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
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
To determine whether a latitude or departurecomputed 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.
Various methods are used in computing areas. Someof the common methods with which the EA should be familiar are discussed below.
Figure 7-19.—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 7-20, the dotted lines indicate the meridiandistances 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 ofCD 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 meridiandistance equals the meridian distance of the preceding
Figure 7-20.—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.
Figure7-21.—Area from double meridian distances.
For convenience, it is customary to usedouble 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 areacontained within a straight-sided 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 7-21 shows entries for the computations of the DMD of the area of the traverse we have been working on. BecauseAB 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 forAB equals the DMD times the latitude or 125.66 x 255.96 = 32,163 .93square feet.
The double area forBC 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 7-22.—Parallel distances.
Figure 7-23.—Area from double parallel distances.
The double area ofDA is 523.62 x 591.64 = 309,794.54 square feet.
The difference between the sum of the minus doubleareas 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 7-22, the parallel distance of a traverse line is the north-to-south distance from the midpoint of the line to a reference parallel. The reference parallel is the parallel passing through the mostsoutherly 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 isDA 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 lineDA 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 7-23 shows entries for the computation of thearea of DPD for the traverse we are working on. Note that the result is identical with that obtained by the computation of the DMD.