The kinetic energy of a drifting tectonic plate…

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

plates
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 thse 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

stall

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.

punters.

A dish designed by nature

Paraboloid 1

When you stir a cup of tea, the surface of the rotating liquid develops a dip in the middle. The faster you stir, the deeper the dip. But the liquid surface is quite uneven; to get a smooth surface, throw the spoon away and spin the whole cup continuously. Once the liquid inside has caught up with the cup, and everything is turning at the same speed, the liquid surface forms a beautiful smooth curve known as a paraboloid of revolution.

Sarah McLeary and I are applying this idea to make thin paraboloidal plaster shells. We spin a bucket on a potter’s wheel (Sarah is a potter), and pour plaster into it. The plaster rapidly flows until its surface forms a deep paraboloidal curve, and then sets. We now use this cast, still spinning, as a mould, and cast a thin layer of plaster inside it, to make a paraboloidal shell.

That’s our first attempt above. It’s about 20 cm across and 3 mm thick. It might look like a part of a sphere, but in a profile view it’s easy to see that the curvature is tightest at the base and gradually decreases up the sides, as you’d expect for a paraboloid.

I love the fact that we didn’t decide what shape this shell was going to be: physics did.

There’s lots of experimentation ahead. When we’ve got the hang of it, I’ll explain our methods in more detail. But as this picture of our third attempt shows, we haven’t quite cracked it yet.

Paraboloid 3 cracked

 

 

 

This is not an illusion

pipe2smallwithcaption-and-copyright

This image is my version of Edward Adelson’s checkershadow illusion (with a little inspiration from Magritte). It’s a photograph of a real, physical scene.

Take a look at the central square of the checkerboard, and the square indicated by the arrow. Which is lighter?  Quite clearly, it’s the central square, isn’t it?

Remarkably, the central square actually emits less light than the square indicated by the arrow!  You could use a light meter to check this claim, but it’s easier to verify it directly by using a piece of card with two holes cut in it to mask off the rest of the image.

Some people will tell you that this image shows you how easy it is to fool your brain. But it does the exact opposite: it shows you what a marvellous piece of equipment your brain is.

Think about the checkerboard itself, and the materials it’s made of. The arrowed square is coated with dark grey paint, and the central square is coated with light grey paint—and that’s exactly what you perceive.

The shadow cast by the pipe means that the light-grey central square is more dimly lit than the dark-grey arrowed square, so much so that it actually reflects less light into your eye than the arrowed square. But your brain cleverly manages to determine the actual lightnesses of the physical surfaces, despite the uneven lighting. Isn’t that a good thing for your brain to do?

If you still don’t believe me, try this thought experiment. Imagine that you live in a forest where there are two kinds of fruit. One is light grey and poisonous, and the other is dark grey and nutritious. Two of these fruits hang next to each other, but in the dappled forest light the (light grey) poisonous fruit is in shadow, and the (dark grey) nutritious fruit is in bright light. Suppose that the depth of the shadow is such that the light-grey poisonous fruit actually reflects slightly less light into your eye than the dark-grey nutritious fruit, just as with the two squares in the picture above. Would you really want your vision to tell you that the poisonous fruit was the dark one and therefore the one to pick? Or would you want it to discount the irrelevant effect of the shadow and tell you which fruit was actually dark and which was actually light (and would kill you)?  I know what I’d want.

I think that it is wrong to call this effect an illusion (and so does Adelson). There is nothing illusory about what you see. You perceive the useful truth about the scene in front of you.

Kettles, cars, and energy

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.

The calculation

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%.

Rock and roll

I made the machine in the video below for no other reason than that I felt like making a vehicle that was propelled in a non-standard way. The idea arose when I wondered about setting my Product Design Engineering students the challenge of making a vehicle that wasn’t driven through its wheels. In the end we did a different project, but I still couldn’t resist having a go myself.

The vehicle is mainly made of corrugated cardboard with other odd bits and pieces. It’s driven by two servos controlled by an Arduino microcontroller board.

Originally there was going to be a separate weight that was shifted back and forth to make the vehicle tip. I was rather pleased when I realised that I could make the batteries, the Arduino, and one of the motors do double duty as the weight.

(I didn’t choose the carpet.)

BEKANT sit-stand table

making a polyhedron croppedI’ve just spent the last day and a half making another irregular polyhedron to use as a lampshade in the room where I work. As I worked, I was reminded how useful it is to have a table with motorised height adjustment. The thing that I was making started off flat, and finished 40 cm in diameter, so a table that was the right height at the beginning would be far too high towards the end of the construction.

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

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

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

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

 

 

 

 

 

Lilac chaser

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

I used an Arduino microcontroller board to control the LEDs.

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

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

How high am I ?

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

Hellaval and Skye from Askival
Am I higher or lower than the summit in the middle distance? (The peak of Hellaval on the island of Rum, taken from the flanks of Askival, with Skye in the background)

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

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

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

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

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

Walking bass

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?

"Vibrating guitar string" by jar [0] is licensed under CC BY 2.0
“Vibrating guitar string” by jar[0]. See end of post for full attribution.

Walking pace. A typical speed for the middle of a guitar string given a good twang is walking pace. And that’s the middle of the string. Near its ends, it’ll be moving much more slowly.

How can that be so? Well, although the string is going back and forth hundreds of times a second, it’s only travelling a few millimetres on each trip. So the distance that it travels in each second isn’t as much as you might expect. It certainly isn’t as far as I expected.

If you think that the string moves slowly, what about the body of the guitar? The string itself radiates very little sound into the air; its job is to set the body of the guitar vibrating. The body of the guitar, with its much larger area, is much more effective than the string at setting air into motion. Yet we can’t even see the body vibrating. At what snail’s pace must it be moving?

Remember also that the air molecules on which the guitar body acts are already travelling at something like 500 metres per second. Isn’t it astonishing that the sub-pedestrian movements of the guitar affect the movement of the air molecules enough to produce a sound that we can easily hear?

The calculation

Suppose that we look near the centre of the string, where its movement is the greatest. Shortly after being plucked, the width of the blur that we see is going to be something like 5 mm. So for every complete oscillation, the string does a round trip of about 10 mm.

The frequencies of the strings on a standard 6-string acoustic guitar are (to the nearest whole number) 82, 110, 147, 196, 247 and 330 hertz (one hertz is one oscillation per second). If we multiply these frequencies by the 10 mm round trip, it tells us how far the centre of each string travels in one second, that is, its average speed. I’ve converted these speeds into metres per second. For comparison, a brisk walk at 4 mph is about 1.8 ms-1.

Note nameMean speed of middle of string in metres per second (to 2 sig. fig.)
E0.82
A1.1
D1.5
G2.0
B2.5
E3.3

Our brisk walk is right in the middle of this range. And remember, we’ve done the calculation for the part of each string that’s moving the most. Near its ends, each string will be moving much more slowly than this.

Complication 1 – how long is the round trip really?

So far, we’ve assumed that each part of the string does a simple back-and-forth movement along a straight line, but if you carefully watch a vibrating guitar string you’ll see that the string often moves in an irregular but roughly elliptical orbit. The wire-wound lower strings show this most clearly; you can see a hint of it in the image at the top. This makes the round-trip distance a bit longer than the 10 mm that we used in the calculation earlier. Does this affect the string’s average speed much?

We’ll take the extreme case where each part of the string moves at constant speed in a circle of diameter 5 mm rather than along a straight line 5 mm long. The circumference of this circle will be 5 \pi 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?

graph cropped 1Consider the function y = \sin \theta, for half a cycle, that is, for theta from 0 to \pi 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 \pi). So we need to work out the area a under the sine curve between 0 and \pi, which we can do by integration:

a = \int\limits_0^\pi \sin \theta \,d\theta

= \left[-\cos \theta\right]_0^\pi

= \left[-(-1)-(-1)\right]

= 2

We constructed the rectangle to have area a. Its width is \pi, so its height, and therefore the mean value of the sine function, is a/\pi = 2/\pi. The height to the peak of the sine curve is 1, so the peak value of the sine function is \pi/2 times its mean value. This means that the peak speed of our guitar string is \pi/2 times, or 50% more than, its average speed. Again, nothing to write home about.

(Rather pleasingly, this ratio of \pi/2 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 r and period t is exactly the same as the (constant) speed of a particle moving in a circular orbit of radius r with period t. 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.)

Credits

Vibrating guitar string” by jar [0] is licensed under CC BY 2.0

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