Talk:2021-04-23 53 9

I am sure you can calculate the probability yourself, but anyway here is my approximation:

Consider a circle of radius r at latitude θ and longitude ϕ. In the latitude/longitude coordinate space, if the circle is small (r ≪ graticule dimensions), it will look like an ellipse with semi-major axis ϕ_r and semi-minor axis θ_r. The semi-major axis will always be longitudinal due to the non-square shapes of the graticules. The lengths of the axes are:

ϕ_r = 1° × r / w_grat,

θ_r = 1° × r / h_grat,

where w_grat and h_grat are the width and the height of the graticule respectively at the centre point of the ellipse. Assuming that the Earth is a sphere with radius R = 6371000 m and taking θ = 53.6401767° from 2021-04-10_53_9, the dimensions of the graticule are approximately

h_grat = π R / 180 = 111195 m,

w_grat = h_grat cos θ = 65922 m.

Now, if r = 384 m, we get the axes of the ellipse:

ϕ_r = 0.00583°,

θ_r = 0.00345°.

The probability P₁ that a random hashpoint hits this ellipse is simply the area of the ellipse divided by the area of the graticule:

P₁ = (π ϕ_r θ_r) / (1° × 1°) = 6.32 × 10⁻⁵ = 1 / 15800.

Finally, the probability P₂ that a hashpoint hits the ellipse at least once in 13 days is

P₂ = 1 − (1 − P₁)¹³ = 8.21 × 10⁻⁴ = 1 / 1220.

An event of this probability will happen on average about once in 3.33 years or three years and four months.

― Kripakko (talk) 13:21, 24 April 2021 (UTC)