Error Propagation

This page will support you in satisfying Writing Learning Outcome: 
ANALYSIS - Analyze lab data by quantifying error.

Learning Objectives

You should be able to

Why Should You Care About Propagation of Error?

In experimental work, increasing the audience's confidence in measured results is important because too much error or uncertainty in a result can render that result useless. Experiments fraught with error are not valuable, but error is inevitable, so both quantifying error and reducing it becomes very important. Computations are often made using values that have error due to measurement precision and other uncertainty. Offering a reader the error bounds for a result that is computed from measured values provides valuable information about the quality of the computed result.

How to Estimate Error in Computed Results

Estimating error is better than not considering error at all. Because engineers are masters of heuristics (tools that use time-saving simplifying assumptions), we will first introduce an easy way to estimate error. 

The relative error in any computed result, expressed as a percentage, must be greater than the individual error of any input value and less than the summation of individual errors in all input values. 

The error in the computed result is closer to the value of the largest error in the input values. There is usually one measurement in any experiment with the greatest error that dominates the error in the computed result. We will see this in the example below. 

How to Rigorously Calculate Error in Computed Results

Propagation of error depends on the mathematical operations performed on input values as well as their standard deviation (s). While calculating error propagation can be relatively complex, if two variables (A and B) are multiplied or divided and their errors are not correlated (meaning the errors are not related in some meaningful way) their standard deviations (sA and sB) are related to the standard deviation (sƒ) of their product (ƒ=AB) or dividend (ƒ=A/B) by Equation 1:

If we assume we are working with normally distributed variables and 3s error bounds, then we can calculate error in a computed result given errors in input values using this procedure:

An Example of Error Propagation

A common task in engineering materials and mechanics courses is to calculate the modulus of elasticity, E, a measure of material stiffness, of a specimen from a tension test involving

The relationship of the quantities to determine the modulus of elasticity is E=σ/ε = (F/A)/(∆/L), where σ is engineering stress (F/A) and ε is engineering strain (∆/L).

Taking a very simplified approach so we can focus on the propagation of error, we will examine a specimen of 6061-T6 aluminum loaded in its elastic range. The test is depicted in Figure 1.

Figure 1. Aluminum sample in tensile tester.

Let us consider the error in the cross-sectional area of the specimen if its width and thickness are measured as 2.000 in. and 0.125 in., respectively, with a measurement precision of 0.001 in. each. The 3s percent error is 0.80% for the thickness and 0.05% for the width. We assume the 0.001 in. is a 3s error, so the standard deviation is 0.001/3, or 0.00033 in. Using Equation 1 to calculate the standard deviation of the cross-sectional area, we determine

The absolute error is then 3sAB or 0.002 in2, which is 0.80% of the 0.25-in2 result. These calculations can be summarized in Table 1. It is worth noting that the relative error (%) in the cross-sectional area is equal to the relative error in the thickness measurement, because the measurement with the greatest error contributes the most significantly to the error in the computed result. It is also worth nothing that the relative error for the thickness measurement is greater than that for the width measurement because we are using the measurement device at the low end of its range. Relative error is reduced, in general, when using a measurement device at the high end of its range, because precision does not generally degrade within the working range of an instrument.

Table 1. Results of a cross-sectional area calculation including propagation of error.

Now that we see how to calculate compounding error through a calculation involving a product of two values, let us perform these same calculations for the value of interest: modulus of elasticity.

Let’s say each input quantity has its own error based on measurement precision or uncertainty:

Using Equation 1 to calculate the standard deviation for computed results,

Table 2. Results of a modulus of elasticity calculation including propagation of error.

If we compute E, the result is 10,666,667 psi, or 10,667 ksi. We can calculate the error in E as ± 12.1% or ± 1,286 ksi. Therefore, the result E is actually somewhere between 9,381 ksi and 11,953 ksi.

The nominal value of modulus of elasticity published for 6061-T6 aluminum is 10,000 ksi. If an error analysis were not conducted, the result of this experiment, 10,667 ksi, would appear to be in error or somehow inconsistent with the 10,000-ksi nominal value. One might compute the percent difference of the experimental value relative to the nominal value and report a 6.67% relative difference.

When accounting for error in the computed result, it can simply be concluded that the modulus of elasticity of the aluminum specimen is somewhere between 9,381 ksi and 11,952 ksi, which is consistent with published values.

If we had used the simplifying assumption of the most significant error in any input value to estimate the error in the modulus of elasticity, we would use 10% of 10,667 ksi or ± 1,067 ksi, for a range of 9,600-11,734 ksi. This much simpler analysis still confirms that the tested specimen has a stiffness consistent with the published nominal value, at least within our ability to measure this value.

These calculations are provided in a spreadsheet available here. Try changing the relative error of the measured values to see how this influences the error in the computed result. 

Propagation of Error Example

Common Mistakes