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.
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.
…is broadly similar to the kinetic energy of me and my bike as I pedal along.
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.
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.
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…
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.
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.
How fast does a plucked guitar string move? It’s a complete blur, so surely it’s travelling at a terrific speed. 50 miles per hour? 100 miles per hour? What do you think?
Walking pace. A typical speed for the middle of a guitar string given a good twang is walking pace. And that’s the middle of the string. Near its ends, it’ll be moving much more slowly.
How can that be so? Well, although the string is going back and forth hundreds of times a second, it’s only travelling a few millimetres on each trip. So the distance that it travels in each second isn’t as much as you might expect. It certainly isn’t as far as I expected.
If you think that the string moves slowly, what about the body of the guitar? The string itself radiates very little sound into the air; its job is to set the body of the guitar vibrating. The body of the guitar, with its much larger area, is much more effective than the string at setting air into motion. Yet we can’t even see the body vibrating. At what snail’s pace must it be moving?
Remember also that the air molecules on which the guitar body acts are already travelling at something like 500 metres per second. Isn’t it astonishing that the sub-pedestrian movements of the guitar affect the movement of the air molecules enough to produce a sound that we can easily hear?
Suppose that we look near the centre of the string, where its movement is the greatest. Shortly after being plucked, the width of the blur that we see is going to be something like 5 mm. So for every complete oscillation, the string does a round trip of about 10 mm.
The frequencies of the strings on a standard 6-string acoustic guitar are (to the nearest whole number) 82, 110, 147, 196, 247 and 330 hertz (one hertz is one oscillation per second). If we multiply these frequencies by the 10 mm round trip, it tells us how far the centre of each string travels in one second, that is, its average speed. I’ve converted these speeds into metres per second. For comparison, a brisk walk at 4 mph is about 1.8 ms-1.
Mean speed of middle of string in metres per second (to 2 sig. fig.)
Our brisk walk is right in the middle of this range. And remember, we’ve done the calculation for the part of each string that’s moving the most. Near its ends, each string will be moving much more slowly than this.
Complication 1 – how long is the round trip really?
So far, we’ve assumed that each part of the string does a simple back-and-forth movement along a straight line, but if you carefully watch a vibrating guitar string you’ll see that the string often moves in an irregular but roughly elliptical orbit. The wire-wound lower strings show this most clearly; you can see a hint of it in the image at the top. This makes the round-trip distance a bit longer than the 10 mm that we used in the calculation earlier. Does this affect the string’s average speed much?
We’ll take the extreme case where each part of the string moves at constant speed in a circle of diameter 5 mm rather than along a straight line 5 mm long. The circumference of this circle will be mm, or about 15 mm. So the speeds of the strings (in this rather unlikely extreme case) will be about 50% greater than the ones listed above. They are still hardly impressive.
Complication 2 – the peak speed
So far, we’ve calculated the mean (average) speed of the string over its round trip. However, unless it’s moving in a perfect circle, its speed changes constantly, and its peak speed will be higher than its mean speed. How much higher?
Imagine that part of the string is vibrating back and forth along a straight line in the simplest possible way. At one end of the movement the string is stationary as it changes direction. It then speeds up, reaching its peak speed at the centre of its range of movement. Then it slows down until it reaches a halt again at the other end of the movement and changes direction again. How do we calculate the peak speed if we know the time taken for the round trip?
You’ll need to know a bit of maths for the next bit. The simplest vibration of the string is where each part undergoes simple harmonic motion, that is, where its position varies sinusoidally with time. This means in turn that the velocity of the string also varies sinusoidally in time. So we need to ask: how does the mean value of a sinusoid compare to its peak value?
Consider the function , for half a cycle, that is, for theta from to radians. We construct a rectangle between these limits, such the rectangle’s area equals the area under the sine curve. The mean value of the sine function is the height of the rectangle, which its area divided by its width (which is ). So we need to work out the area under the sine curve between and , which we can do by integration:
We constructed the rectangle to have area . Its width is , so its height, and therefore the mean value of the sine function, is . The height to the peak of the sine curve is 1, so the peak value of the sine function is times its mean value. This means that the peak speed of our guitar string is times, or 50% more than, its average speed. Again, nothing to write home about.
(Rather pleasingly, this ratio of is the same ratio that we got earlier in Complication 1. It means that the peak speed of a particle doing simple harmonic motion with a given amplitude and period is exactly the same as the (constant) speed of a particle moving in a circular orbit of radius with period . This isn’t a coincidence. It arises because the circular motion can be considered as two linear simple harmonic motions at right angles to each other.)
It’s sometimes said that if you sit an immortal monkey in front of an equally durable typewriter and leave it to tap randomly away at the keys, then eventually it will produce the entire text of Richard III (or any other Shakespeare play of your choice), completely by chance. All you have to do is wait long enough.
I was thinking about this one day, and also thinking about air molecules. In the room I’m sitting in at the moment, there are at least 1,000,000,000,000,000,000,000,000,000 air molecules, all frantically dashing around bumping into each other. How often, I wondered, do little clusters of these molecules fleetingly arrange themselves, by chance, in arrangements that we would regard as being somehow regular or remarkable?
It’s not possible for me to directly observe air molecules, so instead I used my computer to make a 2-dimensional simulation of some atoms of gas doing what atoms of gas do. I set it going, and waited…and waited…and waited…
I promise you that my simulation didn’t involve any secret forces drawing atoms towards certain positions. The movements and collisions of the atoms all occurred in accordance with the laws of mechanics. But I did cheat a little. Can you figure out how?
Molecules and atoms
In case you’re fidgeting on your chair wondering why I started talking about air molecules but finished talking about gas atoms, let me explain.
Nearly all of the air is nitrogen and oxygen. Nitrogen and oxygen atoms are essentially spherical. But in the air, nitrogen atoms are bonded together in pairs to form nitrogen molecules that are, roughly speaking, a stubby rod shape. The same goes for oxygen. Now a collision between two moving rods is much more complicated than a collision between two spheres, because the rods can spin end-over-end in a way that spheres can’t. In fact, I’m not sure that I know how to do the calculations. As the point of the video could be made just as well using atoms rather than molecules, I did the simulation using atoms. If you like, you can think of them as atoms of helium or argon, which do go around on their own.
This post is all about answering the question: “How deep is the atmosphere?”. The question doesn’t actually have a simple answer, because there is no altitude where the atmosphere suddenly stops and space starts. Instead, the air gets progressively thinner and thinner as you get higher, gradually giving way to the vacuum of space. What can be startling is how quickly the air gets thinner as you travel upwards. On this page I’ll be finding some ways of getting to grips with the scale of the atmosphere.
At the end, I hope that you might agree with me that the atmosphere is really awfully shallow, and that we definitely ought to be looking after it much more carefully than we do at the moment.
What do we mean by “the air gets thinner”?
A cubic metre of air at sea level contains about 1.2 kilograms of air. Higher in the atmosphere, a cubic metre would contain less air. For example, at 9,000 metres (just above the summit of Mount Everest) a cubic metre of air would contain only about 0.47 kilograms of air, less than half as much as at sea level. It is this decrease of mass per unit volume (density) that I mean when I write of the air getting “thinner”.
How high is the top of the atmosphere?
To get a feel for the thickness of the atmosphere, we will look at a number of different definitions of the top of the atmosphere. As well as these, we’ll also look at some altitudes with life-and-death significance. I’ve put any calculations at the bottom of the post.
100 km – the Kármán line
The Kármán line is a common definition of the boundary of space. This beautiful and clever idea rests on two points.
The first point: to be in orbit around the Earth (assuming there’s no atmosphere) there’s a certain speed at which you need to travel. If you travel more slowly than this, you’ll fall out of orbit. At the altitudes we’re talking about, the critical speed is nearly 8 kilometres per second.
The second point: an aeroplane stays up in the air because its wings generate lift as they pass through the air. The thinner the air, the faster the aeroplane must fly in order to generate enough lift to stay airborne. So the higher an aeroplane goes, the faster it must fly to stay up there.
Here’s the clever bit: there comes an altitude where the air is so thin that the aeroplane must travel at about 8 kilometres per second for its wings to generate enough lift to stay up. But at this speed, it’s going fast enough to stay in orbit anyway, even if there were no air and it had no wings.
This altitude, which is about 100 km, is the Kármán line. You could say that it’s where an aeronaut becomes an astronaut.
The picture on the right is a scale diagram showing the Earth (grey) and the atmosphere (blue, thickness defined by the Kármán line). 99.99997% of the atmosphere lies in the blue region.
31 km – the 99% line 99% of the atmosphere is below 31 km above the surface of the Earth. The picture on the right shows roughly what part of central London would look like seen from 31 km. It doesn’t look too far away, but from this altitude, you are looking through nearly the entire atmosphere. (Please note that this picture represents roughly how big things would look from 31 km. It doesn’t reflect the optical degradation that viewing through 31 km of air would produce. I scaled the picture from one taken at an unknown altitude.)
19 km – the Armstrong line At this altitude the air pressure is lower than the vapour pressure of the water in your body. Uncontained body fluids (such as saliva) would start boiling at this altitude.
8 km – top of the constant-density atmosphere
Another way to think about the depth of the atmosphere is to ask “How much air are we looking through when we look upwards through the atmosphere?”. To answer this question, imagine that the air in the Earth’s atmosphere is all at sea-level density, instead of getting thinner and thinner with altitude. With the same total amount of air as in the traditional atmosphere, this imaginary atmosphere will come to an abrupt end at a certain altitude. How deep would this constant-density atmosphere be?
The answer is: a little more than 8 kilometres. Mount Everest would just poke out of the top of it. There are details of the calculation later on.
Looked at in this way, the atmosphere is startlingly shallow. You can commonly look horizontally from one place to another place 9 or more kilometres away. When you are doing this, there is more air between you and the not-very-distant object than there is between you and a star overhead.
5.5 km – the habitation line
It appears that no amount of acclimatisation will enable you to survive indefinitely above an altitude of around 6 km. Lambert (1971) reports that in 1961 a team that spent six months at 5,800 m was less fit at the end of this time than newly-arrived people. He also cites an Andean mine at 5,800 m, where the miners chose to walk up daily from 5,300m rather than live at the higher altitude. The current highest permanent human settlement is La Rinconada, at 5,100 m in the Peruvian Andes.
The photograph on the right shows roughly what Buckingham Palace would look like from 6,000 m altitude. It doesn’t look very far away, but at this altitude you wouldn’t last very long. About 50% of the atmosphere is below the habitation limit.
The habitation line on the map The photograph on the right shows an ordinary 1:50,000 Ordnance Survey map, familiar to UK hillwalkers. The grid squares visible in the sea are 1 km across. It shows the town of Aberdeen on the Scottish coast. On this scale, the tip of my thumb is at about 6,500 m, comfortably in the region where the air is too thin to support human life for very long. If you were on the north side of Aberdeen, you’d be closer to the uninhabitable zone than you would be to the south side of Aberdeen. Seen this way, the atmosphere seems very shallow. 55% of the atmosphere is below the level of the tip of my thumb.
About 4 km – the oxygen-mask line
Above this altitude, in an unpressurised cabin, an aeroplane pilot is required to use an oxygen supply. Having said that, thousands of people (myself included) climb 4,000-metre peaks in the Alps and nobody uses oxygen.
12 metres – top of the liquid atmosphere
Finally, suppose we condensed the entire atmosphere to its liquid form. How deep would the resulting “ocean” be? The answer: just under 12 metres.
The alarming thing about the altitudes I’ve listed above is how small they are. Compared to the distances that we regularly travel horizontally across the Earth, these distances are tiny. If you could walk vertically upwards, you’d need an oxygen supply after only an hour. Three hours walking and you’d need a pressure suit. Space itself is only a good day’s bike ride away. If you’re in Birmingham, you’re comfortably closer to space than you are to London.
The tall picture
The tall thin picture on the left at the top of the post is a representation of the way the atmosphere gets thinner with altitude. The density of blue dots is proportional to air density at each altitude, starting with solid blue at sea level. The tick marks on the right-hand side are at 10 km intervals. Note that there is air above 60 km, even though there are no dots. It’s just that the air there is extremely thin, and the dots are so widely spaced that you’d need a much wider picture to have a chance of spotting one.
The picture lets you see at a glance roughly how much of the atmosphere is below any given altitude. The aeroplane silhouette is at a typical cruising altitude for airliners – see how much of the atmosphere is below you when you fly.
You could walk across the bottom of the picture in less than two hours.
The wide picture
The picture below is drawn on the same principle, but to a different scale. It shows Edinburgh (E), London (L), and the atmosphere. The broken line is the boundary of space as defined by the Kármán line.
You are in the vastness of space. In all directions in front of you, almost empty star-studded space stretches out for unimaginable distances, giving an near- overwhelming feeling of exposure. Behind you is an apparently limitless hard surface. A strange force presses your back firmly against this surface, almost as if it were magnetic. A transparent layer of air, only a few miles thick, between you and the void gives you something to breathe and protects you from the cold of space.
But as on most clear nights, it’s a bit chilly for lying on your back on the ground, so after a while you stand up and walk home to warm your toes in front of the fire.
The 99% line
The pressure at any level in the atmosphere must be exactly that required to support the overlying layers of the atmosphere. The pressure at sea level is enough to support the entire weight of the atmosphere. If the pressure at a given altitude is, say, half sea-level pressure, then we know that half of the mass of the atmosphere must be above this level. Therefore with a table of atmospheric pressures we can quickly work out what fraction of the atmosphere is above any given altitude.
The Armstrong line…
is not named after Neil.
The constant-density atmosphere
Imagine a column of air, of cross-sectional area 1 square metre, that extends through the full height of the atmosphere. The air pressure at the bottom of this column, at sea level, is very close to 105 newtons per metre squared. This means that all of the air in the column of the atmosphere has a weight W close to 105 newtons. The acceleration due to gravity is, for our purposes, constant throughout the height of the column – let’s use g = 9.8 ms-2. The mass of air in the column is given by W divided by g, which comes to 10200 kg. The density of air at sea level is about 1.22 kg m-3, and therefore the volume of air in the column is 10200/1.22 = 8360 cubic metres. As the column has a cross-sectional area of 1 square metre, this means its height is 8360 metres.
The liquid atmosphere
When working out the depth of the constant-density atmosphere, we established that the mass of a column of air of cross-sectional area 1 m2, extending the entire height of the atmosphere, is about 10200 kg. The density of liquid air is about 870 kg m-3, and so if we liquefied this amount of air it would have a volume of 11.7 m3. Hence if we liquefied the atmosphere, the resulting ocean would be 11.7 m deep.
This is a minor reworking of a page from my old website. I’m republishing it after receiving an email from Sarah Bush from the Division of Biological Sciences at the University of Missouri, who used the old page as teaching material.
Lambert, D. (1971) Medical appendix in Bonington, C. (1971) Annapurna South Face. Cassell.
West, J.B. (2002) Highest Permanent Human Habitation. High Altitude Medicine & Biology,3, 401-407.
It depends what you mean by see. Single air molecules scatter light (that’s why the sky glows) so with a dark background and an absurdly intense light source you would presumably be able to visually detect a single atom suspended in a vacuum.
But that doesn’t really feel like seeing to me. The question I’m going to answer is: what is the smallest number of atoms that I can quickly assemble using the stuff in my flat, that I can see with my unaided eye by ordinary reflection in typical room lighting?
I’m sure I could look this up somewhere but there’s no fun in that.
My assemblage of atoms was a tiny pencil dot made on white printer paper. There it is, on the right. The dot was definitely visible but so small that I needed to draw marks nearby so that I didn’t lose it.
I estimate that the number of atoms in that minute mark was about 1013, with an uncertainty of at least a factor of 10 in both directions.
In other words, 10 million million, very roughly.
That’s a lot. We talk about atoms very casually, drawing diagrams of chemical structures and so on, and it’s easy to forget how exceedingly tiny they are. It’s useful to do experiments like this one now and again to remind ourselves that atoms really are small beyond our comprehension.
I made the dot by rubbing the end of a propelling pencil to a point and then touching the point lightly against a sheet of white paper.
To estimate the thickness of the layer of pencil lead, I held the pencil perpendicular to some paper and scribbled until the lead had a flat end to it. I then adjusted it so that, as far as I could tell, 1 mm of lead protruded from the pencil. Then, using normal pencil pressure, I drew lines 10cm long until the exposed millimetre of lead had all worn away. I took care to hold the pencil perpendicular to the paper so that the lines I drew were the full width of the lead. I could draw 520 such lines with the millimetre of lead, a total of 52 metres of line.
Calculation 1: volume and mass of the dot
I used the thickness (don’t confuse this with the width) of the lines described in the previous paragraph as a proxy for the thickness of the dot (and in doing so introduced probably the biggest uncertainty in the whole procedure). Propelling pencils leads appear to come in sizes of 0.5 mm, 0.7 mm and 0.9 mm and larger. Holding mine against a ruler showed that it was clearly a 0.7 mm lead. The volume of the initial protruding cylinder of lead was therefore
π × (0.35 mm)² = 0.385 mm³ or 3.85 × 10-10 m³
If the line lines I drew were uniformly 0.7 mm wide (and that’s quite a big if – tilting the pencil will make them narrower) then I can equate the volume of the lines and the volume of the lead cylinder thus:
3.85 × 10-10 m³ = (52 m) × (0.7 × 10-3 m) × t
where t is the average thickness of the layer of lead on the paper in metres. This gives us
t = 1.06 × 10-8 m
It certainly doesn’t deserve 3 significant figures but I’m going to leave more reasonable rounding to the end.
To measure the area of my dot, I took a photograph of it next to the finest scale on my ruler, which is 100ths of an inch (see earlier). Things aren’t made any easier by the non-roundness of the dot, but if I were to say that the dot was 1/300 of an inch in each direction, I don’t think I’d be too far wide of the mark. That makes its area
(1/300 in)² × (25.4 × 10-3 m in-1)² = 7.17 × 10-9 m²
(the 25.4 × 10-3 being the conversion from inches into metres). Using our estimate for the thickness of the pencil layer above, this makes the volume of the dot 7.60 × 10-17 m3.
Next, we need to know the density of the pencil lead. If it was a clay brick, its density would be 2400 kg m-3, and if it was pure graphite its density would be in the range 2090-2230 kg m-3, so it’s a reasonable guess that the density of the graphite/clay mix is about 2300 kg m-3.
So using the volume calculated earlier, the mass of my pencil dot is about
(2300 kg m-3) × (7.60 × 10-17 m3)= 1.75 × 10-13 kg
Calculation 2: how many atoms in a kilogram of pencil lead?
From the Cumberland Pencil Company, cited here, I infer that an HB pencil lead is roughly 50% clay and 50% graphite. They don’t say whether that’s before or after firing (the clay will lose water on firing). I’m going to assume that it’s after firing, but given all the other uncertainties in this calculation, I don’t think it’ll matter much if I’m wrong.
Clay is variable in composition, but a typical constituent of fired clay appears to be various minerals or combinations of minerals of overall composition Al2Si2O7. The relative “molecular” mass of such a compound/mixture will be 220. The relative atomic mass of carbon (in the graphite) is 12, so a 50:50 (by mass) mix of clay and graphite will need about 18 atoms of carbon for every unit of Al2Si2O7, giving a total of 29 atoms per unit of this mixture/compound, and a relative “molecular” mass of 440.
440 g of pencil lead is therefore one mole of pencil lead, and with 29 atoms per elementary entity of this mole, it will contain about 29 × 6.02 × 1023 = 1.75 × 1025 atoms; this is about 3.97 × 1025 atoms per kilogram. (6.02 × 1023 is Avogadro’s number: the number of elementary entities in a mole of a substance)
Calculation 3: number of atoms in the dot
From calculation 1, we know that the dot weighs 1.75 × 10-13 kg, and therefore with 3.97 × 1025 atoms per kilogram, the number of atoms in the dot is
Rounded more reasonably, this is 1013 atoms in the pencil dot.
With the uncertainties in the size of the dot and the composition of the lead, I wouldn’t want to quote the answer any more precisely than this.
We’ve done quite a few steps here. Can we check that this answer looks about right?
Suppose the dot were pure graphite. Its mass would be (2150 kg m-3) × (7.17 × 10-17 m³) kg, which is 1.54 × 10-11 moles and hence about 9.2 × 1012 atoms in the dot. As carbon atoms are smaller than aluminium or silicon atoms, it’s not surprising that this number is a little bit bigger than the unrounded number of atoms calculated in the dot.
Now suppose that it was pure aluminium, with density 2700 kg m-3. Its mass would be (2700 kg m-3)× (7.2 × 10-17 m-3) = 1.94 × 10-13 kg which is 7.46 × 10-12 moles and hence 4.5 × 1012 atoms in the dot. As aluminium atoms are larger than carbon atoms, it’s not surprising that this number is a little bit smaller than the unrounded number of atoms calculated in the dot.
So our clay mineral calculations look at least plausible. The bit I’m really worried about is the thickness of the dot. Making a mark with a pointed lead and drawing a line with a flat end of the lead are likely to involve different pressures and hence different mark thicknesses. My feeling is that I’m most likely to have underestimated the thickness of the dot, and hence the number of atoms in it.