200 Eaztamzples, Showing the Manne7' of Using the Tables.

Second. Cases where the natural smface of the ground has an inclination
at "right angles to the line of the 7‘0ctd.
EXAMPLE 8. An excavation, the base of which is 28 feet, side slope 1 to 1,

and depth throughout 10 feet, has a transverse slope right and left of 12° ; _

required the content.
Fvkrst Method. Here (area (L l E — area E cl m) for 12° in table XXIV

B , 100 , . , .
is -0473, and (10 +  L = (24)? .27: LL L 1n table XXII IS 2133
cubic yards.
Consequently - - - ‘ n 2133 c. yds.
3740 “
multiplied by -0473 - - - - 853 “
149 “
6 (C
gives the correction - - ~ -- 1008 c. yds.
which added to average content from table XI — 14070 “

makes the total content - - - 1507'_8—c._fl:

Second Method, (by equation X). Here,

2

H” (Y + 3/) $3 —-—  equal the content.

 

4 //n
H“ X L = (24)? X  is found in table XXII opposite 24, 2133 c. yds.
. 3740-1 <=
5. (Y + 31,) in table XXIV isgP%1—6 .—. 1-0473 2133-0 «c
853 “
14-9 “
.6 u
t . ' 22338 “
B2 L, :
Subtract 4 m (table XXII) - -» - 726-0 N
and we have for the true content as before 1507-8 “

EXAMPLE 9. An embankment, 25 feet wide on top, having a side slope
of 15 to 1, is 12 feet deep at one end and 4 at the other, and has a transverse
slope right and left of the centre at both ends of 12°; required the content.

First Method, (by formula N)

n I I L ' -
§ H~ + H2 + (H + -H )9 5 (A —-— a)  = correctron for transverse slope.

ioo .
H2 X L = (20-3)” -5-7- = No. cor. to 203 in tab. XXII is 1526 c. yds.
H’? X L = (12-3)? 1-20; = “ ' ’ 12-3 “‘ “ 560 “
. 1 V I
(H +-H )9 X L = (203 + 12-3)’  No. car. to 32-7 “ 3960 “