Personal Blogs

There was an Old Man of St Bees,

Who was stung on the Nose by a Wasp.

When they asked, "Does it hurt?"

He replied. "No, not much."

"I'm so glad it wasn't a Hornet." ^[1]

The modern English name of the insect Vespula vulgaris is of course wasp, but in Old English it was wæps or wæsp, and this goes back via a Germanic line of descent to a Proto-Indo-European (PIE) root *wapso-. This is also the origin of Latin vespa, and the latter may have influenced the form of modern wasp.

Interestingly there is a dialect version wopsie, which I've always been aware of and occasionally use, and this is surprisingly close to the ancestral root, so much so that it's possible to imagine that if we could travel 5,000 years back in time we might agree with people living then what the insect is called.

The French name is guêpe and if we recall that the circumflex over the 'e' originated as a scribal abbreviation for an 's', we can see it is not all that far from Latin vespa after all. But it's thought that it is a kind of amalgam of two words, both coming from PIE *wapso-, one from Frankish (so via the Germanic route) wapsu and the other from Latin vespa.

As a final aside, the name of the Roman Emperor Vespasian might derive from vespa, and there are Italian people with the surname Vespa (and also a famous brand of scooter) and Wasp is an English surname, although quite rare.

[1] I have quoted this Limerick as I remember it but checking the original by W.S. Gilber (of Gilbert and Sullivan), poking fun at Edward Lear, I see my version is a little different, but even more ridiculous so I've left it as is.

Hedgehogs – why can't they just share the hedge?

This was the joke that won Dan Antopolski the Funniest Joke of the Fringe award at the 2009 Edinburgh Festival Fringe. ^[1] 

Hedgehogs were not always called that. In old English, it was an igil, cognate with modern German Igel. But in the early part of the Middle English period this was displaced by yrichon, 'urchin', a word of French origin. This was in turn replaced in the 15^th century by heyghoge and hedgehog it has been ever since.

But where did igil and urchin come from? Well Old French herichun is from Latin ericius and the latter is thought to derive from a Proto-Indo-European (PIE) root *ghers-, 'bristly' or 'spiny', probably the origin also of horror, in the sense of your hair standing on end, bristling, you see.

At this point you may be guessing igil is from the same root, but you'd be wrong ; igil is from a different root *heg-, 'hedgehog'. The hedgehog had a PIE root all to itself, hogging it you might say!

This is also the source of Ancient Greek ekhinos, 'hedgehog' or 'sea urchin'. which of course is where we get echinoderm. It seems people have often seen a parallel between between the spiny land animal and the spiny sea creature, and at one time hedgehogs were sometimes called land urchins to distinguish them.

Modern Greek has a different word for hedgehog which I rather like: skandzóchiros, 'prickly pig', which seems a bit more descriptive than the English equivalent.

[1] BBC News. (2009, August 24). Hedgehog joke is Fringe funniest. bbc.co.uk

Leon Tolstoy, as well as being one of the most famous novelists of all time, was an educationist, who founded a school for peasant children in his country estate, and wrote several textbooks for their use, one of which dealt with arithmetic.

I haven't been able to find the original text of Tolstoy's arithmetic book, but one problem that has become quite famous and appeared in a number of places is reproduced in Yakov Perelman's book Algebra Can be Fun. Here is the wording of the problem as it appears there.

A team of haymakers were assigned the task of scything two meadows, one twice the size of the other. Half a day the team worked on the larger meadow. Then it split into two equal groups: the first remained in the larger meadow and finished it by evening; the second group scythed the smaller meadow, but by evening there still remained a portion to do; this portion was scythed the next day by one haymaker in a single day’s work. How many men were there in the team?

Scroll down to see a solution

Solution: There are doubtless many ways to solve the puzzle but this was my take on it.

Suppose there are people in the team. Then in the first half-day they will put in person-days of work. After the team splits, the people remaining in the first field will work for half a day, so putting in a further person-days, for a total of person-days to mow the first meadow.

Over in the second field the people diverted there will work for half a day, making man-days, and on the second day the lone mower will contribute a further person day, for a total of .

But we know the first is twice the size of the second so

, , which implies and that is the answer.

Imagine your calculator is broken. It won't handle numbers beyond 1,000,000,000,000 but you have been asked to find the last two digits of .

Can you still find the answer. What is it?

Scroll down for Solution and Discussion.

Solution

Even with the broken calculator we can start calculating some small powers of and when we get to we discover the last two digits are . This means all end in (01\).

So will end in also and to find the last two digits of we can just multiply by , which tells us the answer is .

In fact it is possible to prove the rather astonishing fact, that for any prime , has the same last two digits as .

And so for example we can be sure that the last two digits of , a number of some 2,000 digits, are .

If we start with a triangle ABC and erect squares on two of its sides, then join the two outer vertices of the squares it creates a new triangle, called a flank triangle.

Theorem:  If we draw a line passing through the midpoint of the outer edge of the flank triangle and the apex of the original triangle, it will be perpendicular to the base of that triangle. In the diagraml ine HCI is at right angles to AB.

Proof: Rotate the flank triangle through 90° about point C.

Consider the triangle ABD'.

From the construction:

ACD' must be a straight line, and AC = CG, so C is the midpoint of AD'.

BD' is the image of GD and H' the image of midpoint H, so H' is the midpoint of BD'.

Hence CH' joins the midpoints of two sides of ABD' and is therefore parallel to the base AB. CH' is perpendicular to CH and hence to HI.

But this means HI must be perpendicular to AB, because CH' and AB are parallel. //

Exercise: Prove the flank triangle and the original triangle have equal areas.

I got this problem from Cut-the-Knot 'Properties of flank triangles' and drew the graphics in GeoGebra.

If you cut a sheet of A4 paper in half you get two sheets of A5, which have half the area but maintain the same proportions of length to width.

Can we do the same in 3D? Yes we can!! Imagine a block with dimensions , and , with being the longest side.

If we "cut along the dotted line" and let be the longest side of the new blocks so formed, we need the proportions to remain the same,so we can write

or . Solving gives

and . For a proper "A" system we need A0 to have a volume of 1 m³, in the same way as an A0 sheet of paper has an area of 1 m². So

m³ and making the dimensions of the A0 block 1260 mm x 1000 mm x 794 mm.

This extends naturally to higher dimensions, even though we can't physically construct the higher-dimensional blocks we can describe them mathematically. In four dimensions, the A0 hyperbox would have dimensions in the ratio and higher dimensions follow the same pattern.

Another one from Perelman's book.

In each row A - C identify the expression with the largest and smallest values

A. ,   ,   

B. ,   ,   

C. ,   ,   

Scroll down for answers

Answers
A. is only so it is smallest. is clearly bigger than so it is largest.
B. is less than so it is smallest.  is clearly bigger than so it is the largest.
C. is much less than the gigantic , but this time the 'power tower' is the winner because , clearly greater than .

In fact is colossal: it equals

13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096

I found this in another book by Yakov Perelman, Algebra Can Be Fun (1979).

Here's a near solution

Write as . Now notice the first three terms are equal to , so we have .

But we know that for any and we have , the so-called 'Difference of two squares', and applying this to the expression above we get after rearranging slightly, . Providing these are both and thus we have found factors of and it cannot be prime.

I found this puzzle in the 1950 book Figure for Fun, by Yakov Perelman, where it is problem 29.

It concerns two workers who live in the same apartment block and work in the same establishment. Both walk to work each day and the first worker takes 30 minutes for the journey, whereas their faster colleague takes only 20 minutes.

Suppose the slower worker sets off 10 minutes before the faster and they move at their usual speeds. When will the faster worker catch the slower one up?

Scroll down for my solution.

Solution

(This is one way to solve the problem but I expect other approaches are possible.)

Suppose the distance is km. Then the speed of the slower worker must be km/h and that of the faster km/h. After minutes the slower worker will have travelled km.

Now we calculate how far each will have travelled at time hours (after the faster worker sets off.

Slower worker:

Faster worker:

The faster worker will have caught up when these are equal, so , which reduces to  or hours = minutes.

A friend of my daughter's snapped this Brocken Spectre near Abergavenny, in the Bannau Brycheiniog (Brecon Beacons).

When the sun shines from behind an observer onto a bank of mist, the observer's shadow may be cast on the mist and depending on the elevation of the sun the shadow may look like a giant figure, giving rise to the legend of the 'Brocken Spectre', named after the Brocken, a peak in the Harz mountains of northern Germany, where the condition are often favourable for seeing the phenomenon.

In this example the shadow appears tiny but we can clearly see the other feature of a Brocken Spectre: the observer's head is surrounded by a 'glory' – a ring of light like a halo. The ring has colours, in the same order as a rainbow, although distributed somewhat differently, but a glory is much smaller that a rainbow and caused by different mechanisms.

The water droplets in the mist are much smaller than the raindrops responsible for a rainbow, by a factor of about 100. The incoming light is back scattered and the colours are caused by the light waves interfering. But how does this back scattering happen?

Theory 1

The water droplets refract the light back. Unfortunately the water does not bend the light enough for this to be the explanation.

Theory 2

The physics of electromagnetic waves predicts the light can become a wave that travels round the surface of the raindrop and then emerges as a back scattered wave. This seems to happen but does not account for the amount of light actually back scattered.

Theory 3

More recent calculations have shown that energy can tunnel into a water droplet from light waves close by and after travelling round inside the droplet emerge as back scattered light, and that this effect can explain the observed brightness of glories. This is all explained in more detail in the excellent 2012 Scientific American article The Science of the Glory by H. Moysés Nussenzveig.

In this section of the Bayeux Tapestry, which shows Harold landing in Ponthieu, we can see the ship is steered by an oar, as was common at the time. Steering oars did not get replaced by rudders until a couple of centuries later. The steering oar was on the right-side of the ship facing forward,  so that side of the ship was called in Old English steorbord, 'steer side', bord being the term used for the ship's side. I guess if you were a left-handed steerer you just had to live with it.

This has direct cognates in Germanic languages e.g. German Steuerbord, Old Norse stýriborð, and because of the influence these peoples had as seafarers the word was borrowed into Romabce languages, so we have e.g. French tribord.

Tracing the Proto-Germanic *steuro further back seems inconclusive, with more than one explanation appearing in the literature.

Turning now to larboard, the side opposite starboard was at first 'backboard', bæcbord in Old English, but by Middle English we have ladde-borde, which seems to mean the side the ship was loaded or unloaded; think of the modern word laden. Naturally this would be the opposite side from the steering oar.

However having starboard and larboard, although a pleasing rhyming pair, is a recipe for confusion give the similar sounds, and so larboard was been displaced by port, with similar connotations of being the side that comes alongside the quay, so now larboard is rare and archaic, and 'port and starboard' is the normal expression.

Picture credit: Wikimedia, Section of the Bayeux Embroidery showing Harold arriving in Ponthieu, Public Domain.

This iconic picture shows a minister gliding serenely across the ice, but how is thos possib;e. Why is ice slippery enouhgh?

At school we were taught the reason is the pressure of the skates melts the ice and the resulting water act as a lubricant. This is a wonderful explanation, put forward by James Thomson in 1849 based on the fact that pressure lowers the melting point of ice, and it seemed so satisfying that everybody took it for granted that it was correct and it was long taken as received wisdom.

But it can't be quite right; even if the pressure is sufficient to melt the ice when we are talking about ice skating, it is nothing like enough to explain how skis glide along (or why I fall over on icy pavements for that matter!)

So rival theories have been advanced. One is that friction could melt the ice. It's possible this could have an effect but we'd expect melt water from friction to trail behind the sliding object and so not provide much lubrication going forward.

A third idea, based on an observation by Michael Faraday, is premelting. According to this theory the water molecules in the interior of the ice are tightly locked into a crystal lattice that restrict their movement but ones on or close to the surface are bonded to fewer close neighbours, allowing them more mobility, so they can act rather like water. Computer simulations support this as a plausible reason why the surface of ice should act as though it is wet.

It's conceivable all three of the mechanisms outlined above play a part. But very recently a fourth idea has emerged: the skate or ski (or the sole of my shoe) disrupts the crystal lattice of the ice and creates a layer of disorder molecule that are not fixed in a lattice and yet not quite liquid water, but which can act a lubricant. So far this has only been investigated in computer simulations and at present other researchers are not sure how it should be interpreted. 

So we have not finally answered the question 'Why is ice slippery?' but this new research does represent an important advance, and we have edged closer to understanding exactly what is happening.

To read more about this see Quanta Magazine here.

For a good video about the new research see Anton Petrov, 'For 200 Years We Were Wrong About Why Water Ice Is Slippery',

Picture by Sir Henry Raeburn - Reverend Robert Walker (1755 - 1808) Skating on Duddingston Loch. Public Domain.