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:
1.0.0 Indirect Leveling
2.0.0 Level and Traverse Computations SummaryReview Quiz 
Indirect methods of leveling include barometric leveling and trigonometric leveling. These methods are discussed in the following paragraphs.
1.1.0 Barometric Leveling
Barometric leveling makes use of the fact that differences in elevation are proportional to differences in the atmospheric pressure. Therefore, when the atmospheric pressure of a barometer is read at various points on the earth’s surface, you have a measurement of the relative elevation of these points. A mercurial barometer, aneroid barometer, or sensitive altimeter may be used for this purpose. However, the mercurial barometer is too cumbersome to take out into the field. Barometric leveling is used mostly in reconnaissance surveys where differences in elevations are large, such as in mountainous regions. Elevations determined by barometric leveling are typically several feet in error even after they are corrected for the effects of temperature and humidity. These errors are caused by daytoday pressure fluctuations, even by fluctuations from hour to hour. Because of these fluctuations, barometric observations are usually taken at a fixed station during the same period that observations are made on a second barometer that is carried from point to point in the field. The use of two barometers enables you to correct for atmospheric disturbances that could not be readily detected if only one barometer were used. This method is not normally used in construction surveying, except when a construction surveyor may need to run his or her own preliminary topographic control. Barometric or altimeter surveys are run by one of three methods: the singlebase, the twobase, and the leapfrog. The singlebase method requires a minimum number of observers and the least equipment. However, this method needs a series of corrections and is neither as practical nor as accurate as the other two. The twobase method is generally accepted as the standard method for accuracy and is the one most widely used. It requires fewer corrections than the singlebase method. The leapfrog method uses the same type of corrections as the singlebase, but the altimeters are always in close relationship to each other and are operating under reasonably similar atmospheric conditions. The results of the leapfrog method are more accurate than the singlebase method and compare favorably with the twobase method. There are several factors and limitations that must be observed in barometric leveling which are beyond the scope of this training manual. For actual barometric leveling, consult the instruction manual for the instrument being consulted. The theory of twobase barometric leveling is explained below. The twobase method is described here only to provide an idea of how the system works. This method uses at least three altimeters, one at each lower and upper base where elevations are initially known, and one or more roving where elevations are needed between the upper and lower base elevations. In this operation, points of unknown elevations to be determined must lie within the range of the elevations of the lower and upper base stations. The readings of the altimeters at the unknown elevations are taken at the same instant that both the upper and the lower base altimeters are read. If radio communication is not available, a timepiece is needed for each altimeter. 184 All of the timepieces need to be synchronized and the altimeter readings taken at prearranged intervals. Figure 181 – Diagram of a twobase altimeter survey. Figure 181 shows a diagram of the twobase method using three altimeters. The figure shows the known elevations of the lower (Sta. A) and upper (Sta. B) base stations. Altimeter readings at each of the base stations and at field station C are shown. The difference in elevation is computed by direct proportion, using either the lower base or the upper base as reference. For example, to find the differences in elevation between Sta. A and Sta. C, perform the following steps: 105 40 100 = .. ACElDiff ACElDiff 38 feet 105 40 x 100 .. = = Then the result is added to the elevation of Sta. A, as seen below: Elevation C = 100 + 40/105 x 100 = 100 + 38 = 138 feet When the upper base is used as a reference, compute the difference in elevation by using the same method. However, to compute from Sta. B, subtract the result, as shown below: Elevation C = 200 – 65/105 x 100 = 200 – 62 = 138 feet For a more accurate result, do not make altimeter surveys on days when there is not much variation in barometric pressure. Avoid windy days with rapidly moving detached clouds, as the alternating sunlight and shade over the survey area can cause fluctuations in the altimeter reading. Steady barometric pressures generally occur on days with gentle winds and an overcast sky. The recommended time for observations is 2 to 4 hours after sunrise and 2 to 4 hours before sunset. When possible, avoid midday observations. Remember to shade the instrument at all times, and avoid jarring the instrument suddenly during its transfer from one station to another. 185
When the vertical angle and either the horizontal or slope distance between two points is known, the fundamentals of trigonometry can be applied to calculate the difference in elevation between points. This is the basic principle of trigonometric leveling. This method of indirect leveling is particularly adaptable to rough, uneven terrain where direct leveling methods are impracticable or timeconsuming. As in any survey, the equipment used in trigonometric leveling depends on the precision required. For most trigonometricleveling surveys of ordinary precision, angles are measured with a transit, or alidade, and distances are measured either with a tape or by stadia. On reconnaissance surveys the vertical angles may be measured with a clinometer, and distances may be obtained by pacing. The method used in trigonometric leveling is described in the following paragraphs: In Figure 182, a transit is set up and leveled at A, which is also point O. The rodman holds the rod at point B. The instrumentman trains the telescope on C, which is an easily read value (usually a full foot) on the rod. With the telescope trained on C, the vertical angle (a) is read. Then either the horizontal distance or the slope distance between the instrument and the rod is determined. Now one side and one angle of the right triangle (OCD) are known. From your knowledge of trigonometry, you know that the other sides and angle can be computed. However, in trigonometric leveling, you are concerned only with determining the length of the side opposite the measured angle (side CD). The length of this side is the difference in elevation (DE). As seen in Figure 182 the DE is the distance between the height of instrument (HI) and the intersection of the line of sight with the rod (point C). Computing the DE consists of multiplying the measured distance by the proper trigonometric function of the measured angle: sine, when slope distance (OC) is measured; tangent, when horizontal distance (OD) is measured. Figure 182 – Difference in elevation in trigonometry leveling. 186 The following paragraphs discuss typical situations encountered in trigonometric leveling. 1. Depression angle backsight, shown in Figure 183. The rod is on point B below the instrument. The measured vertical angle (a) is a depression (minus) angle. To compute the HI, the rod reading RB and the DE are added to the elevation of B, or HI = RB + DE + Elev. B. 2. Depression angle foresight, shown in Figure 184. The rod is below the instrument, and the vertical angle is minus. The elevation at C equals the HI minus the DE and minus the rod reading RC, or Elev. C = HI – DE – RC. 3. Elevation angle backsight, shown in Figure 185. The rod is above the instrument, and the vertical angle is plus. The HI at F equals the elevation at C plus the rod reading (RC) and minus the DE, or HI = Elev. C + RC – DE. Figure 183 – Backsight Depression (Minus) Angle. Figure 184 – Foresight Depression (Minus) Angle. 187 4. Elevation angle foresight, shown in Figure 186. The rod is above the instrument and the angle is plus. The elevation of G equals the HI plus the DE and minus the rod reading (RG), or Elev. G = HI + DE – RG. As mentioned earlier in this section, the horizontal or slope distances used for calculating the DE may be obtained using various methods. For each method, there are requirements and limitations, which are discussed below. 1. Measured distances obtained by horizontal chaining should be corrected for standard error, temperature, and sag before computing the DE. These corrections are discussed in Chapter 12 of this NRTC. Under ordinary circumstances in the Seabees, corrections for earth curvature and refraction are not necessary. However, methods to perform these corrections can be found in commercial publications, such as Surveying Theory and Practice by Davis, Foote, Anderson, and Mikhail. 2. Measured distances obtained by slope chaining also should be corrected as discussed above. In addition, the slope distance must be converted to a horizontal distance before computing the DE. As an aid in computations, tables have been developed that provide the following data: a. Inclination corrections for 100foot tape Figure 185 – Backsight Elevation (Plus) Angle. Figure 186 – Foresight Elevation (Plus) Angle. 188 b. Differences in elevation for given horizontal distances and gradients from 0° to 45° c. Differences in elevation for given slope distances and gradients from 0° to 45° d. Horizontal distances for given slope distances and gradients from 0° to 45° 3. When using stadia, refer to the stadia procedures and formulas described in Chapter 12 of this NRTC. With practice, these provide a rapid means of determining horizontal distances and elevations. 4. Electronic distancemeasuring devices measure the straightline horizontal or slope distance between instruments. When the same setups slopes is used, replace the electronic equipment with a theodolite and either a target or a rod to measure the vertical angle. The measured vertical angle can be used to convert the measured slope distance to DE by multiplying by the sine of the vertical angle.
Test Your Knowledge 1. Barometric leveling computes the difference in elevations by which of the following measurements?
2. Barometric leveling is used mostly in which of the following types of surveys?
3. Barometric leveling should be performed at what time of day?
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?

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 sixtime 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
In making level computations, be sure to check the notes for a level run by verifying the beginning bench mark (BM). This checking includes making sure the correct BM was used and its correct elevation was properly recorded. Then check the arithmetical accuracy to ensure backlights are added and foresights are subtracted. The difference between the sum of the foresights taken on BMs or turning points (TPs) and the sum of the backlights taken on BMs or TPs should equal the difference in elevation between the initial BM or TP and the final BM or TP. This fact is shown in Figure 187. 1810 This procedure only checks the arithmetic. It does not indicate the accuracy of the measurement made in the field. Figure 187 – Differentiallevel circuit and notes for differential leveling. 2.2.1 Adjusting Intermediate Bench Mark Elevations Level lines that begin and end on points that have fixed elevations, such as bench marks, are often called level circuits. When leveling is accomplished between two previously established bench marks or over a loop that closes back on the starting point, the elevation determined for the final bench mark is seldom equal to its previously established elevation. The difference between these two elevations for the same bench mark is known as the error of closure. The remarks column in Figure 187 indicates the actual elevation of BM 19 is known to be 136.442 feet. The elevation found through differential leveling was 136.457 feet. The error of closure of the level circuit is 136.457 – 136.442 = 0.015 feet. Assume that errors have occurred progressively along the line over which the leveling was conducted. To adjust for this error, the error is equally distributed along the line as shown by the following example. Refer to Figure 187, and notice the total distance between BM 35 and BM 19, over which the line of levels was run, is 2,140 feet. The elevation on the closing BM 19 is found to be 0.015 foot greater than its known elevation. Therefore adjustments must be made for the intermediate BMs 16, 17, and 18. The amount of correction is calculated as follows: Correction = Error of closure [distance between the starting BM and the intermediate BM] distance between the starting and closing BM 1811 BM 16 is 440 feet from the starting BM. The total length distance between the starting and closing BMs is 2,140 feet. The error of closure is 0.015 foot. By substituting these values into the above formula, the correction is as follows: Correction 0.003 ft 2140 440 = 0.015 x −= Since the observed elevation of the closing BM is greater than its known elevation, the adjustments are subtracted from the intermediate BMs. Therefore, for BM 16, the adjusted elevation is 134.851 – 0.003 = 134.848. The adjustments for intermediate BMs 17 and 18 are made in a similar manner. 2.2.2 Calculating the Allowable Error The allowable error of closure depends on the precision required (first, second, or third order). The allowable error of closure in leveling is expressed in terms of a coefficient times the square root of the horizontal length of the actual route over which the leveling was accomplished. Most differential leveling (plane surveying) is thirdorder work. In thirdorder leveling, the closure is usually made on surveys of higher accuracy without doubling back to the bench mark at the original starting point of the level circuit. The length of the level circuit, therefore, is the actual distance leveled. For thirdorder leveling, the allowable error is: 050.0 ft theoflength circuitlevel in miles Refer again to Figure 187. By adding the sight distances in the sixth and seventh columns of the figure, you will find that the length of the level circuit is 2,140 feet (or 0.405 miles). The allowable error of closure, then, is ft = = 032.0)64.0(050.0405.0050.0 ft Since the actual error is only 0.015 foot, the results are sufficiently accurate for thirdorder precision. First and secondorder levels usually close on themselves. This means the leveling party runs a line of levels from an old BM or station to the new BM or station, and then doubles back to the old BM for closure. The actual distance leveled is twice the length of the level circuit. For secondorder leveling, the allowable error is 035.0 ft theoflength circuitlevel in miles Firstorder leveling is even more precise. The allowable error cannot be greater than 017.0 ft theoflength circuitlevel in miles 2.2.3 Adjusting Level Nets When a level survey system covers a large area, you, in turn, adjust the interconnecting network in the whole system. Adjustment of an interconnecting network of level circuits consists of adjusting, in turn, each separate figure in the net, with the adjusted values for each circuit used in the adjustment of adjacent circuits. This process is repeated for as many cycles as necessary to balance the values for the whole net. Within each circuit the error of closure is normally distributed to the various sides in proportion to 1812 their lengths. Figure 188 represents a level net made up of circuits BCDEB, AEDA, and engineering techBE. Along each side of the circuit is shown the length of the side in miles and the observed difference in elevation in feet between terminal BMs. The difference in elevation (plus or minus) is in the direction indicated by the arrows. Within each circuit is shown its total length (L) and the error of closure (Ec) which is determined by summing up the differences in elevation in a clockwise direction. Figure 189 shows the computations required to balance the net. The circuits, sides, distances (expressed in miles and in percentages of the total), and differences in elevation (DE) are listed. Figure 188 – Adjustment of level nets. Figure 189 – Computations required to balance the level net. 1813 For circuit BCDEB, the error of closure is –0.40 foot. This is distributed among the lines in proportion to their lengths. Thus, for the line BC, the correction is 0.07 feet 70 12 0.40 x = Notice that the sign is opposite to that of the error of closure. The correction of +0.07 foot is entered on the first line of the column headed CORR and is added to the difference in elevation (10.94 + 0.07 = +11.01). The sum is entered on the first line under the heading CORR DE (corrected difference in elevation). The same procedure is followed for the remaining lines CD, DE, and EB of circuit BCDEB. The sum of the corrections must have the opposite sign and be equal to the error of closure. The algebraic sum of the corrected differences in elevation must equal zero. The lines in circuit AEDA are corrected in the same manner as BCDEB, except that the corrected value of ED (+27.08 instead of +27.15) is used. The lines of engineering techBE are corrected using the corrected value of engineering tech (+17.97 instead of +17.91) and BE (+5.13 instead of +5.23). In the column Cycle II, the procedure of Cycle I is repeated. You should always list the latest corrected value from previously adjusted circuits before computing the new error of closure. The cycles are continued until the corrections become zero. The sequence in which the circuits are taken is immaterial as long as they are repeated in the same order for each cycle. Computations may be based on corrections rather than differences in elevation.
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.
2.3.3 Adjusting Angles
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 1810 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 deflectionangle 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 1810 – Close traverse by deflectionangle 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 northtosouth meridian running through the point of origin. The departure of a traverse line means the length of the line as projected on the easttowest parallel running through the point of origin, as shown in Figure 1811). 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 1812 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 1813. 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 1812 – Latitude equals length of traverse line times cosine of angle of bearing. Figure 1811 – Latitude and departure. 1816 computations, north latitudes are designated plus and south latitudes minus; east departures are designated plus and west departures minus. Figure 1814 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 1813 – Departure equals length of traverse line times sine of angle of bearing. Figure 1814 – 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 1815 shows a closed traverse with bearings and distances notes. Figure 1815 – Closed traverse by bearings and distance. 1818 Figure 1816 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 1816 – 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 1817 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
In computing latitudes and departures, arithmetical calculations can be greatly expedited by the use of a traverse table. The latitudes and departures for any bearing and distance can be determined mostly by using the tables. Figure 1817 – Form for adjusting latitudes and departures. 1820 Figure 1818 shows sample pages from a table that gives angleofbearing values to the nearest quarter degree (15'). More precise tables give angular values to the nearest 01'. Figure 1818 – Sample pages from traverse table. Under each of the bearing values at the head of the page, a double column gives latitudes and departures for distances of from 1 to 100 feet. For a particular traverse line, you determine the latitudes and departures by breaking down the distance, moving decimal points, and adding up results as in the following example. Suppose you want to determine the latitude and departure for a traverse line 725.32 feet long, bearing N15°30'E. Perform the following steps to find the latitude. In the latitude column 15 1/2°, identify the latitude for 70 feet. You read 67.45 feet. If the latitude for 70 feet is 67.45 feet, the latitude for 700 feet is 674.50 feet. Note this in your notes. Next, you look up the latitude for 25 feet under the same 15 1/2° latitude column, which is 24.09 feet. The latitude for 725 feet, then, is 674.50+ 24.09= 698.59 feet. Finally, for the 0.32 foot, look up the latitude for 32 feet, which is 30.84 feet. If the latitude for 32 feet is 30.84 feet, the latitude for 0.32 foot must be 0.3084 foot, which rounds off at 0.31 foot. The numerical value of the latitude then is 698.59 + 0.31 = 698.90 feet. Because the line AB bears northeast, the latitude is positive. Departure is calculated in the same way by using the departure column. Finally, enter the adjusted latitudes and adjusted departures in the last two columns. Determine the values in each case by applying the correction to the original latitude or departure. Note that the negative latitudes now equal the positive latitudes and the negative departures equal the positive departures. This indicates the errors of closure have been entirely distributed. With the adjusted latitudes and departures, you can now adjust the original bearings and distances by the method called inversing. Inversing simply means computing the bearing and length of a traverse line from the latitude and departure. Again the process is one of simple triangle solution. Figure 1819 shows traverse line AB with the adjusted latitude and departure noted. To determine the adjusted angle of bearing, you solve the triangle AA'B for angle A'AB as follows: 0.06468 583.88 37.70 tan A′ AB = = A′AB °− 243 ′ The adjusted bearing of AB, then, is N3°42'E. For the adjusted distance, solve the triangle for AB as follows: AB 584.22 feet 0.06453 37.70 sin 243 37.70 = = ° ′ = The adjusted length of AB, then, is 584.22 feet.
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 1820 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 1819 – Adjusted bearing and distance from adjusted latitude and departure.
Figure 1820 – Location by plane coordinates.
2.3.6.1 Plane Coordinates from Latitude and Departure
Figure 1820 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 1821 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 1821 – Closed traverse with adjusted latitudes and departures. 1823 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 1822. 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 1822 – Form for computing coordinates. 1824 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 1823, 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 1823 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 onehalf 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 1823 – Meridian distances. 1825 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, divide the double area by 2. Figure 1824 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 1826
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 1825, 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. 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 1826 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 1825 – Parallel distances. Figure 1824 – 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 1827. Figure 1827 – Closed traverse by coordinate method. Figure 1826 – Area from double parallel distances. 1828 Figure 1828 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 1828 – Coordinate entries for computation of Figure 1827. 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. 1829 Figures 1829 and 1830 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 1829, and determine the sum of the products. Then, multiply pairs diagonally in the opposite direction, as shown in Figure 1830, and determine the sum of the products. The difference between the sums (shown in Figure 1829) 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 1829 – First step for tabulated computation of Figure 1827. 1830 Figure 1830 – Second step for tabulated computation of Figure 1827.
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 1831 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 1831 – Area of irregular figure by trapezoidal rule. 1831 To determine the area of this figure, first lay off conveniently equal intervals (in this case, 50.0foot 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 1831, 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 1832 shows the same figure shown in Figure 1831 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. 1832 Figure 1832 – 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 rightofway lines of curved roads. Figure 1833 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 straightline and curvedline boundaries. The data for each of the curves is inscribed on Figure 1833, 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 straightline and chord (also straightline) boundaries. 2. Determine the segmental areas. 3. Subtract the segmental area for Curve 1 from the straightline boundary area. 4. Add the segmental area for Curve 2 to the straightline boundary area. The method of determining a segmental area was explained in the engineering tech Basic Chapter 2. The straightline area may be determined by the coordinate method, as explained in this lesson. For Figure 1833, the segmental area for Curve 1 works out to be 5,151 square feet; for Curve 2, it is 29,276 square feet. 1833 Figure 1833 – Area within straightline and curvedline boundaries (curved segments). Figure 1834 shows a typical computation sheet for the area problem shown in Figure 1833. 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 1834, the area within the straightline 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 1833, the segmental area for Curve 1 lies outside and must be subtracted from the straightline 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. 1834 Figure 1834 – 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 1835. 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 1836 follows the same pattern as Figure 18 34. However, there are two more straightline 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 1836. When the chord bearing is known, the tangent bearing can be computed by adding or subtracting onehalf of delta (∆) as appropriate. The angle between the tangent and the chord equals ∆2. After setting coordinates on the PIs, crossmultiply, accumulate the products, subtract the smaller from the larger, and divide by 2, as before, to get the area of the straightline figure running around the tangents. Then add or subtract each external area as appropriate. In Figure 1835, 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. 1835 Figure 1835 – Area within the curve and its tangents. Figure 1836 – 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 1837 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 1838 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 1837 – 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 1837 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 1839 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 1838 – Plotting traverse lines by parallel method from a single meridian. Figure 1839 – Plotting by tangentoffset method from deflection angles larger than 45 degrees. 1838 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 (tangentoffset 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 1839, 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 1840. Here you measure off a convenient roundfigure 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 1840 – Plotting by tangentoffset 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 1841. 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.
2.3.9.3 Wrong Quadrant
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 1842, 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 1842 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 1841 – 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 sixsided closed traverse has the following measured interior angles: 81612 128101 81147 214154 84118 8190 ° ′ ′ ° ° ′ ° ′ ° ′ ° ′ The interior angles in a sixsided 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 foursided traverse, the difference between the sum of the Rdeflection angles and the sum of the Ldeflection 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 1842 – Proper compass direction of a closed traverse. 1841 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 interiorangle 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 1843, 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 1844 – Graphical method to locate angular mistakes in a closed circuit (angle A). Figure 1843 – Graphical method to locate angular mistakes in a closed circuit (angle C). 1842 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?
6. When you are discussing computations, the term “adjustment” refers to ______.
7. Level lines that begin and end on points that have fixed elevations, such as bench marks, are often called plane coordinates.
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?
9. What is the first step in traverse computations?
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?
11. A positive latitude and a negative departure characterize a traverse line bearing ______.
12. The term “inversing”refers to computing the ____.
13. What is the initial step in finding the area of the figure by counting squares?
14. A bearing that has the correct angular value, but the wrong compass direction is usually caused by which of the following surveying mistakes?
15. When you have an outsized error of closure for latitudes but not departures, what should you check?

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.
1. What is the minimum number of altimeters required for a twobase method of barometric leveling? A. Five B. Two C. Three D. Four 2. The basic principle of trigonometric leveling is when the vertical angle and either the horizontal or slope distance between two points is known, the fundamentals of trigonometry can be applied to calculate the difference in elevation between points. A. True B. False 3. In making level computations, be sure to check the notes for a level run by verifying the ______________. A. turning point B. beginning bench mark C. turning point and beginning bench mark D. none of the above 4. When the horizontal or slope distance is measured by chaining, which of the following corrections 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 5. When you do trigonometric leveling with an electronic distancemeasuring device, what other instrument do you use to measure your angles? A. Transit B. Theodolite C. Level D. Alidade 6. What are you verifying by checking the difference in foresights and backsights? A. Balanced foresights and backsights B. Arithmetic accuracy C. Error of closure D. BM elevation 1846 7. Most differential leveling (plane surveying) is ________ order work. A. First B. Second C. Third D. Fourth 8. When adjusting a level net, you adjust each leg only once, regardless of a leg being in more than one circuit. A. True B. False 9. The terms “elevation angle” and “depression angle” refer to which of the following situations? A. Rod location in relation to the instrument B. Value of the vertical angle C. Instrument location in relation to the rod D. All of the above 10. How is distribution of angular error of closure performed? A. Proportionally by angle size B. Proportionally by distance between angles C. Equally among all angles D. By subtracting the error from all angles 11. Traverse operations are conducted for mapping large construction projects, such as a A. military post B. grocery store C. school D. mall 12. In computing latitudes and departures, calculations can be greatly expedited by the use of a traverse table. A. True B. False 13. The linear error of closure is determined by the __________________. A. sum of the departures. B. difference in the departures and latitudes. C. error of latitude. D. use of the Pythagorean theorem with error of closure in latitude and departure. 1847 14. What is the most common method of balancing a traverse? A. Rule of thumb B. Compass rule C. Closure rule D. Grid rule 15. Plane coordinates describe the location of a point in what manner? A. By a distance north or south and east or west from a point of origin B. By the bearing north or south and east or west from a point of origin C. By the bearing and distance from a point of origin D. By the line of sight from a point of origin 16. What is the major difference between the DMD method of determining area and the DPD method? A. The DPD of the initial traverse line is twice the departure. B. A reference parallel is used instead of a reference meridian. C. The distances are calculated using latitude instead of departure. D. Both B and C 17. After you have determined the segmental areas, what do you do with this information to find the area of the whole parcel? A. Always subtract the areas from the area of the parcel. B. Always add the areas to the area of the parcel. C. Determine if the segmental areas are inside or outside the parcel straightline and chord boundaries, then add or subtract as required. 18. How may you prevent direction error? A. By placing directional arrows on the traverse diagram B. By quickly adding up the angles to ensure the traverse lines were not dropped C. By making both readings and field note entries with deliberation D. By never writing a value without including the appropriate sign 19. Which of the following errors uses the same prevention method as direction error prevention? A. Dropped signs B. Dropped traverse lines C. Wrong azimuths D. Missing decimal points 20. When you have an outsized linear error of closure, what should you check first? A. Accuracy of latitudes and departures B. A dropped traverse line C. Your arithmetic D. Bearings computations 1848 21. After you check computations and ensure that no angle was dropped in the process, you still have a large error of closure. What further check, if any, should you attempt at this time? A. See if an angle is exactly equal to the error of closure; if so, there is a deflection angle error. B. Construct a perpendicular bisector from the line of the linear error of closure to indicate a possible erroneous measurement. C. Construct a line parallel to the suspected erroneous line to find the error. D. None; you must return to the field. 22. When you have an outsized linear error of closure but an acceptable angular error of closure, you should check to see whether you used the _____________________. A. sine of the bearing when finding the latitude of the course. B. sine of the bearing when finding the departure of the course. C. cosine of the bearing when finding the departure of the course. D. tangent of the bearing when finding the latitude of the course. 23. You cannot locate your error and the traverse will not close; therefore, you must rerun the traverse. With which line should you begin? A. Traverse line parallel to the linear error of closure B. Initial traverse line C. Final traverse line D. Line perpendicular to the line of linear closure at its midpoint 1849 Trade Terms Introduced in this Chapter Clinometer An instrument used to establish angles, usually on a hillslope and more rarely along bedding planes.