I could have modelled this just using piece-wise functions with restricted domains. When it was loaded into Geogebra though it was clear that the bottom left section was doing more than just being horizontal. There is a definite negative (but increasing towards zero) gradient… sounds a bit like an exponential decay function to me.

My Geogebra model – feel free to use it and modify it to see if you can make it a better fit to ‘the data’, AKA the wall.

Pro-Tip: The thicker the line, the better the fit. This is a joke.

My standard approach to modelling a problem like this is to load it into Geogebra. Then I look for various parts of the rope which resemble geometric entities I can calculate with. It looks like a straight line will be an adequate model from F to E, and circles for the rest:

I considered using Pythagoras theorem, but this is only worthwhile if you need practice with basic trigonometry.

Position points that you are giong to use for measurements as accurately as possible by zooming in, to minimise errors. Remember that it is the “dead-center’ of points where you will actually measure to. Pro-tip: using crosses instead of circles to mark points is more precise. Next time I will!

3 points well spaced along the circumference in order to define 2 chords: FG and FH. They do not need to be visible as lines for us to be able to bisect them next:

Theorem: The center of a circle always lies along the bisector of any chord.

If you apply this theorem twice, you can find the exact center of any circle! Here I used Geogebra’s Line-Bisector tool twice, to locate the center of the circle.

I used the
“Circle with Center & Radius” tool and colored it orange, also
making it thicker and easier to see. I did this to
compare the shape of the coil to a circle centered on point I. Although not
perfect, it looks circular enough to justify using circles to model it.

Measuring the thickness of a single rope strand by carefully positioning points J & K.

IH = radius of the model circle in this image:

Next we divide the
total radius of the circle by the thickness of one strand to find how many
layers there are in the coil:

Radius / strand
thickness = 2.08 / 0.25 = 8.32

We will round down
and say there are about 8 circles – we could have just counted them of course
but I like to use more general methods when possible, so that we could adapt
this method for more difficult situations. Go and count now to double-check
though…

Yes, there are 8.

We could model the
situation as 8 separate circles, with radiuses beginning at 2.00 and reducing
by 0.25 units each time. The total length of rope would approximately be equal
to the sum of their circumferences.

In fact this is what
I will do. It would of course be more accurate to model the coil as a spiral. In this recreational situation
however it would not make enough difference to make me want to try it… If I
had the actual length to compare my answer to, and I wanted to get as close as
possible, this is what I would do.

Here are the radii
and Circumferences of the 8 circles.

Radius (r)

Circumference (2
*pi *r)

1

2

4*pi

2

1.75

3.5*pi

3

1.5

3*pi

4

1.25

2.5*pi

5

1.00

2*pi

6

0.75

1.5*pi

7

0.5

1 * pi

8

0.25

0.5 * pi

Total

18 * pi

So the total length
of the coiled section is about:

18 * Pi = 56.5 units

Plus the straight
section EF which was 5.93 gives us:

56.5 + 5.9 = 62.4
units

But how long is a unit?

Well a single 1×4
board is 3.5 inches wide (I said you might want to look it up!), so let’s do
this:

3.5 “/ 2.25 =
1.5556 inches per unit

So each unit is
about 1-1/2 inches so we’ll multiply our answer by that conversion factor:

62.4 * 1.5556 = 97.0694 inches

So let’s say about
97 inches.

With approximate
calculations like this you should round your answers to make them easy to work
with. Just be mindful of if you are rounding up or down and try to not drift
too far!

To convert from a
small unit to a larger one, always divide by the conversion factor. You will
always end up with less of the BIG unit!

Next we divide
inches by 12 to find how many feet:

The first controlled flight by humans was courtesy of a large balloon. Filled with warm air from a fire underneath its platform, it became buoyant enough to lift not only itself, but also a large wicker basket slung below it, plus two intrepid (or perhaps well-paid) Frenchmen. Once up in the air away from the fire, the air within soon cooled down, and the balloon descended back to the ground.

For decades balloons ruled the skies, even being shamefully employed to drop bombs on civilians during warfare, but they were soon literally overtaken by aeroplanes, which became a faster & safer way to fly. How would you like to take 111 hours to cruise from the USA to England on an airship? Would that really be so bad?

The modern hot-air balloon shown below is heating the air inside the balloon’s envelope with a gas burner, prior to launch. The burner is attached to the top of the basket, and fired up periodically to keep the air warm, which enables the balloon to stay aloft until it runs out of fuel. I adore the unmistakable throaty, roaring sound these huge propane burners make, and the sight of their billowing luminous flames is quite awe-inspiring for a pyromaniac like me.

“How come the balloons don’t catch on fire I always wondered, having seen images of the explosion of the Hindenberg Zeppelin. It turns out, predictably, that they are made from a fireproof material called – thank you Wikipedia! Which leads me to a question

How hot does it get in the envelope?

What temperature is required for lift-off?

Is the lift-off temperature the same at all altitudes? Imagine launching from a high plateau.

What temperature would be needed at 4000 feet above sea-level?

You have a modest balloon of 100 000 cubic feet

Some balloons are only designed for a couple of passengers, while some can carry up to 25 people at once. Their size varies a lot, so let’s just think about regular sized balloons of 100 000 cubic feet, which could carry about 4 people.

Do you know how to get started with this problem? If not, drop me a comment below and I will send you a hint. If you really don’t have time to try this problem but you want to see a solution, then check this link.

If you quite understandably wish to work in liters (I wish we all would!), then imagine a balloon that is 3 million liters, a teeny bit more than the Imperial value given above.

I have always loved the way Greek prefixes make shape names & units of measurement so easy to remember, and they were my in-road to this problem. It’ll be worth your time to learn them if you don’t already know them!

There are many ways to solve this without any chemistry knowledge, but as a Math teacher I knew heptagons are 7-sided polygons so this was my break-in point. There was only a single 7 in all the formulae so I was in luck! This told me that the prefix of the name is determined by the number of Carbon atoms in the formula.

Heptene = C_{7}H_{14}

There are 3 different types of endings for the chemical names (suffixes) which allowed me to group them into 3 categories:

-enes Heptene & butene

-anes Propane

–ynes Butyne, ethyne & propyne

I tried to draw diagrams of the compounds but did not get very far as I could not remember how their bonds work!

So, if the number of carbons in the compound determines its prefix, it must be the ratio of carbons to hydrogens that determines the suffix (otherwise there would have to be another number in the name (e.g. buhexane)).

Analysing the ratio of carbon to hydrogen atoms in the formulae revealed some patterns:

there were twice as many Hs as Cs in C_{7}H_{14} & C_{4}H_{8}

C_{3}H_{8}, C_{4}H_{6}, C_{3}H_{4} have more Hs than Cs, (but not twice as much)

C_{2}H_{2} has equal numbers of H and C

There are only 2 compounds in the -enes group so I figured that if C_{7}H_{14} is heptene, then C_{4}H_{8}must be butene, the only other -ene This implies that compounds having 4 carbons should be named ‘bu-something’.

I realised I could not group these two compounds together (C_{3}H_{8 & }C_{3}H_{4}), because the grouping was to determine the ratio & hence suffix, (with the 3 Cs giving the prefix). So I put C_{3}H_{8} into its own group because it had more than 2 Hydrogens for every Carbon. Then I put C_{2}H_{2} into the group because it had to go somewhere and I only wanted 3 groups. Not very scientific I know but it worked!

Here are the groups by ratio of C to H, to hopefully work out the suffixes.

Group 1: C_{7}H_{14,} C_{4}H_{8}

Group 2: C_{3}H_{8}

Group 3: C_{4}H_{6}, C_{3}H_{4,} C_{2}H_{2}

So the group with just a single compound must be the only -ane, so thereore: C_{3}H_{8} = propane

Group 3 must be the three “-ynes”.

The prefix ‘pro’ must mean 3 carbons, as propane had 3, so: C_{3}H_{4} = propyne

Above we figured that 4 carbons would be named bu-something so: C_{4}H_{6} = butyne

There is only one left! C_{2}H_{2} = ethyne

The rest was fairly easy. Let me know if you need any help with it. My feet are too cold to write it now – I need to go for a walk!

Match the formulae with their names. Explain your solution.

Write the names of the compounds with the following formulae: C_{2}H_{4}, C_{2}H_{6}, C_{7}H_{12}

Write the formulae for the following elements: propene, butane.

Chemistry was my bugbear science in school – I could not see all the patterns and so I had to memorise stuff. It did make more sense to me when I had learned more Physics however. Maybe our highschool teachers tried to shield us from the complexities of atomic interactions, but by not explaining it deeply, they made the reactions seem somewhat ‘magical’, and that ain’t science.

Most of us are aware that from December 21st to June 20th, days get ever longer. From miserably short snippets of Winter sun, to the seemingly endless days of Summer our planet’s Northern hemisphere slowly tilts towards our Sun. I remember reading a book during a mid-summer trip up to the Isle of Skye which lies at 57.5° latitude, roughly as far North as Hudson’s Bay. It was almost midnight and I did not need any light other than the sun to be able to read. It’s very strange!

I love noticing that it is staying lighter until later in the evenings as spring advances, and this is accompanied by the return of migratory animals and the return to life of trees and flowers. It’s a lovely time of year that brings us hope and the sprouting of healthy food.

This got me thinking: we all love this extra daylight, but how much do we actually get from one day to the next? So I decided to pose this problem, as I really enjoyed solving it, and I intend to make full use of this bonus time from now until…

Using this graph ONLY, how many extra minutes of daylight do we get each day on average from the summer to the winter equinox, in Halifax?

HINT: I used Geogebra to get accurate measurements from which to perform my calculations.

Extra Questions

Do we gain the same number of minutes of daylight per day?

How does this situation change as we go from Summer to winter equinox?

Is this value the same at all places on planet Earth?

If not, where would it be greatest? Where least?

Why does the Ecuadorian tourist board describe Ecuador as: “The land of perpetual spring”?

What do the discontinuities in the graph show? (HINT: Your country may not do this!)

NB: I know we could just Google the distance from the Earth to the Sun, but please do not – who would benefit from that? This is eduation kids!

So to solve this I used ideas I picked up from teaching ‘enlargements’, a subset of ‘geometric transformations’, in highschool Math classes. Some people would call it Thales theorem, to give credit to the chap who originally figured this out about 2500 years hence.

Step 1 I loaded up the image and used the ‘circle through 3 points’ tool in Geogebra to construct a circle which modelled the outlines of both Earth & Moon. I zoomed in a lot to enable me to position the 3 points as close to the circumferences as possible, to minimise errors.

Worked out the ratio of their actual diameters, which is in proportion to their actual radii.

Used the ratio of diameters to divide radius of Earth in photo to get what the moon’s radius would be if it were at the same distance from the camera as the Earth.

Then I looked at the difference and figured it….

the Moon appears 1.327 times larger than it should were it the same distance as the Earth. So it must be….errr. Trouble is, we don’t know where the observer is… think on Dan!

Big thank you to NASA for allowing the use of its images for our education

Okay so we will use the (quickly Googled) ‘fact’ that Earth’s diameter is about 12 742Km. TIDBIT: If you measure around the equator, Earth’s diameter is actually about 21Km greater than measured through the Poles, because the Earth is a huge spinning ball of molten metal that bulges out as it spins like a water-filled balloon!

Also we can use that the Moon’s diameter is about 3474Km

And I’m going to advise you to use my favourite software for this kind of visual modelling problem, Geogebra. Or maybe your brain works better than mine!

FYI this beautiful photo was taken by the Earth Polychromatic Imaging Camera ( truly “EPIC” ) on NASA’s DSCOVR satellite. Nice work everyone!