Aristarchus

Aristarchus of Samos

The Measuring Man of the Ancient World

(In which the distance from the Earth to the Moon can be determined in the comfort of your own home with nothing more than a few easily obtained observations plus some grade school arithmetic which totally eschews more advanced math from middle school such as trigonometry)

Aristarchus at Work

First things first.

Just what the heck did Aristarchus of Samos actually do?

Well, for one thing he was born about 310 BC. And if he was from Samos that's in the Aegean Sea just off the southwest shore of modern Turkey. Little else is known about his life and what we know of his writings are mostly from second hand sources.

However, one of his books has survived. There we learn he calculated the distance from the Earth to the Sun. And we also read he didn't do that bad, that is, if missing the distance by nearly 90,000,000 miles is all right.

But what Aristarchus is really famous for is calculating the distance from the Earth to the Moon. THAT number, we hear, was less than 1% off.

All this we must iterate, is what we read that Aristarchus did. But if you actually go to his book and read what he wrote at the end you'll find:

The diameter of the Earth to the diameter of the Moon is in a ratio greater than 108 has to 43 but less than that which 60 has to 19.

What? (To quote Aristophanes). But ... but ... this is talking about the ratio of the diameters of the Earth and the Moon! And it doesn't even give us an actual value. It just says the ratio is somewhere between 2.5 and 3.2.

And the actual value is 3.7%!

So why do people say Aristarchus calculated the distance from the Earth to the Moon?

With a Little Help from Aristarchus's Friends

Well, although Aristarchus didn't actually calculate the distance from the Earth to the Moon, he could have. But first a bit about the Ancient Greek modus arithematicandi.

One thing that sometimes shocks! shocks! modern readers is that if you crack the ancient mathematical texts you'll see that the Greeks didn't have algebra as we think of it. Instead their proofs were geometric with lots of triangles and circles but never trigonometric functions which they didn't have either.

The Greeks also had a number system without decimal notation. True, they did have unit factions - that is fractions where the numerator is 1 - but usually the only numbers they used were integers and ratios of integers. So the old proofs tend to be rather long and at times difficult to follow.

We saw that Aristarchus actually stopped his calculating after estimating the ratio of the diameters of the Earth and the Moon. But a modern historian looked at Aristarchus's book and keeping with the same method showed that you could relate the ratio of the Earth's radius, d_e, to the Earth-Moon distance, D_EM as somewhere between ⁵⁷/₈ and ²¹⁵/₁₈.

⁵⁷/₈ < ^D_EM/_{d_e} < ²¹⁵/₁₈

... where by taking the diameter of the Earth at 8000 miles - which might be the number the Greeks' used - we have:

57000 miles < D_EM < 95555 miles.

... or averaging the values:

Earth-Moon Distance = 76,277 miles.

With the real value being 240,000 miles, that's 68% error.

Aristarchus Take Umbra-age

Whether being 68% off is good or bad is a matter of how charitable we want to be. But whether the actual numbers were good or bad wasn't really Aristarchus's concern. The Ancient Greek mathematicians were more interested in getting the method right rather than having the numbers spot on.

Figuring out the methods isn't always that easy either, and there are places where even experts aren't quite sure exactly what Aristarchus is doing. Fortunately there are modern equivalents to Aristarchus's calculations which keep to the spirit of his original proofs by avoiding trigonometry. And you'll get numbers that are pretty much spot on.

The first thing, though, is to calculate what Aristarchus actually calculated. That's the ratio of the Earth's and Moon's diameters.

Diameter of Earth / Diameter of Moon

= ^d_e/_{d_m}

To start off we don't bother with anything about the Moon or the Earth. We just suspend a ball in a light so it casts a shadow.

Now it may not be completely obvious but the shadow is divided into three parts: the penumbra, the umbra, and the antumbra.

The umbra is the darkest part of the shadow and is where the ball completely blocks the light. If you're in the umbra, no direct light reaches you at all.

The penumbra is on either side of the umbra. If you're in the penumbra, the ball only partly obscures the light source. Obviously it won't be as dark in the penumbra as in the umbra.

The third part of the shadow, the antumbra, is the part of the shadow that's in line with the umbra but further out. It's also lighter than the umbra.

What's important to know is that any ball suspended in light has an umbra, penumbra, and antumbra. That includes the planets and the Moon.

As everyone knows, when a planet blocks out the view of another planet we have an eclipse. During an eclipse one planet falls within another planet's shadow.

On Earth we see both solar and lunar eclipses. The solar eclipse is when the Moon comes between the Earth and the Sun. And the lunar eclipse is when the Earth comes between the Sun and the Moon.

For a solar eclipse, if the viewers get completely covered by the umbra, it's a total eclipse. If they stay in the penumbra, it will be a partial eclipse. If the viewers are in the antumbra, that's what's called an annular eclipse. Annular solar eclipses are relatively rare and occur only once every year or so.

Shadows and Eclipses

Aristarchus's calculations are based on the geometry of these shadows. The best place to start, though, is to look at the shadows not cast by the Earth, but by the Moon. And yes, although the Moon only shines because of reflected sunlight, the objects lit up by the Moon will also have an umbra, penumbra, and even an antumbra. And at this point you can try an experiment.

Go outside on a clear night and when there's a full moon. Now take a small ball - like a 1" balsa sphere - and attach it to a stick. Place the stick in the ground and then back up so that the ball just obscures the Moon's disk.

Now the point where the ball just blocks the Moon's disk is at the end, or apex, of the umbra. And if you measure how far you've backed up, you'll find that in principle (emphasis added) you're about 9 feet in a direct line from the ball.

[Important Note: This experiment applies only with the Moon. DO NOT, that's NOT! NOT! NOT! try it with the Sun. NEVER EVER look directly at the Sun. Even a brief glance can cause permanent eye damage.]

Once when you've got the measurements down, convert all numbers to inches. Then divide the distance by the diameter of the ball. To avoid excess garrulousness, we'll call the quotient the umbra/distance ratio or the umbra/object ratio. We'll represent this ratio as ^U/_O.

Now if you've been careful in your measurements you'll find the umbra/object ratio is about 108.

U/_O = 108.

And it doesn't matter what size the ball is, the ratio should be close to 108. If you repeat the experiment with a 1/2" ball, then (yes, in principle), the ball will just obscure the moon if you drop back about 4½ feet. A quarter inch ball will obscure the moon at about 2 feet and three inches. A ping pong ball would cast an umbra a bit longer than 14 feet.

You can prove this - geometrically - if you try another experiment - but a thought experiment. First, (mentally) place the different size balls in a line so the centers of the balls and the center of the Moon all form a straight line. Then place the balls so that the smaller ones are just inside the shadow of the larger ones. Then you'll see that the shadow cones will all converge to the same point.

Next note that in the diagram you can see that the diameter of each ball, AD, CD, EF, can be seen as the base of the three triangles: ΔABG, ΔCDG, ΔEFG. For each triangle the two long sides are formed by the edges of the umbra.

Now by the way the balls have been lined up, the two long edges on each triangle are of equal length. Or as mathematicians say, the two sides are congruent. And as everyone learned in grade school, this type of triangle - where two sides are equal - is called an isosceles triangle.

Now isosceles triangles that share a common angle between their congruent sides but which differ in absolute sizes are similar triangles. And with similar triangles all the other geometric properties are proportional to each other.

Since all geometric properties of the triangles are proportional, then all similar ratios are all equal. That means that for all the triangles, the height to base ratios - which we labeled as - ^l_i/_{d_i} - are equal. And all these ratios are also the umbra/object ratio, ^U/_O and have the experimental value of 108.

^l₁/_d₁ = ^l₂/_d₂ = ^l₃/_d₃ = ^U/_O = 108

Making It Easier

The one problem, though, is that trying to measure the umbra/object ratio might be a bit more challenging than described. After all, trying to line up with a 1-inch ball from nine feet away so it just blocks the Moon is a touchy exercise. Repeat measurements may vary quite a bit.

Fortunately, there is another way to measure the umbra to object ratio and you don't even need the umbra or the ball. Instead you measure the image of the full Moon using a pinhole camera.

Pinhole Camera: Simple

A pinhole camera can be as simple or sophisticated as you like. The simplest type is made from two pieces of white cardboard. One board - we call it the screen - is just set up so where the Moon's light will fall flat on it. On the other piece you need to cut a hole about 1/2" square and cover it with a bit of aluminum foil.

Then with a pin you just make a tiny hole in the foil - hence the name. Don't worry if it's not exactly a circle as long as it's small.

Then set up the cardboard with the pinhole so its shadow blocks the moonlight from the screen. Get every thing set up and you'll see that in the middle of the screen is a circle of light.

Pinhole Cameras
(More) Sophisticated

The circle is actually an image of the Moon although it's upside down. You may even see some of the features of the Moon's surface. But all you need is the circle of the moon.

Now take a ruler and simply measure the diameter of the moon's image on the screen. Or you can put some graph paper on the screen. Although the outline of the image is a little fuzzy you should be able to get a decent measurement.

Next measure the distance from the pinhole to the center of the image - that is the distance to the screen. A tape measure works fine.

Do this and now you'll find something verrrrryyyyyy interesting.

Yep. The length from the pinhole to the image divided by the diameter of the image - we'll call it the length to diameter ratio, ^L/_D - is 108. That's the same value as the ^U/_O ratio from the umbra. As to why the ratio of an image from a beam of light is the same as a shadow cast by the moon becomes evident upon mature reflection.

First of all - and this can be another thought experiment - place a ball between the Moon and the pinhole so the apex of the umbra just touches the pinhole. When this is done, the cross section of the umbra defines the isosceles triangle as before. That is, the base of the triangle is the ball's diameter and the edges of the shadow to the pinhole are the sides.

Once the ball is removed, the angle formed by the cone of light inside the camera will be the same as the angle of the umbra at the pinhole. And as with the umbra, the light beam defines an isosceles triangle.

Since the angle at the pinhole of the "light triangle" and the "umbra triangle" are the same, we have similar isosceles triangles. And once more since the triangles are similar, all the proportions are equal. Therefore the ^L/_D ratio of the camera and the U/_O ratio from the umbra are the same and equal to 108.

^L/_D = U/_O = 108.

By the Light of the Silvery Sun

Now this leads us to something stranger. But first a reminder.

Eins mehr you should NEVER! NEVER! NEVER! look directly at the Sun. Making direct measurements of the Sun is extremely dangerous.

However, the image of the Sun projected onto the screen of the pinhole camera can be viewed safely. And you don't have to wait a month to get a good image like you do for the Moon.

And if you do use the pinhole to "capture" an image of the Sun, you can measure the diameter of the Sun's image just like you did for the Moon. And now a really funny thing happens.

You'll find that the diameter of the Sun's image, which we'll call S, is pretty much the same as the diameter of the Moon's image which we called D.

S = D

And this means - with the simplest of grade school math - that the diameter/length ratio of the Sun's image is the same as what we measured for the Moon. And these are the same as the umbra/object ratio of a ball in the moonlight. And they're all equal to 108.

S/_L = D/_L = U/_O = 108

Now it might be tempting to hypothesize that a value of 108 is ubiquitous for all diameter/length ratios whether we're talking about a pinhole camera or the shadow length. That would be most interesting if it were true. But the 108 ratio is not a general case. There's another variable involved and that's how far the ball is from the light source. This is clear if you look at the geometry involved.

At this point we'll pause, not necessarily for refreshment, but to review what we've done so far.

First, we've seen that if you're on the Earth, then spheres of any size held up to the full Moon have the same umbra/object ratio. That makes sense. After all, they're the same distance from the light source, ergo, the Moon.

And we saw that the diameter/length (or distance) ratio of an image in a pinhole camera is the same as the umbra/object ratio. That's just geometry.

But what may not be clear is why the umbra/object ratio of the Moon when the shadow is cast by the Sun is the same as that of a ping pong ball whose shadow is cast by the Moon.

Or why in the camera is the diameter of the Sun's image the same as the Moon's?

Well, once more the answer to the first question is just simple geometry. And the answer to the second question is geometry and a bit of luck.

Ceteris Paribus Coincidentia

First, why is the Moon's umbra/object ratio cast by the Sun, the same as that of a ping pong ball whose shadow is cast by the Moon? That's easily shown in a simple diagram:

... plus a bit of explanation, of course.

To start off we draw the picture of a ball in front of the Sun. Whether it's the Moon or a ping pong ball doesn't matter.

We can now define our two isosceles triangles. Triangle ΔABC is created from the Sun's diameter and the apex of the umbra. Then there's the triangle formed by the Earth's diameter and the apex of the umbra. That's triangle ΔDEC.

Now using the terms:

U	=	Length of Umbra Cast by the Object
O	=	Diameter of Object
S	=	Diameter of Sun
T	=	Distance from Sun to the Apex of Object's Umbra

... and looking at the diagram we immediately see:

^U/O = ^T/S

... since the distances are from similar isosceles triangles.

And since R is the distance from the Sun to the object, then:

R + U

And by substitution:

^U/_O

^{(R + U)}/_S

And with a bit of simple middle school algebra:

^U/_O	=	^{(R + U)}/_S
^U/_O	=	^R/_{S + ^U/_S}
^U/_O - ^U/_S	=	^R/_S
^U(1/_O - ¹/_S)	=	^R/_S
^U	=	^R/_{S(¹/_O - ¹/_S)}
^U	=	^R/(_{^S/_O - ^S/_S)}
^U	=	^R/_{(^S/_O - 1})

So with the last equation:

U = ^R/_{(^S/_O - 1})

We just divide both sides of the equation by the diameter of the object, O:

U/_O = ^R/_{O[(^S/_O - 1)]}

And with a little more twiddling we get:

U/_O = ^R/_{[(^OS/_O - O)]}

^U/_O = ^R/_{(S - O)}

Where U, as we said, is the length of the Earth's umbra, O is the diameter of the object, R is the distance from the Sun to the object, and S is the diameter of the Sun.

Now notice that the objects we are dealing with have diameters smaller than the Sun. Yes, that includes the Moon whose diameter is less than a quarter of a percent of the Sun. So we can readily make the following approximation:

S - O ≅ S

... where ≅ is a mathematician's way of saying "approximately equal to".

What all this brouhaha means is that because S is much greater than O, we can ignore the O in the denominator on the right hand side of the equation.

So instead of:

^U/_O = ^R/_{(S - O)}

... we now have :

^U/_O = ^R/_S

And we learn a couple of things.

First we see that as long as the object is smaller than the Sun and close to the Earth - say no more than a few hundred thousand miles - the ratio of the umbra/object ratio is the same for all objects.

And of course, one of the objects that's - quote - "close to the Earth" - is (what else?) - the Earth. So the umbra/object ratio of the Earth should be 108.

But also notice that if you put everything we've done together, we realize that the umbra/object ratio for things on the Earth is equal to the Sun's distance divided by its own diameter. So the Sun's distance/diameter ratio should be 108.

Is it? Well, we'll get a bit ahead of ourselves and use the most modern and up-to-date numbers. With a Sun-Earth distance of 92,955,807 miles and a diameter of the Sun as 864,938 miles, we get:

^U/_O = ^R/_S = ^92955807/₈₆₄₉₃₈ = 107.5

... which rounds off (using the rules we learned in grade school) to 108.

So geometry shows us that as long as the ball is close to the Earth - whether ping pong, the Moon, or the Earth - the umbra/distance ratio is 108.

But there's still a nagging question. Remember we saw that the theoretical value of the umbra of an object cast by the Sun is the same as the experimental value of an umbra cast by the Moon.

But why should these two numbers be the same?

Well, for this answer, we'll need to ask our second question:

Why in a pinhole camera is the diameter of the Sun's image the same as the Moon's?

THIS question is pretty easy to answer. It's simply coincidence and good luck. The Moon just happens to be the right distance to where the angular diameter of the Sun and the Moon are the same.

And just what is the angular diameter?

The angular diameter according to the dictionary is the apparent angle that an object subtends at the eye of the observer.

Yes, the definition uses the word subtends. And one of several definitions of "subtend" is:

sub^. tend (səb tend' sub-) v. t. 1. Geom. to extend under or be opposite to: a chord subtending an arc. 2. Bot. (of a leaf, bract, etc.) to occur beneath or close to. 3. to form or mark the outline or boundary of. [1560-70; < L subtendere to stretch beneath, equiv. to SUB- + tendere to stretch, see TEND¹]

Subtending

So a chord that's drawn across an arc is said to subtend the arc.

But in plain English if two objects have the same angular diameter, then they have the same apparent size - that's apparent size.

Of course to see if the Sun and the Moon have the same angular diameter, you NEVER look at the Sun. Instead, we use our acquired knowledge of geometry and a new thought experiment.

Put three balls in a row. We put them at points X, Y, and Z and so will call them Balls X, Y, and Z. It should be obvious that we're letting X do duty as the Sun, Y for the Moon, and Z for the Earth.

Angular Diameters

And yes, we're back to (yawn) isosceles triangles.

By now you should know that the distance/diameter ratios are all equal. That is:

XM / AB

YM / CD

But notice we've drawn a "projection" of the middle ball on the surface of ball M. The projection is the line segment EM.

But what exactly is EM?

EM is the apparent size of either Ball X or Y for someone standing on the surface of ball M (ergo, the Earth). And of course it's related to the two other distances by:

XM / AB

YM / CD

ZM / EF

Now suppose that Ball X is 400 times further from Ball M than is Ball Y. That is:

XM = 400YM

And since:

YM / CD

XM / AB

... and that means:

YM / CD

400YM / AB

So with a bit of simple middle school algebra:

AB / CD

400YM / YM

... and that means:

AB / CD

Diameter Ball X / Diameter Ball Y

= 400

So if one object is 400 times larger than another but 400 times further away, then both objects look the same size.

And it just happens that the Moon is indeed 400 times smaller than the Sun and coincidentally the Sun is 400 time further away from the Earth than the Moon. This hasn't always been the case and won't always be the case. The Moon is moving away from the Earth at about 1.5 inches per year.

So we see that since the angular dimensions of the two light sources - the Sun and the Moon - are the same on the Earth, the geometry of calculating the umbra dimensions are also the same. This means that the ^U/_O ratios will be the same for an object regardless of whether it's the Sun or the Moon casting the shadow.

And now as the $500 an hour psychiatrist said, we can begin, yes?

"Methinks it should be now a huge eclipse"
- Othello, William Shakespeare

At this point we've discovered the following tidbits.

The Sun and Moon have the same angular diameter.
The length of an object's umbra to its diameter has a ratio of 108 for all spherical objects close to the Earth whether the light source is the Sun or the Moon.

But this last point brings us to one important conclusion that may not be immediately obvious. And that's:

The length

of the

MOON'S UMBRA

is the

SAME

as the

DISTANCE

from the

EARTH

to the

Moon

Once more everything boils down to geometry where a diagram with isosceles triangles will make things clear.

But it's not just geometry that lets us conclude this point which is crucial to Aristarchus's calculation. A couple of natural phenomenon also supports the notion that the length of the Moon's umbra is the same as the Earth-Moon distance.

The first bit of evidence is that for some solar eclipses the area of totality is tiny. On September 22nd, 1968 the total eclipse lasted only 40 seconds. Historically there was even a shorter eclipse when on February 3, 919 - when Constantine VII was ruling the Byzantine Empire - the period of totality lasted a whopping 9 seconds.

Next there are not just total eclipses but annular eclipses as well. An annular eclipses is when the Moon crosses the path of the Sun but is far enough away from the Earth that the Moon's shadow doesn't quite cover the entire Sun. This means that the apex of the umbra is above the Earth's surface and it's only the antumbra that touches the Earth.

Well, since the apex of the umbra sometimes touches the Earth and sometimes it doesn't, we see the average distance from the Moon to the Earth is certainly close to the length of the Moon's umbra.

Now we come to the point.

Aristarchus's calculation of the Moon-Earth distance is based on the time of a total eclipse - but a total lunar eclipse. But because the Earth's shadow is larger than the Moon's diameter, a total lunar eclipse can last over an hour while a total solar eclipse never lasts much more than 7½ minutes.

As you can guess by now, if you see a partial lunar eclipse, that means the Moon is in the penumbra of the Earth as cast by the Sun. If you see a total eclipse, the Moon is in the Earth's umbra. Of course, if you see a total eclipse, the Moon spends some time in the penumbra as well and so you will also see a partial eclipse.

Like solar eclipses, the duration of the lunar eclipses vary. That's because the Moon isn't always directly in line with the Earth and the Sun. The plane of the orbit slants up and down, and the longer the eclipse, the closer the moon is to being in line. The longer the eclipse the better for the calculation.

Now at this point we'll step back and watch a total lunar eclipse. It starts as the Moon moves into the umbra of the Earth and ends when the Moon finally moves outside the umbra (for simplicity we aren't showing the penumbra).

Yes, we know that the Moon doesn't move in a straight line but actually makes an arc. But we are dealing with the apparent motion of the Moon as seen from the Earth. So having it move in a straight line won't cause any problems.

The first thing is to define how long it takes the Moon to - quote - "cross the shadow (the umbra) of the Earth" - unquote. Note this is not the same as the time of the total eclipse which is the time the Moon is completely covered in the Earth's shadow.

Usually you'd measure the duration by starting the time when the mid-point of the Moon is in shadow and stop timing when half of the Moon has emerged from the shadow. But notice that another way is to start timing as soon as the Moon's edge first touches the shadow of the Earth. Then you stop timing when the other edge just peeks out from behind the shadow.

You'll get the same answer either way, but we'll use the second method since this will come in useful later.

We repeat that the time the Moon crosses the Earth's shadow is not the time of the total eclipse. The total eclipse is when the Moon is completely covered by the Earth's umbra. That is from the time the Moon disappears and to when it just begins to reappear.

At this point everything is simple enough for us to make some definitions and derive some equations.

Although pretty much everything is in the diagram it might be useful to those who are a bit rusty in arithmetic to see how the equations and relations are derived.

First we time how long it takes for the Earth's shadow from when it first touches the Moon's edge until it's completely covered. So if the apparent velocity of the Moon is v, then the distance traveled - which is also the Moon's diameter:

d_m

t_p × v

Next we see that the distance (diameter) of the Earth's shadow is the velocity of the Moon multiplied by the time the Earth to move through both the partial phase and total phase:

d_s

(t_t + t_p) × v

So the diameter of the Earth's shadow compared to the diameter of the Moon is simply the ratio of the two equations:

^d_s/_{d_m}

(t_t + t_p) × v / t_p × v

... which since the velocity numbers cancel out (and yes, we really didn't need v) gives us:

^d_s/_{d_m}

t_t + t_p / t_p

And with some minor twiddling we get:

^d_s/_{d_m}

t_t / t_p

+ 1

With the twiddling complete, we can toss in some actual numbers. There's different ways to proceed.

First you can wait until the next lunar eclipse comes around. Now that is a good thing to do if you're making this a science project. But what you should do even then is to look up some numbers. In particular you'll want to use the most up-to-date data from the longest lunar eclipse you can find. That way the Moon is orbiting nearly above the equator.

Remember that the longer the Eclipse the better. That's because longer eclipses mean the moon is passing through the actual center of the Earth's shadow rather than skimming through near an edge. So the longer eclipses give us the true diameters of the Moon to the Earth's shadow.

Now one of the longest lunar eclipse in history was the one on July 21, 2000. The total eclipse - partial plus total - lasted a whopping 3 hours 56 minutes. That's 236 minutes. The period of totality lasted 1 hour 46 minutes - 106 minutes.

So the entire partial phase at both the beginning and end of the eclipse lasted:

Partial Phase (Total)	=	236 - 106 minutes
	=	130 minutes

So the first part of the partial eclipse is simply half that value:

Partial Phase (Total)	=	130/2 minutes
	=	65 minutes

And this number is the one we're calling t_p from our diagram:

t_p

65 minutes

So we have both the time of the first half of the partial eclipse and the time of the total eclipse. And that's all we need for our equation:

t_p

65 minutes

t_t

106 minutes

So now we pull up our equation:

^d_s/_{d_m}

t_t / t_p

+ 1

... and plug in the numbers:

^d_s/_{d_m}	=	106 / 65 + 1
	=	1.63 + 1
	=	2.63

Now you might read in some places that the ratio of the Moon to the Earth's shadow is constant and doesn't change with distance. So in that case the ratio of the Earth to the Moon would be 2.63.

That, though, isn't correct. If you moved the Moon away from the Earth and to the end of the cone of the shadow, it would still be visible as a disk. But the shadow width would be zero at that point. So the ratio of the Moon's diameter to the Earth's shadow definitely changes and in fact increases without bound as the Moon-Earth distance increases.

What we need, then, is a correction to calculate the Earth's actual diameter to that of the Earth's shadow. Fortunately the correction can be made easily. To this end, we do a very Greek proof.

And yes, we return to the length of the umbra.

As has been mentioned repeatedly, the umbra/diameter ratio is the same for all objects near the Earth. That includes the Moon and the Earth itself.

And as we know the sections of the planet and umbras form similar isosceles triangles. So you can perform a new thought experiment where you move the Moon and its umbra around.

Now as the diagram shows we can take the two umbras of the Earth and the Moon and fit them together in a way that doesn't happen in nature. Then we can draw a quadrilateral ABCD.

Two sides of the quadrilateral are parallel - the diameter lines of the Earth and Moon because we defined them that way. And the Angles ABC and ADC are equal as are angles DAB and DCB.

Again remembering our geometry we can conclude the quadrilateral is in fact a parallelogram. And opposite sides of parallelograms are equal. So line segment AB and line segment DB are the same length.

And line segment AB is in fact the Earth's radius and segment CD is the sum of the diameter of the moon and the Earth's shadow. So lo and behold, we conclude:

AB = d_e

CD = d_m + d_s

... and since:

AB = CD

... we conclude:

d_e = d_m + d_s

... where d_e is the diameter of the Earth, d_s is the diameter of the Earth's shadow where the Moon crosses it, and d_m is the diameter of the Moon.

Remember the shadow of the Earth is 2.63 Moon diameters. So:

d_e = d_m + d_s

... becomes:

d_e = d_m + 2.63d_m

... which is:

d_e = (1 + 2.63)d_m

... and:

d_e = 3.63 × d_m

d_m = ^d_e/_3.63

... or:

d_m = 0.276d_e

So the Moon is a bit more than a quarter the size of the Earth.

You will sometimes read that by looking at the difference of how the Sun shone into a well in Aswan, Egypt, and the shadow cast by an obelisk in Alexandria, the Greek mathematician Eratosthenes measured the diameter of the Earth, d_e as 8000 miles. This is pretty impressive since the modern value is 7926 miles - about 1% error.

But modern scholars have pointed out that Eratosthenes didn't use miles and instead said that the distance from Alexandria to Aswan was 5000 stadia (στάδια, singular: στάδιον). But Eratosthenes did measure the different angles of the shadows cast by the Sun in the two cities. He the concluded that the surface distance between them was ¹/₅₀^th the circumference of a circle. Using modern measurements, the north-south distance from Aswan to Alexandria is 492 miles. That, in turn, gives us a value of 7834 miles for the Earth's diameter.

The 5000 stadia that Eratosthenes mentioned is not likely the actual north south distance from Aswan to Alexandria. Instead it was probably the actual distance traveled along the Nile and was estimated by how long it took a boat to travel the distance. Eratosthenes certainly knew this was only a rough estimate. There's also some disagreement in how long a Greek stadion actually was. Officially it was the unit equal to the length of the stadium - hence the name - at Olympus. Herodotus said this was 600 Greek feet, but of course exactly how long a Greek foot was also varies since each town had its own set of measurements.

Still if you take 5000 stadia as the distance and 600 feet as one stade, you still get a circumference of about 28409 miles. That gives us a diameter of about 9000 miles. A bit large but still not bad.

So accepting there's a margin of error, we'll just go with the modern number for the Earth's diameter. So the diameter of the Moon calculates as:

d_m	=	7,926 miles / 3.63
	=	2183 miles

OK. So where do we go from here?

Remember we have said ad infinitum that the umbra/object, ^U/_O ratio of the Moon is 108.

^U/_O = 108

And O is the diameter of the Moon:

O = d_m

And so we simply plug in the numbers:

D_EM	=	108 × 2183 miles
	=	237,764 miles

And the most up-to-date determination of the distance from the Earth to the Moon is 238,856 with an uncertainty of ¾ of a mile. This gives a calculated error of:

Error	=	\| 100 × (237,764 - 238,855) / 237,764 \|
	=	1.3%

So if we use the method that Aristarchus could have used, when measuring the distance from the Earth to the Moon, we're off by less than 2%

How about that?

References

Aristarchus of Samos - The Ancient Copernicus, Sir Thomas Heath, Dover, 1981.

A History of Pi, Petr Becham, Golem Press, 1970.

"Aristarchus’s On the Sizes and Distances of the Sun and the Moon: Greek and Arabic Texts", J. L. Berggren and Nathan Sidoli, Archives of History of Exact Sciences, 61, 2007, pp. 213–254

"Lunar Eclipses 2011-2029", NASA Eclipse Web Page", National Aeronautics and Space Administration.

Formal Logic: Its Scope and Limits, Richard Jeffrey, McGraw-Hill, 1981.

"How Did Archimedes ACTUALLY Calculate Pi?", Ciaran McEvoy, Video.

"Names of Polygons", Moomoo Math and Science.

"Eratosthenes on the 'Measurement' of the Earth", Bernard Goldstein, Historia Mathematica, Volume 11, Issue 41, November 1984, pp, 411-416.

"Eratosthenes and the Mystery of the Stades", Newlyn Walkup, Mathematical Association of America.

The Eleven Comedies, Aristophanes, Athenian Society, 1912.