(Note: this section requires familiarity with the tangent function. See "The Tangent" in the math refresher section.)

    The picture on the left is meant to represent the astronomer Claudius Ptolemy, who lived around 150 AD. It is an old picture, though not old enough for the artist to have actually known what Ptolemy looked like. But what is that gentleman holding?

    No, it isn't a religious symbol--the proportions are not right, and the marks on the stick do not seem appropriate. It is actually a cross staff (or "Jacob's staff"), a tool widely used by astronomers and navigators before the invention of the telescope, and for a while afterwards (but apparently not yet in Ptolemy's time). It consists of a yardstick with a perpendicular cross stick, attached to the stick in its middle and able to slide up and down.

    Astronomers used the cross-staff for measuring the angle between the directions of two stars. Ships' officers used it to measure the elevation angle of the noontime Sun above the horizon, which allowed them to estimate their latitude (see section on navigation). Various tricks were used to prevent or reduce the dazzling of the eye by sunlight while doing so. Columbus may have used one, and it may also have been part of the navigation equipment aboard the "Mayflower. "

    To measure the angle between two stars, the astronomer would place the staff just below one eye (drawing) and slide the cross-piece up and down. The cross-piece would have a pair of nails sticking out perpendicular to the drawing at symmetric locations such as B and B' (often several pairs of nails). The astronomer would slide the cross-piece up and down, until nail B one of the stars and nail B' the other.

    After that was achieved, he would lower the instrument and measure the distance AC. Then if A was the angle between the staff and the direction of one star, from the definition of the tangent

    The distance BC between the nail and the stick was always the same, and was already known to the astronomer--so, using a table of tangents, he could calculate the angle A. Since the instrument was symmetric, the angle between the directions of the stars was 2A.

Building your own Cross-Staff

    You could paint the nails white and have a friend behind your back illuminating them with a small flashlight (you should not look at any bright light, or else you lose night vision for some minutes). A neat trick is to paint them with glow-in-the-dark paint, or mark them with glow-in-the-dark tape. If you "charge" the glowing substance with a flashlight, then the faint glow is just bright enough to show the positions of the nails.

You will need:

•     A wooden yardstick
•     A strip of wood, 0.5" to 1" wide, about one foot (30 cm) long
•     The side of an empty cereal box, or any piece of cardboard glossy on one side.
•     4 nails about 2" long (thin "finishing nails" if you have them) to go on the strip
•     Several strong rubber bands
•     Glow-in-the-dark tape, or glow-in-the-dark paint.
•     (possibly needed: transparent office tape ("magic tape"),)
•     A pocket calculator with a tan^-1 button.

    The drawing above shows the completed cross-staff, though its proportions have been changed for clarity.

1. Cut a strip of the cardboard, 6" by 2" (15 by 5 centimeters), and wrap it tightly around the yardstick, glossy side to the wood.With the wrapped cardboard near the end of the stick, put several turns of a rubber band around it, to keep it tightly closed. You should now be able to slide the cardboard along the stick, but once in place it should be tight enough to keep its place.

2.     Place a second rubber band around the cardboard, also making several turns. You might add a third one as well.

3.     With pen or pencil, draw a line across the strip of wood at the place which will be its center, and also draw lines across it 1" and 4" (2.5 and 10 centimeters) on each side of it.
       Then draw a line along the strip marking its center, perpendicular to the other lines you have drawn.

4.     Insert the strip of wood under two of the loops of the second rubber band (or two of those of the second band and two of the third, for a firmer grip). The loops should cross the strip in the shape of the letter X. Do not force the strip in, but use your fingers to lift the loops, making it easy to slide the strip to where the middle mark is under the center of the "X" and above the middle of the yardstick. Once you let go of the bands, the strip should be held tight and be perpendicular to the yardstick.

5.     Drive the nails into the strip of wood at the points where the 1" and 4" marks cross the center line. Make sure they are straight, and drive them just deep enough to put them firmly in place.

6.     You want the nails to glow in the dark. If you have glow-in-the-dark paint, you can simply paint the nails. Make sure they are clean, so that the paint sticks. Or if you have glow-in-the dark tape, wrap some of it around each nail.

   If the tape is too thick for wrapping, cut a piece about 1.5" long, and place it lengthwise along one of the nails, facing the side of the yardstick on which its numbering begins, the side from which you will be looking. Then press the sides of the tape around the nail. If it does not stick, put a short length of transparent tape over it, perpendicular to it, and join its ends tightly on the other side (you won't be bothered by tape on the far side of the nail). Repeat with the other 3 nails. Or else, invent your own method!

7.     With a small ruler, measure the distance between the center line on which the nails are and the front of the slider: that distance will have to be added to the distance AC (in the formula above) which you read off the yardstick at the front of the slider.     (Or else, if the "X" and the center line are exactly in the middle of the slider, you can read the yardstick markings both at the front and the back of the slider, than use their average for AC).

   You are now ready to use your cross-staff.

       If you live in a city, that could be the hardest part: the sky may be so bright that only a few stars show. And you may see the nails even if they do not glow in the dark, as dark outlines against the background of the bright sky.

       If however you are lucky enough to get away from bright lights on a dark night (not a moonlit one), enjoy the impressive spectacle you see! For the best view, wait some minutes to let your eyes adapt to the darkness. To use the cross-staff after that, shine the flashlight for a short time at each of the nails, shielding the light from your eyes to preserve their adaptation. Then place the end of the yardstick below one eye, holding it with one hand, while the other adjusts the slide so that one pair of nails covers the two stars whose separation you want to measure.

       The rest is just a calculation of inverse tangents. If you find you cannot easily adjust the cross-staff, you may add another pair of nails, at a different distance from the middle.

Coordinates on a Flat Plane

René Descartes

A more widely used system are cartesian coordinates, based on a set of axes perpendicular to each other. They are named for Rene Descartes ("De-cart"), a French scientist and philosopher who back in the 1600s devised a systematic way of labeling each point on a flat plane by a pair of numbers. You may well be already familiar with it.

The system is based on two straight lines ("axes"), perpendicular to each other, each of them marked with the distances from the point where they meet ("origin")--distances to the right of the origin and above it, the origin being taken as positive and on the other sides as negative (see drawing below).

The distance on one axis is named "x" and on the other axis "y". Given then a point P, one draws from it lines parallel to the axes, and the values of x and y at their intersections completely define the point. In honor of Descartes, this way of labeling points is known as a cartesian system and the two numbers (x,y) that define the position of any point are its cartesian coordinates.

Graphs use this system, as do some maps.

This works well on a flat sheet of paper, but the real world is 3-dimensional and sometimes it is necessary to label points in 3-dimensional space. The cartesian (x,y) labeling can be extended to 3 dimensions by adding a third coordinate z. If (x,y) is a point on the sheet, then the point (x,y,z) in space is reached by moving to (x,y) and then rising a distance z above the paper (points below it have negative z).

Very simple and clear, once a decision is made on which side of the sheet z is positive. By common agreement the positive branches of the (x,y, z) axes, in that order, follow the thumb and the first two fingers of the right hand when extended in a way that they make the largest angles with each other.

What follows uses the trigonometric functions sine and cosine; if these are not familiar to you, either skip the rest of the section, or go learn about them.

Polar Coordinates

Cartesian c oordinates (x,y) are not the only way of labeling a point P on a flat plane by a pair of numbers. Other ways exist, and they can be more useful in special situations.

 One system ("polar coordinates") uses the length r of the line OP from the origin to P (i. e. the distance of P distance to the origin) and the angle that line makes with the x-axis. Angles are often denoted by Greek letters, and here we follow conventions by marking it with f (Greek f). Note that while in the cartesian system x and y play very similar roles, here roles are divided: r gives distance and f direction.

The two representations are closely related. From the definitions of the sine and cosine

x = r cos f
y = r sin f

That allows (x,y) to be derived from polar coordinates. To go in the opposite direction and derive (r,f) from (x,y), note that from the above equations (or from the theorem of Pythagoras) one can derive r:

r² = x² + y²

Once r is known, the rest is easy

cos f = x/r
sin f = y/r

These relations fail only at the origin, where x = y = r = 0. At that point, f is undefined and one can choose for it whatever one pleases.

In three dimensional space, the cartesian labeling (x,y,z) is nicely symmetric, but sometimes it is convenient to follow the style of polar coordinates and label distance and and direction separately. Distance is easy: you take the line OP from the origin to the point and measure its length r. You can even show from the theorem of Pythagoras that in this case

r² = x² + y² + z²

All the points with the same value of r form a sphere of radius r around the origin O. On a sphere we can label each point by latitude l (lambda, small Greek L) and longitude f (phi, small Greek F), so that the position of any point in space is defined by the 3 numbers (r, l, f).

Azimuth and Elevation

An old surveyor's tele-   scope (theodolite).

The surveyor's telescope is designed to measure two such angles. The angle f is measured in a horizontal plane, is known as azimuth and is measured from the north direction. A rotating table allows the telescope to be pointed in any azimuth.

The angle l is called elevation and is the angle by which the telescope is lifted above the horizontal (if it looks down, l is negative). The two angles together can in principle specify any direction: f ranges from 0 to 360⁰, and l from -90⁰ (straight down or "nadir") to +90⁰ (straight up or "zenith").

Again, one needs to decide from what direction is the azimuth measured--that is, where is azimuth zero? The rotation of the heavens (and the fact most humanity lives north of the equator) suggests (for surveyor-type measurements) the northward direction, and this is indeed the usual zero point. The azimuth angle (viewed from the north) is measured counterclockwise.

Mathematicians however prefer their own notation and replace "latitude" (or elevation) l with co-latitude q= 90 - l deg., the angle not to the horizon but to the vertical. T he angle q (theta, one of two t-s in Greek) goes from 0 to 180⁰, not from -90⁰ to + 90⁰. This actually may make more sense, because it is easier to measure an angle between two lines (OP and the vertical) rather than between a line and a flat plane (OP and the horizontal).]

[And in case you have to know: In referring (r, q, f) to cartesian (x,y,z) with the same origin, q is measured from the z-axis and f is measured in the (x,y) plane, counterclockwise from the x-axis. ]

So familiar has the calendar become that people tend to forget that it, too, had to be invented. Early farmers needed to know when to plow and sow ahead of rainy seasons, and to time other seasonal activities. Early priests in Babylonia, Egypt, China and other countries, even among the Maya in America, examined therefore the motions of the Sun, Moons and planets across the sky, and came up with a variety of calendars, some still in use.

The Day

The basic unit is obviously the day: 24 hours, 1440 minutes, 86400 seconds, each second slightly longer than the average heartbeat. The day is defined by the motion of the Sun across the sky, and a convenient benchmark is noon, the time when the Sun is at its highest (i. e. most distant from the horizon) and is also exactly south or north of the observer.

"One day" can therefore be conveniently defined as the time from one noon to the next. A sundial can track the Sun's motion across the sky by the shadow of a rod or fin ("gnomon") pointing to the celestial pole (click here for construction of a folded-paper sundial), allowing the day to be divided into hours and smaller units. Noon is the time when the shadow points exactly south (or north) and is at its shortest.

What then is the period of the Earth's rotation around its axis? A day, you say? Not quite.

Suppose we observe the position of a star in the sky--for instance Sirius, the brightest of the lot. One full rotation of the Earth is the time it takes for the star to return to its original position (of course, we are the ones that move, not the star). That is almost how the day is defined, but with one big difference: for the day, the point of reference is not a star fixed in the firmament, but the Sun, whose position in the sky slowly changes. During the year the Sun traces a full circle around the sky, so that if we keep a separate count of "Sirius days" and "Sun days", at the end of the year the numbers will differ by 1. We will get 366. 2422 "star days" but only 365. 2422 Sun days.

It is the "star day" (sidereal day) which gives the rotation period of the Earth, and it is about 4 minutes shy of 24 hours. A clockwork designed to make a telescope follow the stars makes one full rotation per sidereal day.

The clocks we know and use, though, are based on the solar day--more precisely, on the average solar day, because the time from noon to noon can vary as the Earth moves in its orbit around the Sun. By Kepler's laws (discussed in a later section) that orbit is slightly elliptical. The distance from the Sun therefore varies slightly, and by Kepler's second law, the motion speeds up when nearer to the Sun and slows down when further away. Such variations can make "sun-dial time" fast or slow, by up to about 15 minutes.

Very precise atomic clocks nowadays tell us that the day is gradually getting longer. The culprits are the tides, twin waves raised in the Earth's ocean by (mainly) the Moon's gravitational pull. As the waves travel around the Earth, they break against shorelines and shallow seas, and thus give up their energy: theory suggests that this energy comes out of the (kinetic) energy of the Earth's rotational motion.

The Year

The year is the time needed by the Earth for one full orbit around the Sun. At the end of that time, the Earth is back to the same point in its orbit, and the Sun is therefore back to the same apparent position in the sky.

It takes the Earth 365. 2422 days to complete its circuit (average solar days), and any calendar whose year differs from this number will gradually wander through the seasons. The ancient Roman calendar had 355 days but added a month every 2 or 4 years: it wasn't good enough, and by the time Julius Caesar became ruler of Rome, it had slipped by three months.

In 46 BC Caesar introduced a new calendar, named after him the Julian calendar. It is similar to the one used today: the same 12 months, and an added day at the end of February every 4th year ("leap year"), on years whose number is divisible by 4. Two years afterwards the 5th month of the Roman year was renamed July, in honor of Julius. The name of his successor, Augustus Caesar, was later attached to the month follwing July.

pretty close to the full value 365. 2422. The Metonic calendar is thus more accurate than the Julian one, though less so than the Gregorian. It is still used by Jews, on whose calendar each month begins at or near the new moon, when the Moon's position in the sky is nearest to the Sun's. The traditional Chinese calendar also uses of a formula like Meton's, which was probably invented by the ancient Babylonians.

For more--much more!--see the "Maya Astronomy Page."   About the Maya and Venus, see the chapter "Bringing Culture to the Physicists", p. 313 in "Surely You're Joking, Mr. Feynman!" by Richard Feynman

The priests of ancient Babylonia and Egypt were also pioneer astronomers. They studied the heavens, mapped their constellations, identified the path of the Sun and estimated the periods of the Moon and Sun as they moved across the sky.

But it was a Greek astronomer, Hipparchus of Nicea, who made the first major new discovery in astronomy. Comparing observations more than a century apart, Hipparchus proposed that the axis around which the heavens seemed to rotate shifted gradually, though very slowly.

Viewed from Earth, the Sun moves around the ecliptic, one full circuit each year. Twice a year, at equinox, day and night are equal and the Sun rises exactly in the east and sets exactly in the west. Ancient astronomers had no good clocks and could not tell when the day and night had the same length, but they could identify the equinox by the Sun rising exactly in the east and setting exactly in the west. At those times the Sun's position is at one of the intersections between the ecliptic and the celestial equator.

Around the year 130 BC, Hipparchus compared ancient observations to his own and concluded that in the preceding 169 years those intersections had moved by 2 degrees. How could Hipparchus know the position of the Sun among the stars so exactly, when stars are not visible in the daytime? By using not the Sun but the shadow cast by the Earth on the moon, during an eclipse of the Moon! During an eclipse, Sun, Earth and Moon form a straight line, and therefore the center of the Earth's shadow is at the point on the celestial sphere which is exactly opposite that of the Sun.

"The Dawning of the Age of Aquarius"

Hipparchus concluded that the intersection marking the equinox slowly crept forward along the ecliptic, and called that motion "the precession of the equinoxes. " The rate is about one full circle in 26 000 years. In ancient times the intersection marking the spring equinox was in the constellation of Aries, the ram, and for that reason the intersection (wherever it might be) is still sometimes called "the first point in Aries."

Around the year 1 it moved into the constellation of Pisces (pronounced "pie-sees" in the US) and currently it is again in transition, to the constellation of Aquarius, the water carrier. If you ever heard the song "The dawning of the age of Aquarius" from the musical "Hair," that is what it is all about. To astronomers precession is mainly another factor to be taken into account when aiming a telescope or drawing a star chart; but to believers in astrology, the "dawning of the age of Aquarius" is a great portent and may mark the beginning of a completely new and different era.

The Precession of the Earth's Axis

What does this motion tell us about the Earth's motion in space? If you ever had a spinning top, you know that its axis tends to stay lined up in the same direction--usually, vertically, though in space any direction qualifies.

Precession of a spinning  top: the spin axis traces   the surface of a cone.

Give it a nudge, however, and the axis will start to gyrate wildly around the vertical, its motion tracing a cone (drawing). The spinning Earth moves like that, too, though the time scale is much slower--each spin lasts a year, and each gyration around the cone takes 26 000 years. The axis of the cone is perpendicular to the plane of the ecliptic.

The cause of the precession is the equatorial bulge of the Earth, caused by the centrifugal force of the Earth's rotation (the centrifugal force is discussed in a later section). That rotation changes the Earth from a perfect sphere to a slightly flattened one, thicker across the equator. The attraction of the Moon and Sun on the bulge is then the "nudge" which makes the Earth precess.

Through each 26 000-year cycle, the direction in the sky to which the axis points goes around a big circle, the radius of which covers an angle of about 23.50. The pole star to which the axis points now (within about one degree) used to be distant from the pole, and will be so again in a few thousand years (for your information, the closest approach is in 2017). Indeed, the "pole star" used by ancient Greek sailors was a different one, not nearly as close to the pole.

Because of the discovery mady by Hipparchus, the word "precession" itself no longer means "shift forward" but is now applied to any motion of a spin axis around a cone--for instance, the precession of a gyroscope in an airplane's instrument, or the precession of a spinning satellite in space.

Precession of a spinning scientific payload (also known as its "coning"--from "cone"--or its "nutation") is an unwelcome feature, because it complicates the tracking of its instruments. To eliminate it, such satellites use "nutation dampers," small tubes partially filled with mercury. If the satellite spins as it was designed to do, the mercury merely flows to the part of the tube most distant from the spin axis, and stays there. However, if the axis of rotation precesses, the mercury sloshes back and forth in the tube. Its friction then consumes energy, and since the source of the sloshing is the precession of the spin axis, that precession (very gradually) loses energy and dies down.

[In the section on the calendar, we saw that the Earth's rotation is slowed down very gradually by the tides, raised by the gravity of the Moon. That process is a bit similar to the action of nutation dampers: the energy of the tides is "lost"--that is, converted to heat--when the waves caused by tides break up on the seashore, and that loss is ultimately taken away from the rotational motion (not the precession) of the Earth]

Some say the world will end in fire,
Some say in ice.
From what I've tasted of desire
I hold with those who favor fire.
But if it had to perish twice,
I think I know enough of hate
To say that for destruction ice
Is also great
And would suffice
                                Robert Frost

Some 2000 years after Hipparchus, in the year 1840. Louis Agassiz, a Swiss scientist, published a book on glaciers, a familiar feature of his homeland--huge rivers of ice created by accumulated snowfall, filling valleys and slowly creeping downwards to their end points, lakes of meltwater (or, in some other countries, the sea).

Glaciers leave an imprint on the landscape: they scratch and grind down rocks, and carry loads of gravel, at times even big boulders, from the mountains to the plains, leaving them far from their origins, wherever the ice finally melts. Agassiz, who later became a distinguished professor at Harvard, noted that such imprints existed all over northern Europe, and suggested that the lands now inhabited by Germans, Poles, Russians and others used to be covered by enormous glaciers.

America, too, had its glaciers; Cape Cod, for instance, is a left-over pile of glacial gravel. Later geological studies found evidence that such glaciers advanced and retreated several times in the last million years. The last retreat, a rather abrupt one, occured about 12,000 years ago.

Right now, northern winter occurs in the part of the Earth's orbit where the north end of the axis points away from the Sun. However, since the axis moves around a cone, 13,000 years from now, in this part of the orbit, it will point towards the Sun, putting it in mid-summer just when the Earth is closest to the Sun.

At that time one expects northern climate to be more extreme, and the oceans then have a much smaller effect, since the proportion of land in the northern hemisphere is much larger. Milankovich argued that because winters were colder, more snow fell, feeding the giant glaciers. Furthermore, he said, since snow was white, it reflected sunlight, and with more severe winters, the snow-covered land warmed up less effectively once winter had ended. Climate is maintained by a delicate balance between opposing factors, and Milankovich argued that this effect alone was enough to upset that balance and cause ice ages.

Milankovich was aware that this was just one of several factors, since it turns out that ice ages do not recur every 26,000 year, nor do they seem common in other geological epochs. The eccentricity of the Earth's orbit, which determines the closest approach to the Sun, also changes periodically, as does the inclination of the Earth's axis to the ecliptic. But overall the notion that ice ages may be linked to the motion of the Earth through space may be currently our best guess concerning the causes of ice ages.

Postscript, 28 July 1999.     The magnitude of the "Milankovich effect" depends on the difference between largest and smallest distances from the Sun. That, in its turn, depends on the eccentricity of the Earth's orbit, which varies with a 100,000-year cycle, on which a 413,000-year cycle is superposed. J. Rial (Univ. of North Carolina) found signatures of those cycles in the oxygen isotope content of deep-sea sediments, in full agreement with the Milankovich theory. His work is in "Science," vol. 285, p. 564, 23 July 1999; a non-technical explanation "Why the Ice Ages Don't Keep Time " is on pages 503-504 of the same issue.

Today it is well known that the Earth is a sphere, or very close to one (its equator bulges out a bit because of the Earth's rotation). When Christopher Columbus proposed to reach India by sailing west from Spain, he too knew that the Earth was round. India was the source of precious spices and other rare goods, but reaching it by sailing east was difficult, because Africa blocked the way. On a round globe, however, it should also be possible to reach India by sailing west, and this Columbus proposed to do (he wasn't the first one to suggest this--see below).

Sometimes the claim is made that those who opposed Columbus thought the Earth was flat, but that wasn't the case at all. Even in ancient times sailors knew that the Earth was round and scientists not only suspected it was a sphere, but even estimated its size.

If you stand on the seashore and watch a ship sailing away, it will gradually disappear from view. But the reason cannot be the distance: if a hill or tower are nearby, and you climb to the top after the ship has completely disappeared, it becomes visible again. Furthermore, if on the shore you watch carefully the way the ship disappears from view, you will notice that the hull vanishes first, while the masts and sails (or the bridge and smokestack) disappear last. It is as if the ship was dropping behind a hill, which in a way is exactly the case, the "hill" being the curve of the Earth's surface.
  To find out how the distance to the horizon is calculated, click here

The Greek philosopher Eratosthenes went one step further and actually estimated how large the Earth was. He was told that on midsummer day (June 21) in the town of Syene in southern Egypt (today Aswan, near a huge dam on the river Nile) the noontime Sun was reflected in a deep well, meaning that it was right overhead, at zenith. Eratosthenes himself lived in Alexandria, near the river's mouth, north of Syene, about 5000 stadia north of Syene (the stadium, the size of a sports arena, was a unit of distance used by the Greeks). In Alexandria the Sun on the corresponding date did not quite reach zenith, and vertical objects still threw a short shadow. Eratosthenes established that the direction of the noon Sun differed from the zenith by an angle that was 1/50 of the circle, that is, 7. 2 degrees, and from that he estimated the circumference of the Earth to be 250,000 stadia.

Other estimates of the size of the Earth followed. Some writers reported that the Greek Posidonius used the greatest height of the bright star Canopus above the horizon, as seen from Egypt and from the island of Rhodes further north (near the southwestern tip of Turkey). He obtained a similar value, a bit smaller. The Arab Khalif El Ma'mun, who ruled in Baghdad from 813 to 833, sent out two teams of surveyors to measure a north-south baseline and from it also obtained the radius of the Earth. Compared to the value known today, those estimates were pretty close to the mark.

The idea of sailing westward to India dates back to the early Romans. According to Dr. Irene Fischer, who studied this subject, the Roman writer Strabo, not long after Erathosthenes and Posidonius, reported their results and noted:

"if of the more recent measurements of the Earth, the one which makes the Earth smallest in circumference be introduced--I mean that of Posidonius who estimates its circumference at about 180,000 stadia, then. . . "

and he continues:

"Posidonius suspects that the length of the inhabited world, about 70,000 stadia, is half the entire circle on which it had been taken, so that if you sail from the west in a straight course, you will reach India within 70,000 stadia. "

Notice that Strabo--for unclear reasons--reduced the 250,000 Stadia of Eratosthenes to 180,000, and then stated that half of that distance came to just 70,000 stadia. Handling his numbers in that loose fashion, he could argue that India was not far to the west.

Columbus Again

All these results were known to the panel of experts which King Ferdinand appointed to examine the proposal made by Columbus. They turned Columbus down, because using the original value by Eratosthenes, they calculated how far India was to the west of Spain, and concluded that the distance was far too great.

In the end Queen Isabella overruled the experts, and the rest is history. We may never know whether Columbus knowingly fudged his values to justify an expedition to explore the unknown, or actually believed India was not too far to the west of Spain. He certainly did call the inhabitants of the lands he discovered "Indians," a mislabeling which still persists.

But we do know that if the American continent had not existed, the experts would have been vindicated: Coumbus with his tiny ships could never have crossed an ocean as wide as the Atlantic and Pacific combined. In hindsight the exploration of the unknown may be justification enough!

As for the size of the Earth, it was accurately measured many times since (see item "geodesy" in an encyclopaedia), one notable effort being that of the French Academy of Sciences in the late 1700s. Their aim was to devise a new unit of distance, equal to one part in 10,000,000 of the distance from the pole to the equator (as Eratosthenes showed, it is enough to measure part of that distance). Nowadays that distance is known even more accurately, but the unit introduced by the French academy is still used as the standard of all distance measurements. It is called the meter.

Your line of sight to the horizon is a tangent to the Earth--a line which touches the sphere of the Earth at just one point, marked B in the drawing here. If O is the center of the sphere of the Earth, by a well-known theorem of geometry such a tangent is perpendicular to the radius OB, that is, it makes a 90^o angle with it.

It follows that the triangle OAB obeys the theorem of Pythagoras, which here can be written

(OA)² = (AB)² + (OB)²

or if the length of each line is spelled out

(R + h)² = D² + R²

By an algebraic identity (derived in the "mathematical refresher"), the left-hand side equals R² + 2Rh + h², giving

R² + 2Rh + h² = D² + R²

If now R² is subtracted from both sides and the remaining terms on the left are rearranged

h(2R + h) = D²

The diameter 2R of the Earth is much bigger than h, and therefore the error introduced if (2R+h) is replaced by 2R is very, very small. Carrying out this replacement gives

2Rh = D²

D = SQRT(2Rh)

where SQRT stands here for "square root of". This equation lets one calculate D--in kilometers, if h and R are given in kilometers--but one it is also possible to simplify further:

SQRT(2Rh) = (SQRT(2R)) x (SQRT(h))

with the two square roots multiplied. Using R = 6371 km, SQRT(2R) = 112.88, giving

D = 112.88 km SQRT(h)

If you are standing atop a mountain 1 km high, h = 1 km and your horizon should be 112.88 km away (we neglect the refraction of light in the atmosphere, which may modify this value). From the top of Mauna Kea on Hawaii, an extinct volcano about 4 km high (also the site of important astronomical observatories), the horizon should be about twice as distant, 226 km. On the other hand, standing on the beach with your eyes 2 meters = 0.002 km above the water, since SQRT(0.002) = 0.04472, the horizon is only 5 km distant.

The calculation should also hold the other way around. From a boat on the ocean you should begin seeing the top of Mauna Kea after you pass a distance of 226 km (again, not accounting for refraction). On November 15, 1806, Lieutenant Zebulon Pike of the US Army, leading an exploration party across the plains of the midwestern US, saw through his spyglass the top of a distant peak, just above the horizon. It took his party a week to cover the 100 miles to the mountain, which is now known as Pike's Peak, one of the tallest in Colorado. Pike actually tried to climb to its top, but the snow and the unexpected height of the mountain forced him back.

  "Pre-Trigonometry"

1. The baseline is perpendicular to the line from its middle to the object, so that the triangle ABC is symmetric. We will denote its side by r:
   AC = BC = r

2. The length c of the baseline AB is much less than r. That means that the angle a between AC and BC is small; that angle is known as the parallax of C, as viewed from AB.

3. We do not ask for great accuracy, but are satisfied with an approximate value of the distance--say, within 1%.

(The Greek mathematician Archimedes derived p to about 4-figure accuracy, though he expressed it differently, since decimal fractions appeared in Europe only some 1000 years later.)

we get

and dividing by 2p

Therefore, if we know b, we can deduce r. For instance, if we know that a = 5.73°, 2pa = 36° and we get

Estimating distance outdoors

1.   Stretch your arm forward and extend your thumb, so that your thumbnail faces your eyes. Close one eye (A') and move your thumb so that, looking with your open eye (B'), you see your thumbnail covering the landmark A.

2.   Then open the eye you had closed (A') and close the one (B') with which you looked before, without moving your thumb. It will now appear that your thumbnail has moved: it is no longer in front of landmark A, but in front of some other point at the same distance, marked as B in the drawing.

3.   Estimate the true distance AB, by comparing it to the estimated heights of trees, widths of buildings, distances between power-line poles, lengths of cars etc. The distance to the landmark is 10 times the distance AB.   Why does this work? Because even though people vary in size, the proportions of the average human body are fairly constant, and for most people, the angle between the lines from the eyes (A',B') to the outstretched thumb is about 6°, close enough to the value 5.73° for which the ratio 1:10 was found in an earlier part of this section.   That angle is the parallax of your thumb, viewed from your eyes. The triangle A'B'C has the same proportions as the much larger triangle ABC, and therefore, if the distance B'C to the thumb is 10 times the distance A'B' between the eyes, the distance AC to the far landmark is also 10 times the distance AB.

   How far to a Star?

     When estimating the distance to a very distant object, our "baseline" between the two points of observation better be large, too. The most distant objects our eyes can see are the stars, and they are very far indeed: light which moves at 300,000 kilometers (186,000 miles) per second, would take years, often many years, to reach them. The Sun's light needs 500 seconds to reach Earth, a bit over 8 minutes, and about 5.5 hours to reach the average distance of Pluto, the most distant planet. A "light year" is about 1600 times further, an enormous distance.

     The biggest baseline available for measuring such distances is the diameter of the Earth's orbit, 300,000,000 kilometers. The Earth's motion around the Sun makes it move back and forth in space, so that on dates separated by half a year, its positions are 300,000,000 kilometers apart. In addition, the entire solar system also moves through space, but that motion is not periodic and therefore its effects can be separated.

     And how much do the stars shift when viewed from two points 300,000,000 km apart? Actually, very, very little. For many years astronomers struggled in vain to observe the difference. Only in 1838 were definite parallaxes measured for some of the nearest stars--for Alpha Centauri by Henderson from South Africa, for Vega by Friedrich von Struve and for 61 Cygni by Friedrich Bessel.

     Such observations demand enormous precision. Where a circle is divided into 360 degrees (360°), each degree is divided into 60 minutes (60')--also called "minutes of arc" to distinguish them from minutes of time--and each minute contains 60 seconds of arc (60"). All observed parallaxes are less than 1", at the limit of the resolving power of even large ground-based telescopes.

     In measuring star distances, astronomers frequently use the parsec, the distance to a star whose yearly parallax is 1"--one second of arc. One parsec equals 3.26 light years, but as already noted, no star is that close to us. Alpha Centauri, the sun-like star nearest to our solar system, has a distance of 4.3 years and a parallax of 0.75".

     Alpha Centauri is not a name, but a designation. Astronomers designate stars in each constellation by letters of the Greek alphabet--alpha, beta, gamma, delta and so forth, and "Alpha Centauri" means the brightest star in the constellation of Centaurus, located high in the southern skies. You need to be south of the equator to see it well.

  Aristarchus around 270 BC derived the Moon's distance from the duration of a lunar eclipse (Hipparchus later improved that method).

    It was commonly accepted in those days that the Earth was a sphere (although its size was only calculated a few years later, by Eratosthenes ). Astronomers also believed that the Earth was the center of the universe, and that Sun, Moon, planets and stars all orbited around it. It was only natural, then, that Aristarchus assumed that the Moon moved in a large circle around Earth.

    Let R be the radius of that circle and T the time it takes the Moon to go around once, about one month. In that time the Moon covers a distance of 2pR, where p ~ 3.1415926... (pronounced "pi") is a mathematical constant, the ratio (circumference/diameter) in a circle.

    An eclipse of the Moon occurs when the Moon passes through the shadow of the Earth, on the opposite side from the Sun (therefore, it must be a full Moon). If r is the radius of the Earth, the shadow's width is close to 2r. Let t be the time it takes the mid-point of the Moon to cross the center of the shadow, about 3 hours (in eclipses of the longest duration, when the Moon crosses the center of the shadow).

    If the Moon moves around Earth at a constant speed--and it takes time T to cover 2pR ~ 6.28R, and time t to cover 2r--then

From this Aristarchus obtained

which fits the average distance of the Moon accepted today, 60 Earth radii.

A Few Extra Details

    Viewed from Earth, a "new Moon" (occuring between the time a thin crescent is last seen before sunrise and the time one is seen shortly after sunset) happens when the Moon in its apparent motion around the sky overtakes the Sun. However, by the time of the next new Moon, the Sun's position in the sky has already shifted. If the Sun takes 12 months to go around the sky (or around the ecliptic, or around the zodiac), then in one month it completes 1/12 of its circuit. The Moon must therefore complete [1+(1/12)] circuits to catch up with the Sun again, and the lunar month ("synodic period") is about 1/12 of a month longer than the actual period of 27.32 days.

    Also, the Earth's shadow has only approximately the width of 2r. It would have very nearly a width of 2r if the Sun were a point-like light source (exactly that width if it were infinitely far away). Actually, however, the Sun is large enough to appear as a disk which covers about half a degree of the sky. As a result, the Earth's shadow is not a cylinder but a gradually narrowing cone, and at the Moon's distance it is already about 25% narrower than 2r.

    Here is another way of looking at the same process. Suppose we observe the eclipse from the Moon. Seen from there, the Earth moves from east to west--from A to B in the drawing, assuming the eclipse is of greatest length (i.e. the middle of the Earth passes in front of the Sun).

    The eclipse begins when the last bit of the western edge of the Sun passes point A (bottom of the drawing shown) and ends when the first bit of the eastern edge of the Sun pokes out at point B. That takes less time than it takes for the center of the Sun to pass from A to B (top drawing) which would be the duration of the eclipse if the Sun were a tiny point source, located at its middle.

  In a lunar eclipse, if the width of the shadow of the Earth is twice the width of the Moon, then the width of the Earth itself is (very nearly) three times that of the Moon--not twice, as one might perhaps think. Here is why.

 The Sun is not a point of light but an extended source, with a disk covering a circlular patch in the sky, about 0.5° across. This makes the shadow of the Earth not a cylinder, stretching to infinity without narrowing down, but a cone, with an angle of 0.5° across its apex C (drawing). AB is here the diameter of the Earth, and the directions AC and BC represents rays from opposite edges of the Sun's disk, rays whose directions differ by 0.5°.

 If x is the diameter of the Moon and R its distance, then according to Aristarchus, the width ED of the shadow at distance R equals 2x (actually, 2.5x comes closer to the mark). We add to the drawing points H and K so that HA = KD = x.

 The width of the Moon as seen from point H is KD = x, and since the Moon's size in the sky is about the same as the Sun's, the angle KHD (shaded) should also equal 0.5°. We now extend the line AD = R a further distance R to point F. Then the two shaded triangles HKD and KFD are congruent (= same in size and shape) and have the same 0.5° angle as the angle at C. Indeed, one can prove now that the triangles GFC and AHD are also congruent to the two shaded ones.

It follows then that AC = 3R, and from simple proportions (see drawing) AB = 3x.

 That eclipse was total at the Hellespont--the Dardanelles, part of the narrow strait that separates the European and Asian parts of Turkey--but only 4/5 of the Sun were covered in Alexandria of Egypt, further to the south.

 Hipparchus knew that when the Sun was eclipsed, it and the Moon occupied the same spot on the sphere of the heavens. The reason, he assumed, was that the Moon passed between us and the Sun.

 He believed that the Sun was much more distant than the Moon, as Aristarchus of Samos had concluded, about a century earlier, from observing the time when the Moon was exactly half full (see sections #8c and #9a).. He also assumed that the peak of the eclipse occured at the same time at both locations (not assured, but luckily not too far off), and he then carried out the following calculation.

 Viewed from Alexandria (point A), at that same moment, the point E only overlapped point C on the Sun, about 1/5 solar diameter short of the edge--which was why the eclipse there was not total. One-fifth of the Sun's diameter covers about 0.1° in the sky, so the small angle a (alpha, Greek A) between the two directions measured about 0.1 degrees. That angle is the parallax of the edge of the Moon, viewed from the above two locations.

It is unlikely that Hipparchus knew the distance AB, but he probably knew the latitudes of the Hellespont and Alexandria. The local latitude can be shown to be equal to the elevation of the celestial pole above the horizon and today can be readily deduced by observing the height of Polaris, the pole star above the horizon. In the time of Hipparchus the pole of the heavens wasn't near Polaris (because of the precession of the equinoxes), but Hipparchus, who had mapped the positions of about 850 stars, must have known its position quite well.

The latitude of the Hellespont (from a modern atlas) is about 40° 20' (40 degrees and 20 minutes, 60 minutes per degree), while that of Alexandria is about 31° 20', a difference of 9°. We will also assume Alexandria is exactly due south. If furtermore r is the radius of the Earth, then the circumference of the Earth 2p R, where p = 3.1415926... ("pi", Greek lower-case P) is the ratio between the circumference and diameter of any circle. Since the circumference also equals 360°, we get

where the dot marks multiplication (algebra's equivalent to the x symbol).

The distance R to the Moon

suggesting the Moon's distance is 90 Earth radii, an overestimate of about 50%.

A more accurate calculation

  Then the section cut by the angle a from the circle of radius R around E is not AB but AF (second drawing), which is smaller. Taking this into account reduces the distance.

 We don't know where the Sun was during the 129 BC eclipse, but it must have been on the ecliptic (the words are obviously related!), which places it within 23.5° of the celestial equator, on either side. Assuming it was on the equator (that is, it passed overhead on the Earth's equator) and south of the reported observations (i.e. the eclipse occured near noon) one can make a crude estimate of the correction, using simple trigonometry (see section M-8).

 The Hellespont is around latitude of 40 degrees, and as the drawing shows, that is also the angle between the Moon's direction and the zenith. From the drawing (x marks multiplication)

Repeating the preceding calculation for AF

and in the end

Final comments

  In the absence of accurate timing, the method is almost guaranteed to produce an overestimate. The Earth rotates beneath the shadow spot cast by the Moon, which makes that spot sweep over a long strip, hitting many different locations at different times. The Hellespont was just one of many places where the eclipse was total. Similarly, Alexandria was just one of many locations where 4/5 of the Sun was covered. Randomly selecting point B from the first group and point A from the second may give a much longer baseline AB and a much larger (and incorrect) distance of the Moon. The fact Alexandria is almost exactly south of the Hellespont does not guarantee their peak eclipse times are the same, just that they are not too different.

Does history repeat itself?

1. First, print out this page with its map. If the quality of the print is poor, use a ruler to draw horizontal lines of latitude 30 (through Cairo, Egypt), and 40 (through Ankara, Turkey). Actually, those lines slant and curve a little, but in this crude calculation, the slant can be ignored.
   
   
2. With a ruler measure the distance between the two lines. That distance equals 10 degrees of latitude.
   
   
3. Mark on your map a point on the southern shore of the Black Sea where the line of totality passed. In the lifetime of Hipparchus the city of Heraclea stood there, and Hipparchus might have used it instead of the Hellespont, had his eclipse been like that of 1999.
   
   
4. Measure with the ruler the distance from Alexandria to the point you have marked on the line of totality. Use the scale you calculated to derive the corresponding distance measured in degrees of latitude.
   
   
5. The eclipse occured on 11 August, about halfway between midsummer and the fall equinox. At noon on midsummer day, June 21, the Sun is 23.5° north of the equator; on the autumn equinox, September 21, it is right on the celestial equator.
   
       On August 11 it might be about halfway between those extremes, some 12° north of the equator. The chosen totality site is around latitude 42°, so the angle there at noon, between zenith and the direction to the Sun, is around (42-12) = 30°. For the cosine of that angle, see here.
   
   
6. Given that 71% of the Sun was covered in Alexandria, and assuming that the edge of the Moon then reached 0.71 of the way across the Sun (strictly speaking, not the same), you can now carry out the calculation of Hipparchus for the 1999 eclipse

    The lines of view from the two sites to the edge of the occulting Moon, at the height of the eclipse, therefore made an angle equal to 19% of the visual diameter of the Sun--close enough to the 20% which Hipparchus obtained!

    As the figure above shows, the path of totality on 11 August 1999 passed over Bucharest, the capital city of Romania. The Romanian government commemorated the event by drawing a map of the strip of totality over Romania (continuing the one drawn above) on its 2000-Lei currency note. Click here to see a copy of that note (takes 132K of memory).

  Further details about this eclipse are available at

         http://sunearth.gsfc.nasa.gov/eclipse/TSE1999TSE1999.html.
and at
           http://umbra.nascom.nasa.gov/eclipse/990811/rp.html.

A comic-strip style site about the eclipse of 2.26.98, from "Sky and Telescope," gives many details applicable to solar eclipses in general.

Next Stop:    #9a   Could Earth be Revolving around the Sun?