I think Aristotle solved this one way back in the day.
At a given moment in time, measure the angle of the sun (aristotle did this by looking at the shadow cast in a deep well.) Triangulate. Do this a couple of times in different locations to convince yourself you haven’t accidentaly passed upon, say, a round part of your frisbee, and your done. You can even determine the worlds circumference this way.
Eristosthenes, not Aristotle. And he used two known points, and made both measurements on noon of the Solstice (this is a necessary step, since the position of the Sun in the sky changes both with the time of day and of the year). But while you can determine the radius of the Earth this way (assuming it’s spherical), you can’t rule out a Frisbee unless you travel to locations all around the world (which the questioner ruled out).
There’s a much simpler home-demonstration.
Drop something. Observe that it falls perpendicular to the earth’s surface.
Drop it somewhere else. It still falls perpendicular.
Repeat as often as you like, wherever you like. For the object to always fall perpendicular to the earth’s surface, the earth has to be spherical. If it was a hemisphere, frisbee etc., the centre of gravity would not be in the same place, so the dropped object would only fall perpendicularly(?) at one specific point.
However, I suppose I also now need to prove that the earth doesn’t have some freaky changes in density that make this proof not work. (The boat-behind-horizon effect is the same anywhere, which I believe is enough to prove that the earth’s curvature has a constant radius.)
This proves absolutely nothing except that things always fall perpendicular to the ground, unless you’ve previously demostrated that gravity always pulls towards the center of mass. Do that first, and then you’ve got something.
The Earth’s shadow on the Moon during a Lunar eclipse is the easiest to do. No need to do anything but wait and watch.
I do believe Cecil was slightly off on the answer, though. A round disk straight on would not produce the type of shadow we see (penumbra and umbra). Unless my astronomy books and magazines have been wrong for all these years. The shadow’s shape and characteristics show the Earth to be a near spherical globe.
Unless I’m wrong. But, if I am, show me.
Sure it would. The penumbra/umbra structure is formed because the sun is a finite size. If it was a pinpoint, the shadow of the earth would be sharp and distinct. Because the sun is fairly large, the shadow gets fuzzy.
No, it wouldn’t. The Earth has an atmosphere, which acts somewhat as lens. And the amosphere is full of clouds and dust, which tend to diffract the light even more.
Okay, so it’s not the umbra/penumbra aspect of the shadow.
But, the shadow is always round. So, after viewing sevral eclipses, one could see that, regardless of the starting point of the shadow’s traversing the Moon, it is always round. (The starting points being different at times due to the slight eccentricities of orbits of the main players in relation each other) Only a sphere could cast a round shadow every time.
Is that the proper reasoning on this matter?
Just as a side point:
Wasn’t accepting the Earth as a sphere pretty important to the calulations of Aristarchus when he found the size and distance to the Moon c. 250 BCE?
A side point to the side point:
How did Aristarchus do without Trig? Was the mathematics of Pythagoras enough to work with (was that really trig)? What exactly did Hipparchus add to trig?
Just side I’m wondering about…
[NITPICK] Eratosthenes, not Eristosthenes. [/NITPICK]