354

NICHOLLS'S SEAMANSHIP AND NAUTICAL KNOWLEDGE

13, etc, 3 and placing against the ordinates the figures 1, 3a 3,1, in groups
of four as follows:—

add 1—3

and these are the multipliers for the successive ordinates.

The length of each ordmate is multiplied by its respective multiplier
and the sum of the products multiplied by three-eighths of the
common interval gives the area enclosed by the curves. When half
the section only is treated the result will be half the area of the
total surface enclosed.

Example.—The lengths of the half ordinates of a transverse bulkhead
in feet are 20, 18J, 17, 15,13, 9 and 4 respectively, the common interval
between them being 4 feet. Required the total area of the bulkhead.

01

Ol

OJ

Ol

20 e
Ordinates Multip. Products

181/3 /b
a 20ft. 1 20 b 18£ 3 55 c 17 3 51 d 15 2 30 e 13 3 39 / 9 3 27 04 1 4
Sum of products for half area 226 Total area = 226 Xfx 4x2
CTQ «/^ £4-

17 /C

15 /d

13 /e

9 /

•
4//

Fig. 2.
Example.—Suppose we were asked to find the volume of the vessel,
outlined in Figure 3, enclosed between bulkheads a and e which are each
spaced 30 feet apart. We would, in the first instance, find the half area
of each bulkhead from its figured dimensions, as indicated in the previous
example, either by Simpson's First or Second Rule whichever we found
most convenient. *•
Let these areas be as follows:—
Bulkhead - - - - a b c d e
Area in sq.ft. - 100, 350, 630, 450, 200