Chapter 18 Indirect Leveling/Level and Traverse Computations

Leveling is the surveying operation that determines the difference in elevation between points on the earth’s surface. The operation is divided into two major categories: direct leveling and indirect leveling. From your earlier studies, you should be familiar with the methods and procedures used in direct leveling.

This lesson introduces the theory and basic procedures used in indirect leveling. You also learned in earier lessons that perfect closure in level nets and traverses is seldom, obtained. Typically, a certain amount of linear or angular error occurs. When this error exceeds a prescribed amount, the level net or traverse must be rerun. However, when the error is within the specified allowable limits, then certain adjustments can be made. This lesson provides information on the adjustments and the calculations needed to make the adjustments for the methods used to determine the area of traverses.

When you have completed this lesson, you will be able to:

1. Describe the different types of indirect leveling.
2. Describe the methods for conducting level and traverse computations.
 Contents 1.0.0 Indirect Leveling 2.0.0 Level and Traverse Computations SummaryReview Quiz

1.0.0 INDIRECT LEVELING

Indirect methods of leveling include barometric leveling and trigonometric leveling. These methods are discussed in the following paragraphs.

1.1.0 Barometric Leveling

1.2.0 Trigonometric Leveling

 Test Your Knowledge1. Barometric leveling computes the difference in elevations by which of the following measurements? A. Slope angle B. Atmospheric pressure C. Slope distance D. Vertical offset 2. Barometric leveling is used mostly in which of the following types of surveys? A. Layout B. Route C. Reconnaissance D. Navigational 3. Barometric leveling should be performed at what time of day? A. Midday B. 2 to 4 hours before sunrise or 2 to 4 hours after sunset C. 2 to 4 hours after sunrise or 2 to 4 hours before sunset D. 2 to 4 hours before sunset or 2 to 4 hours after sunset 4. When the horizontal or slope distance is measured by chaining, which of the following must be computed before you determine the difference of elevation? A. Repair adjustment, sag, and temperature B. Standard error, sag, and tension C. Sag, temperature, and standard error D. Tension, heat, and number of movements

2.0.0 LEVEL and TRAVERSE COMPUTATIONS

This section contains information on procedures used in making level and traverse computations. This section also discusses methods of differential leveling, including steps to follow in checking level notes and information on adjusting intermediate bench marks as well as level nets. In addition, this section addresses several methods of plotting horizontal controls used in determining the bearing of the traverses. These methods include plotting angles by protractor and scale, plotting angles from tangents, and plotting by coordinates. Additional information includes the common mistakes an engineering tech may encounter in making or checking computations.

2.1.0 Preliminaries to Computations

Before starting computations, closely check the field data for completeness and accuracy. This includes checking the field notes to ensure they accurately reflect what was actually measured. A field measurement may itself require transformation (called reduction) before it can be applied as a value in computations. For example, field notes may show plate readings for two-, four-, or six-time angles. Each of these must be reduced to the mean angle, as explained earlier. In another example, field notes may show a succession of chained slope distances. Unless the order of precision of the survey permits slope corrections to be ignored, each of these slope distances must be reduced to the corresponding horizontal distance. In a closed traverse you must attain a ratio of linear error of closure and a ratio of angular error of closure that are within the maximums specified for, or implied from, the nature of the survey. An error that is within the maximum allowable is eliminated by “adjustment,” which means the equal distribution of a sum total of allowable error over the separate values that contribute to the total. Suppose, for example, that for a triangular closed traverse with interior angles approximately equal in size, the sum of the measured interior angles comes to 179°57´. The angular error of closure is 03´. Because there are three interior angles about equal in size, 01´ would be added to the measured value of each angle.

2.2.0 Level Computations

2.3.0 Traverse Computations

Traverse operations are conducted for mapping large construction projects, such as a military post or an air base, road railroad and pipeline alignment projects, and for many other projects. A traverse is always classified as either a closed traverse or an open traverse. A closed traverse starts and ends at the same point or at points whose relative horizontal positions are known. An open traverse ends at the station whose relative position is previously unknown and, unlike a closed traverse, provides no check against mistakes and large errors. In the Chapter 12, you studied field procedures for laying out traverses. In this lesson you will study computations that are necessary for adjusting and determining the areas of traverses.

2.3.1 Checking and Reducing Angles

The beginning traverse computations start with checking to make sure all the required angles (including closing angles) were turned and that the notes correctly indicate their sizes. For deflection angles, check to make sure that angles marked L or R were actually turned and have been turned in those directions. Check to make sure sketches agree with your field notes. Next, reduce repeated angles to mean angles using the procedures learned in Chapter 12.

2.3.2 Checking and Reducing Distances

Check to make sure that all required linear distances have been chained. Reduce slope distances when needed. If you broke chain on the slopes, check to make sure that the sums of break distances were correctly added. Finally, apply standard error, tension, and temperature corrections if needed.

From your study of the Chapter 14, you should recall the following three conditions for a closed traverse: 1. The theoretical or geometrical sum of the interior angles is 180° x (n – 2), n being the number of angles measured. 2. The sum of the exterior angles is 180° x (n + 2), where n = number of angles measured. 3. The difference between the sum of the right deflection angles and the sum of the left deflection angles is 360°. Any discrepancy between one of these sums and the actual sum of the angles as turned or measured constitutes the angular error of closure. You adjust the angles in a closed traverse by distributing an angular error of closure that is within the allowable maximum equally among the angles. Figure 18-10 shows a traverse in which one of the deflection angles was turned to the left, all others to the right. The sum of the right deflection angles is 444°59'. By subtracting the left deflection angle (85°01'), the angular error of closure of 02' is determined, which is an average of 20" per deflection angle. This average angular error of closure is then added to each right deflection angle and subtracted from each left deflection angle. After applying this adjustment to each deflection angle in this example, you find, then, that the sum of the adjusted angles to the right equals 445°00'40" and that the sum of the left angles (of which there is only one) is 85°00'40". The difference between these values is 360°00'00", as it should be. Remember that in adjusting the angles in a deflection-angle traverse, apply the adjustments to right and left angles in opposite direction.

2.3.4 Adjusting for Linear Error of Closure

The procedure for distributing a linear error of closure (one within the allowable maximum) over the directions and distances in a closed traverse is called balancing or closing the traverse. However, before this information can be calculated you must understand the concept of latitude and departure. Figure 18-10 – Close traverse by deflection-angle method.

2.3.4.1 Latitude and Departure

Latitude and departure are values used in the method of locating a point horizontally by its plane coordinates. In the plane coordinate system, a point of origin is arbitrarily selected for convenience. The location of a point is given in terms of its distance north or south and its distance east or west of the point of origin. The plane coordinate system will be explained later in this lesson. The latitude of a traverse line means the length of the line as projected on the northto-south meridian running through the point of origin. The departure of a traverse line means the length of the line as projected on the east-to-west parallel running through the point of origin, as shown in Figure 18-11). The point of origin is at O. The line NS is the meridian through the point of origin, and the line EW is the parallel through the point of origin. The latitude of AB is the length of AB as projected on NS and the departure of AB is the length of AB as projected on EW. If a traverse line was running due north and south, the latitude would equal the length of the line and the departure would be zero. For a line running due east and west the departure would equal the length of the line and the latitude would be zero. Now, for a line running other than north to south or east to west, the latitude or departure can be determined by simple triangle solution. Figure 18-12 shows a traverse line 520.16 feet long bearing S61°25'E. To determine the latitude, you solve the triangle ABC for the length of the side AC. From the bearing, you know that the size of angle CAB (the angle of bearing) is 61°25'. The triangle is a right triangle; therefore AC = 520.16 cos 61°25' = 248.86 feet. The latitude of a traverse line, then, equals the product of the length of the line times the cosine of the angle of bearing. To determine the departure, solve the length of the side CB shown in Figure 18-13. CB = 520.16 sin 61°25' = 456.76 feet. The departure of a traverse line, then, equals the length of the line times the sine of the angle of bearing. The latitude of a traverse line is designated north or south and the departure is designated east or west following the compass direction of the bearing of the line. A line bearing northeast, for example, has a north latitude and east departure. In Figure 18-12 – Latitude equals length of traverse line times cosine of angle of bearing. Figure 18-11 – Latitude and departure.  18-16 computations, north latitudes are designated plus and south latitudes minus; east departures are designated plus and west departures minus. Figure 18-14 demonstrates that in a closed traverse, the algebraic sum of the plus and minus latitudes is zero and the algebraic sum of the plus and minus departures is zero. The plus latitude of CA is equal in length to the sum of the two minus latitudes of AB and BC. The minus departure of BC is equal in length to the sum of the two plus departures of CA and AB. Figure 18-13 – Departure equals length of traverse line times sine of angle of bearing. Figure 18-14 – Graphic solution of a closed  traverse by latitude and departure.

2.3.4.2 Linear Error of Closure

In practice, the sum of the north latitudes usually differs from the sum of the south latitudes. The difference is called the error of closure in latitude. Similarly, the sum of the east departures usually differs from the sum of the west departures. The difference is called error of closure in departure. From the error of closure in latitude and the error of closure in departure, the linear error of closure can be determined. This is the horizontal linear distance between the location of the end of the last traverse line (as computed from the measured angles and distances) and the actual point of beginning of the closed traverse. For example, you come up with an error of closure in latitude of 5.23 feet and an error of closure in departure of 3.18 feet. These two linear intervals form the sides of a right triangle. The length of the hypotenuse of this triangle constitutes the linear error of closure in the traverse. By the Pythagorean Theorem, the length of the hypotenuse equals approximately 6.12 feet. Suppose the total length of the traverse was 12,000.00 feet. Then your ratio of linear error of closure would be 6.12:12,000.00, which approximately equates to 1:2,000.

2.3.4.3 Closing a Traverse

A traverse is closed or balanced by distributing the linear error of closure (one within the allowable maximum, of course) over the traverse. There are several methods of doing this, but the one most generally applied is based on the compass rule. This rule adjusts the latitude and departure of each traverse line as follows: 1. Correction in latitude equals the linear error of closure in latitude times the length of the traverse line divided by the total length of traverse. 2. Correction in departure equals the linear error of closure in departure times the length of the traverse line divided by the total length of traverse. Figure 18-15 shows a closed traverse with bearings and distances notes. Figure 18-15 – Closed traverse by bearings and distance.  18-18 Figure 18-16 shows the computation of the latitudes and departures for the traverse entered on the form commonly used for this purpose. As you can see, the error in latitude is +0.33 foot, and the error in departure is +2.24 feet. The linear error of closure, then, is (0.33 2.24 ) 2.26 feet 2 2 =+ Figure 18-16 – Form for computing latitudes and departures. The total length of the traverse is 2614.85 feet; therefore, the ratio of error of closure is 2.26:2614.85, or about 1:1157. Assume this ratio is within the allowable maximum. Proceed now to adjust the latitudes and departures by the compass rule. Set down the latitudes and departures on a form like the one shown in Figure 18-17 with the error of closure in latitude at the foot of the latitudes column and the error of closure in departure at the foot of the departures column. Next, use the compass rule to determine the latitude correction and departure correction for each line. For all, the latitude correction equals 0.07 feet 2614.85 584.21 0.33 x = The error of closure in latitude is plus; therefore, the correction is minus. Note that the sum of the applied latitude corrections equals the error of closure in latitude and the sum of the applied departure corrections equals the error of closure in departure. The corrections, however, are opposite in sign to the error of closure.

2.3.5 Traverse Tables/Adjusting Bearings and Distances

2.3.6 Plane Coordinates

The location of a point by plane coordinates means to describe the location of the point in terms of its distance north or south and east or west from a point of origin. Figure 18-20 shows how coordinate distances are measured on an axis (called the Y axis) running north to south through the point of origin. East to west coordinates are measured on an X axis running east to west through the point of origin. Values on the Y axis north of the point of origin are plus and values south of the point of origin are minus. Values on the X axis east of the point of origin are plus and values west of the point of origin are minus.

Figure 18-19 – Adjusted bearing and distance from adjusted latitude and departure.

Figure 18-20 – Location by plane coordinates.

2.3.6.1 Plane Coordinates from Latitude and Departure

Figure 18-20 also shows the relationship between the plane coordinates of the end stations on a traverse line and the latitude and departure of the line. You can see the difference between the Y coordinate of A and the Y coordinate of B (which is 200.00 feet) equals the latitude of AB. You can also see the difference between the X coordinate of A and the X coordinate of B (which is 600.00 feet) equals the departure of AB. Therefore, if you know the coordinates of one of the stations in a traverse, you can determine the coordinates of the others from the latitudes and departures. Figure 18-21 shows a closed traverse with adjusted latitudes and departures notes. You want to assign plane coordinates to the traverse stations. To avoid the necessity of working with negative coordinates, you select as point of origin a point O that is west of the most westerly traverse station and south of the most southerly traverse station. You determine the bearing and length of dotted line OD and compute from these values the latitude and departure of OD. You can see that the Y coordinate of station D must equal the latitude of OD, or 150.70 feet. Also the X coordinate of D must equal the departure of OD or 556.30 feet. The Y coordinate of station A equals the Y coordinate of D plus the latitude of AD or 150.70 + 591.64 = 742.34 feet. The X coordinate of station A equals the X coordinate of D minus the departure of AD or 556.30 – 523.62 = 32.68 feet. The Y coordinate of station B equals the Y coordinate of station A plus the latitude of AB or 742.34 + 255.96 = 998.30 feet. The X coordinate of station B equals the X coordinate of station A plus the departure of AB or 32.68 + 125.66 = 158.34 feet. The Y coordinate of station C equals the Y coordinate of station B minus the latitude of C or 998.30 – 153.53 = 844.77 feet. The X coordinate of station C equals the X coordinate of station B plus the departure of BC or 158.34 + 590.65 = 748.99 feet. Figure 18-21 – Closed traverse with adjusted latitudes and departures.  18-23 The Y coordinate of station D equals the Y coordinate of station C minus the latitude of CD or 844.77 – 694.07 = 150.70 feet. The X coordinate of station D equals the X coordinate of station C minus the departure of CD or 748.99 – 192.69 = 556.30 feet. These are the same coordinates originally computed for station D, a fact that serves as a check on accuracy. You enter these values on a form that is similar to the one shown in Figure 18-22. In actual practice, however, a wider form is used on which all values and computations from the original station through bearing and distance, latitude and departure, and coordinates can be entered.

2.3.6.2 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 Figure 18-22 – Form for computing coordinates.  18-24 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.

2.3.7 Computing Areas

Various methods are used in computing areas. Some of the common methods are discussed below.

2.3.7.1 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 18-23, 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. Figure 18-23 shows 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 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 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. 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 the line equals its departure. Again, from the rule for meridian distance of the next line, the DMD of that line equals Figure 18-23 – Meridian distances.  18-25 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 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, divide the double area by 2. Figure 18-24 shows entries for the computations of the DMD of the area of the traverse ABCD. 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 the departure of CD (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 = 32,163.93 square 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. 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, 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 .44.7 43560 69.323953 = A  18-26

2.3.7.2 Area by Double Parallel Distance

The accuracy of the area computation of a DMD can be checked by computing the same area from double parallel distances (DPD). As shown in Figure 18-25, the parallel distance of a traverse line is the north-tosouth distance from the midpoint of the line to a reference parallel. The reference parallel is the parallel passing through the most southerly traverse station. The solution for parallel distance is the same as the one used for meridian distance, except that to compute parallel distance, the latitude is used 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 18-26 shows entries for the computation of the area of DPD for the traverse ABCD. Note that the result is identical to that obtained by the computation of the DMD. Figure 18-25 – Parallel distances. Figure 18-24 – Area from double meridian distances.

2.3.7.3 Area from Coordinates B

efore we explain the method of computing area from coordinates, let us set coordinates for the stations of the traverse ABCD. To avoid using negative coordinates, we will measure Y coordinates from an X axis passing through the most southerly station and X coordinates from a Y axis passing through the most westerly station, as shown in Figure 18-27. Figure 18-27 – Closed traverse by coordinate method. Figure 18-26 – Area from double parallel distances.  18-28 Figure 18-28 shows the coordinate entries. It shows that the Y coordinate of A equals the latitude of DA, or 591.64 feet, and the X coordinate of A is zero. The Y coordinate of B equals the Y coordinate of A plus the latitude of AB or 591.64 + 255.96 = 847.60 feet. Figure 18-28 – Coordinate entries for computation of Figure 18-27. The X coordinate of B equals the departure of AB, or 125.66 feet. The Y coordinate of C equals the Y coordinate of B minus the latitude of BC or 847.60 – 153.53 = 694.07 feet. The X coordinate of C equals the X coordinate of B plus the departure of BC or 125.66 + 590.65 = 716.31 feet. The Y coordinate of D obviously is zero; however, it computes as the Y coordinate of C minus the latitude of CD of 694.07 – 694.07, which serves as a check. The X coordinate of D equals the X coordinate of C minus the departure of CD or 716.31 – 192.69 = 523.62 feet. This is the same as the departure of DA, but with an opposite sign—a fact which serves as another check.  18-29 Figures 18-29 and 18-30 show the method of determining the double area from the coordinates. First, multiply pairs of diagonally opposite X and Y coordinates, as shown in Figure 18-29, and determine the sum of the products. Then, multiply pairs diagonally in the opposite direction, as shown in Figure 18-30, and determine the sum of the products. The difference between the sums (shown in Figure 18-29) is the double area or 1,044,918.76 – 397,011.37 = 647,90.39 square feet. The symbol shown beside the sum of the coordinate products is the capital Greek letter (Σ) sigma. In this case, it means sum. Figure 18-29 – First step for tabulated computation of Figure 18-27.  18-30 Figure 18-30 – Second step for tabulated computation of Figure 18-27.

2.3.7.4 Area by Trapezoidal Formula

It is often necessary to compute the area of an irregular figure, when one or more of its sides do not form a straight line. For illustration purposes, assume Figure 18-31 is a parcel of land in which the south, east, and west boundaries are straight lines perpendicular to each other, but the north boundary is a meandering shoreline. Figure 18-31 – Area of irregular figure by trapezoidal rule.  18-31 To determine the area of this figure, first lay off conveniently equal intervals (in this case, 50.0-foot intervals) from the west boundary and erect perpendiculars as shown. Measure the perpendiculars. Call the equal interval “d” and the perpendiculars (beginning with the west boundary and ending with the east boundary) h1 through h6. You can see that for any segment lying between two perpendiculars, the approximate area, by the rule for determining the area of a trapezoid, equals the product of “d” times the average between the perpendiculars. For the most westerly segment, for example, the area is . 2 21   + hh d The total area equals the sum of the areas of the segments. Therefore, since “d” is a factor common to each segment, the formula for the total area may be expressed as follows:   + + + + + + + + + = 22222 hh 21 hhhhhhhh 65544332 dA This works out to   +++++ = 2 22222 hhhhhh 654321 dA In turn, this reduces to   ++++ + = 5432 61 2 hhhh hh dA Substituting in the formula the data from Figure 18-31, you have   ++++ + = 1091219280 2 129105 A 50 When the calculation is completed the result is 25,950 square feet or approximately 0.6 acre.

2.3.7.5 Area by Counting the Squares

Another method of computing the area of an irregular figure is to plot the figure on a sheet of graph paper. The area is then determined by counting the squares within the figure outline and multiplying the result by the area represented by each square. Figure 18-32 shows the same figure shown in Figure 18-31 but plotted to scale on a sheet of graph paper on which each of the small squares is 5 feet x 5 feet or 25 square feet. When squares within the outline are counted, they total 1,038 squares which means 1,038 x 25 = 25,950 square feet.  18-32 Figure 18-32 – Computing area by counting the squares.

2.3.7.6 Parcels That Include Curves

Not all parcels of land are bounded entirely by straight lines. It is often necessary to compute the area of a construction site that is bounded in part by the center lines or edges of curved roads or the right-of-way lines of curved roads. Figure 18-33 shows a construction site with a shape similar to the traverse used in the previous examples. In this site, however, the traverse lines AB and CD are the chords of circular curves, and the boundary lines AB and CD are the arcs intercepted by the chords. The following sections explain the method of determining the area lying within the straight-line and curved-line boundaries. The data for each of the curves is inscribed on Figure 18-33, that is, the radius R, the central angle ∆, the arc length A, the tangent length T and the chord bearing and distance CH. The crosshatched areas lying between the chord and arc are called segmental areas. To determine the area of this parcel, you must take the following steps. 1. Determine the area lying within the straight-line and chord (also straight-line) boundaries. 2. Determine the segmental areas. 3. Subtract the segmental area for Curve 1 from the straight-line boundary area. 4. Add the segmental area for Curve 2 to the straight-line boundary area. The method of determining a segmental area was explained in the engineering tech Basic Chapter 2. The straight-line area may be determined by the coordinate method, as explained in this lesson. For Figure 18-33, the segmental area for Curve 1 works out to be 5,151 square feet; for Curve 2, it is 29,276 square feet.  18-33 Figure 18-33 – Area within straight-line and curved-line boundaries (curved segments). Figure 18-34 shows a typical computation sheet for the area problem shown in Figure 18-33. Included with the station letter designations in the station column are designations (Chord #1 and Chord #2) showing the bearings and distances that constitute the chords of Curves 1 and 2. The remainder of the upper part of the form shows the process (with which you are now familiar) of determining latitudes and departures from the bearings and distances, coordinates from the latitudes and departures, double areas from cross multiplication of coordinates, double areas from the difference between the sums of north and sums of east coordinates, and areas from half of the double areas. As you can see in Figure 18-34, the area within the straight-line boundaries is 324,757 square feet. From this area, segmental area No. 1 is subtracted. Then segmental area No. 2 is added. To calculate the area of the parcel as bounded by the arcs of the curves, add or subtract the segmental areas depending on whether the particular area in question lies inside or outside of the actual curved boundary. In Figure 18-33, the segmental area for Curve 1 lies outside and must be subtracted from the straight-line area, while Curve 2 lies inside and must be added. With the segmental areas accounted for, the area comes to 348,882 square feet or 8.01 acres.  18-34 Figure 18-34 – Computation of area which includes curve segments. The second method of determining a curved boundary area uses the external areas rather than the segmental areas of the curves, as shown in Figure 18-35. The straightline figure is defined by the tangents of the curves, rather than by the chords. This method may be used as an alternative to the chord method or to check the result obtained by the chord method. The computation sheet shown in Figure 18-36 follows the same pattern as Figure 18- 34. However, there are two more straight-line boundaries, because each curve has two tangents rather than a single long chord. The coordinates of A, B, C, and D are the same as in the first example, but the coordinates of the points of intersection (PIs) must be established from the latitudes and departures of the tangents. The computations for determining the tangent bearings are shown in the lower left of Figure 18-36. When the chord bearing is known, the tangent bearing can be computed by adding or subtracting one-half of delta (∆) as appropriate. The angle between the tangent and the chord equals ∆2. After setting coordinates on the PIs, cross-multiply, accumulate the products, subtract the smaller from the larger, and divide by 2, as before, to get the area of the straight-line figure running around the tangents. Then add or subtract each external area as appropriate. In Figure 18-35, the external area for Curve 1 is inside the parcel boundary and must be added, while Curve 2 is outside and must be subtracted. The area comes to 348,881 square feet, which is an acceptable check on the area calculated by using the segmental areas.  18-35 Figure 18-35 – Area within the curve and its tangents. Figure 18-36 – Computation of area which includes external area of curves.

2.3.8 Plotting Horizontal Control

Computations for horizontal control become clarified when a plot or scaled graphic representation of the traverse can be viewed. For example, looking at a plot can tell you whether you should add or subtract the departure or the latitude of a traverse line in computing the departure or latitude of an adjacent line or in computing the coordinates of a station. For linear distances that are given in feet and decimals of feet, use the correct scale on an engineer’s scale for laying off linear distances on a plot. For plotting traverses, there are three common methods: by protractor and scale, by tangents, and by coordinates.

2.3.8.1 Plotting Angles by Protractor and Scale

The adjusted bearings and distances for closed traverse ABCD are as follows: Figure 18-37 shows the plotting method of traverse ABCD using a scale and protractor. First select a scale that will make the plot fit on the size of the paper. Select a convenient point on the paper for station A and draw a light line NS, representing the meridian through the station. AB bears N26°9'E. Set the protractor with the central hole on A and the 00 line at NS, and lay off 26°09'E. The minutes must be estimated. Draw a line in this direction from A, and on the line measure off the length of AB (285.14 feet) to scale. This locates station B on the plot. Draw a light line NS through B parallel to NS through A, representing the meridian through station B. BC bears S75°26'E. Set the protractor with the central hole on B and the 00 line on NS, lay off 75°26' from the S leg of NS to the E, and measure off the length of BC (610.26 feet) to scale to locate C. Proceed to locate D in the same manner. This procedure produces a number of light meridian lines through stations on the plot. Figure 18-38 illustrates a procedure that eliminates these lines. A single meridian NS is drawn clear of the area of the paper on which you intend to plot the Figure 18-37 – Traverse plotted by protractor and scale method. Traverse Line DA CD BC AB Bearing WN WS ES EN 1341 1315 6275 9026 ° ′ ° ′ ° ′ ° ′ Distance feet feet feet feet 96.789 28.720 26.610 14.285  18-37 traverse. From a convenient point O, lay off each of the traverse lines in the proper direction. Then transfer the directions to the plot by one of the methods for drawing parallel lines.

2.3.8.2 Plotting Angles from Tangents

Sometimes instead of having bearing angles to plot from, you might want to plot the traverse from deflection angles turned in the field. The deflection angles for the traverse are as follows: AB to BC 78º25’R BC to CD 90º57’R CD to DA 122º58’R DA to AB 67º40’R You could plot from these angles by protractor. Lay off one of the traverse lines to scale. Then lay off the direction of the next line by turning the deflection angle to the right of the first line extension by protractor and so on. However, the fact that you can read a protractor directly to only the nearest 30 minutes presents a problem. When you plot from bearings, the error in estimation of minutes applies only to a single traverse line. When you plot from deflection angles, however, the error carries on cumulatively all the way around. For this reason, you should use the tangent method when you are plotting deflection angles. Figure 18-39 shows the procedure of plotting deflection angles larger than 45°. The direction of the starting line is called the meridian, following a conventional procedure that the north side of the figure being plotted is situated toward the top of the drawing paper. In doing this, you might have to plot the appropriate traverse to a small scale using a protractor Figure 18-38 – Plotting traverse lines by parallel method from a single meridian. Figure 18-39 – Plotting by tangent-offset method from deflection angles larger than 45 degrees.  18-38 and an engineer’s scale, just to have a general idea of where to start. Make sure that the figure will fit proportionately on the paper of the desired size. Starting at point A, you draw the meridian line lightly. Then you lay off AO, 10 inches (or any convenient roundfigure length) along the referenced meridian. Now, from O you draw a line OP perpendicular to AO. Draw a light line OP as shown. In a trigonometric table, look for the natural tangent of the bearing angle 26°90', which equals to 0.49098. Find the distance OP as follows: OP = AO tan 26°09' = 4.9098, or 4.91 inches. You know that OP is equal to 4.91 inches. Draw AP extended and then lay off the distance AB to scale along AP. Remember that unless a closed traverse is plotted, it is always advantageous to start offsets from the referenced meridian, the reason being that, after you have plotted three or more lines, you can always use this referenced meridian line for checking the bearing of the last line plotted to find any discrepancy. The bearing angle, used as a check, should also be found by the same method (tangent-offset method). To plot the directions of lines from deflection angles larger than 45°, the complementary angle (90° minus the deflection angle) must be used. To plot the direction of line BC in Figure 18-39, draw a light perpendicular line towards the right from point B. Measure off a length of approximately 10 inches, which represents BOJ. The complement of the deflection angle of BC is 90° – 78°25' = 11°35'. The natural tangent value of 11°35’ is equal to 0.20497. From O1 draw O1P1 perpendicular to BO1. Solving for O1P1, you will have O1P1 = BO1 tan 11°35' = 2.0497, or 2.05 inches. Now lay off the distance O1P1. Draw a line from B through P1 extended; lay off the distance BC to scale along this line. The remaining sides, CD and DA, are plotted the same way. Make sure that the angles used for your computations are the correct ones. A rough sketch of your next line will always help to avoid major mistakes. When the deflection angle is less than 45°, the procedure of plotting by tangent is as shown in Figure 18-40. Here you measure off a convenient round-figure length (say 500.00 feet) on the extension of the initial traverse line to locate point O, and from O, draw OP perpendicular to AO. The angle between BO and BC is, in this case, the deflection angle. Assume that 23°21'. The formula for the length of OP is this is OP = BO tan 23°21' = 500 x 0.43170= 215.85 feet. Figure 18-40 – Plotting by tangent-offset method from deflection angles smaller than 45 degrees

2.3.8.3 Plotting by Coordinates

A common and accurate method of plotting by coordinates is shown in Figure 18-41. Each station is located by its coordinate and without angular measurements. To plot station B, for instance, you would lay off from O on the Y axis a distance equal to the Y coordinate of B (847.60 feet). Draw a light line from this point perpendicular to the Y axis, and measure off on this line a distance equal to the X coordinate of B (125.66 feet). The remaining points are plotted in the same way.

2.3.9 Mistakes in Computations

An involved computation, such as determining an area by DMDs, involves a large number of calculations and thus the possibility of a large number of errors. Some of the most common types of mistakes are discussed below.

2.3.9.1 Mistakes with Signs

Be careful to give a value such as a latitude or departure its correct sign (+ or -) and apply the sign correctly in addition, subtraction, multiplication, and division. To prevent not including the correct sign, do not write a value without including the sign. The practice of omitting plus signs is a correct procedure, but it is safer to write in the plus signs.

2.3.9.2 Wrong Column

A wrong column mistake may be an entry made in a wrong column or a reading taken from a wrong column. To avoid column mistakes, make both entries and readings with deliberation.

When you mistake the quadrant in which a line lies, you get a bearing that may have the correct angular value but that has the wrong compass direction. A common cause of this mistake is viewing the direction of a line from the wrong station. In Figure 18-42, the direction of AB is northeast and the direction of BA is southwest. However, AB and BA are the same traverse line. To minimize direction errors, place arrows on the diagram showing the direction of the line.

2.3.9.4 Wrong Azimuth

The same mistake that applies to quadrants also applies to azimuths. Suppose that the bearing of AB in Figure 18-42 is N46°E. Then the azimuth of AB is (measured from north) 46°. While AB and BA are the same traverse line, the azimuth of BA is 226°. Figure 18-41 – Plotting by coordinates.

2.3.9.5 Leaving Out a Traverse Line

A common source of mistakes is leaving out (commonly called dropping) a traverse line, either in the field notes or in computations. If an outsized angular and linear error of closure occurs, check to make sure one of the traverse lines has not been dropped.

2.3.9.6 Wrong Decimal Place

The incorrect placement of a decimal point is a common mistake. Suppose, for example, you are determining an approximate double area by multiplying a DMD of +841.97 feet by latitude of –153.53 feet. If the value of –1535.3 is used instead of the correct –153.53, the result will not be correct.

2.3.10 Locating Mistakes

If a mistake cannot be located and corrected, the whole traverse must be rerun to find the mistake. This may be avoided by using of the following solutions to identifying mistakes.

2.3.10.1 Outsized Angular Error of Closure

The size of an outsized angular error of closure may be a clue to the location of the particular mistake. Suppose, for example, that a six-sided closed traverse has the following measured interior angles: 81612 128101 81147 214154 84118 8190 ° ′ ′ ° ° ′ ° ′ ° ′ ° ′ The interior angles in a six-sided closed traverse should add up to 720°00'. The difference between 720°00' and 612°18' is 107°42'. This large difference suggests that measuring about 107°42' was dropped along the way. You should look for an angle of about this size in the traverse. Suppose that in a four-sided traverse, the difference between the sum of the Rdeflection angles and the sum of the L-deflection angles comes to 180°. For a foursided traverse, this difference should be 360°. This large angle sum difference (180°) suggests one of the angles is the wrong direction. Look for an angle measuring about Figure 18-42 – Proper compass direction of a closed traverse.  18-41 half the error of closure (in this case, measuring half of 180°, or 90°), and see whether this angle is the wrong direction. If an angle has not been dropped, a large interior-angle error of closure probably means a large mistake in measuring or in recording the measuring of one of the angles. You may be able to locate the suspect angle by plotting the traverse from the measured angles. Then draw in the line of the linear error of closure and erect a perpendicular bisector from this line. The bisector may point to the suspect angle. For example, in Figure 18-43, all the bearings are correct except the bearing of CD, which should be S15°31'W for closure, but inadvertently is S05°31'W. Because of this error, the traverse fails to close by the length of the dashed line AA'. A perpendicular bisector from AA' points directly to the faulty angle C. If a perpendicular bisector from the line of linear error of closure does not point at any angle, the faulty angle may lie at the point of the beginning of the traverse. In Figure 18- 44, the bearings of all lines are correct for closure except that of the initial line AB. Line AB should be N29°09'E for closure but was plotted N16°09'E. A perpendicular from AA' does not point at any angle in the traverse.

2.3.10.2 Outsized Latitude and/or Departure Error of Closure

When both the latitudes and departures fail to close by large amounts, there is probably a mistake in an angle or a distance. When one closure is satisfactory and the other is not, a computational mistake is probably the cause of the outsized closure error.

2.3.10.3 Outsized Linear Error of Closure

Check for the following mistakes when an angular error of closure is within allowable limits and there is an outsized linear error of closure. 1. Ensure traverse lines have not been dropped. 2. Ensure each latitude and departure is in the correct column. 3. When computing latitudes and departures, ensure the correct cosine and correct sine are used. The latitude of a traverse line equals the product of the length Figure 18-44 – Graphical method to locate angular mistakes in a closed circuit (angle A). Figure 18-43 – Graphical method to locate angular mistakes in a closed circuit (angle C).  18-42 times the cosine of the bearing. The departure equals the product of the length times the sine of the bearing. 4. Ensure each bearing has the proper compass direction. Ensure the front bearing is used, not the back bearing. 5. Ensure all bearings and distances are copied correctly. 6. Ensure all cosines and sines are copied correctly. 7. Check for arithmetical errors. If these procedures do not identify the mistake, the traverse will have to be rerun. If a rerun is required, examine the direction of the line of linear error of closure on the plot. Often, the traverse line that contains the mistake is parallel to this line. If there is a line that is parallel, start the rerun at that point.

 Test Your Knowledge 5. Before beginning computations based on your field notes, you should perform which of the following actions? A. Reduce slope measurements B. Reduce angles to mean angles C. Check notes for completeness D. All of the above 6. When you are discussing computations, the term “adjustment” refers to ______. A. fudging of the numbers B. alignment of columns C. level computations D. equal distribution of the total error 7.  Level lines that begin and end on points that have fixed elevations, such as bench marks, are often called plane coordinates. A. True B. False 8. You have just completed a level circuit run of 2,640 feet. The error of closure was .021 feet. What order of precision is this leveling work? A. First B. Second C. Third D. Fourth 9. What is the first step in traverse computations? A. Calculating mean angles B. Checking notes for all data C. Correcting distance measurements D. Verifying the crew members 10. You are giving the location by plane coordinates of a point. In what terms should you give the location of that point in relation to the point of origin? A. Its distance east B. Its distance south C. Its distance north or south and the distance east or west D. Its distance south and west 11. A positive latitude and a negative departure characterize a traverse line bearing ______. A. NE B. NW C. SE D. SW 12. The term “inversing”refers to computing the ____. A. latitude and departure of a traverse line from its bearing and length B. bearing and length of a traverse line from its latitude and departure C. ratio of linear error of closure of a traverse from its length and error of closure D. latitude and departure from the corrected angles and distances 13. What is the initial step in finding the area of the figure by counting squares? A. Plot it to scale on graph paper B. Use a planimeter to determine the area C. Divide it into 100-foot squares D. Determine the bearings of the lines 14. A bearing that has the correct angular value, but the wrong compass direction is usually caused by which of the following surveying mistakes? A. Viewing the direction of a traverse line from the wrong station B. Dropping a traverse line C. Taking a reading from a wrong column of a traverse table D. Omitting the plus or minus sign of a written value 15. When you have an outsized error of closure for latitudes but not departures, what should you check? A. Mistake in an angle B. Mistake in a distance C. Error in arithmetic D. Dropped traverse line

Summary

This lesson introduced you to the theory and basic procedures used for indirect leveling, including barometric and trigonometric leveling. This lesson also detailed the procedures used to make level and traverse computations. The section on level computations addressed the adjusting of intermediate bench mark elevations, calculating allowable error, and adjusting level nets. The traverse computations section addressed open and closed traverses and the procedures used for adjusting and determining the areas of traverses. Finally, the lesson detailed some of the mistakes made when calculating the traverse areas.