Puff pastry

I was rather taken by the video above, which I first saw on Core77. I started wondering how many times you have to put the roll of silicone material through the machine to get satisfactory mixing of the two colours of material. The people in the video consider the job done after four passes. What does that mean in terms of the thickness of the red and white layers within the material?

The roll is a rather complicated object, so I worked with an idealised version of the real process, where the sheet emerging from the rollers isn’t rolled up, but cut into several pieces which are stacked up before being passed through the rollers again. I came up with the following:

After only 2 passes, the layers in the slab are too thin to see with the naked eye. And by some margin, too: there are over 600 of them and they’re only a fortieth of a millimetre thick. If you made a perpendicular cut through the slab, it wouldn’t appear to have red and white layers in it.

After only 4 passes, a standard compound microscope operating in visible light wouldn’t be able to resolve the layers in the slab.

After only 6 passes, the layers would be thinner than the width of the molecules of the silicone material. At this stage the concept of red and white layers no longer makes sense.

These results will only apply to material near the centre of the roll. It’s easy to see from the video that material near the edges is not mixed so well.

The calculation

From the video, it looks like there are about 9 turns in the roll. Each time the roll is flattened by the rollers, those 9 turns are converted into 18 layers. The resulting sheet is rolled up and passed through the rollers again, multiplying the number of layers by 18, and so on.

This doesn’t work at the sides of the roll. We’ll ignore that complication, and work with a flat analogue of the actual situation. We’ll assume that we start with two long rectangular flat sheets of material, a white one and a red one, laid on top of each other. We’ll cut this assembly into 18 identical pieces, and make a stack of them; this stack will have 36 layers. We now flatten this stack in the rollers, cut it into 18 pieces, stack them up (giving us 648 layers), and repeat.

On emerging from the roller, the sheet appears, by eye, about 1.5 cm thick. We’ll assume that we start with two layers of half this thickness. The table below shows the number of layers and the thickness of each layer after 0, 1, 2, 3… passes through the rollers.

Number of passesNumber of layersLayer thickness (m)
027.50 × 10-3
1364.17 × 10-4
26482.31 × 10-5
311 6641.29 × 10-6
4209 9527.14 × 10-8
53 779 1363.97 × 10-9
668 024 4482.21 × 10-10

We can identify various milestones, as follows:

Limit of visual acuity. A person with clinically normal vision can resolve detail that subtends roughly 1 minute of arc at the eye. At a viewing distance of 30 cm, this corresponds to about 0.1 mm (10-4 m). The layers of material are much thinner than this after only 2 passes. If you made a perpendicular cut through the slab of material, after two passes you wouldn’t be able to see the layered structure. (This might not be true if the cut was oblique.)

Limit of standard light microscopy. A compound microscope working in visible light can resolve detail down to about 200 nm (2 × 10-7 m). The layers become thinner than this after only 4 passes.

Single-molecule layers. The question here is the number of passes needed before the layers are less than a molecule thick (at which point the idea of layers fails). The difficulty is that molecules of silicones are long chains, and these chains are almost certainly bent, so their size is ill-defined. This part of the calculation will be hugely approximate. We’ll be as pessimistic as possible, assuming that the molecules are roughly straight and that they lie parallel to the layers in the slab of material.


A common silicone material is polydimethylsiloxane or PDMS. This consists of a silicon-oxygen backbone with methyl groups attached. The lengths of carbon-silicon and carbon-hydrogen bonds are 1.86 × 10-10 m and 1.09 × 10-10 m respectively. So the width of the molecule is going to be, very, very approximately, of the order of 4 × 10-10 m. The layers are thinner than this after only 6 passes.





Faces in the snow

A hollow impression of my face in the snow.

The hollow face illusion is a wonderful visual effect in which a hollow mask of a face appears to be convex, like the face itself. Making a hollow mould of your face (for example using plaster) is difficult and potentially dangerous. However, last weekend my attention was drawn to an easier and safer way.

I was walking down from Coire an Lochain in the Scottish Highlands with a group from the Red Rope club, when I saw my friend Maia standing on the path ahead, chuckling. She’d been making face imprints in a steep snowdrift, and they showed the hollow face illusion beautifully.

Ben face plantThe procedure needs no explanation (see right). The snow needs to be fresh and soft; you’d be surprised how hard it is to push your face into what feels to your hand like very soft snow. The tip of my nose is noticeably flattened in the picture above.

Ben and Matthew on skyline
Near Coire an Lochain on the day in question. (Readers familiar with Highland place names will realise that I’m not giving much away here.)



The kinetic energy of a drifting tectonic plate…

…is broadly similar to the kinetic energy of me and my bike as I pedal along.

Map of tectonic plates (United States Geological Survey) http://pubs.usgs.gov/publications/text/slabs.html

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.

Me struggling up one of the many steep roads in NW Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably less than that of a drifting tectonic plate.
Me struggling up one of the many steep roads in north-west Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably much less than that of a drifting tectonic plate ;-).

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.

The calculation

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…

…1500 joules!

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!

Edinburgh Mini Maker Faire 2017


On 16th April Sarah McLeary and I had a busy day at the 2017 Mini Maker Faire in Edinburgh (part of the science festival). We showed the paraboloidal castings that we’ve been working on together, and I showed developments in my irregular polyhedra since I showed them at the 2015 Mini Maker Faire.

As well as plaster casts, we’ve tried slipcasting porcelain using our paraboloidal moulds, squirting the slip on as the mould spins to try and get a lacy structure. You can see one of these near the front left of the table. I’ve also been having a go at making irregular puckered plane-based tilings (standing up at the back of the table). I will write more about both of these projects before too long.