My Great Grandfather and the Great War

My Great Grandfather, John Wrigley, 1915

My Great Grandfather, John Wrigley, 1915

Over the summer vacation, I was rummaging around some old family photos at my Dad's house and he showed me two articles from 1915.  It turns out that my great grandfather not only fought in the First World War, but also survived the Battle of Hill 60 at Ypres in 1915.  His story is incredible, he was shot in the arm during the battle.  As the Germans hurled grenades at his platoon he dived back into his trench, found some cardboard packaging to use as a makeshift bandage for his arm and walk 5km back to Ypres to receive medical treatment.   He survived the ordeal, and had to live with a plate in his arm for the remainder of his life.  It was taken out a few years before he died.  A typical Wrigley, he asked to keep it as a souvenir, but the doctors refused and it ended up as a museum piece.

Below are his accounts written in a letter to his father, published in the Sheffield Telegraph in 1915.  His words do it far more justice than I ever could. 

J. Wrigley's arm with plate.

J. Wrigley's arm with plate.

The Sight Around Hill 60.

Sheffield Telegraph 1st May 1915

My Great Grandfather, J. Wrigley

My Great Grandfather, J. Wrigley

Private J. Wrigley, of the 2nd King’s Own Yorkshire Light Infantry, who was wounded in the fierce fighting around Hill 60, sends an interesting letter to his father, Mr. John Wrigley, of 425 Langsett Road, Sheffield, contract manager of the Post Office telephones.

“I never want to witness another sight like that I saw on Sunday night.” he writes.  “There were hundreds of dead and dying all around, men blown to bits, shells falling every yard.  We were only a few yards from the Germans, and they were throwing hand grenades and bombs among us.  They also had two or three machine guns turned on us, in fact it was hell.  I was going forward with the remainder of my platoon when my arm fell, as if somebody had given me a terrible kick.  I looked down, and it was bleeding fast, so I turned back, and got into our trenches, put my arm on a cardboard box, and walked with it this way to Ypres, three miles away.”

Mr. Wrigley has in all four sons in the service.  He has another son in the 2nd K.O.Y.L.I., one in the 1st Field Company Sheffield Engineers, and another in training with the Sheffield Engineers at Doncaster.


Father and Five Sons - Sheffield Family’s Fine Army Record

Sheffield Daily Independent 27th November 2015

1. Father, John Wrigley 2. Richard Wrigley 3. Ernest Wrigley 4. John Wrigley (My Great Grandfather) 5. Joseph Wrigley 6. Robert Wrigley 7. Mother Mrs. Wrigley

1. Father, John Wrigley 2. Richard Wrigley 3. Ernest Wrigley 4. John Wrigley (My Great Grandfather) 5. Joseph Wrigley 6. Robert Wrigley 7. Mother Mrs. Wrigley

A striking instance of family patriotism, possibly a record for the Sheffield district is furnished in the case of Sergt. John Wrigley, R.A.M.C., and his wife.  In civilian life Sergt Wrigley, who is now stationed at the Third Northern Base, is well-known as telephone contracts manager.  He served with the Gordon Relief Expedition and holds the medal and Khedive’s clasp.  In addition to “doing his bit” he has five sons now serving King and Country.  In order they are:-

Private Ernest Wrigley, 2nd K.O.Y.L.I., now in France:  Pte. John Wrigley, of the same regiment, wounded in France on Hill 60, and has been in hospital since:  Second Lient. Joseph Wrigley, 3/5 Y. and L. Regt., Rotherham Territorial unit, now in training:  Sapper Richard Wrigley, 1st Field Co., Sheffield Engineers, who was one of the first to land with the Engineers in the Dardanelles, was wounded in both legs, in the eyes and shot across the top if the head, since invalided home with dysentery:  Sapper Robert Wrigley, now with the Sheffield Engineers in training.

Throwback Thursday: Superhero Science

I wrote this post a couple of years back on my old website, which has long gone.  I thought it could do with a shameless repost. 

Hawkeye should be as far away from danger as possible!

Hawkeye gets a bad press.  One of the 'normal' humans in the Avengers.  No superpowers, except he's really good at darts.  Which unless he's in a showdown with Phil 'The Power' Taylor, isn't probably very helpful to anyone.  If I was him, I'd want to be shooting that bow from as far away as possible from Thanos, Ultron, or whoever.  So what's the best angle he should shoot his bow at, in order to avoid danger?

Projectile Motion

Once Hawkeye shoots the arrow, it becomes a projectile.  And to work out how Hawkeye can get maximum distance, we're going to start by looking at how fast the arrow is going, or its velocity.

A projectile is basically any object that has been dropped or thrown (projected).  It will continue to move by its own inertia and is influenced only by the pull of gravity.  It will move an a parabola, a symmetrical curve (as long as we ignore air resistance and drag, which we will).

Because Hawkeye will most likely have to launch his bow at an angle to the horizontal, it will have a speed (or more correctly velocity) that can be broken into two components.  A vertical velocity, and a horizontal velocity.

Both the horizontal component and vertical component can be considered independent from each other.

Don't Panic!

Now I'm about to use a term called 'cosθ', and in a bit I'm going to use a term called 'sinθ', which unless you are a maths nerd like me will result in painful flashbacks to maths lessons, and you reaching for a paper bag to breathe heavily into.  Just hold on, it will be explained hopefully as painlessly as possible.

The magnitude of the velocities are represented to scale by the length of their arrows.  The longer the arrow, the greater the velocity.  When combined with the initial velocity, all three lengths can be combined to make a right angled triangle (a triangle that has one angle of 90°).

Because the three velocities together form a triangle when drawn to scale it means that we can use trigonometry to find the magnitude of the horizontal and vertical velocities.  And that's it, you can stop breathing heavily into that paper bag now.

Horizontal Velocity

Horizontally, once the arrow has left the bow, there are no external forces acting on it.  Meaning that from the left of from the right there is nothing pushing or pulling on it.  Because of this the velocity horizontally will be the same magnitude (size) throughout the journey of the arrow, as there is nothing pushing on the arrow, making it slow down or speed up.

Vertical Velocity

Because gravity is always pulling directly downwards on the arrow, it will cause the arrow to change its velocity vertically.  Once launched, the arrow's vertical velocity will gradually slow down until it reaches its highest point and will have a velocity of zero.  Once it reaches it's highest point it will gain velocity in the opposite direction and start to speed up.  Gravity will cause the arrow to decelerate when it is going upwards, and acclerate when it is going downwards at a rate of 9.81 metres per second, every second.  In any equation using the acceleration due to gravity, we label that value 'g'

This means that if it was going up at a vertical velocity of say 30 metres per second (m/s), after a second the vertical velocity would be 20.19 m/s (30 - 9.81).  After two seconds the vertical velocity would be 10.38 m/s (30 - 9.81 - 9.81).

It also means that after the arrow reaches its highest point and as it starts to 'fall' the arrow will gain a vertical velocity of 9.8 m/s every second that it is falling.

Now instead of using - 9.81 - 9.81 - 9.81 - ... for every second of time, it's easier to multiply 'g' by the number of seconds and subtract that from the starting vertical velocity.

We're going to use these two equations as a starting point to work out everything else.  How long the arrow is in the air, distance traveled and finally launch angle.

How far does the arrow go?

We're going to use the two equations above to work out a few things, first off we are going to figure out what the distance the arrow would travel horizontally in terms of the arrow's velocity.  The next time you're in a car look at the speedometer, you'll see the units in mph (miles per hour) or kph or km/h (kilometres per hour).  This means that:

So we have range in terms of launch angle and time taken.  We have two unknowns, so how can we 'get rid of time'?

What do we know about the time of flight?

We know that if there is no air resistance, the path of the arrow is symmetrical.  That means that when it is at its highest point, the arrow is halfway through its journey time.  We also know that at its highest point, the vertical velocity is 0.  So let's go back to the equation for vertical velocity and see if we can relate it to time in any way:

Now this 't' in the equation above is the time taken for the arrow to reach its highest point.  This is half the amount of time the arrow spends in the air overall.

We can put this expression for time in the air into our equation for the range of the arrow.

If Hawkeye always fires his arrow at the same launch speed no matter what angle he aims at we can label that a constant (it doesn't change).  Gravity 'g' also doesn't change.  This means that the only thing that can change is sinθ and cosθ.

Every other value in the equation (velocity, gravity...) has a constant value, and they will be multiplied by sinθ multiplied by cosθ.  This means that to get the biggest range, we need sinθ multiplied by cosθ to produce the biggest value possible.

Crunching the numbers

We find the biggest value of sinθ multiplied by cosθ by trial and error.  But I love a good spreadsheet, so I decided to plot the values of sinθ multiplied by cosθ for all angles from 0 degrees (completely horizontal) to 90 degrees (shooting straight up), to see which would give us the biggest value.

When we look at the chart, we see that the biggest value of sinθ multiplied by cosθ is when θ is 45 degrees.

This means?

This that for Hawkeye to be as far away from villains with actual real powers he needs to shoot his arrow with an angle of 45 degrees.  Or he could just let Hulk smash them, which is what I'd do in his position.

Brazil's flag is a starchart

I'm a space nut.  The classroom I teach in is decorated with numerous posters of the planets, models of spacecraft, an awesome inflatable star globe. There's also a star chart, and surrounding the star chart are cutouts from the middle Brazilian flag.

Before I moved to Brazil, I just assumed  that the stars were most likely in some order, but I barely paid attention to the flag so I couldn't remember how the stars were arranged within the flag itself.

When I arrived in Brazil, I was told that each star represented an individual state, and that apparently these stars were visible the night that Brazil became an independent country.  Some time trawling through Wikipedia confirms that this is the case, and that these were the stars visible from "Rio de Janeiro at 8:30 in the morning on 15 November 1889, the moment at which the constellation of the Southern Cross was on the meridian of Rio de Janeiro and the longer arm [of the cross] was vertical." 

The cool thing for me is that as I learn more and more about the astronomy of the Southern Hemisphere, the more Iearn about the Brazillian flag.  For example the three southern most states  Rio Grande do Sul, Santa Catarina and Paraná are represented by 'The Southern Triangle' (Triangulum Australe) constellation (which, sidenote is quite possiblely the most boring constellation ever imagined)

The Southern Triangle - I know, I couldn't contain my excitement either when I first saw it.

The Southern Triangle - I know, I couldn't contain my excitement either when I first saw it.

The most boring constellation, immortalised in one of Earth's most important nation's flag...

The most boring constellation, immortalised in one of Earth's most important nation's flag...

I also learnt how to find my way in the Southern Hemisphere.  As a life long Northern Hemisphereian, I know that the pan of the Big Dipper points to the North Star, Polaris.  It's easy to identify and once you spot it, you know where north is. 

The Big Dipper, and Polaris are represented on the state flag of Alaska

The Big Dipper, and Polaris are represented on the state flag of Alaska

This method for orientating yourself doesn't work in the Southern Hemisphere, as these stars are never visiible south of the equator.  However, there is a constellation, The Southern Cross (Crux) that points directly to the Southern Pole Star. 

The Southern Cross, Crux.  Image 

The Southern Cross, Crux.  Image 

Crux, pointing to the Southern Pole Star.

Crux, pointing to the Southern Pole Star.

This is shown on the flag of Brazil.  There is also deeper meaning to the use of the Southern Pole Star.  The Federal District, Brasilia is represented by the Southern Pole Star.  In the night sky, because of the Earth's rotation all stars appear to move around a fixed point.

This symbolism was used on purpose in the flag of Brazil, it represents that all of the States revolve around the capital.

Swriling star trails caused by the Earth's rotation.

Swriling star trails caused by the Earth's rotation.

I love this flag, I love flags and I love astronomy, so that's not really much of a surprise. The main reason I love it is because it's a lot like the country itself.  It's complicated, with a lot of deeper meanings.  It can appear messy when you first look at it, but as you learn more and more about it, you start to appreciate it and love it. 

Candy Physics - Marshmallows and Boyle's Law

I love trying to relate physics to the everyday world, it makes the concepts that you cannot see like atomic physics and electricity and magnetism easier to visualise and understand.   One of the best ways I find to do this is with food.  Either by making ice cream to discuss latent heat, using M&Ms to explain radioactive decay, or marshmallows and air pressure...

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