Drawing a graph by hand

Builds on: Sections S2 and S3
Reading time: 1 h
Exercises: 6
Olympiad problems: 1

ζ Geminorum, a star of the fourth magnitude in the middle of Gemini, is easy to see without a telescope from a dark site. Every 10.15 days it brightens and fades a little, because it is a Cepheid, a yellow supergiant star that swells and shrinks with a steady period. In 1912 Henrietta Leavitt plotted the periods of 25 such stars in the Small Magellanic Cloud against their brightness and found that the points fell along a straight line when she used the logarithm of the period. That graph, drawn by hand, gave astronomers a way to measure the distances of galaxies. In the data-analysis round the graph is also drawn by hand, on graph paper, and a large share of the marks goes to drawing it well. In this section we learn how to choose the axes and scales, plot the points with their error bars, rule the best line and read its slope, and we draw the graph for five Cepheids near the Sun ourselves.

D2.1What a graph is for

Before you readMake a guess first

A student has ten measurements that should lie on a straight line. Which gives the most reliable slope?

A graph does four jobs at once. It shows whether the data follow the model we expect, because points that should lie on a straight line either do or visibly curve away. It averages out the random errors of single measurements, because the line follows all the points together. It shows up a point that does not belong, such as a misread value or a star of a different kind, which a table of numbers hides. And the slope and intercept of the line give the quantities the problem asks for. A data-analysis problem usually asks for the graph by name, with words such as "plot a suitable linear graph", and the marks for it are given for each feature that a marker can check on the paper, from the labels of the axes to the slope.

Every graph in this section is a straight line. When the model is a curve, such as a power law or an exponential, we first change the variables so that the curve becomes a straight line, as Leavitt did by taking the logarithm of the period. Section D5 shows how to choose the variables, and the drawing is then exactly the same.

D2.2Axes and scales

Before you readMake a guess first

The values to be plotted along an axis run from 2.1 to 3.0, and the axis is 12 large squares long. What should one large square stand for?

The quantity we choose or know precisely, such as the period of a star, usually goes along the horizontal axis, and the quantity it controls, such as the brightness, goes up the vertical axis. When a problem says "plot $y$ against $x$ ", it means $y$ up and $x$ across. Each axis is labelled with its quantity and its unit, such as " $P$ (day)" or " $M_{V}$ (mag)". A logarithm has no unit, so for the logarithm of a period we write the unit inside it, as $\log (P / day)$ , which says that the period was measured in days before its logarithm was taken.

The scale of an axis is the value that one large square stands for. Graph paper is ruled in large squares of 1 cm, each divided into small squares, and a scale of 1, 2 or 5 times a power of ten per large square makes every value quick to place. A scale of 0.1 per large square puts the value 2.37 at 3.7 large squares from 2.0, a step of mental arithmetic, while a scale of 0.3 would turn it into a division by 0.3 for every point. The axis should start at a round value, a whole number of large squares from zero, but it does not need to start at zero itself. Finally the points should spread over more than half of each axis, so that each one can be read and plotted precisely. The way to meet all three rules is to take the smallest easy scale at which all the points still fit.

The graph we draw in this section plots the absolute magnitude of each Cepheid, the magnitude it would have if it were 10 parsecs away, against the logarithm of its period. The absolute magnitude measures how luminous the star really is, whatever its distance.

Example D.1Scales for the Cepheid graph

The five Cepheids in the table of Figure D.1 have $\log (P / day)$ from 0.571 to 1.551, and absolute magnitudes $M_{V}$ that reach from −2.66 to −5.49 once their error bars are included. The sheet is 12 large squares across and 16 up. Choose the two scales.

Solution

Across the sheet the points span $1.551 - 0.571 = 0.980$ . A scale of 0.05 per large square covers only $12 \times 0.05 = 0.6$ , which is too little, so we take 0.1 per large square, which covers 1.2. Starting at 0.5, the axis runs to 1.7 and holds every point, which fill $0.980 / 1.2 = 82 %$ of it.

Up the sheet the points span $5.49 - 2.66 = 2.83$ magnitudes. A scale of 0.1 per large square covers only 1.6, so we take 0.2, which covers $16 \times 0.2 = 3.2$ , and the points fill $2.83 / 3.2 = 88 %$ of the axis. Brighter stars have more negative magnitudes, and astronomers usually put them at the top, so we let the axis start at −2.6 at the bottom and fall to −5.8 at the top.

AnswerOne large square stands for 0.1 in

\log (P / day)

, from 0.5 to 1.7, and for 0.2 magnitudes in

M_{V}

, from −2.6 at the bottom to −5.8 at the top.

D2.3Points and error bars

Before you readMake a guess first

A point has $M_{V} = - 3.02 \pm 0.14$ , on a scale where one small square is 0.04 magnitudes. How far does its error bar reach?

Every point in the table is plotted, because a missing point loses marks even when it would not change the line. Each point is a small, sharp mark, such as a cross or a dot with a small circle round it, so that its centre is clear. When the table gives the uncertainty of a value, the point gets an error bar, a short line from the value minus its uncertainty to the value plus its uncertainty, with a small crossbar at each end. A value of $- 3.02 \pm 0.14$ has an error bar from −3.16 to −2.88. If the values along the other axis are uncertain too, the point gets a horizontal error bar as well. The periods of the Cepheids are known very precisely, so their points have vertical error bars only.

The absolute magnitudes in the table come from the parallaxes of the stars, and each error bar shows how far the parallax lets the magnitude be wrong. A star with a less certain parallax has a longer bar, and the line should follow the points with short bars more closely than those with long ones. Section D4 shows how the uncertainty of the parallax gives the uncertainty of the magnitude.

D2.4The best line

Before you readMake a guess first

Which line through a set of points with error bars is the best one?

The best line is one straight line, ruled with a ruler and drawn across the whole range of the data. It does not join the points from one to the next, and it does not have to pass through any of them. It should pass through the error bars of all the points, or of nearly all, with the points balanced above and below it, and it should pass through the centre of the points or very near it. A transparent ruler helps, because it lets us see the points on both sides of the line while we rule it.

ResultThe centre of the points(D.1)

\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}, \bar{y} = \frac{1}{n} \sum_{i = 1}^{n} y_{i}

$x_{i}, y_{i}$	the coordinates of point $i$
$n$	the number of points
$\bar{x}, \bar{y}$	the coordinates of the centre of the points

The best straight line through points with similar uncertainties passes through the centre of the points. The line found by the method of least squares, which Section D6 teaches, always does.

For the five Cepheids the centre of the points is at $\bar{x} = 0.925$ and $\bar{y} = - 3.74$ . The line is not forced through the origin, or through any other point, unless the model says that it must pass there. When one point lies far from the line that all the others follow, we draw the line through the others, circle the odd point and write beside it why it was left out. A point is left out only for a reason, never just because it spoils the line.

Figure D.1The period and luminosity of five Cepheidsinteractive

Star	P (day)	log P	M_V (mag)	σ (mag)
RT Aur	3.728	0.571	−2.83	0.17
FF Aql	4.471	0.650	−3.02	0.14
X Sgr	7.013	0.846	−3.64	0.13
ζ Gem	10.151	1.007	−3.93	0.14
l Car	35.551	1.551	−5.27	0.22

1 Scales
2 Points
3 Line
4 Slope

The figure shows a sheet of graph paper and the data for five Cepheids in our Galaxy, whose distances were measured from their parallaxes. The table gives each star's period

P

, its logarithm, its absolute magnitude

M_{V}

in the V band and the uncertainty

σ

M_{V}

. The four steps above the sheet go through choosing the scales, plotting each point and its error bar, ruling the best line and reading its slope, and each step is checked as a marker would check it. The data are from IOAA 2017 (Phuket), data-analysis problem 1.

D2.5Reading the slope

The slope of the line is read from two points on the line itself, never from two of the data points, which lie off the line by their errors. The two points should be far apart, at least as far as the first and last of the data, so that small errors in reading them hardly change the slope, and it helps to choose them where the line crosses a grid line, so that one of their coordinates is a round number. We write the coordinates of both points beside them on the graph, draw the triangle between them, and divide the rise by the run in the units of the axes. The slope therefore has the units of the vertical axis divided by those of the horizontal axis. Counting squares instead of values gives the right answer only when both axes happen to have the same scale, which they rarely do.

Example D.2The slope of the Cepheid line

A line drawn through the five Cepheids passes through the points (0.60, −2.93) and (1.50, −5.16). Find its slope, and say how much more luminous a Cepheid with a period ten times as long is.

Solution

The slope is the rise divided by the run,

\begin{aligned} slope & = \frac{- 5.16 - (- 2.93)}{1.50 - 0.60} \\ = \frac{- 2.23}{0.90} = - 2.48 mag . \end{aligned}

A period ten times as long has a logarithm larger by exactly 1, so the star's absolute magnitude is smaller by 2.48. Each magnitude is a factor of 2.512 in luminosity (Section O1.1), so the star is ${2.512}^{2.48} = 9.8$ times as luminous.

AnswerThe slope is −2.48 magnitudes for each unit of

\log (P / day)

. A Cepheid with ten times the period is about 9.8 times as luminous, so the luminosity of a Cepheid is nearly proportional to its period.

Example D.3A Cepheid of 20 daysfinish it

A Cepheid in our Galaxy has a period of 20.0 days. Use the line through (0.60, −2.93) with slope −2.48 to find its absolute magnitude $M_{V}$ .

The start

The logarithm of the period is $\log 20.0 = 1.301$ . Along the line, $M_{V}$ changes by −2.48 for each unit of $\log (P / day)$ , starting from −2.93 at 0.60.

Your stepNow find

M_{V}

mag

Common errorsSection D2

A scale of 3, 4 or 7 per large square turns the plotting of every point into a long division, and it is where most misplaced points come from. A scale of 1, 2 or 5 times a power of ten avoids this.
The points are not joined from one to the next. The graph needs one ruled straight line that follows all of them.
A line forced through the origin, when the model does not say that it passes there, gives the wrong slope.
The slope is read from two points on the line, far apart. Two data points, or two points close together, give a slope that carries their errors.
An axis labelled with only a symbol, such as " $P$ ", loses the mark for its label. The unit belongs on the axis too, as in " $P$ (day)".
A graph squeezed into one corner of the sheet, with the points covering less than half of each axis, cannot be read precisely and loses marks for its scale.

ExercisesSection D26 questions

D2.1
Values from 12.3 to 47.9 are to be plotted along an axis 10 large squares long. Which scale is best?
D2.2
ζ Gem has $M_{V} = - 3.93 \pm 0.14$ . On a scale of 0.2 magnitudes per large square, with 5 small squares to each large one, how many small squares long is its whole error bar?
small squares
D2.3
Find $\bar{y}$ , the mean of the five values of $M_{V}$ in the table of Figure D.1.
mag
D2.4
A student's line passes through (0.55, −2.80) and (1.55, −5.30). What is its slope?
mag
D2.5
With a slope of −2.48 magnitudes per unit of $\log (P / day)$ , how many times as luminous is a Cepheid with a period of 40 days as one with a period of 5 days?
times
D2.6
One point lies far below the line that the other nine follow. What should the student do?

Olympiad problemsPast IOAA problems, restated1 problem

2017The Cepheid relation in the infrared36.5 points

Adapted from IOAA 2017 (Phuket), data-analysis problem 1, part 1. We have restated it in our own words, kept the original data, and kept only the K band, since Section D2 works through the V band. The solution was last checked on 19 October 2026.

Five Cepheids in our Galaxy have parallaxes measured with the Hubble Space Telescope. The table gives each star's period $P$ , its mean apparent magnitude $⟨ K ⟩$ in the infrared K band, the extinction $A_{K}$ , which is the dimming of the star by dust in the K band, and its parallax $ϖ$ with the uncertainty $σ$ of the parallax. The magnitudes have negligible uncertainty.

Star	$P$ (day)	$⟨ K ⟩$ (mag)	$A_{K}$ (mag)	$ϖ$ (mas)	$σ$ (mas)
RT Aur	3.728	3.925	0.02	2.40	0.19
FF Aql	4.471	3.465	0.08	2.81	0.18
X Sgr	7.013	2.557	0.07	3.00	0.18
ζ Gem	10.151	2.097	0.01	2.78	0.18
l Car	35.551	1.071	0.06	2.01	0.20

Find the absolute magnitude $M_{K}$ of each star and its uncertainty, plot a suitable linear graph of the period and luminosity relation in the K band with error bars, and find the slope of the relation. In the original paper this part, together with the same graph for the V band, was worth 36.5 marks.

I solved it without the solution

AnswerThe absolute magnitudes $M_{K}$ are −4.19, −4.37, −5.13, −5.69 and −7.47, each uncertain by the same amount as $M_{V}$ , from 0.13 to 0.22. The slope of the relation in the K band is about −3.4 magnitudes for each unit of $\log (P / day)$ , steeper than the −2.5 of the V band. Section D3 shows how to find its uncertainty from the steepest and shallowest lines.