Chapter 7: Total Uncertainty, Bias Error, and Propagation of Error

Bias Error: Measuring Volume in a cylinder

What measurements to we need to measure the volume of a cylinder? Do you remember geometry?

The equation for the volume of a cylinder is

\begin{equation}
V = \pi r^2 l
\end{equation}

where r is the radius of the cylinder and l is the length of the cylinder. When thinking about the uncertainty of these measurements there are many things that may come to mind but looking at the equation above we want to identify the number of independent variables as that is where uncertainty will originate from. So in the above equation we will have uncertain in r and l. Now of course when we calculate volume we will have to measure r and l and there may be precision error in these measurements due to temperature fluctuation and thermal expansion and we know how to quantify precision error. However, there will also be bias error present as well. Bias error will manifest itself in the type of device utilized to measure these quantities. For example if you are using a yard stick with 1 yard intervals that will have a much larger uncertainty than a micrometer, or some type of quantum laser technique...we unfortunately do not have that available in lab.

Propagating Uncertainty

Usually we measure several quantities and use those measurements to calculate a desired quantity just like we saw in volume. Each measurement has some associated uncertainty which will create an uncertainty in the calculated result. How do we quantify that uncertainty?

We propagate the uncertainty!

We will utilize a statistical theorem to propagate uncertainty in independent variables. Let’s consider a linear function y of several several independent variables x_i with standard deviation σ_i.

The standard deviation of this equation would be

\begin{equation}
\sigma_{y} = \sqrt{\bigg (\frac{\partial y}{\partial x_{1}} \sigma_{1} \bigg )^{2} + \bigg (\frac{\partial y}{\partial x_{2}} \sigma_{2}\bigg )^{2} + ... + \bigg (\frac{\partial y}{\partial x_{n}} \sigma_{n}\bigg )^{2} }
\end{equation}

Each variable x_i will have some associated uncertainty u_i which will then contribute to the total bias uncertainty in y which we will call u_y. In order to estimate u_y we will make a key assumption that each uncertainty is relatively small (reasonable) so that we can perform our favorite operation! A first order Taylor expansion of y(x_i)

\begin{equation}
y(x_{1} + u_{1}, x_{2} + u_{2}, ..., x_{n} + u_{n}) \approx y(x_{1}, x_{2}, ..., x_{n}) + \frac{\partial y}{\partial x_{1}}u_{1} + \frac{\partial y}{\partial x_{2}}u_{2} + ... + \frac{\partial y}{\partial x_{n}}u_{n}
\end{equation}

Thus y is a linear function of independent variables and the uncertainty will behave like standard deviations:

\begin{equation}
u_{y} = \sqrt{\bigg (\frac{\partial y}{\partial x_{1}} u_{1} \bigg )^{2} + \bigg (\frac{\partial y}{\partial x_{2}} u_{2}\bigg )^{2} + ... + \bigg (\frac{\partial y}{\partial x_{n}} u_{n}\bigg )^{2} }
\end{equation}

The key thing for this theorem to hold true and so that we can utilize this equation is that all uncertainties must be independent of each other and have the same odds.

Let’s do some examples to make this clear

Volume of a Cylinder

A tube of circular cross section has a nominal length of 50 m ± 0.5 m and a radius of 15m ± 0.1m. Determine the uncertainty in the volume.

So when we first approach this problem we must identify if there is precision error, bias error, or both. So when we read this problem if we want to identify precision error do we have any information about the number of samples? A list of measurements? Or anything else?

Well the answer in this problem is no so we have no precision error contribution, something to note when you are reporting your results in a lab report.

For identifying the bias error do we know that tool we were using to measure the independent variables, manufacturer’s information, or if we are really luckily we are just given the uncertainties.

Well in this problem we are really lucky we are just given the uncertainties so we just have deal with bias error in calculating the total uncertainty!

The next step we should do is to identify what we need to measure, the equation required, and the number of independent variables.

In this problem

\begin{equation}
V = \pi r^2 l
\end{equation}

So we have a function V (r,l) that is a function of two independent variables r and l both of which will have some uncertainty associated with the measurement as seen below

So let’s proceed with our solution to calculate the bias error in V , B_V we have to first take our differential

\begin{equation}
B_{V} = \sqrt{\bigg(\frac{\partial V}{\partial r}u_{r}\bigg)^{2} + \bigg(\frac{\partial V}{\partial l}u_{l}\bigg)^{2}}\\
\end{equation}

where we know from our problem statement that u_r = 0.1m and that u_l = 0.5m.

And we also know that

\begin{eqnarray}
\frac{\partial V}{\partial r}= 2 \pi r l\\
\frac{\partial V}{\partial l}= \pi r^2\\
\end{eqnarray}

We can then calculate B_V as seen below by plugging in the nominal or mean values for r and l as well as the uncertainties associated with them.

So we find

\begin{equation}
B_V = 589.05 m^3
\end{equation}

and since in this problem we have no precision uncertainty we find that the total uncertainty in volume, U_V is just B_V as seen below

\begin{equation}
U_V = \sqrt{B^2_V + P^2_r + P^2_l} = B_V
\end{equation}

where P_r = P_l = 0 since there is no precision error in this experiment.

It is critical when performing this analysis to keep your units consistent, this is where most people will make mistakes always always always convert to SI units.

Now often it is more useful to express your answers in terms of a fraction or a percent we can do this simply by dividing our differentials and B_V equation above and re-write as

\begin{equation}
\frac{B_{V}}{V} = \sqrt{\bigg(\frac{\frac{\partial V}{\partial r}u_{r}}{V}\bigg)^{2} + \bigg(\frac{\frac{\partial V}{\partial l}u_{l}}{V}\bigg)^{2}}\\
\end{equation}

We can clearly see here that

\begin{equation}
\frac{\frac{\partial V}{\partial r}u_{r}}{V}= \frac{2 u_r}{r}
\end{equation}

and that

\begin{equation}
\frac{\frac{\partial V}{\partial l}u_{l}}{V} = \frac{u_l}{l}
\end{equation}

Plugging this back into our equation we will find that now

\begin{equation}
B_V = 0.01667
\end{equation}

This is a unitless fraction and you should check that this is the case for yourself if we want to convert this fraction to a percent we must simply multiply by 100 in this problem the nominal volume is 35343 m³ so to double check our answer on the bias uncertainty we can find that if we multiply our fraction by the nominal volume we indeed get that the uncertainty is 589.05 m³. This is a nice way to double check your answer as the problems get more complex from here.

Let’s do another example problem focusing on material mechanics

Uniaxial Tension Test

I am trying to measure the stress in a rectangular block of aluminum under uniaxial tension. I repeated the experiment 14 times and find the force to exhibit a mean of 48.9N and a standard deviation of 3.24. I also measured the length of the block and width 14 times and found a mean width of 0.1m and a standard deviation of 0.01m, the length had a mean of 0.5m and a standard deviatin of 0.02m. I measured the length and width with a micrometer and the force applied is ± 1N. What is the total uncertainty in stress?

So lets break down this problem will we have precision error, bias error, or both?

We will have both precision error and bias error in this problem. Precision error comes from the multiple measurements made in this problem and we are also given standard deviations as well. We will also have bias error because we are given an uncertainty in force plus the tool used to measure length and width.

What is the equation we are working with?

Well since we are dealing with uniaxial tension we can use Hooke’s law, we will talk about tensor notation much much more in depth soon so

\begin{equation}
\sigma = \frac{F}{A} = \frac{F}{l \cdot w}
\end{equation}

So we have a function σ(F,l,w) that is a function of 3 independent variables so we will have to take 3 partials.

So at this point we have some options in terms of how to approach this problem. You can go directly to the partials but when you have precision error I would suggest calculating precision error first, expressing this uncertainty in percent, calculating bias error in percent, and then calculating total uncertainty in percent.

So to start with precision we have precision error in all 3 independent variables so we can calculate the precision error as we have done previously

\begin{eqnarray}
P_F = t_{0.025,13}\frac{S_F}{\sqrt{n}}\\
P_l = t_{0.025,13}\frac{S_l}{\sqrt{n}}\\
P_w = t_{0.025,13}\frac{S_w}{\sqrt{n}}
\end{eqnarray}

But as I mentioned we want to work with percentages so we need to divide these values by the nominal value i.e. \(\frac{P_F}{F}\), \(\frac{P_l}{l}\), \(\frac{P_w}{w}\).

So those values can be expressed as a dimensionless fraction as

\begin{eqnarray}
\frac{P_F}{F} = 0.04\\
\frac{P_l}{l} = 0.023\\
\frac{P_w}{w} = 0.06
\end{eqnarray}

Now we can move on to our partial equations so to calculate the bias error, B_σ we must use the following equation

\begin{equation}
B_{\sigma} = \sqrt{\bigg(\frac{\partial \sigma}{\partial F}u_{F}\bigg)^{2} + \bigg(\frac{\partial \sigma}{\partial l}u_{l}\bigg)^{2}+ \bigg(\frac{\partial \sigma}{\partial w}u_{w}\bigg)^{2}}
\end{equation}

Now here you can plug and check and get an uncertainty in units of stress, Pa, but I would highly suggest that you convert to dimensionless units and a fraction and you will find your answer looks much nicer as seen below

\begin{equation}
\frac{B_\sigma}{\sigma} = \sqrt{\bigg(\frac{u_F}{F}\bigg)^2+\bigg(\frac{u_l}{l}\bigg)^2)+\bigg(\frac{u_w}{w}\bigg)^2}
\end{equation}

Now you can plug in chug to solve this problem but wait I know that F,l,w are the nominal or mean values but what is the uncertainty? Well the uncertainty in F is given and we know the tool to measure length is a micrometer so the bias uncertainty for that tool is 1µm so we have our uncertainties.

Plugging in you will find that

\begin{equation}
\frac{B_\sigma}{\sigma} = \sqrt{\bigg(\frac{u_F}{F}\bigg)^2+\bigg(\frac{u_l}{l}\bigg)^2)+\bigg(\frac{u_w}{w}\bigg)^2} = 0.02
\end{equation}

And now we can calculate the total uncertainty as a percent

\begin{equation}
U_\sigma = \sqrt{\bigg(\frac{B_\sigma}{\sigma}\bigg)^2 + \bigg(\frac{P_F}{F}\bigg)^2 + \bigg(\frac{P_l}{l}\bigg)^2 +\bigg(\frac{P_w}{w}\bigg)^2} \cdot 100 = 7.93 \%
\end{equation}

You will get much more practice with these problems during class, on YouTube, and also in Psets and exams but this is generally how you should approach these problems.

Suggested Procedure for Calculating Total Uncertainty

(1) Determine if problem includes bias error, precision error, or both

(2) Write out fundamental equation, i.e. what you are trying to calculate and identify independent variables

(3) Write out known quantities, i.e. means, uncertainties, standard deviation, number of trials, etc, etc

(4) Start with Precision Error first and convert to percentages or fractions

(5) Take Partials and Calculate Bias Error and convert to percentages or fractions

(6) Combine Precision and Bias error to calculate Total Uncertainty

Minimizing Error in Designing Experiments

Best time to minimize experimental error is in the design stage. You should perform a single-sample uncertainty analysis of the proposed experimental arrangement prior to building the apparatus. This allows one to make a decision on whether the expected uncertainty is acceptable and to identify the sources of uncertainty. So when designing an experiment:

• Avoid approaches that require two large numbers to be measured to determine the small difference between them

• Design experiments or sensors that amplify the signal strength to improve sensitivity

• Build null designs in which the output is measured as a change from zero rather than as a change in nonzero value

• Avoid experiments with large correction factors must be applied as part of the data-reduction procedure

• Attempt to minimize the influence of the measuring system on the measured variable

• Calibrate entire system to minimize calibration related bias errors.

When identifying outliers and a particular point is in doubt, temporarily exclude the data point and fit a new line through the remaining data. If the fit is better the point can possibly be dropped however it could indicate that the relationship is not linear it may indicate another scaling regime or a transition in behavior.