The video above is a night-time time-lapse video taken from Portobello, Edinburgh looking roughly north over the Firth of Forth towards the coast of Fife, a few miles away. I made the video in March 2022.
Look at the vertical relative movement of the distant lights. I believe that this is caused by atmospheric refraction as bodies of warm or cool air rise or fall (I’m not sure which). Light rays are refracted (bent) as they pass between bodies of air at different temperatures, because air’s density depends upon temperature. It’s like the shimmering haze that you see above hot ground on a very hot day, but on a grander scale and unfolding at a more leisurely pace.
The frames were taken at 5-second intervals, which means that the video represents about half an hour of real time.
I don’t believe that we’re looking at camera shake, because that would move the entire image as a piece, rather than causing some parts to move relative to others.
In the previous post I looked at how coloured shadows are formed. As I wrote it, I realised how much there is to learn from the coloured shadows demonstration; that’s what this post is about. The image above shows coloured shadows cast by a white paper disc with a grey surround.
Mixing red, green, and blue lights
In the image at the top, we’re mixing shadows. If we were mixing lights in the normal way, it would look like the picture on the right.
So what do we learn from the coloured shadows image?
Red, green and blue add to make white
The white paper disc has all three lights shining on it, and it appears white. Mixing lights shows us this too.
The three lights still exist independently when they are mixed
Some descriptions of colour light mixing could leave you with the impression that when we mix red, green, and blue lights together to make white, they combine to make a new kind of light, a bit like the way butter, eggs, flour and sugar combine to create something completely different: a cake.
But if that were so, we’d only get a black shadow. The fact that we get three coloured shadows show that the three coloured lights maintain their independence even though they’re passing through the same region of space. It’s very like ripples on a pond: if you throw two pebbles into a pond, the two sets of ripples spread through the same region of water, each one travelling through the water as if the other pebble’s ripples weren’t there.
When we mix coloured lights, additive colour mixing rules apply:
Red and blue make magenta.
Red and green make (surprisingly) yellow.
Blue and green make cyan.
All three colours add to make white.
Mixing magenta, cyan, and yellow shadows
In the coloured shadows image, it looks at first glance as if we are adding together coloured lights. But if we were, we’d expect the centre of the pattern, where all the lights overlap, to be white, as it is in the light-mixing image. Instead, the centre of the coloured shadows pattern is black.
The reason is that we aren’t adding coloured lights, we’re adding coloured shadows, and now subtractive colour mixing rules – the rules of mixing paints – apply.
In the cyan shadow, red has been blocked by the disc, leaving green and blue. In the yellow shadow, blue has been blocked by the disc, leaving red and green. Where the cyan and yellow shadows overlap, the only colour that has not been blocked by one disc or the other is green, so that’s the colour we see. We get the same result when we mix blue and yellow paints: the only colour that both paints reflect well is green. (If the blue and yellow paints reflected only blue and only yellow respectively, the mixture would appear black.)
In the black centre of the pattern, all three lights are blocked by the disc. Something similar happens when you mix every colour in your paint box together.
Colour printing is based on these subtractive colour mixing rules.
Brightness matters, and blue isn’t very bright
In either of the light- or shadow-mixing images, the boundaries between the regions aren’t all equally distinct. The least distinct ones are:
magenta and red
green and cyan
blue and black
yellow and white
In each case, the difference in the colours is the presence or absence of blue light.
There are two things at work here. Firstly, more than we might think, our vision is based on brightness, not on colour. We happily watch black-and-white movies; after a while we hardly notice the absence of colour.
Secondly, our sensation of brightness is largely due to the red-yellow-green end of the spectrum – blue makes a very small contribution, if any. So although the presence or absence of blue light can have a strong effect on colour, it has a weak effect on brightness. So boundaries defined by the presence or absence of blue light tend to be relatively indistinct compared to those defined by the presence or absence of red or green light.
This is a photograph that I took as a response to a challenge that was set by photographer Kim Ayres as part of his weekly podcast Understanding Photography. The challenge was to produce a photo where the main interest was provided by shadows. I lit a rose cutting using red, green, and blue lights that were about 3 metres away and 30 centimetres apart from each other. The result is a gorgeous display of coloured shadows. Coloured shadows are nothing new, but they are always lovely.
Kim suggested that I do a blog post to explain more about how coloured shadows arise. To do this I set up an arrangement for creating simple coloured shadows. One part of the arrangement is three lights: red, green, and blue arranged in a triangle.
The lights shine upon a white screen set up about 3 metres away. In front of the screen, a wire rod supports a small black disc.
First of all, let’s turn on the red light only. The screen appears red, and we can see the shadow of the disc on it. The shadow occupies the parts of the screen that the red light can’t reach because the disc is in the way.
Next, we’ll turn on the green light only. Now the screen appears green, and for the same reasons as before, there’s a shadow on it. The shadow is further to the left than it was with the red light; this is because the green light is to the right of the red light as you face the screen.
Next, we’ll turn on the blue light only, with the expected result. The blue light is lower than the red and green ones, so the shadow appears higher on the screen. (The shadow is less sharp than the previous two. This is because my blue light happens to be larger than the red or green lights).
Now we’re going to turn on both the red and green lights. Perhaps unsurprisingly, we see two shadows. They are in the same places as the shadows we got with the red and green lights on their own. But now they are coloured. The shadow cast by the red light is green. This is because, although the disc blocks red light from this part of the screen, it doesn’t block green light, so the green light fills in the red light’s shadow. Similarly, the shadow cast by the green light is red.
The screen itself appears yellow. This is because, by the rules of mixing coloured lights (which aren’t the same as the rules for mixing coloured paints), red light added to green light gives yellow light.
We can do the same with the other possible pairs of lights: red & blue, and green & blue. (The green shadow looks yellowish here. It does in real life too. I think this is because it’s being seen against the bluish background.)
We’re now going to turn on all three lights. As you might expect, we get three shadows. The colours of the shadows are more complicated now. The shadow cast by the red light is filled in with light from both of the other lights – green and blue – so it has the greeny-blue colour traditionally referred to as cyan. The shadow cast by the green light is filled in with light from the red and blue lights, so it is the colour traditionally called magenta. And the shadow cast by the blue light is filled in with light from the red and green lights, and thus appears yellow.
The rest of the screen, which is illuminated by all three lights, is white, because the laws for mixing coloured lights tell us that red + green + blue = white. The white is uneven because my lights had rather narrow and uneven beams.
Finally, let’s add further richness by using a larger disc, so that the shadows of the three lights overlap. Now we get shadows in seven colours, as follows.
Where the disc blocks one light and allows two lights to illuminate the screen, we see the colours of the three pairwise mixtures of the lights: yellow (red+green), magenta (red+blue), and cyan (green+blue).
Where the disc blocks two lights and allows only one light to illuminate the screen, we see the colours of the three individual lights: red, green, or blue.
And in the middle, there’s a region where the disc blocks the light from all three lights, so here we get a good old-fashioned black shadow.
If it’s a bit hard to wrap your head around this, let’s trying looking at things from the screen’s point of view. Here I’ve replaced the screen with a thin piece of paper so that the shadows are visible from both sides. I’ve made holes in the screen in the middle of each of the coloured regions, so that we can look back through the screen towards the lights.
Here’s what you see when we look back through the magenta shadow. We can see the red light and the blue light, but not the green one – it’s hidden behind the disc.
This is the view looking back through the green shadow. We can see only the green light. The red and blue lights are hidden behind the disc.
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)
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.
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.
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
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.
Exploring the coast at Rerwick Point.
Showery weather meant that we were treated to many magnificent rainbows, like this one seen at Tankerness.
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.
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 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.
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.
…is broadly similar to the kinetic energy of me and my bike as I pedal along.
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 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!
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.
Be careful about how much water you put in your kettle and ‘do your bit’ to save energy!
That’s what we’re often told, and boiling only as much water as you need for your cup of tea can only be a good thing. But how much impact does it really have on your overall energy consumption?
Consider this: if you’re driving along in a car at 50 mph, you’re using enough energy to make a cup of tea every few seconds.
1, 2, 3, tea, 1, 2, 3, tea, 1, 2, 3, tea…
As most people don’t think twice about driving their cars for an extra few minutes, never mind an extra few seconds, it’s clear that the energy savings made by being frugal with your kettle are very small compared to your overall energy use.
Does this mean that we shouldn’t be careful with our kettles after all? No. It all matters. But having boiled exactly one cup’s worth of water for your refreshing cuppa, you shouldn’t put your feet up and think that your energy consumption issues are sorted. You’ve much bigger fish to fry.
The calculation
Imagine that you’re travelling in a car that’s doing 50 miles per hour, with a rate of fuel consumption of 50 miles to the gallon (the units are customary in a UK motoring context). This is a realistic situation; if anything it’s optimistic in terms of energy consumption.
In this scenario, your car will burn a gallon of petrol, that’s 4.5 litres approximately, every hour.
With 3600 seconds in an hour, that’s 4.5L/3600s = 0.00125 litres per second. The density of petrol is about 0.8 kg per litre, so the rate of petrol consumption comes to 0.00125 L s-1 × 0.8 kg L-1 = 0.001 kilograms per second.
The energy density of hydrocarbon fuels is about 46 megajoules per kilogram, so the rate of use of energy by your car is 0.001 kg s-1 × 46×106 J kg-1 = 46000 joules per second (watts).
Now let’s think about the tea. The energy E required to produce a temperature change of ΔT in a mass m of a substance whose specific heat capacity is C is
E = m C ΔT
A typical mug of tea contains about 250 ml, or 0.25 kg, of water. For water, C = 4200 J kg-1 K-1. The temperature change is about 90 K, if the water comes out of the tap at about 10°C and is raised to boiling point. So the amount of energy required to heat the water for a cup of tea is 0.25 kg × 4200 J kg-1 K-1× 90 K = 94500 joules.
As the car is using 46000 joules per second, it follows that the car is using enough energy to heat the water for a cup of tea every two seconds.
Now to be fair to the car, we need to recognise that in real life there is more energy used to heat the water for the tea than actually goes into the water. There are heat losses in energy generation and transmission, and from the kettle itself.
Things get very complicated here. A thermal electricity generating plant (coal, oil, gas, or nuclear) converts energy in the fuel to electrical energy with about 35-45% efficiency. But in the UK, about 20% of our electricity is generated from renewables, where the concept of efficiency is harder to pin down. For example, I could use some oil to heat my water directly instead of doing it indirectly by generating electricity, but I couldn’t do the same with the wind. I’m going to assume that 80% of the electricity that I used was generated in thermal plants at 40% efficiency, and 20% of it was generated at 100% efficiency, giving an overall efficiency of 52%. But be aware that this is a very rough figure, and that there is no right answer.
According to this document, about 93% of the energy in generated electricity makes it unscathed through the transmission and distribution systems to the end user.
I did an experiment and estimated the efficiency of my kettle to be about 88%. I give details later.
This gives an overall efficiency of boiling water as 52% × 93% × 88% = 43%. So instead of using 94500 joules to boil water for a cup of tea, we’re actually using 94500J/0.43 = 220000 joules, which means that the car’s rate of energy use is equivalent to a cup of tea roughly every 5 seconds. Because of all the uncertainties involved (not least the size of a cup of tea), we should treat this figure as very approximate.
Efficiency of my kettle
My kettle took 180 s to raise 1 litre of water from 17°C to boiling.
The kettle is rated at 1850-2200 W for supply voltages in the range 220-240 V. When the kettle was running, I measured the supply voltage to be 242 V, so I assumed that the kettle was operating at the top of its power range, ie 2200 W.
The heat supplied by the kettle element was therefore 2200 W × 180 s = 396000 J.
Using the same equation as earlier, the energy required to raise the temperature of 1 kg of water by 83 K is 1 kg × 4200 J kg-1 K-1 × 83 K = 348600 J.
The kettle was therefore heating the water with efficiency 348600/396000 = 0.88 or 88%.
