Galactic greenhouse

Before Christmas, my enterprising friend Clare decided to brighten the dark nights of a Scottish winter by turning her greenhouse into an illuminated art gallery. She asked friends to produce translucent artworks that could be hung in the greenhouse and lit from within.

My contribution is a representation of the movement across the sky of Jupiter and Saturn (and some smaller planets) in the two years bracketing the recent Great Conjunction. It’s made from a sheet of wallpaper, painted black, with holes cut out and with coloured filters placed behind the holes.

A photograph of the piece. It’s just over 50 cm wide. The variations in brightness of the discs aren’t part of the plan; the illumination wasn’t perfectly even.

The piece is divided into 30 rows. All but one of these rows contain a large red disc (representing Jupiter) and a large yellow disc (representing Saturn). A row may also contain smaller discs, representing Mars in pink, Venus in white and Mercury in blue. The purple discs represent the ex-planet Pluto. All of the discs hugely exaggerate the size of their planets.

Each row represents the same strip of the sky, in the sense that if I had included stars on the piece, the same stars would appear in the same positions on every row. From top to bottom, the rows show that strip of sky at 25-day intervals, covering a period roughly from roughly one year before the Great Conjunction to one year after. The discs in each row indicate the positions of any planets that are in that strip of sky at the time.

Concentrating on Jupiter (red) and Saturn (yellow) first, we see that they have a general leftward motion, but with periods of rightward motion. Jupiter’s overall leftward motion is faster than Saturn’s: it starts to the right of Saturn and finishes to the left. Because Jupiter overtakes Saturn, there comes a point where they are at the same place in the sky. This is the Great Conjunction: in this row, both Jupiter and Saturn are represented by a single large white disc.

Mars, Venus and Mercury move much faster. Mars crosses our field of view in only 4 rows (roughly 100 days) and Venus and Mercury make repeat visits. Pluto wavers back and forth without appearing to make much leftward progress at all.

 

The FAQ

Why do the planets move along the same line? They don’t exactly, but it’s pretty close. All of the planets, including the Earth, move around the Sun in roughly circular orbits. Except for Pluto’s, these orbits are more or less in the same plane (like circular stripes on a dinner plate). Because our viewpoint (the Earth) is in this plane, we look at all the orbits edge on, and the planets appear to follow very similar straightish paths across the sky. I have chosen to neglect the slight variations in path and depict the planets as following one another along exactly the same straight line

Why do Jupiter and Saturn move mainly right to left? Looking down from the North, all of the planets orbit anticlockwise. Mars, Jupiter, and Saturn have bigger orbits than the Earth, we’re observing them from inside their orbits (and from the Earth’s northern hemisphere). Thus their general movement is leftwards. (If you don’t get it, whirl a conker around your head on a string, so that it moves anticlockwise for someone looking down. The conker will move leftwards from your point of view.) The orbits of Venus and Mercury are inside the Earth’s orbit; their movements as seen from the Earth are rather complicated.

Why do Jupiter, Saturn, and Pluto sometimes move from left to right? Earth is in orbit too, so we’re observing the planets from a moving viewpoint. If you move your head from side to side, nearby objects appear to move back and forth against the background of distant objects. Exactly the same effect happens with our view of the outer planets as the Earth moves around its orbit from one side the Sun to the other – they appear to move back and forth once a year against the background of distant stars. But at the same time, they are also really moving leftwards (as we look at them). The sum of the planet’s real motion with their apparent back-and-forth motion gives the lurching movement that we see: mainly leftwards but with episodes of rightward motion. Note that the planets never actually move backwards: they just appear to. The same thing happens to Mars, but none of its periods of retrograde motion coincided with its visit to our strip of the sky.

Why do some planets move faster across the sky than others? The larger a planet’s orbit, the more slowly it moves. For the outer planets, a larger orbit also  means that we’re watching it from a greater distance, so it appears to move more slowly still. Saturn’s orbit is about twice as big as Jupiter’s, so it moves more slowly across the sky than Jupiter. Jupiter “laps” Saturn about once every 20 years: these are the Great Conjunctions. Mars’ orbit is smaller than Jupiter’s, so it moves more quickly across the sky. Meanwhile lonely Pluto plods around its enormous orbit so slowly that the leftward trend of its motion is barely discernible; all we see is the side-to-side wobble caused by our own moving viewpoint. As for Mercury and Venus: it’s complicated.

Please could you stop being evasive about the movements of Venus and Mercury? It really is complicated. The orbits of Venus and Mercury are smaller than the Earth’s: we observe them from the outside. If the Earth was stationary, we’d see Venus and Mercury moving back and forth from one side of the Sun to the other. Returning to our conker-whirling experiment, it’s like watching a conker being whirled by somebody else rather than whirling it yourself. But the Earth is moving around its orbit too. And then Venus and Mercury are also moving rather fast: Mercury orbits the Sun 4 times for each single orbit made by the Earth. Combine all of these things and it becomes very confusing. Whereas the outer planets’ episodes of retrograde (backwards) movement across the sky occur less than once a year, Mercury is retrograde about three times a year.

Do the planets really follow a horizontal path across the sky? This question doesn’t have an answer. We’re using the pattern of stars, all inconceivably distant compared to the planets, as the fixed background against which we view the movement of the planets. You may have noticed that the stars move in arcs across the sky during the night; this is due to the Earth’s rotation on its axis. So our strip of sky moves in an arc too, and turns as it moves. So if it ever is horizontal, it is only briefly so, and when and if it is ever horizontal will depend upon your latitude.

Jupiter and Saturn never exactly lined up, did they? No, they didn’t (see the answer to the first question). On this scale, at the Great Conjunction the discs representing Jupiter and Saturn should be misaligned vertically by about a millimetre. With our hugely over-sized planets, this means almost total overlap, which still misrepresents the actual event, where the planets were separated by many times their own diameter. And for all other rows, where the two discs don’t overlap, a millimetre’s misalignment would be imperceptible. A final and maybe more compelling reason for my neglect of the misalignment of the planets’ paths is that I don’t know how to calculate it.

Anything else to confess?  Yes. There’s a major element of fiction about the piece in that it’s not physically possible to see all of these arrangements of the planets. The reason is that for some of these snapshots, the Earth is on the opposite side of the Sun from most or all of the planets, and Sun’s light would drown out the light from the planets. In other words, it would be daytime when the planets are above the horizon, and therefore in practice they would be invisible. This was almost the case for the Great Conjunction, where there was only a short period of time between it becoming dark enough for Jupiter and Saturn to be visible, and them disappearing over the horizon.

A further element of fiction is that, even in the depths of a Scottish winter’s night, Pluto is far too faint to be seen with the naked eye, not to mention not being regarded by the authorities as a planet any more. But it was passing at the time of the Great Conjunction and it seemed a pity to miss it out.

 

Bradycardia

Here again is the processor package from my old laptop. The processor has a clock in it that delivers electric pulses that trigger the events in the processor. The clock on this processor “ticks” at 2.2 gigahertz, that is, it sends out 2.2 billion pulses per second.

Over two thousand million pulses every second! How can we make sense of such a huge number?

In this post, I’m going to do with time what I did with space in the previous post. I’m going to ask the question:

Suppose that we slow down the processor so that you could just hear the individual “ticks” of the the processor clock (if we were to connect it to a loudspeaker), and suppose that we slow down my bodily processes by the same amount. How often would you hear my heart beat?

Answer: My heart would beat about once every year and a half.

The calculation

How slow would the processor clock need to tick for me to be able to hear the individual ticks? A sequence of clicks at the rate of 10 per second clearly sounds like a series of separate clicks. Raise the frequency to 100 per second, and it sounds like a rather harsh tone; the clicks have lost their individual identity. Along the way, the change from sounding like a click-sequence to sounding like a tone is rather gradual; there’s no clear cutoff.

You can try it yourself using this online tone generator. Choose the “sawtooth” waveform. This delivers a sharp transition once per cycle, which is roughly what a train of very short clicks would do, and play around with the value in the “hertz” box. (Hertz is the unit of frequency; for example, 20 hertz is 20 cycles per second.)

I found that a 40 hertz sawtooth definitely sounds like a series of pulses, and that a 60 hertz sawtooth has a distinct tone-like quality. So let’s say that the critical frequency is 50 hertz, that is, 50 ticks per second. I don’t expect you to agree with me exactly.

If I can hear individual pulses at a repetition rate of 50 hertz, then to hear the ticks of a 2.2 gigahertz clock I need to slow down the clock by a factor of

(1)   \begin{equation*}   \frac{2.2 \times 10^9}{50} = 44 \times 10^6 \end{equation*}

At rest, my heart beats about once per second, so if it was slowed down by the same factor as the processor clock, it would beat every 44 × 106 seconds, which is about every 17 months.

Or should it be twice as long?

The signal from the processor clock is usually a square wave with 50% duty cycle. Try the square wave option on the online signal generator with a 1 hertz frequency (one cycle per second). You’ll hear two clicks per second, because in each cycle of the wave, there are two abrupt transitions, a rising one and a falling one.

This means that if we did connect a suitably slowed-down processor clock to a loudspeaker, we’d hear clicks at twice the nominal clock rate. Looked at this way, we’d need to slow down the clock, and my heart, twice as much as we’ve calculated above. My heart would beat once every three years.

However, most processors don’t respond to both transitions of the clock signal. Some processors respond to the rising transition, others to the falling transition. To assume that we hear both of these transitions is to lose the spirit of what we mean by one “tick” of the processor clock.

 

 

Anticrepuscular rays

Converging rays

I took this photograph at dusk recently from the beach at Portobello, where Edinburgh meets the sea. As sunset pictures go, it’s not much to look at. But what caught my attention was the faint radiating pattern of light and dark in the sky.  The light areas are where the sun’s rays are illuminating suspended particles in the air. The dark areas are where the air is unlit, because a cloud is casting a shadow.  You may have seen similar crepuscular rays when the sun has disappeared behind the skyline and the landscape features on the skyline cast shadows in the air.

The rays in my picture appear to radiate from a point below the horizon, because that’s where the sun is…isn’t it?

No! Portobello beach faces north-east, not west. The sun is actually just about to set behind me! So why do the rays appear to come from a point in front of me? Shouldn’t they appear to diverge from the unseen sun behind me?

To understand why, we need to realise that the rays aren’t really diverging at all. The Sun is a very long way away (about 150 million kilometres), so its rays are to all intents and purposes parallel. But just as a pair of parallel railway tracks appear to diverge from a point in the distance, so the parallel rays of light appear to diverge from a point near the horizon.

The point from which the rays seem to diverge is the antisolar point, the point in the sky exactly opposite the sun, from my point of view. It’s where the shadow of my head would be. When I took the photograph, the sun was just above the horizon in the sky behind me, so the antisolar point, and hence the point of apparent divergence, is just below the horizon in the sky ahead of me.

For normal crepuscular rays, the (obscured) sun is ahead, and the light is travelling generally* towards the observer. The rays in the picture are anticrepuscular rays, because the light is generally travelling away from me. This was the first time that I had knowingly seen anticrepuscular rays.

*I say “generally” because the almost all of the rays aren’t travelling directly towards the observer. An analogy would be standing on a railway station platform as a train approaches: you’d say that it was travelling generally towards you even though it isn’t actually going to hit you.

 

“I’m deuterawhat?” – colour vision at Orkney Science Festival

No need to look so sad, Garry. You're special.
No need to look so sad, Garry. You’re special.

You’re deuteranomalous, Garry.

The distressed man on the right is Garry McDougall. Garry’s just found out that his colour vision is not the standard-issue colour vision that most of us have. He made this discovery while watching my talk on the science of colour vision, in Kirkwall as part of the Orkney International Science Festival 2018.

Garry and I were part of a team funded by the Institute of Physics to perform at the festival.  Also on the team were Siân Hickson (IOP Public Engagement Manager for Scotland) and Beth Godfrey.

Garry needn’t look quite so woebegone: he’s not colour blind, and he’s in plentiful company – about 1 in 20 men have colour vision like his.

Normal metameric lights
To Garry, these two lights looked different.

How did Garry’s unusual colour vision come to light? In one of the demos in my talk, I compare two coloured lights. One (at the bottom in the picture on the right) is made only of light from the yellow part of the spectrum. The other (at the top) is made of a mixture of light from the red and green parts of the spectrum. If I adjust the proportions of red and green correctly, the red/green mixture at the top appears identical to the “pure” yellow light at the bottom.

Except that to Garry it didn’t. The mixture (the top light) looked far too red. By turning the red light down, I could get a mixture that matched the “pure” yellow light as far as Garry was concerned. But it no longer matched for the rest of us!  To us, the mixture looked much greener than the “pure” yellow

Garry metameric lights
To Garry, these two lights looked the same.

light; the lower picture on the right shows roughly how big the difference was. This gives us an insight into how different the original pair of lights (that we saw as identical) may have appeared to Garry. It’s not a subtle difference.

We can learn a lot from this experiment.

Firstly, we’re all colour blind. The red/green mixture and the “pure” yellow light are physically very different, but we can’t tell them apart. “Colour normal” people are just one step less colour blind than the people we call colour blind.

Secondly, it shows that there’s no objective reality to colour. People can disagree about how to adjust two lights to look the same colour, and there’s no reason to say who’s right.

Thirdly, it shows that Garry has unusual colour vision. Our colour vision is based on three kinds of light-sensitive cell in our eyes. They’re called cones. The three kinds of cone are sensitive to light from three (overlapping) bands of the spectrum. Comparison of the strengths of the signals from the three cone types is the basis of our ability to tell colours apart. Garry is unusual in that the sensitivity band of one of his three cones is slightly shifted along the spectrum compared to the “normal” version of the cone. This makes him less sensitive to green than the rest of us, which is why the red/green mixture that matches the “pure” yellow to Garry looks distinctly green to nearly everyone else.

Garry isn’t colour blind. He’s colour anomalous. A truly red-green colour blind person has only two types of cone in their eyes. Garry’s kind of colour anomaly is quite common, affecting about 6% of men and 0.4% of women. It’s called deuteranomaly, the deuter- indicating that it’s the second of the three cone types that’s affected, ie the middle one if you think of their sensitivity bands arranged along the spectrum.

My thanks to Siân Hickson for the photographs.

Ben on rocks
Exploring the coast at Rerwick Point.
rainbow
Showery weather meant that we were treated to many magnificent rainbows, like this one seen at Tankerness.

A note to deuteranomalous readersNormal metameric lights

Please don’t expect the illustrations of the colour matches/mismatches above to work for you as they would have done if you’d seen them live. A computer monitor provides only one way to produce any particular colour, so the lights that appear identical to colour “normal” people (image duplicated on the right) will also appear identical to you, because, in this illustration, they are physically identical.

Puff pastry

I was rather taken by the video above, which I first saw on Core77. I started wondering how many times you have to put the roll of silicone material through the machine to get satisfactory mixing of the two colours of material. The people in the video consider the job done after four passes. What does that mean in terms of the thickness of the red and white layers within the material?

The roll is a rather complicated object, so I worked with an idealised version of the real process, where the sheet emerging from the rollers isn’t rolled up, but cut into several pieces which are stacked up before being passed through the rollers again. I came up with the following:

After only 2 passes, the layers in the slab are too thin to see with the naked eye. And by some margin, too: there are over 600 of them and they’re only a fortieth of a millimetre thick. If you made a perpendicular cut through the slab, it wouldn’t appear to have red and white layers in it.

After only 4 passes, a standard compound microscope operating in visible light wouldn’t be able to resolve the layers in the slab.

After only 6 passes, the layers would be thinner than the width of the molecules of the silicone material. At this stage the concept of red and white layers no longer makes sense.

These results will only apply to material near the centre of the roll. It’s easy to see from the video that material near the edges is not mixed so well.

The calculation

From the video, it looks like there are about 9 turns in the roll. Each time the roll is flattened by the rollers, those 9 turns are converted into 18 layers. The resulting sheet is rolled up and passed through the rollers again, multiplying the number of layers by 18, and so on.

This doesn’t work at the sides of the roll. We’ll ignore that complication, and work with a flat analogue of the actual situation. We’ll assume that we start with two long rectangular flat sheets of material, a white one and a red one, laid on top of each other. We’ll cut this assembly into 18 identical pieces, and make a stack of them; this stack will have 36 layers. We now flatten this stack in the rollers, cut it into 18 pieces, stack them up (giving us 648 layers), and repeat.

On emerging from the roller, the sheet appears, by eye, about 1.5 cm thick. We’ll assume that we start with two layers of half this thickness. The table below shows the number of layers and the thickness of each layer after 0, 1, 2, 3… passes through the rollers.

Number of passesNumber of layersLayer thickness (m)
027.50 × 10-3
1364.17 × 10-4
26482.31 × 10-5
311 6641.29 × 10-6
4209 9527.14 × 10-8
53 779 1363.97 × 10-9
668 024 4482.21 × 10-10

We can identify various milestones, as follows:

Limit of visual acuity. A person with clinically normal vision can resolve detail that subtends roughly 1 minute of arc at the eye. At a viewing distance of 30 cm, this corresponds to about 0.1 mm (10-4 m). The layers of material are much thinner than this after only 2 passes. If you made a perpendicular cut through the slab of material, after two passes you wouldn’t be able to see the layered structure. (This might not be true if the cut was oblique.)

Limit of standard light microscopy. A compound microscope working in visible light can resolve detail down to about 200 nm (2 × 10-7 m). The layers become thinner than this after only 4 passes.

Single-molecule layers. The question here is the number of passes needed before the layers are less than a molecule thick (at which point the idea of layers fails). The difficulty is that molecules of silicones are long chains, and these chains are almost certainly bent, so their size is ill-defined. This part of the calculation will be hugely approximate. We’ll be as pessimistic as possible, assuming that the molecules are roughly straight and that they lie parallel to the layers in the slab of material.

PDMS
Polydimethylsiloxane

A common silicone material is polydimethylsiloxane or PDMS. This consists of a silicon-oxygen backbone with methyl groups attached. The lengths of carbon-silicon and carbon-hydrogen bonds are 1.86 × 10-10 m and 1.09 × 10-10 m respectively. So the width of the molecule is going to be, very, very approximately, of the order of 4 × 10-10 m. The layers are thinner than this after only 6 passes.

 

 

 

 

My longest shortest shadow

Longest shortest shadow small

In the morning, as the sun climbs in the sky, my shadow gets shorter and shorter. In the afternoon, as sun gets lower and lower in the sky, my shadow grows again. At local noon, when the sun is at its highest, my shadow is as short as it’s going to be that day. My noon-time shadow is my shortest shadow.

How high the sun gets at local noon depends on the time of year. In the summer, it gets much higher than in the winter. On the day of the winter solstice, the noon-time sun is lower than the noon-time sun on any other day of the year.  Therefore, of all of my shortest (noon-time) shadows, the one on the winter solstice will be the longest. It’s my longest shortest shadow.

Today is the winter solstice, and I’d hoped to photograph my longest shortest shadow on the beach at Portobello in Edinburgh. But implicit in all of the above is that the sun will be visible in the sky at the critical moment in order to cast the shadow. In Scotland in December this can be a tricky condition to meet, and it wasn’t quite met today. My shadow was barely visible. Since this idea came to me, I’ve only had one chance to photograph my longest shortest shadow, back in 2010.

 

Pedal-powered geology

Graham's bike revised
What on earth am I doing here? Read on to find out. Many thanks to Simon Gage for the idea and to Graham Rose for the wonderful illustration.

In the previous post, we discovered that the kinetic energy of a drifting continent is of the same general magnitude as that of a moving bicycle and its rider – 1500 joules would be a typical figure.

I went on to calculate that, whereas it takes me only about 10 seconds to get my bike up to full speed, it would take me hundreds of years to get the continent up to its tiny full speed were I to put my shoulder against it and push (assuming that it was perfectly free to move). How can this be, when the amount of energy that I’m giving each of these objects is the same?

The problem is that when I push the continent, I am, effectively, in the wrong gear.

On a bike with gears, you’ve got a range of choices about how you power it: you can ride in a high gear, pedalling slowly but pushing hard on the pedals, or ride in a low gear, pedalling more quickly but pushing less hard on the pedals. There’s a simple tradeoff: if you want to pedal half as fast, you’ve got to push twice as hard for the same effect.

But there’s a limit to how hard you can push on the pedals, which means that if you move up too far up through the gears, there comes a point where you can no longer make up for the decreased pedalling rate by pushing harder on the pedals, and the power that you can supply to the bicycle falls.

Anyone who’s tried to accelerate a bicycle when they are in too high a gear will have experienced this problem, and it’s what I experience when I try to push the continent directly. Because the top speed of the continent is extremely low (about the speed of a growing fingernail), I’m necessarily pushing it very slowly as I accelerate it. This means that to give it energy at the rate that I want to (1500 joules in 10 seconds, like the bike) I would have to push it impossibly hard – the force needed is about the same as the weight of a 300-metre cube of solid rock.

Is there a way that we can put me into a lower gear, so that I can push with a force that suits me, over a longer distance, and still apply the very high force over a short distance to the continent?

Yes. Just as we’ve all used a screwdriver as a lever to get the lid of a tin of paint off, so I could use a lever to move the continent. Similarly to the bike gears, the lever allows me to exchange pushing hard over a small distance with pushing less hard over a longer distance. To do the job, the lever would need to be long enough to allow me to push, with all my might, through a distance of about 2.5 metres, with the short arm of the lever pushing the continent. We’d need an imaginary immoveable place for me to stand, and we could use the edge of the neighbouring continent as the pivot (just as we use the rim of a paint tin as the pivot). The catch is the length of the lever: if the short arm was 1 metre long, the long arm would be about 1.5 million kilometres long.

Simon Gage of Edinburgh International Science Festival suggested a more compact arrangement: a bicycle with an extremely low gear ratio, with the front wheel immobilised on the neighbouring Graham's bike extractcontinent (assumed immoveable), and the back wheel resting on the continent we’re trying to accelerate. A transmission giving 17 successive 4:1 speed reductions would do the job nicely. Ten seconds of hard pedalling would get the continent up to full speed. To me on the saddle, it shouldn’t feel any different to accelerating my bike away from the lights.

A wee caveat. This is a thought experiment, and we’ve swept some fairly significant engineering issues under the carpet. The rearmost parts of the power train would be moving at speeds that are literally geological, so in reality it would take me years of pedalling to take all of the slack and stretch out of the system. These parts would also be transmitting mountainous forces, and so they’d need to be supernaturally strong. There will be frictional losses. And then there’s the issue of transmitting a gigantic force to the continent through the contact of a bike tyre on the ground.

The calculations

What force is required to accelerate the Eurasian plate to top speed in 10 seconds?

The top speed of the plate is 3.2 × 10-10 ms-1. If I accelerate it uniformly, its average speed will be half of this, and so in the 10 seconds over which I hope to accelerate it, it will travel 1.6 × 10-9 m.

Now W = fd

where W is the work that I do on the plate (ie the kinetic energy that I give it), f is the force that I apply to it, and d is the distance through which I push the plate. Rearranging gives us

f = W/d

We know W from the previous post (it’s 1500 joules) and we’ve just calculated d. Thus f works out at about 9.4 × 1011 newtons.

For comparison, a 300-metre cube of rock of density 2700 kg m-3 will have a weight of (300 m)3 × 2700 kg m-3 × 9.81 m s-2 = 7 × 1011 newtons roughly.

The lever

When a lever is used to amplify a force, the ratio of the lengths of the arms of the lever needs to be the same as the ratio of the two forces. Suppose that I can push with a force equal to my own body weight, about 600 newtons. If I’m to use a lever to amplify my push of 600 N to a force of 9.4 × 1011 N, the ratio of the lengths of the arms needs to be (9.4 × 1011)/600, or roughly 1.5 × 109. So if the short arm of the lever is 1 metre long, the long arm needs to be about 1.5 × 109 metres long, which is 1.5 million kilometres. For comparison, the Moon is about 400,000 kilometres away.

To do 1500 joules of work with a force of 600 N, I’d need to push over a distance of 2.5 metres (because 600 × 2.5 = 1500).

The bicycle gearing

I estimated that it takes me 15 pedal revolutions to get my bike up to full speed. Knowing the length of the pedal cranks, I know the total distance that I have pushed the pedals through, and I know how much work I have done on the bicycle – 1500 joules. (I’m ignoring energy losses here, because they are small at low speeds on a bike and the calculation is highly approximate anyway). Using work done = force × distance, this gives an average force on the pedals of about 94 newtons.

The 17 stages of 4:1 reduction mean that the back wheel is rotating 417 = 1.7 × 1010 times slower than I’m pedalling. The pedalling force is amplified in the same ratio, to give a force on the teeth of the rearmost gear of 1.6 × 1012 newtons. We now have to allow for the fact that the radius of the rear wheel is about twice the length of the pedal crank. This roughly halves the force available at the rim of the rear wheel, giving a force of about 8 × 1011 newtons, which is close to what we need.

 

The kinetic energy of a drifting tectonic plate…

…is broadly similar to the kinetic energy of me and my bike as I pedal along.

plates
Map of tectonic plates (United States Geological Survey) http://pubs.usgs.gov/publications/text/slabs.html

According the the theory of plate tectonics, the outer layer of the Earth is divided into a number of separate plates, which very slowly drift around, opening and closing oceans, causing earthquakes, and thrusting up mountain ranges.

A moving body has energy by virtue of its motion: kinetic energy. Kinetic energy is proportional to a body’s mass and to the square of its speed.

Now tectonic plates move extremely slowly: the usual comparison is with a growing fingernail. But they are also extremely heavy: tens of millions of square kilometres in area, over 100 km thick, and made of rock. I wondered how the minute speed and colossal mass play out against each other: what’s the kinetic energy of a drifting tectonic plate?

There are so many variables, that vary such a lot, that this calculation is going to be extremely approximate. But the answer is delightfully small: the kinetic energy of the tectonic plate on which I live, as observed from one of the plates next door, is about the same as the kinetic energy of me and my bike when I’m going at a reasonable pace: about 1500 joules.

Me struggling up one of the many steep roads in NW Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably less than that of a drifting tectonic plate.
Me struggling up one of the many steep roads in north-west Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably much less than that of a drifting tectonic plate ;-).

This is a fun calculation to do, but we shouldn’t get carried away thinking about the kinetic energy of tectonic plates. Plates are driven by huge forces, and their motion is resisted by equally large forces. The mechanical work done by and against these forces will dominate a plate’s energy budget in comparison to its kinetic energy.

But the calculation does provoke an interesting thought about forces and motion. I can get my bike up to full speed in, say, 10 seconds. If the Eurasian plate were as free to move as my bike, and I were to put my shoulder against it and shove as hard as I could, it would take me about 500 years to get it up to its (very tiny) full speed.

In both cases, I’m giving the moving object roughly 1500 joules of kinetic energy. How come I can give that energy to my bike in a few seconds, but to give it to the plate would take me centuries?

I’ll return to that thought in a later post.

The calculation

Depending on how you count them, there are 6-7 major tectonic plates, 10 minor plates, and many more microplates. The plates vary hugely in size, from the giant Pacific Plate with an area of 100 million km2, to the dinky New Hebridean plate, which is a hundred times smaller. The microplates are smaller still. Plates also vary a lot in speed: 10-40 mm is typical.

I’m going to be parochial, and choose the Eurasian plate for this calculation.

Let’s call the area of the plate a and its mean thickness t. Its volume is then given by at, and if its mean density is ρ, then its mass m is ρat.

A body of mass m moving at a speed v has kinetic energy ½mv2. So our plate will have kinetic energy ½ρatv2.

The area of the Eurasian plate is 67,800,000 km2 or 6.78 × 1013 m2, and its speed relative to the African plate is (the only speed I have) is given as 7-14 mm per year. We’ll use 10 mm per year, which is 3.2 × 10-10 ms-1. The thickness of tectonic plates in general varies roughly in the range 100-200 km depending upon whether we are talking about oceanic or continental lithosphere; let’s call it 150 km or 1.5× 105 m. The density of lithospheric material varies in the range 2700-2900 kg m-3; we’ll use 2800 kg m-3.

Putting all of these numbers into our formula for kinetic energy, we get a value of 1500 joules (to 2 significant figures, which the precision of the input data certainly doesn’t warrant).

Now for me and my bike. I weigh about 57 kg, my bike is probably about 10 kg. Suppose I’m riding at 15 mph, which is 6.7 ms-1. My kinetic energy is almost exactly…

…1500 joules!

The closeness of these two values is unmitigated luck*, and we shouldn’t be  seduced by the coincidence. Just varying the speed of the plate in the range 7-14 mm would cause a 4-fold change in kinetic energy, and there’s the variability in plate thickness and rock density to take into account as well. The choice of bike speed was arbitrary, I guessed the mass of the bike, and I’ve since realised that I didn’t account for the fact that the wheels of my bike rotate as well as translate.

However, what we can say is that the kinetic energy of a drifting continent is definitely on a human scale, which leads to a new question:

Suppose the Eurasian plate were as free to move as my bicycle, and that I put my shoulder against it and shoved, how long would it take me to get it up to speed?

From the figures above, the mass of the plate is 2.85 × 1022 kg. If I can push with a force equal to my own weight (about 560 newtons) then by Newton’s 2nd Law I can give it an acceleration of about 1.96 × 10-20 ms-2. Rearranging the equation of motion v = at, where v is the final speed, a is the acceleration, and t is the time, then t = v/a. Inserting the values for v and a, we get t = 1.6 × 1010 seconds, or about 500 years.

 

* I didn’t tweak my assumptions: what you see above really is the very first version of the calculation!

A dish designed by nature

Paraboloid 1

When you stir a cup of tea, the surface of the rotating liquid develops a dip in the middle. The faster you stir, the deeper the dip. But the liquid surface is quite uneven; to get a smooth surface, throw the spoon away and spin the whole cup continuously. Once the liquid inside has caught up with the cup, and everything is turning at the same speed, the liquid surface forms a beautiful smooth curve known as a paraboloid of revolution.

Sarah McLeary and I are applying this idea to make thin paraboloidal plaster shells. We spin a bucket on a potter’s wheel (Sarah is a potter), and pour plaster into it. The plaster rapidly flows until its surface forms a deep paraboloidal curve, and then sets. We now use this cast, still spinning, as a mould, and cast a thin layer of plaster inside it, to make a paraboloidal shell.

That’s our first attempt above. It’s about 20 cm across and 3 mm thick. It might look like a part of a sphere, but in a profile view it’s easy to see that the curvature is tightest at the base and gradually decreases up the sides, as you’d expect for a paraboloid.

I love the fact that we didn’t decide what shape this shell was going to be: physics did.

There’s lots of experimentation ahead. When we’ve got the hang of it, I’ll explain our methods in more detail. But as this picture of our third attempt shows, we haven’t quite cracked it yet.

Paraboloid 3 cracked

 

 

 

This is not an illusion

pipe2smallwithcaption-and-copyright

This image is my version of Edward Adelson’s checkershadow illusion (with a little inspiration from Magritte). It’s a photograph of a real, physical scene.

Take a look at the central square of the checkerboard, and the square indicated by the arrow. Which is lighter?  Quite clearly, it’s the central square, isn’t it?

Remarkably, the central square actually emits less light than the square indicated by the arrow!  You could use a light meter to check this claim, but it’s easier to verify it directly by using a piece of card with two holes cut in it to mask off the rest of the image.

Some people will tell you that this image shows you how easy it is to fool your brain. But it does the exact opposite: it shows you what a marvellous piece of equipment your brain is.

Think about the checkerboard itself, and the materials it’s made of. The arrowed square is coated with dark grey paint, and the central square is coated with light grey paint—and that’s exactly what you perceive.

The shadow cast by the pipe means that the light-grey central square is more dimly lit than the dark-grey arrowed square, so much so that it actually reflects less light into your eye than the arrowed square. But your brain cleverly manages to determine the actual lightnesses of the physical surfaces, despite the uneven lighting. Isn’t that a good thing for your brain to do?

If you still don’t believe me, try this thought experiment. Imagine that you live in a forest where there are two kinds of fruit. One is light grey and poisonous, and the other is dark grey and nutritious. Two of these fruits hang next to each other, but in the dappled forest light the (light grey) poisonous fruit is in shadow, and the (dark grey) nutritious fruit is in bright light. Suppose that the depth of the shadow is such that the light-grey poisonous fruit actually reflects slightly less light into your eye than the dark-grey nutritious fruit, just as with the two squares in the picture above. Would you really want your vision to tell you that the poisonous fruit was the dark one and therefore the one to pick? Or would you want it to discount the irrelevant effect of the shadow and tell you which fruit was actually dark and which was actually light (and would kill you)?  I know what I’d want.

I think that it is wrong to call this effect an illusion (and so does Adelson). There is nothing illusory about what you see. You perceive the useful truth about the scene in front of you.