Edwin Pickstone, a colleague at Glasgow School of Art, appointed me as a maths consultant recently. His project was to produce a book which had a black square on each page. The size of the square and the number of pages had to be such that the envelope of the resulting block of ink was a cube containing 1 kilogram of ink.
His printer provided a sample run so that we knew what the areal density of the ink film would be, and I did the sums to work out how big the squares would need to be and how many pages we’d need. It turned out that we needed squares of side 19.28 cm, and (coincidentally) 1928 pages.
The responsibility weighed heavily on me, and I was a little nervous as the order went off to the printer. So I was relieved and delighted when Edwin told me that when he’d walked into the printers to pick the job up, the printer said ‘bloody hell, it took a whole kilo tin of ink to print your job’.
Edwin is Lecturer, Typography Technician and Designer in Residence at Glasgow School of Art.
In the previous post, we discovered that the kinetic energy of a drifting continent is of the same general magnitude as that of a moving bicycle and its rider – 1500 joules would be a typical figure.
I went on to calculate that, whereas it takes me only about 10 seconds to get my bike up to full speed, it would take me hundreds of years to get the continent up to its tiny full speed were I to put my shoulder against it and push (assuming that it was perfectly free to move). How can this be, when the amount of energy that I’m giving each of these objects is the same?
The problem is that when I push the continent, I am, effectively, in the wrong gear.
On a bike with gears, you’ve got a range of choices about how you power it: you can ride in a high gear, pedalling slowly but pushing hard on the pedals, or ride in a low gear, pedalling more quickly but pushing less hard on the pedals. There’s a simple tradeoff: if you want to pedal half as fast, you’ve got to push twice as hard for the same effect.
But there’s a limit to how hard you can push on the pedals, which means that if you move up too far up through the gears, there comes a point where you can no longer make up for the decreased pedalling rate by pushing harder on the pedals, and the power that you can supply to the bicycle falls.
Anyone who’s tried to accelerate a bicycle when they are in too high a gear will have experienced this problem, and it’s what I experience when I try to push the continent directly. Because the top speed of the continent is extremely low (about the speed of a growing fingernail), I’m necessarily pushing it very slowly as I accelerate it. This means that to give it energy at the rate that I want to (1500 joules in 10 seconds, like the bike) I would have to push it impossibly hard – the force needed is about the same as the weight of a 300-metre cube of solid rock.
Is there a way that we can put me into a lower gear, so that I can push with a force that suits me, over a longer distance, and still apply the very high force over a short distance to the continent?
Yes. Just as we’ve all used a screwdriver as a lever to get the lid of a tin of paint off, so I could use a lever to move the continent. Similarly to the bike gears, the lever allows me to exchange pushing hard over a small distance with pushing less hard over a longer distance. To do the job, the lever would need to be long enough to allow me to push, with all my might, through a distance of about 2.5 metres, with the short arm of the lever pushing the continent. We’d need an imaginary immoveable place for me to stand, and we could use the edge of the neighbouring continent as the pivot (just as we use the rim of a paint tin as the pivot). The catch is the length of the lever: if the short arm was 1 metre long, the long arm would be about 1.5 million kilometres long.
Simon Gage of Edinburgh International Science Festival suggested a more compact arrangement: a bicycle with an extremely low gear ratio, with the front wheel immobilised on the neighbouring continent (assumed immoveable), and the back wheel resting on the continent we’re trying to accelerate. A transmission giving 17 successive 4:1 speed reductions would do the job nicely. Ten seconds of hard pedalling would get the continent up to full speed. To me on the saddle, it shouldn’t feel any different to accelerating my bike away from the lights.
A wee caveat. This is a thought experiment, and we’ve swept some fairly significant engineering issues under the carpet. The rearmost parts of the power train would be moving at speeds that are literally geological, so in reality it would take me years of pedalling to take all of the slack and stretch out of the system. These parts would also be transmitting mountainous forces, and so they’d need to be supernaturally strong. There will be frictional losses. And then there’s the issue of transmitting a gigantic force to the continent through the contact of a bike tyre on the ground.
What force is required to accelerate the Eurasian plate to top speed in 10 seconds?
The top speed of the plate is 3.2 × 10-10 ms-1. If I accelerate it uniformly, its average speed will be half of this, and so in the 10 seconds over which I hope to accelerate it, it will travel 1.6 × 10-9 m.
Now W = fd
where W is the work that I do on the plate (ie the kinetic energy that I give it), f is the force that I apply to it, and d is the distance through which I push the plate. Rearranging gives us
f = W/d
We know W from the previous post (it’s 1500 joules) and we’ve just calculated d. Thus f works out at about 9.4 × 1011 newtons.
For comparison, a 300-metre cube of rock of density 2700 kg m-3 will have a weight of (300 m)3 × 2700 kg m-3 × 9.81 m s-2 = 7 × 1011 newtons roughly.
When a lever is used to amplify a force, the ratio of the lengths of the arms of the lever needs to be the same as the ratio of the two forces. Suppose that I can push with a force equal to my own body weight, about 600 newtons. If I’m to use a lever to amplify my push of 600 N to a force of 9.4 × 1011 N, the ratio of the lengths of the arms needs to be (9.4 × 1011)/600, or roughly 1.5 × 109. So if the short arm of the lever is 1 metre long, the long arm needs to be about 1.5 × 109 metres long, which is 1.5 million kilometres. For comparison, the Moon is about 400,000 kilometres away.
To do 1500 joules of work with a force of 600 N, I’d need to push over a distance of 2.5 metres (because 600 × 2.5 = 1500).
The bicycle gearing
I estimated that it takes me 15 pedal revolutions to get my bike up to full speed. Knowing the length of the pedal cranks, I know the total distance that I have pushed the pedals through, and I know how much work I have done on the bicycle – 1500 joules. (I’m ignoring energy losses here, because they are small at low speeds on a bike and the calculation is highly approximate anyway). Using work done = force × distance, this gives an average force on the pedals of about 94 newtons.
The 17 stages of 4:1 reduction mean that the back wheel is rotating 417 = 1.7 × 1010 times slower than I’m pedalling. The pedalling force is amplified in the same ratio, to give a force on the teeth of the rearmost gear of 1.6 × 1012 newtons. We now have to allow for the fact that the radius of the rear wheel is about twice the length of the pedal crank. This roughly halves the force available at the rim of the rear wheel, giving a force of about 8 × 1011 newtons, which is close to what we need.
…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!
Be careful about how much water you put in your kettle and ‘do your bit’ to save energy!
That’s what we’re often told, and boiling only as much water as you need for your cup of tea can only be a good thing. But how much impact does it really have on your overall energy consumption?
Consider this: if you’re driving along in a car at 50 mph, you’re using enough energy to make a cup of tea every few seconds.
1, 2, 3, tea, 1, 2, 3, tea, 1, 2, 3, tea…
As most people don’t think twice about driving their cars for an extra few minutes, never mind an extra few seconds, it’s clear that the energy savings made by being frugal with your kettle are very small compared to your overall energy use.
Does this mean that we shouldn’t be careful with our kettles after all? No. It all matters. But having boiled exactly one cup’s worth of water for your refreshing cuppa, you shouldn’t put your feet up and think that your energy consumption issues are sorted. You’ve much bigger fish to fry.
Imagine that you’re travelling in a car that’s doing 50 miles per hour, with a rate of fuel consumption of 50 miles to the gallon (the units are customary in a UK motoring context). This is a realistic situation; if anything it’s optimistic in terms of energy consumption.
In this scenario, your car will burn a gallon of petrol, that’s 4.5 litres approximately, every hour.
With 3600 seconds in an hour, that’s 4.5L/3600s = 0.00125 litres per second. The density of petrol is about 0.8 kg per litre, so the rate of petrol consumption comes to 0.00125 L s-1 × 0.8 kg L-1 = 0.001 kilograms per second.
The energy density of hydrocarbon fuels is about 46 megajoules per kilogram, so the rate of use of energy by your car is 0.001 kg s-1 × 46×106 J kg-1 = 46000 joules per second (watts).
Now let’s think about the tea. The energy E required to produce a temperature change of ΔT in a mass m of a substance whose specific heat capacity is C is
E = m C ΔT
A typical mug of tea contains about 250 ml, or 0.25 kg, of water. For water, C = 4200 J kg-1 K-1. The temperature change is about 90 K, if the water comes out of the tap at about 10°C and is raised to boiling point. So the amount of energy required to heat the water for a cup of tea is 0.25 kg × 4200 J kg-1 K-1× 90 K = 94500 joules.
As the car is using 46000 joules per second, it follows that the car is using enough energy to heat the water for a cup of tea every two seconds.
Now to be fair to the car, we need to recognise that in real life there is more energy used to heat the water for the tea than actually goes into the water. There are heat losses in energy generation and transmission, and from the kettle itself.
Things get very complicated here. A thermal electricity generating plant (coal, oil, gas, or nuclear) converts energy in the fuel to electrical energy with about 35-45% efficiency. But in the UK, about 20% of our electricity is generated from renewables, where the concept of efficiency is harder to pin down. For example, I could use some oil to heat my water directly instead of doing it indirectly by generating electricity, but I couldn’t do the same with the wind. I’m going to assume that 80% of the electricity that I used was generated in thermal plants at 40% efficiency, and 20% of it was generated at 100% efficiency, giving an overall efficiency of 52%. But be aware that this is a very rough figure, and that there is no right answer.
According to this document, about 93% of the energy in generated electricity makes it unscathed through the transmission and distribution systems to the end user.
I did an experiment and estimated the efficiency of my kettle to be about 88%. I give details later.
This gives an overall efficiency of boiling water as 52% × 93% × 88% = 43%. So instead of using 94500 joules to boil water for a cup of tea, we’re actually using 94500J/0.43 = 220000 joules, which means that the car’s rate of energy use is equivalent to a cup of tea roughly every 5 seconds. Because of all the uncertainties involved (not least the size of a cup of tea), we should treat this figure as very approximate.
Efficiency of my kettle
My kettle took 180 s to raise 1 litre of water from 17°C to boiling.
The kettle is rated at 1850-2200 W for supply voltages in the range 220-240 V. When the kettle was running, I measured the supply voltage to be 242 V, so I assumed that the kettle was operating at the top of its power range, ie 2200 W.
The heat supplied by the kettle element was therefore 2200 W × 180 s = 396000 J.
Using the same equation as earlier, the energy required to raise the temperature of 1 kg of water by 83 K is 1 kg × 4200 J kg-1 K-1 × 83 K = 348600 J.
The kettle was therefore heating the water with efficiency 348600/396000 = 0.88 or 88%.
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 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.
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 big margin, 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.
What about the other exit routes for carbon?
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.
Not the Lake District
The picture at the top of the post wasn’t actually taken in the Lake District. It was taken in Glen Feshie, in the Cairngorms. To keep the weight down, I didn’t take my camera on the trip to the Lakes.