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File: RadiusOfTheEarthPartTwo

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********
!! 2017/09/16 - Radius of the Earth - Part 2

Some nine and a bit weeks ago I posted about a method of calculating
(or estimating) TheRadiusOfTheEarth using a stopwatch and watching the
Sun at sunrise (or sunset).  When "Mike the Sundial" told me the idea
I was just stunned at the simplicity and elegance.  Colin Beveridge
took up the challenge, and you can read his account of his method
here:

* http://www.flyingcoloursmaths.co.uk/stab-colins-puzzle/

[[[>50
In a moment you'll say "Hang on - you're supposed to be calculating
the size of the Earth, but here you are assuming you know it!"

What I'm doing here is working forward from the radius and calculating
the time.  Then we can reverse the calculation - if we know the time we
can calculate the radius.  So the initial calculation is just to set
the scene, and familiarise ourselves with the calculations we'll be
doing.
]]]
When I read that I was interested to see just how different his take
was from mine, but then I realised - it comes down to the phenomenon
of "chunking".  ColinB was working - as most people would - from first
principles.

But I have a piece of knowledge that ColinB doesn't have, and I hadn't
realised just how much it affected my approach.  I know the formula
for the distance to the horizon.

Well, actually, I don't /know/ the formula, but I do know that when
you are 5 metres above sea level, the distance to the horizon is about
8000 metres.  What's more, I know how to compute that value, so if I
need to know it more accurately, I can just sketch the diagram and
compute the answer.

[[[> |>> https://www.solipsys.co.uk/images/BeforeSunrise.png ---- Before Sunrise <<| ]]]
How does that help?

Consider the minute or two before sunrise.  Let's suppose we are at
the equator, and it's the equinox. Later we can correct for all these
things, but it makes our initial calculations easier.  So the picture
here shows the moment before sunrise.  The Sun is just illuminating
the top of a 5 metre tall wall, and the sunlight is now descending
the wall.  We're timing how long it will take to get to the base

[[[< |>> https://www.solipsys.co.uk/images/AtSunrise.png ---- Sunrise for the base _ of the wall <<| ]]]
And now we see the situation as the sunlight reaches the base of the
wall.  The obvious thing to do is now consider the angle at the centre
of the Earth, /etc,/ but there is something else we can do.  We know
that the point of contact with the Earth of the Sun's rays has rolled
over the Earth's surface, and we know that it's come from 8000 metres
away.  So the time it's taken the sunlight to descend our wall is the
time taken for the tangent to roll 8000 metres around the Earth.

But we /also/ know that the Earth's circumference is 40 million metres,
because that's the original definition of the metre, so the distance
that point of contact has moved is 8000/40,000,000 or $1/5000^{th}$ of
the Earth's circumference.  So the time taken is $1/5000^{th}$ of a day.

Which is about 17.28 seconds.

"Ah!" you say, "But you're /assuming/ we know the size of the Earth!"

Yes, in all the above I've been assuming I know how big the Earth is,
and I've calculated an approximation of the time taken for light to
descend a 5 metre wall.  But now we can go back and replace all the
figures with letters and follow the calculation through.  When we do
that we get a relationship between the Earth's radius/circumference
and the time taken to descend a wall.  Instead of using 8000 we use
$\sqrt{2.R.h}$ where $R$ is the Earth's radius, and $h$ is the height
of the wall.  Instead of 40 million metres we use $C,$ which is
$2\pi{R}.$  So using $C$ as the Earth's circumference, $R=C/(2\pi),$
$D=\sqrt{2hR}=\sqrt{2hC/(2\pi)}=\sqrt{hC/\pi}.$

That means the distance the sunlight rolls as a fraction of a day is
$D/C$ which comes out as $\left(\sqrt{hC/\pi}\right)/C = \sqrt{h/(\pi{C})}.$  For
brevity we write $S=86400$, we get:

|>> $T = S\sqrt{\frac{h}{\pi{C}}}$ <<|

We can turn that around, and we get:

|>> $C = \left(\frac{S}{T}\right)^2 \frac{h}{\pi}$ <<|

We're on the final straight.  We've computed the size given the time
to descend from some height $h$ to the bottom of the wall, which is
assumed to be at sea level.  Suppose instead, as ColinB did, we go
from some height $2h$ to the lower height of $h$ - what then?

We go back to the equation telling us $T$ from $h$ and $C.$  If we
double $h$ then $T$ goes up by a factor of $\sqrt{2}.$  So that means
the time from $2h$ to $h$ is $\sqrt{2}-1$ times the time from $h$ to
0.  So take the time measured, divide by $\sqrt{2}-1,$ and use that.

So if $T$ is the time taken from $2h$ to $h$ then we use $\frac{T}{(\sqrt{2}-1)}$
and put that in the above formula.  So now we get:

|>> $C = \frac{S^2.h}{\pi{(T/(\sqrt{2}-1))^2}}$ <<|

which finally simplifies to:

|>> $C = \left(\frac{S}{T}\right)^2\frac{(\sqrt{2}-1)^2}{\pi}h$ <<|

We've used $T$ as the time in seconds.  If instead we use $t$ as the
time in days then the $S$ disappears, and if we set $h$ to be 2 then
we get:

|>> $C = \frac{0.109}{t^2}$ <<|

!! Correcting for latitude ...

[[[>50 Well, it turns out that this is wrong!  We certainly do
need to make a correction for the latitude, but it's not as
simple as dividing by $\cos(theta)$

We look at finding our latitude here:
  * LatitudeCorrection
The actual calculation for the correction is here:
  * EarthRadiusRefined
]]]
At a latitude of $\theta$ the radius we measure is a factor of
$\cos(\theta)$ too small, so we need to divide by $\cos(\theta)$.
A latitude of 53.13 degrees corresponds to a correction of 0.6.
So with $t$ in days we get:

|>> $C = \frac{0.182}{t^2}$ <<|

!! Sanity check ...

We /know/ the Earth's circumference is 40 million metres, and plugging
that in we get a time of 5.828 seconds for sunrise to descend from the
4 metre height to 2 metre height.

Better be quick!
----

|>>
| |>> <<<< Prev <<<< ---- GrepTimingAnomaly <<| | : | |>> >>>> Next >>>> ---- TheFourPointsPuzzle ... <<| |
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