I’ve just spent the last day and a half making another irregular polyhedron to use as a lampshade in the room where I work. As I worked, I was reminded of how glad I am that I bought a table with motorised height adjustment. The thing I’m making starts off flat and finishes 40cm in diameter, so a table that would be the right height at the beginning would be far too low when I’m nearly finished.

But it’s not just about workpieces that grow. Different assembly operations are best done at different heights: cutting and folding need a lower table than glueing, for example, and if I need to take a close look at something, it’s useful to be able to raise it as close as possible to eye level. (I do nearly all making tasks standing up, by the way.)

There are many different sit-stand desks out there. Mine is an Ikea BEKANT electric sit-stand desk. The height adjusts electrically from 65cm to 120 cm, and takes about 20 seconds to cover the full range. It seems to be solidly built, though I wouldn’t try to do woodwork on it.

It seemed a bit of an extravagance when I bought it, but I love it. By enabling me to keep a better posture, it’s much more comfortable to work at than a fixed-height desk, and the ability to move the workpiece to the best height for any given operation materially improves the quality of the things I make.

How could it be improved? Foot switches (or even better, speech control!) would be handy for those times when you need to change the height of the desk quickly, but both hands are occupied holding something together. And being able to tilt the tabletop would be wonderful – a project for the future, maybe…

The lilac chaser is a remarkable visual phenomenon that is normally seen as a computer animation. Dr Rob Jenkins of York University wanted to show people that the effect works with real, honest-to-goodness, physical lights, so he asked me to make him the equipment to do this. The video below shows you how the apparatus works. Note that the limitations of my camera mean that the effect is not as strong in the video as it is in real life.

One useful technique that I developed here was a way of producing an an even spot of light from an LED. Diffused LEDs give an even spread of light but send light in all directions, which is wasteful if you want only a small bright spot. Clear LEDs are available which direct the light in quite a narrow beam, but the distribution of light is very uneven. I found that shining the light from a clear LED down a short white tube, about 10 mm internal diameter and 60 mm long, did a very good job of producing a sharp-edged even spot on a piece of tracing paper placed at the end of the tube. I assume that the many reflections inside the tube thoroughly mix up the light. I found the tubing in the plumbing section of a hardware shop, and lightly roughened the inside of it using fine sandpaper.

To get a spot with a blurred edge, I placed a second tracing-paper screen a short distance away from the end of the tube. By varying the distance of this screen I was able to vary how blurred the patch of light on it was.

Last week I was on holiday in Wales. It wasn’t the driest of weeks, and while I was inside not climbing mountains, I finally got round to doing some mountain-related geometry that I’ve been putting off for the last 30 years or so. It’s about knowing how high you are.

If you’re climbing or descending a mountain, you sometimes want to know roughly how far (vertically) there is still to go. One way to get an idea of your altitude is to use nearby peaks or other points of known altitude as reference points. But how do you judge whether you are above or below another point? It’s not always obvious, and without some sort of rule, the worry is that you’ll make optimistic judgements, leading to disappointment in the long run.

My friend Malcolm once told me that, as a rule of thumb, you should look at the reference point relative to the distant skyline. If the point appears to be above the skyline, you are lower than it, and if it appears to be below the skyline, you are higher than it. I’ve used this skyline-rule ever since, but I’ve never checked how accurate it is.

The fact that there’s any doubt about the rule is because the Earth is not flat. If it was flat, then your line of sight to the (infinitely distant) sea-level horizon would be exactly horizontal, and the rule would work perfectly. If the skyline was made up of mountains, the rule would work perfectly as long as they were as high as your reference point.

But the Earth isn’t flat: it’s a big ball. How does this affect the accuracy of the rule? I used a wet Welsh Wednesday afternoon to find out.

It turns out that the rule is good enough for general hillwalking purposes as long as the reference point is no more than two or three kilometres away (as it usually will be). The errors are smaller if the skyline is distant mountains rather than the horizon at sea level. The rule consistently underestimates your altitude, which, in ascent at least, is probably better than the alternative.

Some more detail

It’s fairly easy to see that the skyline rule will underestimate your altitude if the skyline is the horizon at sea level. Imagine you’re standing on a peak at . For you, the horizon is at point . Any reference point that is in a direct line between you and must clearly be lower than you are, so using its altitude as an estimate of your altitude is going to underestimate your altitude.

This result is good in ascent: you aren’t going to give yourself false hope. Conversely, it could let you down in descent. But how big is the error? Is it worth allowing for?

The answer is: maybe. The following table shows you how much higher you are, in metres, than the reference point, if the reference point is aligned with with the sea-level horizon, for different heights and distances of reference point. For example, if the reference point is at 1000 m altitude and is 2 km distant (a fairly typical UK situation), then when it’s aligned with the horizon, you’re actually 36 m higher than it. This isn’t huge in terms of climbing effort, but it’s not completely trivial either. For more distant reference points, the errors are larger; the rule would be seriously misleading for a reference point 10 km away.

Distance of reference point

1 km

2 km

5 km

10 km

Height of ref. point

250 m

9

18

46

96

500 m

13

25

64

133

750 m

15

31

78

161

1000 m

18

36

90

185

Table showing the overestimation of your altitude (in metres) if you use the skyline-rule when the skyline is the sea-level horizon.

A possibly more realistic situation is that the distant skyline is filled with mountains. How does this change the results? We have too much choice now, because the height and distance of the skyline now become variables in our equations. I chose to calculate the case where the skyline is (a) of the same altitude as the reference point, and (b) as far away as the sea-level horizon would be if you could see it.

Distance of reference point

1 km

2 km

5 km

10 km

Height of ref. point

250 m

4

9

23

48

500 m

6

13

32

67

750 m

8

15

39

81

1000 m

9

18

45

92

Table showing the overestimation of your altitude (in metres) if you use the skyline-rule when the skyline is distant mountains.

Not surprisingly (because the skyline is now higher) the underestimation of your altitude is quite a lot smaller, about half of what it was before. In fact if the reference point is less than 2 km away, the error really isn’t worth worrying about.

Even more detail

Case 1: The skyline is the horizon

In the diagram, the Earth (sea level) is represented by the grey arc of a circle. Its centre is at and its radius is . You are standing at point , at an altitude above sea level. The line is your line of sight to the horizon. Your reference point is , at an altitude above sea level. The distance between you and the horizon is and that between you and the reference point is .

Applying Pythagoras’ Theorem in triangle , we get

which we can rearrange to

For any realistic altitude, will be a thousand or more times larger than , so , and we can safely neglect the term to give

(1)

Now apply Pythagoras in triangle , giving

which, if (as it will be) becomes

Now substituting for from equation (1) and rearranging, we get

This tells you how high you are () when another point of known altitude and distance is lined up with the horizon.

Case 2: The skyline is distant peaks

Out of all of the various possibilities, I’m going to choose the one where the skyline is as high as the reference point, and as far away as the sea-level horizon would be if you could see it.

The notation is largely as in the previous case. Point is the sea-level horizon as it would be seen from where you are at . Point is the distant skyline; for convenience, the line is the y-axis. The angle is denoted by .

We know that must lie on the dashed line (because is the horizon seen from ), and we know that it must lie on the straight line passing through and (because, seen from , appears aligned with ). We also specify how far it is from the reference point . We know also that the reference point and the distance skyline point have equal altitudes above sea level. At this point my geometry failed me, so I resorted to coordinate geometry and the goal-seeking feature of Libre Office Calc. We will treat as a parameter that we can vary.

The origin of my coordinate system is at .

The coordinates of are and those of are . The gradient of the line joining them is therefore

and as the line passes through its equation is

At point , , so

and so the coordinates of point are

The final requirement is that be a specified distance from the reference point (1000 m, 2000 m, 5000 m… as in the tables above). We do this by using the goal seek feature of the spreadsheet to adjust until the distance (calculated using Pythagoras’ Theorem) is what we want it to be. We can then determine the altitude of point from the coordinates of point .

A complication: atmospheric refraction

I’ve done the calculations on the assumption that light travels in straight lines. But the density (and hence refractive index) of the atmosphere changes with height, which causes light to travel in a slight downward curve, making distant objects appear slightly higher than they really are. The effect varies depending upon atmospheric conditions, but an approximate calculation is that the apparent altitude (in degrees) of a mountain will exceed its true altitude by an amount equal to its distance in kilometres divided by 1500. If you’re at an altitude of 1000 m, the sea-level horizon (and hence the skylines in my calculations) will be 113 km away, and thus the skyline would appear 0.075 degrees higher than it really should. When sighting a reference point 2 km away against this skyline, this would result in a height error of only 2.6 m, which isn’t worth worrying about, especially as it reduces the underestimation of your altitude.

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?

The calculation

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}.

Note name

Mean speed of middle of string in metres per second (to 2 sig. fig.)

E

0.82

A

1.1

D

1.5

G

2.0

B

2.5

E

3.3

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.

In a previous post I showed some examples of irregular polyhedra that I’ve been making out of paper. These polyhedra were all based on points distributed quasi-randomly over a sphere. At each point, my program placed a plane tangent to the sphere. The result of the intersection of all of those planes was an irregular polyhedron.

The 47-hedron in the picture above was made in a rather different way. It’s certainly irregular, but the process that created it was not at all random. As in the previous polyhedra, the starting point was a sphere with a number of points distributed over it. But this time, the points were placed according to a simple and perfectly regular rule.

Imagine you you draw a line from one pole of a sphere to the opposite pole, winding sort-of-helically around the sphere. Then place a number of points at exactly equal spacings along this line. In the examples below, there are 43 points.

If your quasi-helix had 2.5 turns, the result would look like the diagram on the right. The numbers are where the points are placed on the imaginary sphere; the resulting polyhedron is also drawn. The helical structure is very clear, but notice the irregularity of the faces: for example, the lower edges of faces 26, 27, and 28 are all different. This is because the radius of the quasi-helix is not constant, so turns of the helix near the poles will have fewer faces on them than turns near the equator, which means that the way the numbers on one turn line up with the numbers on the previous and subsequent turns keeps changing.

If we increase the number of turns in the quasi-helix to 5, the helical structure is still clear:With 7.5 turns in the quasi-helix (below), adjacent-numbered faces are now so far apart that the faces on the turns above and below them are starting to intrude into the spaces between them. See, for example, how faces 27 and 28 are almost pushed apart by faces 19 and 35.By the time we have 10 turns (below), consecutively-numbered faces are so far apart that faces from the turns above and below often meet between them, making the helical structure hard to discern (see, for example, how faces 34 and 22 squeeze in between faces 28 and 29). The polyhedron is beginning to look quite irregular, but note that consecutively-numbered faces are still more similar to each other than they are to the other faces.

The 47-hedron on the right (same as at the top of the post) was based on a quasi-helix with about 12.5 turns. At first sight, it looks quite random, and it’s certainly irregular, but there are still visible similarities between faces if you look hard enough. Compare, for example, the mid-blue and light-blue faces at roughly 10 o’clock and 2 o’clock respectively. And what about the dark-blue and very dark-blue faces just below the centre? One looks suspiciously like a rotated version of the other – and it is! In fact the whole polyhedron has an axis of 180° rotational symmetry. This suprised me when I spotted it, but it shouldn’t have. A helix has 180° rotational symmetry (turn a corkscrew upside down and it looks the same) so it’s inevitable that shapes based on a helix will have that symmetry also.

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.

So what?
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.

Afterthought
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 10^{5} newtons per metre squared. This means that all of the air in the column of the atmosphere has a weight W close to 10^{5} 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 m^{2}, 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 m^{3}. Hence if we liquefied the atmosphere, the resulting ocean would be 11.7 m deep.

Acknowledgement
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.

References
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.

This summer Chris Wallace and I spent a week renovating the chemistry-themed Chain Reactor machine that we built for the Edinburgh International Science Festival in 2011. Since then the machine has appeared every year, not just in Edinburgh but in Abu Dhabi, where it was installed next to a beach.

Unfortunately the salt air in Abu Dhabi meant that we got more chemistry going on than we’d intended, as the rusty condition of the nuts and bolts above shows (pictured not in Abu Dhabi but instead on a chilly November morning on the beach at Portobello in Edinburgh, where I live). Thus a big part of the job was replacing hundreds of zinc-plated nuts and bolts with stainless steel equivalents.

As we did this, I noticed a clear pattern to the corrosion. Surfaces facing upwards suffered much more badly than surfaces facing downwards, and if a bolt was sheltered from above by an overhanging part of the machine, it didn’t rust very much, even though it was completely open to the air.

This would make sense if the corrosive agent (tiny droplets of brine or particles of salt, maybe) was predominantly falling downwards through the air, which such droplets/particles must do if the air is reasonably calm.

You can see evidence of something similar in kitchens, where frying fills the air with an invisible mist of fat particles, each slowly falling through the air. You may not realise this until you see the tacky layer that gradually develops on the top sides of your shelves and cupboards, but not on the undersides. And if you wear glasses, it’s the inside of the lenses that you need to clean after frying; this is the side that faces upwards as you look down to cook.

This makes me wonder whether we’d need to clean or redecorate our kitchen walls less often if they were slightly overhanging so that grease particles were less likely to land on them. A slope of a degree or so from the vertical wouldn’t be too conspicuous and might make all the difference.

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 10^{13}, 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.

The experiment

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

(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} m^{3}.

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} m^{3})= 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 Al_{2}Si_{2}O_{7}. 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 Al_{2}Si_{2}O_{7}, 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 × 10^{23} = 1.75 × 10^{25 }atoms; this is about 3.97 × 10^{25} atoms per kilogram. (6.02 × 10^{23} 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 × 10^{25} atoms per kilogram, the number of atoms in the dot is

Rounded more reasonably, this is 10^{13} 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.

Checking

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 × 10^{12} 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 × 10^{12} 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.

It’s a remarkable fact that there are only five regular convex polyhedra, that is, solid shapes whose flat faces are all identical regular polygons. The most familiar example is the cube, with 6 identical square faces. There are other, less regular but still orderly polyhedra, such as the truncated icosahedron, seen in a bloated form in some footballs.

But what, I wondered, about truly irregular polyhedra, where every face is a different irregular polygon? What would they look like? Last January I set out to make some and find out.

My irregular polyhedra are all based on spheres. Imagine placing a number of dots on a sphere. At each dot, let there be a plane just touching the sphere. These planes will all intersect each other, and if we remove the parts of each plane cut off by the neighbouring planes, we’re left with a polyhedron, with one face for each dot that we placed on the sphere. The nature of this polyhedron depends upon how the points are distributed over the sphere. In the polyhedra shown here, the placement of dots was random but subject to certain constraints.

I wrote some software that allowed me to choose how many faces I wanted, and to regulate how evenly spread over the sphere the dots were. I could preview the resulting polyhedron, and when I saw one that I liked, the program produced a set of images of the faces of the polyhedron, with numbered tabs on them. Then all I had to do was print them, cut them out, fold the tabs, and glue them all together. This is not a task for the impatient: the 83-hedron at the top took about 3 days.

The polyhedra shown here differ principally in how evenly the dots were spread over the imaginary internal sphere. After randomly placing the dots, the program simulated repulsion between them (as if they were electric charges). The longer this repulsion process went on, the more evenly distributed the dots became. For the 29-hedron, no repulsion process was done, and for the 43-hedron, the process was allowed to continue until the dots stopped moving; presumably this is now (in some undefined sense) as regular as a 43-hedron can get.

For the 29-hedron (right), I took advantage of the four-colour map theorem, which tells me that if I want to colour the faces such that no two neighbouring faces have the same shade, 4 shades of card are all I need. I leave it to you to convince yourself that if this theorem is true for flat maps, it must be true for maps on balls too. (If you don’t see where maps come into it, think of each face of the polyhedron as a country on a political map.)

None of this would have happened had it not been for a visit to Jenny Dockett to talk about her Illuminating Geometry project. It was while talking to Jenny that the idea for making irregular polyhedra came to me.

I wrote the software in Python. The internally-illuminated shapes are made out of layout paper (very thin paper), and the 29-hedron is made of thin card. I used UHU Office Pen glue (not to be confused with UHU Pen glue). This glue doesn’t wrinkle the thin paper, and dries at about the right speed, but has proved to be somewhat unreliable in humid weather.