How does this not disprove a flat earth?
Erastothenes alone doesn't disprove a FE.
Earth was known to be round before this experiment, with this experiment just determining the size of Earth.
By using just 2 measurements you have an unsolved problem.
You need to determine the height of the sun and the curvature of Earth.
In general (and using radians):
You have a point directly below the sun, where the angle of elevation to the sun is pi/2, with the sun a distance of h above it.
You have another point, some distance l along the surface of Earth, where the sun is at an angle of elevation a.
This allows us to construct a right angle triangle where one side is of height h2, and you have an angle of b.
From this we can also get angle c, such that a+c=b.
Note that for an inside out Earth (with us on the inside of Earth), c is negative.
There is also the horizontal distance d between the 2 points.
And the difference in height dh of the sun between the 2 points.
c is also the angle at the centre of Earth.
That means c*R=l.
Or to express it in terms of curvature where C=1/r, c/C=l, c=l*C
This means b=a+l*C.
The horizontal distance is given as d=R*sin(c), noting that if c is negative and R is negative, this is still positive.
Again, to express in terms of curvature this is d=sin(l*C)/C.
If C is 0, then it is simply d=l (otherwise you have pesky divide by 0 errors, but this is also the limit as C approaches 0)).
dh can also be given with a similar term, dh=R*(1-cos(c)).
Note that in this case if R is negative and c is negative, then dh is negative, as cos(c) will still be positive and less than 1.
Again, to express this in terms of the curvature we have dh=(1-cos(l*C))/C.
This now means that h2=h+dh=h+(1-cos(l*C))/C
So now we have our right angle triangle, with an angle of b, the opposite side of h2 and the adjacent side of d.
i.e. tan(b)=h2/d
i.e. tan(a+l*C)=[h+(1-cos(l*C))/C]/[sin(l*C)/C]
Noting that if C=0, then the equation is better expressed as:
tan(a)=h/l, again to avoid the divide by 0 errors and also is the limit as C approaches 0.
And if h is very large, such that the 2 lines are basically parallel, it instead simplifies as b=~pi/2, thus from
b=a+l*C
we get pi/2=a+l*C
This is an equation with 4 terms, a, l, C and h.
From the experiment 2 are know, a and l.
C and h are unknown and varying one will vary the other.
So in order to solve it, you need to know one of them.
And if you have limits on one you can place limits on the other.
It was known that h is very large, many times the size of Earth and thus practically infinite. This allows us to use h=infinity as an approximation and get pi/2=a+l*C
So you either need to use something else, like observations of the sun remaining roughly the same size throughout the day and regardless of location to establish the sun is many times the distance to Earth and thus h is approximately infinity.
Or alternatively, you need more points, which are not symmetric (e.g. if you had a point 100 m away from the sub-solar point, it is pointless to go to another point 100 m below the sub solar point, as they give you the same a and l).
With just a single point you have a single equation, with 2 unknowns, h and C, and 2 knowns, a1 and l1.
With 2 points, you have 2 equations, with 2 unknowns, h and C, and 4 knowns, a1, a2, l1 and l2.
You can then solve those 2 simultaneous equations and determine what h and C are, or at least put limits on them.
So Eratosthenes doesn't prove a round Earth, but the observed angle of elevation to the sun or any celestial object does.