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What will be the correction for curvature of three level sections, if the areas are given as 43.56 sq. m, 22.54sq.m and h = 56.43 sq. m, b= 2, n= 2 which are having faces at a unity distance having 59.87 m radius?

What will be the correction for curvature of three level sections, if the areas are given as 43.56 sq. m, 22.54sq.m and h = 56.43 sq. m, b= 2, n= 2 which are having faces at a unity distance having 59.87 m radius? Correct Answer 220.19 sq. m

The formula of correction for curvature can be given as C = d*(w2 – w12)*(h + b/2n) / 6R. On substitution, we get C = 1*(43.562 – 22.542)*(56.43 + 2/2*2) / 6*59.87 C = 220.19 sq. m.

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