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CHAPTER VI.

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Quantities of Materials.

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I. Volumes from Cross Sections.

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2. Volumes of Single Prismoids.

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Application of the Prismoidal Formula.

As generally written the Prismoidal formula is:
V = L ÷ 6 (A_{0} + 4 A_{m} + A_{1})
... (1)
in which
V = Vol. of Prismoid in Cu. Ft.

L = Dist. in Ft. bet. End Sections.

A_{0} = Area in Sq. Ft. of one End Area or Section.

A_{1} = Area in Sq. Ft. of other end Area or Section.

A_{m} = Area in Sq. Ft. of a Section of Prismoid midway bet.,
the two sections, the lengths of the sides of this middle Section being
the means of the lengths of the corresponding sides of the two End Sections.

When the volume of the given Prismoid is required in Cu. Yds. Eq.
1 becomes:

V Cu. Yds. = L ÷ (6 × 27) (A_{0} + 4 A_{m}
+ A_{1}) ... (2)
If L be taken as 100 ft., which it usually is in the calculation of
railroad excavation, Eq. 2 becomes:

V Cu. Yds. = 100 ÷ (6 × 27) (A_{0} + 4 A_{m}
+ A_{1}) ... (3)
In Fig. 1 of Plate IV, let us suppose that
we have plotted the two End Sections A_{0} and A_{1} as
there shown; Section A_{0} being ABCPD and Section A_{1}
being ABHNG, both sections being plotted to a scale of say 1 inch = 8 ft.,
and both sections having, as in all railroad sections, the same width of
road bed and the same ratio of side slopes.

Since these sections are plotted to a lineal scale of 1 in. = 8 ft.,
1 sq. in. of section will represent 8² = 64 sq. ft., and we have as
the Volume represented by the two sections:

V Cu. Yds. = (100 × 64) ÷ (6 × 27) (A_{0}
+ 4 A_{m} + A_{1}) ... (4)
or by reduction
V Cu. Yds. = 39.5061 (A_{0} + 4 A_{m} + A_{1})
... (5)