How to Combine Absolute Uncertainties When Subtracting Measurements

Two metal strips are measured separately. Strip A has a measured length of \(62.4\pm0.1\ \mathrm{cm}\), while strip B has a measured length of \(12.4\pm0.1\ \mathrm{cm}\).

Their measured length difference is \(50.0\ \mathrm{cm}\). But what uncertainty should accompany that result?

When subtracting measurements, add their absolute uncertainties. Under the usual worst-case uncertainty convention used in HSC-style practical calculations:

\[ \boxed{\Delta D=\Delta A+\Delta B} \]

Therefore,

\[ \Delta D=0.1+0.1=0.2\ \mathrm{cm} \]

The final result is:

\[ \boxed{D=50.0\pm0.2\ \mathrm{cm}} \]

This might seem strange. If you subtract the measurements, why don’t you subtract their uncertainties as well?

The answer comes from understanding what an uncertainty actually represents.

01Why do absolute uncertainties add when subtracting?

Imagine that you repeat both measurements. Neither reading is necessarily exactly the same because each measurement has some uncertainty.

For strip A:

\[ 62.4-0.1\leq A\leq62.4+0.1 \]

Therefore:

\[ 62.3\leq A\leq62.5 \]

Similarly, for strip B:

\[ 12.3\leq B\leq12.5 \]

These ranges are called measurement intervals. They describe the lowest and highest values allowed by the stated uncertainty bounds.

Now consider the difference:

\[ D=A-B \]

Before doing any calculations, predict which combination gives the smallest possible difference.

• A is at its lowest possible value.
• B is at its highest possible value.

Both changes make the difference smaller.

The reverse combination gives the largest possible difference:

• A is at its highest possible value.
• B is at its lowest possible value.

This is the important idea: both measurement uncertainties can push the final difference in the same direction.

Finding the minimum and maximum difference

The minimum possible difference is:

\[ D_{\min}=(A-\Delta A)-(B+\Delta B) \]

Expanding:

\[ D_{\min}=A-B-(\Delta A+\Delta B) \]

The maximum possible difference is:

\[ D_{\max}=(A+\Delta A)-(B-\Delta B) \]

Expanding:

\[ D_{\max}=A-B+(\Delta A+\Delta B) \]

Therefore, the absolute uncertainty in the difference is:

\[ \boxed{\Delta D=\Delta A+\Delta B} \]

For our metal strips:

The difference can range from 49.8 cm to 50.2 cm.

Its central value is 50.0 cm, and its absolute uncertainty is 0.2 cm.

The common mistake: subtracting the uncertainties

A student might reason:

\[ \Delta D=0.1-0.1=0 \]

This is incorrect.

The student has confused the subtraction of the measured values with the combination of their uncertainties.

Uncertainties are not signed quantities. They describe how far each measurement might vary from its reported value.

For example, A could be 0.1 cm lower than reported while B is 0.1 cm higher. These two changes reduce the difference by 0.2 cm, not zero.

Likewise, if A is higher and B is lower, the difference increases by 0.2 cm.

The uncertainty must account for both possibilities.

02Understand which uncertainty convention you’re using

Before combining uncertainties, check what the quoted \(\pm\) values actually represent.

There are different ways to describe measurement uncertainty, and they do not always use the same mathematical rule.

For ordinary HSC-style maximum-minimum uncertainty calculations, use the first rule unless the question specifies a different statistical or measurement model.

Uncertainty versus error

These terms are related, but they are not interchangeable.

Measurement error is the difference between a measured value and the true value.

\[ \text{Error}=\text{Measured value}-\text{True value} \]

The actual error is usually unknown because the true value is not known exactly.

Absolute uncertainty describes the estimated range or maximum possible deviation associated with a measurement.

For example, \(15.2\pm0.1\ \mathrm{cm}\) means the measurement is reported as 15.2 cm with an absolute uncertainty of 0.1 cm.

The absolute uncertainty is expressed in the same units as the measurement.

Standard deviation describes the spread of repeated measurements around their mean.

It is not automatically the maximum uncertainty of an individual measurement. It can be used to quantify random variation, but its meaning must be identified before applying an uncertainty propagation rule.

For independent standard uncertainties, the uncertainty in a difference is:

\[ \sigma_D=\sqrt{\sigma_A^2+\sigma_B^2} \]

Here, \(\sigma_A\) and \(\sigma_B\) are the standard uncertainties of the two measurements.

Notice that this is different from adding maximum absolute uncertainties.

For the mean of \(n\) independent repeated measurements, the standard error is commonly estimated using \(s/\sqrt n\), where \(s\) is the sample standard deviation.

Do not switch between these conventions halfway through a calculation.

A quick decision process

When you encounter a subtraction uncertainty question:

1. Write the quantity being calculated, including its subtraction order.
2. Calculate the central value using the measured values.
3. Identify whether the quoted uncertainties are maximum bounds or standard uncertainties.
4. For maximum absolute uncertainties, add their magnitudes.
5. Check the minimum and maximum possible results if you are unsure.
6. Report the answer with appropriate units and rounding.

One important qualification: if a question explicitly identifies a shared calibration error affecting both measurements, you may need to account for that correlation rather than automatically adding every contribution. We will examine this more carefully in Question 5.

03Worked examples

04What does the uncertainty tell you about the result?

Calculating an uncertainty is only half the job. You also need to understand what the resulting interval allows you to conclude.

Consider a difference reported as:

\[ D=0.3\pm0.4\ \mathrm{mm} \]

Its possible interval is:

\[ -0.1\leq D\leq0.7\ \mathrm{mm} \]

The interval includes zero.

This means the measurements cannot establish whether the first quantity is larger or smaller than the second.

It does not prove that the two quantities are equal. It means the measurement uncertainty is too large to distinguish their difference confidently using the stated bounds.

There are three useful cases:

What about percentage uncertainty?

After finding the absolute uncertainty, you might also need the percentage uncertainty.

For a non-zero difference:

\[ \%\text{ uncertainty} = \frac{\Delta D}{|D|}\times100 \]

For example, if:

\[ D=7.3\pm0.3\ \mathrm{cm} \]

Then:

\[ \%\text{ uncertainty} = \frac{0.3}{7.3}\times100 \]

\[ \boxed{\%\text{ uncertainty}=4.1\%} \]

Notice that a relatively small difference can have a large percentage uncertainty, even when the individual measurements are quite precise.

This is particularly important when subtracting two measurements that are close together.

For a more detailed explanation of the distinction between error and uncertainty, see Percentage Error vs Percentage Uncertainty: HSC Physics Guide.

05Questions and solutions

Question 1

Two independently measured metal wires have lengths:

\[ L_A=14.8\pm0.2\ \mathrm{cm} \]

\[ L_B=9.6\pm0.1\ \mathrm{cm} \]

Treat the uncertainties as maximum absolute bounds.

Calculate:

a) The difference in their lengths, \(L_A-L_B\).

b) The maximum absolute uncertainty in this difference.

c) The minimum and maximum possible differences.

Solution 1

The difference is \(5.2\pm0.3\ \mathrm{cm}\), with a possible interval from 4.9 cm to 5.5 cm.

Part a: Calculate the difference.

\[ D=L_A-L_B \]

\[ D=14.8-9.6 \]

\[ D=5.2\ \mathrm{cm} \]

Part b: Calculate the absolute uncertainty.

For subtraction, add the absolute uncertainties:

\[ \Delta D=\Delta L_A+\Delta L_B \]

\[ \Delta D=0.2+0.1 \]

\[ \boxed{\Delta D=0.3\ \mathrm{cm}} \]

Part c: Find the measurement interval.

The minimum difference occurs when A is lowest and B is highest:

\[ D_{\min}=14.6-9.7 \]

\[ D_{\min}=4.9\ \mathrm{cm} \]

The maximum difference occurs when A is highest and B is lowest:

\[ D_{\max}=15.0-9.5 \]

\[ D_{\max}=5.5\ \mathrm{cm} \]

Therefore:

\[ \boxed{D=5.2\pm0.3\ \mathrm{cm}} \]

The result is consistent with a difference between 4.9 cm and 5.5 cm under the stated uncertainty bounds.

Question 2

A trolley is pulled horizontally by two opposing forces.

The measured rightward force is:

\[ F_R=8.3\pm0.3\ \mathrm{N} \]

The measured leftward force is:

\[ F_L=8.0\pm0.3\ \mathrm{N} \]

Take right as the positive direction. Treat both uncertainties as maximum absolute bounds.

a) Calculate the net force and its absolute uncertainty.

b) Determine the minimum and maximum possible net forces.

c) Can you conclude that the net force is definitely directed to the right? Explain.

Solution 2

The estimated net force is \(+0.3\pm0.6\ \mathrm{N}\). Its possible interval includes both negative and positive values, so the direction of the net force cannot be established.

Part a: Calculate the net force.

Taking right as positive:

\[ F_{\mathrm{net}}=F_R-F_L \]

Substituting:

\[ F_{\mathrm{net}}=8.3-8.0 \]

\[ F_{\mathrm{net}}=+0.3\ \mathrm{N} \]

Now combine the absolute uncertainties:

\[ \Delta F_{\mathrm{net}}=0.3+0.3 \]

\[ \boxed{\Delta F_{\mathrm{net}}=0.6\ \mathrm{N}} \]

Therefore:

\[ \boxed{F_{\mathrm{net}}=+0.3\pm0.6\ \mathrm{N}} \]

Part b: Find the possible range.

The minimum net force occurs when the rightward force is smallest and the leftward force is largest.

\[ F_{\min}=8.0-8.3 \]

\[ F_{\min}=-0.3\ \mathrm{N} \]

The maximum net force occurs when the rightward force is largest and the leftward force is smallest.

\[ F_{\max}=8.6-7.7 \]

\[ F_{\max}=+0.9\ \mathrm{N} \]

Therefore:

\[ \boxed{-0.3\leq F_{\mathrm{net}}\leq+0.9\ \mathrm{N}} \]

Part c: Interpret the result.

No, the net force is not definitely directed to the right.

Although the central estimate is positive, the possible interval includes negative values and zero.

The trolley could therefore experience a leftward net force, a zero net force, or a rightward net force within the stated measurement bounds.

The tempting mistake is to look only at the central measurements and conclude that 8.3 N is greater than 8.0 N.

However, the uncertainties are large enough that this small difference cannot establish the direction.

Question 3

Two independent experimental groups measure the same difference between two lengths.

Their reported mean measurements are:

\[ A=25.00\ \mathrm{mm} \]

\[ B=24.70\ \mathrm{mm} \]

The standard uncertainties of the two mean measurements are both 0.20 mm.

Assume the uncertainties are independent random uncertainties, and that the quoted values represent one standard deviation rather than maximum absolute bounds.

a) Calculate the difference between the mean measurements.

b) Calculate the standard uncertainty of the difference.

c) Explain why adding the two uncertainties directly would use a different convention.

Solution 3

The difference is 0.30 mm, with a standard uncertainty of approximately 0.28 mm.

Part a: Calculate the difference.

\[ D=A-B \]

\[ D=25.00-24.70 \]

\[ \boxed{D=0.30\ \mathrm{mm}} \]

Part b: Calculate the standard uncertainty.

Because the uncertainties are independent standard uncertainties, combine them using the square root of the sum of their squares.

\[ \sigma_D=\sqrt{\sigma_A^2+\sigma_B^2} \]

Substituting:

\[ \sigma_D=\sqrt{0.20^2+0.20^2} \]

\[ \sigma_D=\sqrt{0.04+0.04} \]

\[ \sigma_D=\sqrt{0.08} \]

\[ \boxed{\sigma_D=0.28\ \mathrm{mm}} \]

The result is therefore:

\[ \boxed{D=0.30\pm0.28\ \mathrm{mm}} \]

Here, the \(\pm0.28\ \mathrm{mm}\) represents one standard deviation, not a guaranteed maximum interval.

Part c: Explain the different convention.

If the two quoted uncertainties represented maximum absolute bounds, the correct result would instead be:

\[ \Delta D=0.20+0.20=0.40\ \mathrm{mm} \]

However, the question explicitly identifies them as independent standard uncertainties.

For independent random uncertainties, the variances add, which is why we use the square root of the sum of the squares.

The important lesson is to identify what the quoted uncertainty represents before selecting a formula. The two methods answer different questions.

Question 4

An experiment measures the energy entering a device and the useful energy transferred out.

The measured input energy is:

\[ E_{\mathrm{in}}=48.6\pm0.4\ \mathrm{J} \]

The measured useful output energy is:

\[ E_{\mathrm{useful}}=46.9\pm0.3\ \mathrm{J} \]

Assume both uncertainties are maximum absolute bounds, with no shared systematic uncertainty.

a) Calculate the energy dissipated.

b) Determine its maximum absolute uncertainty.

c) Calculate its percentage uncertainty.

d) Can you establish that some energy was dissipated within the stated uncertainty bounds? Justify your answer.

Solution 4

The estimated dissipated energy is \(1.7\pm0.7\ \mathrm{J}\), with a percentage uncertainty of approximately 41%. The entire possible interval is positive, so the measurements support a positive energy loss.

Part a: Calculate the dissipated energy.

The energy balance gives:

\[ E_{\mathrm{diss}}=E_{\mathrm{in}}-E_{\mathrm{useful}} \]

Substituting:

\[ E_{\mathrm{diss}}=48.6-46.9 \]

\[ \boxed{E_{\mathrm{diss}}=1.7\ \mathrm{J}} \]

Part b: Calculate the absolute uncertainty.

Because the energy values are being subtracted, add their absolute uncertainties.

\[ \Delta E_{\mathrm{diss}} = \Delta E_{\mathrm{in}}+\Delta E_{\mathrm{useful}} \]

\[ \Delta E_{\mathrm{diss}}=0.4+0.3 \]

\[ \boxed{\Delta E_{\mathrm{diss}}=0.7\ \mathrm{J}} \]

Therefore:

\[ \boxed{E_{\mathrm{diss}}=1.7\pm0.7\ \mathrm{J}} \]

Part c: Calculate the percentage uncertainty.

Use:

\[ \%\text{ uncertainty} = \frac{\Delta E_{\mathrm{diss}}}{E_{\mathrm{diss}}} \times100 \]

Substituting:

\[ \%\text{ uncertainty} = \frac{0.7}{1.7}\times100 \]

\[ \boxed{\%\text{ uncertainty}=41\%} \]

This is a relatively large percentage uncertainty because the dissipated energy is much smaller than either of the original energy measurements.

Part d: Interpret the result.

The possible energy interval is:

\[ E_{\min}=1.7-0.7=1.0\ \mathrm{J} \]

\[ E_{\max}=1.7+0.7=2.4\ \mathrm{J} \]

Therefore:

\[ 1.0\leq E_{\mathrm{diss}}\leq2.4\ \mathrm{J} \]

The lower limit is still positive.

This means that, within the stated measurement bounds, the experiment supports a positive energy loss.

If the interval had included zero, the measurements would not have been sufficient to establish that the dissipated energy was definitely positive.

Question 5

A student wants to measure the length of a metal rod by recording the coordinates of its two endpoints on the same ruler.

The displayed readings are:

\[ r_1=12.40\ \mathrm{cm} \]

\[ r_2=62.40\ \mathrm{cm} \]

However, the ruler has three sources of measurement uncertainty.

Its displayed reading \(r\) is modelled by:

\[ r=(1+s)x+c+e \]

where:

• \(x\) is the true position.
• \(c\) is a common additive zero offset, with \(|c|\leq0.20\ \mathrm{cm}\).
• \(s\) is a common scale-factor error, with \(|s|\leq0.004\), equivalent to 0.4%.
• \(e\) is an independent endpoint reading error, with \(|e_1|,|e_2|\leq0.03\ \mathrm{cm}\).

The same values of \(c\) and \(s\) apply to both endpoint readings, while the endpoint reading errors can vary independently.

Assume all errors are small, use a first-order uncertainty approximation, and treat the stated limits as maximum bounds.

Two students disagree:

• Student A argues that the common zero offset must be added twice because it affects both readings.
• Student B argues that all calibration uncertainty cancels because the same ruler is used.

Determine the rod’s measured length and its conservative maximum absolute uncertainty. Explain why neither student has the complete answer.

Solution 5

The estimated rod length is \(50.00\ \mathrm{cm}\), with a first-order maximum absolute uncertainty of approximately \(0.26\ \mathrm{cm}\).

Neither student is completely correct. The common zero offset cancels, but the scale-factor error does not.

This is an important distinction: subtracting measurements can eliminate some shared systematic effects without eliminating every source of uncertainty.

Step 1: Calculate the measured rod length.

The rod length is the difference between the two displayed coordinates.

\[ L=r_2-r_1 \]

\[ L=62.40-12.40 \]

\[ \boxed{L=50.00\ \mathrm{cm}} \]

Step 2: Examine the common zero offset.

Write the two measurement equations:

\[ r_1=(1+s)x_1+c+e_1 \]

\[ r_2=(1+s)x_2+c+e_2 \]

Subtract the first equation from the second:

\[ r_2-r_1 = (1+s)(x_2-x_1)+(e_2-e_1) \]

Notice that:

\[ c-c=0 \]

The common zero offset disappears completely.

Both readings are shifted by exactly the same amount, so their difference is unaffected.

Student A is therefore incorrect to add the common zero offset twice.

Step 3: Account for the scale-factor error.

The scale-factor error is different.

It changes the spacing between the ruler markings, rather than shifting every reading by the same amount.

Because the two endpoints are separated by 50.00 cm, the approximate absolute uncertainty contributed by the scale factor is:

\[ \Delta L_s=|s|L \]

Substituting:

\[ \Delta L_s=0.004\times50.00 \]

\[ \boxed{\Delta L_s=0.20\ \mathrm{cm}} \]

The scale-factor uncertainty remains in the difference because it affects the measured separation itself.

Student B is therefore incorrect to assume that all calibration uncertainty cancels.

Step 4: Account for the independent endpoint reading errors.

The two endpoint reading errors can vary independently.

The maximum uncertainty in their difference is:

\[ \Delta L_e=\Delta e_1+\Delta e_2 \]

Substituting:

\[ \Delta L_e=0.03+0.03 \]

\[ \boxed{\Delta L_e=0.06\ \mathrm{cm}} \]

Step 5: Combine the remaining uncertainty contributions.

Using the conservative maximum-bound convention, add the scale-factor and endpoint reading contributions:

\[ \Delta L=\Delta L_s+\Delta L_e \]

\[ \Delta L=0.20+0.06 \]

\[ \boxed{\Delta L\approx0.26\ \mathrm{cm}} \]

The common zero offset contributes nothing because it cancels exactly.

Therefore, the result before final rounding is:

\[ \boxed{L\approx50.00\pm0.26\ \mathrm{cm}} \]

Using one significant figure for the uncertainty, the conventionally rounded result is:

\[ \boxed{L\approx50.0\pm0.3\ \mathrm{cm}} \]

This problem reveals why blindly adding every uncertainty can sometimes overestimate the uncertainty in a difference.

The ordinary HSC maximum-minimum rule assumes that the quoted uncertainties can vary independently to their allowed extremes. When a question explicitly identifies a shared systematic component, you must examine how that component affects both measurements.

The key distinction is:

• A shared additive offset cancels when two readings are subtracted.
• A shared multiplicative scale-factor error generally remains.
• Independent endpoint reading errors still combine through the worst-case addition rule.

The measurement model tells you which contributions survive. You should never assume that every uncertainty cancels simply because the same instrument was used.

The central rule is straightforward: when subtracting measurements with maximum absolute uncertainty bounds, add the absolute uncertainties.

The more interesting question comes afterwards: does the resulting interval allow you to distinguish the two quantities?

That is the next step in interpreting experimental results. It connects directly to Do Overlapping Error Bars Mean Two Results Agree? and, when working with experimental graphs, to Error Bars and Maximum-Minimum Gradients: HSC Physics Guide.