Middle Eastern corrosion problem leads to kitchen cleaning insight

bolts on beach 1This 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.



What’s the smallest number of atoms that I can see?

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.

pencil dot
The dot is where the two pencil lines would meet. The ruler scale is 100ths of an inch.

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.

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

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

(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

(1.75 × 10-13 kg) × (3.97 × 1025 kg-1) = 6.95 × 1012

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.

Most irregular

83-hedron (about 35 cm across)

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.

43-hedron (about 35 cm across)

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.

very irregular small
29-hedron (about 30 cm across)

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.

art walk polyhedra.
Exhibition at Porty Art Walk

In 2015 I exhibited these polyhedra, and others, at Mini Maker Faire and Porty Art Walk in Edinburgh.

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.

On the fate of my food

076 Glen Feshie campRecently 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 sums

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.

065 Sardines

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.

If I suddenly became weightless, what would happen to me?

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.

Diagram of the Earth, looking straight down on the north pole.

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.

The geometry

Let the centre of the Earth be at O, the origin, and let the Earth’s radius be r and its angular velocity about its own axis be \omega. You are standing at latitude \phi and are therefore a distance r \cos \phi from the Earth’s axis. Your linear velocity as you stand still on the rotating Earth will be \omega r \cos \phi, tangential to the Earth’s surface.

nogravity3Suppose that gravity stops acting on you at time zero, when you are at point Q. 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 t, you will have travelled a distance \omega t r \cos \phi, to point P.

Your altitude is the distance PR, where R is the point on the Earth’s surface for which P is directly overhead.  R lies on OP, the line from P to the centre of the Earth.  The length of OP is given directly by Pythagoras’ Theorem in triangle OPQ: it’s \sqrt{(\omega t r \cos \phi)^2 + r^2}. As OP=PR+r, the altitude of point P is OP - r. So

PR = ((\omega t r \cos \phi)^2 + r^2)^{1/2} - r

All we need now is \omega = 2\pi/86400 \text{ s}^{-1}, because the Earth does one full rotation in 86400 seconds, and r=6.4\times10^6 \text{ m}, because that’s how big the Earth is. We can now choose \phi and calculate PR for any value of t.


nogravity4We can check this answer in two ways. Firstly, we can use the very useful intersecting chords theorem to calculate the distance marked h in the diagram on the right. For small values of t, where h \ll r, then h should be approximately equal to the your altitude PR. For values of t of 1, 10, or 60 seconds, h and PR agree to 4 significant figures. As we expect, as t increases, the agreement gets less good: h and PR differ by about 2% after 1 hour.

The second check is to differentiate the expression for PR twice with respect to time. The first differentation gives us an expression for the rate of change of PR with respect to time, that is, your rate of gain of altitude:

\frac{dPR}{dt} = \frac{\omega^2rt}{  (1+\omega^2t^2)^{1/2}}

(Note: to keep things clear, I’ve omitted \cos \phi, which merely accompanies r everywhere and doesn’t change the conclusions.) Where t=0, and your motion is purely tangential, this expression for your speed away from the ground should be zero, and where t is very large (\omega t \gg 1), and your motion is purely radial, your speed away from the ground should be \omega r. Both are true.

The second differentiation gives us an expression for your upward radial acceleration:

\frac{d^2PR}{dt^2} = \frac{\omega^2r}{(1+\omega^2t^2)^{3/2}}

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 t=0 and your path is tangential to the surface, become the same as that for the centripetal acceleration of the ground, \omega^2 r, which it does. In addition, when \omega t r \gg r, and your path is almost radial, the expression for your acceleration should approach zero, which it does.