Uncertainty Propagation in Calculations
What if you want to determine the uncertainty for a quanitity that was calculated from one or more measurements? There are complicated and less complicated methods of doing this. For this course, we will use the simple one. The Upper-Lower Bounds method of uncertainty in calculations is not as formally correct, but will do. The basic idea of this method is to use the uncertainty ranges of each variable to calculate the maximum and minimum values of the function. You can also think of this procedure as exmining the best and worst case scenarios. For example, if you want to find the area of a square and measure one side as a length of 1.2 +/- 0.2 m and the other length as 1.3 +/- 0.3 meters, then the area would be:
A = l * w = 1.2 * 1.3 = 1.56 m^2
The minimum area would be using the "minimum" measurements so l = 1.2 - 0.2 = 1.0 and w = 1.3 - 0.3 = 1.0
So the "minimum' area is A-min = 1.0 * 1.0 = 1.0 m^2
Likewise for the maximum area, l = 1.2 + 0.2 = 1.4 and w = 1.3 + 0.3 = 1.6
So the Maximum area is A-max = 1.4 * 1.6 = 2.24 m^2
Thus, we can say the area is A = 1.5 +/- 0.6 m^2
From: http://www2.southeastern.edu/Academics/Faculty/rallain/plab194/error.html
What if you want to determine the uncertainty for a quanitity that was calculated from one or more measurements? There are complicated and less complicated methods of doing this. For this course, we will use the simple one. The Upper-Lower Bounds method of uncertainty in calculations is not as formally correct, but will do. The basic idea of this method is to use the uncertainty ranges of each variable to calculate the maximum and minimum values of the function. You can also think of this procedure as exmining the best and worst case scenarios. For example, if you want to find the area of a square and measure one side as a length of 1.2 +/- 0.2 m and the other length as 1.3 +/- 0.3 meters, then the area would be:
A = l * w = 1.2 * 1.3 = 1.56 m^2
The minimum area would be using the "minimum" measurements so l = 1.2 - 0.2 = 1.0 and w = 1.3 - 0.3 = 1.0
So the "minimum' area is A-min = 1.0 * 1.0 = 1.0 m^2
Likewise for the maximum area, l = 1.2 + 0.2 = 1.4 and w = 1.3 + 0.3 = 1.6
So the Maximum area is A-max = 1.4 * 1.6 = 2.24 m^2
Thus, we can say the area is A = 1.5 +/- 0.6 m^2
From: http://www2.southeastern.edu/Academics/Faculty/rallain/plab194/error.html
