Category Archives: Phenomena

Old Lady-Day

At length it was the eve of Old Lady-Day, and the agricultural world was in a fever of mobility such as only occurs at that particular date of the year. It is a day of fulfilment; agreements for outdoor service during the ensuing year, entered into at Candlemas, are to be now carried out. The labourers—or “work-folk”, as they used to call themselves immemorially till the other word was introduced from without—who wish to remain no longer in old places are removing to the new farms.

Thomas Hardy Tess Of The d’Urbervilles (1891)

Yesterday (as this post goes live) was Old Lady-Day, once a significant day in the English agricultural calendar, as Thomas Hardy describes above. And today (April 6th), a new tax year begins in the UK. These dates are not unrelated to each other, and are also linked to the Christian Feast of the Annunciation, which commemorates the Biblical event depicted in the Leonardo painting at the head of this post—the arrival of the Angel Gabriel to inform the Virgin Mary that she was to conceive a miraculous child. As the King James Version of the Bible tells the story:

And the angel came in unto her, and said, Hail, thou that art highly favoured, the Lord is with thee: blessed art thou among women.
And when she saw him, she was troubled at his saying, and cast in her mind what manner of salutation this should be.
And the angel said unto her, Fear not, Mary: for thou hast found favour with God.
And, behold, thou shalt conceive in thy womb, and bring forth a son, and shalt call his name JESUS.

Luke 1:28-31

This event was called “The Annunciation To Mary”, or”The Annunciation” for short, annunciation being the act of announcing something. When it came to nailing these events to the Christian calendar, it made sense for the Feast of the Annunciation to fall exactly nine months before the Feast of the Nativity, which celebrates the birth of Jesus. Since that festival, Christmas Day, falls on December 25th in the Western Christian tradition, the Feast of the Annunciation occurs on March 25th. In Britain, that day is commonly called “Lady Day”, a reference to “Our Lady”, the Virgin Mary.

As well as being a religious feast-day, Lady Day was a significant secular date, too. As one of the four English quarter days*, it was a time when payments fell due and contracts were honoured. Farm labourers were usually indentured to work for a year at a time, and if they wanted to change jobs they all did so on Lady Day.

In fact, Lady Day was such an important date in the calendar, it marked the start of the New Year in English-speaking parts of the world for almost 600 years. While it may seem very strange to us now, under what was called “The Custom of the English Church” the year number would increment on March 25th each year, rather than on January 1st.

Scotland switched to using January 1st for New Year’s Day in 1600, but England didn’t make the change until it adopted the Gregorian calendar reform in 1752. I’ve written before about how eleven days were dropped from September that year, to bring the calendar back into alignment with the seasons.

Gregorian Calendar 1752 (Great Britain)

So 1752 was famously a short year in the English-speaking world. But it’s probably less well-known that 1751 was an even shorter year in England, since it began on March 25th, but ended on December 31st.

The missing eleven days in 1752 created a problem for all the legal stuff relating to contracts and debts that fell due on Lady Day. All the famous protests about “Give Us Back Our Eleven Days” were not from ignorant people who thought their lives had been shortened, but from people who were being paid daily wages, but were settling their debts monthly or quarterly or yearly.

One solution to this problem was to shift the dates of contracts and payments to compensate—so instead of changing jobs on Lady Day 1753, farm labourers worked until eleven days later, April 5th, and continued to renew their contracts on that day in subsequent years. April 5th therefore became known as “Old Lady-Day”, an expression that was still in use in 1891, when Hardy wrote about it in the quotation at the head of this post.

Similarly, the date of the new tax year moved from March 25th to April 5th—so the workers were given back their eleven days, at least by the tax authorities, if not by their landlords and other creditors.

But wait. I told you that the tax year in the UK begins on April 6th, not on Old Lady-Day. Why has it shifted by another day? Because under the old Julian calendar, the year 1800 was due to be a leap year, but under the new Gregorian calendar, it was not. So workers were being done out of a day’s wages in 1800 because of the calendar reform, and the tax authorities duly shifted the date of the new tax year by a day, to compensate. By 1900, which also dropped a leap day, these calendrical niceties seem to have been forgotten, and no shift in the tax year occurred, so the date has remained the same ever since.

So there you are. People in Britain used to start a new tax year at the start of a new calendar year, back when Lady Day was also New Year’s Day. Now, thanks to a centuries-old calendar reform and a surprising impulse of fairness from the tax authorities of the time, we calculate our taxes from what seems (to the uninitiated) like a random date in April.


* The other quarter days were Midsummer Day (June 24th), Michaelmas (September 29th) and Christmas (December 25th).
Oddly enough, the same date was also used in Florence, which earned that system of year reckoning its alternative name, Calculus Florentinus.

Falling Through The Earth

Falling through the Earth
Earth image prepared using Celestia

If the alien cyborgs have constructed this miraculous planet-coring device with the precision I would expect of them, I predict we shall plunge entirely through the center and out to the other side.

Gregory Benford Tides Of Light (1989)

Cover of Tides of Light by Gregory Benford

There’s an old puzzle in physics, to work out how long it would take a person to fall right through the centre of the Earth to the other side, along a tunnel constructed through its core. Gregory Benford is the only science fiction writer I’ve ever seen attempt to incorporate that scenario into a story—a particularly striking bit of audacity that made it on to the cover of some editions of his novel.

For simplicity, the puzzle stipulates that the tunnel contains no air to slow the faller, and usually also specifies that it has a frictionless lining—any trajectory that doesn’t follow the rotation axis will result in the faller being pressed against the side of the tunnel by the Coriolis effect. (Benford had his protagonist Killeen fall through an evacuated tunnel from pole to pole, thereby avoiding Coriolis.)

So our unfortunate faller drops through a hole in the surface of the Earth, accelerates all the way to the centre, and then decelerates as she rises towards the antipodal point, coming to a momentary halt just as she pops out of the hole on the far side of the planet—hopefully not too flustered to remember to grab a hold of something at that point, so as to avoid embarking on the return journey.

We can set a lower bound to the duration of that journey by working out how long it would take to fall from the surface of the Earth to the centre, assuming all the Earth’s mass is concentrated at that point. Because the journey is one that involves symmetrical acceleration and deceleration, doubling this number will give us the duration of the complete traverse of the tunnel.

This means our faller moves under the influence of an inverse-square gravitational field throughout her fall. The acceleration at any given distance from the centre of such a field is given by:

\Large
a=\frac{GM}{r^2} 

where a is the acceleration, G is the Gravitational Constant, M is the central mass and r the radial distance. That’s an important equation and I’ll invoke it again, but for the moment it’s more useful to know that the potential energy of an object in such a field varies inversely with r. The gravitational potential energy per unit mass is given by:

\large U_{m}=-\frac{GM}{r}

If we drop an object starting from rest at some distance R, we can figure out its kinetic energy per unit mass at some distance r by subtracting one potential energy from the other. And from that, we can figure out the velocity at any given point during the fall:

\large v=\sqrt{2GM\left (\frac{1}{r}-\frac{1}{R}\right)}

Finally, integrating* inverse velocity against distance gives us the time it takes to fall from a stationary starting point at R to some lesser distance r. For the special case of r=0, this comes out to be

\large t=\frac{\pi }{2\sqrt{2}}\sqrt{\frac{R^3}{GM}}

This calculation, incidentally and remarkably, is a major plot element of Arthur C. Clarke’s 1953 short science-fiction story “Jupiter V”.

Plugging in values for the mass and mean radius of the Earth, we find that t turns out to be 895 seconds. Doubling that value gives a total time to fall from one side of the Earth to the other of 29.8 minutes. That’s our lower bound for the journey time, since it assumes the faller is subject to the gravitational effect of Earth’s entire mass throughout the journey. (We draw a veil over what would actually happen at the centre point, where the faller would encounter a point of infinite density and infinite gravity.)

We can also put an upper bound on the journey time by assuming the Earth is of uniform density throughout. Under those circumstance, instead of the gravitational acceleration getting every higher as our faller approaches the centre of the Earth, the acceleration gets steadily lower, and reaches zero at the centre. This is because of something called Newton’s Shell Theorem, which shows that the gravitational force experienced by a particle inside a uniform spherical shell is zero. (Which rather undermines the premise of Edgar Rice Burroughs’s “Hollow Earth” novels.)

So as our faller descends into the Earth, she is accelerated only by the gravity of the spherical region of the Earth that is closer to the centre than she is. For any given radial distance r, the mass m of this interior sphere will be

\large m=\rho \frac {4}{3} \pi r^3

where ρ is the density.

Plugging that into our equation for the acceleration due to gravity (the first in the post) we get:

\large a=\frac {4} {3} \pi \frac {G \rho r^3} {r^2}=\frac {4} {3} \pi G \rho r

So the acceleration is directly proportional to the distance from the centre. This is the defining property of a simple harmonic oscillator, like a pendulum or spring, for which the restoring force increases steadily the farther from the neutral position we displace the oscillating mass.

Which is handy, because there’s a little toolbox of equations that apply to simple harmonic motion (they’re at the other end of my link), and with a bit of fiddling we can derive our journey time . The basic time parameter for oscillators is the period of oscillation, which I’ll call P. But that would be the time taken to fall from one side of the Earth to the other and back again. So the time we’re interested in is just half of that:

\large \frac {P} {2}=\pi \sqrt {\frac {3} {4 \pi G \rho}}

And plugging in the value for the mean density of the Earth, that shakes down to 42.2 minutes.

Notice how this length of time depends only on the density—it actually doesn’t matter how big our uniform sphere is, the time to fall from one side to the other remains the same so long as the density is the same. This is analogous to the fact that the period of oscillation of a pendulum doesn’t depend on how far we displace it—which is why we have pendulum clocks.

And because the acceleration due to gravity, g, at the surface of a spherical mass of radius R is given by

\large g=\frac {GM} {R^2}=\frac {4 \pi G \rho R^3} {3R^2}=\frac {4 \pi G \rho R} {3}

we can also derive our half-period P/2 as

\large \frac {P} {2}=\pi \sqrt{\frac{R}{g}}

A nice compact formula depending on radius and surface gravity, which is often quoted in this context.

And this is the gateway to another interesting result for spheres of uniform density, if you’ll permit me a brief digression. Suppose we dig a straight (evacuated, frictionless) tunnel between any two points on the surface of the sphere, and allow an object to slide along the tunnel under the influence of gravity alone—a concept called a gravity train. How long will such an object take to make its journey from one point on the surface to the other? It turns out that this time is exactly the same as the special case of a fall through the centre. We can see why by constructing the following diagram:

Geometry of gravity train tunnel

By constructing similar triangles, we see that the ratio of R (the distance to the centre of the earth) to d (the distance to the centre of the tunnel), is always locally the same as the ratio of g (the local gravitational acceleration) to a (the component of g that accelerates the gravity train along the tunnel. So for any straight tunnel at all, d/a is always equal to R/g, which (from the equation above) we know determines the period of oscillation through a central tunnel.

Remarkably then, we can take a sphere of uniform density of any size, and drill a straight hole between any two points on its surface, and the time it takes to fall from one surface point to the other will be exactly the same, and determined entirely by the density of the sphere.

But back to the original problem. I’ve determined that the fall time is somewhere between 29.8 and 42.2 minutes. Here are plots of the velocity and time profiles for the two scenarios I’ve discussed so far:

Falling to the centre of the Earth, two limits
Click to enlarge

Can I be more precise? I can indeed, by using the Preliminary Reference Earth Model (PREM), which uses seismological data to estimate how the density of the Earth varies with distance from its centre.

Taking those figures and Newton’s Shell Theorem, I can chart how the acceleration due to gravity will vary as our faller descends into the Earth. Here’s the result, with density in blue plotted against the left axis, and gravity in red against the right axis:

PREM density and gravity chart
Click to enlarge

As our faller descends through the low-density crust and mantle of the Earth, and approaches the high-density core, she actually finds herself descending into regions of higher gravity, reaching a maximum about 9% higher than the gravity at the Earth’s surface when she reaches the boundary between the mantle and the core.

If I take the data from the PREM as representing a succession of shells, across which acceleration rises or falls linearly within each shell, I can integrate my way through each shell in turn, deriving velocity and elapsed time. For each shell of thickness s, with initial velocity v0, initial acceleration a0 and final acceleration a1, the final velocity is given by

\large v=\sqrt{v_{0}^{2}+(a_{0}+a_{1})s}

and the time taken to traverse the shell is

\large t=\int_{0}^{s}\frac{dx}{\sqrt{v_{0}^{2}+2a_{0}x+\frac{a_{1}-a_{0}}{s}x^{2}}}

Handing off v as v0 to the next shell and summing all the t‘s once I reach the centre of the Earth will give me the answer I want.

The solution to the time integral is a bit messy, coming out as an arcsin equation when a0 > a1, and a natural log when a1 > a0.

But it’s soluble, and with steely nerves and a large spreadsheet, the graphs of the solution fall neatly between the two extremes I figured out earlier:

Falling to the centre of the Earth, PREM curves
Click to enlarge

And the summed times for the full journey come out to be 38.2 minutes. And that’s my best estimate of the answer to the question posed at the head of this post.


* Integrating this expression turned out to be a little tricky, at least for me.

\small t=\sqrt{\frac{R}{2GM}}\int_{r}^{R}\sqrt{\frac{r}{R-r}} dr

After mauling it around and substituting sin²θ for r/R, then mauling it around some more, I ended up with this eye-watering equality as the general solution:

\small t=\sqrt{\frac{R^3}{2GM}}\cdot \left [( \frac{\pi }{2}-asin\left ( \sqrt{\frac{r}{R}} \right )+ \sqrt{\frac{r}{R}}\cdot \sqrt{1-\frac{r}{R}}\right ]

Exact solutions for this integral look like this:
For a0>a1:

\tiny t=\sqrt{\frac{s}{a_{0}-a_{1}}}\cdot \left [ asin\left ( \frac{a_{0}}{\sqrt{a_{0}^2+v_{0}^2\left ( \frac{a_{0}-a_{1}}{s}\right )}} \right )-asin\left ( \frac{a_{1}}{\sqrt{a_{0}^2+v_{0}^2\left ( \frac{a_{0}-a_{1}}{s}\right )}} \right ) \right ]

For a1>a0:

\small t=\sqrt{\frac{s}{a_{1}-a_{0}}}\cdot ln\left [ \frac{\sqrt{\frac{a_{1}-a_{0}}{s}\left ( v_{0}^2+sa_{0}+sa_{1} \right )}+a_{1}}{\sqrt{\frac{a_{1}-a_{0}}{s} v_{0}^2}+a_{0}} \right ]

(If you can simplify any of these, be sure to let me know.)

Long-Exposure Bicycle Spokes

Blurred cyclist
Click to enlarge
© The Boon Companion, 2020

The Boon Companion has been experimenting with long exposure times and intentional camera movement, of late. She was just about to discard the motion-blurred cyclist above as a failed experiment when something about the image caught my eye.

Blurred bicycle wheel

In the thirtieth-of-a-second exposure, the bicycle wheel has rolled a short distance. But why do the spokes look curved? Why don’t the curves point towards the centre of the wheel? And why is the effect only visible in the lower half of the wheel?

So I sat down to figure out the trajectory of a bicycle spoke as the wheel rolls along the ground. As you do.

Any point on wheel rolling across a flat surface without slipping follows a curve called a trochoid (from Greek trochos, “wheel”). I won’t pester you with the relevant equations (they’re at the other end of the link above). Here’s what the trochoid curves look like for the two ends of a radial spoke, spanning the distance between a wheel hub and a thick wheel rim:

Trochoid trajectories of a radial wheel spoke
Click to enlarge

The shape of the wheel is plotted in grey dashes, with a single vertical spoke marked. As the wheel rolls left or right, the ends of the spoke follow the curved trochoid trajectories, with successive positions marked at 20º intervals.

But bicycle wheels don’t (usually) have radial spokes, and I felt obliged, going into the problem, to look at the position of real bicycle spokes. Here’s a very common pattern:

Bicycle spoke pattern, 36 spokesThirty-six spokes, laid out in what’s called a “three cross” pattern. Fundamentally, there are only two different spoke alignments in this pattern—leading and trailing.

Bicycle spoke pattern, 36 spokes, leading and trailing spokesFor a wheel rolling from right to left, the spoke I’ve highlighted in red is leading, and the blue spoke is trailing. They’re simply mirror images of each other. One side of the wheel has nine leading spokes, spaced 40º apart, and nine trailing spokes in the same pattern. The other side of the wheel is laid out exactly the same way, but with the pattern rotated by 20º. The final result is called a “three cross” pattern because each trailing spoke crosses three leading spokes on its side of the wheel (and vice versa).

In this pattern, the anchor point for the spoke at the hub is offset 60º relative to its attachment at the rim. So to see the trajectory of a representative bicycle spoke, I need to slide the trochoid curve for the hub 60º out of alignment with the rim curve. Here its, with the spoke drawn in at 20º intervals:

Trochoid trajectory of a bicycle wheel spoke
Click to enlarge

This is the trajectory of a trailing spoke for a wheel rolling right to left, and a leading spoke for a wheel rolling left to right.

We can already get a hint of why some sort of spoke pattern shows up in the lower half of the wheel, but not the upper. In the upper half, the spoke is moving rapidly sideways, as it pivots across the top of the wheel; in the lower half it performs a sort of dipping motion, arcing downwards towards the point at which the wheel contacts the road, and then arcing back up again.

Now, I figure a cyclist moving at a reasonable speed for a shared-use path will rotate the wheels through about 30º during a thirtieth-of-a-second exposure. Here’s a more detailed trajectory for a trailing spoke (for a bicycle moving right-to-left) during a 30º rotation:

30-degree trajectory of low trailing spoke
Click to enlarge

You can see that the spoke slides along itself, to some extent—different parts of the spoke occupy the same spatial position at different times. These are the only parts of the spoke that will show up as a dense “shadow” during a prolonged exposure that blurs the other parts. In the final photograph we’ll therefore see something like the arc I’ve sketched in red, while the rest of the spoke is smeared into a blur:

30-degree trajectory of low trailing spoke, superimpositions marked
Click to enlarge

Something similar happens for a leading spoke as it passes through the same position:

30-degree trajectory of low leading spoke, superimpositions marked
Click to enlarge

So, actually, while the leading/trailing distinction slightly changes the details, both kinds of spoke produce a curved shadow during a prolonged exposure.

If we catch a spoke that’s higher in its trajectory, we get another arc:

30-degree trajectory of high trailing spoke, superimpositions marked
Click to enlarge

Now we can put these images together, showing the curved shadows of several spokes at once. Each of them will lie on a different set of trochoid arcs, shifted laterally according to how far the spoke lies from the lowest point of its trajectory. Like this:

Overall effect of spoke blur during 30-degree rotation
Click to enlarge

I’ve marked the visual centre of the wheel, for reference. Notice how the partial shadow arcs formed by the lower spokes seem to point below the centre, while the arcs of the higher spokes point above the centre. It happens because none of the spokes are radial, and because the centre of the wheel is never stationary, but shifts horizontally as the spokes sketch out their curved arcs.

So. Although I was baffled by the photograph when I first saw it, I began to get an inkling of what was going on as I made the first trochoid sketches, and was then pleasantly surprised by how things began to fall neatly into place as I added more detail. I’m hoping you’re as pleasantly surprised as I am.

Why Do Mirrors Reverse Left And Right But Not Up And Down?

Still from the BBC's QI R Series: Reflections
Click to enlarge

Reflection: A transformation under which each point in a shape appears at an equal distance on the opposite side of a given line—the line of reflection.

It’s not often I have occasion to shout at the television, but a recent episode of the BBC’s long-running television series QI precipitated just such an outburst. The cause of my vexation was their answer to the question that forms the title of this post. The offending episode was the R Series: Reflections, and the explanation was an excellent approximation to gibberish, involving as it did some business about “The mirror doesn’t flip things around; we flip things around,” intoned by Sandi Toksvig as she stood in front of a mirror fiddled with a bit of card with the word BOSS written on it (see above). To be fair to Toksvig, it probably wasn’t her idea, and she did manage to deliver the entire farrago while wearing the sort of anxious expression people wear when they’re not entirely convinced by their own argument.

The answer to the question is really that it is ill-posed. Mirrors actually don’t reverse left and right, for the simple reason that mirrors have no way of telling left from right. They have no left-right asymmetry, in other words. The only asymmetry they do possess is in the plane of reflection—stuff in front of the mirror is the real world; stuff “behind” the mirror is the reflected world.

That’s what’s being described in the definition at the head of this post, which refers to a two-dimensional reflection, like this:

Reflected B

Here we have a “line of reflection”, corresponding to the mirror; a letter “B” in front of the mirror, representing the real world; and a reflected letter “B” behind the mirror. Every point on the reflected “B” is the same distance from the mirror as the corresponding point on the original “B”. So because the spine of the “B” is the farthest part from the mirror, its reflection also lies farthest from the mirror. Conversely, the curved parts of the letter lie closer to the mirror on both sides. And it’s that preservation of “near” and “far” on either side of the mirror plane which causes the reflection to be a reversed image of the original. If we travel from one side of the mirror to the other, we encounter in turn spine-curves-mirror-curves-spine. So, actually (and pace Toksvig), the mirror very much does “flip things around”. Indeed, many introductory geometry texts gloss the word “reflection” as “a flip”.

The same thing happens when a three-dimensional person stands in front of a real mirror:

Reflected person

The reflected image’s left and right (and head and feet) are pointing in the same direction as the real person’s. What has been flipped in the mirror image is the direction in which the nose and toes are pointing. So the mirror has reversed front and back, not left and right. If you lie down with your feet pointing at the mirror, the reflection will also have its feet pointing at the mirror—so on this occasion the mirror leaves your front and back, and left and right, in the same positions, but reverses you top to bottom. Only if you stand sideways on to the mirror does it truly reverse your left and right—but that’s because you’ve chosen to place one side of your body close to the mirror and the other far from it, not because the mirror has some magical ability to tell left from right.

So why do we always think the mirror image has reversed left and right, no matter how we orientate ourselves before the mirror? The answer, I think, lies in the single plane of symmetry in our own bodies. Our fronts are very different from our backs, our heads are very different from our feet, but our left side is very similar to our right side. So it’s very difficult for us to see the mirror reflection as having reversed front and back—instead, we see it as another person who has turned around to face us. In which case, their right hand now moves when we move our left hand, and vice versa. No matter what the orientation of the reflection, we always interpret it as a left-right reversed person, because a head-foot reversed person or a front-back reversed person is harder to conceptualize.

Remove the left-right symmetry, and we stop talking about mirrors reversing left and right.

Reflected barber's pole

This barber’s pole lacks a clear left-right distinction. So instead, we find ourselves saying that it “spirals the opposite way”. We’re still, apparently, unable to discern that what has been reversed is the front and back, but now we can’t blame a left-right switch either.

So if anyone ever asks you the title question, permit yourself the slightest of headshakes and the faintest of smiles, and say: “But mirrors don’t reverse left and right; they preserve near and far.”

Strange Moon

Cover of We Landed By Moonlight by Hugh Verity
A couple of weeks ago, I reviewed three books about the activities of 161 (Special Duties) Squadron, RAF, during the Second World War. For this post, I want to talk specifically about the cover of Hugh Verity’s memoir and personal history of 161 Squadron, We Landed By Moonlight (Revised Edition), published in 2000 by Crécy. It’s a marvellous book, as a source of both anecdote and historical record, and we should all be grateful to Crécy for keeping it in print—but it’s an odd cover.

The first thing that struck me about it is that the Westland Lysander on the cover is sporting South-East Asia Command roundels and flashes, putting it a very long way from 161 Squadron’s base in the south of England. The image in fact comes from this photograph of Lysander V9289, of 357 (Special Duties) Squadron, RAF, operating in Burma. Right kind of aircraft, right kind of duties, wrong continent. But that’s fair enough, given that 161 Sq. Lysanders flew almost entirely at night on secret missions, so tended to go unphotographed unless they actually crashed.

The Lysander image has been composited with a fine full moon, to produce an atmospheric cover image. (In fact, there are two versions of this cover from Crécy, both using the same Lysander and moon images—you can find the other in my link from the book title, above.) And it’s that moon image that really got me puzzling, and inspired this post. Here it is, in a larger and more contrasty version:

Moon from cover of We Landed By MoonlightIt definitely doesn’t look like our own familiar full moon:

Full moon
Photo via Good Free Photos

But it doesn’t look like a random painting, either. And I found it naggingly familiar. At first, I wondered if it was a photograph of some other moon in our solar system, but then I began to recognize significant features. The curve of Mare Nectaris below the three linked blotches of Mare Serenitatis, Mare Tranquilitatis and Mare Fecunditatis settled it—it is a photograph of our Moon.

But there are three things wrong with it.

One is that it is mirror-reversed. The real Moon looks like this:

Reversed Moon from cover of We Landed By MoonlightThe second is that it is pretty much lying on its side. By my estimate, the north-south axis is tilted at about twenty degrees to the horizontal:

Axis of reversed Moon from cover of We Landed By MoonlightNow, you can see the Moon in this orientation, if you catch it just after moonrise in the tropics, when it’s moving almost vertically towards the zenith. But the farther north or south you go, the more tilted is the Moon’s trajectory as it rises, and the closer to vertical is its axis. Even when the Moon is as far into the northern sky as it ever gets, it can never be seen in that orientation anywhere in France, which was the 161 Sq. stamping ground.

But that’s a nitpick, really, because the striking thing about this view is that you can never see it from Earth. The paired dark blotches about halfway towards the upper rim of the Moon, above, are Mare Marginis and Mare Smythii, which (as the former name implies) sit right on the edge of the Moon’s disc when seen from Earth. The photograph used on Crécy’s cover has actually been taken by a spacecraft, somewhere over about 60ºE lunar longitude.

Once I’d figured all this out, I realized why the image looked naggingly familiar. This view is a classic, because it’s what successive Apollo astronauts saw as they departed the Moon towards Earth. At the conclusion of their lunar mission, they fired their main engine as they orbited over the far side of the Moon, and then came looping around the eastern hemisphere, pulling away in a long orbit back towards Earth. The same view was photographed several times, by several relieved astronauts, but I think the Moon on Crécy’s cover is this one:

AS11-44-6667
Click to enlarge
AS11-44-6667

It’s one of a series of departure photographs taken by Apollo 11 on 22 July 1969. If Verity landed his Lysander by that moonlight, he was a very long way from home!


Note: Crécy’s current edition of this classic book features the Lysander from the Shuttleworth Collection, which is painted in early 161 Sq. markings, landing in a field of poppies.

Harvest Moon

The Harvest Moon exhibited 1872 by George Mason 1818-1872
Click to enlarge
The Harvest Moon by George Mason, exhibited 1872, bequeathed by Lord Faringdon, 1934
Photo © Tate, used under Creative Commons CC-BY-NC-ND 3.0 (Unported) licence.

In the northern hemisphere, the Harvest Moon falls on 1 October in 2020, which is what provokes this post. The Harvest Moon is defined as the full moon that occurs closest to the autumnal equinox, which fell on 22 September (in the northern hemisphere, in 2020). You can find many lists of “names of the full moons” on-line (there’s a rather marvellous compilation of lists here), but the Harvest Moon is the only one that’s defined by the date of the equinox, rather than the month in which it falls—about three times in four it occurs in September, but the rest of the time (as on this occasion) it drifts into October.

The other thing about the Harvest Moon is that it has real astronomical and historical significance. Like many other full moon names, it obviously derives from what’s going on in the seasonal cycle at the time it appears—but there’s a deeper significance, which is what I want to write about here.

To understand what’s special about the full moon around the autumnal equinox, and its relevance to harvesting crops, we need to talk a bit about the orbit of the moon.

As is well known, the Earth’s rotation axis is inclined to the plane of its orbit around the sun, by about 23½º. So the Earth’s rotation and its orbit define two planes, tilted relative to each other—the celestial equator, which is the extension of the Earth’s equatorial plane; and the ecliptic, which is the plane of the Earth’s orbit. So from the vantage point of the Earth, the sun moves around the sky along the ecliptic plane, from west to east, completing one revolution per year. Like this:

Relationship between celestial equator and ecliptic
Click to enlarge

The two points at which the celestial equator and the ecliptic intersect have names with complicated astrological origins. The point on the celestial equator which the sun crosses when heading north is called the First Point of Aries. The point opposite that is The First Point of Libra. Both are symbolized by the zodiacal symbols for their corresponding constellations. These are the locations of the sun at the times of the equinoxes—it crosses the First Point of Aries at the March equinox, spends six months bringing summer to the northern hemisphere, and then crosses the First Point of Libra at the September equinox, on its way south for the southern hemisphere summer.

The moon orbits close to the ecliptic plane. For the purposes of this discussion, we can treat it as travelling in the ecliptic plane, and come back to the slight deviation later. So the moon moves (roughly) along the ecliptic from west to east, taking a month to make a full revolution. It also passes through the First Points of Aries and Libra, spending two week over the northern hemisphere, and two weeks over the south.

The moon makes one complete circuit of the celestial sphere every 27.3 days. If it moved at a constant rate along the celestial equator, it would therefore be about 13º farther west every day. The Earth would need to rotate correspondingly farther between successive moonrises and moonsets, making each moonrise and moonset occur about fifty minutes later than its predecessor. And that’s true on average for the real moon. But the fact that the moon’s orbit follows the ecliptic, and not the equator, introduces a subtle variation.

Here’s what happens to successive moonrises at 50ºN latitude, when the moon is passing through the First Point of Aries.

Harvest moonrise in northern hemisphere
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Its 13º displacement along the ecliptic has a northward component in this part of its orbit, which means that it lies closer below the horizon on successive nights than it would do if it were moving parallel to the equator. So the Earth has to rotate less far between successive moonrises, and the moon rises only slightly later each night. (The effect becomes more pronounced at higher latitudes, and less so at lower latitudes.)

But at moonset, the northward movement at the First Point of Aries serves to lift the moon farther above the horizon than it would otherwise be. So successive moonsets show longer delays than the average 50 minutes when the moon is in this part of its orbit.

Harvest moonset in northern hemisphere
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The situation is reversed two weeks later, as the moon passes through the First Point of Libra. Now each successive moonrise at northern latitudes is delayed more than 50 minutes, like this:

"Antiharvest" moonrise in northern hemisphere
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And it should come as no surprise that the delay between successive moonsets is correspondingly shortened at this point in the moon’s orbit.

So although it all averages out over the course of a month, there’s a regular variation in the delay between successive moonrises (and moonsets) during that period. Here’s a chart of the delays for a representative period (September and October 2018) at 50ºN; I’ve marked the passages through the First Point of Aries:

Delay between successive moonrises/moonsets, Sep-Oct 2018
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So this happens every month. Why is it particularly relevant only once a year, on the Harvest Moon? Because full moons occur only when the sun is on the opposite side of the sky from the moon. Which means the only time we see a full moon passing through the First Point of Aries is when the sun is in the vicinity of the First Point of Libra—which, you’ll recall from the top of this post, happens during the September equinox. So in the northern hemisphere, at the time of the autumnal equinox, the full moon rises at almost the same time for several successive evenings, but sets more than an hour later each morning. And if you’re bringing in the harvest (as you do in temperate latitudes in the autumn), and you don’t have access to artificial outdoor illumination, then that’s hugely advantageous. For several nights, the full moon rises before twilight fades, and sets only when the morning sky is already bright. You can work around the clock, day and night, to get the crops in, in other words. Which is what’s going on in Mason’s painting at the head of this post.

Does the southern hemisphere have a Harvest Moon? It surely does. All the geometry flips over, so the First Point of Libra assumes the role that the First Point of Aries does in the northern hemisphere. Like this:

Harvest moonrise in southern hemisphere
Click to enlarge

Full moons occur at this point when the sun is passing through the First Point of Aries—the March equinox, which is the autumnal equinox for the southern hemisphere. Isn’t that neat? (Well, I think it’s neat.)

There are a couple of subtleties, which mean not every Harvest Moon is the same. The first complicating factor is that the moon’s orbit does not lie perfectly in the ecliptic, but inclined to it at about 5º. The inclined orbit of the moon twists continuously in the ecliptic plane, completing one rotation every 18.6 years, under the influence of the sun’s gravity. This means that the tilt of the moon’s orbit sometimes subtracts from the angle between the ecliptic and the celestial equator, and sometimes adds to it, like this:

Effect of moon's inclination to ecliptic on Harvest Moon
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So we have “seasons” when the Harvest Moon is delayed even less than average on successive nights, and seasons (nine years later) when it is delayed more than average.

The other complicating factor is that the moon doesn’t orbit in a perfect circle—it moves in an ellipse, and it crosses the sky more slowly when it’s farthest from the Earth (its apogee), and more quickly when it’s closest (perigee). This ellipse twists around in the plane of the moon’s orbit with a rotation period of about 8.8 years. When the apogee aligns with the First Point of Aries, the delay between successive Harvest Moon moonrises is shortened even farther. Conversely, a few years later, the perigee will prolong the delay between successive moonrises.

It so happens that apogee is passing through the First Point of Aries in 2020, and we can see a noticeable effect on moonrises and moonsets. Here’s the delay graph for September and October 2020, again at 50ºN.

Delay between successive moonrises/moonsets, Sep-Oct 2020
Click to enlarge

The slow movement of the moon at the First Point of Aries shortens the delay between successive moonrises, and between successive moonsets. Conversely, the fast movement at the First Point of Libra lengthens these same time periods. So the delay graphs are dented downwards at Aries, and shoved upwards at Libra—you can see this most clearly in the flattened tops on the moonset curve.

There are few places left in the world where any of this is relevant to the life of farmers, of course. But it’s a fine astronomical curiosity, I hope you’ll agree.


Note: Moonrise and moonset times used in the graphs were taken from the calculator at timeanddate.com.

Tertiary Rainbows, etc

Rainbow enclosing bright sky
Click to enlarge
Photo by Binyamin Mellish via Good Free Photos

In my last two posts about rainbows, I discussed the formation of the primary and secondary rainbows, respectively, tracing their origins to specific light paths through falling raindrops.

The primary rainbow ray follows a path like this:
Green rainbow rayFor a raindrop at the apex of the rainbow arc, sunlight enters near the top of the drop, bounces once off the back, and then exits the bottom, descending towards the observer’s eye, making an angle of around 41.5º for the green ray shown.

For the secondary rainbow, sunlight enters near the bottom of the drop, is reflected internally twice, and then exits the front of the drop, descending towards the observer at an angle of around 51º.

Green secondary rainbow rayThese rainbow rays are special, representing the maximum or minimum angles of deflection of the incoming ray. And they are associated with a particular offset of the incoming ray in its interaction with the raindrop. I measured offset like this:Rainbow offset parameterThe rainbow ray for the primary enters the raindrop at an offset of about 0.86; the secondary rainbow ray at about 0.95.

For more about these topics, in particular the importance of maximum and minimum deflections, I direct you back to my previous posts.

I finished my most recent post on this topic with a question: If a single internal reflection produces a primary rainbow, and two internal reflections produce a secondary rainbow, is there such a thing as a tertiary rainbow? And if so, where is it?

The path of the rainbow ray for a tertiary bow looks like this:

Path of rainbow ray for tertiary rainbowThe three reflections carry it almost all the way around the raindrop, so that (in contrast to the primary and secondary bows) it leaves the drop heading away from the sun. This tells us that, to see a tertiary bow, we’d need to look towards the sun, rather than (as for the primary and secondary) towards the antisolar point.

Plotting my usual graph of light deflection and transmission against the full range of ray offsets with three internal reflections, I get this:

Deflection and tranmission in formation of tertiary rainbow
Click to enlarge

In contrast to my graphs for the primary and secondary, this one shows the deflection from the “solar point”—the position of the sun. The maximum occurs at an offset of about 0.97, reaching 42.5º for red light, and 37.7º for violet. The angular distance between red and violet is therefore about two-and-a-half times what we see in the primary bow. The light transmission is scaled to match my previous two graphs, and it shows that, because so much light is lost during multiple internal reflections, only about 1% of the tertiary rainbow ray survives to exit the drop. (But that’s equivalent to about 24% of the transmission for the primary rainbow ray, so not catastrophically dim.)

So it seems straightforward enough. We should see a broad, faint tertiary bow, about the same diameter as the primary but centred on the sun. Have you seen that? No, me neither.

The problem is that there’s a lot of light in the sky around the sun, particularly when rain is falling through the line of sight. Les Cowley at Atmospheric Optics calls this the “zero-order glow”, because it is formed from sunlight that passes through the raindrop without being reflected. So there’s always a directly transmitted, zero-order light ray parallel to the tertiary rainbow ray, like this:Path of rainbow ray for tertiary rainbow, and zero-order glowMuch more light pours through the raindrop in the zero-order glow than survives through three internal reflections. This makes the tertiary rainbow very difficult to see.

But not impossible. There have been sporadic reports over the last few decades, by careful observers who knew what they were looking for. And finally, in 2011, a paper* appeared in the “Light And Color In The Open Air” edition of Applied Optics, entitled “Photographic evidence for the third-order rainbow”. The authors describe taking a photograph under favourable conditions (the sun obscured, a dark cloud in the region that would be occupied by the tertiary bow). The photographer could barely discern a hint of the tertiary rainbow—“only a faint trace of it at the limit of visibility for about 30 seconds”. But after a bit of image processing, a rainbow arc appeared in the resulting photograph. And, after careful analysis, the authors confirmed they had taken the first known photograph of a tertiary rainbow.

You’ll have discerned a pattern. Each successive rainbow (primary, secondary, tertiary) is produced by a rainbow ray which is increasingly offset from the centre of the drop, undergoing one more reflection that its predecessor. The increasing offset means that refraction is greater, with a larger difference between red and violet rays, leading to a broader rainbow. Each additional reflection along the light path means more light lost, and so a fainter rainbow.

Rather than bore you with light paths for higher-order rainbows, I’ll show you them all on one diagram. (The design is based on Jearl Walker’s “Amateur Scientist” column in the July 1977 issue of Scientific American—the calculations and drawing are all my own.)

First twenty rainbows for light entering top of raindrop
Click to enlarge

The rosette shows the first twenty rainbows, produced by the first twenty internal reflections of light that enters the upper half of a water droplet. Light bounces around the drop clockwise, and the difference in deflection for successive rainbows quite quickly converges to settle at a little less than a right angle. So you can see that the primary (labelled “1” in the lower left quadrant) is followed by the secondary in the upper left quadrant, the tertiary in the upper right quadrant, and the quaternary in the lower right. The quinary rainbow (number five) brings the progression almost full circle, and appear just a little anticlockwise of the primary. I won’t bore you with the names for higher orders of reflection, but you should be able to pick out how they go around again, and again, and again, becoming successively wider and fainter.

Because violet is refracted more than red, all the rainbows have the same layout, with violet always lying clockwise of red.

Rainbows in the lower left quadrant are being reflected downwards, and back towards the light source. So an observer would look for them as circles around the antisolar point. Because the red light is descending more steeply than the violet, red will appear on the outer edges of all these rainbows. Rainbows in the lower right quadrant are formed from light that has been reflected downwards and away from the light source—so the observer must look towards the sun to see them. In this position, violet light is descending more steeply than red light, so all these rainbows have violet on their outer edges.

Light in the top half of the diagram is all spraying upwards—these rainbows would not be visible to an observer standing below the drop. But we’ve neglected the light that enters the lower half of the raindrop, and bounces anticlockwise. It produces its own rosette of rainbows, identical to the one above, except mirrored in the horizontal plane, like this:

First twenty rainbows for light entering bottom of raindrop
Click to enlarge

The reversed light path means that violet always lies anticlockwise of red for this family of rainbows. So antisolar rainbows with this light path have violet as their outer colour, and solar rainbows have red outermost.

So the rainbows we see in the sky come from a mixture of the two sets of possible light paths in the two diagrams above. So let’s stack them together, and switch from a view centred on the water drop to a view centred on the observer:

The first twenty rainbows
Click to enlarge

The sky is full of overlapping rainbows! You should think of the upper and lower “copies” of each rainbow as being linked by a vertical circular arc sticking out at right angles to your screen, which the observer sees as a (potential) circular rainbow.

My diagram has the sun directly behind the observer, in which case the rainbows in the lower part of the diagram would be invisible under normal circumstances, superimposed on the ground, and each rainbow would form a semicircular arc against the sky. But the sun is usually some distance above the horizon—as it climbs higher, it pushes the antisolar rainbows lower in the sky, but carries the solar rainbows higher. So antisolar rainbows  generally form less than a semicircular arc, and when the sun is higher above the horizon than the radius of the rainbow they will drop entirely below the horizon. But solar rainbows will be lifted above the horizon by the sun, forming more than a semicircular arc, and at the extreme when the sun is higher above the horizon than the radius of the rainbow they can form complete circles. (Complete circles are also possible for antisolar rainbows, but only when the observer is looking down on water droplets suspended below the horizon, as from a plane flying over clouds.)

Are any of them visible to the naked eye? We know that the tertiary has been spotted very rarely. The quaternary sits right next to it, farther out in the zero-order glow, but additional light losses mean its rainbow ray transmission is just 15% of the primary. I don’t know of anyone who has seen it, but it has been photographed. In fact, the article entitled “Photographic observation of a natural fourth-order rainbow” appeared in the same themed edition of Applied Optics as the tertiary rainbow report. (Early reports of the tertiary photographs had inspired the author to search out the quaternary using image stacking.)

The quinary bow has also been photographed. With rainbow-ray transmission sitting at just 10% of the primary, it is also partially obscured by the brighter secondary bow. But its green, blue and violet portion sit within Alexander’s Dark Band, giving reasonable hope of detecting those colours. And Harald Edens managed to (just) pick out the green stripe of the quinary rainbow in a photograph taken in 2012.

The sixth-order rainbow sits within the bright sky at the inner edge of the primary bow, and seems like a poor candidate for detection. The seventh is better separated from the zero-order glow than the third and fourth, but its rainbow ray transmission is only 6% of the primary. However, it seems likely that some of the higher order bows will yield to the sort of photographic techniques commonly employed in astrophotography these days. Watch this space. Meanwhile, for your delectation, I’ve appended some basic descriptive data for the first twenty rainbows, as featured in my diagrams above.


* Großmann M, Schmidt E, Haußmann A. Applied Optics 50(28): F134-41
Oh, alright, I will. Beyond quinary, five, the sequence goes senary, septenary, octonary, nonary, denary, undenary, duodenary … at which point we need to start making up names until we get to vigenary, twenty.
Theusner M. Applied Optics 50(28): F129-F133

Note: All values, here and in previous posts, are calculated using a refractive index for red light (wavelength 700nm) of 1.33141, and for violet light (wavelength 400nm) of 1.34451. See the page dealing with refractive index on Philip Laven’s excellent website, The Optics Of A Water Drop, for further information and references.


The First Twenty Rainbows
OrderOuter
Radius (º)
Width (º)Outer
Colour
Solar/
Antisolar
Transmission
(% Primary)
142.41.9RedAntisolar100
253.83.4VioletAntisolar42.6
342.54.8RedSolar23.5
448.96.1VioletSolar14.9
552.97.4RedAntisolar10.3
639.68.7VioletAntisolar7.6
765.610.0RedSolar5.8
829.011.2VioletSolar4.5
979.112.5RedAntisolar3.7
1017.813.8VioletAntisolar3.0
1193.1*15.1RedSolar2.6
1210.110.1RedSolar2.2
1390.3*17.6VioletSolar1.8
1424.418.9RedAntisolar1.6
1578.520.1VioletAntisolar1.4
1638.821.4RedSolar1.3
1766.622.7VioletSolar1.2
1853.323.9RedAntisolar1.0
1954.625.2VioletAntisolar0.9
2067.926.5RedSolar0.8

* The order-11 and order-13 rainbows lie predominantly in the solar hemisphere, but extend slightly into the antisolar
The order-12 rainbow spans the solar point, and therefore overlaps itself—a small disc of blue-violet rainbow (radius 6.1º) is superimposed on a larger disc of green-yellow-red rainbow (radius 10.1º)

Secondary Rainbows

Rainbow enclosing bright sky
Click to enlarge
Photo by Binyamin Mellish via Good Free Photos

In my previous post about rainbows, I described how the light of the rainbow was reflected back to our eyes by falling water droplets. For a raindrop at the top of the rainbow arc, light follows a path that enters near the top of the raindrop, bounces off the back, and then exits from the bottom:

Green rainbow rayThe angle between the incoming light ray and the reflected ray ranges from 42.4º for red light to 40.5º for violet light. All other light rays, entering the drop either closer to its centre or closer to its edge, are reflected back at smaller angles—and it’s their smeared and superimposed light which accounts for the white glow visible within the arc of the rainbow, above. I used the name “Offset” for the parameter that measures how close to central the incoming light ray hits the droplet. It’s measured like this:

Rainbow offset parameterAnd by plotting the deflection of light rays at various offsets, along with light transmission along the reflected pathway, I showed how the rainbow forms at the point of maximum deflection, corresponding to an offset of about 0.86:

Click to enlarge

I called the ray that follows this maximum-deflection route the “rainbow ray”. (See my previous post for much more detail.)

I also produced a little diagram of how light is lost from the rainbow ray each time it encounters a surface at which it is either reflected or transmitted, like this:

Light losses from rainbow raySo the rainbow ray arrives at your eye containing only about 4.5% of the light that entered the water drop.

What I want to talk about this time is the fate of the light that undergoes a second internal reflection (labelled “0.5%” in the diagram above).

It’s possible for light from this second reflection to exit the drop when it next encounters the water-air interface, like this:

Green two reflection rainbow rayThe second reflection takes the light to the front of the raindrop, and it exits on an upward course, crossing the path of the incoming ray. For this ray to reach the eyes of an observer looking towards the antisolar point (the centre of the rainbow arc), we have to flip the geometry upside-down, like this:

Green secondary rainbow raySo now the incoming light enters near the bottom of the raindrop, and the reflected light is deflected downwards, towards an observer on the ground.

This light is the source of the secondary rainbow, which is larger and fainter than the primary rainbow I’ve been describing so far. (A secondary rainbow is dimly and partially visible in the photograph at the head of this post.)

Like the singly reflected light that forms the primary bow, the doubly reflected light that forms the secondary bow has its own characteristic angle of deflection, but this time (because of the flipped light-path described above) it’s the minimum angle of deflection from the antisolar point where light is concentrated to form the secondary rainbow ray:

Deflection and tranmission in formation of secondary rainbow
Click to enlarge

I’ve kept the scale of the transmission curve the same, to allow comparison with that of the primary rainbow. And the range of the angle-of-deflection axis is the same, but it spans from 45º to 90º this time, rather than the 0º to 45º of the primary plot.

The minimum deflection occurs at an offset of about 0.95. The deflection is 50.4º for red light, and 53.8º for violet.

So there are several things going on with this secondary rainbow. Firstly, because the light enters closer to the edge of the raindrop, the refraction is greater, and that causes the separation between red and violet light to be greater—so the secondary rainbow has a width of 3.4º, compared to just 1.9º for the primary rainbow. Secondly, the fact the light-path is flipped over compared to the primary reverses the colour sequence—red is on the inside of the secondary bow, but on the outside of the primary. Thirdly, the additional reflection means more light is lost from the rainbow ray as it passes through the drop. This is partially offset by the fact that the rainbow-ray offset is closer to the offset of maximum light transmission—so the light transmitted into the secondary bow works out to be about 43% of that transmitted into the primary. Finally, because the rainbow ray occurs at a minimum deflection from the antisolar point, the rest of the sunlight entering the drop is reflected to greater angles than the rainbow ray—it lights up the sky outside the secondary bow.

So if we imagine a raindrop falling vertically towards the antisolar point, it at first sends a doubly reflected mixture of light, appearing white, towards the observer’s eye. When it has fallen to 53.8º from the antisolar point, it sends a relatively pure, doubly reflected violet light to the observer—the top of the secondary bow. As it falls from 53.8º to 50.4º, it reflects all the spectral colours in sequence, ending with red light at the inner rim of the secondary bow. Then, it quite literally goes dark. It is too low to send any doubly reflected light in the observer’s direction, but too high to send any singly reflected light. The region of sky between the secondary and primary bow is therefore noticeably darker than either the region outside the secondary, or inside the primary.

Primary and secondary rainbows
Click to enlarge
Original photograph used under Creative Commons Attribution-ShareAlike 3.0 Licence

This dark region is called Alexander’s Dark Band*, in honour of the Greek philosopher Alexander of Aphrodisias, who described it in about 200AD.

After passing through the Dark Band, the raindrop lights up with singly reflected red wavelengths when it reaches 42.4º, and then runs through the spectral colours until it reaches violet at 40.5º. Below that point, it reflects a mixture of wavelengths (white light again) until it hits the ground.

The whole sequence looks like this:

Primary and secondary rainbow formation
Click to enlarge

The obvious question now is this: if the primary rainbow forms from singly reflected light, and the secondary from doubly reflected light, what happens to triply reflected light? Is there a tertiary rainbow? And if so, where is it?

That’s the topic for next time.


* Not to be confused with the 1911 Irving Berlin hit, Alexander’s Ragtime Band.

Rainbow Rays

Chalked lockdown rainbow
Click to enlarge

The COVID-19 lockdown, in my part of the world, has produced an outpouring of children’s rainbow art—often stuck up in people’s windows, but sometimes sketched on the pavements, too.

I’ve been struck by the generally good command of spectral colours on display, with red on the outside and an appropriate progression towards violet on the inside. I was amused by the one above, which is a pretty flawless piece of artistry, undermined only by the position of the sun.

It reminded me that I had been planning to write about rainbow colours for years, ever since I wrote about converging rainbows back in 2015.

That was when I posted this diagram, showing the relationship between the sun’s position and the rainbow arc:

Formation of rainbow
Click to enlarge

The rainbow is a complete circle of coloured light, centred on the antisolar point—the direction exactly opposite the position of the sun, which is marked by the shadow of your head. We usually only see an upper arc, because the area below the horizon usually contains too few raindrops along the line of sight to generate bright colours.

Every raindrop 42½º away from the antisolar point reflects red light towards our eyes; every raindrop at 40½º reflects violet light in the same way. And between those extremes, raindrops reflect all the spectral colours in turn, changing their apparent colour as they fall. But quite why that happens is a little complicated, and that’s what I want to write about this time.

Here’s the route that a ray of green light takes when it passes through a rainbow-forming raindrop and bounces back towards our eyes. Let’s call this particular trajectory the “rainbow ray” for short:

Green rainbow ray

If the raindrop is falling through the top of the rainbow arc, the light enters near the top of the drop, is refracted as it crosses the air-water interface, then reflected from the back of the drop, then exits through the bottom of the drop, being refracted again as it moves from water back to air. The angle between the incoming ray and the outgoing is about 41½º for green light. At the sides of the rainbow, you have to imagine the diagram above lying horizontal—the ray enters the outward-facing side of the drop, and is reflected towards you sideways. And, obviously, the rest of the rainbow is formed by light paths that are more or less tilted between the horizontal and the vertical.

Violet light is refracted more than green light, which is in turn refracted more than red light. So a ray of white light (drawn in black below) is split into a fan of coloured rays as it enters the raindrop. These follow slightly different courses within the drop, and exit at different angles (exaggerated here for clarity):

Violet, green and red rainbow rays, exaggerated

So we see red at 42½º, and violet at 40½º, with green between. Simple.

But why choose to consider just those rainbow rays? What about all the light that enters the drop closer to its rim, or closer to its centre? I’ll call this general group of light rays “reflected rays”, of which the rainbow ray is only one example.

If I plot a couple of examples, you can see that rays entering the drop farther out than the rainbow ray are refracted more strongly, and end up exiting the drop at an angle less than the rainbow ray. Those that enter the drop nearer the centre are refracted less, but bounce back at a narrower angle from the back of the drop, and also exit the drop at an angle less than the rainbow ray. So for a given colour of light, it turns out that the rainbow ray is the light path that results in the maximum deflection angle from the antisolar point. All other reflected rays exit at narrower angles, and so should appear within the visible coloured arc of the rainbow itself.

Rainbow ray and other rays
How does that help, though? Why is the rainbow ray’s role as the maximum angle of deflection important?

To show why, I’m going to make a plot of what happens to all the reflected rays, using a parameter I’ll call Offset*, which works like this:

Rainbow offset parameterIt’s just the proportional distance from the centre of the raindrop at which the ray enters—a zero offset means that the ray spears straight into the centre of the drop; an offset of one means the ray just grazes the edge of the drop.

So here’s how red and violet reflected rays are deflected, for the full range of offsets:

Deflection of red and violet light relative to antisolar point for various ray offsets
Click to enlarge

Deflection peaks at an offset of about 0.86, the location of the rainbow ray, with lesser deflection occurring on either side, as shown in the diagram above. Red and violet are at their maximum separation at the peak, and the rounded peak of the curves means that a lot of rays close to the rainbow ray end up reflected in the same part of the sky as the rainbow ray. You can see from the chart that all the rays with offsets from 0.75 to 0.95 end up within two degrees of the rainbow ray; a similar span from 0.2 to 0.4 is spread over fifteen degrees.

So, in the vicinity of the rainbow ray, there’s a lot of light in a small area of sky, and the spectral colours are well separated. Farther from the rainbow ray, the deflected light is smeared over a large area of sky within the rainbow arc, and the spectral colours are not well separated—all these other rays average out to a patch of white light filling the curve of the rainbow, with no colour separation. This bright area within the rainbow can often be strikingly visible, if the rainbow has dark clouds behind it.

Rainbow enclosing bright sky
Click to enlarge
Photo by Binyamin Mellish via Good Free Photos

One other thing contributes to the colour intensity of the rainbow—oddly, that’s that fact that some light is lost from the reflected rays every time they interact with an air/water or water/air interface. Here’s a diagram of how much light goes missing from the green rainbow ray as it passes through the raindrop:

Light losses from rainbow ray

The large proportion of light that shoots straight out the back of the drop, doubly refracted but without being internally reflected, creates a bright patch around the sun that appears whenever the solar disc is viewed through falling rain. Les Cowley at the excellent Atmospheric Optics site has dubbed this the “zero-order glow”.

Interestingly, for a given ray each interaction with an interface results in exactly the same ratio of reflection to transmission (though not quite in my diagram, which features rounded figures). This is unexpected (at least to me), because the reflective properties of a water/air interface are generally different from those of an air/water interface; the former features the phenomenon of total internal reflection, for instance. But it turns out that the first passage through the air/water interface changes the angle of the ray just enough to make it interact with subsequent water/air interfaces in exactly the same way as its initial air/water encounter.

If I plot the final amount of transmitted light for all the different offset rays, and add it to my previous graph, it looks like this:

Transmission, and deflection of red and violet light relative to antisolar point for various ray offsets
Click to enlarge

The transmission data are in brown, and refer to the new, brown axis on the right side of the chart. You can see that transmission starts to ramp up just as we get into the vicinity of the rainbow ray, boosting the brightness in the rainbow’s part of the sky. The peak of transmission does occur at very high offsets, beyond the rainbow ray, but in that region the angle of deflection changes very rapidly with slight changes in offset, which diffuses that light over a large arc.

The calculations I did to produce the transmission graph above involved Fresnel’s equations, so I had to track two different polarizations of light independently. For light reflecting from a surface between two transparent mediums, there’s a critical angle of incidence called Brewster’s angle, at which the reflected light becomes totally polarized. At that angle, the reflected light is entirely s-polarized; light polarized at right angles to this (p-polarized) is completely transmitted through the reflective surface. (Your polarizing sunglasses are designed to filter out s-polarized light reflected from horizontal surfaces, to reduce glare.)

The Brewster angle for an air/water interface is around 53º; for water/air it is about 37º. And it turns out that any light entering the water at an angle of incidence of 53º has its angle changed to 37º by refraction. So in the case of our raindrop, a ray that strikes the surface of the drop at 53º (corresponding to an offset of about 0.8) will continue through the drop and strike the water/air interface at 37º—it hits two Brewster angles in succession! This means that p-polarized light that hits the drop at Brewster’s angle is entirely refracted into the drop—none of it escapes by reflection from the air/water interface. But then it hits the back of the drop, and now none of it is re-reflected—it is all transmitted, again. So at an offset of 0.8, no p-polarized light gets into the reflected ray—it is all lost out the back of the raindrop.

So now if I mark up the total amount of transmitted light with its s- and p-polarized components, you can see that the light making up the rainbow will be strongly s-polarized, because the rainbow rays are pretty close to Brewster’s angle:

Polarization of light transmitted by the rainbow
Click to enlarge

The resulting polarization follows the curve of the rainbow. Your polarizing sunglasses will largely block the light coming from the top curve of the rainbow, but will let light through from the sides. However, if you tilt your head, you’ll remove light from the sides of the rainbow, and bring the upper curve into view.

Here’s a nice short little YouTube video, by James Sheils, demonstrating how to make the lower curve of a rainbow appear and disappear using a polarizing filter:

And that’s it, for now. Some time in the future I’ll get around to discussing the secondary rainbow.


* The measure I’ve called Offset is sometimes called the “impact parameter”, a term borrowed from nuclear physics. While the analogy is strong, if you know its original application, I’m not sure the phrase itself helps with visualization, so I’m sticking with Offset in this and subsequent posts.

Does The Sun Set On The British Empire?

British Empire map, 1886
Click to enlarge
Walter Crane’s map of the British Empire, 1886

In short, taking every thing into consideration, the British empire in power and strength may be stated as the greatest that ever existed on earth, as it far surpasses them all in knowledge, moral character, and worth. On her dominions the sun never sets. Before his evening rays leave the spires of Quebec, his morning beams have shone three hours on Port Jackson, and, while sinking from the waters of Lake Superior, his eye opens upon the mouth of the Ganges.

Caledonian Mercury, 15 October 1821, page 4: “The British Empire”

It’s noticeable, when reading the above, that none of the places it mentions by name still belong to the United Kingdom. The British empire is now much reduced in size; in fact, its overseas possessions are confined to a scatter of places that few people could reliably place on a map:

Overseas Territories
● Anguilla
● Bermuda
● British Virgin Islands
● Cayman Islands
● Falkland Islands
● Gibraltar
● Montserrat
● Pitcairn Islands
● Saint Helena (with Ascension & Tristan da Cunha)
● Turks and Caicos Islands
● British Indian Ocean Territory
● South Georgia and South Sandwich Islands
● British Antarctic Territory (in abeyance under Antarctic Treaty)

Dependent territory
● Sovereign Base Areas of Dhekelia & Akrotiri (Cyprus)

If you’re one of the people who would have trouble placing these names on a map, here’s a map:

British overseas territories
Click to enlarge
(Source of base map)

Cover of Britain's Treasure Islands by Stewart McPhersonAnd if you’d like to know more about all these places, I heartily recommend Stewart McPherson’s marvellous book, Britain’s Treasure Islands: A Journey To The UK Overseas Territories, as well as the accompanying BBC television series.

What stands out from the map above is that the UK still has the Atlantic, Caribbean and Mediterranean pretty well covered. There’s a solitary (and I do mean solitary) British possession in the Pacific, the Pitcairn Island group. (I’ve written about Pitcairn and its neighbouring islands a couple of years ago, when we were lucky enough to visit them.) And there’s another single possession in the Indian Ocean, the catchily named British Indian Ocean Territory (BIOT). BIOT occupies the whole of the Chagos Archipelago, and is inhabited entirely by British and American military personnel and contractors, based on the largest island, Diego Garcia. It used to be home to 2000 Chagossians, who were chucked out around 1970 to make way for the UK/US military installations. The poor Chagossians are still grinding through the courts attempting to get their homeland returned to them.

Anyway. Pitcairn and BIOT, which are a long way west and east of most UK territories, look like the key locations to examine when it comes to deciding whether the sun still “never sets on the British empire”. With Pitcairn’s time zone of GMT-8, and BIOT’s of GMT+6, there’s only ten hours of difference between the two territories, which should mean that the sun is visible from both locations for a couple of hours a day. But there’s a potential problem with the seasonal variation in day length—while BIOT sits close to the equator and won’t have much variation in the times at which the sun rises and sets, Pitcairn is south of the tropics, and so we can expect its sunsets to be noticeably earlier in June than they are in December.

So we’re going to need to plot daylight charts for the whole year. Here’s one for Greenwich:

Sunrise and sunset in Greenwich
Click to enlarge

Along the x-axis we have the months of the year, numbered from 1 to 12. On the y-axis, Greenwich Mean Time. The lower curve marks the time of sunrise, throughout the year, at Greenwich. The upper curve is sunset. The yellow area between the curves therefore represents the totality of daylight seen in Greenwich throughout the course of a year.

OK. Let’s superimpose the sunrise and sunset curves for Adamstown on Pitcairn, giving times in GMT:

Sunrise and sunset in Greenwich & Pitcairn
Click to enlarge

The Pitcairn sunrise and sunset curves are in red, and Pitcairn daylight extends a long way through the Greenwich night. But sunset on Pitcairn always occurs before sunrise in Greenwich, so there’s a brief period when the sun is shining in neither location.

Will BIOT, with its sunrise earlier than Greenwich, fill the gap? Here are the BIOT curves (calculated for Diego Garcia) added in green:

Sunrise and sunset in Greenwich, Pitcairn & BIOT
Click to enlarge

It’s a close-run thing. Pitcairn’s midwinter sunset on 21 June 2020 comes just 38 minutes after BIOT’s sunrise. Here’s a south polar view of the Earth on that date, capturing the brief period when both territories are in sunlight:

Diego Garcia and Adamstown both in sunlight on 21 June 2020
Click to enlarge
(Prepared using Celestia)

But there’s no doubt the chart is full of daylight, and the sun still never sets on the British empire!