# How did ancient physicists around 500 BC decide that the earth is spherical?

One of the standard stories that is given for this (and is mentioned in Aristotle) is that the hull of a ship disappears first as it sails towards the horizon on a calm sunny day. Is this a myth, or is it possible? After all, at the time no telescopes were available.

Scenario: Assume, for instance the hull of the ship rides 1 metre above the surface of the sea, with a sail whose height is 3 metres (I am going by 'Min of the Desert', a replica of a 3800 years old oceangoing Egyptian ship, so about the right time period.)

1. How far must it travel before the curvature of the earth completely hides the hull?
2. How far must it travel (assuming a flat earth) before visual acuity of a normal naked eye can no longer discern it?

I get an answer for question 1 of around 120km, this seems too distant (by my unaided intuition) for the naked eye to resolve. Also it seems wrong from general observation of the horizon from the coast, as one of the commenters has pointed out it looks like only a few kilometers. And surely at that kind of distance atmospheric efects begin to kick in, especially in the mediterrean where its generally warm.

Its been a very long time since I've done geometrical optics, so the answer to question 2 is beyond me. Can someone help. Also a verification of 1 would be nice (I've used a radius of the earth as 6000km).

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"I get an answer for question 1 of around 120km, this seems too distant" Much too far. If you are standing near sea level it's only a handful of kilometers. –  dmckee Jun 10 '12 at 20:15
There's another way, the shadow of Earth on the moon. –  stupidity Jun 10 '12 at 20:16
And the famous differences in sun shadows down vertical wells, separated some distance apart. –  ja72 Jun 11 '12 at 17:50

Use:

$L = R(\arccos{(\frac{R}{R+h})} +\arccos{(\frac{R}{R+p})})$

Where $h$ is the height of the hull above the surface of the earth and $p$ is the height of the observer.

$L = 6.378\times10^{6}m\times (\arccos{\frac{6.378\times10^{6}m}{6.378\times10^{6}m+1m}} + \arccos{\frac{6.378\times10^{6}m}{6.378\times10^{6}m+1.776m}}) \approx 8331.309m$

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For the hull question, make this construction:

• We draw a triangle CAB where C is the center of the earth, A is the top of the observer head (which will make 2 meters over the water) and B is the top of the hull (1 meter over the water) at the last point it can be seen (meaning that AB just grazes the water).

• Draw the altitude from C to the point D where AB is tangent to the water).

We have two right triangles there, and the Pythagorean theorem gives us the lengths

$$L_{AD}^2 + R^2 = (R + a)^2$$

where I have written $L_{AD}$ for the length of segment $AD$, $R$ for the radius of the Earth and $a$ for the height of the observer. From that

$$L_{AD} = a\sqrt{1 + 2\frac{R}{a}}$$

and

$$L_{BD} = b\sqrt{1 + 2\frac{R}{b}} .$$

Punch the numbers and you get a bit over 8 km for the length of $AB$.

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Great. My mistake was that I didn't factor in the height of the observer, I assumed that the observer was level with the surface of the earth. And your answer tallies with general observation of the horizon. –  Mozibur Ullah Jun 10 '12 at 20:39