# Why aren't calculation results in error propagation at the center of the range?

We have two copper rods, with $L_1$ and $L_2$ as their lengths respectively, and we want to glue the two bars together, with glue that's infinitesimally thin. \begin{align} L_1 &= 20 ± 0.2\ \mathrm{cm} \\ L_2 &= 30 ± 0.5\ \mathrm{cm} \end{align} To calculate the length of the composite bar, $L$, as well as its uncertainty, we can do the following (which I admit is a rather crude method, but is done for completeness): \begin{align} L_\text{MAX} &= 20.2 + 30.5 = 50.7\ \mathrm{cm} \\ L_\text{MIN} &= 19.8 + 29.5 = 49.3\ \mathrm{cm} \end{align}

Therefore, $L = 50 ± 0.7\ \mathrm{cm}$.

This, although is a long method, is a correct as the length $L$ is just a sum of the values of $L_1$ and $L_2$ with an uncertainty of the range of values possible divided by two.

Now, if we want to calculate the area with the following length and width, as well as the uncertainty, we could use a method similar to the one described above: \begin{align} W &= 20 ± 0.2\ \mathrm{cm} \\ L &= 10 ± 0.2\ \mathrm{cm} \\ A_\text{MAX} &= (20.2\ \mathrm{cm})(10.2\ \mathrm{cm}) = 206.04\ \mathrm{cm}^2 \\ A_\text{MIN} &= (19.8\ \mathrm{cm})(9.8\ \mathrm{cm}) = 194.04\ \mathrm{cm}^2 \end{align} In this case, the answer without the uncertainty, $10\ \mathrm{cm} \times 20\ \mathrm{cm} = 200\ \mathrm{cm}^2$, is not the center of our range of values. Although 200 isn't the smack center, there does exist a center, which in this case is 200.04.

The actual uncertainty of the area is 6, which is in fact half of the range of the maximum and minimum, giving us a final answer of $200 ± 6\ \mathrm{cm}$.

The way we have defined the propagation of uncertainties in physics is such that the answer is not necessarily the smack center of the minimum and maximum value range, but is instead the product of the two measurements, the two lengths in this case. This approach to the first problem made a lot of intuitive sense, however, I cannot understand why the final answer is 200 (which is not the center of the range) ± 6, and why this answer gives a different range of values than the range calculated using the long, crude method.

I am a high school student who has not covered calculus yet, which is what prevented me from understanding the proof of adding fractional or percentage uncertainties when we multiply or divide quantities. Any help will be greatly appreciated, thanks in advance.

Your question is valid and is very good you are thinking this way so young.

First, error propagation is a whole area of its own and there is not a unique way to do so, nor there is an absolute best way to be provided. So much so that the ISO document used for having a consensus on this, is called "Guide to the Expression of Uncertainty in Measurement", not a manual.

This is the case mainly because determining uncertainties is very much a compromise (depends on how much confident you want to have on your value) and model guided (depends on your assumptions of what you don't know).

What does this means?

Well first it is impossible to have absolute confidence on a value, so when you say a rod's length $20.0\pm0.2cm$ (and please note the number of significant digits should match) you are assuming that the value can be anything between $19.8cm$ and $20.2cm$ but there is no chance it will be lower nor higher.

Secondly, you are assuming that all values between $19.8cm$ and $20.2cm$ are equally likely, which is not usually the case.

Now don't get me wrong: your reasoning is not flawed, is completely legitimate, although not the most practical, and therefore not what is generally used. All this was to clarify what it really means your error propagation and your expectations of them falling inside a range, etc.

How we do things

From experience we have seen that many values, when measured exhibit a Normal distribution, which means in few words, that repeated measurements on the same conditions will be concentrated around the mean value, and the further from this value the less you find. Since this is very often, we commonly assume that this is the case in general. So in this sense you are correct in expecting that your rod's value $20cm$ should be in the center of the interval.

As for the interval to report, when assuming normal distribution, it is common practice to report the interval whose center is the mean value of the measurements, and which will contain roughly 63% of all possible values. The logical reasons for this are numerous and lengthy to cover here, but still this is a convention.

Now you see the information behind a value: $20.0\pm0.2cm$ means generally that we assume the Normal distribution for the values, we estimated $20.0cm$ to be the mean, and 63% of the measurements will be between $19.8cm$ and $20.2cm$.

What does it all have to do with error propagation?

Well if it all depends on our definition of a value and its uncertainty, then it has everything to do.

To propagate errors we don't just sum $L_1$ and $L_2$ max and min values, and this is precisely because we know that these values are not 100% inside their intervals, they are 63% inside. And so the values of $L_t = L_1 + L_2$ will not be 63% inside the interval between $L_1min+L_2min$ and $L_1max+L_2max$. This is not easy to prove in concise manner with your math skills, but you seem curious so you will find it out.

So the way we propagate errors in this case is $\delta L_t = \sqrt{(\delta L_1)^2 + (\delta L_2)^2}$ and although its advantages can be argued logically, this is still based on the Normal distributed model, and the 63% interval requirement.

So what about the area? Well as you saw above, your idea of propagating by using the minimum and maximum values changes the coverage of your interval. Above, when a sum is involved, it will not in general yield a 63%. Here the same happens, you can not expect to have a correct $63%$ interval because, for example $W_{min}*L_{min}$ gives a value of the area that actually is outside of the 63% interval for the area. The prove of this involves probability and again you will have to take my word here.

The formula used is in fact $\delta A / A = \sqrt{(\delta W / W)^2 + (\delta L / L)^2}$ which needs some basic calculus knowledge to deduce, but I will try to give some sense to it from your point of view. See how similar it is to the previous one? Is like the same as before, only you are using $\delta value/value$ which you can think of as a relative uncertainty: is how important is the uncertainty in relation to the value it characterizes.

Indeed, if you would like to compare the uncertainty between your height, say $700\pm2mm$ and say the Earth's radius $696342\pm65km$. Now which one is measured more precisely? It would be hard to say here since $2mm$ is much smaller than $65km$, but the values are very different too. However looking at the relative uncertainties you will see that Earth's value is indeed more precise!

Finally, let me say that although I only spoke of the general case encountered in physics, it is not always correct to assume the Normal distribution, nor even expect that the mean is in the center of the interval. Under some conditions you may encounter Poisson distribution in which the mean is usually closer to the smallest values of the integral. Also in other cases, particularly in research, we expect to have values much more rigorously measured, and intervals are chosen to contain as much as 95% of the measured values, and sometimes more.

When proving addition of fractional uncertainties, one neglects the product of uncertainties. In your case the product of fractional uncertainties is (0.01)(0.02) = 0.0002 which is considerably less than the sum of them 0.01 + 0.02 = 0.03. This is the discrepancy you see in multiplication example. Nevertheless if you take significant figures only in your example, then 200.04 ± 6 becomes 200 ± 6 and the answers are the same.

I think you should have a look in this http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf it should clear up how to do error estimation in a more rigorous manner. I would only worry about the first two sections on addition and multiplication if you haven't covered calculus yet.

This method of taking the maximum and minimum value and halving the difference will only result in a symmetric error if the operations you are dealing with are linear. So for example if I was to work out the value and error of some quantity $y=x^2$ and I know the value of $x$ and its error $\delta x$ can you see why the error in $y$ isn't symmetric under your method? Put in some numbers and try it for yourself.

If you have any follow up post it as a comment and I will try and answer you.