LATITUDE AND DEPARTURE FROM PLANE COORDINATES.— The
numerical values of latitude and departure of a traverse line are easily
computed from the coordinates of the end stations of the line. For traverse line
AB, for
example, the numerical value of latitude equals the difference between the Y
coordinate of A
and the Y
coordinate of B,
while the numerical
value of departure equals the difference between the X
coordinate of A and
the X coordinate
of B. To determine whether a latitude or departure computed
this way is positive or negative, the best method
is to examine a sketch of the traverse to determine
the compass direction of the bearing of the line in question. If the line bears
northeast, the latitude is positive, or north, and the departure is positive, or
east. If the line bears southwest, both latitude and departure are negative.
Computing Areas
Various methods are used in computing areas. Some of
the common methods with which the EA should be familiar
are discussed below.
Figure 719.—Form for
computing coordinates
AREA BY DOUBLE MERIDIAN DISTANCE.—
The
meridian
distance of a traverse
line is equal to the length of a line running east to west from the midpoint of
the traverse line to a reference meridian. The reference meridian is the
meridian that passes through the most westerly
traverse station.
In figure 720, the dotted lines indicate the meridian distances
of the traverse lines to which they extend from the
reference meridians. You can see that the meridian distance
of the initial line AB equals
one half of the departure of AB.
The meridian distance of
the next line BC equals
the meridian distance of AB,
plus one half of the
departure of AB, plus
one half of the departure of
BC.
You can also see that the meridian distance of CD
equals the meridian distance of BC,
plus one half of the
departure of BC,
minus one
half of the departure of DC.
Similarly, the meridian distance of
AD equals
the meridian distance of DC,
minus one
half of the departure of DC,
minus one
half of the departure of AD.
You should now be able to
understand the basis for the
following rules for determining meridian distance:
1. For the initial traverse line in a closed traverse, the
meridian distance equals one half of the departure.
2. For each subsequent traverse line, the meridian distance
equals the meridian distance of the preceding
Figure 720.—Meridian distances.
line, plus one half of the departure of the preceding line, plus
one half of the departure of the line itself. However, it
is the algebraic sum
that results—meaning that plus departures
are added but minus departures are subtracted.
Figure721.—Area from double
meridian distances.
For convenience, it is customary to use double
meridian distance (DMD)
rather than meridian distance
in calculations. When the meridian distance of the
initial traverse line in a closed traverse equals one half
of the departure of the line, the DMD of this line equals its departure. Again,
from the rule for meridian distance of the next line, the DMD of that line
equals the DMD of the preceding line, plus the departure of the preceding line,
plus the departure of the line itself.
It can be shown geometrically that the area contained
within a straightsided closed traverse equals the
sum of the areas obtained by multiplying the meridian
distance of each traverse line by the latitude of that
line. Again the result is the algebraic sum. If you multiply
a positive meridian distance (when the reference
meridian runs through the most westerly station,
all meridian distances are positive) by a plus or north latitude, you get a plus
result that you add. If you multiply a positive meridian distance by a minus or
south latitude, however, you get a minus result that you subtract.
Therefore, if you multiply for each traverse line the double meridian
distance by latitude instead of meridian distance by latitude, the sum of the
results will equal twice the area, or the double area. To get the area, you
simply divide the double area by 2.
Figure 721 shows entries for the computations of the DMD of the area of the
traverse we have been working on. Because AB
is the initial traverse
line, the DMD of AB equals
the departure. The DMD of BC
equals the DMD of AB (125.66),
plus the departure of AB (125.66), plus the departure of BC
(590.65), or 841.97
feet. The DMD of CD equals
the DMD of BC
(841.97), plus the departure of BC
(590.65), plus theCD (which
is minus 192.69,
and therefore is subtracted), or 1239.93 feet. The DMD of DA
equals the DMD of CD
(1239.93), plus the
departure of CD (–192.69), plus the departure of DA (–523.62), or 523.62
feet. Note that the DMD of this last traverse line equals the
departure of the line, but with an opposite sign. This fact
serves as a check on the computations.
The double area for AB
equals the DMD times the
latitude or 125.66
x 255.96 = 32,163 .93square feet.
The double area for BC
equals 841.97 (the DMD)
times minus 153.53
(the latitude), or minus 129,267.65
square feet.
The double area of CD
is 1,239.93
x (694.07) = –860,598.21 square
feet.
Figure 722.—Parallel distances.
Figure 723.—Area from
double parallel distances.
The double area of DA
is 523.62
x 591.64 = 309,794.54 square
feet.
The difference between the sum of the minus double areas
and the sum of the plus double areas is the double area which is 647,907.39
square feet. The area is one half of this, or 323,953.69 square feet. Land area
is generally expressed in acres. There are 43,560 square feet in 1 acre;
therefore, the area in acres is
AREA BY DOUBLE PARALLEL DISTANCE.—
You can check the accuracy
of the area computation of a
DMD by computing the same area from
double
parallel distances (DPD).
As shown in figure 722, the parallel distance of a traverse line is the
northtosouth distance from the midpoint of the line to a reference parallel.
The reference parallel is the parallel passing through the most
southerly traverse
station.
You can see that the solution for parallel distance is the same as the one
used for meridian distance, except that to compute parallel distance you use
latitude instead of departure. The parallel distance of the initial traverse
line (which is DA in
this case) equals one half of the latitude. The parallel distance of the next
line, AB, equals
the parallel distance of the preceding line, DA,
plus one half of the
latitude of the preceding line DA,
plus one half of the
latitude of line AB itself.
It follows from the above that the DPD of the initial traverse line DA
equals the latitude of
the line. The DPD of the next line, AB,
equals the DPD of the
preceding line, DA, plus
the latitude of the preceding line, DA,
plus the latitude of the
line AB itself.
The solution for area is the same as for area by meridian distance except that,
for the double area of each traverse line, you multiply the DPD by the departure
instead of multiplying the DMD by the latitude.
Figure 723 shows entries for the computation of the area
of DPD for the traverse we are working on. Note that
the result is identical with that obtained by the computation
of the DMD.
