Maximum Circles on Rectangular Field Diagonals

Diagonal length = $\sqrt{40^2 + 30^2} = 50$ m. For a circle of radius $r = 1.25$ m touching two adjacent sides of the corner (angle $\theta$ at corner where $\tan \theta = 30/40 = 3/4$), $\sin(\theta/2)$ determines distance from vertex to center = $1.25 \times \sqrt{1 + (3/4)^2}/... = 1.25 \sqrt{2}$ roughly. Usable diagonal length is $50 - 2 \times 0.51 = 48.98$ m. Max circles per diagonal = $\lfloor 48.98 / 2.5 \rfloor = 19$. Since 1 circle can be placed exactly at the intersection of the two diagonals (common to both), each half diagonal gets 9 circles. Total circles = $4 \times 9 + 1 = 37$.