Altitude and azimuth

Builds on: Section 1.1
Reading time: 20 min
Exercises: 5
Olympiad problems: 0

On the evening of 15 February in Ankara, at nine o'clock, Orion stands high in the south. Sirius, the brightest star of the night sky, shines below it and a little to the left, about a third of the way up from the horizon. To show it to a friend we would say which way to face and how far up to look. Astronomers give the same two instructions as two angles, the azimuth and the altitude, and for Sirius at that moment they are 165° and 32°. These two angles are the most natural way to say where something is in the sky, and many telescopes are pointed with them. In this section we define them and see why they belong to one observer at one moment.

Figure 1.4The evening sky in altitude and azimuthinteractive

Seen from

Facing

Local time21:00

Star
Altitude
Azimuth

The sky over Ankara (39.9° N) on the evening of 15 February, drawn from a star catalogue as it looks when we face one direction. Altitude is marked up the left side and azimuth along the horizon, with grid lines every 10°. Sirius is ringed, and the dashed line drops from it straight down to the horizon. Tap any star, or choose one from the list, to read its altitude and azimuth, then move the time and watch both numbers change. In Sydney, at the same local time, Sirius is high in the north-east.

1.2.1The horizon and the zenith

Before you readMake a guess first

Out on a flat plain, how much of the celestial sphere is above the horizon?

Each way of giving positions on the sky in this chapter starts from a great circle on the celestial sphere and the two points that are 90° from every point of it, which act as the poles of the circle. On Earth the circle is the equator and its poles are the North and South Poles. For an observer standing on the ground, the natural circle to start from is the horizon.

A plumb line, a weight hanging on a string, shows which way is straight down. Extended upwards, the line meets the celestial sphere at the zenith Z, the point directly overhead, and extended downwards it meets the sphere at the nadir, directly below our feet. The plane through the observer at right angles to the plumb line cuts the sphere along a great circle, the horizon. At sea or on a wide plain the edge of the sky is very nearly this circle. Hills and buildings hide parts of the sky above it, but when astronomers speak of the horizon they mean the ideal circle 90° from the zenith. The horizon is the base circle of this system, and the zenith and the nadir are its poles.

Any great circle through the zenith and the nadir is a vertical circle. It meets the horizon at right angles, and our eyes follow one when we look straight up from a point of the horizon to the zenith. The most important vertical circle is the meridian, which passes through the celestial poles. It meets the horizon at the north point N, the point of the horizon nearest to the north celestial pole, and at the south point S opposite it. The east point E and the west point W lie halfway between them, 90° from each, with east on our right as we face north. These four points are the cardinal points, and the vertical circle through E and W is the prime vertical. Figure 1.5 shows all of them.

Quick checkUse the definitions above

The meridian and the prime vertical are both great circles, so they cross at two points. Which two?

1.2.2Altitude and azimuth

Before you readMake a guess first

Azimuth is measured around the horizon starting from north. In which direction is azimuth 90°?

Through any star X, unless it is at the zenith or the nadir, there passes exactly one vertical circle. It meets the horizon at the point F directly below the star, or directly above it if the star is below the horizon, and we call F the foot of the star's vertical circle. Two angles then fix the direction of X. The altitude $h$ is the arc FX, the angle of the star above the horizon. It runs from 0° on the horizon to 90° at the zenith, and it is negative for an object below the horizon, down to −90° at the nadir. The azimuth $A$ is the arc of the horizon from the north point to F, measured through the east, so that north is 0°, east 90°, south 180° and west 270°. In place of the altitude we sometimes give the zenith distance $z$ , the arc ZX from the zenith down to the star.

ResultZenith distance(1.4)

z = 90 ° - h

$z$	zenith distance, from 0° at the zenith to 180° at the nadir
$h$	altitude, from −90° at the nadir to +90° at the zenith

Some problems call the zenith distance the zenith angle.

Figure 1.5Altitude and azimuth on the celestial sphereinteractive

Azimuth

A

Altitude

h

Where to look
Zenith distance

The celestial sphere around the observer O, seen from outside. The horizon carries the cardinal points N, E, S and W, Z is the zenith, and the thin circle through N, Z and S is the meridian. Parts of circles on the far side of the sphere are dashed. The star X lies on its own vertical circle, the thin circle through Z and X, which meets the horizon at F. The heavy arcs are the azimuth

A

, measured along the horizon from N through E to F, and the altitude

h

, measured up from F to X. Try , , , and .

A telescope on the simplest kind of mount, an altazimuth mount, turns about a vertical axis and tilts about a horizontal one, so it moves in azimuth and in altitude. A ship's navigator who measures the height of a star above the sea with a sextant is measuring its altitude, apart from small corrections for the dip of the horizon, which we meet at the end of this section, and for refraction.

Quick checkUse the definitions above

A star is due south-east, halfway up from the horizon to the zenith. What are its azimuth and altitude?

NoteSome books, and some olympiad papers, measure azimuth from the south point through west instead, so that south is 0° and north 180°. The planetarium round of IOAA 2011 did this, and so did one theory problem at IOAA 2013, while most IOAA papers measure from north through east and say so. Always use the convention the problem states.

1.2.3The horizon system belongs to the observer

Before you readMake a guess first

At the same moment, one friend in Istanbul and another in Ankara both look at Sirius. Do they see it at the same altitude and azimuth?

The horizon, the zenith and the cardinal points are tied to the observer. Move to another place on Earth and the zenith points in a new direction, taking the whole grid of altitude and azimuth with it. Stay in one place and the grid stays put, but the sky turns past it, so the altitude and azimuth of every star change from minute to minute, as Figure 1.4 shows. The horizon system is the right one for pointing a telescope or telling a friend where to look. It is no use for a star catalogue, which needs coordinates that are the same for every observer on every night, and Section 1.3 builds such coordinates from the celestial poles and the celestial equator.

NotePanoramas like Figure 1.4 have appeared in the observation round. At IOAA 2017 in Phuket, one station showed a 360° panorama of the night sky over an unknown place, and the task was to estimate the local sidereal time, which is defined in Section 1.3.

Altitude and azimuth also behave differently from the two coordinates of a flat map. The lines of equal altitude, apart from the horizon itself, are small circles, and they shrink towards the zenith in the same way as the parallels of latitude shrink towards Earth's poles. Near the zenith, two stars can differ in azimuth by a large angle and still be close together in the sky.

Example 1.4Two stars, opposite in azimuth

Two stars both have an altitude of 60°. One is due east and the other due west. How far apart are they in the sky?

Solution

Both stars lie on the prime vertical, the vertical circle through E, Z and W. Each is $90 ° - 60 ° = 30 °$ from the zenith, one on the eastern side and the other on the western side. The arc between them along the prime vertical passes through the zenith and is $30 ° + 30 ° = 60 °$ long. The prime vertical is a great circle, so this arc is the angular distance between the stars.

Answer60°, although their azimuths differ by 180°.

Quick checkUse the example above

Two stars both have an altitude of 80°, one due north and the other due south. What is the angular distance between them?

degrees

1.2.4The horizon seen from a height

Before you readMake a guess first

Standing on a cliff 100 m above the sea, how far away is the horizon?

So far the horizon has been the great circle 90° from the zenith, and on flat ground this is what we see. From a mountain, a tower or an aircraft, the edge of the land or sea lies below this circle, because Earth curves away beneath us. To tell the two apart, the circle 90° from the zenith is called the astronomical horizon, and the edge of the land or sea is the visible horizon. The angle by which the visible horizon lies below the astronomical horizon is the dip of the horizon.

Figure 1.6The dip of the horizon

A section through Earth, with centre C and radius R. The observer O is at a height H above the surface, greatly exaggerated in the drawing. The horizontal through O is at right angles to the vertical line CO. The line of sight from O to the visible horizon just grazes the surface at T, a distance

ℓ

from O, so it is at right angles to the radius CT. The dip

θ

is the angle between the horizontal and the line of sight, and the same angle appears at C.

In Figure 1.6 the line of sight touches Earth at T, and a line that just touches a circle is at right angles to the radius there, so the triangle OTC has a right angle at T. The horizontal at O is at right angles to CO, and the line of sight lies $θ$ below it, towards C, so the angle at O between the line of sight and CO is $90 ° - θ$ . The three angles of the triangle add up to 180°, which leaves an angle $θ$ at C. In the same triangle the hypotenuse is CO $= R + H$ and the side next to the angle at C is CT $= R$ , which gives the dip.

ResultDip of the horizon(1.5)

\cos θ = \frac{R}{R + H}

$θ$	dip of the horizon, the angle of the visible horizon below the horizontal
$H$	height of the eye above the ground or the sea
$R$	radius of Earth, 6371 km

When

H

is much less than

R

, which is true for any mountain or aircraft, the dip in radians is very nearly

θ = \sqrt{2 H / R}

Show the derivation

For a small angle $\cos θ \approx 1 - θ^{2} / 2$ , with $θ$ in radians. When $H$ is much less than $R$ ,

\frac{R}{R + H} = \frac{1}{1 + H / R} \approx 1 - \frac{H}{R} .

Setting the two equal gives $θ^{2} / 2 = H / R$ , so $θ = \sqrt{2 H / R}$ .

The distance from the eye to the visible horizon is the side OT of the same triangle. By Pythagoras' theorem $ℓ^{2} = (R + H)^{2} - R^{2} = 2 R H + H^{2}$ , and for any height we can reach the last term is tiny compared with the first.

ResultDistance to the horizon(1.6)

ℓ = \sqrt{2 R H + H^{2}} \approx \sqrt{2 R H}

$ℓ$	distance from the eye to the visible horizon

Both results ignore refraction, which bends light from near the horizon so that it follows Earth's curve a little. In practice the dip is slightly smaller and the horizon slightly further away. Olympiad problems usually say to ignore this.

Example 1.5The horizon from an airliner

An airliner cruises 11 km above the sea. How far below the horizontal is the horizon seen from a window, and how far away is it? Take $R = 6371$ km.

Solution

By (1.5),

\cos θ = \frac{6371}{6371 + 11} = 0.99828,

so $θ = 3.36 °$ . The approximate form gives almost the same, $θ = \sqrt{2 \times 11 / 6371} = 0.0588$ rad $= 3.37 °$ . By (1.6) the horizon is

ℓ = \sqrt{2 \times 6371 \times 11 + 11^{2}} = 375 km

away.

AnswerThe horizon is 3.4° below the horizontal and 375 km away. From the aircraft we can see stars down to 3.4° below the astronomical horizon.

Quick checkUse the result above

What is the dip of the horizon from the top of a cliff 100 m above the sea, in arcminutes?

arcminutes

Example 1.6A dip of one degreefinish it

How high must we climb for the horizon to dip by 1°?

The start

Rearranging the approximate form of (1.5) gives $H = R θ^{2} / 2$ , with $θ$ in radians. One degree is $π / 180 = 0.017453$ rad.

Your stepFinish it: find

H

in metres.

NoteOlympiad problems use the dip of the horizon when a sunset is timed from a height, as in “Height of a hill” at the end of Section 1.3, and when a mountain shows extra stars, as in “Stars from Mount Damavand” in Section 1.4.

Common errorsSection 1.2

Measuring azimuth from the wrong point or in the wrong direction. In this course, as in most IOAA problems, azimuth runs from north through east, so west is 270°, not 90°. If a problem measures azimuth from the south point through west, each of its azimuths differs from ours by 180°.
Mixing up altitude and zenith distance. A star 20° above the horizon has a zenith distance of 70°.
Treating a difference in azimuth as an angular distance. Even for two stars at the same altitude, the two are equal only on the horizon. Higher up, the same difference in azimuth is a smaller distance on the sky, and near the zenith a very small one, as Example 1.4 shows.
Putting $H$ and $R$ in different units in (1.5), or forgetting that the approximate form $\sqrt{2 H / R}$ gives the dip in radians.

ExercisesSection 1.25 questions

1.2.1
What is the azimuth of the north-west point of the horizon?
degrees
1.2.2
A vertical stick 1.00 m tall stands on level ground and casts a shadow 1.50 m long. What is the zenith distance of the Sun?
degrees
1.2.3
Two stars on the horizon have azimuths of 350° and 10°. How far apart are they?
1.2.4
The highest viewing deck of the Burj Khalifa in Dubai is 555 m above the ground, which there is close to sea level. What is the dip of the horizon seen from the deck, in arcminutes?
arcminutes
1.2.5
Seen from Ankara through a whole night, which coordinates of Polaris stay almost the same?