The city of "Aswan, Egypt" lies around 24° Latitude and, in the time of Eratosthenes, there was a well in this city (called Syene at that time) and it was known that the Noon sun on the Summer Solstice would shine directly down the well, casting no shadow.
NOTE: the exact latitude for the Tropic of Cancer (the point where the Summer Solstice sun is directly overhead) changes slightly over the years. So to repeat this measurement today you would need a slightly different location along ~23°26′13.7″N (where does Flat Earth model explain that? Spherical Earth model explains it due to precession of rotating Earth).
Eratosthenes thought to measure the angle of a shadow cast in a distant city (Alexandria, Egypt) during the Solstice - and, using basic geometry, to therefore measure the curvature of the Earth between the two locations (both are within a few degrees longitude of each other).
He found the angle cast by the sun in Alexandria was about 7°14’ from true, and he estimated the distance between Syene and Alexandria at about 5000 stadia, which meant the circumference of the Earth was about 250,000 stadia (or about 50 times the distance between the two cities). This is because 7°14’ is just about 1/50th of a full circle (360°).
That distance is approximately 522 miles which means it would be about 26,100 miles around - the polar (north-south) circumference of the Earth is actually 24,874 miles so that is pretty amazing since Eratosthenes would have had to have measured the distance of 522 miles the hard way (surveying is very subject to measurement error, even little bitty tiny errors in measuring an angle can throw off the final result by many miles).
Now, there are TWO possible ways to account for the increasing angle of the shadow...
#1 the sun is far away and the angle is (mostly) a product of a spherical Earth.
#2 the sun is close and the angle is just because of the position of the sun in the sky.
However, we find that Option #2 is ruled out because the angles measured at different latitudes do not agree with the linear angle model (they get stepper exponentially with distance). This is impossible on a flat earth (see details).
What about Sun Size
Earth radius is ~3959 miles. The sun is 92.96 million miles away. That is about 23,480 Earth radii away -- there is no reasonable way I can make a picture that wide.. This great distance is why we says the Sun's rays are almost parallel. But actually the Sun is also very large so those rays are not exactly parallel and in future posts we will probably touch on this as it is why shadows are fuzzy on the edges unless the object is very close to the ground. When you measure actual shadow lengths you get the results above that match a spheroid Earth with a radius of ~3959 miles.
When we look at the sun (very carefully so we don't damage our eyes) we find that the angular size of the sun (how big it LOOKS to us) is 30 arc minutes across the sky. This is true all day, the angular size of the sun in the sky does NOT get larger and smaller throughout the day (this is another way we know the flat earth model is incorrect).
If the Sun is 92.96 million miles away and has a radius of 432287.95 miles how big should it appear to us when we view it in the sky?
Well, if we made a right triangle from our viewing point to the center of the sun (92.96 million miles) then a leg out to the edge then we could solve for that angle, knowing that the tangent of the angle Θ = sun radius / distance to sun, then we just need to double that.
tan Θ = 432287.95 / 92960000
So we can take the arctan of that value to get our angle.
arctan(432287.95 / 92960000)*2 in arcminutes = 31.97' or 0.5329° (one-half degree)
This is how you compute apparent angular size of a distant object if you know how big it is.
You can also use this Angular size calculator - but note that it doesn't double the result.
However, since the sun is VERY far away this doesn't matter - you can enter distance = 92960000 and Size = 864575.9 and you still get 0.53288 degrees. But if you are measuring things nearby you need to be VERY accurate in your measurements.
What other experiments can you think of to do that would prove the sun is very distant?
Using your "Flat Earth" model (however far away and large or small you imagine the sun to be) how would the angular size of your 'Sun' change through a Flat Earth day?
How can you measure the actual angular size of the sun without hurting your eyes?
Maybe try this method: http://www.eaae-astronomy.org/WG3-SS/WorkShops/Sundiam.html
Since the Earth orbits the Sun in a slight ellipse can you measure the change in angular size of the sun? Does that measurement agree with the heliocentric model? (yes, yes it does)
Educated people have known for a very long time the Earth is not flat
Earth radius is ~3959 miles. The sun is 92.96 million miles away. That is about 23,480 Earth radii away -- there is no reasonable way I can make a picture that wide.. This great distance is why we says the Sun's rays are almost parallel. But actually the Sun is also very large so those rays are not exactly parallel and in future posts we will probably touch on this as it is why shadows are fuzzy on the edges unless the object is very close to the ground. When you measure actual shadow lengths you get the results above that match a spheroid Earth with a radius of ~3959 miles.
When we look at the sun (very carefully so we don't damage our eyes) we find that the angular size of the sun (how big it LOOKS to us) is 30 arc minutes across the sky. This is true all day, the angular size of the sun in the sky does NOT get larger and smaller throughout the day (this is another way we know the flat earth model is incorrect).
If the Sun is 92.96 million miles away and has a radius of 432287.95 miles how big should it appear to us when we view it in the sky?
Well, if we made a right triangle from our viewing point to the center of the sun (92.96 million miles) then a leg out to the edge then we could solve for that angle, knowing that the tangent of the angle Θ = sun radius / distance to sun, then we just need to double that.
tan Θ = 432287.95 / 92960000
So we can take the arctan of that value to get our angle.
arctan(432287.95 / 92960000)*2 in arcminutes = 31.97' or 0.5329° (one-half degree)
This is how you compute apparent angular size of a distant object if you know how big it is.
You can also use this Angular size calculator - but note that it doesn't double the result.
However, since the sun is VERY far away this doesn't matter - you can enter distance = 92960000 and Size = 864575.9 and you still get 0.53288 degrees. But if you are measuring things nearby you need to be VERY accurate in your measurements.
What other experiments can you think of to do that would prove the sun is very distant?
Using your "Flat Earth" model (however far away and large or small you imagine the sun to be) how would the angular size of your 'Sun' change through a Flat Earth day?
How can you measure the actual angular size of the sun without hurting your eyes?
Maybe try this method: http://www.eaae-astronomy.org/WG3-SS/WorkShops/Sundiam.html
Since the Earth orbits the Sun in a slight ellipse can you measure the change in angular size of the sun? Does that measurement agree with the heliocentric model? (yes, yes it does)
Educated people have known for a very long time the Earth is not flat
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