'm sure you lost me by now...

er. yes...

What do you mean by "an uncertainly of 0.03 each"? Do you mean the uncertainty is 3% of the measured mass? Or an uncertainty of 0.03Kg?

If you show your complete calculation for one of two points, somebody might be able to help more. Using ESP to guess what you did is even harder than doing physics :smile:
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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