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D1

The data-analysis round

Builds on
Sections S2 and S3
Reading time
35 min
Exercises
6
Olympiad problems
0

The Sun sometimes throws a cloud of magnetised gas out into space, a coronal mass ejection, and when one reaches Earth a day or a few days later it can disturb satellites and power grids. At IOAA 2025 in Mumbai, one data-analysis problem gave the speeds of ten such clouds as they left the Sun and as they arrived at Earth, together with the time each took to cross. From these numbers the contestants built a simple model that predicts when a storm will arrive, the kind of forecast that space-weather services make every day. That one problem has everything the data-analysis round asks for, with real measurements, a model, a graph and a result with its uncertainty. In this section we describe the round, set out a plan that works for almost every problem in it, and look at the errors that every measurement carries.

D1.1The round

Before you readMake a guess first

A data-analysis paper of four hours is worth 150 marks. Roughly how long is there for each mark?

Section S1 describes the round in outline. In recent years the marks of its paper have added up to 150, as in 2019, 2022, 2024 and 2025, which leaves about 1.6 minutes for each mark, and Section S2 shows how to plan the time by marks. Each problem gives a table of measurements, or an image or a spectrum to measure, and asks for a chain of results, usually a quantity worked out for every row of the table, a graph, the slope and intercept of a line with their uncertainties, and a final value with its uncertainty.

The marks are given step by step, as in the theory round. A careful table and graph earn their marks even when the final number comes out wrong, and a final number with no graph or working earns very few. A data-analysis problem rewards an orderly way of working more than a clever idea.

D1.2A plan for every problem

Before you readMake a guess first

What should come first when we start a data-analysis problem?

Almost every data-analysis problem can be worked through with the same plan, and each step earns its own marks.

  1. We read the whole problem, find the quantity that the last part asks for, and write down the relation that links the data to it.
  2. We choose the variables that make the relation a straight line, which Section D5 explains.
  3. We add columns to the table for these variables and their uncertainties, with the units in the headings.
  4. We draw the graph by hand, as in Section D2.
  5. We read the slope and intercept of the line and their uncertainties, as in Section D3.
  6. We turn the slope and intercept into the quantity asked for and carry the uncertainties through, as in Section D4.
  7. We finish with a sentence on whether the result is reasonable and what limits its accuracy.

The plan is the method of Section S2 applied to data. Its first step is the one most often skipped under time pressure, and skipping it costs the most, because the rest of the work depends on it.

D1.3Errors in a measurement

Before you readMake a guess first

A stopwatch is started two seconds late every time, so that every time it gives is two seconds short. What kind of error is this?

Every measurement differs a little from the true value, and the difference has two kinds of cause. A random error pushes a reading up one time and down the next, like the varying reaction time of the person pressing a stopwatch, so repeating the measurement and averaging, or drawing a line through many points, makes it smaller. A systematic error pushes every reading the same way, like a stopwatch started late, and no amount of repetition removes it. It shows up only when we compare with something we trust, such as a standard star of known brightness.

A graph treats the two kinds differently. Random errors scatter the points on both sides of the line, and the line averages them out. A systematic error that adds the same amount to every value plotted up the graph moves the whole line up or down, so it changes the intercept and leaves the slope alone, while one that multiplies every such value by the same factor changes the slope as well. This is one reason why many problems are set up so that the answer comes from a slope.

Reading a scale adds a small error of its own. The reading error of a scale is usually taken as half its smallest division, so a length read from a ruler marked in millimetres is uncertain by 0.5 mm. Finally, a result can only be as precise as the data. A product or a quotient keeps as many significant figures as the least precise value in it, while a sum or a difference keeps as many decimal places, so a difference of two close values can lose most of its figures. In the problem below, CME-F leaves the Sun at 347 km/s and reaches Earth at 350 km/s, a change of only 3 km/s that is known to a single figure, so its acceleration is far less certain than the others. In the working we keep a figure or two more than this, so that rounding does not pile up (Section S2).

Quick checkA reading error

The distance between two stars on a photograph is measured with a ruler marked in millimetres. How uncertain is the reading?

mm

D1.4From textbook to olympiad

The first part of the problem on solar storms from IOAA 2025 asks for the average acceleration of each coronal mass ejection, its change of speed divided by the time it took. It looks like one line of school physics, and the only real difficulty lies in the units.

Example D.1The acceleration of a solar storm

A coronal mass ejection, CME-A, leaves the Sun at 804 km/s and reaches Earth 74.5 hours later at 470 km/s. What is its average acceleration in m s⁻²?

Solution

The change of speed is 470−804=−334 km/s, which is −334000 m/s, and the time is 74.5×3600=268200 s. The average acceleration is therefore

a=−334000 m/s268200 s=−1.25 m s−2.
AnswerThe average acceleration is −1.25 m s⁻². It is negative because the cloud slows down on its way, held back by the slower wind of gas that streams out from the Sun.

The same arithmetic for all ten clouds gives the last column of the table below.

Cloudu (km/s)v (km/s)τ (h)a (m s⁻²)
CME-A80447074.5−1.25
CME-B247360127.50.246
CME-C523396103.5−0.341
CME-D83041571.0−1.62
CME-E665400104.5−0.704
CME-F347350101.50.008
CME-G44637599.5−0.198
CME-H15536097.00.587
CME-I101651567.0−2.08
CME-J68341054.0−1.40

The fast clouds slow down and the slow ones speed up, which is what we expect if the solar wind drags every cloud towards its own speed. The problem's model, a=m(u/u0)+α, measures the speed u in units of u0=1000 km/s, so that CME-A, for example, has u/u0=0.804. The model is a straight line in a graph of a against u/u0, so the plan of this section leads straight to a graph, its slope m and its intercept α. Section D6 restates the whole problem and fits the line by least squares.

Example D.2A cloud that speeds upfinish it

CME-H leaves the Sun at 155 km/s and reaches Earth 97.0 hours later at 360 km/s. Find its average acceleration.

The start

The change of speed is 360−155=205 km/s, or 205000 m/s, and the time is 97.0×3600=349200 s.

Your stepNow find a in m s⁻².
m s⁻²
Common errorsSection D1
  1. A change of speed in km/s divided by a time in hours gives an acceleration in km s⁻¹ h⁻¹, which is not m s⁻². Converting both to SI units first, metres per second and seconds, avoids the mistake.
  2. Rounding each intermediate result to two or three figures lets the rounding errors add up. The working keeps a figure or two more than the data, and only the final answer is rounded.
  3. Averaging many readings does nothing about a systematic error. A stopwatch started late gives the same wrong time however often it is used.
  4. A column in a table without its unit in the heading cannot be checked, and it loses the mark that the table carries.
  5. A graph drawn before the relation is worked out often plots the wrong variables, and the time spent on it is lost.
ExercisesSection D16 questions
  1. D1.1

    A part of a data-analysis problem is worth 12 marks. At 1.6 minutes for each mark, about how long does it deserve?

    minutes
  2. D1.2

    CME-I leaves the Sun at 1016 km/s and reaches Earth 67.0 hours later at 515 km/s. What is its average acceleration?

    m s⁻²
  3. D1.3

    How many m s⁻² is an acceleration of 1 km/s per hour?

    m s⁻²
  4. D1.4

    Every distance in a table is too large by the same 2 km, because of an error in the instrument. Which part of a straight-line graph of distance against time does this change?

  5. D1.5

    Repeated measurements of the brightness of a star scatter by a few hundredths of a magnitude about their mean. What kind of error is this?

  6. D1.6

    A distance and a time are each given to three significant figures. To how many figures should the speed worked out from them be given?