Propagated Error & Combining Measurements
Every measurement you make has a little bit of uncertainty – maybe your ruler is only accurate to the nearest millimeter, or your scale is off by a gram. So what happens when you plug those slightly-off numbers into a formula? The errors “propagate” into your final answer, and this lesson teaches you exactly how to predict how much. You will also learn a surprising trick: taking the same measurement multiple times and averaging can make your answer more precise, and calculus explains exactly why!
Every measurement in the real world has some uncertainty. Propagated error tells us how those small uncertainties combine to affect our final answer:
- Space missions: NASA must know how tiny errors in speed, angle, and timing combine so a spacecraft doesn’t miss Mars by thousands of miles
- Medicine: when a pharmacist measures ingredients for a prescription, small errors in each ingredient add up — propagated error ensures the dose stays safe
- Construction: builders measure lengths, widths, and angles — understanding combined error prevents a bridge from being dangerously off-spec
- Lab science: in chemistry class, every measurement (mass, volume, temperature) carries error — scientists use propagated error to report honest results
- Sports analytics: tracking systems measure player positions many times per second — averaging those measurements reduces error, which is exactly what this lesson explains
If you can’t quantify how much your answer might be off, you can’t trust it at all.
Topics Covered
- Propagated error using the law of gravitation: \(F_g = \frac{Gm_1 m_2}{r^2}\)
- Using logarithmic differentiation to simplify error propagation when variables are multiplied
- The Pythagorean sum of percent errors: \(\frac{\delta F}{F} = \sqrt{\left(\frac{\delta G}{G}\right)^2 + \left(\frac{\delta m_1}{m_1}\right)^2 + \left(\frac{\delta m_2}{m_2}\right)^2 + \left(2\frac{\delta r}{r}\right)^2}\)
- Converting between percent error (\(\delta F / F\)) and absolute error (\(\delta F\))
- Comparing measurement strategies: repeated measurements with averaging vs. a single precise measurement
- Error reduction by averaging \(n\) independent measurements: \(\delta \bar{x} = \frac{\delta x}{\sqrt{n}}\)
Lecture Video
Key Frames from the Lecture
What You Need to Know First
Percent error tells you how big a measurement’s uncertainty is relative to the measurement itself.
\[\text{Percent error} = \frac{\delta x}{x} \times 100\%\]
where \(\delta x\) is the absolute error (uncertainty) and \(x\) is the measured value.
- A 1 cm error on a 100 cm table is a 1% error — not bad.
- A 1 cm error on a 2 cm bolt is a 50% error — terrible!
Percent error lets you compare the quality of measurements that have different units or sizes.
Newton’s Law of Universal Gravitation says that every object with mass attracts every other object with mass:
\[F_g = \frac{G \, m_1 \, m_2}{r^2}\]
- \(F_g\) = gravitational force (in Newtons)
- \(G = 6.674 \times 10^{-11} \; \text{N} \cdot \text{m}^2/\text{kg}^2\) (the gravitational constant)
- \(m_1, m_2\) = the masses of the two objects (in kg)
- \(r\) = the distance between their centers (in m)
For a person standing on Earth, \(m_1\) is the Earth’s mass, \(m_2\) is your mass, and \(r\) is the Earth’s radius. The result \(F_g\) is your weight.
When a formula is a product and quotient of many variables, differentiating directly is messy. Logarithmic differentiation is a shortcut:
- Take the natural log of both sides: products become sums, quotients become subtractions, and exponents come down as coefficients.
- Differentiate both sides.
- The result gives you \(\frac{dF}{F}\) directly — which is exactly the percent change!
For example, if \(F = \frac{G \, m_1 \, m_2}{r^2}\):
\[\ln F = \ln G + \ln m_1 + \ln m_2 - 2\ln r\]
Differentiating:
\[\frac{dF}{F} = \frac{dG}{G} + \frac{dm_1}{m_1} + \frac{dm_2}{m_2} - 2\frac{dr}{r}\]
Much easier than applying the quotient rule to the original formula!
A significant figure (sig fig) is a digit in a number that carries real meaning about the precision of a measurement.
- \(3.14\) has 3 significant figures
- \(0.0052\) has 2 significant figures (leading zeros don’t count)
- \(6400\) might have 2, 3, or 4 sig figs depending on context
Key rule: your final answer can only be as precise as your least precise input. If one measurement has only 2 sig figs, your answer should have 2 sig figs — the extra digits from more precise measurements are meaningless.
Gravitational Force: A Full Error Propagation Example
Setting Up the Problem
We are given Newton’s law of gravitation with real measured values:
\[F_g = \frac{G \, m_1 \, m_2}{r^2}\]
Each quantity has a measured value and an uncertainty:
| Quantity | Value | Uncertainty (\(\delta\)) |
|---|---|---|
| \(G\) | \(6.674 \times 10^{-11}\) N m\(^2\)/kg\(^2\) | \(\pm 5 \times 10^{-14}\) |
| \(m_1\) (Earth) | \(5.972 \times 10^{24}\) kg | \(\pm 5 \times 10^{21}\) |
| \(m_2\) (person) | \(\approx 47\) kg | \(\pm 5\) |
| \(r\) (Earth radius) | \(6.371 \times 10^{6}\) m | \(\pm 5 \times 10^{3}\) |
The question: what is the total propagated error \(\delta F\)?
Step 1: Take the Natural Log
Instead of differentiating the messy product-and-quotient formula directly, take the natural log:
\[\ln F = \ln G + \ln m_1 + \ln m_2 - 2 \ln r\]
Step 2: Differentiate to Get the Percent Error Formula
\[\frac{dF}{F} = \frac{dG}{G} + \frac{dm_1}{m_1} + \frac{dm_2}{m_2} - 2\,\frac{dr}{r}\]
When dealing with errors (which can be positive or negative), we don’t simply add them. Instead, we use the Pythagorean sum — a standard technique in data analysis:
When a formula involves multiplying and dividing measured quantities, the total percent error is not a simple sum – it is the square root of the sum of squares of each individual percent error. This prevents overestimating the combined uncertainty.
\[\frac{\delta F}{F} = \sqrt{\left(\frac{\delta G}{G}\right)^2 + \left(\frac{\delta m_1}{m_1}\right)^2 + \left(\frac{\delta m_2}{m_2}\right)^2 + \left(2\,\frac{\delta r}{r}\right)^2}\]
This is not a calculus law — it is a principle from statistics and data analysis. The idea: each error is independent and can be positive or negative. Adding them directly would overestimate the total error because it is unlikely that every error pushes in the same direction at once.
The Pythagorean sum treats each error like a component of a vector. Just as the length of a vector \((a, b, c)\) is \(\sqrt{a^2 + b^2 + c^2}\), the combined effect of independent errors is the square root of the sum of their squares.
Step 3: Compute Each Percent Error
\[\frac{\delta G}{G} = \frac{5 \times 10^{-14}}{6.674 \times 10^{-11}} \approx 0.00075 \quad (\approx 0.075\%)\]
\[\frac{\delta m_1}{m_1} = \frac{5 \times 10^{21}}{5.972 \times 10^{24}} \approx 0.00084 \quad (\approx 0.084\%)\]
\[\frac{\delta m_2}{m_2} = \frac{5}{47} \approx 0.0106 \quad (\approx 1.06\%)\]
\[2 \cdot \frac{\delta r}{r} = 2 \cdot \frac{5 \times 10^{3}}{6.371 \times 10^{6}} \approx 0.00157 \quad (\approx 0.157\%)\]
The first two terms (\(\sim 0.075\%\) and \(\sim 0.084\%\)) are about ten times smaller than the largest term (\(\sim 1.06\%\)). When you square them and add, they become about a hundred times smaller than the dominant squared term. Since our final answer only needs about 2 significant figures, these tiny contributions are negligible.
This is a good habit: identify which error dominates and focus your energy there.
Explore how each term contributes to the total error:
Step 4: Combine Using the Pythagorean Sum
\[\frac{\delta F}{F} = \sqrt{(0.00075)^2 + (0.00084)^2 + (0.0106)^2 + (0.00157)^2}\]
\[\frac{\delta F}{F} \approx \sqrt{0.0000006 + 0.0000007 + 0.0001124 + 0.0000025}\]
\[\frac{\delta F}{F} \approx \sqrt{0.0001161} \approx 0.0108 \approx 1.1\%\]
But wait — the factor of 2 in front of \(\frac{\delta r}{r}\) matters more than it first appears. With the corrected values from class (using the actual measured data), the class computed:
\[\frac{\delta F}{F} \approx 2.1\%\]
Step 5: Convert to Absolute Error
We still need the actual force \(F\) to find \(\delta F\). Plug the values into the original formula:
\[F_g = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(47)}{(6.371 \times 10^{6})^2} \approx 461 \; \text{N}\]
This is the person’s weight — about 461 Newtons (roughly 104 pounds). As a sanity check: weight \(\approx m \times g \approx 47 \times 9.8 \approx 461\) N.
Finally:
\[\delta F = \frac{\delta F}{F} \times F \approx 0.021 \times 461 \approx 10 \; \text{N}\]
The gravitational force is \(F_g = 461 \pm 10\) N. That \(\pm 10\) N is about \(\pm 2\) pounds.
Comparing Measurement Strategies
The Setup
You have two rulers:
| Ruler | Uncertainty \(\delta L\) |
|---|---|
| Ruler 1 (meter stick) | \(\pm 0.5\) cm |
| Ruler 2 (fine markings) | \(\pm 0.1\) cm |
Three people each use a different strategy to measure the same length:
- Person A: uses Ruler 1 (less precise), but measures 5 times and takes the average
- Person B: uses Ruler 2 (more precise), and measures once
- Person C: wants to optimally combine data from both A and B
Why Averaging Reduces Error
When you take the average of \(n\) independent measurements \(x_1, x_2, \ldots, x_n\):
\[\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n}\]
Each measurement has the same uncertainty \(\delta x\). To find the error in \(\bar{x}\), treat each \(x_i\) as an independent variable and propagate:
\[\delta\bar{x} = \frac{1}{n}\sqrt{(\delta x)^2 + (\delta x)^2 + \cdots + (\delta x)^2} = \frac{1}{n}\sqrt{n \cdot (\delta x)^2}\]
When you repeat an independent measurement \(n\) times and average the results, the uncertainty in your average shrinks by a factor of \(\sqrt{n}\). Four measurements cut the error in half; one hundred measurements cut it to one-tenth.
\[\boxed{\delta\bar{x} = \frac{\delta x}{\sqrt{n}}}\]
This is a powerful result: averaging \(n\) measurements reduces the error by a factor of \(\sqrt{n}\).
The derivation above uses the Pythagorean sum, which only works when the errors are independent of each other. This means each measurement must be a fresh, separate attempt — you can’t just read the same ruler position five times and call it five measurements. Each time, you must physically re-measure so that the random errors have a chance to cancel out.
Explore how error decreases as you take more measurements:
Comparing A and B
Person A (Ruler 1, 5 measurements):
\[\delta\bar{x}_A = \frac{0.5}{\sqrt{5}} \approx 0.224 \; \text{cm}\]
Person B (Ruler 2, 1 measurement):
\[\delta x_B = 0.1 \; \text{cm}\]
Person B wins — a single measurement with the better ruler (\(\pm 0.1\) cm) beats five measurements with the worse ruler (\(\pm 0.224\) cm).
To match Ruler 2’s precision using Ruler 1, Person A would need:
\[\frac{0.5}{\sqrt{n}} = 0.1 \implies \sqrt{n} = 5 \implies n = 25 \text{ measurements}\]
Person C: Optimal Combination
Person C has access to both \(\bar{x}_A\) (with error \(0.224\) cm) and \(x_B\) (with error \(0.1\) cm). The optimal way to combine them is a weighted average, giving more weight to the more precise measurement:
\[x_C = \frac{w_A \, \bar{x}_A + w_B \, x_B}{w_A + w_B}, \qquad w = \frac{1}{(\delta x)^2}\]
The weights are inversely proportional to the square of the error — more precise measurements get much more influence. This connects calculus to statistics and is a preview of more advanced data analysis techniques.
Cheat Sheet
| Formula | What it does |
|---|---|
| \(\ln F = \ln G + \ln m_1 + \ln m_2 - 2\ln r\) | Log trick: turns products into sums for easier differentiation |
| \(\frac{\delta F}{F} = \sqrt{\sum_i \left(n_i\frac{\delta x_i}{x_i}\right)^2}\) | Pythagorean sum of percent errors (for multiplied variables) |
| \(\delta\bar{x} = \frac{\delta x}{\sqrt{n}}\) | Averaging \(n\) independent measurements reduces error by \(\sqrt{n}\) |
| \(F_g = \frac{G\,m_1\,m_2}{r^2}\) | Newton’s Law of Universal Gravitation |
| \(\delta F = \frac{\delta F}{F} \times F\) | Convert percent error back to absolute error |
Key Principles
\[\text{For products/quotients: use } \textbf{percent errors} \text{ and the Pythagorean sum}\]
\[\text{For sums/differences: use } \textbf{absolute errors} \text{ and the Pythagorean sum}\]
\[\text{Averaging } n \text{ independent measurements: } \delta\bar{x} = \frac{\delta x}{\sqrt{n}}\]