First off, this measurement was not made at the north pole, but much further south (in Egypt. Also degrees, as a unit of measure of an angles, have no definition along a straight line. They only have a definition as fraction of the circumference of a circular. (The full circumference represents an angle of 360 degrees, half circle 180 degrees, etc.). They are also defined by the intersection of two lines, because you can construct a circular arc of radius 1 originating at the intersection point which begins at one line at ends at the other. You can do this physically with a protractor. So I'm not sure how you can justify labelling the north pole as 90 degrees (or any other degree measurement, really), assuming a flat earth model.
I also do not see the intersection of any lines, or anything that represents an angle, which correspond to the red line that is supposed to represent 1/25 of the radius, so I'm not how the measurement of 7°12' is coming into play here.
7°12' is 1/25th of 180 degrees.
If you move North to South from one side to the Sun's area of light to the other side the Sun will rise from the horizon and make 180 degrees arc over your head.
When you say "move North to South", I presume you mean from a given point towards the closest point on the outer edge of the disc (which would be away from the North Pole).
Herein lies the problem. It will be impossible for an observer on a flat disc to ever observe the Sun at/below the horizon, at the same time another observer at another point of the disc observes it directly overhead, (or really, any point substantially above the horizon). Let me illustrate. Keep in mind this is a side view.
Observer O observe the sun directly overhead (90 degrees). Observer A will observe at some angle < 90 degrees. Observer B will observe the sun at a some angle < 90 degrees. As we imagine observer B getting farther and farther away from A and O, the angle keeps diminishing, this is true. But you cannot imagine even getting close to the horizon (an angle of 0 degrees), until observer B is very far away.
But we can better than just intuitive observations - we can actually do the math to determine how far B must be from A or O, in order to observe the sun at some given angle.
I'll use the following convention. ang(X) is the angle formed at point X, between the line from X to the sun, and line of the ground. In terms of Erathosnes, it would be 90 - the angle of the shadow formed in his stick experiment. For example, ang(O) = 90 degrees. dist(X, Y) is the distance from point X to Y. Also, though I didn't label it, I'll use S for the sun.
This following requires trigonmetry. Since we have a right triangle (one where one of the angles is 90 degrees), we can easily solve for lengths if we know one of them and angle between one of the sides.
In a right triangle, there will two line segments which are short, and a third line segment which is longer than the other two. This is called the hypotenuse.
There is a cool function, called "sine", which is defined for a given angle, assuming the hypotenuse is 1. It is writen as sin(X), reading as "the sine of X" sin(X) would be the length of the side opposite that angle. Since the hypotenuse is 1, sine can never assume a value > 1. In a real right triangle, where the hypotenuse > 1, we can simply multiply sin(X) by the hypotenuse to get the length of the opposite side.
There is also a similar function called "cosine", similar to sin. This gives you the length of the adjacent side to the angle. It is written as cos(X). cos(X) * length of hypotenuse gives the length of the side
Also I shoudl note, when using angles in the sine and cosine functions, you need to plug them in as "radians". Radians are defined as the length of the circular arc formed by the angle assuming a radius of 1. The circumfrence of a half circle is defined to be PI, which is roughly 3.141527. This correspondes to 180 degrees. So the conversion from degrees to radians is radians = PI * degrees / 180.
Looking at the diagram again, we can see both triangles share a common side, which is dist(O, S). We can use trigonometry to solve for the dist(B, O), if we have enough information.
In this case, we know ang(A), we know ang(O), we know dist(A, O), and we are presuming ang(B) The issue is to determine the dist(B, O) which satisfies the observations.
Now lets write down some equations, just going by the trigonometry
1) dist(B, S)*sin(B) = dist(O, S)
2) dist(B, S)*cos(B) = dist(B, O)
3) dist(A, S)*sin(A) = dist(O, S)
4) dist(A, S)*cos(A) = dist(A, O)
Equation (2) is the one we want to solve for, dist(B, O). However, we do not know dist(B, S). But we can use equation (1) to solve for dist(B,S). Dividing both sides by sin(B), we get
5) dist(B,S) = dist(O,S)/sin(B).
Now the problem is we don't know dist(O,S). But we can use equations 3) and 4) to solve for it. Equation 3) gives us the exac trelation - the only problem is that we don't know dist(A,S). But we can use equation 4, dividing both sides by cos(A), we get
6) dist(A, S) = dist(A,O)/cos(A)
Plugging this back into (3), we get
3) dist(A, S) * sin(A) = dist(O, S)
7) (dist(A,O)/cos(A)) * sin(A) = dist(O, S)
Now plugging this back into equation (5)
5) dist(B,S) = dist(O,S)/sin(B).
9) dist(B,S) = [ dist(A,O)/cos(A)) * sin(A) ] /sin(B).
Now we plug this back into (2) to solve for dist(B, O)
2) dist(B, S)*cos(B) = dist(B, O)
10) [[ dist(A,O)/cos(A)) * sin(A) ] /sin(B) ] * cos(B) = dist(B.O)
And we ended up with a mess, however a mess that can be calculated.
Now let's plug in Erathosnes numbers, and a reasonable number for ang(B), like lets say 5 degrees. He observe an angle of 7.2 degrees, which means that ang(A) = 90 - 7.2 = 82.8 degrees angle of the sun above the horizon. The distance, dist(A, O) was 500 miles.
So dist(B,O) = [[ (dist(A,O)/cos(A)) * sin(A) ] /sin(B) ] * cos(B)
= [[ (500 / cos(A * PI/180)) * sin(A * PI/ 180) ] / sin (B * PI / 180) ] * cos ( B*PI/180)
= 45239.09 miles.
The claimed diameter of the Earth in the wiki was 25,000 miles. And we weren't actually measuring the diameter, but something less than the radius. What that means the diameter woud have to be more than twice our number, so > 90,000 miles.
And this is assuming an angle of 5 degrees, as a I felt that was reasonable. But lets see what happens to our numbers as they get closer to 0 degrees.
For 2 degrees, we have dist(B, O) = 113339 miles ( to the nearest mile)
For 1 degrees, we have dist(B, O) = 226748 miles (to the nearest mile )
For 0.1 degrees we have dist(B, O = 2267711 miles (to the nearest mile)
So we can see whats going on, as we get closer and closer to 0, the distance is increasing very quickly. In fact looking back out our equation, since we are dividing by sin(B), it cannot be zero otherwise the value is undefined. sin(B) is only zero when B itself is 0, i.e, the sun is at the horizon. The model predicts that we can never get far enough away, to actually observe this.
Now, I suppose you could introduce that the idea that the light from the sun is doing something funny which causes it not to move in a straight line (refracting, accelerating, etc.) causing the observation you get. But, when do that, it trashes the basic geometry of the problem, such that he observations themselves worthless without knowing what the light is doing/ why it is doing that.
Therefore we can take the known distance and multiply by 25 to get that distance across the sunlight area. If we think that the Sun is moving in a circle like the Monopole model we can double our figure for the total diameter.
It's really just measuring the sunlight area on one side and then multiplying by two to get dimensions for a model.
I determined that the sun goes around the Monopole model because I read the FAQ.
I determined that the sun would rise and eventually set when you move towards it from one side of its area of light to the other based the sun's rising and setting over the course of a day.
I determined that the Sun would move consistently or near-consistently in the sky by observation.
The sun moving consistently in the sky at a given point, over the course of the year, is another problem (it doesn't, it varies over the course of a year at any given point), but I'll put that issue aside for now. Right now I want to focus on the basic geometrical reasoning.
So, in conclusion, it is my position that, assuming a flat Earth, Eratosthenes experiment can tell you nothing about the diameter of the Earth.
As a addendum, I'll note that I am not trying to necessarily argue for an alternate mode such as RET. I am simply working out the claims of FET, as claimed in that section of the Wiki, to its rational conclusions given basic geometrical reasoning.