I wasn't aware of the other thread on the same subject. Looked at it now and totally disagree with the fragment you posted there in #1. Using ##A_\text {max} - A_\text {min} ## as estimate for the uncertainty in the area is a gross exaggeration. There are several aspects.
Let me try to explain a little bit using drawings. Suppose we have measured ##W = 3.0 \pm 0.5## and ##L = 5.0 \pm 1.0## for the blue rectangle. We have
##A = 15##, ##A_\text {max}=21## and ##A_\text {min} = 10##.
So ##A_\text {max}-A =6## (the green shaded area) and ##A - A_\text {min} = 5## (the red shaded area).
And ##A_\text {max} - A_\text {min} =11##. Simple math so far.
According to the fragment you copied, the area would be quoted as ##A = 15\pm 11##, so the biggest rectangle that would still be within error margin would be like the dashed green one with area 26 in the second picture and the smallest would be like the dashed red one with area 4.
View attachment 325057
You can check that the sum of the green and red shaded areas is the difference beteween the area of the blue rectangle and the green dashed rectangle.
And also the difference beteween the area of the blue rectangle and the red dashed rectangle.
But the largest deviation from ##A=15## is only 6 and the average deviation is 5.5. To me it is clear that the dashed rectangles are not justfiable from the measurements, so quoting ##A = 15 \pm 6## is the best one can do (not ##15\pm 5.5## because that suggests an accurate error estimate that is nonexistent).
If you agree and want to dive deeper, read on. If not, stop here
.
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There is more: error handling is an exact science up to a point. Then judgment and common sense come in.
Errors are usually small and are treated in first order. So in a formula like ##A = L * W## we do have $$\begin{align*}A_{max} = (L+\Delta L)*(W+\Delta W) = L*W + \Delta L*W +L* \Delta W +\Delta L*\Delta W\\A_{min} = (L-\Delta L)*(W-\Delta W) = L*W - \Delta L*W -L* \Delta W +\Delta L*\Delta W \end{align*}$$ we ignore the last term in each because it is second order. The little rectangles shown here in the first picture:
View attachment 325059
Well, not so little in this case with such large errors, but the concept is clear, I hope.
That way we slightly underestimate ##A_{max}-A## and overestimate ##A-A_{min}##, but for
the average deviation we have ##{1\over 2}(A_{max}-A_{min}) = \Delta L*W +L* \Delta W ##.
Taking the average deviation as ##\Delta A## and rearranging: ##{\Delta A\over A} = {\Delta L\over L} + {\Delta W\over W} ##
As in post #4. (the expressions for a product and a quotient are the same).
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There is still more:
Error handling is based on statistics. Convention is to silently assume a few things unless stated otherwise (or clear from the context for experts
) :
- Errors have a Gaussian distribution; the uncertainty reported is the standard deviation
- Errors in separate measurements are uncorrelated (independent)
The first assumption means that ##L+\Delta L## is less likely than ##L##: the shading density should be lower when going further away from ##L## Same for W.
The second assumption means that probabilities (represented by shading density) can be multiplied. I'm not good at creating suitable graphics, so I have to steal a picture (from
here) that shows the probability for the position of the top right point of a rectangle
View attachment 325060
The consequence is that independent errors must be added in quadrature: for
example ##\sigma_{a+b}^2 = sigma _a^2 + \sigma_b^2## (I did this with the 40% and 10% in post #3).
If you are not bored to death yet, do some googling and pick up what you like and can understand
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