Last update: 31 May 2013

Quick note on the questioner: their identity is not revealed but think of him/her as a friendly alien, who may or may not be more intelligent than us. Or possibly a pet.

**So what do we see in the sky?**

Stars, sun, moon and planets - we can see the planets out to Saturn without a telescope (Mercury, Venus, Mars, Jupiter, Saturn). There are some other things, especially once you start using telescopes. For now, though, we're concentrating on using the movement that we see in the stars, sun, moon and planets to understand what's happening in the solar system. We'll start with the stars.

**Does it matter where we're looking from?**

Yes. You'll see different behaviour at different latitudes; and also longitudes in terms of the time at which things happen, but the longitude doesn't affect the general behaviour. When we want a specific example we'll take Winchester, UK - approximate latitude 51°N, approximate longitude 1.3degW (or 51.06 and 1.32 to be more accurate). You can also see more with clear air and low light pollution which is why the best observatories are on top of remote mountains. Now if you go out into space then it's even clearer and you don't have the atmosphere and the Earth isn't in the way of half the sky ... but leave that for now, we're talking about what people have seen for thousands of years.

**Since you mention them, can you explain latitude and longitude please?**

Sure. Greek philosophers seem to have believed from two and a half thousand years ago that the Earth is spherical, and it was as long ago as the first century that Ptolemy published an atlas with places located by latitude and longitude co-ordinates. There is a myth that Columbus 'proved' the world was round by his four voyages between 1492 and 1503. He was attempting to reach the East Indies (SE Asia, east of India) by sailing west from Spain to save going round Africa on the eastward route. He actually found the Americas (which he insisted was the East Indies). The first circumnavigation of the world was led by Magellan from 1519 to 1522, although he died on the way; of the five ships that set out only the Victoria made it all the way - the voyage is a fascinating tale of politics, mutiny, battles, shipwrecks and pursuits. However, the sphericity of the earth was widely accepted by scientists well before this - for example deduced by the way ships slowly disappear over the horizon or by the circular shadow of the Earth on the moon during a lunar eclipse (we'll discuss that when we talk about the moon).

**Latitude lines** are parallel to each other and range in value from 90°N at the North Pole to 0° at the equator to 90°S at the South Pole, the angles being measured from the centre of the Earth to the N or S of the equator. The 90° latitude (at either pole) is a single point, the length of the 0° latitude line (the equator) is about 40,000 km (25,000 miles), and the other latitudes have lengths somewhere inbetween. As we'll find out the Earth rotates about an axis through the earth between the N and S Poles, and the Equator is the line of latitude half way between the two poles. Now even without knowing this, there are good astronomical reasons to draw latitudes as they are - for example, on any day the length of day or the highest altitude of the sun is the same for all places with the same latitude. The equator has all days of length 12 hours (they're a few minutes longer actually due to various physical effects), whereas everywhere else has longer days in summer and shorter days in winter; it also experiences the quickest sunrise and sunset because the sun rises and sets more or less vertically. There are four other special lines of latitude - the Arctic Circle, the Tropic of Cancer, the Tropic of Capricorn, and the Antarctic Circle (approximately 66.5°N, 23.5°N, 23.5°S and 66.5°S respectively). Anyway, more on the significance of specific latitudes later when we talk about the sun.

**Lines of longitude (or meridians)** run from N Pole to S Pole perpendicular to the lines of latitude, and define the east-west position. Longitudes range between 180°E and 180°W, with the central 0° longitude line placed near the Royal Observatory, Greenwich, London (this is an arbitrary point, chosen through history - there is nothing special about 0° longitude). The angles are from the centre of the Earth to E or W of the Greenwich meridian. Each line of longitude has the same **local time** - that is, the instant that the sun reaches its highest altitude (which is the definition of local noon) is the same for all places on that longitude. As you travel each 15° of longitude east, the local time will be one hour later. Hourly **time zones** are thus defined in 15° longitude blocks...although it doesn't work out exactly like that. The time zone boundaries are adjusted by countries to ensure the time is the same in a country or a region; in addition some countries define **daylight savings** by advancing clocks an hour in spring and then back in autumn - this ensures more daylight in the evening and less in the morning. Now if you go 360° right round the globe back to where you are standing the time will be 360/15 = 24 hours later - which of course doesn't make sense and this is why there is an **International Date Line** at 180° of longitude. Similar to time zones, the International Date Line weaves it way between countries rather than being exactly on 180°. Crossing to the east means a day is subtracted, whereas a day is added crossing to the west. In the UK the time zone is Greenwich Mean Time (GMT), with British Summer Time (BST) an hour ahead between 1:00 AM the last Sunday of March and 1:00 AM the last Sunday of October. GMT provides the 'mean solar time' at Greenwich: that is, on average noon will be when the sun is at its highest at Greenwich in this time zone. It is only an average because various effects like the variable speed of the Earth (we'll discuss that later) mean the time of the sun's highest altitude at Greenwich has to vary slightly from noon on a day to day basis or else some days would be longer than others. GMT is also very close to Coordinated Universal Time (UCT) which is defined as a standard time for Earth (regardless of time zone) and is calculated using atomic clocks.

Now for accurate mapping it's more complicated than that, because the Earth is not a perfect sphere; in fact it's not even a perfect ellipsoid. To measure real latitude and longitude on the Earth a **geodetic datum** is defined. A datum is a set of values used to define a coordinate system. For instance WGS84 is used by GPS and is the most widely used global datum: it is defined by a reference ellipsoid (precise parameters of the ellipsoid are defined to best fit the Earth's shape); reference positions on the Earth defined with specific latitude and longitude values to 'anchor' the ellipsoid to the Earth; and a reference surface to model sea level (so that a third coordinate of height above sea level can be defined to measure depths or altitudes). Different datums will have slightly different coordinates for the same place - software exists to convert coordinates between datums (you can find programs online). Local datums can offer a better fit to the Earth than a global datum because they just have to be accurate over a smaller area. For instance Ordnance Survey maps used in the UK use the OSGB36 datum. The same coordinates in WGS84 and OSGB36 will be different places - up to about 100m apart.

Right I'm getting a bit carried away now, but these diversions allow us to explore side issues that catch our interest before I get back to finding the secret of the universe. Let me finish with a mention of **mapping**, and **northings and eastings**. Positions on the curved surface of the Earth can't be exactly drawn onto a flat two dimensional map. A mathematical projection is defined and this will distort the picture - either shapes will be inaccurate (for example, of a country) or the scale will be inaccurate (for distance and area - it will be different at different parts of the map) or directions (the bearing or angle) between points will be inaccurate; or possibly all three. As an example, this is why Greenland looks massive on a world map. However, you can be quite accurate if you restrict your map to a reasonably narrow area - so an accurate projection is defined in one area, and a different projection in a neighbouring area. The most standard way of doing this is to use a **Universal Transverse Mercator (UTM) projection**. This is a series of 60 projections each with a 6° wide longitude band; between them they cover the Earth (except for high latitudes near the Poles). The **central meridian** of each projection defines the projection and is the central longitude line of the area covered, for instance at 9°W, 3°W, 3°E, 9°E, etc. Maps are then drawn for a particular area, using the relevant projection, e.g. UTM with central meridian 3°W (the origin of the mathematical projection is then 3°W, at the equator). The maps are drawn with gridlines to pinpoint positions - to locate a position eastings (a value in metres east of the origin) and northings (the value north of the origin) are used, the eastings being given first. The origin is the central meridian (which is given a value of 500,000 eastings - so you don't get negative numbers as you go to the west) and the equator (given a value of 0 for points in the northern hemisphere; and 10,000,000 in the southern hemisphere - again so you don't get negative numbers in the south). UTM projections preserve shape and direction fairly closely but distort distance and area; they are more accurate near the central meridian; and are not applicable near the Poles (the UTM is valid from 80°S to 84°N - a different type of projection is generally used for mapping polar regions).

**Grid north** is the (up) direction of the northing grid lines on the map and this will vary slightly from **true north** (which points at the N Pole) except at the central meridian where they are equal. The difference is called **grid convergence** (or just convergence) which is defined as the bearing from true north to grid north - this will vary from point to point. This is measured as a bearing (or angle) which is positive to the east (or clockwise). For the UTM projection convergence is positive to the east of the central meridian and negative to the west; this means true north is to the west (or left) of grid north on the eastern side of the central meridian and to the east (or right) on the western side. A grid bearing is a bearing measured clockwise from the northing grid lines on a map with 0° being in the direction of grid north; for instance a direction of SW on the map is equivalent to a grid bearing of 135°. Since true north deviates from grid north by the convergence, then the true bearing at a point will deviate by the same amount from the grid bearing and they will be related by the formula *true bearing = grid bearing + convergence* (e.g. the true bearing is SW or 135°, convergence is +2° and therefore the grid bearing as measured on the map is 137° - slightly south of SW).

In addition to the variance between grid north and true north, if you are trying to navigate using a compass there is a different and third north - **magnetic north**, which is the direction a compass points. There is a Magnetic North Pole - somewhere in the Canadian Arctic at the moment - and a Magnetic South Pole, but a compass needle doesn't point exactly to the Magnetic North Pole; it varies according to the local magnetic field. Both the Magnetic North Pole and the local direction of a compass needle change over time. We are not going to get into the details of the Earth's magnetic field now! Suffice to say that at any point and date there is a **magnetic declination (or magnetic variation)** which is the bearing from true north to magnetic north, positive to the east (or clockwise). You can look this up on various websites - for instance, the Mag Dec Calculator at the US National Geophysical Data Center tells us that the magnetic declination for Winchester on 4 Mar 2013 is 1.62°W changing
by 0.15°E per year. Now if you are trying to navigate accurately by map and compass, you don't need the difference between magnetic north and true north, you need the difference between magnetic north and grid north - which you can work out by the formula * magnetic grid variation (also called grid magnetic angle) = magnetic declination - convergence* - this gives you the bearing from grid north to magnetic north. As ever with bearings this is positive to the east (or clockwise). I'm going to reiterate the ins and outs of this formula because it's easy to mix the signs up: magnetic grid variation and magnetic declination and convergence are all bearings (which means they are positive to the east and negative to the west); magnetic declination is from true north to magnetic north, convergence is from true north to grid north and magnetic grid variation is from grid north to magnetic north. You have to get the signs right in the formula! Maps will usually tell you the magnetic grid variation for the year the map is published and the rate of change (the rate of change will be the same as for magnetic declination), so you don't need to worry about working it out from the magnetic declination and convergence. Quite often you don't need to worry about the difference between grid and compass (or magnetic) bearings because the difference may only be a few degrees, but sometimes this will be important...in which case, here's how you swap between magnetic bearings and grid bearings when you are trying to navigate by map and compass:

Finally a brief word about the UK and the **Ordnance Survey** which uses the **National Grid** to cover the whole of Great Britain with its maps. The Ordance Survey uses a Transverse Mercator projection for its maps, but it is not a UTM projection because its central meridian is non-standard at 2°W and the latitude of the origin is 49°N. To avoid negative numbers a 'false origin' is defined 400km to the east of 2°W and 100km to the north of 49°N - that is, somewhere to the SW of the Scilly Isles. All of Great Britain and its islands (not Ireland though) are covered by the Ordnance Survey maps which extend 700km to the east and 1300km to the north of the false origin. As an example, Winchester Cathedral has **grid reference** approximately 448500 129500. Now that's to the nearest metre, so unless you were being very accurate you might truncate this to 44850 12950 (which defines a square with 10mx10m square) or 4485 1295 (which defines a 100m square) - using less than six digits tells you how accurate you are being (you have to use the same number of digits for the easting as the northing). The Ordnance Survey allow a further abbreviation by defining 'lettered' 100km squares which are specified on maps - e.g. the square with bottom left corner at 400000 100000 is SU, so the 4 and 1 can be left off the grid reference and Winchester Cathedral becomes SU485295 (to the nearest 100m) - note you typically don't put a space between the easting and northing values. The grid convergence for Winchester is approximately 0.5°, so true north is very slightly west of the north grid lines. You will see this grid convergence value stated on the Winchester Ordnance Survey map. As above we saw the magnetic declination for Winchester is 1.62°W (in 2013), so the magnetic grid variation will be -1.62 - 0.5 = -2.12, or 2.12°W (moving east by 0.15° per year). And indeed this is what is says on the map - well actually it was published in 2006 and it says 'magnetic north is 3.15°W of grid north moving 0.15°E per year', which works out more or less the same. So if you walk directly north according to your compass, you will be walking approximately 2° to the west on the map; and if you try and walk exactly along a north grid line, you will need to set your compass to 2°E. Try not to get lost.

As a summary, you're likely to use latitude and longitude for global activities such as global navigation or relating your position to the stars you are likely to see; and northings and eastings for local activities such as travel within a country or finding your way on a mountain. You may also need to take into account the variations between magnetic, grid and true north for accurate navigation. Now the technicalities and mathematics of goedetic datums, map projections, coordinate transformations, accurate mapping, GPS positioning, the Earth's magnetic field and general manipulation and storage of geographic data is a large subject and many companies work in these areas; for instance in GIS (Geographic Information Systems) software, mapping, surveying or GPS applications. So we'll leave it right there and get back to the main subject!

**Zzzzz. Only kidding - shall we get back to the stars then? What do we see and how do they move?**

**Ok, a question: the stars rise in the east and set in the west - is that exactly east and west?**

No. To answer this best I'm going to introduce **celestial coordinates**, the coordinate system we use for stars. Stars are considered as being on the underside of the fictitious celestial sphere I just mentioned; that is they are on the interior of a sphere that is centred on the Earth and has enormous (or infinite) radius. Note that this implies all the stars are the same distance from us, which is not true; however they are so far away we don't have any depth perception to tell which are closer or further from us - we'll get to how to estimate star distances later. We define a star's location by two celestial coordinates: **1.** **The declination (dec)**. This is the angle of the star (from the centre of the Earth) north of the celestial equator, where the celestial equator is the projection of the Earth's equator on the celestial sphere. So if the star is on the celestial equator the declination is 0, if it is directly above the N Pole it is +90 and if it is directly above the S Pole it is -90. **2.** **The right ascension (RA)** is the angle eastwards from the **vernal equinox**, although it is usually measured in time units - from 0 hours to 24 hours (equivalent to 0° to 360°, so 1h is 15° and for example, an RA of 6^{h} 30^{m} is the same as 15 * 6.5 = 97.5°.). We'll get to what the vernal equinox is later - it is only the declination that is important to answer the question. Celestial coordinates are analogous to latitude and longitude on the Earth. For any visible star you can look it up in a star catalogue - e.g. the Bright Star Catalogue - and you'll find its RA and declination.

**Whoa there! How do I find a star with, say, declination 25° and RA 11 ^{h}? How do I locate the celestial equator and the vernal equinox (whatever it is), and then find the star?**

Let's give some **terms and definitions (and consequences and analysis)** first, in a sensible order, then get back to your question (and the previous one). I'll repeat a couple we've already mentioned to put these all in one place. This actually ends up considerably longer than I intended, but it gives a lot of background for our subsequent discussions on observational astronomy and will keep those shorter (in theory). And it's mostly fascinating!

**A note on units - degrees, minutes, seconds and hours, minutes, seconds:**The angular measure of degrees specifies that 360° is one full rotation. Divisions of one degree are minutes (') equivalent to 1/60° and seconds (") equivalent to 1/60'. You can also specify the angle as a decimal, so for example 24.5° is equivalent to 24° 30'. Daily time is measured in hours, minutes (1/60 an hour) and seconds (1/60 minute) with one day equal to 24 hours, as everyone is aware. These units can be abbreviated hr, min, sec or h, m ,s - we'll use the latter, e.g. 3h 5m 6s. Now hours, minutes and seconds can also be used as an angular measure in astronomy, for example for right ascension. In this case 24 hours is equivalent to 360° and hence one hour is equivalent to 360/24 = 15° , one minute is equivalent to 15/60 = 0.25° and one second is equivalent to 0.25/60 = 0.0042° (roughly). The abbreviations for hours, minutes and seconds when they are used as angles are superscripts, so for example a right ascension is expressed as 6^{h}30^{m}30^{s}(which is the same as 6^{h}30.5^{m}or even 6.508^{h}(roughly)); converting this to degrees would give 97.625° or 97° 37.5' or 97° 37' 30".**Spherical Astronomy:**This is the discipline that is used to derive the positions of the stars and other celestial objects on the celestial sphere, as seen from the Earth. It is used to calculate what stars can be seen from a particular point on the Earth, how they move around the sky, the times that they rise and set; and similarly for the Sun and other celestial objects, including the timing of events like eclipses. Initially assume that the Earth is spherical and rotates in relation to the fixed stars; and that the stars, sun, moon and other bodies can be observed and their future positions predicted. Star positions are published in star catalogues and the positions of other bodies are published in what is called an astronomical ephemeris which can give both past and predicted future positions. Once the celestial position of a star is observed and catalogued, then the positional properties of that star as seen from Earth - when it rises, where you can see it from, etc. - can be derived using straightforward geometry and trigonometry. Similarly for the Sun and other bodies given an ephemeris which provides their positions at particular times and a way of predicting their movement. The basic point I want to make here is that the calculation of sunrises, what stars can be seen and when, and similar details, uses fairly simple mathematics that has been around for a long time (although it may get very involved sometimes!). So when I say that e.g. the altitude above the horizon of the fixed North Celestial Pole is equal to the observer's latitude you can work this out by drawing the picture and doing some trigonometry, or you can look up the derivation on the web...or you can take my word for it! It can be a bit difficult to visualise sometimes because your local horizon is a tangential plane on the Earth's surface. A big 3D model or even better a holographic projection that you could manipulate would make it easier - but if you try and visualise yourself standing on a rotating Earth or hold a tennis ball and spin it while moving it around a football you can hopefully see the gist of what's happening. I won't very often try and write the mathematical derivations out because that will distract us from finding the secrets of the universe, mostly I'll just give a nod and say 'from spherical astronomy' to indicate this is based on established mathematical disciplines. Now the results quoted won't usually be exact since they won't take a number of subtle factors into account - for example, that the Earth is roughly ellipsoidal or the effects of parallax or the effects of the Earth's atmosphere or oddities in the orbits of the Earth or other bodies or starlight aberration (and more!). Taking all this into account can make the maths much more complicated (especially concerning orbital motions where the full power of modern numerical methods is employed). We'll mention all these by the by.**Great circle:**This is a mathematical term defined as the intersection between a sphere and a plane that passes through the centre of the sphere. Any great circle is a circle with the same radius as the sphere it is defined in relation to: for instance the equator and any circle of longitude (from the North Pole to the south pole and back up to the North Pole - this actually equates to two lines of longitude, separated by 180° - e.g. 10°E and 170°W) are great circles on the Earth - or they would be if the Earth was a perfect sphere, but they're close approximations. Lines of latitude are circles but are not great circles, apart from the equator. The shortest distance between two points on the surface of a sphere is always an arc that is part of a great circle, and hence long distance flights or shipping will plot a route (more or less) on great circles (there will be adjustments for currents or wind conditions). A small circle is a circle on a sphere formed by the intersection with a plane that does not pass through the centre of the circle - for instance latitudes (apart from the equator) as just mentioned. Note that the intersection of a plane and a sphere always forms a circle (from trigonometry).**Celestial sphere:**An imaginary sphere of arbitrarily large radius with centre at the centre of the Earth (actually the centre is arbitrary if the sphere is considered of infinite radius - it could be at the observer or the sun, or in fact anywhere). All objects in the observer's sky are thought of as projected onto the underside of the celestial sphere, which appears to rotate from east to west about the NCP-SCP axis. All objects seem equally far away - distance is considered unknown or unimportant. The celestial sphere can be considered to be infinite in radius and this is important for geometric calculations. This means any point within it can be considered the centre (as noted above), that all parallel lines will intersect the sphere at a single point, and that all parallel planes will intersect the sphere in the same great circle. The essential point is that all observers see the same things in the same direction, so for example I can measure the declination of a star from where I am on the Earth rather than having to offset it to the centre of the Earth (which is the official centre of the celestial sphere). Now the celestial sphere (even if it existed and all the stars were the same distance away) does not have an infinite radius. There are two cases where we need to take this into account as follows - and because of this we do consider the celestial sphere centre to be the centre of the Earth:**For close objects like the moon**which will appear to change their position on the celestial sphere if you quickly run to the other side of the Earth. This is known as**parallax**. This is the apparent displacement of objects as you move to a different line of sight. For example, as you walk into town, that tree on your right is in front of the pub and you can't see the pub; when you walk a bit further you can see the pub (thank goodness) and the tree is in front of something else from the new angle (or line of sight) you are looking from. Back to the moon, it's declination and RA on the celestial sphere will thus be slightly different on one side of the Earth to the other at the same time. Now its coordinates differ over time anyway, since it (like the sun and the planets) is not fixed on the celestial sphere; but the parallax is a different effect applying at the same time for two different observer locations. The parallax is the difference in the angle to the object (e.g. the moon) from the centre of the Earth to the angle from your local position on the Earth's surface. The parallax will be zero if the moon is overhead because it will be in line with the centre of the Earth, and it will be at a maximum when the moon is on the horizon (rising or setting). The lunar horizontal parallax is defined as the angle between the lines from*centre of Earth to moon*and*surface position to moon where the moon is on the horizon*- this will give the maximum parallax and is about 1°, although it varies by about 10% because the distance from Earth to moon varies. To account for this, celestial coordinates for the moon are calculated (and published in astronomical almanacs) as**geocentric**coordinates (i.e. from the centre of the Earth) and corrections need to be applied to take parallax into account, depending on your local position on the Earth. The local coordinates are called**topocentric**coordinates and provide the apparent RA and declination coordinates for your location - for casual observation the 1° is not that significant, although it is for calculation of eclipses and exact moonrise and moonset times. Note that the parallax will resolve into differences in both RA and declination for the topocentric position - a maximum of about 1° in either case (but not both that large at the same time), depending on the configuration of the moon and Earth and your position on the Earth. The Topocentric Positions of Major Solar System Objects and Bright Stars is a website that provides the topocentric positions of the moon, sun and planets if you input your latitude and longitude and the time and date. We also see a bit of parallax for the closest stars (they are in slightly different directions between summer and winter as the Earth moves round the sun) - we'll discuss this more later as it allows us to work out star distances for the closest stars.**If you travel to the other side of the Milky Way**you'll find the star directions and patterns are very different, so the infinite radius approximation is not valid if you travel those kind of distances. It'll take 100,000 years at the speed of light though, so I wouldn't worry too much.

**North Celestial Pole (NCP) and South Celestial Pole (SCP):**The two points where the Earth's axis of rotation intersects the celestial sphere. These points appear to stay in a fixed position and all the other stars seem to rotate about these points - that is, the NCP in the northern hemisphere and the SCP in the southern hemisphere. To be more precise the stars rotate about the axis from NCP to SCP. You can't see both NCP and SCP, but this is the same as saying the stars rotate about the axis from the observer to the NCP (in the northern hemisphere) or the axis from the observer to the SCP (in the southern hemisphere) - because the celestial sphere has more or less infinite radius, you don't need to be at the centre of the Earth for this to be true (as just discussed in the section on the celestial sphere). Polaris is within a degree of the NCP and hence is used as a navigational aid to find north; there is not a bright star very close to the SCP. The altitude above the horizon of the NCP (facing north) is equal to your latitude in the northern hemispshere; similarly for the SCP (facing south) if you are in the southern hemisphere (from spherical astronomy).**Celestial equator:**The projection of the Earth's equator out into space, onto the celestial sphere. It is a great circle on the celestial sphere. An observer can visualise the celestial equator as a semicircle (you can see half of it from any particular location) that meets the horizon at due east and due west. It's maximum altitude will be give by*altitude = 90 - latitude*(from spherical astronomy), where the altitude is the angle upwards from the southern horizon when in the northern hemisphere (and from the northern horizon in the southern hemisphere). For example in Winchester at 51°N, the celestial equator can be located by facing south and visualising a semicircle from the eastern to western horizons that reaches an altitude of 39° at due south. At the equator the celestial equator will be directly overhead passing through the zenith; and at the poles the celestial equator will be on the horizon (this is the exception to the rule you can visualise half the celestial equator in the sky - at the poles you can see it all, albeit right on the horizon).**Hour circle:**This is a great circle on the celestial sphere through the north and south celestial poles and equates to a line of right ascension, just as a longitude circle on the Earth is a great circle through the north and south poles. In fact it equates to two lines of right ascension separated by 12h, e.g. 2h and 14h (in the same way that a longitude circle is two lines of longitude separated by 180°). It will be perpendicular to the celestial equator. The declination of an object is measured along its hour circle: that is, you locate the appropriate hour circle by moving the RA angle from the vernal equinox (e.g. move 90° from the vernal equinox in the plane of the celestial equator if RA = 6^{h}- we will see how to do this using sidereal time later) and then measure the declination up or down that hour circle.**Zenith (and nadir):**The zenith is the local point directly above an observer's head on the celestial sphere - that is, it is an extrapolation of the line through the centre of the Earth, through the observer and onto the celestial sphere. The nadir is the extrapolation in the opposite direction, out the other side of the Earth - the observer can't see the nadir, of course, because the Earth is in the way. The zenith has a declination equal to the observer's latitude (for example, +51° in Winchester - that is, 51° above the celestial equator). The right ascension of the zenith will vary because the celestial sphere appears to rotate each day (and be about 4 minutes earlier each night) - and because of this, the stars at and near the observer's zenith vary through the night and the year although they will always be those stars in a narrow band of declinations (that is, stars with a declination near 51° for Winchester).**The fixed stars:**This is a term used to describe the distant stars which keep the same relative arrangement in the sky - this usually refers to any sky objects that are part of this pattern, so for example distant galaxies are included. The original theory was that the stars are on a fixed pattern on the celestial sphere which rotates round the Earth; now we know the Earth rotates and the stars really are in fixed positions. Either way, the fixed stars are differentiated from other objects such as planets, the sun or the moon which appear to change their positions over a short period - these objects were sometimes called wandering stars. Now I said the understanding is that the stars really are in fixed positions - so ok, that's not quite true, we understand and have detected various independent motions of the 'fixed' stars. The point is that any motion is small relative to how very far away the stars are and so is very difficult to detect (and won't be detected with the naked eye). The fixed stars are sometimes quoted as a reference frame for measuring motion - for example you would describe the motion of the objects in the solar system (like the Earth revolving around the Sun) relative to the fixed stars. That's physics, and we'll get to that in some way or other - the reference frames you use and whether they can really be fixed ends up a large subject!**Constellations and Asterisms:**The stars appear to be in the same place night after night and year after year, and therefore fixed relative to each other (as we've just mentioned). They do appear to rotate around the sky each night until they get back to the same positions the next night (four minutes earlier, as previously mentioned). However the distance between stars and the shape and size of groups of stars are preserved as they rotate, which is why we can recognise different star groupings throughout the night. From ancient times, prominent star shapes have been named by the shapes they appear to make - these are informally called constellations, with names such as Scorpius or Orion (The Hunter - Orion was a hunter in Greek mythology). Ptolemy (a Greco-Roman inhabitant of Egypt, which was under Roman rule then) published a list of 48 constellations in the 2nd century, many of which have much earlier references - right back to cave paintings of the star shapes. Today we're a bit more formal: there's an official list of 88 constellations that was created by the International Astronomical Union in 1922 which together cover the whole sky - they are defined by precise boundaries of right ascension and declination, and include Ptolemy's 48 constellatons. Celestial objects are often referred to by the constellation they are in - for example a planet like Mars moves through the sky, so it is helpful to locate it by noting it is (at a particular time) in, say, Saggitarius. A few well known stars are named by historical names, but mostly they are named by one of several logical systems (they kept changing as the number of stas discovered rapidly increased) which usually includes an abbreviation of the constellation they are in - star catalogues list the stars and their locations. There are so many stars identified by modern telescopes that there are now ways of buying your own star name! A pattern of stars (if it's not a constellation in its own right) is called an asterism. An asterism could be composed of stars from one or several constellations. For example, the Plough (called the Big Dipper in the US) is a prominent asterism of 7 stars that's in the Ursa Major constellation. As a note, Polaris can be found from the Plough - follow a line between the two end stars of the rectangle part and extend it 5 times (from the bottom of the Plough upwards), where you will get to Polaris which is one of the stars in the Little Dipper asterism in the Ursa Minor constellation. Asterisms are generally not stars that are really grouped together - they just appear so since we cannot tell how far away they are, in reality stars within them are likely to have very different distances from the Earth.**True and Visible Horizon:**Loosely speaking the horizon is the skyline - that is the line that separates the land or sea from the sky. The true horizon is the horizon you would see on the (almost) spherical Earth in the absence of obstructions like buildings or mountains; the visible horizon is what you actually see. The distance to the (true) horizon is approximately 5km (3 miles) for someone 2m tall at ground level - to be precise for someone whose eyes are at 2m, which makes them quite tall; or 40km (25 miles) for someone on top a 100m hill. To be accurate, 'ground level' strictly means 'sea level' and the distance is correct when you are looking out to sea since this removes the effects of uneven ground - but if the area around you is flat it's a good estimate. This calculation uses the formula*d = $\sqrt{\mathrm{2Rh\; +h2}}$*, where d is distance to the horizon, R is the radius of the Earth and h is the height the observer. An approximate formula close to the Earth is*d = 3570$\sqrt{h}$*, with d and h in metres - this approximation is ok for mountain tops or aeroplanes, but not something really high like a satellite. The formula is derived by applying Pythgoras' theorem assuming a spherical Earth, and the approximation assumes the h^{2}term is small. Note that the formula measures the distance to the ground-level horizon. If there are, e.g. tall buildings or trees or ship masts over the horizon, you will see the top of these objects when their base is over the calculated horizon - for example if a tower 100m high is 40km away, you will just see the top of the tower if you are at ground level (this is the reverse of the above example that the (ground level) horizon is 40km away if you are at a height of 100m). It is worth mentioning dip as well - this is the angle to the horizon from the observer, or from the line perpendicular to their eyes to be precise. For the 2m high observer the angle is approximately 0.02° and for the observer on the 100m hill it is 0.14° - again this is straightforward trigonometry. The formula doesn't take into account atmospheric refraction which can increase the distance to the horizon by about 10% in normal conditions. The actual dip is reduced slightly by refraction in normal conditions: this seems an odd statement because you can see further so you would think the dip is greater. However because the light bends the observer sees the object at a tangent to the (curving) light that enters their eyes and it appears higher up - the overall effect is to slightly reduce the dip, where dip is the angle to the farthest object they can apparently see. The actual object is not really where they see it, it is lower down. Atmospheric refraction depends on the relative densities of the air near the surface and air higher up, which depends on temperature and pressure differences - in certain conditions it can increase the horizon distance considerably or in fact reduce it or create mirages. We'll go into more detail on atmospheric refraction in What we see in the Sky: The Atmosphere.**Celestial Horizon:**The celestial horizon is also called the**rational horizon**and is defined as "a great circle on the celestial sphere having a plane that passes through the centre of the Earth at a right angle to the line formed by an observer's zenith and nadir". There is also a related concept called the**sensible horizon**which is defined as "the plane of an observer's position lying at a right angle to the line formed by the observer's zenith and nadir". These are used to define an alternative astronomical coordinate system based on the observer's horizon. The sensible horizon is the plane tangential to the Earth's surface where the observer is located, and is parallel to the plane of the celestial horizon (which is through the centre of the Earth). From our definition of the celestial sphere we saw that since it is considered of infinite raduius (at least as far as the stars are concerned) parallel planes intersect the celestial sphere at the same great circle ... and therefore the celestial and sensible horizons are effectively one and the same in most circumstances. They divide the celestial sphere into two hemispheres - one hemisphere representing the half the sky visible to the observer at that location and the other representing the half that is hidden from the observer below the Earth's horizon. That's not quite true: unless the observer is standing in a hole they will see a bit extra because of their height as explained in the section above on True and Visible Horizon - an observer 2m high will see approximately an extra 0.02°, or if they are on a 100m hill it is approximately an extra 0.14°. So the person on the hill will see objects that have altitudes from -0.14° to 90°, slightly more than half the celestial sky. We're about to define altitude in the next definition but one! The horizon extent will also increase slightly due to atmospheric refraction - as we just discussed the observer won't see it like that; it will seem as if refraction slightly reduces the extent of the horizon based on where they see things, but in reality they see objects like stars higher than they really are and they can actually see objects which are at a greater depth below the horizon. A final point is that there will be a difference between measurements from the celestial and sensible horizon for close objects like the moon due to parallax, as we discussed in the definition for the celestial sphere above.**Celestial meridian (or local meridian):**The celestial meridian is local to the observer and is the great circle on the celestial sphere passing through the NCP and the observer's zenith. It will also pass through the SCP and the nadir since these points are their opposites on the celestial sphere. The celestial meridian intersects the horizon at north and south -**in fact this is what defines north and south**. We've spoken about magnetic north and grid north, and that true north is the direction to the North Pole. True north can only be defined by astronomical means - it is the fixed point of the NCP (and SCP) that determines the axis of rotation of the Earth and hence true north (and south). The plane of the celestial meridian is perpendicular to the plane of the celestial horizon. You would normally refer to the celestial or local meridian as the upper half of it - that is, the half that is in the sky you can see and that goes through your zenith and intersects your horizon at north and south. An object reaches it highest altitude as it crosses (or**transits**) the local meridian - this is called the**upper culmination**of the star (or celestial object - the same applies to the Sun, moon, planets, etc.). The object will actually transit the meridian twice - once it will achieve maximum altitude at the upper culmination and once (about 12 hours later) it will achieve minimum altitude or**lower culmination**. Depending on your latitude and the declination of the object you may see both upper and lower culmination (for a circumpolar star), just the upper culmination (the star rises and sets and you see the higher altitudes), or neither (the object is not visible at your latitude). In the northern hemisphere the upper culmination on the meridian is given by*altitude of upper culmination = 90 - (latitude - declination)*, where the altitude is from the southern horizon. Note that if the altitude of culmination is greater than 90°, then the upper culmination on the meridian will take place to the north of the zenith (so you should face north to see it) - this will apply if the declination is greater than the latitude; the altitude from the northern horizon is then*180 - southern altitude*(or*90 + (latitude - declination*). The lower culmination is given by*altitude of lower culmination = (latitude + declination) - 90*, where the altitude is measured from the northern horizon - note, the lower culmination (if it is visible) will always be to north of the zenith. If this is less than zero (*lat + dec < 90*) then the lower culmination will not be seen, on the other hand if this is greater than zero (*dec > 90 - lat*) then the star is circumpolar. The equivalent formulas for the southern hemisphere are*altitude of upper culmination = 90 - (latitude + declination)*(from the northern horizon; if this is greater than 90 then subtract from 180 to get the southern altitude of*90 + (latitude + declination)*) and*altitude of lower culmination = (latitude - declination) - 90*(from the southern horizon). In these southern hemisphere formulae the southern latitude is considered positive, e.g. 45°S is +45°. The term culmination on its own will generally refer to upper culmination. The results here are from spherical astronomy. You can perhaps visualise why objects should reach their maximum altitude on the meridian in terms of symmetry - objects rise in the easterly half of the sky, gradually get higher until they reach the 'middle' point - i.e. the north-south meridian line, then get lower until they set in the westerly half of the sky; for circumpolar objects it is similar except they don't get low enough to set.**Hour Angle (HA):**The hour angle to a point is the angle between an observer's celestial meridian and the hour circle through the point. It is measured along the celestial equator and is positive to the west. It can be measured in degrees (from 0 to 360) or hours, minutes and seconds (from 0 to 24 hours). Together with the declination this forms an alternative system of equatorial coordinates - in this case one of the coordinates (the declination) is fixed with respect to the stars and the other (the hour angle) is local to the observer so this will continually change as time goes by (and the stars appear to rotate). When the hour angle of an object is 0 it will be on your local meridian. The hour angle is very useful because of its relation to sidereal time (later in our list of definitions) and the right ascension. It also tells you how long it was since the object was at your meridian (if positive - e.g. 2h means it was there 2 hours ago) or how long until it gets there if negative (e.g. 23h is the same as -1h, so 1 hour until it gets there) - the times are in sidereal hours which are slightly shorter than standard hours which we'll discuss shortly. The RA/declination coordinates I have been referring to as 'celestial coordinates' are actually also equatorial coordinates: there is a choice between HA and declination or RA and declination for an equatorial system. For formal or catalogued coordinates you are going to want RA/dec of course, since these are fixed. I'll keep calling RA/dec celestial coordinates, even though there are other types of celestial coordinates (e.g. galactic coordinates to measure objects in relation to the Milky Way's plane and centre).**Altitude and azimuth:**These provide an alternative coordinate system (**the horizontal or alt/az coordinate system**) to locate celestial objects. Altitude (alt) (also called elevation) is the angle from the celestial horizon to the object and is between 0° and 90° above the horizon; or between 0° to -90° below the horizon - you can't see it in that case unless it is very slightly less than 0°, when effects of your height or atmospheric refraction may mean it is visible as previously discussed. Azimuth (az) is the bearing from north on the celestial horizon and is between 0° and 360° (e.g. east would be 90°). Since the horizon is rotating, the alt/az coordinates of a celestial object are changing all the time. However once you have the alt/az it is easier to locate the star than if you have declination and RA (unless your telescope works it out for you). In addition horizontal coordinates are used to calculate rising and setting times - these equate to the altitude being zero. There are online calculators that convert dec/RA celestial star coordinates to alt/az if you input the precise date, time and latitude and longitude - for example Celestial to Horizon Co-ordinates Calculator, which also explains the calculation (note that atmospheric refraction is not taken into account). Note again that for close objects, in particular the moon, parallax will apply - we are measuring from the celestial horizon not the sensible horizon. The same discussion we had when we described the celestial sphere will apply. There is a maximum parallax of about 1° to the moon and that will resolve into a difference for altitude and (possibly!) azimuth when considering them from a local position - that is, relative to the sensible horizon as opposed to the published figures based on the goecentric (celestial horizon) coordinates. The moon will be lower in the local sky than the goecentric coordinates suggest - varying from about 1° when the moon is near the horizon to 0 when it is overhead. This effect is lunar parallax in altitude. There is no lunar parallax in the azimuth for a spherical Earth because the plane of the horizon at the surface is parallel to the plane through the centre and*the azimuth is measured on that plane*- so you will get the same values; for the altitude, this is measured out of the plane so you do get the parallax with the displacement between the centre and surface of the Earth. You may have to draw this out to reassure yourself! For higher accuracy measurements - when the Earth is considered as an ellipsoid - there is a small amount of azimuth parallax because the two planes won't be exactly parallel. The parallax does leads to different local (topocentric) coordinates for both RA and declination as discussed in the description of the celestial sphere, as an altitude difference will resolve into a difference in both of these celestial coordinates.**Ecliptic and ecliptic plane:**The ecliptic is the apparent path of the Sun on the celestial sphere (as seen from the centre of the Earth); and the ecliptic plane, sometimes also called just the ecliptic, is the plane of this path (the path is more or less in a two dimensional plane). At any time the Sun is in a particular area of the sky - for example in late June it is in the same direction as the constellation of Gemini. Through the year the Sun appears to move eastwards through the fixed stars, approximately 1° per day, until it gets back to the same area of sky after one year. This is as opposed to the daily east to west motion of the Sun across the sky; the stars appear to be moving westwards across the sky a bit faster than the Sun is moving westwards and hence the Sun appears to be moving eastwards amongst the stars. An obvious question: since during the day you can't see the stars and at night you can't see the Sun, how do you know where the Sun is in relation to the stars? Well a fairly obvious answer - look at where the Sun sets and see what stars appear in that direction just afterwards, or alternatively look where the Sun rises and match this to the stars that have just disappeared in that direction. The ecliptic plane is at an angle of approximately 23.5° to the celestial equator. This angle is known as the obliquity of the ecliptic and leads to some interesting astronomical events: specifically the Sun on its ecliptic will intersect the celestial equator twice a year, and it will also reach a maximum and a minimum altitude above the celestial equator once each per year. The moon and the planets all seem to move on a narrow band centred on the ecliptic that's about 20° wide. What is really happening is that the ecliptic defines the Earth's orbit around the Sun. The planets also orbit the Sun in approximate planes that are within a few degrees of the ecliptic; similarly the moon orbits the Earth (roughly) in a plane inclined at about 5° to the ecliptic.**Vernal equinox (and autumnal equinox):**Ok, so I will say what the vernal equinox is! The vernal equinox is the point on the celestial sphere where the Sun crosses the celestial equator northwards. The date at which the Sun makes this crossing defines the first day of spring in the northern hemisphere (or the first day of autumn in the southern hemisphere) and is generally 20 or 21 March. The variation in the date (and exact time) of the vernal equinox is mostly due to the effect of leap years but also due to some orbital subtleties. The vernal equinox can refer to either the precise point on the celestial sphere or the date and time that the Sun is actually at that position. The vernal equinox point always exists as a location throughout the year (not just when the Sun is there!). The position of the vernal equinox very slowly changes over the years due to an effect called the precession of the equinoxes which means the Earth's orientation rotates relative to the stars approximately once every 26,000 years which we won't delve into right now: star coordinates are updated or corrected for periodically because of this (they will shift about .01° per year relative to the vernal equinox). On the vernal equinox the Sun will rise exactly in the east and set exactly in the west (more or less) and the hours of day and night will be roughly 12 hours each. The autumnal equinox is the point where the Sun crosses the celestial equator southwards and similarly defines the start of autumn (northern hemisphere) or spring (southern hemisphere), and the date is generally 22 or 23 September. The same points apply about equal night and day and the sun rising and setting due east and west. The reality of the vernal and autumnal equinoxes is that this is the time when the tilt of the Earth is neither towards or away from the Sun and the Sun is in the same plane as the equator. Now the position of the vernal equinox on the celestial equator is the zero point of right ascension, so rotation to the east from this point defines the RA in hours, minutes and seconds as we have mentioned (with 1 hour equivalent to 15°). The autumnal equinox is opposite the vernal equinox on the Sun's path (or in reality on the Earth's ellipse around the Sun) and hence the RA of the autumnal equinox is 12h. The axis of the RA is the line from the centre of Earth to the vernal equinox which is a line in the plane of the celestial equator.**Summer and Winter Solstices:**In the northern hemisphere the summer solstice occurs when the Sun reaches its maximum altitude above the celestial equator and is generally on 20 or 21 June; the winter solstice occurs when the Sun reaches its minimum altitude below the celestial equator and is generally on 21 or 22 December. The solstices are precise moments in time, but the terms are also used to refer to the day on which they occur. The days are also called midsummer and midwinter. The names are reversed in the southern hemisphere to reflect the fact that the June solstice is winter and the December solstice is summer. The solstices are the longest and shortest days of the year (well, there are some subtleties in the Arctic and Antarctic Circles and the tropics - we'll visit this in our section on the Sun). At the solstices the sun stops its movement to the north or south in readiness to reverse; the name solstice derives from Latin words meaning*sun*and*to stand still*.**Zodiac:**The zodiac is the name of the band about 20° wide around the ecliptic on which the Sun, the moon and all the planets appear to move. The ecliptic currently passes through the following constellations in order, starting from the vernal equinox (approximately 21 March) when the Sun will be in Pisces: Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Ophiuchus, Sagittarius, Capricornus, Aquarius. The period of time spent in each constellation varies from 7 days to 45 days. Note that these are the constellations as defined by the International Astronomical Union. The Sun will slowly change the constellation it is in at a particular time of year; for example, 2,000 years ago it would have been in Aries at the vernal equinox - its vernal equinox position is rotating around the constellations roughly every 26,000 years due to the precession of the equinoxes. Now you're probably going to look at those constellations and say they don't match the astological ones: there's one extra (Ophiuchus), the dates don't match and they all last different periods. True. The zodiac is not actually the list of constellations I've just given. It is a division of 12 30° 'zodiac' or 'star signs' centred on the ecliptic, with the first point of Aries (the beginning of Aries) defined as the location of the vernal equinox (at an RA of 0). So the Sun will be in Aries (the fictional zodiacal sign as opposed to the real constellation) from approximately 21 March - 19 April, and similarly dates are provided for the rest of the signs up to Pisces from 19 February - 20 March. The dates are not exact as the vernal equinox will vary by a day or so due to the effects of leap years - if you were born on the cusp (close to the boundary between two star signs) then you will need to double check which sign you were born under (or ask an astrologer). In theory the boundary of Aries should start at the exact time of the vernal equinox but I don't think this is usually taken into account when working out your star sign, it is just the day. In astrology your personality and life events are influenced by what sign you are born under and (being more precise) by the configuration of the planets and moon at your exact time of birth. In ancient times, zodiac signs did more closely match the constellations as the vernal equinox was in Aries 2,000 years ago although the 12 zodiacal constellations (or asterisms) varied in width; the 30° division was made to split the year equally rather than to exactly match the constellations (probably by both astrologers and astronomers of the time). I'm a Gemini by the way.**Solar Time: Apparent Solar Time and Mean Solar Time:**We need to discuss time here because we need to understand sidereal time to determine the right ascension, and thus how to find a star with a particular RA. To understand sidereal time we should first address the more familiar solar time.**Apparent solar time**- also called local apparent solar time or LAST - is the time measured directly by the Sun, for example by a sundial. This is defined as apparent solar time = hour angle of the Sun + 12 hours, or LAST = HA + 12. Remember the hour angle is zero when the object (in this case the Sun) is on the local meridian. Therefore the definition means that it is apparent noon when the Sun is on the local meridian, and that it is 1:00pm (or 13:00) when the Sun has moved 15° West (and therefore HA=1). This is called*apparent*solar time because it is based on the apparent motion of the Sun when it is really the Earth that is moving. Note that the apparent solar time is local and will differ for two people at different longitudes, even if it is only a few metres - at the equator there will be 1 second of time difference for approximately each 460m of distance, or at Winchester it will be 1 second for about 300m.**Mean solar time**- also called local mean solar time or LMST. Although the average length of the apparent solar day is 24 hours over the year, the actual length of individual days is up to 20s shorter (on about 16 Sept at the moment) or 30s longer (about 22 Dec) than the 'average' 24 hour day. Therefore an average value of the time is defined called the mean solar time based on the (fictitious) position of a mean Sun that travels round the Earth in exactly 24 hours every day. This is the time your watch will show, once we correct for the time zone you are in. The differences will be cumulative - that is, if you have several days that are all short or all long, then the daily time differences will add together to differ by several minutes. If the apparent solar days are longer than 24 hours, then mean solar time will reach noon before the sundial and the mean time will be ahead of the sundial time; the fictitious mean Sun will be ahead (to the west) of the real (apparent) Sun. The converse will apply if the apparent solar days are shorter than 24 hours - the mean time will be behind sundial time and the mean Sun will be to the east of the apparent Sun. The maximum deviations of the mean solar time from the apparent solar time turns out to be about 16 minutes earlier (i.e. the Sun will show noon and the watch will show 11:44am - on about 3 November) or 14 minutes later (the watch shows 12:14pm at apparent noon - about 11 February). The two times are equal on 4 different dates through the year - approximately 15 Apr, 13 Jun, 1 Sept and 25 Dec.**The difference is defined by the equation of time**(sometimes abbreviated E or EOT) which is a published tabulation or graph of the difference between apparent and mean solar time - and therefore can be used to convert apparent solar time to mean solar time (or vice versa). Now let's say we have the apparent solar time and have converted to mean solar time - this still won't give your wrist watch time because you are in a time zone which has standardised time across a range of latitudes. For standard time zones this is a band 15° of latitude wide and the time for all of that band will be the mean solar time at the central longitude - for example, for Greenwich Mean Time, the time is the mean solar time at 0° longitude and this will apply to the band 7.5° either side. As we discussed in our diversion on longitude it doesn't work this neatly because time zones are based on political boundaries, but as long as you know the central meridian of your time zone you can use this to calculate the time.**To find your wrist watch time in Winchester from the Sun, here's what you do**: read the sundial time to get apparent time, let's make it noon for simplicity; apply the equation of time to get mean time, say this is 2 minutes later than solar time for today's date, so we are at 12:02pm; correct for longitude so we measure the time at the central longitude of our time zone at 0° (Greenwich) - Winchester is at 1.3°W so the time at 0° will be later by 1.3/15 * 60 minutes = 5.2 minutes or 5m 12s (remember each 15° of longitude is an hour, or equivalently each 360° is 24 hours or each 1° is 4 minutes). And hence GMT is 12:07:12. Finally, if it's British Summer Time then you need to add an hour for daylight saving. Note that Greenwich Mean Time - the mean solar time at Greenwich, which could also be described as Greenwich Mean Solar Time - is adopted as the worldwide time standard known as Universal Time (UT). You can find the equation of time values easily enough on the web - one example is Equation of Time. One thing to be careful of is the sign of the equation of time - the convention is that*E = apparent solar time - mean solar time*, so to get mean time you would subtract E from the solar time, but there does seem to be some variation in publications. The source you use should make the sign clear, but as a check the mean solar time will be later than the apparent solar time for Jan, Feb and Mar so you would expect to add a few minutes to get from apparent to mean time in these months (for the next hundred years plus at least!). You can also reverse the instructions and do the calculaton the other way to get from watch time to Sun time. The equation of time corrections have been recorded since accurate clocks were available and the effect became obvious because of the discrepancy between these and the Sun, and tables date from the late 1600s. The difference was determined initially by observations which could be assumed to apply for future years since the equation of time doesn't vary much short term; apart from a correction based on how near the year is to a leap year - the graph shifts by 1/4 day each year until the next leap year when it shifts back. Latterly the equation of time is determined by calculation - which we're just about to discuss. The two main terms are as below. As ever there are additional subtleties if you want to be supremely accurate, like gravitational effects on the motion of the Sun from the planets - but the two effects below are by far the main components of the equation of time.**The Earth's elliptical orbit:**The Earth orbits the Sun in an ellipse. The Earth's orbit has low 'eccentricity' which means it's close to circular - there's a difference of approximately 5 milliom kilometres (or 3 million miles) between the closest and furthest points of the Earth to the Sun. The closest point of the Earth to the Sun occurs at a point called perihelion which currently happens on approximately 2 January; the furthest point is called aphelion and happens on approximately 4 July. As we'll find out later, one property of gravitational orbits means that the Earth moves faster when it's closer to the Sun. This means that the day (the solar day, as measured by the Sun) will be longer. The reason is that for a day to pass, the Earth has to rotate once (which takes roughly 23h 56m 4s) and then has to rotate a bit further to 'catch up' because the Earth has moved about 1° round the Sun (so the direction of the Sun from the same point on Earth - e.g. South at noon in the northern hemisphere - is the same again). [The main thing to visualise here is whether the Earth has to rotate a bit further or a bit less so the Sun is back in the same compass direction. Hopefully you can see that it is further (play with a tennis ball), given that the Earth rotates anticlockwise and revolves round the Sun anticlockwise] On average the length of day comes to 24 hours, but when the Earth is moving faster it will have moved that bit further round the Sun and the 'extra rotation' to catch up will therefore take a bit longer - about 8 seconds at perihelion. Similarly the solar day will be about 8 seconds shorter at aphelion. The cumulative time difference due to this effect adds up to the mean solar time reaching a maximum of approximately 8 minutes later (than apparent solar time) in April, 8 minutes earlier in October, and the same at perihelion and aphelion. The equivalence of the two times at perihelion and aphelion is a choice which makes the difference symmetrical through the year (half the time mean solar time is ahead of apparent solar time and half the time it is behind), and this also minimises the difference between apparent and mean solar time. If we had made them equal in October, then the difference would go up to 16 minutes in April and back to zero the next October - note though, that wherever we chose the zero difference point it would always cancel out over a year. Over long periods this effect will vary because of changes in the Earth's eccentricity - that is, the difference between closest and furthest approaches of the Earth to the Sun (and hence variations in the Earth's speed) will vary. This variation has a cycle of approximately 100,000 years. [The eccentricity varies between about 0.005 and 0.058 and is currently about 0.0167.] The dates of the perihelion and aphelion also change - they advance by about 1 day every 60 years (the complete cycle to bring them back to the same dates takes about 22,000 years), which means the equation of time corrections due to this effect will also shift one day each 60 years.**The obliquity of the Earth's orbit:**This effect is purely one of geometry due to the Earth's rotation in the plane of the equator (which defines the east/west directions) varying from the ecliptic plane in which the Earth revolves around the Sun. Note that the Earth both rotates from west to east and revolves around the Sun from west to east (as it both rotates anticlockwise and revolves around the Sun anti-clockwise). The Sun appears to move (more or less) eastwards through the stars on the ecliptic plane which is at approximately 23.5° to the celestial equator. This motion can be split into two components. The majority of the Sun's apparent motion is from west to east along the celestial equator - this is the projection of the Sun (which moves on the ecliptic at 23.5° to the celestial equator) onto the celestial equator. The Sun also appears to move vertically (increasing or decreasing its altitude): at the summer solstice (in June, using northern hemisphere terminlology) it is at its highest altitude above the celestial equator, at midwinter it is at its lowest altitude below the celestial equator, and at the vernal and autumnal equinoxes it is on the equator. It is the east-west direction along the celestial equator that defines the time and that a sundial will measure (because we measure the easterly or westerly bearing along the equator (or a circle of latitude parallel to the equator); to put it another way, if the Sun is at its highest altitude (so it is noon) at one longitude, then it will be an hour later (at that location) when the Sun is at its highest altitude at a longitude 15° to the west (time is measuring the east-west passage of the Sun). In terms of measuring shadows on sundials, the altitude determines the length of a shadow not the direction of the shadow. The vertical motion is slowest when it reaches the peak values at midsummer or midwinter (it actually becomes zero at these exact points because it comes to a stop and reverses direction) and is fastest at the equinoxes when it crosses the equator. The total speed of the Sun across the sky (or in reality the Earth around the Sun) is constant (ignoring the effect of the elliptical orbit just discussed); so when it is slower in the vertical direction it will be faster in the east/west direction. Imagine the Sun is in a certain direction (e.g. south at noon in the northern hemisphere) and let the Earth rotate once (in roughly 23h 56m 4s); for the Sun to be in the same direction on the next day the Earth has to rotate a bit further*to catch up with the extra motion of the Earth eastward around the Sun*. Now since the Earth moves faster eastwards near the solstices (because it moves slower vertically) it will take longer to catch up and hence the day will be longer. Similarly the day will be shorter near the equinoxes since the east/west motion is slightly slower. We can explain this slightly more mathematically as: the apparent solar time is defined as the hour angle of the Sun + 12 hours; the hour angle is measured on the celestial equator; we therefore have to resolve the motion of the Sun into equatorial motion (which gives the time, as shown by a sundial) and vertical motion (which leads to changes in the Sun's altitude); the equatorial motion is faster at the solstices (and the vertical motion slower) so the day is slightly shorter. The overall effect is that the day is about 20 seconds shorter at the solstices and 20 seconds longer at the equinoxes. The mean solar time (due to this effect) is chosen to be equal to apparent solar time at the vernal equinox; again this choice is to make the difference symetrical and to minimise the maxium difference between mean and apparent solar time. The mean and apparent solar times (due to this effect) are also equal at the autumnal equinox and the summer and winter solstices. The maximum cumulative difference will be about 10 minutes: the mean solar time will be about 10 minutes later than apparent solar time in early February and August; and about 10 minutes earlier in early May and November. Over long periods this effect will vary because of changes in the Earth's obliquity - which seems to vary between about 22° and 24.5° over a 41,000 year cycle. You then need to add these two effects up to get the overall equation of time.

**Sidereal Time:**Sidereal time or local sidereal time (LST) is the hour angle of the vernal equinox. Remember the hour angle is the angle between an observer's celestial (or local) meridian and the hour circle through the object in question and is positive to the west. The positive sign to the west makes sense - since all the objects in the sky, including the vernal equinox, appear to move to the west (because the Earth is rotating anti-clockwise), then LST will increase as time goes on, which is what you expect for time - you don't want it to decrease! Also recall that apparent solar time (LAST) is the hour angle of the Sun + 12 hours. A sidereal day is thus the time taken for the fixed (background) stars to be in the same position again which is the time for the Earth to rotate; while a solar day is the time for the Sun to be in the same position again (on average) which is a bit longer, as the Earth has to rotate a bit further to catch up. The length of the solar day is 24 hours - technically you could call these solar hours, but they are what we think of as normal 'wrist watch hours' and we just call them hours. The length of the sidereal day is approximately 23h 56m 4.09s (in solar hours, minutes and seconds). Note that just as the solar day length is an average due to the effects defined by the equation of time, the sidereal day length is also an average due to some orbital subtleties (specifically**nutation**, which we'll discuss when we talk about orbits). There is an eqivalent equation called the**equation of the equinoxes**that converts 'apparent' sidereal time (the hour angle to the actual vernal equinox) to 'mean' sidereal time (the hour angle to an 'average' vernal equinox) - the effect is small (the difference is never much more than one second). A sidereal clock measures mean sidereal time and when we refer to sidereal time we will be talking about mean sidereal time. Note also, that because of precession of the equinoxes (which we've mentioned before and will discuss properly when we discuss orbits), a sidereal day is not exactly the time for the Earth to rotate because the vernal equinox will move slightly in that time - in fact the sidereal day is about 0.0084s shorter than the rotation period, and therefore the stars do not come back to exactly the same posiitons after one sidereal day. However, we measure right ascension relative to the vernal equinox so the sidereal time turns out to be very useful; because of this though, we will actually see the star positions rotate very slowly (over about 26,000 years) - this is why star charts are updated every few decades to take account of the new positions relative to our RA/dec coordinate axes, which are moving relative to the fixed stars because of the moving vernal equinox. There is actually a**stellar day**defined that tracks the day exactly relative to the fixed stars (and is therefore about 0.0084s longer than the sidereal day) - we probably won't need to mention this again. Now since sidereal time is defined in terms of hour angles (to the vernal equinox), one sidereal day will still take 24 'hours' even though the hours are smaller than solar hours - these are sidereal hours and if you do the maths a sidereal hour is approximately 59m 50.2s (of solar time). Similarly a sidereal minute is approximately 59.84 solar seconds and a sidereal second is approximately 0.997 solar seconds. Here are a few more facts about sidereal time, its relationship to solar time and crucially the link to RA:- A solar day is approximately 1.003 sidereal days, and a sidereal day is approximately 0.997 solar days. The same ratio applies for hours, minutes and seconds.
- One solar day is, in sidereal time, 24h 3m 56.56s. This means that after one solar day, a sidereal clock that started at the same time as a normal (solar) clock will gain 3m 56.56s on the solar clock. It gains roughly 10s per hour or 1s per 6 minutes. After one solar year (approximately 365.24 days) it will have gained 24 hours - and hence a sidereal year is about 366.24 sidereal days. Since it gains 24 hours a year, it will gain approximately 2 hours per month.
- One sidereal day is, in solar time, 23h 56m 4.09s. This means that after one sidereal day, the solar clock reads 3m 55.91s short of the sidereal clock. And of course after 366.24 sidereal days they are back in sync (for a moment, at least).
- Sidereal time is a local measure. Sidereal time at Greenwich is used as a standard and published in almanacs - it is refered to as
**Greenwich Sidereal Time (or GST)**(which is Grenwich mean sidereal time - although you can also have Greenwich apparent sidereal time). Sidereal time for defined time zones is not used as it is for solar time, as there is no need for it. To convert from GST to your local sidereal time (LST), you add an hour for each 15° of longitude you are to the east of Greenwich. For example if it is 14:30 GST, then the LST at Winchester (1.3°W) will be 1.3/15 hours before 14:30 - which is .0866 hours or 5.2 minutes (multiplying it by 60) or 5 minutes 12s before 14:30, and so the LST is about 14:24:48 (just take 5 minutes off in Winchester, unless you need to be really accurate). - There is also a sidereal date defined, but first I have to mention the Julian Day (or Julian Date) (either way, JD). This is the number of days (in decimal) since noon GMT on 1 January 4713 BC. Astronomers use Julian Days because it makes it easier to calculate the time between two dates, and it also avoids confusion about different calendars (our calendars have changed at various times through history, sometimes skipping days to get it back on track with the seasons). The current Julian Day on 14 April 2013 is 2,456,397. The Greenwich Sidereal Date (GSD) is the number of sidereal days since 1 January 4713 BC. Since there is an extra sidereal day for each year, then the GSD will be greater than the Julian Day by the number of years since 4713 BC. The actual definition of GSD is "the number of sidereal days elapsed at Greenwich since the beginning of the Greenwich sidereal day that was in progress at the Julian date (JD) 0.0". This means that the start point was not noon on 1 January 4713 BC as for the Julian Day, but likely a few hours earlier (whenever the sidereal day in progress at that point started) - exactly when that was I don't know but let's not get hung up on it.
- We need to know when sidereal time and mean solar time are equal since if we know this we can convert mean solar time to sidereal time because we know the rate at which they diverge. At the exact moment of the vernal equinox the LST and the local mean apparent time (LAST) will be exactly 12 hours apart because the Sun and the vernal equinox are in the same place and LST is the hour angle of the vernal equinox while LAST is the hour angle of the Sun + 12 hours. At the exact moment of the autumnal equinox the Sun and the vernal equinox will be opposite each other and therefore
**the LST and LAST will be equal at the autumnal equinox**(since the hour angles of the vernal and autumnal equinoxes are 12 hours apart). Now we know that a sidereal day gains 3m 56.56s on the solar day and therefore we can calculate sidereal time by working out the number of days (including fractional days) since the solar and sidereal times were equal and multiplying that by 3m 56.56 to give the discrepancy. Since our current time is given in mean solar time (rather than apparent solar time) we need to know the date and time when the mean solar time and sidereal time are equal which will be slightly different than the autumnal equinox (since that is based on apparent solar time). Looking at the equation of time this is approximately +7m on 22 Sept, and hence mean solar time is approximately 7 minutes earlier than apparent solar time (and than LST) at the autumnal equinox. One day earlier mean solar time would be only 7m - 3m 56.56s = 3m 3.44s behind LST, since in a solar day 24h 3m 56.56s of sidereal time passes. To find when they were equal you have to go back 7m/(3m 56.56s) = 420s/236.56s which is approximately 1.77 days earlier. To be very accurate here you will need to know the exact value of the equation of time at the autumnal equinox and the exact time of the autumnal equinox. Similarly, the date when the LST is 12 hours apart from the mean solar time is approximately 1.77 days after the vernal equinox (since the equation of time is about -7m close to the vernal equinox). - The Right Ascension of an object is the angle eastwards from the vernal equinox. Local sidereal time is the hour angle of the vernal equinox - that is, the angle westwards from the local meridian to the vernal equinox. When the object is on your local meridian it's right ascension will therefore be the angle eastwards from the vernal equinox to your meridian - which is the same as the angle westwards from your meridian to the vernal equinox. And therefore
**the right ascension of an object that is on your local meridian is equal to the local sidereal time**. As such, sometimes the LST is quoted as the right ascension of the local meridian which is equivalent to its more usual definition of the hour angle of the vernal equinox. More generally the**local sidereal time = right ascension of an object plus the hour angle of the object**. That is*LST = RA + HA*. The reason is that: since the LST is the angle west to the vernal equinox, this is the same as the angle west to the object (it's hour angle) plus the angle west from the object to the vernal equinox (the same as the angle east from the vernal equinox to the object, which is the RA). So if you know the RA of an object and the LST, you can find the object at an hour angle of*LST - RA*, that is at an angle to the west of your local meridian of*LST - RA*. Conversely if you can see an object at a particular hour angle and you know the LST, then you can calculate the right ascension as*RA = LST - HA*. This is what we've been waiting for in our quest to understand how to locate a star or object at a particular RA (or how to measure the RA of an object). Astronomers have been doing this for a long time - nowadays they do it incredibly accurately with modern instruments, but in older times a catalogue of RA/dec coordinates for stars was slowly built up and made more accurate over the years. Similarly RA/dec observations for the Sun, moon and planets were built up - in this case, the co-ordinates change over time, so the observations mapped coordinate values over time and various patterns emerged, for instance that the Sun's values repeat over a year or that the moon's repeat over about 27 days (very roughly, there are lots of subtleties with the moon). These observations and catalogued coordinates were used to formulate and refine theories of how these objects all move - the more accurately we measure, the more subtle effects we discover, for instance orbital 'wobbles' due to the gravitational influence of other planets. Anyway, we're heading off topic here - the key point is the link between LST and the RA. **Converting between Sidereal Time and Mean Solar Time:**We saw how to convert from sundial time to wrist watch time earlier. Now we need to convert from wrist watch time to sidereal time (LST). To do this let's take the example of it being 20:00 in Winchester on 13 April 2013. First we need to change this from daylight saving time to GMT (since BST applies on 13 April) - so it's 19:00 GMT. Now we need to convert this to local mean solar time in Winchester - this will be 1.3*15 * 60 minutes = 5.2 minutes earlier - so call it 5 minutes earlier and the LMST will be roughly 18:55. For a quick estimate, note that mean solar time and sidereal time are equal round about 21 September - so 13 April is about 6.7 months later. Since the sidereal time gains roughly 2 hours in a month we would expect the time to be about 6.7*2 hours = 13.4 hours = 13 hours 24 minutes later than 18:55 - which is 08:19. To do it more accurately the principle is simple but the numbers are fiddly: you need to know how many days it is since the times were equal (which was roughly at the autumnal equinox) until now and multiply that by 3m 56.56s to see how much later sidereal time is from 18:54:48. You will have to look up the date of the previous autumnal equinox, check the equation of time and work out when the mean solar and sidereal times really were equal, then work out how many days between then and now (make sure both times are in the same time zone - usually GMT because equinox dates and times will generally be published in GMT) to several decimal places, multiply that by 3m 56.56s and add that to your current local (Winchester) time of 18:54:48. Alternatively you can use an online calculator to do the calculation for you, e.g. this Sidereal Clock which gives the sidereal time as 08:23 at the Winchester longitude of 1.3° W at 19:00 GMT. Which is within 4 minutes of our estimate anyway.

**Fantastic. I can see you've set the scene to answer my question about how to locate a star. Can you go for it?**

Ok. Let's take your example of a star with dec = 25° and RA = 11^{h}, and assume we're in Winchester (51°N, 1.3°W) on 13 April. All new results here not fully explained are from spherical astronomy (and assume a spherical Earth and no refraction or other 'subtleties')! We'll do this step by step:

- Since we're at latitude 51°N any star with
*dec < latitude - 90 = 39°S*will be invisible (it will never rise). Also any star with*dec > 90 - latitude = 39°N*will be circumpolar (always up). Our star with dec = 25°N fits neither of these so it will rise and set. - The next thing to check is that the star is visible at night. Since it rises and sets there may be a time of year when it is only up in the daytime, and hence we won't be able to see it. The star will reach its maximum altitude on the local meridian and it will reach this when the local sidereal time is equal to the star's RA of 11
^{h}. We've just worked out that the sidereal time is 08:23 at 20:00 on 13 April - so when the sidereal time is 11:00, it will be 22:37 (more accurately, since the sidereal clock runs faster, in 2h 37m of sidereal time only .997 times 2h 37m = approximately 2h 36m 32s of solar time passes, so the time will be 22:36:32). And therefore the star will be on the meridian at 10:37pm - which means we will have no problem seeing it during the night. - The declination is 25° and hence the star is 25° above the celestial equator. The celestial equator can be visualised from Winchester by facing south and imagining a semicircle from the eastern to western horizons that reaches an altitude of
*90 - latitude = 39°*from the south. We want to define a new circle of constant declination = 25°. This will be at an angle 25° higher than the celestial equator at the due south point - so it will be at an elevation of 64° here. The circle will no longer intersect the horizon at due east and west though (this is only the case for a star with dec = 0), and is no longer a great circle (again this only applies if dec = 0, when the star moves on the celestial equator) but describes a small circle on the celestial sphere. Stars move in circles parallel to the celestial equator - this is because since a star has constant declination it maintains the same angle to the celestial equator (from the centre of the Earth, or effectively from the observer since the difference is negligible on the celestial sphere) for its whole journey. If the dec > 0 then the star will intersect the horizon north of east (where the star rises) and north of west (where it sets), and the visible circle arc is larger than a semicircle and the star is up the majority of the time; if the dec < 0, the star will intersect the horizon south of east and south of west, and the visible circle arc is smaller than a semicircle and the star is down the majority of the time. Once the*dec > 90 - latitude*the circle will miss the northerly horizon entirely and the star is cirumpolar; similarly if the*dec < latitude -90*the circle will disappear over the southerly horizon and the star will be invisible. So in our case the circle of declination 25° has a maximum altitude of 64° to the south and is parallel to the celestial equator which goes from due east to due west with altitude 39°. Now you can visualize this, but it might be easier if you know the rising and setting azimuth of the star - and helpfully there is a formula for that,*cos(rising azimuth) = sin(declination)/cos(latitude)*(assuming a spherical Earth and no extra 'complications' like refraction). Plugging in the numbers, our*rising azimuth = cos*(roughly). The^{-1}(sin(25)/cos(51) = 47.8°*setting azimuth of a star = 360 - rising azimuth*, so the star sets at 312.2°. So roughly the star rises NE and sets NW; it is up for approximately (360 - (2 * 47.8))/360 = 73.4% of the time. So our star roughly traverses a three-quarters circle from the NE to the NW horizon points reaching an altitude of 64° at due south. The centre of this circle is its intersection with a line between the observer and the NCP. All we need to know is where the star is on this arc. - The RA is 11
^{h}and so it will be on our meridian - the circle through our zenith and the north and south horizon points - at 22:37. At other times we need the hour angle of the star, which equals LST - RA, to know how far west of the local meridian the star is. Let's say the time is 2:00am (on the 14 April now), then the LST will be roughly 14:23 (adding 6 hours to 08:23 - or we can look it up on the sidereal time website which gives 14:24:24). So the hour angle is roughly 14^{h}23^{m}- 11^{h}= 3^{h}23^{m}(or 3.38^{h}) which means the star is 3.38 * 15° = 50° (roughly) to the west of the meridian. The star will therefore be on the hour circle with hour angle approximately 50° at 2:00am. The hour circle of 50° is the circle that goes through the NCP and a point 50° west of your zenith point. Note that the hour circle with hour angle 50° together with the hour circle with hour angle 330° (the 'other side' of it) defines a great circle on the celestial sphere which you will be able to see half of (a semicircle) above the horizon. You are only interested in the westerly part of that though (where the hour angle is 50°) - the star will reach an hour angle of 330° in another 12 hours time. - So ... you can find the star at 02:00am by placing it at the intersection of: a three-quarters circle from the NE to NW horizon at an altitude of 64° at due south and a semicircle through the NCP and a point 50° west of the zenith. Probably an easier way is: to visualise the celestial equator semicircle and take its intersection with the meridian (i.e. the point due south at altitude of 39°); trace a path 50° westerly along the celestial equator; then trace a path of 25° (the declination) in the direction of the NCP to reach the star. Note that if the hour angle is greater than 90° (and less than 180°) then the path along the celestial equator will go over the horizon - and then come back into view when you trace up the declination (assuming it is visible); if the hour angle is between 180° and 360° then you will be traversing to the east along the celestial equator (and going over the horizon between 180° and 270°). Yet another alternative is to take the circle of constant declination we defined (a three-quarters circle from the NE to the NW horizon points reaching an altitude of 64° at due south) and measure 50° from its central due south point along the circle to reach the star. Use a compass to find north, south, east and west (and correct for mag dec if that much accuracy is needed), or alternatively use the NCP to identify north from the stars! If you're just using your eyes this will give you a fair idea of the star's whereabouts and if it's a distinctive bright star you could probably find it - but you'd have to be pretty skillful at estimating angles and visualising circles in the sky to locate the point accurately! Luckily telescopes have viewfinders, crosshairs and motorised control that allows much more accurate locating and tracking of stars by RA and declination - these go from sophisticated professional observatory telescopes controlled by software that allow the measurement of RA/dec to very high accuracy to amateur telescopes with lesser accuracy but still allowing much better positioning than the 'eyeballing' just sketched - and of course you can see amazingly more detail. I don't have a telescope or know much about them, so I'll leave this subject for now.
**Note that a final and easier method is to wait for the star to be on your central meridian, in this case at 10:37 - then all you have to do is face south and look up at an angle of 64°.** - An alternative is to convert to altitude/azimuth co-ordinates. Use the Celestial to Horizon Co-ordinates Calculator mentioned earlier (or equivalent). Put in the values for lat/long, RA/dec and date/time and you will find that the star we have been discussing has alt/az coordinates of approximately 43° altitude and 256° azimuth at Winchester at 2:00am on 14 April. So the star is at a bearing of 256° (14° south of west) at 43° altitude. As a test, if I put in the time of 22:37 on the 13 April for the star I get approximately an azimuth of 180° and an altitude of 64° - which is as expected since the star is on the meridian at this time (with a bearing of due south) with a maximum angle of 64° as already seen. In theory this method seems easier and it is for a 'one-time' look up of the star (as long as you have access to a web application to work it out for you - note, the coordinate conversion calculation is do-able and is 'just spherical astronomy', but you wouldn't want to do it with pen and paper!). The drawback is this doesn't give you a good picture of how the star will move as the night progresses, whereas the circles of declination and the increasing hour angle does give a much better picture of the ongoing motion.
- Another alternative is to use a reference star fairly close to the values of your star, which you are able to easily find and whose coordinates you know. Then offset from those coordinates to the required star coordinates,.

**Right, I see you managed to slip in details of my previous question about which directions the stars rise and set. Are we done here?**

More or less. **I want to finish with a few notes on star rising and setting, focussing on what happens at the equator, the poles and intermediate latitudes.** So ... here you go (with the usual caveat that results are approximate, assuming a spherical Earth, etc.):

**The equator:**The equator is the only place where all the stars it is possible to see from Earth can be seen. There are no invisible stars and no circumpolar stars. Except ... in theory the NCP and the SCP with declinations of 90° (N or S) are always on the horizon (to the north and south) and are therefore circumpolar although there are not visible stars at those exact points; in practise due to atmospheric refraction (or if you are at altitude) some stars close to these points may be circumpolar and be always visible right on the horizon. The celestial equator is seen as a semicircle from due east to due west directly over the observer's head (through the zenith). All stars travel in semicircles parallel to the celestial equator and are up for 12 hours and down for 12 hours. After one day an observer will have seen all the stars that can be seen go through such a semicircle. Stars rise vertically and set vertically. [To explain this: they travel on a semicircle and the diameter of the semicircle is the line between the rising and setting points on the horizon. The apparent direction of motion of the star at any time is the tangent to the semicircle it is travelling on. At the rising and setting points the tangent will be perpendicular to the diameter (i.e. the ground) and hence at the moment of rising (or setting) the apparent motion is vertical.] The stars rise at an azimuth away from east equal to their declination and set at an angle of*360° - rising azimuth*, so stars with a declination of 0° rise in the east and set in the west, and for example stars with a declination of 40° rise at an azimuth of 50° (just south of north east) and set at 310°. The highest altitude of a star is achieved when the semicircle of travel intersects the local meridian which will be 6 hours after rising (and 6 hours before setting). The altitude achieved at this point will be*90 - dec*, varying from an altitude of 90° (that is, at the zenith) for stars with dec = 0° to an altitude of 0° for stars with a dec of 90° (the NCP/SCP points as discussed).**The poles:**At the poles, exactly half the stars can be seen and these are all circumpolar. At the North Pole the stars that can be seen are those with dec greater than 0° and at the South Pole those with dec less than 0° (those with dec = 0 will be right on the horizon, for either pole). The celestial equator will be the circle of the horizon. The stars travel around horizontal circles (one circuit each day) at a constant altitude equal to the star's declination; the NCP (at the North Pole) will be stationary at 90° (the observer's zenith), and similarly for the SCP at the South Pole. Note that the stars travel parallel to the celestial equator, as always.**Intermediate latitudes**(e.g. Winchester 51°N): At intermediate latitudes - where to be honest, most of us live - some stars are circumpolar, some rise and set, and some are never visible. Here's the lowdown:**Stars are circumpolar**if*dec > 90 - latitude*in the northern hemisphere; for the southern hemisphere the formula is*dec < latitude -90*. Note both northern and southern latitudes are treated as positive here, so e.g. 45°N or 45°S are both considered as +45° - this shouldn't be confusing because we only apply the first formula to northern latitudes and the second one to southern latitudes. For Winchester, this is all stars with declination greater than 39°. For a southern equivalent of Winchester (at latitude 51°S) the circumpolar stars are those with declination less than -39.**Stars are invisible**(never rise) if*dec < latitude - 90*in the northern hemisphere or if*dec > 90 - latitude*in the southern hemisphere. In Winchester then, the invisible stars are those with declination less than -39; and for our southern equivalent at 51°S, it is stars with declination greater than 39.**The celestial equator**intersects the horizon at due east and due west and reaches its highest altitude at an angle of*90 - latitude*when facing south in the northern hemisphere (39° in Winchester); or when facing north in the southern hemisphere.**Stars travel in circles parallel to the celestial equator**, about the axis from the observer to the NCP (in the northern hemisphere) or the axis from the observer to the SCP (in the southern hemisphere). For both northern and southern hemipsheres, the stars travel across the sky in circles from the eastern half the sky to the western half: which is anticlockwise if you're facing north and clockwise if you're facing south. So if you are in the northern hemisphere facing the equator (south) the stars will be moving clockwise, while if you are in the southern hemisphere facing the equator (north) the stars will be moving anticlockwise. Similarly if you face the pole of the hemisphere you are in, the stars will move anticlockwise in the north and clockwise in the south.**Rising and setting azimuths and angles: northern hemisphere**Stars with a declination of 0 rise due east and set due west, and are up for 12 hours and down for 12 hours. In the northern hemisphere, stars with a declination greater than 0° rise in a direction (or azimuth) north of east and set north of west. They will rise at an angle south of the perpendicular (the perpendicular to the ground at their rising point, that is) equal to the latitude and make their way across the sky, reach the meridian half way through their journey and then set at an angle south of the perpendicular again equal to the latitude. The stars will be up longer than 12 hours, increasing towards 24 hours as the declination increases. Once the declination equals*90 - latitude*then the star touches the northern horizon but is otherwise always up and for any greater declinations the star is circumpolar (that is, up 24 hours). Those with declination less than 0° rise at an azimuth south of east and set south of west. Again they rise and set at angles south of the perpendicular equal to the latitude. The stars will be up less than 12 hours, decreasing towards 0 hours as the declination decreases. Once the declination equals*latitude - 90*the star touches the southern horizon but is otherwise not visible, and for any lesser declinations the star will be permanently invisible (that is, down 24 hours). Note that the stars will traverse a path through the sky that is always south of the rising and setting azimuths (except, of course at the actual rising and setting points on the horizon). Note also that stars that rise north of east (and therefore set north of west) are up longer than 12 hours and those rising south of east (and setting south of west) are up less than 12 hours. There is a formula for the rising azimuth as mentioned earlier,*cos(rising azimuth) = sin(declination)/cos(latitude)*- this formula is consistent with the earlier statement that at the equator (where*cos(latitude) = 1*) the rising azimuth is equal to the angle away from east equal to the star's declination. The*setting azimuth of a star = 360 - rising azimuth*, so the rising and setting points are symmetrical about the north-south axis.**Rising and setting azimuths and angles: southern hemisphere**Again, stars with a declination of 0 rise due east and set due west, and are up for 12 hours and down for 12 hours. In the southern hemisphere stars with a declination less than 0° will rise at an azimuth south of east and set south of west. They will rise at an angle north of the perpendicular equal to the latitude and make their way across the sky, reach the meridian half way through their journey and then set at an angle north of the perpendicular again equal to the latitude. The stars will be up longer than 12 hours, increasing towards 24 hours as the declination becomes more negative. Once the declination equals*latitude - 90*then the star touches the southern horizon but is otherwise always up and for any more negative declinations the star is circumpolar (that is, up 24 hours). Those with declination greater than 0° rise at an azimuth north of east and set north of west. Again they rise and set at angles north of the perpendicular equal to the latitude. The stars will be up less than 12 hours, decreasing towards 0 hours as the declination increases. Once the declination equals*90 - latitude*the star touches the northern horizon but is otherwise not visible, and for any greater declinations the star will be permanently invisible (that is, down 24 hours). Note that the stars will traverse a path through the sky that is always north of the rising and setting azimuths (except, of course at the actual rising and setting points on the horizon). Note also that stars that rise south of east (and therefore set south of west) are up longer than 12 hours and those rising north of east (and setting north of west) are up less than 12 hours.**To calculate the time that a star is up:**this is equal to twice the hour angle of the setting point. The reason is: first, we actually need the hour angle (remember this is measured from the celestial meridian) east to the rising point plus the hour angle west to the setting point - these are symmetric about the meridian so we simply need twice the hour angle to the setting point (hour angles are positive to the west, so we use that the setting point to get the sign right). Hour angles are measured along the celestial equator and stars travel parallel to the celestial equator, so the sum of these two hour angles will give the time for the star to move (along the axis of the celestial equator) from the rising point to the meridian and then to the setting point; it will be below the horizon for*24 - time above horizon*, of course. Take the example of stars on the equator: here the celestial equator is over the observers' heads from east to west and all stars travel parallel to this; the meridian is over the observers' head from north to south, so the direction of star travel is at right angles to the meridian; so the hour angle from meridian to the setting points will be on a quarter of a circle or 6^{h}whatever the star's rising and setting azimuth may be - and hence all stars are up for 12 hours at the equator, as we have already said. The question then is, how do we calculate the hour angle of the setting point away from the equator? Well spherical astronomy has the answer and to save us the trouble*hour angle of star setting point = - (sin(latitude)*sin(declination))/(cos(latitude)*sin(latitude))*. This is in degrees, so we need to divide by 15 to convert to hours and then double it since we need twice the hour angle. This will give you the time the star is up, more or less. To be slightly more accurate, note that the time you have will be in sidereal hours, so multiply by 0.997 to get the proper value in solar hours. We could also adjust for refraction, the non-spherical nature of the Earth, anomalies in the Earth's orbit, etc. (as ever), but we're pretty close here. We may need to adjust the sign of the formula for southern hemisphere latitudes (I haven't checked!).**Stars reach their highest and lowest altitudes**(upper and lower culmination) as they cross (or transit) the local meridian. We discussed this in the section on the celestial meridian, but for summary: in the northern hemisphere*altitude of upper culmination = 90 - (latitude - declination)*, where the altitude is measured from the southern horizon; and*altitude of lower culmination = (latitude + declination) - 90*, where the altitude is measured from the northern horizon. In the southern hemisphere*altitude of upper culmination = 90 - (latitude + declination)*where the altitude is measured from the northern horizon; and*altitude of lower culmination = (latitude - declination) - 90*where the altitude is measured from the southern horizon. You will only see the lower culmination if the star is circumpolar, and the upper culmination will be to the north (in the northern hemisphere) or the south (in the southern hemisphere) if the calculated altitude exceeds 90° (which happens if*dec > latitude*).**The altitude of the North Celestial Pole**(facing north, in the northern hemisphere) gives the latitude of the observer. At Winchester the NCP will be at an angle of 51° above the northern horizon. The altitude of the South Celestial Pole (facing south, in the southern hemisphere) similarly gives the southern latitude.

And to finally finish this chapter, **here's a list of the websites we've referenced:**

- Mag Dec Calculator at the US National Geophysical Data Center
- Bright Star Catalogue
- Topocentric Positions of Major Solar System Objects and Bright Stars
- Celestial to Horizon Co-ordinates Calculator
- Equation of Time
- Sidereal Clock

**Thank you.**

No problem. Right, I'm going to move on now. There is of course vastly more to cover about the stars - for instance aberration of starlight, motions of individual stars and galaxies, star catalogues, orbital influences on star positions (like the precession of the equinoxes, nutation), the whole field of spectroscopy (analyzing starlight), etc.. But we'll cover that in future sections. It's the Sun next. However ... I need to take a hiatus, as my next story Culture Man needs to be written before the summer's out. Can you give me a hand with that?

**Sure. Do you have any ice cream?**