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V cu. ft. = (2 × L) ÷ 3 (A0 + 2 A1 + A2 + 2 A3 + A4 + ... 2 An–1 + An – (A0 + An) ÷ 2) ... (19)
In the particular problem we are using for illustration there are 10 small constituent prismoids each 2 ft. long.
In applying Eq. 20 to this particular case it is evident that the areas ABC and ARSTV of Fig. 1 Plate X are the A0 and An of that Equation and since there are 10 of these small prismoids each 2 ft. long n will be 10 and L = 2.0.
Designating the areas of the bases of these 10 constituent prismoids by the numbers of the contours on the diagram we have the areas 18, 14, 10, 6, 2 as the values of the quantities A1, A3, A5, A7 and A9, respectively or the odd A's of Eq. 20, while the areas 20, 16, 12, 8, 4 and 0 are the corresponding values of A0, A2, A4, A6, A8 and A10 or the even A's of that Equation.
Substituting all these values in Eq. 20 we have as the total volume of excavation
Since our diagram is plotted to a scale of 1" = 10', 1 Sq.. Inch = 10² = 100 sq. Ft., which being introduced into Eq. (a) gives for the total volume after reduction.
By assuming the expression within the parenthesis of Eq. (c) to be 1 Sq. inch of actual area we have
Eq. (d) shows that after having performed all of the operations indicated by the expression within the parenthesis of Eq. (a) each square inch of actual area indicated by the result of those operations when plotted to a scale of 10 ft. to 1 inch will represent a volume of 4.9382 Cu. Yds.
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