and embankment,

(18 + 1½h)h
0.27,

the figure being inverted for embankment. For a prism of ten or of one thousand feet in length, we have only to move the decimal point. In forming a table, proceed as follows:—

hB + 1½hB + 1½h × h(B + 1½h) × h
0.27
abcd
a′b′c′d′
anbncndn.

Fig. 55.

It is evident from inspection of fig. 55, that c exceeds co by h × 2r; and that c″ exceeds c′ by h′ × 2r′; and so on as far as we go; this increase being constant, we have then to find the area of c, and for the area c + c′ double c, and add the increment; whence the rule:—

Having found the increase (which varies with the angle of the slope) for the second section, add the increase to twice the first. For the third, add twice the increase to three times the first; and for the nth, add n – 1 times the increment to n times the first area, or algebraically calling a the first area, a′ the second, a″ the third, an the nth area, and we have

The first areaa= a;
The second area2a + i= a′;
The third area3a + 2i= a″;
The nth areana + (n – 1)i= an.

We might operate at once upon the cubic contents, but for the length to which some decimals run; some indeed circulating.