Why More Decimal Places Do Not Always Mean Better Measurements

A digital calliper measures a steel rod and displays \(126.84\ \mathrm{mm}\). An analogue instrument measures the same rod and gives \(126.8\ \mathrm{mm}\). Which result is better?

The digital reading has more decimal places, so it looks more precise. But suppose the digital calliper has a stated uncertainty of \(\pm 0.20\ \mathrm{mm}\), while the analogue instrument has an uncertainty of just \(\pm 0.05\ \mathrm{mm}\).

The analogue measurement is actually more tightly constrained.

More decimal places indicate finer measurement resolution, not necessarily greater accuracy or lower uncertainty. A measurement is only as useful as the uncertainty attached to it, the instrument’s calibration, and the method used to obtain it.

This distinction matters when you evaluate experimental results in HSC Physics. A result such as \(126.84\ \mathrm{mm}\) is not automatically better than \(126.8\ \mathrm{mm}\). You need to know what those numbers can genuinely tell you.

01What do the extra decimal places actually tell you?

Start by separating two ideas: resolution and uncertainty.

Resolution is the smallest change an instrument can display or distinguish.

For example:

• A ruler marked in millimetres has a resolution of \(1\ \mathrm{mm}\).
• A digital calliper displaying hundredths of a millimetre has a resolution of \(0.01\ \mathrm{mm}\).
• A digital thermometer displaying two decimal places in degrees Celsius has a resolution of \(0.01^\circ\mathrm{C}\).

A digital instrument can therefore display a much finer reading than an ordinary ruler.

However, resolution does not tell you everything about the measurement.

Imagine that a digital calliper has a small calibration problem. Its display might consistently read \(0.20\ \mathrm{mm}\) above the actual length.

The instrument can still display \(126.84\ \mathrm{mm}\). Its extra digits have not disappeared, but the measurement is affected by a much larger uncertainty than the display resolution suggests.

This is why you should distinguish:

The important point is that a fine resolution does not guarantee accuracy.

For a closer look at this distinction, see How to Distinguish Instrument Resolution from Measurement Uncertainty.

02A measurement is an interval, not just a number

Suppose you measure a length and obtain:

\[ x = 12.5 \pm 0.2\ \mathrm{cm} \]

The central measurement is \(12.5\ \mathrm{cm}\), and the absolute uncertainty is \(0.2\ \mathrm{cm}\).

This means the measurement is represented by the interval:

\[ x_{\min}=12.5-0.2=12.3\ \mathrm{cm} \]

\[ x_{\max}=12.5+0.2=12.7\ \mathrm{cm} \]

So the reported measurement interval is approximately \(12.3\) to \(12.7\ \mathrm{cm}\).

The uncertainty determines the width of this interval.

A smaller uncertainty produces a narrower interval. A larger uncertainty produces a wider interval.

Return to the two instruments measuring the steel rod.

Notice something interesting.

The analogue measurement has fewer decimal places, but its uncertainty interval is considerably narrower.

The two intervals overlap, meaning the measurements are compatible with each other given their stated uncertainties. However, overlapping intervals do not prove that the measurements are identical or that either instrument is accurate.

An uncertainty interval is an estimate based on a particular measurement method or convention. It is not automatically a guarantee that the true value must lie inside it.

That distinction becomes especially important when an instrument has a calibration problem.

03An experiment that exposes the problem

Consider a hypothetical investigation designed to test whether a digital calliper displaying two decimal places actually measures length to that level of accuracy.

The investigation uses:

• A digital calliper with a display resolution of \(0.01\ \mathrm{mm}\).
• A calibrated gauge block with a known length of \(50.00 \pm 0.01\ \mathrm{mm}\).
• A flat working surface.

The gauge block provides a reference measurement against which the calliper can be tested.

The original experimental design

The gauge block is placed between the calliper’s external measuring jaws.

One jaw is fixed to the main beam, while the other slides along it. An internal position sensor detects the separation between the jaws and sends the reading to the digital display.

The jaws are brought gently into contact with opposite faces of the gauge block.

The initial prediction is straightforward.

If the calliper is properly calibrated, repeated measurements should produce values close to \(50.00\ \mathrm{mm}\).

Because the display changes in increments of \(0.01\ \mathrm{mm}\), a student might also assume that the measurement uncertainty is approximately \(0.01\ \mathrm{mm}\).

The block is measured five times, with the following illustrative results:

These readings are remarkably close together.

The range is:

\[ \text{Range}=50.38-50.35 \]

\[ \text{Range}=0.03\ \mathrm{mm} \]

The half-range is:

\[ \Delta x=\frac{0.03}{2}=0.015\ \mathrm{mm} \]

The mean is:

\[ \bar{x}= \frac{50.35+50.36+50.36+50.37+50.38}{5} \]

\[ \bar{x}=50.364\ \mathrm{mm} \]

But compare this with the accepted reference length:

\[ x_{\mathrm{reference}}=50.00\ \mathrm{mm} \]

The mean measurement is approximately \(0.36\ \mathrm{mm}\) too high.

The instrument produces consistent readings, but those readings are systematically displaced from the reference value.

What went wrong?

The problem is a systematic error.

A systematic error shifts measurements in a consistent direction. It might arise from incorrect calibration, a zero offset, or a consistent problem with the measurement method.

In this example, the calliper is consistently reading too high.

A student might suggest taking more measurements and calculating a more accurate mean.

That won’t solve the problem.

If every measurement is affected by approximately the same offset, averaging them simply produces a more precise estimate of the wrong value.

Think about a kitchen scale that has not been zeroed properly. If it always adds \(200\ \mathrm{g}\), weighing the same object ten times won’t remove those extra \(200\ \mathrm{g}\).

The repeated measurements can be extremely consistent while remaining inaccurate.

Revising the experiment

The experimental design needs an additional calibration check.

Before measuring the gauge block, close the calliper jaws without an object between them.

The display should read:

\[ 0.00\ \mathrm{mm} \]

Suppose it instead reads:

\[ +0.36\ \mathrm{mm} \]

This suggests a zero offset of approximately \(+0.36\ \mathrm{mm}\).

If the offset is stable, the correction is:

\[ x_{\mathrm{corrected}} = x_{\mathrm{display}}-0.36 \]

The corrected readings become:

The corrected mean is:

\[ \bar{x}_{\mathrm{corrected}}=50.004\ \mathrm{mm} \]

This is much closer to the accepted reference value of \(50.00\ \mathrm{mm}\).

The investigation has revealed something important: the original display resolution was not a reliable indication of the overall measurement quality.

There is one limitation. A zero check detects a constant offset, but it cannot establish that the instrument measures correctly across its entire range. A proper calibration check would use reference lengths at several points to identify possible scale-factor errors as well.

04Uncertainty, error, and standard deviation are different

These three terms are easy to mix up in experimental questions.

They describe different features of your measurements.

1. Uncertainty

Uncertainty describes the estimated range associated with a measurement.

For example:

\[ L=50.00\pm0.05\ \mathrm{mm} \]

The absolute uncertainty is \(0.05\ \mathrm{mm}\).

It tells you how tightly the measurement is constrained under the stated uncertainty convention.

You can estimate uncertainty from instrument limitations, repeated measurements, or other known sources of variation.

Importantly, uncertainty does not tell you the actual difference between your measurement and the true value.

2. Error

Error is the difference between a measured value and an accepted reference value.

For this article, use the sign convention:

\[ e=x_{\mathrm{measured}}-x_{\mathrm{reference}} \]

A positive error means the measurement is higher than the reference value.

A negative error means the measurement is lower.

For the original calliper experiment:

\[ e=50.364-50.00 \]

\[ e=+0.364\ \mathrm{mm} \]

The calliper has a positive error of \(0.364\ \mathrm{mm}\).

You cannot normally calculate the actual error of an unknown measurement unless you have an appropriate reference value.

3. Standard deviation

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

A small standard deviation means the readings are clustered closely together.

A large standard deviation means the readings are more widely spread.

For a sample of repeated measurements, the standard deviation is:

\[ s= \sqrt{ \frac{\sum(x_i-\bar{x})^2}{n-1} } \]

Where:

• \(s\) is the sample standard deviation.
• \(x_i\) is an individual measurement.
• \(\bar{x}\) is the mean.
• \(n\) is the number of measurements.

For the five calliper readings, the sample standard deviation is approximately:

\[ s=0.011\ \mathrm{mm} \]

This is small compared with the \(0.364\ \mathrm{mm}\) error.

That is not a contradiction.

The readings are closely grouped, but they are grouped around an incorrect value.

Standard deviation measures the spread of the dataset. It does not automatically represent the total measurement uncertainty, the uncertainty of the mean, or the size of a systematic error.

Comparing the three ideas

A useful way to remember this:

• Uncertainty asks how tightly you can specify the measurement.
• Error asks how far the measurement is from a reference.
• Standard deviation asks how spread out your repeated readings are.

05Which uncertainty convention should you use?

You cannot calculate an uncertainty properly without knowing how it is being estimated.

For HSC-style experimental questions, pay attention to whether the instrument is analogue or digital, whether a manufacturer specification is supplied, and whether repeated measurements are involved.

Analogue instruments

For a single analogue reading, a common school convention is:

\[ \Delta x= \frac{\text{smallest scale division}}{2} \]

For example, a ruler has markings every \(1\ \mathrm{mm}\).

Its estimated reading uncertainty is:

\[ \Delta x=\pm0.5\ \mathrm{mm} \]

The uncertainty is half the smallest scale division.

This assumes the scale can be read properly and that the measurement method does not introduce a larger limitation.

Digital instruments

Digital instruments require a little more care.

The display resolution tells you the smallest displayed increment, but the instrument may have a larger stated accuracy limitation.

For example:

• Display resolution: \(0.01\ \mathrm{mm}\).
• Stated instrument accuracy: \(\pm0.20\ \mathrm{mm}\).

The stated accuracy limitation is much larger than the display increment.

For the digital examples in this article, when no instrument specification is supplied, we adopt a classroom convention of using \(\pm1\) smallest displayed increment.

There is another convention worth recognising: if a digital display simply rounds to the nearest increment, the rounding interval alone is half an increment. For a display changing in steps of \(0.01\ \mathrm{mm}\), this is approximately \(\pm0.005\ \mathrm{mm}\).

That rounding interval does not account for calibration, drift, or other instrument limitations.

If a question explicitly supplies an instrument accuracy or uncertainty, use that information rather than assuming the display resolution represents the whole uncertainty.

Repeated measurements

Repeated measurements help reveal random variation.

A simple school-level estimate of uncertainty from repeated measurements is half the range:

\[ \Delta x= \frac{x_{\max}-x_{\min}}{2} \]

For example, suppose repeated readings are:

\[ 12.4,\ 12.6,\ 12.5,\ 12.7,\ 12.5 \]

The range is:

\[ 12.7-12.4=0.3 \]

Therefore:

\[ \Delta x=\frac{0.3}{2}=0.15 \]

The mean is:

\[ \bar{x}=12.54 \]

The result can be expressed as:

\[ x=12.54\pm0.15 \]

The half-range is an estimate of the spread of the readings. It is not the same calculation as standard deviation.

Also, repeating a measurement does not automatically remove systematic errors.

For more practice with repeated readings and relative uncertainty, see How to Find Percentage Uncertainty from Repeated Measurements.

06Worked examples

07Questions and solutions

Question 1

A student measures a metal strip using a ruler with markings every \(1\ \mathrm{mm}\).

One end of the strip is aligned with the zero mark, and the other end gives a reading of \(73\ \mathrm{mm}\).

Assume the uncertainty is half the smallest scale division.

Calculate:

a) The absolute uncertainty.

b) The measurement interval.

c) The percentage uncertainty.

Solution 1

The measurement is \(73\pm0.5\ \mathrm{mm}\), giving an interval of \(72.5\) to \(73.5\ \mathrm{mm}\) and a percentage uncertainty of approximately \(0.68\%\).

Step 1: Calculate the absolute uncertainty.

The smallest scale division is \(1\ \mathrm{mm}\).

Therefore:

\[ \Delta L=\frac{1}{2} \]

\[ \Delta L=0.5\ \mathrm{mm} \]

Step 2: Write the measurement interval.

The lower boundary is:

\[ 73-0.5=72.5\ \mathrm{mm} \]

The upper boundary is:

\[ 73+0.5=73.5\ \mathrm{mm} \]

Therefore:

\[ 72.5\leq L\leq73.5\ \mathrm{mm} \]

Step 3: Calculate percentage uncertainty.

\[ \%\text{ uncertainty} = \frac{0.5}{73}\times100 \]

\[ \boxed{0.68\%} \]

The ruler’s resolution limits how finely the length can be read. Reporting the measurement as \(73.000\ \mathrm{mm}\) would not make it more accurate.

Question 2

Two instruments are used to measure the same metal rail.

Instrument A is a digital sensor that gives:

\[ L_A=2.483\pm0.020\ \mathrm{m} \]

Instrument B is an analogue instrument that gives a reading of \(2.48\ \mathrm{m}\). Its smallest scale division is \(0.01\ \mathrm{m}\).

Assume the analogue uncertainty is half the smallest division.

a) Calculate the uncertainty interval for each instrument.

b) Calculate the percentage uncertainty for each measurement.

c) Identify which measurement has the smaller relative uncertainty.

d) Can you conclude that the instrument with the smaller uncertainty is more accurate? Explain.

Solution 2

Instrument B has the smaller absolute and percentage uncertainty. However, this does not prove it is more accurate because accuracy requires comparison with an accepted reference value.

Step 1: Calculate Instrument A’s interval.

Its measurement is:

\[ L_A=2.483\pm0.020\ \mathrm{m} \]

Lower boundary:

\[ 2.483-0.020=2.463\ \mathrm{m} \]

Upper boundary:

\[ 2.483+0.020=2.503\ \mathrm{m} \]

Therefore:

\[ 2.463\leq L_A\leq2.503\ \mathrm{m} \]

Step 2: Calculate Instrument B’s uncertainty.

Its smallest division is \(0.01\ \mathrm{m}\).

Therefore:

\[ \Delta L_B=\frac{0.01}{2} \]

\[ \Delta L_B=0.005\ \mathrm{m} \]

Its measurement interval is:

\[ 2.48-0.005=2.475\ \mathrm{m} \]

\[ 2.48+0.005=2.485\ \mathrm{m} \]

Therefore:

\[ 2.475\leq L_B\leq2.485\ \mathrm{m} \]

Step 3: Calculate Instrument A’s percentage uncertainty.

\[ \%\text{ uncertainty}_A = \frac{0.020}{2.483}\times100 \]

\[ \%\text{ uncertainty}_A\approx0.81\% \]

Step 4: Calculate Instrument B’s percentage uncertainty.

\[ \%\text{ uncertainty}_B = \frac{0.005}{2.48}\times100 \]

\[ \%\text{ uncertainty}_B\approx0.20\% \]

Step 5: Compare the instruments.

Instrument B has a percentage uncertainty of approximately \(0.20\%\), compared with \(0.81\%\) for Instrument A.

Its measurement is therefore more tightly constrained relative to its measured value.

The intervals also overlap:

\[ [2.475,\ 2.485]\ \mathrm{m} \]

lies entirely within Instrument A’s interval.

This means the results are compatible within their stated uncertainties.

However, you cannot conclude that Instrument B is more accurate without knowing an accepted reference value.

A narrow uncertainty interval and a small error are not necessarily the same thing.

Question 3

A digital balance is tested using a reference mass with an accepted value of:

\[ m_{\mathrm{reference}}=10.00\pm0.01\ \mathrm{g} \]

The balance displays readings in increments of \(0.01\ \mathrm{g}\).

Five measurements produce:

Calculate:

a) The mean mass.

b) The sample standard deviation.

c) The signed error relative to the reference value, using measured minus reference.

d) Explain whether the small spread of readings demonstrates that the balance is accurate.

Solution 3

The mean is \(10.24\ \mathrm{g}\), the sample standard deviation is approximately \(0.016\ \mathrm{g}\), and the signed error is \(+0.24\ \mathrm{g}\).

The balance produces relatively consistent readings but is not accurate relative to the reference mass.

Step 1: Calculate the mean.

\[ \bar{m}= \frac{10.22+10.24+10.25+10.23+10.26}{5} \]

\[ \bar{m}=\frac{51.20}{5} \]

\[ \bar{m}=10.24\ \mathrm{g} \]

Step 2: Calculate the sample standard deviation.

The deviations from the mean are:

\[ -0.02,\ 0,\ 0.01,\ -0.01,\ 0.02 \]

Therefore:

\[ s= \sqrt{ \frac{ 0.0004+0+0.0001+0.0001+0.0004 }{4} } \]

\[ s=\sqrt{0.00025} \]

\[ \boxed{s\approx0.016\ \mathrm{g}} \]

Step 3: Calculate the signed error.

Using the specified sign convention:

\[ e=m_{\mathrm{measured}}-m_{\mathrm{reference}} \]

\[ e=10.24-10.00 \]

\[ \boxed{e=+0.24\ \mathrm{g}} \]

The positive sign means the balance consistently overestimates the reference mass.

Step 4: Interpret the results.

The standard deviation is small compared with the mean, showing that the repeated readings are relatively closely grouped.

However, the mean is \(0.24\ \mathrm{g}\) above the reference value.

The reference mass itself has an uncertainty of only \(\pm0.01\ \mathrm{g}\).

The balance is therefore producing a systematic overestimate.

It would be incorrect to claim that the total measurement uncertainty is simply \(\pm0.016\ \mathrm{g}\) because that number describes the spread of the readings, not the calibration error.

The instrument needs to be checked and recalibrated before its measurements can be trusted.

Question 4

A student investigates the period of a pendulum by timing ten complete oscillations.

Five measurements produce the following times:

The stopwatch displays increments of \(0.01\ \mathrm{s}\).

An independent assessment of the student’s start-stop timing method estimates an additional uncertainty of \(\pm0.20\ \mathrm{s}\) for the full timing interval. Assume this contribution is independent of the observed trial-to-trial spread.

For this question, combine these two absolute uncertainty estimates conservatively by addition. The \(\pm0.20\ \mathrm{s}\) estimate already includes the stopwatch display limitation.

Calculate:

a) The mean time for ten oscillations.

b) The mean period of one oscillation.

c) The half-range uncertainty in the period.

d) The final period and uncertainty, including the additional timing uncertainty.

Solution 4

The mean period is \(1.926\ \mathrm{s}\), and the final conservatively estimated result is:

\[ \boxed{T=1.93\pm0.03\ \mathrm{s}} \]

The stopwatch displays hundredths of a second, but the student’s timing method introduces a much larger uncertainty.

Step 1: Calculate the mean time for ten oscillations.

\[ \bar{t}= \frac{19.20+19.34+19.26+19.28+19.22}{5} \]

\[ \bar{t}=\frac{96.30}{5} \]

\[ \bar{t}=19.26\ \mathrm{s} \]

Step 2: Calculate the mean period.

The period is the time taken for one complete oscillation.

Since ten oscillations were timed:

\[ T=\frac{\bar{t}}{10} \]

\[ T=\frac{19.26}{10} \]

\[ T=1.926\ \mathrm{s} \]

Step 3: Calculate the half-range uncertainty.

The maximum time is \(19.34\ \mathrm{s}\).

The minimum time is \(19.20\ \mathrm{s}\).

\[ \Delta t_{\mathrm{spread}} = \frac{19.34-19.20}{2} \]

\[ \Delta t_{\mathrm{spread}}=0.07\ \mathrm{s} \]

This uncertainty applies to the full ten-oscillation timing.

For one oscillation:

\[ \Delta T_{\mathrm{spread}}= \frac{0.07}{10} \]

\[ \Delta T_{\mathrm{spread}}=0.007\ \mathrm{s} \]

Step 4: Include the additional timing uncertainty.

The independent timing-method uncertainty for ten oscillations is:

\[ \Delta t_{\mathrm{method}}=0.20\ \mathrm{s} \]

Using conservative addition:

\[ \Delta t_{\mathrm{total}}=0.07+0.20 \]

\[ \Delta t_{\mathrm{total}}=0.27\ \mathrm{s} \]

For one oscillation:

\[ \Delta T_{\mathrm{total}}= \frac{0.27}{10} \]

\[ \Delta T_{\mathrm{total}}=0.027\ \mathrm{s} \]

Step 5: Report the final result.

Rounding the uncertainty appropriately:

\[ \boxed{T=1.93\pm0.03\ \mathrm{s}} \]

The hundredths displayed by the stopwatch do not determine the final uncertainty.

The student’s reaction time and timing method impose a much larger limitation.

Timing more carefully, or timing more oscillations and dividing by their number, can help reduce the relative effect of reaction time. However, simply displaying more decimal places cannot fix it.

Question 5

A digital displacement sensor is used to measure the thickness of a thin metal plate.

The plate is placed between two rigid measuring jaws. The student first measures the gap between the jaws without the plate, then repeats the measurement with the plate inserted.

The sensor has the following response:

\[ R=1.002L+0.30 \]

Where:

• \(R\) is the displayed reading in millimetres.
• \(L\) is the actual separation of the jaws in millimetres.
• \(0.30\ \mathrm{mm}\) is a constant zero offset.
• \(1.002\) is a scale factor.

The two displayed readings are:

The display resolution is \(0.01\ \mathrm{mm}\).

Use an uncertainty of \(\pm0.01\ \mathrm{mm}\) for each reading. Assume the rounding uncertainties of the two readings are independent and combine their absolute bounds conservatively by addition.

Also assume:

• The same sensor and measuring orientation are used for both readings.
• The zero offset remains constant between measurements.
• The scale factor of \(1.002\) is exact for this problem.
• The plate increases the jaw separation by exactly its thickness.

Calculate the actual plate thickness and its uncertainty.

Explain why the zero offset does not simply make the calculated thickness \(0.60\ \mathrm{mm}\) too large, and why subtracting the two displayed readings without correcting the scale factor is still not exact.

Solution 5

The corrected plate thickness is:

\[ \boxed{t=5.09\pm0.02\ \mathrm{mm}} \]

The key is that the constant zero offset cancels when the two readings are subtracted, but the scale factor does not.

Step 1: Calculate the difference between the displayed readings.

Let:

• \(R_0\) be the reading without the plate.
• \(R_1\) be the reading with the plate.
• \(t\) be the actual plate thickness.

The displayed difference is:

\[ R_1-R_0=52.40-47.30 \]

\[ R_1-R_0=5.10\ \mathrm{mm} \]

A tempting first answer is therefore \(5.10\ \mathrm{mm}\).

But this ignores the scale factor.

Step 2: Apply the sensor’s response equation.

For the measurement without the plate:

\[ R_0=1.002L_0+0.30 \]

For the measurement with the plate:

\[ R_1=1.002L_1+0.30 \]

Subtract the first equation from the second:

\[ R_1-R_0 = 1.002(L_1-L_0) \]

Notice that:

\[ 0.30-0.30=0 \]

The constant zero offset cancels.

Since the plate increases the jaw separation by its thickness:

\[ L_1-L_0=t \]

Therefore:

\[ R_1-R_0=1.002t \]

Step 3: Calculate the actual plate thickness.

Rearrange:

\[ t=\frac{R_1-R_0}{1.002} \]

Substitute:

\[ t=\frac{5.10}{1.002} \]

\[ t=5.0898\ldots\ \mathrm{mm} \]

Therefore:

\[ t\approx5.09\ \mathrm{mm} \]

Step 4: Calculate the uncertainty in the displayed difference.

Each displayed reading has an uncertainty of:

\[ \pm0.01\ \mathrm{mm} \]

Using conservative addition when subtracting two independent measurements:

\[ \Delta R=0.01+0.01 \]

\[ \Delta R=0.02\ \mathrm{mm} \]

Step 5: Propagate the uncertainty through the scale correction.

The scale factor is exact in this question, so divide the absolute uncertainty by the same factor:

\[ \Delta t=\frac{0.02}{1.002} \]

\[ \Delta t\approx0.01996\ \mathrm{mm} \]

Therefore:

\[ \Delta t\approx0.02\ \mathrm{mm} \]

The final result is:

\[ \boxed{t=5.09\pm0.02\ \mathrm{mm}} \]

Why is this question more complicated than it first appears?

There are two different calibration effects.

The constant zero offset adds the same amount to both measurements. Because the same instrument is used without its offset changing, that contribution disappears when the readings are subtracted.

The scale factor behaves differently. It multiplies both actual lengths, so it also multiplies their difference. It must therefore be corrected.

If the student simply reports \(5.10\ \mathrm{mm}\), the result is approximately \(0.0102\ \mathrm{mm}\) too high.

This difference is small, but the underlying reasoning matters.

The cancellation would also fail if the zero offset changed between the two measurements. Similarly, if the scale factor were uncertain rather than exact, its uncertainty would need to be included.

The important lesson is that measurement quality depends on how the instrument behaves across the entire experiment, not just how many digits appear on its display.

08Where to go next

Once you can assign an uncertainty to each measurement, the next challenge is carrying those uncertainties through calculations.

This becomes especially important when you subtract two measurements to find a displacement, thickness, or change in length.

See How to Combine Absolute Uncertainties When Subtracting Measurements for the next step.