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.
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.
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.
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 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.
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.
Recently I spent a few days walking and wild camping among the mountains of the English Lake District. I was moving on every day and carrying everything I needed for the trip on my back. My rucksack got lighter and lighter as I worked my way through my food supply, and as there was no evidence that my body was getting any heavier, I started wondering where all that mass goes.
The food was largely very dry, so I’m not going to concern myself with water (though some water will be generated as the food is metabolised). Of the dry mass, I will excrete some as faeces, and some in my urine. I’ll secrete some skin oils and will also shed some skin. Then there’s hair, toenails and fingernails, and we mustn’t forget the odd bit of snot and earwax. Quite a trail of debris, really.
But what I started to wonder as I walked was: how much of this dry mass do I breathe out? I must lose some mass with each breath, because outbreaths are poorer in oxygen and richer in carbon dioxide than inbreaths, and carbon dioxide is heavier than oxygen. Can we put some numbers to this loss of mass?
It turns out that we can, and the result surprised me: by a factor of several, the most important exit route for carbon appears not to be my bottom but my lungs.
The volume of gas exchanged on each breath in ordinary breathing is about 0.5 litres. On the way in, the concentration of carbon dioxide is practically nothing (about 0.04%), but on the way out, it’s about 5%, So every 20 breaths results in me breathing out the equivalent of half a litre of pure CO2, which means 40 breaths to breathe out a litre.
Assuming that for every molecule of carbon dioxide breathed out, there is one molecule of oxygen breathed in, then the extra (non-water) mass breathed out is just the mass of the carbon in the carbon dioxide; we can ignore the oxygen.
How much carbon is there in one litre of CO2? One mole of carbon is 12 grams, and one mole of a perfect gas occupies roughly 24 litres in everyday conditions. So one litre of carbon dioxide contains one-24th of a mole of carbon, or about 0.5 grams.
So every 40 breaths, I breathe out 0.5 grams of carbon that originally entered my body as food. At rest, I breathe about once every 4 seconds, so it takes 2 × 40 × 4 = 320 seconds to breathe out a gram of carbon. There are 86400 seconds in a day, so over the course of a day, I’ll breathe out 86400/320 = 270 grams of carbon.
I’ve made all sorts of approximations, so let’s say that I breathe out 200-300 grams of carbon daily.
Do I believe this answer?
This is a much bigger number than I expected. How does it compare to the other output routes for carbon, and do the numbers stack up when we consider how much carbon I ingest?
From various internet sources, fat is something like 75% carbon, carbohydrate is about 40% carbon, and protein is about 50% carbon. On this basis, I’m going to estimate that the dry matter of faeces is in the region of 50% carbon. A typical person produces about 130 grams of faeces a day, of which 75% is water. So the amount of carbon excreted daily in faeces is about 130 × 0.25 × 0.5 = 16 grams.
Wikipedia tells me that we typically produce about 1.4 litres of urine a day, with 6.9 grams per litre of carbon, which gives about 10 grams of carbon excreted by this route per day.
Several internet sources (which may be equally wrong) suggest that we shed about 10 grams of dead skin every day. I don’t know what the moisture content of dead skin is, but it looks like we’re not going to lose more than 5g of carbon by this route every day.
Neglecting the other minor carbon loss processes, this gives a total of about 30 grams of carbon leaving the body daily by routes other than the lungs, compared to 200-300 grams via the lungs.
If these sums are right, I must ingest an equal amount of carbon in my food. This isn’t easy to estimate at all precisely. My diet is largely a mixture of fat, protein, and carbohydrate, but I don’t know what the balance is, and it’ll vary a lot between people. So let’s go to an extreme and suppose that I meet a daily energy requirement of 2500 kilocalories by eating fat only. At 9 kilocalories per gram I’d need to eat 280 grams of grease a day. At 75% carbon, this would be 210 grams of carbon. Doing a similar calculation assuming a diet of undiluted protein or carbohydrate gives a daily carbon intake of 310 grams or 250 grams respectively. So maybe I eat about 250 grams of carbon per day, which tallies reasonably well with the total figure for carbon output that I calculated earlier.
Suppose that you were standing perfectly still, and gravity suddenly stopped operating on your body. What would happen? Nothing much, you might think, apart from a queasy feeling of weightlessness. After all, an object won’t start to move unless a force acts on it, and no force is acting on your body.
However, when you stand “still”, in fact you’re travelling in a very large circle at rather high speed, as the Earth turns on its axis and carries you with it. Newton’s 1st Law tells us that things travel in straight lines unless a sideways force acts upon them. The force that keeps tugging you to make you travel in a circle rather than in a straight line is gravity. This means that the moment gravity stops acting on you, you’ll start moving along a straight line (the red line in the diagram) while the ground continues to move in a circular path underneath you (the blue line).
The consequence is that you’ll lose contact with the Earth and float upwards, serenely or otherwise. At least, that’s what it will look like to earthbound observers. But what’s really happening is that the ground is accelerating downwards away from you as it moves on its curved path. Try to remember that as you watch your footprints receding beneath you.
I wondered how fast this would all happen. The answer is: remarkably quickly. I give the geometry later, for those who are interested, but here are some example results for a person standing in Edinburgh, on a latitude of 56° N. Just for now, we’ll pretend that there isn’t any air.
After 1 second your feet will be 5 millimetres off the ground.
After 10 seconds you’ll be 53 centimetres off the ground.
After a minute, you’ll be 19 metres up
After an hour you’ll be at an altitude of over 68 km (though, being half frozen to death by now, you may be losing interest).
The lower your latitude, the quicker your ascent. At the equator, you’ll rise about three times as fast as in Edinburgh, and at the poles, you won’t lose contact with the ground at all.
Why on earth should I be interested in a situation that is contrived and physically impossible? It’s because it brings home the fact that each of us is constantly moving along a curved path as the Earth rotates. On the equator, it takes only 10 seconds for our trajectory to deviate from a straight line by 1.7 metres (during which time we’ve travelled 465 metres).
Why did I pretend that there isn’t any air? Because the presence of air muddies the waters by adding another force: the upward force of our buoyancy in the air. At low altitudes this force isn’t negligible: it slightly more than doubles the first three figures above. As you rise further and the air gets thinner, it matters less and less. I left it out because I wanted to make the effect of the Earth’s rotation clear.
The question that I’ve just answered is a trimmed-down version of a question that my friend Malcolm and I occupied ourselves with once when we were on a rather long and boring tramp along a glen at the end of a camping trip in the Cairngorms in Scotland. The question we asked then was: what would we observe if gravity suddenly stopped operating altogether? I may return to that subject in a future post.
Let the centre of the Earth be at , the origin, and let the Earth’s radius be and its angular velocity about its own axis be . You are standing at latitude and are therefore a distance from the Earth’s axis. Your linear velocity as you stand still on the rotating Earth will be , tangential to the Earth’s surface.
Suppose that gravity stops acting on you at time zero, when you are at point . With no gravity acting on you, will now travel in a straight line tangential to the Earth’s surface. The diagram shows the situation from a suitable vantage point, looking sideways on to your direction of travel. We are not looking down on the north pole.
After a time , you will have travelled a distance , to point .
Your altitude is the distance , where is the point on the Earth’s surface for which is directly overhead. lies on , the line from to the centre of the Earth. The length of is given directly by Pythagoras’ Theorem in triangle : it’s . As , the altitude of point is . So
All we need now is , because the Earth does one full rotation in 86400 seconds, and , because that’s how big the Earth is. We can now choose and calculate for any value of .
We can check this answer in two ways. Firstly, we can use the very useful intersecting chords theorem to calculate the distance marked in the diagram on the right. For small values of , where , then should be approximately equal to the your altitude . For values of t of 1, 10, or 60 seconds, and agree to 4 significant figures. As we expect, as increases, the agreement gets less good: and differ by about 2% after 1 hour.
The second check is to differentiate the expression for twice with respect to time. The first differentation gives us an expression for the rate of change of with respect to time, that is, your rate of gain of altitude:
(Note: to keep things clear, I’ve omitted , which merely accompanies everywhere and doesn’t change the conclusions.) Where , and your motion is purely tangential, this expression for your speed away from the ground should be zero, and where is very large , and your motion is purely radial, your speed away from the ground should be . Both are true.
The second differentiation gives us an expression for your upward radial acceleration:
As it’s not you accelerating, but the ground that is accelerating away from you as it continues on its circular path, this expression for your upwards acceleration should, where and your path is tangential to the surface, become the same as that for the centripetal acceleration of the ground, , which it does. In addition, when , and your path is almost radial, the expression for your acceleration should approach zero, which it does.
This blog replaces my old website. I’m planning to use it to write about my science and design thoughts and activities. I’ll probably also recycle some items from my old website so that they remain present on the web.