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.
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.
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
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.
A note to deuteranomalous readers
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.
Sarah Kenchington and I made this machine for the Full of Noises festival in Barrow-in-Furness in August 2018.
Sarah designed and made the bicycly bits that raise the table-tennis balls from the pit into the hoppers at the top, and I made the two devices that the balls descend through on their way to the cow bells and glockenspiel.
We shot the video in this post in a hurry on a dark damp Tuesday morning before packing the machine up to take it to Barrow, so it comes with apologies for the poor lighting in places.
The peg board (Galton board) that appears from 1:13 to 1:31 is an established classic (see below if you want to make one). The swinging-ramp ball-feeding device (2:09 to 2:18) is a revival of something I designed for the Chain Reactor.
What’s new from me is the arrangement for feeding the balls from the wire chute into the swinging-ramp assembly (1:56 to 2:18). Its operation should be clear from the video, except perhaps for one detail. Because this device may jam if it tries to collect a ball that has not quite arrived at the bottom of the wire chute, and because the timing of the arrival of the balls is erratic, it’s necessary to maintain a queue of balls in the chute to guarantee that there’s always a ball in place at the bottom to be collected. To achieve this, we arranged that the average rate of ball delivery into the chute (determined by the number of spoons on the bicycle chain) was greater than the rate of collection of balls out of the chute, and had an overflow route for the excess balls. Once three balls have accumulated in the chute, any further balls are diverted back into the ball pit (2:30-2:40).
Chris Wallace and I discovered while making the Chain Reactor that the horizontal spacing of the pegs on a Galton Board is important. If the spacing is too great, a ball that sets off rightwards will tend to keep going rightwards, and vice versa. To get good randomisation, the ball should rattle between each pair of pegs, and to get this to happen, the gap between the pegs should be only slightly greater than the diameter of the balls. This in turn means that the pegs need to be precisely placed to avoid there being pairs of pegs that don’t let the balls through at all.
In that project we achieved the necessary precision by making the position of each peg (a bolt) adjustable, but with something like 100 bolts, this difficult job was very tedious and sorely tried Chris’s patience.
This time round, I developed a system that let me get every hole in the right place first time. Firstly, I cut the board into four strips so that all parts of it were accessible to a pillar drill. This guaranteed that every hole was accurately perpendicular. Secondly, I made a drilling jig (top right) to get the hole spacing correct. After drilling each hole, I put the peg (the bolt on the right-hand part of the jig) into the just-drilled hole, and the drill for the next hole into the drill hole on the left-hand part of the jig. The spacing between the peg and drill hole is adjustable using the long bolt. Thirdly, I made a large custom table for the pillar drill (bottom right), with a fence arrangement so that each row of holes was straight.
When I was doing the drilling, the only measurements I had to make were to get the first hole in each row in the right place with respect to the previous row. It took me a few hours to perfect the drilling arrangements, but then only an hour or so to drill 90 holes, all exactly where I wanted them.
No, this post isn’t about wealth inequality. It’s about daylight inequality.
Today is the day of the spring equinox. For the past three months, the days have gradually been getting longer, and from tomorrow, the sun finally starts to spend more time above the horizon than below it*.
It’s also the day when everyone in the world enjoys a day of approximately equal duration. But from tomorrow onwards until the autumnal equinox in September, the further north you are, the longer your days will be, in the sense that the sun will spend more time above the horizon.
I wondered where Edinburgh, where I live, fits into this scale of day lengths. For the next six months, we’ll have longer days than anyone living south of us. What fraction of the world’s population is that?
I estimate that, over the coming summer, we’ll have longer days than roughly 99.3% of the world’s population.
I was very surprised at how large this number is. And delighted too: it somehow seems to make up for the seemingly endless dark dreich dampness of the Scottish winter.
Rather than counting the number of people who live south of Edinburgh, it’s easier to count the much smaller number who live further north.
There are a few countries that are wholly north of Edinburgh, namely Finland, Norway, Latvia, Estonia, Iceland, the Faroes, and Greenland. There are some countries that are partially north of Edinburgh: Sweden, Denmark, and Lithuania. And then there is Russia, which spans a vast range of latitudes, but which has relatively few cities north of Edinburgh, of which the largest and/or most well known are St Petersburg, Nishny Novgorod, Perm, Yekaterinburg, Tomsk, Archangelsk, and Murmansk (though I counted a few more). There’s one state of the USA: Alaska. Finally we have Aberdeen, Inverness, Dundee and Perth, the main centres of population further north in Scotland.
To roughly compensate for the fact that these counts don’t cover minor centres of population, and that the northern part of Moscow probably overlaps somewhat with Edinburgh, I included the whole of Sweden, Denmark and Lithuania in the sum, despite the fact that the countries have major towns that are south of Edinburgh.
When I added it all up, it came to just under 44 million people living north of Edinburgh, against a world population at the time (I actually did the sums a few years ago) of 6.8 billion. Expressed as a percentage, 99.35% of the world’s population live south of Edinburgh, which I’ll round to 99.3% to avoid overstating my case.
*Actually, the sun appears to be above the horizon even at times when, geometrically speaking, it is slightly below it. This is because the light rays are refracted by the atmosphere and so travel in slight curves rather than straight lines.
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.
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 passes
Number of layers
Layer thickness (m)
7.50 × 10-3
4.17 × 10-4
2.31 × 10-5
1.29 × 10-6
7.14 × 10-8
3 779 136
3.97 × 10-9
68 024 448
2.21 × 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.
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.
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.
The hollow face illusion is a wonderful visual effect in which a hollow mask of a face appears to be convex, like the face itself. Making a hollow mould of your face (for example using plaster) is difficult and potentially dangerous. However, last weekend my attention was drawn to an easier and safer way.
I was walking down from Coire an Lochain in the Scottish Highlands with a group from the Red Rope club, when I saw my friend Maia standing on the path ahead, chuckling. She’d been making face imprints in a steep snowdrift, and they showed the hollow face illusion beautifully.
The procedure needs no explanation (see right). The snow needs to be fresh and soft; you’d be surprised how hard it is to push your face into what feels to your hand like very soft snow. The tip of my nose is noticeably flattened in the picture above.
You’re walking along a woodland path. Suddenly you hear a recipe for yogurt being recited somewhere in the bushes. A short while later, you hear an apology for kicking you in a tender place emerging from the trees. What is going on?
Cut Adrift is an installation by Edinburgh artist Mark Haddon. Over the past 20 years or so, Mark has collected handwritten notes that he has found lying on the ground. The notes are a varied mix: letters, recipes, instructions, apologies… Mark recorded people reading these notes out loud. As part of Sanctuary, a 24-hour art event in southern Scotland, he and I arranged that the passing of people along a path would trigger the playing of these recordings from a sound system hidden in the undergrowth.
I helped Mark with the technical side of the project. We used a Teensy microcontroller board with an audio adaptor board audio shield to store, sequence, and play the audio clips. The signal went to a 12 V audio amplifier and thence to a pair of loudspeakers. We used a PIR (passive infra-red) motion detector to detect people walking by. The output from the detector was connected to one of the input pins of the Teensy. The whole thing was powered using a 12 V lead-acid battery. The battery, Teensy, and amplifier all went in a plastic storage crate to protect them from the weather.
The passage of a person triggered one audio clip. The choice of which clip to play was random, but subject to a rule that made a clip more and more likely to be chosen the longer it was since it was last played. My program held the clips in a queue. When a clip was needed, there was a 50% probability that the first clip in the queue would be chosen, a 25% probability that the second clip would be chosen, and so on as far as the 5th clip. Clips lower in the queue than this would never be chosen. Once the chosen clip had been played, it was put at the bottom of the queue and all the other clips moved up one place.
I’m grateful to Jen Sykes of Glasgow School of Art for pointing me in the direction of the Teensy and its audio board.
Edwin Pickstone, a colleague at Glasgow School of Art, appointed me as a maths consultant recently. His project was to produce a book which had a black square on each page. The size of the square and the number of pages had to be such that the envelope of the resulting block of ink was a cube containing 1 kilogram of ink.
His printer provided a sample run so that we knew what the areal density of the ink film would be, and I did the sums to work out how big the squares would need to be and how many pages we’d need. It turned out that we needed squares of side 19.28 cm, and (coincidentally) 1928 pages.
The responsibility weighed heavily on me, and I was a little nervous as the order went off to the printer. So I was relieved and delighted when Edwin told me that when he’d walked into the printers to pick the job up, the printer said ‘bloody hell, it took a whole kilo tin of ink to print your job’.
Edwin is Lecturer, Typography Technician and Designer in Residence at Glasgow School of Art.
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 continent (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.
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.
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.