Eratosthenes' experiment on a flat earth versus on a spherical earth

Question

On a flat earth we assume that the sun is a point and that it is a distance $H$ from the ground. We put three sticks in the ground, all of height $h$. The rays of the sun will be lines from the sun to the tip of the stick and a shadow is defined as a line segment from the bottom of the stick where it was put in the ground to the intersection point of the sun ray with the ground.

On a spherical earth the sticks are stuck in the ground such that they are normal to the earth. The rays are parallel lines that go through the tips of the sticks, and the shadow is defined as the arc between the bottom point of the stick to the intersection point of the sun ray with the ground

Neil Tyson demonstrated this with wells in this video: https://youtu.be/hLPPE3_DVCw?t=318

He said that the angles somehow aren't the same when we add a third stick on a spherical and on a flat earth. I've tried to convince myself that these two models wouldn't produce the same angles this way:

Consider a stick stuck in one place on a spherical earth. The angle between the stick and the ray is $x_2$. Now consider a stick stuck at the same place the first one was stuck, but longer. The angle between the new stick and the new ray would still be $x_2$ since the two rays are parallel.

Now consider the same setup on a flat earth. The angles formed by two sticks and two rays wouldn't be the same.

Therefore, these two models aren't equivalent.

Another way I tried was to calculate those angles. On a spherical earth the angle $x$ formed by any stick and a ray would be $x = d/r$, where $d$ is the distance from that stick to a point where the ray falls on the earth at a right angle and $r$ is the radius of the earth. On a flat earth the angle would be $tan^{-1}(\frac{d}{H-h})$.

Since on a spherical earth the angle depends only on $d$ (since $r$ is constant), and on a flat earth the angle depends on the height of the stick $h$, these two models cannot be equivalent.

Are these two justifications correct? If not, what would be the correct one?

Answer 1

Both justifications are correct.

(It took me a while to understand what you meant by $d$ in the spherical model calculation though : better to say something like: $d$ is the distance—following the curvature of the earth— from the point where the stick is planted to the point where sun ray falls at a 90° angle on the earth)

But the second justification where you observe that the angle doesn't depend on $h$ is just a proof of the intuition you gained with the first justification. So I would would not call them separate.

Here's another justification: calculating the length of the shadows.

• First, flat earth (using the same parameters as you, and calling the length of the shadow $s$): Starting with $\frac{s}{s+d}=\frac{h}{H}$ (Thales' theorem) we get $s=\frac{d}{H-h}$. If the stick is as tall as the sun's height, we get an infinite shadow, as expected. (Assume $h<H$)
• Second, spherical earth: I'm using angles here (see the diagram), but remember $d=r\alpha$ and $s=r\beta$. For clarity the shadow $s$ is in red, the sun rays come from the right.

First calculate $k = (r+h)\sin\alpha$, then use that for $\alpha + \beta = \sin^{-1}\left(\frac{k}{r}\right)$, and finally $s=r\left((\alpha+\beta)-\alpha\right)$ ; plugging all back in with your parameters: $s=r\left(\sin^{-1}\left(\frac{r+h}{r}\sin\frac{d}{r}\right)\right)-d$

When the top of the stick isn't even projected onto the earth (it's too close to the earth's shadow), the above isn't defined, as expected. Just use $s=r\frac{\pi}{2}-d$ instead. When $d$ is too great (the stick, or at least its base, is in the earth's shadow), set the result to $0$.

Two very different functions for calculating the shadows, hence two very different models.

And what about the case where we have a spherical earth and a sun at distance $H$? Left as an exercise to the reader, of course.