Circumference of the earth at the equator (latitude 0) C_{E} = 40 075 km

In order to simplify we will take 40 000 km.

Circumference of a given latitude circle =
C_{l} = C_{E} * cos l

Thus the circumference of a latitude circle is greatest at the equator (cos 0 = 1) and zero at the poles (cos 90° = 0).

The circles of longitude all have the same length. If we consider the earth as perfectly spherical, these circles will have the same circumference as that of the earth at the equator, namely 40 000 km.

Please look at the visibility curve for 7^{th} August 2013 as well as the diagram.

We have chosen as the point of reference for our calculation of the extension of the observational zone the point 40° S and 9° E. This point is situated in the blue zone of the visibility curve for 7^{th} August 2013. The blue zone represents naked eye visibility under perfect conditions of observation. Like all the other zones, the blue zone is a parabola. In the diagram, R represents our point of reference.

The tip of the blue zone in the visibility curve is situated at 42° S and 12° E. We have represented this point by P in the diagram.

In order to evaluate in kilometres the distance between the tip of the blue zone at P and the point of reference R, we will use latitude 41° (intermediate between the latitude of the tip of the blue zone P and that of the reference point R).

Circumference of the circle of latitude at 41° = 40 000 * cos 41° = 40 000 * 0.755

= 30 200 km

On this circle, each step of one degree in longitude is = 30 200/360° = 84 km

The distance between the points P and R is 3° in longitude. In kilometres, this corresponds to 84 * 3 = 252 km.

We follow the longitude at the point of reference R for our calculations, that is to say, 9°E. We then find that the blue zone of visibility terminates at 33° latitude towards the north and 49° latitude towards the south. Thus, in latitudinal degrees, the extension of this zone is 16°. These points are marked as N and S on the diagram.

Each circle of longitude has the same circumference, 40 000 km approximately. Thus a step of 1 degree in latitude on any circle of longitude represents:

40000/360 = 111 km

Consequently, the distance between N and S = 16° in latitude = 1776 km

Our zone of visibility for the crescent of 7^{th} August 2013 comprises the parabola defined by the points P, N and S.

Using a mathematical technique known as approximating an integral by the method of rectangles, which we do not develop here, it can be shown that the area of the parabola defined by the points P, N and S =

4/3 * the area of the triangle PNS

Now the area of the triangle PNS = ½ * NS * PR

= ½ * 1776 * 252

= 223 776 km^{2}

Thus the area of the parabola (zone of observation)

= 4/3 * 223 776 = 298 368 km^{2}

We will retain, by way of approximation, for the surface of our observational zone the value of 300 000 km^{2}.

*Some comparisons: *

- The surface of the earth is 510 000 000 km^{2}. Thus our zone of observation covers 0.06 % of the terrestrial surface.

- The surface of France is 674 846 km^{2}. Thus our zone of observation covers 44 % of the surface of France.

- The surface of France is 674 846 km

Our visibility predictions are calculated according to the visibility of the young crescent at the point R in the blue zone of visibility. It is important to note that we are speaking about naked eye visibility. The visibility is real and not supposed so.

If, in a given country, the Fajr prayer is **after** the visibility of the crescent at the point of reference R, then Shawwal can begin on that day. Otherwise, it will begin on the next.

But we have seen that the blue zone in which our reference point R is situated represents a naked eye visibility of the crescent under perfect observational conditions. It is not evident that these conditions will be realised for a unique point like R. It is sometimes estimated that, for a unique point, the probability of really observing the crescent is of the order of 70 %.

However, we have to note an important point: to the **east** of the line NS, the crescent will be visible **before** the visibility at the point R itself. Thus, if the Fajr prayer is after visibility at the point R, it will be even more so if this visibility occurs earlier than at the point R. Thus, we can consider that if our calculations are right for the point R, they will also be right for the totality of the zone of observation represented by the parabola comprised between the points P, N and S.

What does this change? The probability of observing the crescent with the naked eye is no longer limited to a unique point R, but takes into account the entire zone of observation, which means a surface of 300 000 km^{2}. Thus, even if we base our calculations on the visibility of the crescent at the point R, in reality, this visibility relates to a very large surface area. We thus have almost complete certainty that the crescent shall be seen as per our prediction.