If the elongation factor during rolling of an ingot is 1.22.

If the elongation factor during rolling of an ingot is 1.22. The minimum number of passes needed to produce a section 250 x 250 mm from an ingot of 750 × 750 mm are

Elongation factor =	A_i	= 1.22
	A_r

Let n → number of passes

750 × 750	= 250 × 250
(1.11)ⁿ

(1 – 22)ⁿ = 9
n/n 1.22 = 9
n = 11.04