Measuring the Earth with camels and sticks

[Before I begin, I strongly recommend that you read [Calculating the Size of the Earth](https://brilliant.org/assessment/techniques-trainer/calculating-the-size-of-the-earth/) in the techniques trainer section. The image below is taken from there (with slight modifications)]

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I'll use this image as a reference. We're calling $PQ$ the lighthouse of Alexandria which is $130$ meters tall. $QS$ is the distance between Syene and Alexandria [ $928$ kilometers]. And $QR$ is the shadow of the lighthouse.

Let's determine $\angle A$ .

Since $QR$ is really small compared to the circumference of the earth, we can consider $QR$ to be a straight line.

So in the right triangle $PQR$ , $\angle A=\tan^{-1} \frac{16}{130}=0.1225$ radians approximately.

$\angle A$ is equal to $\angle B$ because they're alternate angles.

And the circumference of the earth is $\frac{2\pi}{0.1225}\times QS= \frac{2\pi}{0.1225}\times 928=\boxed{47726}$ kilometers approximately.

Let $θ_{lighthouse}$ be the angle created by the lighthouse and its shadow. So, computing for $θ_{lighthouse}$ :

• tan $θ_{lighthouse}$ = $\frac{16}{130}$
• arc tan ( $\frac{16}{130}$ )= $θ_{lighthouse}$
• $θ_{lighthouse}$ = 7.016501745

Taking for account that the rays are parallel, we can say that the angle formed by the lighthouse-shadow is the alternate interior angle of the earth formed from the distance between Alexandria and Syene. Thus, $θ_{earth}$ - which is the central angle formed by the distance from Alexandria and Syene is equal to the angle formed in the lighthouse-shadow.

$θ_{lighthouse}$ = $θ_{earth}$

So computing for earth's circumference : (In my solution, I used means and extreme) - Let x be the circumference of the earth

$θ_{earth}$ : $d_{Alexandria->Syene }$ , 360 : x

x = $d_{Alexandria->Syene } \times 360$ / $θ_{earth}$

Substituing the values :

x = $928 km \times 360$ / 7.016501745

x = $\boxed{47726 km}$

The angle made by the sun rays at the lighthouse is $\theta = \arctan(130/16)$ . The angle made by the sun rays above Syene is $90^\circ$ . Since we can assume the Earth is spherical, therefore it's cross-sections are circles. Syene and the lighthouse lie on the edge of a circle, the distance between them, $L$ , is $928m$ and the angle between them is $\phi = 90^\circ - \theta = (90 - 82.98)^\circ = 7.02^\circ$

Now we can get the radius of circle like this:

Convert $\phi$ to radians: $\phi_{r} = 0.122$

$r = \frac{L}{\phi_{r}} = \frac{928}{0.122} = 7606 km$

$\therefore$ Circumference = $2\pi r= 2 \times 3.14 \times 7606km = 47765km$

Edit: This is not very accurate but hey! we are in ancient Greece!

In Alexandria the Sun makes a angle of $\theta=tan^-1(\frac{16}{130})=0.12246$ with the normal with the surface.

Because we consider earth as a sphere we can represent the angle $\theta$ as the angle between two points on a circunference with a distance of $928 km$ between them.

Now we just make a proportion $\frac{\theta}{2\pi}=\frac{928}{C}$ where $C$ is the circumference of the Earth. Solving we have $C=47726$ .