. The elements of railroad engineering . es a -^ ^? = —~^— X 9 = sq. ft. 31 8 111Area of triangles r + ^/ = —-—^— X 8 = sq. ft. Total area of section A ~ sq. ft. Taking the dimensions of the section B given in Fig. 485,we have 1020 RAILROAD CONSTRUCTION. 9 7 -j_ 2 2Area of triangles a-\- b — —-—^—-— X 0 = sq. ft. Area of triangles <:+ d— ^ t — X 5 = sq. ft. Total area of section B = sq. ft. ;r r • 1 1 71 4- ,^„ ^Mean area of sections A and h = = sq. ft. Contents of the included prismoid = x 50 = 8,790cu. ft. = cu. yd. In
. The elements of railroad engineering . es a -^ ^? = —~^— X 9 = sq. ft. 31 8 111Area of triangles r + ^/ = —-—^— X 8 = sq. ft. Total area of section A ~ sq. ft. Taking the dimensions of the section B given in Fig. 485,we have 1020 RAILROAD CONSTRUCTION. 9 7 -j_ 2 2Area of triangles a-\- b — —-—^—-— X 0 = sq. ft. Area of triangles <:+ d— ^ t — X 5 = sq. ft. Total area of section B = sq. ft. ;r r • 1 1 71 4- ,^„ ^Mean area of sections A and h = = sq. ft. Contents of the included prismoid = x 50 = 8,790cu. ft. = cu. yd. In apulying the prismoidal formula we calculate the areaof a section midway between the given sections, and for itsdimensions we take the mean of the dimensions of the givensections. These dimensions will be as follows: Center cut, 8 + 5 ft. 14+ side distance, ^^ = ft. ^ , ., ,. + ,Lett side distance, — ft. With these dimensions, construct the section J/shown inFig. 486. 20Ji-. FlG. 486. The area of section Jl/ is computed l)y the same methodas that used with sections A and B in Figs. 484 and 485,and is as follows: RAILROAD CONSTRUCTION. 1021 Area of triangles a + b = lL^±Ji^ X 9 = sq. ft. Area of triangles c+d ,^^^!:1+1M^ x = «q. area of section J/= sq. ft. Denoting the distance between the sections by L, and thecubical contents of the prismoid by 5, we have, by applyingthe prismoidal formula 111, S^^{A+^B). Substituting known values in the formula, we have 5 = -^ ( + 4 X + ) = 8,703 cu. ft. = cu. Ans. Comparing the results, we have By averaging end areas, contents = cu. prismoidal formula, contents = cu. difference of about 1 per cent. Fig. 487 represents a mixed section of which the parta b c \?, solid rock, the part^rtr/is loose rock and thepart d c g Ji is earth. The slope a c in solid rock is \ hori-zontal to 1 vertical. In a s
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Keywords: ., bookcentury1800, bookdecade1890, booksubjectrailroadengineering
