Gheorghe M. T. Radulescu
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Gheorghe M. T. Radulescu Document Transcript

  • 1. GHEORGHE M.T.RADULESCUGENERAL TOPOGRAPHY TUTORIALS
  • 2. FOREWORD The practical applications represent the basis for learning General Topography. Irecommend to all those who wish to be initiated and to improve themselves in this area,to use the three manuals (lecture notes, tutorials, problems) in parallel, by chapter, in thepresented order. I ensure them that, if they respect this suggestion, the results will be as expected. I want to thank my two colleagues from the department within the PolytechnicInstitute from Cluj-Napoca: Mrs. Viorica Balan and Mr. Gheorghe Bendea, for thesuggestions they have given me concerning the presentation of practical applications. The Author II
  • 3. TABLE OF CONTENTSFOREWORD....................................................................................................................IITABLE OF CONTENTS................................................................................................III1. TOPOGRAPHIC ELEMENTS OF THE TERRAIN – MEASURING UNITSAND COMPUTATIONAL MEANS IN TOPOGRAPHY.............................................1 1.1. THE TOPOGRAPHIC ELEMENTS OF THE TERRAIN.......................................1 1.1.1. CLASSIFICATION............................................................................................1 1.2.2. DISTANCE MEASURING UNITS.....................................................................3 1.2. TOPOGRAPHIC SURFACES.................................................................................3 1.2.1. ACTUAL SURFACES AND HORIZONTAL SURFACES.................................3 1.2.2. SURFACE MEASURING UNITS......................................................................4 1.3. ANGULAR TOPOGRAPHIC ELEMENTS............................................................4 1.3.1. ANGLES MEASURED IN TOPOGRAPHY.......................................................4 1.3.2. THE ANGLE IN GEOMETRY AND TOPOGRAPHY.......................................5 1.3.3. ANGLE MEASURING UNITS...........................................................................5 1.3.4. REVISION OF TRIGONOMETRIC NOTIONS, THE TRIGONOMETRIC CIRCLE.......................................................................................................................7 1.3.5. ORIENTATIONS, THE RELATION BETWEEN COORDINATES AND ORIENTATIONS........................................................................................................11 1.4. PROBLEMS FOR TUTORIAL 1...........................................................................13 1.4.1. SOLVED PROBLEMS.....................................................................................13 1.5. EXAMPLE FOR SOLVING THE HOMEWORK (FOR (N) = 0)..........................152. STUDYING THE THEODOLITE.............................................................................19 2.1. THE GENERAL CONSTRUCTION SCHEMA OF A THEODOLITE...............19 2.1.1. GENERAL NOTIONS, CLASSIFICATIONS...................................................19 2.1.2. THE PRINCIPLE SCHEMA – AXES AND COMPONENT PARTS................20 2.1.3. THE DETAILED SCHEMA OF THE THEODOLITE (figure 2.4.)................22 2.1.4. THE DETAILED SCHEMA OF OTHER TYPES OF THEODOLITES...........24 2.2. WORKING PROCEDURE FOR THE THEODOLITE.........................................25 2.2.1. WORKING PRINCIPLES................................................................................25 2.2.2. VERIFYING THE DEVICE.............................................................................25 2.2.3. PLACING INTO THE STATION.....................................................................26 2.2.4. AIMING AND POINTING...............................................................................28 2.2.5. DEVICES FOR READING ANGULAR VALUES ON THE THEODOLITE...29 2.2. THE HOMEWORK OF THE TUTORIAL............................................................313. MEASURING ANGLES WITH THE THEODOLITE...........................................32 III
  • 4. 3.1. THE NATURE OF TOPOGRAPHIC ANGLES....................................................32 3.2. ANGLE MEASURING METHODS......................................................................34 3.2.1. THE CASE OF MEASURING ONE ANGLE...................................................35 3.2.2. MEASURING MORE ANGLES FROM ONE THEODOLITE STATION........39 3.3. THE HOMEWORK OF THE TUTORIAL............................................................42 3.4. EXAMPLE FOR SOLVING THE HOMEWORK.................................................434. DIRECT AND INDIRECT DISTANCE MEASURING METHODS....................47 4.1. MEASURING DISTANCES DIRECTLY.............................................................47 4.1.1. INSTRUMENTS FOR THE DIRECT MEASUREMENT OF DISTANCES.....47 4.1.2. PREPAIRING THE TERRAIN FOR MEASUREMENTS................................48 4.1.3. CORRECTIONS APPLIED TO LENGTHS MEASURED DIRECTLY...........50 4.2. MEASURING DISTANCES INDIRECTLY.........................................................52 4.2.1. TACHEOMETRIC METHODS AND INSTRUMENTS...................................52 4.3. THE HOMEWORK OF THE TUTORIAL............................................................575. PLANIMETRIC TRAVERSE....................................................................................61 5.1. GENERAL PROBLEMS........................................................................................61 5.1.1. CLASSIFICATIONS........................................................................................61 5.1.2. CONDITIONS THAT CHARACTERIZE A TRAVERSE:................................64 5.2. DESIGNING THE TRAVERSE............................................................................65 5.2.1. MEASUREMENT PREPARING WORKS........................................................65 5.3. FIELD WORKS......................................................................................................65 5.3.1. PERFORMING MEASUREMENTS................................................................65 5.4. OFFICE WORKS...................................................................................................66 5.5. THE HOMEWORK OF THE TUTORIAL............................................................66 5.6. THE SOLUTION OF THE HOMEWORK FOR N = 0...........................................666. SURVEYING PLANIMETRIC DETAILS...............................................................78 6.1. SURVEYING DETAILS THROUGH RADIATION (POLAR COORDINATES) ........................................................................................................................................78 6.2. SURVEYING DETAILS THROUGH SQUARING (SQUARE COORDINATES) ........................................................................................................................................83 6.3. SURVEYING DETAILS THROUGH THE METHOD OF ALIGNMENTS.......84 6.4. THE HOMEWORK OF THE TUTORIAL............................................................857. PROBLEMS SOLVED ON MAPS AND PLANS.....................................................87 7.1. PROBLEMS CONCERNING USING MAPS AND PLANS................................88 7.1.1. SYMBOLS........................................................................................................88 7.1.2. THE GRATICULE OF MAPS AND PLANS....................................................88 7.1.3. THE SCALE OF MAPS AND PLANS.............................................................92 7.1.4. ORIENTING MAPS AND PLANS IN THE FIELD.........................................93 7.2. SOLVING SOME PLANIMETRY AND LELVELING PROBLEMS ON A TOPOGRAPHIC PLAN................................................................................................94 7.2.1. THE HOMEWORK OF THE TUTORIAL.......................................................94 7.2.2. SOLVING METHOD EXAMPLE....................................................................968. THE STUDY OF LEVELING INSTRUMENTS....................................................107 IV
  • 5. 8.1. SIMPLE INSTRUMENTS – WITHOUT TELESCOPE.....................................107 8.8.1. THE LEVEL HOSE........................................................................................107 8.1.2. THE LEVELING LONG BOARD AND THE AIR-BUBBLE LEVEL............109 8.2. LEVELING INSTRUMENTS WITH TELESCOPE...........................................110 8.2.1. RIGID INSTRUMENTS FOR GEOMETRIC LEVELING.............................110 8.2.2. AUTOMATIC INSTRUMENTS FOR GEOMETRIC LEVELING.................114 8.2.3. ACCESSORIES FOR GEOMETRIC LEVELING DEVICES........................116 8.3. THE HOMEWORK OF THE TUTORIAL..........................................................1219. METHODS FOR MEASURING ALTITUDE DIFFERENCES...........................122 9.1. GENERAL PRINCIPLES....................................................................................122 9.1.1. THE PRINCIPLES OF GEOMETRIC LEVELING.......................................122 9.1.2. THE PRINCIPLES OF TRIGONOMETRIC AND TACHEOMETRIC LEVELING..............................................................................................................125 9.2. APPLICATIONS..................................................................................................12910. GEOMETRIC LEVELING TRAVERSE WITH RADIATIONS......................138 10.1 THE GENERAL CONDITIONS OF A GEOMETRIC LEVELING TRAVERSE ......................................................................................................................................138 10.2. RECOGNIZING AND PREPARING THE ROUTE OF THE TRAVERSE.....139 10.3. FIELD WORKS – MEASUREMENTS THAT ARE PERFORMED IN THE CASE OF SIMPLE MIDDLE GEOMETRIC LEVELING TRAVERSE, SUPPORTED AT THE ENDS............................................................................................................139 10.4. OFFICE OPERATIONS.....................................................................................14411. LEVELING OF PROFILES AND SURFACES...................................................148 11.1 LONGITUDINAL AND TRANSVERSAL LEVELING THROUGH PROFILES ......................................................................................................................................148 11.1.1 THE CONDITIONS OF THE ROUTE.........................................................148 11.1.2. RECOGNIZING THE TERRAIN AND PREPARING THE ROUTE...........149 11.1.3. PERFORMING FIELD MEASUREMENTS................................................149 11.1.4. OFFICE OPERATIONS..............................................................................150 11.2. SURFACE LEVELING......................................................................................152 11.2.1. SURFACE LEVELING THROUGH THE METHOD OF SMALL SQUARES .................................................................................................................................152 11.2.2. SURFACE LEVELING THROUGH THE METHOD OD LARGE SQUARES .................................................................................................................................154 11.3. USING THE DATA OBTAINED THROUGH SURFACE LEVELING..........155 11.4. THE HOMEWORK OF THE TUTORIAL........................................................157 V
  • 6. 1. TOPOGRAPHIC ELEMENTS OF THE TERRAIN – MEASURING UNITS AND COMPUTATIONAL MEANS IN TOPOGRAPHY The content of the tutorial: The topographic surveys needed for drafting plansand maps consist in measuring the relation in which the topographic points that define asurface are, either using a control network (the planimetric problem), or using ahorizontal datum level (the leveling problem). Actually – the linear elements (horizontal and vertical distances) and the angularelements (horizontal and vertical angles) are measured in the field, formed bytopographic points and reference elements. The purpose of this work is to determine thetopographic elements of the terrain, the relation between them: what does measuringthem represent, which are the measuring units that are used and which are the auxiliarymeans used for computations.1.1. THE TOPOGRAPHIC ELEMENTS OF THE TERRAIN1.1.1. CLASSIFICATION a. The nature of topographic elements Consider two topographic points A and B in the field, materialized in some form(wood or metal stake, concrete boundary marks, etc.). In what concerns these points, we can identify the following topographicelements: - THE AB ALIGNMENT – which represents the intersection of the topographic surface of the terrain with a vertical plan that passes through the given points. In practice, the sinuous line is geometrized (approximated) by a right line, which represents the direction materialized in the field by the points A and B. - THE INCLINED DISTANCE LAB – represents the line segment limited by the points A and B on the direction mentioned above. 1
  • 7. B L AB ΔZ AB Z ZB A ϕ DAB ZA Vertical datum Figure 1.1. The linear elements measured in the field - THE HORIZONTAL DISTANCE DAB – represents the projection of the slanted distance on the horizontal plan, having as value the horizontal segment between the verticals of the given points. - THE HEIGHTS ZA and ZB – of the points A and B – represent the value of the vertical segment between the vertical datum and that point. - THE ALTITUDE DIFFERENCE ∆ZAB – between the given points – represents the vertical distance measured between the horizontal plans that pass through these points ∆ZAB = ZB – ZA (1.3’). b. Relations between the topographic elements The relation in which are the elements presented above results from theexpression of the trigonometric functions of the angle ϕ – called slope angle (being theangle formed by the distances LAB and DAB). ∆ZAB sin ϕ = -------- (1.1) LAB DAB cos ϕ = ------- (1.2) LAB 2
  • 8. ∆ZAB tg ϕ = --------- (1.3) DAB L²AB = D²AB + ∆Z²AB (1.4) Using these formulas we can determine the unknown elements based on theknown (measured) ones. Measuring the linear elements presented above consists in comparing their size,using a chosen etalon (measuring unit).1.2.2. DISTANCE MEASURING UNITS Most of the countries use the meter (m) as distance measuring unit. Determined in 1799 by the French DELAMBRE and initially considered as beingthe 40,000,000th part of the length of the terrestrial meridian, and by more recentcomputations, as the 40,000,000.42th part, now it is defined (since 1960) as being equal to1,650,763.73 wavelengths of the orange radiation produced by the KRYPTON 86 gas. The multiples of the meter are: 1 km = 10 hm = 100 dam = 1000 m, and thesubmultiples: 1 m = 10 dm = 100 cm = 1000 mm. The measuring units of the Anglo-Saxon system are given the appendix tables inGENERAL TOPOGRAPHY – lecture notes.1.2. TOPOGRAPHIC SURFACES1.2.1. ACTUAL SURFACES AND HORIZONTAL SURFACES The topographic surface (St) is the actual surface of the terrain, which, not havinga regular form, cannot be mathematically expresses, and therefore, it cannot bemathematically represented. For this reason a geometric schematization of the terrain is performed by choosingthe characteristic points. It should be mentioned that, since the constructions are 3
  • 9. performed with horizontal foundations, the horizontal projection of the surface of theterrain (S in figure 1.2.) is represented on all topographic maps and plans. St S Figure 1.2. Surfaces in topography1.2.2. SURFACE MEASURING UNITS Derived from the metric system, the measuring unit for surfaces is the squaremeter (m2) with the following multiples and submultiples: 1 km² = 100 ha, 1 ha = 100 ari = 10,000 m²; 1 m² = 100 dm² = 10.000 cm² = 1,000,000 mm². In the appendix tables from GENERAL TOPOGRAPHY – lecture notes there arepresented the measuring units of the Anglo-Saxon system, too.1.3. ANGULAR TOPOGRAPHIC ELEMENTS1.3.1. ANGLES MEASURED IN TOPOGRAPHY In topography there are measured horizontal and vertical angles. In figure 1.3, the angle α is horizontal, being the angle formed by the horizontalprojections of two aiming lines. The vertical angles (ϕ) are formed by some direction with its horizontalprojection. 4
  • 10. (V AB) (VBC ) A ZA ZB B CA Z ϕA ϕB HC ∆ α D CA DCB A0 B0 Figure 1.3. The angular elements measured in the field The vertical angle formed by a straight line, which represents the support of aslanted distance, between two points, with its horizontal projection, is called slope angle(figure 1.3. the angles ϕA and ϕB). Usually, the theodolites (topographic devices used for measuring angles) recordthe angle Z, called zenithal angle, and the vertical angles result by computation.1.3.2. THE ANGLE IN GEOMETRY AND TOPOGRAPHY The geometric notion of angle – as shape formed by two half-lines having thesame origin – is incomplete for topographic use, knowing the sign and the sense ofmeasuring the angle being necessary. Thus, the topographic angles are oriented angles, the first side of the angle and thesense of measuring being known. By measuring an angle one can understand comparingit with another angle, chosen as unit.1.3.3. ANGLE MEASURING UNITS In topography the new (centesimal) degrees are usually used as measuring units. 5
  • 11. One centesimal degree (1g) represents the hundredth part of the right angle (D) orthe 400th part of the entire circle (C). D C 1g = ----- = ------ (1.5) 100 400 Submultiples: 1g = 100c (centesimal minutes); 1c = 100cc (centesimal seconds). Most of the measuring instruments in topography are graduated in centesimaldegrees. The advantage of this system consists in the simplicity of the operations, thedivision of degrees being done in the decimal system. E.g. 123g32c17cc = 123g.3217 Other measuring units: Sexagesimal degrees (10), which represent the 90th part of the right angle (D) orthe 360th part of the entire circle (C): D C 0 1 = ----- = ------ (1.6) 90 360 Submultiples: 10 = 60’ (sexa minutes); 1’ = 60” (sexa seconds). The radian (1 RAD) is the central angle corresponding to the arc of circle equal tothe circle radius. It is known that a circle has 2πRAD. For various computations is required to pass from one gradation system toanother, this transformation being performed through one of the equivalence relations: π C 100g = 900 = ------ RAD = 1D = ------ (1.7) 2 4 α0 αg a(RAD) ------- = ------ = ------------- (1.8) 1800 2000 π 1 RAD ≅ 63g66c20cc≅ 57017’5” (1.9) 10 = 1g.111… 1g = 54’ 1’ = 1c85cc.2 1c = 52”.4 (1.10) 1” = 3cc.09 1cc= 0”.34 6
  • 12. For transforming in radians, with the use of formula (1.4), the followingcoefficients are obtained for the centesimal gradation: 200g ρg = -------- = 63.661977… π 200g x 100c ρ = -------------- = 6366.1977… c π 200g x 100c x 100cc ρcc = ------------------------- = 636619.77… π and for the sexa gradation: (1.11) 1800 ρ0 = -------- = 57.295779… π 1800 x 60’ ρ = ------------- = 3437.7467… ’ π 1800 x 60’x 60” ρ = -------------------- = 206264.80… ,, π Letting π = 3.14159265… It can be said that: 1g = 0.015708 RAD (1.12) 10 = 0.017453 RAD1.3.4. REVISION OF TRIGONOMETRIC NOTIONS, THE TRIGONOMETRIC CIRCLE a. The trigonometric circle and the topographic circle The computations performed in topography need a thorough knowledge oftrigonometric functions, of the trigonometric circle, which, in topography, is transformedin the topographic circle. We define the trigonometric circle as the circle having the center in a pointdenoted by 0, the radius equal to the unit, having the origin for measuring arcs in thepoint A and the measuring sense in left-handed direction (figure 1.4.a). 7
  • 13. Y II -ctg B +ctg V I ctg P +cos N tg -cos R=1α +sin α +sin A -sin A’ 0 M -sin α -tg X α +cos - cos III IV B’ Figure 1.4.a. The trigonometric circle A V X II - tg A +tg I tg T M +cos N +ctg -cos θ’ =0 c α cos cos + + α θ” P B’ 0 θ” cos B cos - -ctg Y α - θ ”’ α +sin -sin III IV A’ Figure 1.4.b. The topographic circle In topography, the trigonometric circle is replaced by the topographic circle(figure 1.4.b), for the following reasons: 8
  • 14. - The reference direction in the field, and therefore in topography, is the direction of the topographic North, which coincides with the ordinate axis (this is why this axis is denoted here by 0X); - The sense of measuring angles, in topography, is the right-handed direction. Consequently, it can be seen, comparing the two circles, that in the topographiccircle, the quadrants II and IV are switched, and the quadrants I and III stay in the sameposition as in the trigonometric circle. Therefore, the order of the quadrants is given by the sense of measuring angles.Since one of the characteristics of the trigonometric circle is that the origin and the senseof measuring arcs can be changed without changing the rules and formulas established onthe quadrants, in the two circles, the formulas and the signs of the trigonometric functionsare identical. Hence: defining the trigonometric functions and the variation oftrigonometric lines is equivalent in both circles (see tables 1.1 and 1.2). b. Reduction to the first quadrant, determining the values of trigonometric functions The trigonometric functions of some given angles θ, located in the quadrants II –IV, can be determined as functions of some corresponding angles in the first quadrant –α. The transformation formulas for passing to the first quadrant presented in table 1.3 arebuilt in the following way:Table 1. 1 – The corresponding sign and line of the trigonometric functions, in the four quadrants The function The trigonometric The sign in the quadrants line I II III IV sin MN + + - - cos OM + - - + tg AT + - + - ctg BV + - + - - The sign of the function for the four quadrants is the one specified in table 1.1; - For quadrant I, the functions have the significance that results from figure 1.4.a and 1.4.b and from table 1.1; 9
  • 15. - For quadrant III, the function of the angle from this quadrant is equivalent to the function of the angle from the first quadrant, obtained by subtracting 200 g from the initial angle;Table 1. 2 – The variation of trigonometric lines The quadrant I II III IV θ The trigonometric line 0g 100g 200g 300g 400g sinθ 0 +1 0 -1 0 cosθ +1 0 -1 0 +1 g g tgθ +∞ 100 +ε 0 +∞ 300 +ε +1 g g 0 100 -ε -∞ 300 -ε -∞ g ctgθ +∞ 200 -ε +∞ 0 -∞ 200g+ε 0 - For quadrants II and IV, the function is equivalent to the cofunctions of the angles from the first quadrant (obtained by subtracting 100g and 300g, respectively, from the initial angle); These rules lead to establishing table 1.3, which is used in the following way:Table 1. 3 – The values of the trigonometric functions in the four quadrants Quadrant I II III IV θ = given angle θ* =α θ” =α + 100g θ” =α + 200g θIV =α + 300gα= reduced angle sinθ + sinα + cosα - sinα - cosα cosθ + cosα - sinα - cosα + sinα tgθ + tgα - ctgα + tgα - ctgα ctgθ + ctgα - tgα + ctgα - tgα - Having an angle θ, which can be found in one of the four quadrants and knowing the fact that there exist tables of natural values of trigonometric functions only for the angles situated in the first quadrant, it becomes necessary to transform the function of the angle θ into that corresponding to quadrant I. Depending on the quadrant in which is situated the angle θ, it can be expressed as: 10
  • 16. θI = α (1.13) θII = α + 100g θIII = α + 200g θIV = α + 300g corresponding to the quadrants I, II, III, and IV. Extract the trigonometric function, from the mentioned table, from theintersection of the line corresponding to the initial function (of the angle θ) with thecolumn corresponding to the quadrant in which θ is found.1.3.5. ORIENTATIONS, THE RELATION BETWEEN COORDINATES AND ORIENTATIONS The orientation is the horizontal angle formed by some direction in the field, or onthe plan (map) with the direction of the topographic North, parallel to the 0X axis of thecoordinate system, and it is denoted by θ. We specify that the orientation is an oriented angle, measured in right-handeddirection, starting from the direction of the North, until the given direction isencountered. The Cartesian coordinates of a point A represent the distances from this point tothe rectangular axes of the chosen system, and are denoted by (XA, YA). As it has already been mentioned, the axis of ordinates in rectangular topographicsystems is denoted by 0x, and the axis of abscissas is denoted by 0y. In figure 1.5.a, b, c, d there are presented the four possible situations concerningthe relative position of points in the field. N The relations between orientations and coordinates result by expressing thetrigonometric functions of the angle θ, the computation of the unknown elements being N N B N B the known elements.possible depending on θAB ∆XAB X θAB A ∆YAB ∆XAB A ∆YAB A ∆XAB ∆XAB θAB θAB ∆YAB ∆YAB B A B Y 11 a) QUADRANT I b) QUADRANT II c) QUADRANT III d) QUADRANT IV Figure 1.5.
  • 17. Hence we have: ∆YAB sinθAB = -------- (1.14) DAB ∆XAB cosθAB = -------- (1.15) DAB ∆YAB YB - YA tgθAB = ------- = ----------- => θAB from tables (1.16) ∆XAB XB - XA ∆XAB ctgθAB = -------- from tables (1.17) ∆YAB Knowing the orientation of the direction formed by two points A and B, θAB, thedistance between the points DAB and the coordinates XA and YA of one of the points, wecan compute from the relations (1.14) and (1.15) the relative coordinates ∆XAB and ∆YABof the second point (B), with respect to the known one (A). Thus: ∆XAB = DAB · cosθAB (1.15’) ∆YAB = DAB · sinθAB (1.14’) Since: ∆XAB = XB - XA (1.18) ∆YAB = YB - YA (1.19) 12
  • 18. It will result that: XB = XA + ∆XAB (1.18’) YB = YA + ∆YAB (1.19’) From the relations (1.16) and (1.17) we can determine the orientation θAB,depending on the known coordinates of two points. Returning to figure 1.5.a, b, c, and d it can be seen that the sign of the relativecoordinates ∆XAB and ∆YAB indicates the position of the orientation in the topographiccircle. Therefore: in quadrant I + ∆X, + ∆Y, θI = α in quadrant II + ∆X, + ∆Y, θII = α + 100g (1.20) in quadrant III - ∆X, - ∆Y, θIII = α + 200g in quadrant IV - ∆X, + ∆Y, θIV = α + 300g α being the angle reduced to the first quadrant.1.4. PROBLEMS FOR TUTORIAL 11.4.1. SOLVED PROBLEMS Problem #1: The following data concerning the topographic points A and B wascollected as result of field measurements: LAB = 147.32 m; Z = 97g 31c; Also, the height of the point A is known: ZA = 300.53 m + n (mm); Determine: DAB, ∆ZAB, ZB. Remark: (n) represents the number of the student from the half-group. Problem #2: The following data is known concerning two points A and B: LAB = 121.56 m – n (m); 13
  • 19. ∆ZAB = 2.454 m. Compute DAB and ϕ. Problem #3: Transform the following angles, from the given basis into therequired one: a) From centesimal degrees into sexagesimal degrees: 32g 43c36cc + nc 121g 52c37cc + ncc 237g 82c58cc + ng 321g 52c84cc - nc b) From sexagesimal degrees into centesimal degrees: 52°36’28” – n’ 131°52’42” + n” 236°58’36” – n” 321°31’43” + n’ Problem #4: Express the functions of the following angles through trigonometricfunctions of the angles from the first quadrant: 121g 36c42cc + ng 237g 52c38cc - nc 346g 82c56cc + nc 98°52’36” - n’ 231°36’48” + n” 303°21’52” + n’ Problem #5: Determine the angles θ corresponding to the following values: ∆XAB = 148.05 m + n (m); ∆YAB = - 136.21 m - n (m); ∆XAB = - 121.37 m + n (m); 14
  • 20. ∆YAB = - 111.66 m + n (m); in centesimal and sexagesimal degrees. Problem #6: The coordinates of two points A and B are known: XA = 1321.52 m + n (m); YA = 3436.48 m; XB = 1464.49 m; YB = 3542.64 – n (m); Compute DAB and θAB. Problem #7: The coordinates of a point A, the distance to the point B and theorientation of the direction formed by the two points are known. Compute the coordinatesof the point B. XA = 1336.92 m ; YA = 2438.84 m; DAB = 184.52 m + n (m); θAB = 236g 51c36cc.1.5. EXAMPLE FOR SOLVING THE HOMEWORK (for (n) = 0) Problem #1: See figure 1.1, the relations (1.2), (1.1), (1.3), (1.3’). DAB = LAB cosϕ It can be seen that ϕ = 100g 00c00cc – Z = 100g 00c00cc – 97g 31c00cc = 2g 69c, Hence DAB = 147.32 m x cos2g 69c = 147.32 x 0.999107 = 147.19 m. ∆ZAB = LAB sinϕ = 147.32 m x sin2g 69c = 147.32 m x 0.042242 = 6.223; ZB = ZA + ∆ZAB = 300.53 m + 6.223 = 306.753 m. Problem #2: DAB = √L²AB - ∆Z²AB 15
  • 21. Hence DAB = √1477.8017 = 121.54 m ∆ZAB 2.454 m sinϕ = ------- = ------------- = 0.020187562 => ϕ = 1g 28c00cc LAB 121.56 mProblem #3:a) 32g 43c36cc = see the appendix table 2, part one 32g = 28048’ 43c = 23’13”.2 36cc = 11”.66 32g 43c36cc = 29011’24”.86 ≅ 29011’26”b) 52036’28” = see the appendix table 2, part two 520 = 57g 77c77cc.00 36’ = 66c66cc.70 28” = 86cc.40 52036’28” = 58g 45c30cc.90 ≅ 58g 45c31cc.The other transformations will be solved similarly.Problem #4: sin 121g 36c42cc = cos 21g 36c42cc cos 121g 36c42cc = sin 21g 36c42cc tg 121g 36c42cc = - ctg 21g 36c42cc ctg 121g 36c42cc = - tg 21g 36c42cc sin 237g 52c38cc = - sin 37g 52c38cc cos 237g 52c38cc = - cos 37g 52c38cc tg 237g 52c38cc = tg 37g 52c38cc ctg 237g 52c38cc = ctg 37g 52c38cc 16
  • 22. sin 346g 82c52cc = - cos 46g 82c52cc cos 346g 82c52cc = sin 46g 82c52cc tg 346g 82c52cc = - ctg 46g 82c52cc ctg 346g 82c52cc = - tg 46g 82c52cc For the angles expressed in sexagesimal gradation, we shall proceed in a similarmanner. Problem #5: DAB = √∆X2AB + ∆Y2AB = √(XB - XA) 2 + (YB - YA) 2 = = √(1464.49 – 1321.52)2 + (3542.64 – 3436.48)2 =√ 142.972+ 106.972 = 178.07 m; ∆YAB 106.16 tgθAB = ---------- = ----------- = 0.742533 ∆XAB 142.97 θAB = arctg 0.742533 = 40g 66c12cc Problem #6: ∆XAB = DAB cosθAB = 184.52 cos236g 51c36cc = 184.52 · (-0.939978) = -154.99 m; ∆YAB = DAB sinθAB = 184.52 sin 236g 51c36cc = 184.52 · (-0.542621) = -100.12 m; XB = XA - ∆XAB = 1336.92 – 154.99 = 1181.93 m; YB = YA - ∆YAB = 2438.84 – 100.12 = 2338.72 m. Problem #7: ∆YAB - 136.21 tgθAB = ------- = ----------- = - 0.920027 ∆XAB 148.05 - ∆YAB HavingtgθAB =---------- we are in quadrant II. ∆XAB Hence, we are looking for the angle β, for which ctgβ = 0.920027. We have for 0.920001 – ctg52g 65c10cc 17
  • 23. 0.920027 – 0.920001 = 27 units 1cc …………………… 2.91 units Xcc …………………… 27 27 · 1cc Xcc.= ----------- ≅ 9cc 2.91 Hence ctg (52g65c10cc - 9cc) = 0.920027 => β = 52g65c01cc; => tgθAB = β + 100g = 152g65c01cc; The other computations are performed similarly. Proposed problems: Problem #1’: The following data is known concerning two topographic points Aand B: LAB = 136,54 m – n (m); ϕ = 2g51c32cc - nc; Compute DAB and ∆ZAB. Problem #2’: The following data was determined concerning two points C and D,as result of field measurements: LAB = 243.76 m + n (m); ∆ZAB = 12.345 m; Compute DAB and ϕ (the slope angle of the terrain). Problem #3’: Transform the following angles in the required basis: a) From centesimal degrees into sexagesimal degrees: 64g31c12cc + nc; 356g17c24cc - ncc; b) From sexagesimal degrees into centesimal degrees: 18
  • 24. 126°31’15” + ng; 223°17’38” – nc.2. STUDYING THE THEODOLITE The content of the tutorial: In different topographic works there appearsfrequently the need to measure the angular elements of the terrain – horizontal andvertical angles – the theodolites being the optical instruments that serve this purpose. We shall study some types of these devices that are more frequently used in ourcountry.2.1. THE GENERAL CONSTRUCTION SCHEMA OF A THEODOLITE2.1.1. GENERAL NOTIONS, CLASSIFICATIONS So, the theodolites are optical devices that allow measuring angles with anapproximation of minutes (or seconds), being used in topographic measurements. The tacheometer-theodolites (the tacheometers) are optical devices that can beused to measure optically (hence, indirectly) both angles and distances, by the use ofstadia hairs traced on the reticule of the devices. B V(V’) L SB Clinome ter L AB DSB ϕB A B’ DAB α S ϕA D SA Be aring circle A’ Horiz ontal projection plan V(V’) S: station point; DiJ : the horizontal distance; VV: the vertical axis ofthedevice; A, B: aimed points; ϕA, ϕB : vertical angles; V’V’: the verticalofthestation point; LiJ : the slanted distance; α: horizontal angle. 19 Figure 2.1. Measuring angles with theodolites
  • 25. Types of theodolites: - Classical type, with metallic graduated circles, equipped with decentralized reading devices; - Modern type, with crystal graduated circles and centralized reading devices; Measuring angles is performed with these devices placed in geodetic ortopographic station points, obtaining (figure 2.1): - Horizontal angles αi (the plan dihedral angle formed by the station point as angle apex and the directions that unite it with other topographic points); - Vertical angles (formed by some direction with its horizontal projection).2.1.2. THE PRINCIPLE SCHEMA – AXES AND COMPONENT PARTS Theoretical axes: - MAIN VERTICAL (VV) AXIS – The device can be rotated around this axis (rotation r1). During measurements, the VV axis coincides with the vertical of the station point (V’V’); - SECONDARY HORIZONTAL (HH) AXIS – The complex telescope- clinometer is rotated around this axis, on the vertical plan (rotation r2); - AIMING AXIS OF THE TELESCOPE (Γ0) – Materializes the aiming line of the topographic points. These three axes meet in one point (C v), called the aiming center, in the followingway: HH ⊥ VV, Γ0 ⊥ HH (figure 2.2). The main parts of the device are the following (figure 2.3): THE GRADUATED HORIZONTAL CIRCLE (or the bearing circle) ((1) infigure 2.3) is a metallic disc (in the case of classical theodolites) or a crystal disc (in thecase of modern theodolites), having a diameter of φ 70 ÷250 mm, and being graduated onthe entire circumference with centesimal (sexagesimal) degrees, increasing towards right-handed direction, being used for determining horizontal angles. 20
  • 26. THE GRADUATED VERTICAL CIRCLE (or the clinometer), having the samecharacteristics as the previous one, being used for determining vertical angles ((2) infigure 2.3). 0 V Γ2 H H Cv Γ Γ1 V Figure 2.2. The main axis and rotations of the theo-tacheometer THE ALIDADE ((3) in figure 2.3) is a plate in the interior of the bearing circle,which bears two diametrically opposed reading indexes (i1 and i2). With the use of a distaff ((a) and (b)), the alidade supports the complexclinometer-telescope. THE TELESCOPE ((4) in figure 2.3) is the optical device with the use of whichaiming the topographic points is possible. THE BASE ((5) in figure 2.3) consists of a metallic support, equipped with threefoot screws (C) and a threaded inlet (d), which allows fixing the device on the trivet, bythe use of a screw. 21
  • 27. 0 4 H H 2 Cv r a b i2 i1 1 5 c d c Figure 2.3. The main parts of the theodolite Operating the foot screws, the device can be brought to horizontal. One difference between measuring horizontal angles and vertical angles is thatwhile in the first case, the bearing circle is fixed and the reading indexes are mobile (withthe alidade and the telescope), in the case of vertical angles, the clinometer is mobile(with the telescope) and the indexes are fixed.2.1.3. THE DETAILED SCHEMA OF THE THEODOLITE (figure 2.4.) 1. THE TELESCOPE (1’: LENS, 1”: OCULAR, 1”’: APPROXIMATE AIMING DEVICE, 1IV: IMAGE FOCUSING – CLARIFYING MUFF, 1V: RETICULE CLARIFYING MUFF, 1VI: RETICULE). 2. VERTICAL GRADUATED CIRCLE – CLINOMETER (2’: INSCRIPTION THAT INDICATES THE POSITION OF THE DEVICE). 3. HORIZONTAL GRADUATED CIRCLE – BEARING CIRCLE (the area in which it can be found is indicated). 22
  • 28. 4. THE ALIDADE (4’: DISTAFFS FOR SUPPORTING THE TELESCOPE WITH CLINOMETER). 5. THE BASE (5’: FOOT SCREWS, 5”: THE TENSION PLATE OF THE BASE). Parts that ensure the correct position of the device for measurements: 6. THE AIR-BUBBLE LEVEL FROM THE HORIZONTAL CIRCLE (6’: ADJUSTING SCREW OF THE LEVEL). 7. THE SPHERICAL LEVEL (FOR APPROXIMATE HORIZONTAL SETTING). 8. THE AIR-BUBBLE LEVEL FROM THE VERTICAL CIRCLE (8’: THE REFLECTING MIRROR, 8”: MICROMETRIC SCREW). Parts that ensure reading angular values: 9. MICROSCOPE (9’: THE OCULAR OF THE MICROSCOPE, 9”: READING CLARIFYING MUFF, 9”’: RELFECTING MIRROR). Parts that ensure the motion of the device, the motion of the telescope withclinometer around the horizontal axis: 10. LOCKING SCREW (CLAMP) (10’: SLOW MOTION SCREW). Parts that ensure the motion of the device around the vertical axis: 11. LOCKING SCREW (CLAMP) (11’: SLOW MOTION SCREW). Parts that ensure the motion for recording angular values: 12. THE CLAMP FOR FASTENING THE BEARING CIRCLE TO THE ALIDADE AND FOR LOCKING THE RECORDING MOTION. Parts that ensure fastening the device to the base: 13. CLAMPING SCREW. TRIVET (t) PLATE (t’) FEET (t”) SCREW FOR FASTENING THE BASE TO THE TRIVET (t”’) PLUMB-BOB WIRE THAT ENSURES CENTERING THE DEVICE ON THE STATION POINT (tIV). 23
  • 29. 2 1 8’ 1’ 8 10 2’ 4’ 1IV IV 10’ 1 8” 6 1’’’ 9, 9” 1IV 1” 11 6’ 1 V 9’’’ 11’ 4 9’ 7 13 3,4 12 5 t,t’ 5’ 5” t” t’’’ tIV Figure 2.4. The detailed schema of the tacheometer-theodolite THEO 030 Carl Zeiss Jena (Germany) – position I2.1.4. THE DETAILED SCHEMA OF OTHER TYPES OF THEODOLITES In the volume GENERAL TOPOGRAPHY – problems and practical applicationsII, we present the detailed schema for other types of theodolites, frequently used onconstruction sites. 24
  • 30. 2.2. WORKING PROCEDURE FOR THE THEODOLITE2.2.1. WORKING PRINCIPLES It should be specified that the theodolite, being a very sensitive optical device,should be kept in certain conditions, being carefully and gently handled, but not beforethoroughly knowing the purpose of each part. The device should be kept away from: hits, shocks, moisture, high temperaturedifferences, high pressures, dust, vibrations, etc. After operating the screws (clamps) 10 and 11, only small amplitude motion ofthe device will be performed, with the use of the screws 10’ and 11’.2.2.2. VERIFYING THE DEVICE In order to ensure the precision of measurements, the theodolites must satisfy aseries of required conditions. The errors during angle measurement, due to the device, can be classified into thefollowing categories: ERRORS DUE TO: 1) Construction imperfections of the device; 2) The degradation, wrong-going, destruction of some component parts; 3) The alteration of component parts. The errors from the first category can be significantly improved by choosing someadequate working methods. When the device is found in a state characteristic to thesecond category, the operator will turn to shops specialized in repairing optical devices.The errors from the third category can be identified and reduced by adjustment. All problems presented above are discussed in detail in the lecture and we do notthink it is necessary to treat them here. It should be specified that the device should be periodically verified-rectified, andannually, the devices are presented to specialized shops for verification. 25
  • 31. 2.2.3. PLACING INTO THE STATION It is performed through some operations that have the purpose to ensure thecoincidence of the vertical axis of the device (VV) with the vertical of the station point(V’V’). The working steps are the following: 1) Centering the device, on the vertical of the station point, is performed in the following way: - Place the trivet above the station point, watching the plumb-bob wire to be as close to the point as possible, the plate of the trivet to be horizontal, in the same time ensuring the stability of the trivet, by successively pushing the foot on the shoes of the trivet; - Remove the device from the casing, place it on the plate of the trivet and fasten it temporarily with the screw of the trivet; - Moving the device on the plate (or, if this is not possible, adjusting the height of the feet of the trivet), the plumb-bob wire is brought to the vertical of the station point. V Instrument Plate n Clamping screw Plumb- bob wire Trivet Shoe V’ A V’ Figure 2.5. Centering 26
  • 32. 2) The horizontal setting of the device is ensured in two steps (figure 2.6). 2 2 1 I 1 II Figure 2.6. Horizontal setting Therefore, the air-bubble level is brought to be parallel, with the use of two footscrews, which are operated in separate directions (1 or 2), so that the bubble of the levelis brought between the benchmarks. Then, we rotate the device such that the level tobecome normal to the previous position, and operating the third foot screw, the air-bubbleis brought again between the benchmarks of the level. If the level is adjusted, after these operations the device is set horizontally, that is,its horizontal axis should be parallel to the horizontal of the location. This can be checkedbringing the level in different positions, the bubble remaining between the benchmarks.Otherwise, the level does not operate adequately, therefore its verification andrectification is necessary, after which the horizontal setting operation should be doneagain. We mention that in the case of devices that are equipped with spherical level, anapproximate horizontal setting is possible to be previously performed, bringing thebubble of this level in the benchmark circle. 27
  • 33. After the device was horizontally set, the centering operation is performed again,maybe optically, and then the horizontal setting is verified again, and so on, until the twooperations have satisfactory results. In this moment, the device is ready for measurements, its vertical axis (VV) beingidentical with the vertical of the station point (V’V’). Centering can be performed with the use of the centering stick or optically, withthe so-called “optical-plumb wire” (which is an optical device, inserted in the base ofmedium and high precision devices).2.2.4. AIMING AND POINTING Their purpose of to bring the image of the point (of the aimed signal) in the centerof the reticule and they are performed in the following way: a) Clarify the cross-hairs (figure 2.7.a), which can be done aiming a bright background with the telescope and operating the clarifying muff of the reticule (1V figure 2.4); b) Aim approximately the signal (figure 2.7.b), overlapping the device 1”’ (figure 2.4) on the free image of the signal; c) Lock the motion of the device, with the screws (clamps) 10 and 11 (figure 2.4); d) Clarify (focusing) the image of the aimed signal, operating the muff 1 IV (figure 2.4); the image of the aimed signal will appear in the center of the reticule; e) Point the signal operating the screw 10’ (obtaining the image (2)) and then 11’ (obtaining the image (3) of the pointed signal). Then, read and record the angular values. Handling conveniently the reflecting mirror 9”’ (figure 2.4), the image ofgradations is lit up. The muff 9” (figure 2.4) is used to clarify these gradations, and thenthe reading is performed. 28
  • 34. Operating muff IV a. Clarifying cross-hairs Aimed signal Aimers Telescope b. Approximate aiming with various devices 10’ 11’ 1 2 c. Pointing 3 Figure 2.7. Aiming and pointing2.2.5. DEVICES FOR READING ANGULAR VALUES ON THE THEODOLITE We shall discuss only those devices that the medium precision topographicinstruments that are frequently used by the construction engineer are equipped with. a. The microscope with lines has a fix hair carved on its reticule, whichoverlaps on the image of the graduated circles, ensuring the reading. The precision of the devices is equal to the smallest division: 1g 100c P = ------- = ---------- = 10c 10 div 10 div The reading consists of an integer part (PI) and an approximate part (PII). On the horizontal circle (Hz) we have: PI = 274g30c 29
  • 35. PII = 3c CHz = 274g33c On the vertical circle (V) we have: PI = 302g50c PII = 7c CV = 302g57c V 302 303 READINGS V = 302g54c HZ = 274 33c g 274 Hz 275 Figure 2.8. The microscope with lines b. The scale microscope has a graduated scale carved on its reticule, equal invalue to the apparent value of a division on the graduated circle. The scale is divided into 100 divisions (minutes) grouped by ten. The precision of the device (and the smallest division) is: 1g 100c P = --------- = ----------- = 1c 100 div 100 div The readings are (figure 2.9.a and b): On the horizontal circle (Hz): PI = 372G67C PII = 50CC CHZ = 372G67C50CC On the vertical circle (V): 30
  • 36. PI = 267G14C PII = 50CC CHZ = 267G14C50CC 1 267 2 V 267 V = 26714c50 cc g 266 0 1 2 3 4 5 6 7 8 9 10 6 372 7 Hz = 372 67 40cc g c b- reading details 0 1 2 3 4 5 6 7 8 9 10 373 Hz 372 a - readings V = 267g14c 50cc Hz = 372g67c 40cc Figure 2.9. The scale microscope2.2. THE HOMEWORK OF THE TUTORIAL Problem #1: Draft the detailed schema of a theodolite, explaining the purpose ofeach component part and the conditions that the theoretical axes must satisfy. Problem #2: Set up a station, parsing all the steps from centering until reading theangular values, explaining the purpose of each operation, the method and the parts thatare used for it. Problem #3: Draft the schemas for the following readings on the microscope withlines (V = 321g32c + nc; Hz = 268g52c + nc) and on the scale microscope (V = 321g32c20cc+ ncncc; Hz = 268g52c 60cc + ncncc). Remark: Problems 1 and 3 represent the topic of the paper, and problem 2 will beperformed practically. 31
  • 37. 3. MEASURING ANGLES WITH THE THEODOLITE The content of the tutorial: In the previous tutorial we have studied thetheodolite and its usage, therefore, we shall see in the sequel how this device is used intopographic measurements.3.1. THE NATURE OF TOPOGRAPHIC ANGLES We call theodolite station placing it on the vertical of a topographic point (calledstation point) and performing some topographic measurements from this position. In what follows we shall present the angles that can be measured from such apoint: a) HORIZONTAL AIMS (c): formed as horizontal angles between the origin on the bearing circle and the vertical aiming plan towards a given point (figure 3.1.a), being the reading performed on the bearing circle. b) HORIZONTAL DIRECTIONS (ω): in the case when the horizontal graduated circle (the bearing circle) is placed with the origin towards one of the given points, then they are measured on the direction of the second point, reading the angle formed by the station as angle apex, with the directions towards the given points (figure 3.1.b). The horizontal angles (α) are formed between the vertical plans for aiming two given points. It results in value the difference of the aims performed to the two points. c) ZENITHAL ANGLES (Z): are the angles read usually on the vertical graduated circle, being formed by the vertical of the location (of the station) with the aimed direction (figure 3.1.e). d) VERTICAL ANGLES (V): are formed by some direction with its horizontal projection (figure 3.1.c). e) SLOPE ANGLES (ϕ): of the terrain are vertical angles, obtained by aiming a given point, at the height i of the device, in the station; they represent the 32
  • 38. angle formed by the direction determined by the station and the aimed point, with the horizontal of the station point. In figure 3.1.d. it can be seen that in this case we measure an angle ϕ’ equal to the slope angle ϕ of the terrain, which are angles with parallel sides. In the general case, there are measured the angles from the categories a and e, theother ones being computed based on the measured values. 1 α 100g 1 ω 100 g 2 C 0g : Origin on the 0g bearing circle; S S: Station point; S 200g 1: Aim ed point; 300 g 200 g ω : Horizontal 1, 2: Aimed points; direction 300g a) α: Horizontal angle b) Aimed direction Aimed direction s UV Horizontal s= ϕ’ Device i AP ϕ horizon i i Station point ϕ horizontal i: The height of the instrument in the station; ϕ: Angle measured with the theodolite; s: The aim ing height; ϕ’ : Slope angle of the terrain; UV:Vertical angle i ≠ s ϕ’ = ϕ . c) d) V ZI i ϕ’ i ZII ϕ S V Z : zenithal angle read in position I I Z : zenithal angle readin position II II ϕ’I + ϕ’ II 100 g - ZI+ ZII - 300 g e) ϕ= ϕ’ = ------- = --------------------------- 2 2 Figure 3.1. Topographic angles 33
  • 39. We mention that a zenithal angle is the angle of type Z I, the angle ZII beingassimilated as angle measured in the second position of the telescope (the clinometer onthe right of the telescope).3.2. ANGLE MEASURING METHODS As result of the operations of placing the theodolite in the station, presented in theprevious tutorial, the device is ready for measurements, satisfying the followingconditions: - The main (VV) and secondary (HH) axes of the device become vertical, and horizontal, respectively; - The (VV) axis coincides with the vertical (V’V’) of the station point; - The device placed on the trivet is stable, in completely fixed position, little sensitive to the touches during handling; - As result of the verification and rectification procedures of the device, the correctness and the precision needed for using it, the integrity and sound functioning of each component part have been ensured. The topographic angles can be measured through various methods, chosendepending on the following factors: - The precision needed and the purpose of the measurement; - The existing equipment; - The number of points and measured angles; - The distance to the aimed points; - The condition of the atmosphere, and in general, the conditions in which the measurement is performed; - The existence or the lack of vegetation that could prevent or make difficult to perform some aims; - The available time, etc. 34
  • 40. 3.2.1. THE CASE OF MEASURING ONE ANGLE a. Angle computed as difference of readings We mention that for all angle-measuring procedures listed here, we assume that the device is set up for measurement, in the theodolite station. In this case, we have the following steps (figure 3.2): 1) Aiming and pointing the first point (1) in the position I of the telescope (the clinometer on the left of the telescope); 2) Recording the reading on the bearing circle (CI1) and maybe on the clinometer (ZI1); 3) Rotating the device in right handed direction and aiming the second point in position I; 4) Recording the readings (CI2, ZI2); 5) Turning the device in position II and aiming – pointing the second point; 6) Recording the readings (CII2, ZII2); 7) Rotating the device in left handed direction and aiming the first point in position II; 8) Recording the readings (CII1, ZII1); 9) Verifying the readings performed and recorded in table (3.1), as it follows: - CI1 with CII1, and CI2 with CII2, respectively, must differ with 200g to which the permissible error specified for each type of device is added; - ZI1 with ZII1, and ZI2 with ZII2, respectively should sum up to 400g plus the error specified above. Only after these checkings have been performed we can leave the station. Table 3.1Station Aimed Readings on the Mean Horizontal Readings on the Vertical point point bearing circle directions angle clinometer angle Horizontal directions Position Position Position Position I II I II 1 2 3 4 5 6 7 8 9 1 CI1 CII1 C1M ZI1 ZII1 V1 S1 α= 2 CI2 CII2 C2M ZI2 ZII2 V2 35
  • 41. The bearing circle of The origin on the the theodolite bearing circle V(V’) 1” g Direction 1 V(V’) ≡0The bearing 0g 100g Direction 1 1”circle of the S’ 1” 300g 0gtheodolite 2” 1’’’ 300g 200g Direction 2 S’ 2”’ 200 2” g 1 100g 2” 1 1’ 1’ 2 α α S 2’ S 2 Horizontal projection Horizontal a) plan V(V’) projection plan V(V’) 2’ b)VV:The vertical of the device VV = V’V’(measuring condition) α = Direction 2V’V’:The vertical of the point SS: Station point S’1’’’ The horizontal formed by the S’2’’’ center of the bearing circle and theS’: The center of the bearing circle1,2 - Measured points signals1’,2’- The projection of the points on an imaginary horizontal plan1”,2”-Aims on the signals placed in these pointsIt can be seen: ∠1’’’ ’2’’’ = ∠ 1”S’2” = ∠ 1S2 =α = Direction 2 – Direction 1 S 4” V(V’) 1” d) Position II 4 Position I 1 i = the height of the device in 3” the station; 4” V3 1’ s = the aiming S i 2” 3’’’ height = 11”= 22” =….. α If i = s => V =ϕ 2 ϕ3 δ i = s => V ϕ ≠ 2’ 3’ V(V’) Figure 3.2. Angle measuring methods 36
  • 42. 1 100g 2 CI2 C I1 0g 200g The origin on the bearing S1 C II1 CII2 circle Bearing circle 300g Figure 3.3. Data processing is performed in the following way: (CI2) + (CII2)² 1 I g - Column (5): C M = (C ) + ------------------ 1 2 Therefore, the degrees from position I are recorded and is computed the mean ofthe minutes from the two positions. - C2M is computed similarly; - Column (6): C2M – C1M, with the remark that if C 2M < C1M then 400 g are added to the first one (α= C2M + 400 g – C1M); 100g – ZI1+ ZII1 - 300g 1 - Column (9): V = -----------------------------; 2 - V2 is computed similarly. In table 3.2 we present another method for processing the measured values, bycomputing the mean of the horizontal angles (the vertical ones being computed as in thefirst table). Measuring the angles in the two positions of the telescope has the followingconsequences: - Removing (or reducing) the instrumental errors; - Increasing the measuring precision; 37
  • 43. - The mutual control of the values measured in the two positions.Table 3.2 Readings on Horizontal angles Readings on the Vertical angle Station point Aimed point the bearing clinometer V circle (horizontal directions) Position Position αI αII α Position I Position I II II 100g-ZI1+ZII1-300g I II I II 1 C1 C Z Z V1=------------------- 1 αI + αII 1 1 2 S1 I C 2-C I 1 II C 2-C II 1 α= -------- 100g-ZI2+ZII2-300g 2 CI2 CII2 2 ZI2 ZII2 V2=------------------- 2 b. Angle measured through the method of “zeros in coincidence” (with zero origin on the bearing circle, towards the first point) Between the operation of setting up the device in the station and the firstoperation of measurement presented in the previous case, we insert the operation ofbringing the reading index of the alidade (in position I) in coincidence with zero, on thebearing circle. This can be obtained in the following way: - Rotate the device around the VV axis, watching the reading microscope on the horizontal circle; - When the gradation 0g appears in the microscope, lock the motion by the use of clamp 11 (figure 2.4); - Operating the slow motion screw in horizontal plan (11’), bring the zero gradation in coincidence with the reading index (the line of the microscope or 0 on the scale); - Lock the recording motion by the use of clamp 12 (figure 3.4), the bearing circle being locked now by the alidade; - Aim and point the first point (step I); - Unlock clamp 12; - Parse the steps 2-8 presented previously; 38
  • 44. - The verification (9) in this case is that C II1 = 200g ± e, where e = the permissible error for the device used for measuring angles. In this case: αI = CI2; αII = CII2 - 200g αI + αII And α = ---------- 2 Or: C1M = 0g ± e/2; (CI2)c + (CII2)g 2 1 g C M = (C 2) + ------------------- 2 And: α = C2M - C1M The table to be used can be chosen in this case, too, between tables 3.1 or 3.2.3.2.2. MEASURING MORE ANGLES FROM ONE THEODOLITE STATION THE METHOD OF THE HORIZON TOUR Let S be the station point and 1, 2, 3, and 4 the points the have to be aimed fromthis station (figure 3.4). 4 The origin on the 1 The origin of aims bearing circle 0 g 10 0g 30 0g 200g 2 Position I Position II 3 Figure 3.4. Measuring by horizon tour 39
  • 45. In this figure it can be seen that usually the ORIGIN OF AIMS does not overlapon the ORIGIN OF THE BEARING CIRCLE. But we know from the previous case that this can be done by applying themeasuring method with “zeros in coincidence”. The horizon tour in position I is performed in right handed direction, on the route1-2-3-4-1, and in position II it is performed in left handed direction, on the route 1-4-3-2-1. After the device is set up in the station, the following operations are performed: - Aiming and pointing the signal from point 1 (maybe with “zeros in coincidence”); - Recording the values from the bearing circle (CIi1) and from the clinometer (ZI1); - Unlock the alidade and the telescope with the vertical circle (the bearing circle stays fixed) and aim the point 2; - Record the values (CI2), (ZI2); - The values (CI3, ZI3), (CI4, ZI4) are obtained similarly; - Aim again the point 1, obtaining (CIf1). 1) The measured data are inserted in the columns 3 and 4 (readings on the bearing circle) and 11 and 12 (readings on the clinometer); 2) The computation for the mean of the directions: (CIi1)c + (CIIi1)c i I g C M1 = (C ) + -------------------- i1 2 3) The computation of the error (e) is: e = CfM1 – CiM1 4) The computation of the corrections: - The total correction: Ct = - e Ct - The unitary correction: Cu = ---- n (n = the number of measured points); 40
  • 46. Table 3.3 FIELD BOOKFor measuring directions in the station……………………………Theodolite type…………………………………………………… Mean corrected directions reduced to zero C0MY Aiming point i = ? Vertical angles Vi Correction CCMY Y = 1 - 4Mean corrected directions Aimed point CMY Y = 1-4Mean directions Horizontal Horizontal Readings directions angles on the read on the clinometer bearing circle Position I ZIY Position I ZIIY Position I CIJ Y = 1-4 CIIJ Y = 1-4Position I Notation Value g c cc g c cc 1 2 3 4 5 6 7 8 9 10 11 12 13 1 CIi1 CIIi1 CiM1 - C C M1 00.00.0 α 0 C M2 – ZI1 ZII1 V1 0 C0M2 2 CI2 CII2 CM2 1xCu CCM2 C0M2 β C0M2 – ZI2 ZII2 V2 3 CI3 CII3 CM3 2xCu CCM3 C0M3 C0M1 ZI3 ZII3 V3 4 CI4 CII4 CM4 3xCu CCM4 C0M4 γ C0M1+400- ZI4 ZII4 V4 1 CIf1 CIIf1 CfM1 4xCu CCM1 00.00.0 C0M4 - - - 0 C0M3 – δ C0M1 γ 1 Remarks 4 Schema α S1 β 2 δ 3 - The initial direction 1: Ci1 = 0 x Cu; - The direction 2: C2 = 1 x Cu; - The direction 3: C3 = 2 x Cu; - The direction 4: C4 = 3 x Cu; 41
  • 47. - The final direction 1: Cf1 = 4 x Cu; 5) The correction of mean directions: CciM1 = CiM1 + Ci1; CcM2 = CM2 + C2; …………………; CcfM1 = CfM1 + Cf1; CciM1 = CcfM1 (verification) 6) The computation of directions reduced to zero: C0M1 = 00.00.00 (It is considered that the bearing circle had its origin in the coincidence with the first aim); 7) The computation of the horizontal angles: = C0M2 – C0M1 = forward direction – backward direction; 8) The computation of the vertical angles: 100g - ZIi + ZIIi - 300 Vi = ------------------------------ 2 Measuring angles (directions) from a traverse point is performed similarly tothe case presented previously, with the following remarks: - The height of the instrument i in the station is recorded in each station; - In order to obtain the declivity angle of the terrain, the aim is performed similarly to the one presented in figure 3.1.d. The methods for improving the measuring precision are used in precisiontopographic measurements, being presented in detail in the topography lecture.3.3. THE HOMEWORK OF THE TUTORIAL Let Si be a station point and let (101, 102, …) be a series of points that have to bemeasured: 1. Measure the angle formed by the points 101 and 102 with the station point and the corresponding vertical directions. 42
  • 48. The data will be written in table 3.1 and will be processed. 2. Repeat the previous measurement using the method of “zeros in coincidence”. The data will be written in table 3.2, will be processed and compared to the data obtained in the previous case. 3. Applying the method of the horizon tour, measure five specified points from the given station. The data will be written and processed in table 3.3. 3.4. EXAMPLE FOR SOLVING THE HOMEWORK The method for executing the field works was presented in this tutorial, and we shall not return to this subject. We want to present in the sequel the method for processing real data, and to this end we shall give one computation example for each particular case. Table 3.4Station point Aimed point Readings on the Mean Horizontal Readings on the Vertical bearing circle directions angle clinometer angle Position Position Position Position I II I II 101 37g21c00cc 237g22c00cc 37g21c50cc 92g26c00cc 307g74c00cc 7g74c00ccS4 47g26c50cc 102 84g47c00cc 284g49c00cc 84g48c00cc 96g51c00cc 303g47c00cc 3g48c00cc Hence 21c + 22c C101M = (37g) + ---------------- = 37g21c50cc 2 47c + 49c C102M = (84g) + ---------------- = 84g48c00cc 2 = 84g48c00cc - 37g21c50cc = 47g21c50cc 100g00c00cc - 92g26c00cc + 307g74c00cc - 300g00c00cc V01 = ------------------------------------------------------------------ = 7g74c00cc 2 43
  • 49. 100g00c00cc - 96g51c00cc + 303g47c00cc - 300g00c00cc V02 = ------------------------------------------------------------------ = 3g48c00cc 2Station point Aimed point Readings on Horizontal angles Readings on Vertical The the bearing the clinometer angle measurement circle schema (horizontal directions) Position Position αI α II α III Position Position I II I II 101 00g00c 200g00c 47g26c 47g27c 47g26c50cc 96g26c 307g74c 7g74c S 1 S4 α 102 47g26c 247g27c 96g51c 303g47c 3g48c 2 I = 47g26c - 00g00c = 47g26c II = 247g27c - 200g00c = 47g27c C0MJMean directions reduced to zero I + II = ------------- = 47g26c50cc 2 FIELD BOOK For measuring directions in the station S5 Theodolite THEO 080 Horizontal Readings on Mean corrected directions CCMJ Horizontal directions read on the angles the bearing circle clinometer Z Mean directions CMJ Schema Remarks CJ = correction Station point i= Vertical angles Aimed point Position Position Position Position I II Value I II left right left right 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 31g22c 231g28c 31g25c - 31g25c 00.00.00 92.53 307.45 7.46 α 56.27.20 2 87g51c 287g53c 87.52 20cc 87.52.20 56.27.20 96.46 303.52 3.52 132.8 132.84.4 101.59.4 i = 1.52 mS5 3 132g83c 332g85c 40cc 98.12 301.90 +1.89 4 0 0 β 241.35.60 246.5 246.53.6 215.28.6 4 246g52c 46g54c 60cc 101.32 298.66 -1.33 3 0 0 328.8 328.87.8 297.62.8 - 5 328g87c 128g89c 80cc 106.51 293.48 7 0 0 γ 184.71.60 6.51.50 6 31g26c 231g22c 31.24 1cc 31.25 00.00.00 - - - 44
  • 50. 1) On the field there were recorded the unitary corrections, the values from the columns 3, 4, 11, 12;2) Column 5: 22c + 28c 31 + -------------- = 31g25c g 2 51c + 53c 87 + -------------- = 87g52c g 2 83c + 85c 132 + -------------- = 132g84c g 2 87c + 89c 328 + -------------- = 328g88c g 2 26c + 22c 31 + -------------- = 31g24c g 23) The error e = 31g24c - c31g25c = 1c4) The total correction: Ct = - e = +1c Cu = Ct / n = 1c/5 = + 205) The corrections on directions: C21 = 0 x (+ 20cc) = 0 C2 = 1 x (+ 20cc) = + 20cc C3 = 2 x (+ 20cc) = + 40cc C4 = 3 x (+ 20cc) = + 60cc C5 = 4 x (+ 20cc) = + 80cc Cf1 = 5 x (+ 20cc) = + 1c6) The computation of the mean corrected directions: CCM1 = 31g25c + 0cc = 31g25c CCM2 = 87g52c + 20cc = 87g52c20cc CCM3 = 132g84c + 40cc = 132g84c40cc CCM4 = 246g53c + 60cc = 246g53c60cc 45
  • 51. CCM5 = 328g87c + 80cc = 328g87c80cc CCM1 = 31g24c + 1cc = 31g25c7) The computation of the directions reduced to zero: 31g25c - 31g25c = 00.00.00 - 31g25c = 56.27.20 132.84.40 - 31g25c = 101.59.40 - 31g25c = 215.28.60 - 31g25c = 297.62.80 31g25c - 31g25c = 00.00.008) The computation of horizontal angles: = 56.27.20 – 00.00.00 = 56.27.20 = 297.62.80 – 56.27.20 = 241.35.60 = 00.00.00 – 215.28.60 + 400.00.00 = 184.71.409) The computation of vertical angles: 100g – 92.53 + 307.45 – 300 ---------------------------------- = 7.46 2 100g – 96.46 + 303.52 – 300 ----------------------------------- = 3.52 2 100g – 98.12 + 301.90 – 300 ----------------------------------- = 1.89 2 100g – 101.32 + 298.66 – 300 ----------------------------------- = - 1.33 2 100g – 106.46 + 293.48 – 300 ------------------------------------- = - 6.51.50 2 46
  • 52. 4. DIRECT AND INDIRECT DISTANCE MEASURING METHODS The contents of the tutorial: Distances in topography can be measured directly,by comparison to a length-measuring unit, applying some instruments (tapes, measuringreels) in the field; or indirectly, measuring some elements in the field and then computingthe distance.4.1. MEASURING DISTANCES DIRECTLY In the construction activity, the direct method is frequently used, both for draftingthe necessary documentation for the design, and especially for applying the projects inthe field.4.1.1. INSTRUMENTS FOR THE DIRECT MEASUREMENT OF DISTANCES The kit for measuring distances consists of a tape, a measuring reel, tensioners,steel pickets, thermometer and dynamometer. The tape is a steel tape having the following characteristics: - Length of 20, 25, 50 or 100 m; - Width of 1.5 ÷ 2 cm; - Thickness of 0.4 ÷ 0.8 mm; 47
  • 53. - It is equipped at the ends with two rings (handles) that are used to tension it; - It is wrapped on a metallic frame, equipped or not with a distaff; - It is divided by each decimeter (wholes punched in the axis); - Half meters are marked by rivets; - Meters are marked by brass plates, having the corresponding value written on them. The measuring reel is a steel (plastic, textile) tape having the followingcharacteristics: - Length of 10, 15, 20, 25 and 50 m; - Width of 10 ÷ 13 mm; - Thickness of 0.2 mm; - Calibrated at +20ºC and tensioning force of 3-10 daN; - Divided in cm, and the first and last decimeter in mm; - Wrapped in metallic or leather casings, or on frames equipped with handles. Annex instruments: The tensioner, made of wood, is used for measurements performed with the tape– one at each end – having the purpose to ensure its perfect stretching. The metallic steel pickets with length of 30 cm and diameter of 5-6 mm, havethe purpose to mark each applied end of the tape (measuring reel), being grouped 11 on ametallic ring. The dynamometer ensures the stretching of the tape (measuring reel) with aforce equal to the calibrating one. The thermometer is used to determine the atmospheric temperature duringmeasurements.4.1.2. PREPAIRING THE TERRAIN FOR MEASUREMENTS Direct measurement of distances assumes a previous preparation of the terrain,which consists in: - Cleaning the measured alignment of vegetation and obstacles; - Setting out and pegging out the alignment, between the measured points; 48
  • 54. - Designating the points of declivity change. a) The direct measurement of a horizontal distance is performed in the terrain with slope ≤ 2%. The total length D will results from the relation: D = nl + l’ (4.1) Where: l = the length of a tape (measuring reel); l’ = the distance measured on the last tape applied; n = the number of whole tapes applied. b) Measuring a distance on a slanted terrain b.1) Measuring the slanted length After designating the points of declivity change (1, 2, …), the slanted distances(l1, l2, …) and the slope angle (ϕ1, ϕ2, …) are measured, or, eventually, the altitudedifferences (δh1, δh2, …). - li is measured as in the previous case; - ϕi is measured with the theodolite; - δhi is determined using one of the methods of leveling. Therefore, we have: di = li cosϕI (4.2) Or di =√l2i - δh2i (4.3) In the end: D = ∑di (4.4) b.2) Measuring distances in horizontal projection The method of the LONG BOARD and the LEVEL WATER The LONG BOARD is a right board (length l = 3 ÷ 5m, width 15 cm, thickness 5cm). The LEVEL WATER is a wooden bar, having mounted a level with air-bubble. Thelong board is brought to the horizontal with the use of the level water, and having aknown length, applying it step by step, for a certain number (n) of times, the point n isreached, near B. 49
  • 55. The distance nB is determined by measurement with the measuring reel appliedon the long board (l1). The total distance: D = nl x l1. The method of culteratie, less precise, consists in measuring the partial distanceswith the use of the tape (measuring reel) kept at horizontal – at the eyes – partialdistances that by summing up give the total distance. Hence: D = D = ∑di4.1.3. CORRECTIONS APPLIED TO LENGTHS MEASURED DIRECTLY MISTAKES that can appear during direct measuring of distances: - On large distances – the number of entire tapes that are applied can be mistaken; - Sometimes, the steel pickets are forgotten to be changed; - The tape can twist at the final reading (instead of reading the increasing gradation, the decreasing one is read – for example 28 m instead of 22 m, for a 50 m long tape). The MISTAKES are avoided by performing measurements with increased careand by measuring the distance back and forth (that is, from A to B and from B to A), thusincreasing the precision, too. In this case: DFORTH + DBACK D = ---------------------------- (4.8) 2 With the condition: ∆D = DFORTH - DBACK ≤ T (4.8) Where T = the measuring tolerance. The ERRORS that appear in the case of direct measuring of distances havesystematic or accidental nature, and can be significantly reduced applying somecorrections. SYSTEMATIC ERRORS 50
  • 56. 1. The CALIBRATION ERROR is determined by the fact that between the moment of calibration and the moment of use, the tape can change its size because of the action of various factors. DIt is corrected by: Ck = ∆lk ---- (4.9) LWhere Ck = calibration correction; ∆lk = lr - l (lr = actual length, l = nominal length of the tape)(4.10)2. The TENSIONING ERROR is determined by the different tensioning force in the moment of measurement, compared to that in the moment of calibration. 1000It is corrected by: CP = ------- D (P – Pe) (4.11) E·SWhere CP = calibration correction; E = 2.1 · 106 daN/cm² (the longitudinal elasticity coefficient of steel); S = the section of the tape in cm²; P = the actual tensioning force of the tape; Pe = the tensioning force of the tape at calibration (3daN/mm²).3. The ALIGNMENT ERROR and4. The ERROR OF NOT SETTING THE INSTRUMENT ON THE HORIZONTAL are reduced by rigorously respecting the instructions that concern direct measurements.5. The ERROR OF REDUCING TO THE SEA LEVEL is corrected only in the case of geodetic measurements.ACCIDENTAL ERRORS6. The TEMPERATURE ERROR is determined by the different temperature during measurements (tº), compared to the calibration temperature (tº 0 = 20ºC).It is corrected by: D 51
  • 57. Ct = ∆lt ---- (4.12) l Where Ct = temperature correction; ∆lt = lα (t° - t°0) (4.13) Where α = 0.0115 mm/1°C, 1m the thermal dilatation coefficient of steel. In conclusion, a distance D directly measured, is computed using the formula: DCORRECTED = Dmeas + Ck + CP + Ct (4.14) With the permissible error: eD = ± 0.003√D4.2. MEASURING DISTANCES INDIRECTLY There are multiple possibilities for measuring distances indirectly – from amongwhich, we shall discuss only those that are currently used. Indirect measurements offerthe advantage of reduced measuring time, and some methods bring an improved precisionin comparison to direct measurements.4.2.1. TACHEOMETRIC METHODS AND INSTRUMENTS Tacheometers are theodolites that, besides measuring horizontal and verticalangles, allow determining distances using optical methods. a. Stadimetric tachometry The reticule of these devices (figure 4.1) has, besides the cross hairs (1 and 2needed for measuring angles), stadia hairs 3 with the constant K = 100 and 4 with theconstant K = 50, that are used for determining distances. 2 4 3 1 3 4 52 Figure 4.1. The reticule of the tachometer
  • 58. The main accessory is the STAFF (figure 4.2), on the image of which the stadiahairs overlap, determining the necessary data for determining the distance. The reading on the measuring staff is performed downward, in the following way: - Read the numbers which are exactly below that hair (which represent meters and decimeters), count the entire divisions from the line that is below the mentioned numbers until the hair (which represent centimeters) and transform into millimeters. Each reading will thus consist of four numbers (figure 4.2); - On the upper stadia hair 0330, (CS); - On the horizontal cross hair 0200, (CM); - On the lower stadia hair 0070, (CJ). For indirect measuring of distances there are two cases: - Measurements on plan terrain; - Measurements on slanted terrain. On plan terrain (figure 4.3) the distance is determined in the following way: - The tacheometer is centered at one end of the 04 measured panel; - The measuring staff is placed perfectly vertical on 0322 03 the point that designates the other end of the panel; - Aim the measuring staff and perform the three readings (CS, CM, CJ); 02 0195 - Verify the readings using the following relation: CS + C J CM = ---------- (4.16) 2 01 - Compute the distance using the following formula: 0068 D=K·H (4.17) 00 Where K = 50 or 100 and it represent the constant of the device; 53 Figure 4.2. Reading on the measuring staff
  • 59. H = C S – CJ (4.18) Represents the generator number. Z = 100g 0322 H 0195 0068 B A D Figure 4.3. Tachometry on plan terrain On slanted terrain (figure 4.4) ϕ C’S CS In this case it is necessary to determine the vertical angle (slope angle). H H’ CM Z CJ C’J i ϕ Bi LAB ϕ DAB 54 A Figure 4.4 Tachometry on slanted terrain
  • 60. The distance LAB is deduced using the formula: LAB = KH’ (4.19) H’ = Hcosϕ (4.20) In the triangle CS, CM, CS’ it represents a fictive generator number. The horizontal distance DAB will be: DAB = KH cos²ϕ (4.21) (since it can be seen that DAB = LAB cosϕ). The precision of determining the distance using the stadimetric method variesbetween ± 5 and ± 20 cm/100 m. b. Diagram tacheometers These devices are self-reducing, giving directly the distances between the stationpoint and the aimed point, in which a special vertical measuring staff is placed,centimetrically graduated. The device is equipped with a diagram engraved on glass and mountedconcentrically with the vertical circle. For the distance, the diagram has two curves, themark curve (zero) and the distance curve. The distance between the two curves isproportional to cosϕ, where ϕ is the declivity angle of the speed with respect to thehorizontal, and therefore, it is also the rotation angle of the diagram with respect to thevertical image of the measuring staff. In the plan of the image (figure 4.5), the curves overlap on the image of themeasuring staff, and only one reading (C d) is performed for determining the distance, atthe intersection of the distance curve with the gradations of the measuring staff. 55
  • 61. We specify that the telescope of the device is fixed with the zero curve on themark of the measuring staff (which is at 1.4 m from the base), and C d represents thedistance on the measuring staff between the projections of the distance curve and of thezero curve. In this case, the horizontal distance DAB will be: DAB = Kd Cd (4.22) Where Kd = the constant of the device = 100. The precision is the same as in the previous case. 3 Cd = 0.271 2 1 Mark Zero curve Figure 4.5. Reading on the diagram c. Telemeter tacheometer and d. Self-reducing tacheometer with double refraction, are studied in the lecture and we shall not discuss them here. 56
  • 62. e. Parallactic tachometry consists in measuring an angle (called parallactic angle), which is formed by the aims performed with a precision theodolite (1cc – 5cc), at the ends of a 2 m invar horizontal measuring staff (figure 4.6) placed perpendicularly on the direction of the aim. It results: DAB = b/2 ctg γ/2 (4.23) Where:b is the length of the measuring staff (2 m). We specify that this method can be applied in many ways: with the measuringstaff at the end (the studied case), in the middle, with helping base, etc. γ γ/2 A B b=2m DAB Figure 4.6. Parallactic tachometry4.3. THE HOMEWORK OF THE TUTORIAL Problem #1: After performing measurements between various points in the field,the following data was recorded (the type of the measurement is also specified – throughthe paragraph). Direct measurements: 4.1.2.a n = 26, l = 50 m, l1 = 21.46 m; 4.1.2.b n1= 12, l1= 12.46 ϕ1 = 6g32c; n2= 6, l2= 31.28 ϕ2 = ng31c; l = 50m. 57
  • 63. n3= 8, l3= 18.52 ϕ3 = 8g56c; 4.1.2.b2 n = 32, l = 5 m, l1= 2.46 m; (long board and level water) d1= 36.52 m d2= 15.46 m, d3= 6.21 m (culteraţie). Correct the distances from the points 4.1.2.a and 4.1.2 b2 Culteraţie, knowing thefollowing data: lr= 50.02 m, p = 4 daN/ mm², s = 2 cm x 0.5 mm, tº = 10º. The following data was obtained through indirect measurement: 4.2.1.a - Between the points A and B (constant K = 100; ϕ = 0; CS = 0322, CM = 0195, CJ =0068; - Between the points A and C the following data was obtained: K = 50; ϕ = 8g31c, CS = 1631, CM = 1402, CJ = 1173; 4.2.1.b Kd = 100, Cd1 = 0,321 + 0,n, Cd2 = 1,671 + 0,n; 4.2.1.c b = 2 m, γ1 = 2g31c + n c, γ2 = 4g56c + n c (base at the end); b = 2 m, γ1 = 1g52c + n c, γ2 = 1g48c + n c (base in the middle). Example of solving the homework: 1. 4.1.2.a. D1 = n · l + l1 = 26 x 50 + 21.46 = ………………… = 1321.46m; 4.1.2.b d1 = (n1 · l + l1) cos 6g32c = (12 x 50 + 12.46) · 0.995076 = 609.44m; d2 = (n2 · l + l2) cos 0g31c = (6 x 50 + 31.28) · 0.999988 = 331.28m; d3 = (n3 · l + l3) cos 8g56c = (8 x 50 + 18.52) · 0.990974 = 414.74m; 3 D2 = ∑di = …………………………………… = 1355.46m; i=1 4.1.2.b2 D3 = n · l + l1 = 32 · 5 + 2.46 = 162.46 m (long board and levelwater); 3 D2 = ∑d2 = 36.52 + 15.46 + 6 · 21 = 58.19 m (culteraţie); i=1 Computing corrections: For D1 = 1321.46 m D1 1321.46 1321.46 Ck1 = ∆lk ---- = (lr1 – l)---------- = (50,02-50) ----------- = 0.53 m l 50 50 1000 100 58
  • 64. Cp1 = ------ D1(P1 – Pe) = ----------------- = 1321.46 (4-3 daN/mm²) = 0.06 m E · S1 2.1 · 106 · 2 0.05 D1 D1 1321.46 m Ct1 = ∆lt ---- =lα (t°1 – t°0) ---- = 50m 0.0115mm/1°C, 1m(10°-20°) ------------- l l 50 = - 0.15 m D1CORRECTED = D1MEASURED + Ck1 + Cp1 + Ct1 = 1321.46 + 0.53 + 0.06 – 0.15 = = 1321.90 m. The corrections for D4 will be computed similarly. Problem #2: 4.2.1.a. The distance DAB: - Verifying the readings on the measuring staff: CS + C J 0322 + 0068 CM = ---------: 0195 = ------------------- 2 2 Computing the distance: DAB = KH = 100 · 0.254 m = 25.4 m; H = CS - CJ = 0322 – 0068 = 0.254m. The distance DAC: - Verifying the readings on the measuring staff: CS + C J 1631 + 1173 CM = ---------. 1402 = ----------------- 2 2 Computing the distance: DAC = K H cos²ϕ = 50 · 0.458 m cos ²8g31c = 22.51 m; H = CS - CJ = 1631 – 1173 = 0.458 m. b = 2mA γ1 γ2 B D1 D2 D59 Figure 4.7. Parallactic tachometry with base in the middle
  • 65. 4.2.1.b. Kd = 100, Cd1 = 0.321 m D8 = Kd · Cd1 = 100 · 0.321 = 32.1 m D9 = Kd · Cd2 = 100 · 1.671 = 167.1 m.4.2.1.e. b = 2 m, γ1 = 2g31c (base at the end): γ1 2g31c00 D10= ctg ---- = ----------- = ctg1g15c50cc 2 2 Similarly D11 (γ2 = 4g56c) b = 2m, γ1 = 1g52c, γ2 = 1g48c (base in the middle): D12 = D’12 = D”12 = ctg γ1 / 2 + ctg γ2 / 2 = ctg0g76 + ctg0g74c = 169.79m. 60
  • 66. 5. PLANIMETRIC TRAVERSE The content of the tutorial: Basic method for thickening the control networks,the planimetric traverse has the purpose to create a basis of points of known coordinatesin a given area, points that will be used either for surveying the details from that area, orfor applying a project in the field. In the sequel, the main categories of the class of planimetric traverses will bestudied, by practical examples.5.1. GENERAL PROBLEMS Technically, traverses are supported on points of known coordinates, determiningthe coordinates of other points situated in the neighborhood of the detail points. The traverse is a sequence of alignments (sides) that intersect in points calledvertices.5.1.1. CLASSIFICATIONS A. Depending on the importance and the supporting method (figure 5.1): 61
  • 67. 1. Main traverses – are supported on superior order points (triangulate or intersection points); 2. Secondary traverses – are supported on a superior order point and a traverse point; 3. Tertiary traverses – are supported on traverse points.The set of traverses from figure 5.1 is called polygonation.B. Depending on the method of determining the orientations. 1. Traverses with orientations determined by computation, or known orientations; 2. Traverses with magnetic orientations (determined using the compass); 3. Traverses with gyroscopic orientations (determined using the GYROTHEODOLITE). 1 101 103 102 2 106 108 107 109 3 4 104 105 Figure 5.1. Traverse supporting method Main traverse, 1-101-102-103- 2 and 3-104-105-4 Secondary traverse, 1-106-107-105 Tertiary traverse, 102-108-109-105 1,2,3,4 - superior order points (known) 1-101, 101-102 etc - the sides of the traverse 101,102,103, … - new points (unknown) – the vertices of the traverse 62
  • 68. C. Depending on the shape of the route (figure 5.2) 1. Supported traverses (figure 5.2) a) With two ends and two orientations (figure 5.2.a); b) With two ends and one orientation (figure 5.2.b); c) With one end and one orientation (suspended traverses) (figure 5.2.c). 2. Traverses in closed circuit (figure 5.3) 3. Traverses with nodal point (figure 5.3) 1 2 N N θi2 θ31 105 3 104 a. N θ56 6 5 111 110 b. 70 9 N θ89 112 c. 113 8 114 Figure 5.2. Supported traverses orientations θ 31a - with two ends (3 and 4) and two 63b – with two ends (5 and 70) and one orientationθc – with one end and one orientation θ 89 56
  • 69. N 17 19 N θ16.17 θ18.19 123 1611 N 117 18 Node 118 120 N 122 15 116 13 119 θ14.15 121 12 115 θ12.13 14 θ10.11 a. b. Figure 5.3.a. Traverse in closed circuit; b. Traverse with nodal point5.1.2. CONDITIONS THAT CHARACTERIZE A TRAVERSE: - The angle apexes should be chosen in such a way that: 64
  • 70. - They are protected from motion, damages, settling; - There exists visibility between the points; - The direct measuring of distances is possible; - They are in the neighborhood of details that are to be surveyed; - They are well and solidly designated – materialized; - They will be numbered from 100 upwards; - The sides will satisfy the following conditions: - The length of a side 30 – 300 m (80 – 120 m on average); - The total length of the traverse 2 –4 km; - The number of sides – at most 30.5.2. DESIGNING THE TRAVERSE5.2.1. MEASUREMENT PREPARING WORKS a. Analyzing starting data and problems that have to be solved. b. Designing the traverse in general; it is performed on a plan on the scale 1:10.000 – 1:25.000 and it consists in choosing the route. c. Recognizing the terrain: the general project is compared to the actual situation in the field, the support points and their state are identified. d. Finishing the traverse project. e. Designating the points and drafting their marking schemas.5.3. FIELD WORKS5.3.1. PERFORMING MEASUREMENTS a. Measuring directly the sides back and forth – with the tape or measuring reel, either the horizontal distance, or the slanted one, and computing the horizontal distance. 65
  • 71. The following condition is to be satisfied: l = lforth – lback ≤ Tl (5.1) Where:Tl = ± 0.003√ l is the permissible tolerance in m (5.2). b. Measuring angles on the slope ϕ i of the traverse sides It is performed in the two positions of the theodolite, both back and forth. c. Measuring horizontal angles ω i, formed by the sides of the traverse (breaking angles), is performed in both positions of the telescope (see methods for measuring angles). d. Surveying details (if the traverse has radiations) – this problem will be discussed separately, in the next tutorial.5.4. OFFICE WORKS These will be treated based on the example given in the Homework of theTutorial.5.5. THE HOMEWORK OF THE TUTORIAL The coordinates of the support points 71, 72, 73, and 74, in the case of asupported traverse (the case A1, B1 with computed orientations, C1 a) are given and inthe table 5.1 there is given the data measured in the field (the columns 4, 6, 7, 10, 11). Compute the traverse and repeat it on the scale 1:1000 (on an A3 sheet of paper). Index of coordinates of the support points Point X(m) Y(m) 71 8306.43 + n (m) 4418.54 + n (m) 72 8284.56 + n (m) 4406.32 + n (m) 73 8255.67 + n (m) 4704.40 + n (m) 74 8334.76 + n (m) 4747.94 + n (m)5.6. THE SOLUTION OF THE HOMEWORK for n = 0 66
  • 72. 1. The data recorded in the field book (table 5.1) is processed, and the slanted distance (LiJ), the horizontal angle (ωiY) and the vertical angle (ϕiY) are obtained – see the remark of the table. 2. The processed data is written in the office book (table 5.2). 3. The computation of the traverse has the purpose to compute the coordinates of the vertices of the traverse, and the following steps are parsed: A. Processing the field book (table 5.1, page 75) After performing the field operations, the columns 1, 2, 3, 4, 6, 7, 10, and 11 werefilled in the book. Column 5 is obtained by computing the average of column 4, slanted distancesback and forth. For columns 8, 9, and 12 see tutorial 4 – concerning column 12, having thevertical angles back and forth, their mean was computed. B. The computation of the coordinates of the traverse points (table 5.2, page 76) a. Reducing the distances at the horizon It is performed using the relation: D = L cos ϕ 72 Where L is the slanted distance (column 4) ϕ is theX N vertical slope angle of the terrain (column 5). In column 6 cosϕ was written. ∆X 4 Hence, the product of column71.72 and 6 will be written in column 10. θ 71.72 b. Computing the support orientations ∆ Y71.72 71 0 Y 74 X N θ 73.74 ∆ X 73.74 ∆ Y73.74 73 0 67 Y Figure 5.4.
  • 73. The starting orientation θ71.72 and the ending orientation θ73.74 are computed fromtheir tangent, in the following way: ∆Y71.72 Y72 - Y71 4406.32 – 4418.54 - 12.22 tg θ71.72 = ---------- = ----------- = ------------------------ = --------- ∆X71.72 X72 - X71 8284.56 – 8206.43 78.13 = - 0.156406 (6.4) - ∆Y71.72 Hence, we have: tg θ71.72= ------------ => θ71.72 is situated in the quadrant IV. + ∆X71.72 We will search the trigonometric tables for the angle β Where:β = tg θ71.72 – 300g which has ctgβ = 0.156406: We find: β =90g12c29cc Thus: tg θ 71.72 = β + 300g =390g12c29cc ∆Y73.74 Y74 - Y73 4747.94 – 4704.40 43.54 tg θ73.74 = ---------- = ----------- = ------------------------ = --------- ∆X73.74 X74 - X73 8334.76 – 8255.67 79.09 = 0.550512 ∆Y73.74 Thus, we have: tg θ73.74= ----------- =>θ73.74 is situated in the quadrant I. ∆Y73.74 In the trigonometric tables we find for this values of the tangent: θ 73.74=32g03c00cc c. Computing the orientations of sides 68
  • 74. Computing coarse orientations (transmitting orientations) θ71.72 = β + 300g =390g12c29cc (computed from the coordinates) θ71.101 = tg θ71.72 + ω1 = 390g12c29cc + 121g37c50cc = 511g49c89cc =111g49c89cc (wespecify that from the value 511g we subtracted 400g – an entire circle, which has nosignificance). θ101.71 = θ71.101 + 200g = 111g49c89cc + 200g = 311g49c89cc θ101.102 = θ101.71+ ω2 = 311g49c89cc + 185g30c50cc = 496g80c39cc = 96g80c39cc θ102.101 = θ101.102 + 200g = 96g80c39cc + 200g = 296g80c39cc θ102.103 = θ102.101 + ω3 = 296g80c39cc + 214g38c00cc = 511g18c39cc =111g18c39cc θ103.102 = θ102.103 + 200g = 111g18c39cc + 200g = 311g18c39cc θ103.104 = θ103.102 + ω4 = 311g18c39cc + 191g57c50cc = 502g75c89cc =102g75c89cc θ104.103 = θ103.104 + 200g = 102g75c89cc + 200g = 302g75c89cc θ103.73 = θ104.103 + ω5 = 302g75c89cc + 62g31c50cc = 365g07c39cc θ73.104 = θ104.73 + 200g = 365g07c39cc + 200g = 565g07c39cc = 165g07c39cc θ73.74 = θ73.104 + ω6 = 165g07c39cc + 266g95c00cc = 432g02c39cc =32g02c39cc N 7472 N θ73.74 θ73.104 73 ω6 θ71.101 θ102.103 θ101.102 ω3 ω1 ω2 θ103.104 102 ω4 ω5 71 101 θ102.101 θ71.72 103 104 θ101.71 θ104.105 θ103.102 θ104.73 Detail station 101 N N θ101.102 θ71.101 102 θ71.101 ω2 71 69 101 200g θ101.71 Figure 5.5. Transmitting orientations
  • 75. The last orientation must coincide with the orientation θ74.74 computed from thecoordinates, within the margins of the tolerance T θ, which is computed using thefollowing relation: Tθ = 2.5 · 1c √ n Where n = the number of stations = 6; 1 c = the reading approximation on thedevices. It will result: Tθ = 6c12cc Establishing errors, respecting the tolerance condition, total and unitarycorrections: x – the computation of the error: eθ = θ73.74 COMPUTED - θ73.74 GIVEN FRIM COORDINATES We mention that the first orientation is affected by a series of errors for measuringthe horizontal angles ω. The second orientation is correct. It will result: eθ = 32g02c39cc - 32g03c70cc = - 1c31cc • It is compared to Tθ = eθ  < Tθ (1c31cc < 6c12cc) The error is permissible, and it will be corrected by the operation calledcompensation. • The computation of the total correction Cθ is done using the relation: Cθ = - eθ; Hence Cθ = - (- 1c31cc) = 1c31cc (5.7) • The computation of the unitary correction Cuθ: Cθ Cuθ = -------- (5.8) n In this case: 70
  • 76. 1c31cc Cuθ = -------- ≅ 22cc 6 Establishing the corrections on orientations: θ71.72 is considered correct, it is not corrected C0 = 0cc for θ71.101 there is applied the correction C1 = Cuθ x 1 = 22cc for θ101.102 there is applied the correction C2 = Cuθ x 2 = 44cc for θ102.103 there is applied the correction C3 = Cuθ x 3 = 66cc (6.9) for θ103.104 there is applied the correction C4 = Cuθ x 4 = 88cc for θ104.73 there is applied the correction C5 = Cuθ x 5 = 1c10cc for θ73.74 there is applied the correction C6 = Cuθ x 6 = 1c31cc = Cθ Compensating the orientation is done adding the corresponding correction to eachorientation. Thus θ71.101 CORRECTED = θ71.101 COARSE + C1 and so on (5.10) (e.g.: θ71.101 CORRECTED = 111g49c89cc + 22cc = 111g50c11cc). d. Computing relative coordinates 7472 N θ73.74 71. 101 N X 102 ∆Y104.73 ∆ 101 .10 103 73 104 .10 θ71.101 .10 θ101.102 3 θ102.103 .10 5 2 X 4 X X ∆Y101.102 ∆ X θ103.104 ∆ ∆ ∆ ∆Y.71.101 102 D ∆Y103.104 D 101.102 102.103 104 101 71 D71.101 ∆Y102.103 103 D103.104 θ104.73 Figure 5.6. Computing relative coordinates The relative coordinates are obtained using the relations: ∆XiJ = DiJ cos θiJ (5.11) 71
  • 77. ∆YiJ = DiJ sin θiJ (5.12) For example: ∆X71.101 = 70.09 cos111g50c11cc = - 12.59 m; ∆Y71.101 = 70.09 sin111g50c11cc = 68.95 m; Having the corrected orientations θiJ (in column 8), the values sinθiJ and cosθiYwere extracted for them (in column 9). The (horizontal) distances θiY have been written in column 10. • The coarse relative coordinates (affected by a series of errors) are obtained by applying the relations (5.11), (5.12), which were written in the columns 11 and 12, in the upper part of the rows. • The coarse relative coordinates are summed up on columns, obtaining: ∑∆XCOMPUTED = 49.12 (m) (5.13) ∑∆YCOMPUTED = 285.74 (m) (5.14) These basically represent the coordinate increase from 71 to 73, but affected byerrors. • These corrected coordinate increases are obtained with the relations: ∆X71.73 = X73 - X71 = 8255.67 – 8206.43 = 49.24 m (5.15) ∆Y71.73 = Y73 - Y71 = 4704.40 – 4418.54 = 285.86 m (5.16) • The closing discrepancy errors on coordinates will be: eX = ∑∆X - ∆X71.73 = 49.12 – 49.24 = - 0.12 m (5.17) eY = ∑∆Y - ∆Y71.73 = 285.74 – 285.86 = - 0.12 m (6.18) • Computing the total error is done in the following way: eX,Y =±√ e²X + e²Y = ±√( - 0.12)² + (- 0.12)² = ± 0.17 m (6.19) • Computing the tolerance on distances: 1 TD = ± (0.003√ ∑D + -------- ∑D) (5.20) 2500 Where D represents the sum of the distances from column 10; We obtain: 72
  • 78. 424.41 TD = ± (0.003√ 424.41 + --------) = ± 0.23 m 2500 • The error is compared with the tolerance: lX,Y < TD (± 0.17 < ± 0.23) (5.21) • Since the condition that the error should be smaller (at most equal with the correction) is satisfied, we can move on to computing the corrections: CX = - eX = 0.12 m (5.22) CY = - eY = 0.12 m (5.23) • Compute the unitary corrections: CX 12 cm CuX = ------------ = --------------- = 0.107 cm/m (5.24) ∆X 112.28 m CY 12 cm CuY = ------------ = --------------- = 0.032 cm/m (5.25) ∆Y 378.96 m Where ∑ ∆X and ∑ ∆X represent the sum of coarse relative coordinates,taken in absolute value. • Compute the corrections of relative coordinates: CXiJ = CuX∆XiJ are written in column 11, and 12, respectively CYiJ =CuY∆YiJ the second line of the row. (5.26), (5.27) For example: CX71.101 = 0.107 x 12.59 ≅ 1 cm CY71.101 = 0.032 x 68.95 ≅ 2 cm The corrections CXiJ and CYiJ are rounded to cm, taking care that: ∑ CXiJ = CX; (5.28) ∑ CYiJ = CY; (5.29) • The corrected relative coordinates are obtained adding the corrections to the coarse relative coordinates and are written in the last line of the rows, on columns 11 and 12. e. Computing absolute coordinates 73
  • 79. Starting from the point 71, step by step (from the top of the columns 13 and 14downwards) applying the formulas: Xi = Xi – 1 + ∆Xi-1, i (5.30) Yi = Yi – 1 + ∆Yi-1, i (5.31) Thus: X71 = 8206.43 m X101 = X71 + ∆X71.101 = 8206.43 – 12.58 = 8193.88 X102 = X101 + ∆X101.102 = 8193.85 + 4.40 = 8198.27 X103 = X102 + ∆X102.103 = 8198.27 – 15.02 = 8183.25 X104 = X103 + ∆X103.104 = 8183.25 – 3.95 = 8179.30 X73 = X101 + ∆X101.73 = 8179.30 + 76.37 = 8255.67 X73 DAT = 8255.67 (for verification) Y71 = 4418.54 m Y101 = Y71 + ∆Y71.101 = 4418.54 + 68.87 = 4487.51 Y102 = Y101 + ∆Y101.102 = 4487.51 + 87.82 = 4575.33 Y103 = Y102 + ∆Y102.103 = 4575.33 + 84.68 = 4660.01 Y104 = Y103 + ∆Y103.104 = 4660.01 + 90.99 = 4751.00 Y73 = Y101 + ∆Y101.73 = 4751.00 – 46.60 = 4704.40 X73 DAT = 4704.40. Having the coordinates of the vertices of the traverse, the computation is done. C. The computation of a planimetric traverse in closed circuit (table 5.3, page 77) We present this type of computation, too, in order to ensure a better understandingof processing planimetric traverses. The computation in this case is performed in the same way as for the traversepresented before, with the particularity that the ending orientation is the same as thestarting (initial) orientation, the traverse beginning and ending on the same point. We have presented in table 5.1 the data recorded in the field, processed (in a tableof the type 5.1, which we do not consider to be necessary to present), in the columns 4, 5,7. 74
  • 80. In column 1 are given: the schema of the measurement and the starting data. In column 7 was performed the correction of horizontal angles, in the followingway: e = ∑ωMEASURED - ∑ωTHEORETICAL = 200g (n + 2) (5.32) n = the number of sides of the polygon formed by traversing. The error is compared to the tolerance, and if the condition is satisfied, then theunitary correction is computed: Cω Cuω = ------ (5.34) N Having the compensated angles, the permanent orientations will be computeddirectly, in the same way as in the previous traverse. The other computations are similarto those performed for the traverse supported at the ends. It should be mentioned that, in general, the computation and the compensation ofplanimetric traverses is performed directly in the table – having the type 5.2 or 5.3. 75
  • 81. 76
  • 82. 77
  • 83. 6. SURVEYING PLANIMETRIC DETAILS The content of the tutorial: By the topographic survey of planimetric details weunderstand the set of operations of measuring, computing and representing on the planthe situation from the terrain. A minimal number of points will be surveyed in the plan, which are needed toreproduce the details from the terrain and that surface on the whole: building corners,traffic routes, details concerning various installations, property and usage limits, as wellas natural details: forests, swamps, river sides, etc. We specify that the survey of planimetric details is supported on a series ofknown points, which form the control network of the survey, usually a planimetrictraverse.6.1. SURVEYING DETAILS THROUGH RADIATION (POLAR COORDINATES) 78
  • 84. Using this method, there are surveyed the details that are situated radially near astation point, measuring for each of them the distance (directly or indirectly) and theangle formed by the direction station-details with a known direction (figure 6.1). 71 501 502 ω 1 ω L 2 101- 103 503 D 3 101 Figure 6.1. The method of radiation The order of measurements is the following: - Draw a field schema of the measured elements; - Measure the elements needed for the traverse (see tutorial 5); - Measure the elements needed for the radiated points; - The horizontal distance di, the horizontal angle ωι; - The slanted distance li, from the station to the radiated point, once, directly (with the measuring tape or reel) or indirectly (using one of the methods of tachometry, or electronically): - The horizontal angle ωi; - The vertical angle, which will be needed for reducing the distance to the horizontal. These elements are recorded in a table of type 6.1 (see the homework of thetutorial). Also, in each station, a schema will be drafted, containing all the measureddetails, photographed – with respect to the known direction. 79
  • 85. For verification, the aiming begins with the support point and ends with it, thus closing the horizon tour. We mention that in the case of radiation, the horizon tour is not compensated anymore, but the possible closure discrepancy must be with the margins of tolerance. BOOK FOR TRAVERSES WITH RADIATIONS Table 6.1Team …………… Date …………Operator ……… WORK ……………………… Weather ………Instrument ……… Aimed point Station Schema Distances Horizontal Horizontal Slanted Vertical (L) (D) directions angles directions angles Slanted Horizontal Position I α Position I V m cm m cm g c cc g c cc g c cc g c cc 1 2 3 4 5 6 7 8 9 72 - - 30 60 00 00 00 00 - - 501 - 38 90 174 61 00 144 31 00 - - 502 - 36 12 179 68 00 149 38 00 - - 503 - 27 60 196 21 00 165 61 00 - - 71 504 - 32 44 203 01 00 172 41 00 - - ∆ 532 - 65 70 205 61 00 175 01 00 - - 508 - 24 05 209 71 00 179 11 00 - - 80
  • 86. 507 - 31 93 216 96 00 186 36 00 - - 506 - 38 40 220 81 00 190 21 00 - - 505 - 57 92 226 61 00 196 01 00 - - 509 - 34 24 242 23 00 211 63 00 - - 510 - 24 63 244 31 00 213 71 00 - - 511 - 26 80 262 36 00 231 76 00 - - 512 - 40 05 263 61 00 233 01 00 - - 513 - 26 12 277 33 00 246 73 00 - - 514 26 10 26 00 278 21 00 218 61 00 94 32 00 5 68 00 515 21 05 20 97 292 71 00 232 11 00 94 34 00 5 66 00 72 - - 30 60 00 00 00 00 - - 101 ○ 71 - - 0 00 00 00 00 00 - - 516 - 94 50 100 32 00 100 32 00 - - 517 - 57 20 109 53 00 109 53 00 - - 518 - 55 12 115 42 00 115 42 00 - - 519 - 30 16 149 05 00 149 05 00 - - 520 - 21 83 170 52 00 170 52 00 - - 522 - 21 37 174 49 00 174 49 00 - - 503 - 21 53 195 56 00 195 56 00 - - 504 - 34 86 193 37 00 193 37 00 - - BOOK FOR TRAVERSES WITH RADIATIONS Table 6.1 (cont.)Team …………… Date …………Operator ……… WORK ……………………… Weather ………Instrument ……… Aimed point Station Schema Distances Horizontal Horizontal Slanted Vertical (L) (D) directions angles directions angles Slanted Horizontal Position I α Position I V m cm m cm g c cc g c cc g c cc g c cc 1 2 3 4 5 6 7 8 9 continued101 525 - 40 37 220 37 00 220 37 00 - - 526 - 6 12 240 83 00 240 83 00 - - 527 - 21 43 273 62 00 273 62 00 - - 529 - 17 11 295 12 00 295 12 00 - - 528 - 6 63 311 52 00 311 52 00 - - 530 - 9 15 341 06 00 341 02 00 - - 81
  • 87. 533 - 22 12 342 06 00 342 06 00 - - 531 - 9 83 351 07 00 351 07 00 - - 101 533 - 18 12 363 12 00 363 12 00 - - 534 - 25 12 369 72 00 369 72 00 - - 501 - 36 35 376 03 00 376 03 00 - - 71 - - 00 00 00 00 00 00 - - 102 101 - - 0 00 00 00 00 00 - - 538 - 17 52 12 51 00 12 51 00 - - 537 - 18 36 17 82 00 18 82 00 - - 537 - 16 92 27 31 00 27 31 00 - - 536 - 54 12 80 73 00 80 73 00 - - 535 - 92 83 89 12 00 89 12 00 - - 535 - 92 80 90 17 00 90 17 00 - - 539 - 6 25 333 91 00 333 91 00 - - 540 - 8 83 351 92 00 351 92 00 - - 541 - 13 27 370 73 00 371 73 00 - - 543 - 14 72 375 32 00 375 32 00 - - 542 - 13 78 376 41 00 376 41 00 - - 544 - 23 21 385 12 00 385 12 00 - - 545 - 26 12 388 34 00 388 34 00 - - 524 - 55 33 394 35 00 00 00 00 - - A - 7 93 0 00 00 0 00 00 - - 101 - 00 00 00 00 00 00 - - BOOK FOR TRAVERSES WITH RADIATIONSTeam …………… Date …………Operator ……… WORK ……………………… Weather ………Instrument ……… Aimed point Station Schema Distances Horizontal Horizontal Slanted Vertical (L) (D) directions angles directions angles Slanted Horizontal Position I α Position I V m cm m cm g c cc g c cc g c cc g c cc 1 2 3 4 5 6 7 8 9 74 - - 00 00 00 0 00 00 - - 104 - - 133 36 00 133 36 00 - - 546 - 69 12 170 91 00 170 91 00 - - 103 - - 202 12 00 202 12 00 - - 73 821 - 83 31 209 53 00 209 53 00 - - ∆ 551 - 53 94 229 61 00 229 61 00 - - 550 - 101 51 255 11 00 190 11 00 - - 549 - 100 72 277 77 00 277 77 00 - - 82
  • 88. 548 - 111 37 296 21 00 297 21 00 - - 547 - 49 23 366 43 00 366 43 00 - - 74 - - 00 00 00 00 00 00 - -6.2. SURVEYING DETAILS THROUGH SQUARING (SQUARE COORDINATES) In this case, the details are placed longitudinally with respect to a known axis(102 – 103 in figure 6.2) and close to that. The order of the operations is the following: - Perform a schema of the measured elements; - Choose (or create) a repeating axis (which will be identical to the 0Y’ axis), and the 0X axis will be normal to the 0Y’ axis in one of the support points; - Tension the tape, with the origin in the end 102, towards 103, aligning it either by setting out or with the theodolite; - Draw a perpendicular with the topographic square, on the 0Y’ axis, from wach detail; - The foot of the perpendicular (e.g. 801-801’) will give the rectangular coordinates Xi’, Yi’, which are measured with the measuring reel, or with the tape; - Proceed in the same way for each surveyed point, also drafting a detailed schema of the measurements; - It is necessary to determine the perimeter (by measuring the sides a, b, c, d – figure 6.2) of all measured buildings. X’ Y’801 801’ 102 801 X’801 802 103 Y’ a d 801-802 c - X’ b Figure 6.2. The method of squaring 83
  • 89. The measured values will be written in a table of type 6.2 (see the homework ofthe tutorial). SQUARE COORDINATESTable 6.2 POINT X’ Y’ POINT X’ Y’ 801 5.52 12.31 814 5.20 22.75 802 7.21 15.32 815 16.31 23.02 803 7.21 29.72 816 5.17 33.21 804 5.48 36.91 817 16.36 33.21 805 8.37 43.62 818 5.12 64.23 806 8.37 56.03 819 16.34 64.23 807 5.46 65.32 820 16.30 72.35 808 9.36 69.33 821 5.10 77.51 809 9.34 78.83 810 5.42 83.87 811 5.21 8.36 812 7.13 9.52 813 7.15 19.366.3. SURVEYING DETAILS THROUGH THE METHOD OF ALIGNMENTS The details situated in alignment (figure 6.3) can be measured using this method,by determining directly the distances li. 101 D (L ) AB AB βAB βAB l2 l1 A c B a b Figure 6.3. The method of alignments 84
  • 90. It should be mentioned that, for repeating the measured details on the plan usingthis method, it is necessary that the support points A and B from the alignment to bemeasured with respect to the control network of the survey (∠βAB and DAB are measured). Also, the perimeter of the measured buildings is determined and a detailedschema of the measurements is drafted (see the homework of the tutorial).6.4. THE HOMEWORK OF THE TUTORIAL Supported on the traverse (72-71.101-102-103.104.73.74), a area of the terrainwas surveyed using various methods. The data recorded in the field is presented in table 6.1, table 6.2 and the schema6.1. Process this data in order to repeat the measured details on the plan and draft thetopographic plan of the measured area on the scale 1:1000. Remark: We specify that, for recording the data from the radiation method, table6.1 can be used, too, but in this case, in the field only the columns 1, 2, 3, 4, 6, and 10 arefilled in, and columns 9 and 12 are filled in by processing the previous ones (therefore,only position I of the telescope is used). Explanations – procedure: 1. Processing the table 6.1 - In column 1 draft a schema of the measurement (in this case, since we have presented the location plan, these schemas were not recorded in the table anymore); - In column 5 are presented the horizontal distances measured directly (in the case of the distance 71.514, it was measured in value – slanted distance, the horizontal distance being determined based on this); - Column 6 contains the readings recorded on the bearing circle, which will be used to compute the horizontal angles (α); These angles (that are written in column 7) were computed in the following way: 85
  • 91. - In the case when the bearing point (e.g. 71 from station 101) is aimed using the method of zeros in coincidence, that is, with the origin on the bearing circle, the columns 6 and 7 are identical. - When the bearing point is aimed with some value on the bearing circle, the angle α will be computed subtracting this value from each direction from column 6. The slanted directions were recorded only in the case when the value on theclinometer circle was not equal to 100g (that is, when the telescope was not horizontal). 2. Repeating the details on the topographic plan is performed in the following steps: - Trace the rectangular graticule; - Repeat the bearing points (71, 72, 101, …) through coordinates (see the tutorial 6); - Repeat the detail points measured through distances (column 5) and angle (column 7), from each station point (in the case of measurements by radiation). The distances are scaled down using the relation: Diy diy = -------- N Where:diy = the distance between two points on the plan; Diy = the corresponding distance in the field; N = the denominator of the plan scale (here N = 1000). If we do not have a centesimal protractor, the angles from column 7 will betransformed into sexagesimal gradation using the method discussed in tutorial 1. Repeating is performed through the angle traced with the protractor and the 516distance traced with the bar (see figure 6.4). 71 d 10 g 0 3 c 2 0g 86 101 Figure 6.4. Repeating radiated details
  • 92. Then, the details surveyed through square coordinates and through the method ofalignments are repeated, initially scaling down the coordinates or the distances. The last step is finishing the plan, which consists in drawing the planimetricdetails, through the contour (buildings, property limits, roads, etc.).7. PROBLEMS SOLVED ON MAPS AND PLANS THE CONTENT OF THE TUTORIAL The TOPOGRAPHIC MAP is a scaled down, generalized and standardized,representation of a large area of the terrestrial topographic surface on a plan surface,taking into account the terrestrial curvature. The TOPOGRAPHIC PLAN gives a scaled down, standardized image, similar tothe scaled down horizontal projection of some small surfaces of terrain, not taking intoaccount the terrestrial curvature. In design and execution works, the engineer often faces the situation of exploitingtopographic maps or plans, precious tools only for those who know how to use them. In their study, the planimetry and leveling problems that appear (being studiedwithin this tutorial), are concerned with knowing the symbols, the geometric andgeographic graticule, and the numerical or graphical scales. 87
  • 93. 7.1. PROBLEMS CONCERNING USING MAPS AND PLANS7.1.1. SYMBOLS The symbols express graphically the content of topographic maps and plans,being a schematical representation, of standard chosen shape and size, suggesting theimage of the topographic object that they represent. When choosing symbols, these should suggestively indicate the details theyrepresent, they should be distinct, represented on scale, not with respect to the size of therepresented object, but based on graphs called scale diapasons. The symbols can be divided into: - Symbols for planimetry; - Symbols for leveling. In Appendix 4, there are presented the most important symbols used intopography.7.1.2. THE GRATICULE OF MAPS AND PLANS The graticule of maps is formed of the geometric and geographic graticule. a. The geographic graticule is formed by the meridians and parallels that bound the map – such that, each corner of the topographic sheet has the geographic coordinates written upon it (figure 7.1). 45°50’ 45°45’ difference 21°37’30” 21°37’30” 21°30’ 21°30’ 45°50’ 45°45’ 88 Figure 7.1. The geographic graticule
  • 94. In figure 7.2 we have presented in detail a corner of the graticule, and also themethod for computing the geographic coordinates of a point (A) situated on the mapsheet. From here it can be deduced that the coordinates of each point are determinedthrough interpolation, with respect to the closest graticule corner. b. The rectangular geometric graticule is specific to each meantime zone of the GAUSS projection, being formed of the X axis (representing the axial meridian of the meantime zone) and the Y axis (representing the equator) and a network of squares with 1 km sides, parallel to the coordinate axes. In order to make the coordinates positive on the Y axis, the X axis was translatedtowards west with 500 km (figure 7.3). 74 The number of the meantime zone is added to the value of the Y coordinate. N 60” A’ A A”’ ∆Y ” ∆ϕ 50 S 72 45 ° 45’ 24° 30’ 30” 52 97 A” R 98 M 30” ∆λ ” 60” 89 Figure 7.2. The rectangular geographic graticule and the computation of coordinates
  • 95. Our country is situated within the meantime zones 34 and 35, having the axialmeridians 21º and 27º. Usually, it is simplified, saying that it is in the meantime zones 4and 5. For example, if: X = 5372.363 km; Y = 4290.572 km; X represents the distances until the Equator, and the digit 4 from Y represents thenumber of the meantime zone, 290…. Being the distance until the 0x axis, given by the21º meridian. (figure 7.3). X 24° 27° 30° 19° 21° YA A XA Equator Y Translation with 500 km Axial meridians Figure 7.3. The Gauss graphic graticule It should be mentioned that in our country, since 1970, there is used thestereographic projection system on secant plan, with the center of the Cartesian 90
  • 96. coordinate system at north from Fagaras. The 0x axis is taken on the north direction ofthe coordinate system, and the 0y axis on the east direction. In order to make all the coordinates positive, the axes are translated towards southand west, respectively, with 500 km. The computation method of the geographic coordinates of the point A: - Trace perpendiculars on the geographic frame, obtaining A’ and A”; - Measure the distances MA’, MN, PA”, PR; - It can be seen that: ϕA = ϕM + ∆ϕ” = 45°45’ + ∆ϕ” λA = λM + 30” + ∆λ” = 24°30’30” + 30” +∆λ” = 24°31” + ∆λ” from M to P - ∆ϕ” and ∆λ” are obtained through the rule of three: bo” ………………………MN ∆ϕ”………………………. MA’ MA PA” ∆ϕ”= -------- bo” (similarly ∆λ” = --------- bo”) MN PR - Replacing ∆ϕ” and ∆λ” in the relations (7.1), (ϕA and λA) are obtained. The computation method of the Cartesian coordinates of the point A: - Determine de Cartesian coordinates of the point S: XS = 5072 km YS = 5297 km - Measure δXSA and δYSA in graphical value: δX = 20.5 m δY = 16.1 m - Measure ∆XSA and ∆YSA in field values: ∆XSA = δXSA 25.000 = 512,500 mm = 0.512500 km ∆YSA = δYSA 25.000 = 402,500 mm = 0.402500 km - Compute XA and YA: XA = XS + ∆XSA = 5072 + 0.512500 = 5,072,512.50 m 91
  • 97. YA = YS + ∆YSA = 5297 + 0.402500 = 5,297,402.50 m7.1.3. THE SCALE OF MAPS AND PLANS The constant ratio between a distance d on the plan or map and its correspondentin the field D, expressed in the same measuring unit, is called scale. Numerical scales are expressed as the ratio: d 1 --- = --- D n Thus, from this relation, one of the terms can be determined from the other threeterms, depending on their values. Usually, the scale ratio is given as 1:n, d is measured on the plan, and D = n · d. The graphical scale can be done in two ways: - Simple (or linear) – with the precision 1:10 from the value of the base (figure 7.4) Scale: 1:5000 Base = 2 cm = 100 m Precision = Base/10 = 2 mm = 10 m Distance = 300 m on base 70 m on talor 370 m 100 0 100 200 300 400 500 Figure 7.4. The simple linear graphical scale The following problems can be solved using graphical scales: - Computing the distance D – from the field, by graphically measuring on the mal and comparing with the field equivalent, read on the scale (figure 7.4); 92
  • 98. - Repeating a distance D on the plan – by graphically applying the value d determined from the graphical scale (figure 7.4).7.1.4. ORIENTING MAPS AND PLANS IN THE FIELD This problem represents placing the plan or the map for a possible confrontationwith the terrain, such that each line on the plan to be parallel to its correspondent from thefield. There are two possibilities: - The lines from the plan that are known in the field, too (road axis or railway) are placed parallel and in the same direction (figure 7.5); 96°20’ 0 Figure 7.5. Orientation based on field elements 270 90 180 93 Figure 7.6. Orientation with the compass
  • 99. - Orientation with the compass, by the north direction, known on the map, rotating the map until the magnetic indicator of the compass becomes the extension of the meridian traced on the map (figure 7.6).7.2. SOLVING SOME PLANIMETRY AND LELVELING PROBLEMS ON A TOPOGRAPHIC PLAN We shall discuss this problem in a specific example, stating the homework of thetutorial and then solving each point and presenting the necessary theoretical notions.7.2.1. THE HOMEWORK OF THE TUTORIAL A. PLANIMETRY a) Determine the Cartesian coordinates (XA, YA, XB, YB) of the points A and B, respectively; b) Repeat two points C and D of known coordinates on the plan [XC = 2334.36 + n(m), YC = 1408.52 + n(m), XD = 2358.46 + n(m), YD = 1456.34 + n(m)]; c) Determine the distance DAB both graphically and analytically, verifying it to be within the margins; d) Compute the surface SABCD analytically; 94
  • 100. e) Determine the orientation of the distance AB: θAB, both analytically and graphically B. LEVELING a) Determine the heights of the points A, B, C, D and of a point E situated on the AB alignment, at 90 m from A; b) Compute the slopes of the terrain on the AB alignment; c) Corresponding to the required slope p0 % = (7 = 0,n)%, trace a line having this slope between the points A and B, on two variants; d) Draw the longitudinal profile of the AB alignment; e) Trace the transversal profile of the terrain in the point E, on a length of 30 m to the left and 30 to the right of the alignment. NX= 2400 B 9 8 Bra 18 7 t esr 17 iv er T 6 16X= 2350 P 15 5 14 E 12 13 4 11 10 3 2 S 1 AX= 2300 M Y= 1400 Y= 1450 1:500 Y= 1500 Plate 7.1. 95
  • 101. 7.2.2. SOLVING METHOD EXAMPLE A. PLANIMETRY a. The coordinates will be determined with respect to the closest graticule corner to the given point (figure 7.7); - Draw normals from the points A, B, C, D on the axes of the graticule; - Measure the distance from the points to these axes, determining the relative coordinates (by multiplying by the scale factor) ∆X and ∆Y; - Compute the coordinates of the points by modifying the coordinates of the mentioned graticule corner, with the values of the relative coordinates. Thus, for A: δXMA = 17.2 mm values measured on the plan δYMA = 20.0 mm The relative coordinates: ∆XMA = XMA x 500= 17.2mm x 500 = 8600.0mm = 8.6m; ∆YMA = YMA x 500= -20.0mm x 500= -10,000.0mm = -10.0m; The absolute coordinates: XA = XM + ∆XMA = 2300 + 8.6 = 2308.6 m YA = YM + ∆YMA = 1500 + 10.0 = 1490.0 m X 2400 2350 δ Y MA A δ X MA M Y 2300 0 1400 1350 1500 Figure 7.7. Graphical96 determination of coordinates
  • 102. For the point B we obtain similarly: XB = 2380.1 m YB = 1433.4 m Repeating the points on the plan is the inverse operation of the previous one. For point C we have: XC = 2334.36 m YC = 1408.52 m - The closest graticule corner is the one having the coordinates: XP = 2350.00 m YC = 1400.00 m We compute the relative coordinates: ∆XPC = 2350.00 – 2334.36 = 15.64 m ∆YPC = 1400.00 – 1408.52 = - 8.52 m Scaled down: ∆XPC 15.64 δXPC = ----------- = --------------- = 31.3 mm N 500 ∆YPC - 8.52 δYPC = ----------- = --------------- = 17.0 mm N 500 Applying these values with respect to the point P, we obtain the position on theplan of the searched point. We proceed similarly for the point D. b. The distance DAB graphically determined is obtained by measuring it on the plan and then multiplying it by the scale denominator; thus: DAB = dAB x N = 182.2 mm x 500 = 91,100 mm = 91.10 m; c. The distance DAB determined analytically is obtained through the coordinates of the points, with the formula (figure 7.8): 97
  • 103. DAB = √∆X²AB +∆Y²AB (7.2) ∆XAB = XB - XA = 2380.10 – 2308.60 = 71.50 m; ∆YAB = YB - YA = 1433.40 – 1490.00 = - 56.60 m; DAB = √(71.50) ² + (-56.60) ² = 91.19 m d. There exist various procedures for computing surfaces, which we shall not present here, because they are the topic of topography courses. We have chosen the most general and precise method for computing the givensurface: the analytical method (from coordinates). Computation formulas: ∑ Xi (Yi +1 – Yi –1) S’ = ----------------------------- (7.3) 2 ∑ Yi (Xi -1 – Xi +1) S” = ----------------------------- (7.4) 2where i represents the order number of a point representing one of the corners of thegiven surface, i+1 being the point after it, i-1 being the point before it, in right-handeddirection (figure 7.9). X B N XD B D XB X DAB XC ∆XAB C A θ AB XA ∆YAB 0 YC YB YD YA Y Y D AB = √ ∆ X 2 AB + ∆ Y 2 AB 1 0 ∆YAB S = -- ∑ X i ( Y i +1 - Y i -1 ) tg θ = ------- 2 i=A AB ∆XAB 1 0 S = -- ∑ Y i ( X i -1 - X i +1 ) 2 i=A 98Figure 7.8. The analytical computation Figure 7.9. The analytical computation of distance and orientation of the surface
  • 104. Having: XA = 2308.60 m, XB = 2380.10 m, XC = 2334.36 m, XD = 2358.46 YA = 1490.00 m, YB = 1433.40m, YC = 1408.52 m, XD = 1456.34 Thus: 1 S’ = ---[XA(YB - YD) + XB(YC - YA) + XC(YD - YB) + XD(YA - YC)] = 586.147 m2 2 1 S” = ---[YA(XD - XB) + YB(XA - XC) + YC(XB - XD) + YD(XC - XA)] = 586.146 m2 2 S’ + S” S = ---------- = 586.147 m2 2 e. Graphical determining: θAB = 357g50c (with the centesimal protractor) ∆YAB -56.60 Analytical determining: tg θAB = -------- = ----------= - 0.791608 ∆XAB 71.50 - ∆Y Since tg θAB = --------- => θAB is in the quadrant IV: β = θAB - 300g is searched for + ∆X ctgβ = 0.791608 => β = 300 g + 57 g 37 c18 cc = 357 g 37 c18 cc. B. LEVELING a. The point A is situated on the contour 453, therefore ZA= 453 N 444 443 m. M B E=1m444 d d’ 443 B M d’ N d 99 Figure 7.10. Determining the heights with respect to contours
  • 105. For other points that are between two contours, apply the rule of three or thetheorem of THALES. - Measure d and d’: d = 16.3 m; d’ = 11.4 m. According to figure 7.10.b: d’ d ----- = ------ ∆h 1m d’ => ∆h = ------- 1m (7.5) d 11.4 ∆h = -------- 1m = 0.7 m 16.3 B L AB ∆ Z AB A ϕ AB b D Figure7.11. The slope 100
  • 106. ZB = ZM + ∆h => ZB = 443 + 0.7 = 443.7 m.The heights of the points C, D, and E are determined similarly.b. Determining the slopes on a given alignment presents the following problems: 1) Determining the slopes on the AB alignment of the slope of the direction AB (the mean slope); 2) Determining the (maximal) slope of the terrain in the area of the points A and B (figure 7.12); 3) Determining the maximal and minimal slope of the terrain on the given alignment AB (figure 7.13). ∆ZiJ pij = tg ϕiJ = ------- DiJ N 444 B E = 1m M ϕB 443 DMN =d·N Figure 7.12. The slope of the terrain in the area of a point 5 44 0 44 7 8 9 10 5 6 4 2 3 d min . . 1 A d max . 453 452 451 450 p min 449 448 447 446 p max 445 444 443 101 Figure 7.13. The maximal and minimal slope
  • 107. For example: -E -1m pA1 = --------- = ---------, hence: DA1 dA1N -1m - 1000 mm pA1 = ---------------------- = --------------- = - 0.084 23.8 mm x 500 11,900 mm ∆ZAB ZB - ZA 443.7 – 453 9.3 pAB = -------- = ----------- = ---------------- = - --------- = - 0.1020 DAB DAB 91.19 91.19 The slope expressed as percentage: p%A1 = pA1 x 100 = 0.084 x 100= - 8.4%; p%AB = pAB x 100 = 0.1020 x 100= - 10.20%; 2) E 1m 100 cm pA = tg ϕA = ------ = -------- = ------------- = 0.0870 DA d x N 2.3 x 500 pA% = 0.0870 x 100 = 8.7 % E 1m 100 cm pB = tg ϕA = ------ = -------- = ------------- = 12.27% = 0.1227 DB d x N 1.63x 500 3) As it results from figure 7.13, the maximal slope (p max) on a given alignment isthere where the distance between two consecutive contours is minimal (dmin). By analogy, for (dmax) it corresponds (pmin). On the AB alignment: E 100 cm 1cm pmax = ----------- = ------------ = --------------- = 0.1786 (pmax% = 17.86%) dmin x N d16 x 500 1.12 cm x 5 E 100 cm 1cm 102
  • 108. pmin = ------------ = ------------ = --------------- = 0.0694 (pmin% = 6.94%) dmax x N d34 x 500 2.88 cm x 5 c. When designing the axis of a traffic route, it is required to perform it with constant slope, on various sections. In our example, between the points A and B, there will be traced a line withconstant slope p = 7%, using contours (figure 7.14). Therefore: 1m 100 cm 100 1 1 pmin = 100 ------------ = 100 ---------- = ------ x ------ = 20 ---- (cm) d0 x N d0 x 500 5 d0 d0 1 20 d0 = 20 ------- = -------- = 2.9 cm; p0 % 7 Having this distance in the compass (called design step), with the point of thecompass in A, an arc of circle is drawn, which will intersect the next contour (452) in twopoints (α and β). Practically, the number of variants will double at each intersection with Ethe contours.100 --- p0 %= D0 From among these, the designer will choose the optimal solution. Y 451 E =1 m i ϕ D0=d0 · N 450 Figure 7.14. Determining the length of the line with constant slope d0 B d0 d0 d0 445 d0 d0 d0 100 d0 d0 450 D0 = E d0 d0 d0 p0 % d0 d0 α 10 d0 = E d0 d0 d0 d0 p0%⋅ N β A 103 Figure 7.15.Tracing the line with given slope
  • 109. d. The longitudinal profile of the terrain between two points (A and B) is obtained intersecting the terrain with a vertical plan that passes through those two points (figure 7.16). Considering the two points A and B, the longitudinal profile of the terrainbetween those points is obtained as follows: - Unite the points A and B with a right line; - Number (1, 2, 3, …) the points where the line intersects the contours; - Measure the distances between these points, transforming them into field values (DiJ = diY x N); - Draft the profile (see figure 7.17) in the following way: - Draw the frame and the ruling; - Write the headers, the scale, etc.; - Transform the points of the profile on the scale (d iJ = DiJ / N’) in the first box; - Write the heights of each point, the distances between the points; - Compute the cumulated distances (the distances from the left end of the profile – the point A, to some point); - Compute the slope (either the slope of the given direction, or on sections or between the contours, etc.); - Draw normals from each point, intersecting the contour that has the height of the point written on it; - Unite these points, obtaining the longitudinal profile. 104
  • 110. Remark: Usually, in order to increase the expressivity of the profile, the heightscale is chosen to be 10, 20 or 50 times larger than the distance scale. As origin for theheights is chosen 0D, which is denoted by a height placed 1-2 m lower than the smallestheight from the project. e. The transversal profile is obtained similarly with the previous one, representing the intersection of the terrain with a vertical plan and transversal to an alignment. Horizontal sections and vertical sections The profile contains only the boxes presented in figure 7.18, and the height scale Vertical sectionis usually taken equal to the distance scale. 3 3 E 2 2 4 4 E 1 1 5 5 E A 350 B A B 350 ă Vertical datum Scaling A’ 1’ 2’ 3’ 4’ 5’ B’ down Scaling Longitudinal profile Length scale 1:1000 down 354 Height scale 1: 200 353 352 351 3” 5” 350 A” 1” 2” 4” B” 348 349 Point A 1 2 3 4 5 B 350 351 352 353 352 351 350 Height partial 6.25 6.97 11.94 7.21 7.83 4.96 Distances 13.22 25.16 97.37 105.20 110.16 6.25 0.0 cumulated Slope p% Figure 7.16. The principles of relief representation on plans and maps with contours and in longitudinal profile 105
  • 111. Distance scale 1:500 Z Longitudinal profile Height scale 1:200 454 453 452 Z Transversal profile Distance scale 1:500 451Remark 450 Height scale 1:100 449 454Usually, the height 448 447scale is equal to the 446 453distance scale. 445 444 452 443In order to442 44 present 441 451 D the drafting way, 0Point number A for this profile, the 1 450 2 3 4 5 6 7 8 9Point height was height scale 453 B 452 451 450 449 448 447 446 445 444 443.7 enlarged. 449Partial distances 11.9 9.10 5.6 14.4 7.7 12.9 5.6 10.5 11.0 2.4Cumulated distances 0 11.9 21.0 26.6 448 41.0 48.7 61.6 67.2 77.7 88.7 91.10 p41% p12% p34% p98Slopes % p23% p43% p56% p67% p78% p89% % 447 Figure 7.17. The longitudinal profile 446 D 445 Point number 0 S9 10 11 12 13 E 14 15 16 17 18 T 448.00 448.00 450.13 450.00 449.00 448.00 448.26 448.00 449.00 450.00 450.20 Point height 2.5 Distances 1 7.75 5.50 2.4 10.45 4.75 4.05 4.15 6.4 7.25 3.4 Figure 7.18. The transversal profile 106
  • 112. 8. THE STUDY OF LEVELING INSTRUMENTS The content of the tutorial: Within this topic, the main simple instruments orinstruments with telescope used in geometric leveling will be described, in the same time,being presented the working procedure, through practical examples. Theoretical problems and working procedure8.1. SIMPLE INSTRUMENTS – WITHOUT TELESCOPE We shall present only those instruments that have a broader usage.8.8.1. THE LEVEL HOSE Or, also called – the level with water and hose – it consists of two tubes made ofglass (or glass in metallic fitting), connected by a rubber tube long of 5 up to 50 m (figure8.1). 107
  • 113. 3 CA 0 0’ A 1 ∆Z 1 B 0 000 2 Figure 8.1. The usage of the level hose 1 – glass tube, 2 – rubber tube, 3 – graduated bar Based on the principle of communicating vessels, if the instrument is filled upwith water, then, in the two glass tubes (1), the water will rise up to the same level 00.Thus, with the use of the graduated bar (3), there can be determined the altitudedifference (∆ZAB) between two points A and B, situated at a distance of up to 50 m. The working procedure is the following: - Place the first glass tube in front of the point A, at the level of the water from the tube; - On the vertical of point B, place a graduated bar (3) with the origin CB = 0.000, in front of the point B; - The gradation CA, situated on the horizontal of the point A (00’), is determined with the second glass tube, brought in front of the graduated bar. The altitude difference (∆ZAB) will be: ∆ZAB = CA - CB (8.1) It is considered that the precision that can be obtained is of order of ± 1 cm, for adistance between the points of up to 50 m, enough for some construction works. However, it is recommended that this instrument to be used only in the absence ofdevices with telescope. 108
  • 114. 8.1.2. THE LEVELING LONG BOARD AND THE AIR-BUBBLE LEVEL Currently called – the long board and the level water – they are also used todetermine some altitude differences, especially on rough terrains. The long board (1) is a wooden bar, perfectly right, having 2-5 m length (therecan be used a measuring staff, too). The level water (2) is a level with an air bubble, assembled on a wooden support. As accessories, there are used: a graduated bar (3), (or a measuring staff) which isused to determine the altitude differences, and a plumb-bob wire, which is used to bringthe graduated bar to vertical (figure 8.2). N N 3 2 A 1 ∆Z B Figure 8.2. The long board and the level water 1 – the long board, 2 – the level water, 3 – the graduated bar Working procedure: - One end of the long boards is placed on a point (A) and the second end is elevated (or lowered) until the air-bubble of the level (placed on the long board) is situated between the benchmarks; 109
  • 115. - In this moment, with the use of the long board (1), the horizontal of the point A was transmitted in front of the point B and on the graduated bar (3) we can determine the altitude difference between the two points.8.2. LEVELING INSTRUMENTS WITH TELESCOPE These are special instruments used to perform geometric leveling, the aiming axisof the telescope being perfectly horizontal. Depending on the way in which this operation is performed, we can distinguishbetween two types of devices: - The rigid leveling device – for which the axis is brought to the horizontal with the use of an air-bubble level rigidly connected to the telescope (through the operation called horizontal setting) and - The automatic leveling device (also called the instrument with compensator), for which the aiming axis is automatically brought to the horizontal – with the use of a compensator – after an approximate horizontal setting of the spherical level. We shall study a device from each category, from among the most used devices.8.2.1. RIGID INSTRUMENTS FOR GEOMETRIC LEVELING In our country, the most known and used device from this category is the level Ni030 produced by Carl Zeiss Company from Jena, Germany. Usage: It is used in technical and precision leveling, and by attaching a dived witha drum, it is used in precision leveling, too. Construction characteristics and device description (figure 8.3): 1 – the telescope – is an optical device that allows: - Materializing the optical aiming axis of the level; - Obtaining and observing the enlarged image of the aimed objective. 110
  • 116. Also, by the use of stadia cross hairs, horizontal distances can be measuredoptically (indirectly), up to 200 m. The telescope consists of the following parts: 2 – the lens – which forms the real, reduced and turned upside-down image of theaimed objective; 3 – the ocular – which forms the enlarged and virtual images of the image formedon the lens; the clarification of the cross hairs can be performed by the use of the ocular; 4 – the reticule of the telescope – formed of a circular metallic support thatsupports a glass slide on which are engraved the reticular traces (figure 8.4). At the intersection of the vertical cross hair (1) and the horizontal cross hair (2) isthe center of the reticule (r), which, together with the optical center of the lens,materializes the aiming optical axis of the telescope LL. a. The focusing screw – by the use of which is operated the focusing lens of the telescope, ensuring the sharp focus of the image of the aimed objective. 11 12 1 10 13 2 7 8 4 3 5 9 6 Figure 8.3. The Zeiss Ni 030 level – view from the focusing button 1 – telescope, 2 – air-bubble level, 3 – base, 4 – spherical level, 5 – foot screw, 6 – tension plate, 7 – clamp for locking the tensioning motion, 8 – horizontal slow motion screw, 9 – gradienter screw of the telescope, 10 – lens, 11 – ocular, 12 – telescope focusing button, 13 – the ocular of the microscope for reading the horizontal circle 111
  • 117. 1 4 3 4 2 3 Figure 8.4. The reticule of the telescope 1, 2 – main cross hairs, 3 – symmetric hairs, 4 – bisector cross hairs for reading on invar measuring staffsb. The air-bubble level – with coincidence centering (figure 8.5), when the bubble of the level is brought to coincidence – the level is perfectly set horizontally and therefore the aiming axis of the telescope (L-L) is horizontal – operation that is performed with the use of the slow motion foot screw (7). a. b. Figure 8.5. The image of the bubble of the air-bubble level a. not horizontally set level, b. horizontally set level 112
  • 118. 8 – graduated horizontal circle – which is used to determine the horizontal angles(with low precision); 9 – scale microscope – accessory used for reading the horizontal angles, equippedwith an ocular (9’), which is used to clarify the image in the microscope; 10 – clamp for locking the motion of the device in the horizontal plan; 11 – horizontal slow motion screw – which is used to fix the image of the aimedmeasuring staff in the plan of the cross hairs; 12 – the spherical level – used for the approximate horizontal setting, operatedfrom the horizontal setting screws of the level; 13 – the base of the device; 14 – foot screws; 15 – the tension plate for fixing the device on the trivet. To the parts mentioned above, we can add an optical micrometer with planparallel plate, which can be fixed in front of the objective and with the use of which themeasuring precision is improved – using invar measuring staffs. It is assumed that under normal measuring circumstances, a 2-5 mm/km of doubleleveling precision (given by the squared error) can be ensured; and in what concernsmeasuring horizontal angles, a precision of ± 2c can be ensured by approximation. Concerning the working procedure of this device in each station are performed thefollowing operations: a) Centering the device on the station point (only in the case of end geometric leveling), consisting in: - Placing and fixing the trivet above the station point; - Taking the device out of its casing and verifying the integrity of component parts; - Temporarily fixing the device on the trivet – by operating its clamping screw; - Centering the device, by the use of the plumb-bob wire that materializes the vertical axis – operation that is performed by moving the device on the plate of the trivet, or by varying the height of the feet of the trivet, until coincidence with the plumb-bob wire of the station point is obtained. 113
  • 119. These operations are necessary in the case of middle geometric leveling, in thiscase the device and the trivet being placed arbitrarily, in a point located approximately inthe middle of the distance between the two measured points. 2 b) Approximate horizontal setting – with the use of the spherical level – by operating the foot screws, the spherical bubble being brought in the interior of the circle – benchmark (figure 8.6.1); c) Clarifying the cross hairs – by operating the ocular, performing an aim on a bright 1 background – in the end the wires should be Figure 8.6.1. The horizontally seen as some lines, finely traced with china ink; set spherical level 1 – d) The approximate aiming of the measuring staff that is situated on the first the air bubble, 2 – the benchmarkmeasured point; circle e) Clarifying the image of the measuring staff – by operating the focusing screw; f) Ensuring permanent horizontal setting – bringing the half images of the bubble of the air-bubble level into coincidence – by operating the slow motion screw (7), operation that is performed for each aim that is performed; g) Reading on the measuring staff – the values that interest us in front of the cross hairs (and stadia hairs, if it is necessary) Operations d – f are repeated for each measured point.8.2.2. AUTOMATIC INSTRUMENTS FOR GEOMETRIC LEVELING These instruments present the particularity that within certain limits of inclinationof the vertical axis, the aiming axis is automatically brought to the horizontal andtherefore, the aiming axis is not brought into this position manually (operating the slowmotion foot screw) before each rod reading. The automatic levels are equipped with a spherical level that is horizontally set ineach station, at the beginning of measurements, obtaining a almost vertical position of the 114
  • 120. vertical axis – the automatic horizontal setting of the aiming axis being performed withthe use of an optical-mechanical element, called compensator. The advantage of these devices is that under the same circumstances of precision,time is saved, since there is removed the manual horizontal setting for each aim, whichwas necessary for rigid levels. From among the automatic technical leveling instruments, the most used inRomania, we have chosen for presentation the automatic level Zeiss Ni 025, produced bythe Carl Zeiss Company from Jena, too. The description of the device: 1 – the telescope of the level consists of the following parts: 2 – the lens of the telescope; 3 – the ocular of the telescope; the clarity of the cross hairs is obtained by rotatingthe ocular rosette; 6 5 1 3 2 9 8 10 Figure 8.6. The compensator level Ni 025 Carl Zeiss Jena 4 – the focusing screw, by operating on which the clear image of the aimed objectis obtained. The other parts of the level are: 5 – the spherical level – whose purpose we mentioned above; 115
  • 121. 6 – the folding aiming mirror for observing the spherical level; 7 – foot screw; 8 – slow motion screw in the horizontal plan (“no end screw”); 9 – horizontal circle, having the same purpose as in the case of rigid levels; 10 – tension plate. Concerning the operations for performing measurements with this instrument,these are the same as in the case of rigid level, except operation (f) – the slow horizontalsetting – which is not necessary here, as we have seen, being performed automatically.8.2.3. ACCESSORIES FOR GEOMETRIC LEVELING DEVICES8.2.3.1. Leveling measuring staffs They represent the main accessory for measurements performed in geometricleveling. The measuring staffs are wooden bars, graduated in cm (and those with invar strip– used in precision leveling – have the smallest gradation equal to 0.5 cm), which areplaced vertically in the points between which the altitude difference is measured. 32 08 We shall discuss only the study of measuring staffs used in technical leveling –those with centimetric gradations.31 09 1115m The regular leveling measuring staffs are wooden boards, well dried, long of 2 – 4 m 30m, wide of 1 – 14 cm, and thick of 2 – 2.5 cm, made of one piece (those having 2 and 3 10m) or of two joined pieces (those having 3 and 4 m) (figure 8.7). 29 The gradation of the measuring staff is performed centimeter by centimeter, in 11contrasting colors (white and red or white and black). The centimetric gradations aregrouped by five (or ten). 28 12 2888m 27 13 m 26 14 25 a. b. Figure 8.7. Technical 116 leveling measuring staffs a – for devices with right image b – for devices with image turned upside down.
  • 122. In figure 8.8 we have presented an example of reading on the centimetricmeasuring staff. After the vertical cross hair is brought over the image of the measuring staff byoperating the slow motion screw in the horizontal plan, the three readings are performed(at the upper stadia hair, at the level hair – horizontal cross hair and at the lower stadiahair) shortly CS, CM and CJ. 24 2388 23 2305 2202 22 117 Figure 8.8. Performing rod readings
  • 123. Each reading is performed in the following way: - Meters and decimeters are read directly on the measuring staff, being the digits written right below (for upside down images – it is above) the hair at which the reading is performed; - Centimeters are counted – being the number of entire gradations that can be found between the line that is below the digits read before and the mentioned hair; - Millimeters are approximated, between 0 and 9 – assuming that a gradation has a centimeter, that is ten millimeters – as the division fragment situated between the last gradation previously counted and the hair. As a conclusion – for example, at the horizontal cross hair (also called the levelhair) – in figure 8.8: 2.3 m are read 0 cm are counted 5 mm are approximated visually 2.305 m (in general, we don’t put the dot after the meter, we readdirectly 2305 mm). The verification that is performed in order to determine whether a rod reading isgood or not is the following: CS + C J ------------- = CM, or CM, CS - CM = CM – CJ; 2 the permissible deviation being of 1-2 mm. In order to increase the measuring precision, the measuring staffs can be placed inperfectly vertical position, with the use of a spherical level assembled on the measuringstaff or with the use of a plumb-bob wire. 118
  • 124. It is recommended that in the case when the measuring staffs are placed inintermediary points, without height marks – then leveling broaste to be used – made ofiron or cast iron (figure 8.9). Figure 8.9. Leveling broaste In what follows we shall present some examples of execution and control ofmeasurements – specifying that in leveling we are interested only in the middle reading –at the horizontal cross hair, the other two being used only for control (and for computingthe distance device – measuring staff, when it is necessary). Example 1 (figure 8.10) – Reading on the device with right image; Readings: CS = 1713 CM = 1664 CJ = 1614 Reading control: CS + C J With the condition ----------- - CM ≤ 2 mm 2 1713 + 1614 ---------------- = 1663.5 2 1.7 1713 1663.5 – 1664 = - 0.5 mm. 1664 1.6 1614 119 Figure 8.10 Reading on the device with right image
  • 125. Example 2 (figure 8.11) – Reading on the device with upside down image CS = 2316 CM = 2241 CJ = 2164 2316 + 2164 ----------------- = 2240 (verification) 2 2240 – 2241 = - 1 mm, therefore the tolerance condition is satisfied. 21 2164 22 2241 23 2316 Figure 8.11. Readings on the device with upside down image 120
  • 126. 8.3. THE HOMEWORK OF THE TUTORIAL Problem #1: Draw the two types of topographic levels that have been studied,emphasizing the similarities and the differences between them, and the purpose of thedevices in measurements and their working procedure. Problem #2: Draw the schemas of the following rod readings: CS = 1962 + n; CS = 3184 – n; a) Device with right image CM = 1860 + n; CM = 3076 – n; CJ = 1758 + n; CJ = 2968 – n. CS = 1472 + n; CS = 2546 – n; b) Device with upside down image CM = 1394 + n; CM = 2512 – n; CJ = 1316 + n; CJ = 2478 – n . Problem #3: Exemplify some works for which the level differences (or theheights) can be determined using the level hose and some works for which the levelingdevices with telescope are indispensable. 121
  • 127. 9. METHODS FOR MEASURING ALTITUDE DIFFERENCES The content of the tutorial: Determining the heights of points – by the use of altitude differences is a frequentoperation in performing and investment, from the phase of technical study, design, untilthe field application of the project; tracing step by step the investment works and beinguseful even after the construction is done, for monitoring its time behavior. This is why we consider that the future engineer should know this problem, andwe present the various possibilities to determine altitude differences within this tutorial.9.1. GENERAL PRINCIPLES9.1.1. THE PRINCIPLES OF GEOMETRIC LEVELING We have seen in the previous tutorial that the geometric leveling instruments withor without telescope are used to make a horizontal direction that passes through thevertical plan determined by the two measured points. Comparing the vertical distancesbetween each point and the horizontal, the altitude difference between these points is 122
  • 128. obtained, and then, knowing the height of one point, we can use the altitude difference tocompute the height of the other point. There are two types of geometric leveling, which we shall present in the sequel,but we recommend that the second method to be avoided, since the first one solves anysituation with increased precision.9.1.1.1. Middle geometric leveling Having two points A and B in the field (A with known height), the problem ofdetermining the altitude difference between these points has risen (figure 9.1). Figure 9.1. Determining the altitude differences by means of middlegeometric leveling (reading backward, backward measuring staff, level, span, span,forward measuring staff, reading forward, the vertical datum surface) There are two methods for computing the height of a point: a) Using the altitude difference δZAB – which is given by the formula: δZ AB = a − b (9.1) For the case from figure 9.1: δZ AB = 2.084 − 1664 = +0.420 m. Knowing the height of the point A (for example Z A = 351.430 m), the height ofthe point B can be computed applying the formula: Z B = Z A + δZ AB (9.2) 123
  • 129. For our case: Z B = 351.430 + 0.420 = 351.850 m b) Using the altitude of the aiming plan (Zi): It can be seen that: Zi = Z A + a (9.3) Hence: Z i = 351.430 + 2.084 = 353.514 m Or, the altitude of the aiming plan (or the horizon of the instrument) is equal tothe known height of a point, to which the rod reading from that point, from the levelingstation, is added. But in the same time, for the point B: Zi = Z B + b (9.4.) and knowing ZA, reading a and b, by confronting the formulas (9.3) and (9.4), itcan be obtained: Z B = Zi − b (9.5) or ZB = ZA + a − b (9.6) through computation obtaining the same value ZB as in case (a). It should be mentioned that there are cases when the level cannot be placed on thealignment of the aimed points (as in figure 9.2.a); in this case, another position can bechosen on the mediatrix of the segment AB, therefore maintaining equal distances pointA – station and station – point B (figure 9.2.b). Figure 9.2.9.1.1.2. End geometric leveling Also called “forward geometric leveling”, it consists in placing the level in one ofthe measured points (figure 9.3). 124
  • 130. Figure 9.3. (Level, span, forward rod reading, level hair, vertical datum) In this point, the height (i) of the instrument in the station is measured (with themeasuring staff or a reel), and in point B the value (b) is determined on the measuringstaff. There are two computational methods, too, similar to those studied above. a) Using the altitude difference (δZAB) Z AB = i − b (9.7) In our case: Z AB = 1.640 −1.040 = 0.600 m Then, knowing Z A = 350.053 m, ZB is determined with the formula: Z B = Z A + δZ AB (9.8) Hence: Z B = 350.053 + 0.600 = 350.653 m b) Using the height of the aiming plan (Zi): It can be seen that: Zi = Z A + i (9.9) And then: Z B + b = Z i , hence Z B = Z i − b (9.10) with the use of which, the same value is determined for the height of the point B.9.1.2. THE PRINCIPLES OF TRIGONOMETRIC AND TACHEOMETRIC LEVELING 125
  • 131. Trigonometric leveling, as well as the tacheometric leveling, consists incomputing the altitude difference based on the horizontal distance between two pointsand a vertical angle, recorded with a theodolite-tacheometer. This is formed by theperformed slanted aim and the horizontal that passes through the optical center of thedevice, and is situated in the same plan with the slanted aim and the measured points.9.1.2.1. Trigonometric leveling Depending on the distance between the station and the aimed point, wedistinguish: a) Near trigonometric leveling In the case when the distance from the station point to the aimed point is small,the aim on the signal (range pole) placed in the measured point can be performed at thesame height as the height of the device in the station (figure 9.4). The condition is that the aim should not be covered by vegetation or otherobstacles. It can be seen that the aim, being parallel to AB, the angle α determined usingthe device is equal to the slope angle ϕ. Figure 9.4. (vertical datum) From the triangle formed by the sides AB and δZAB, we can write: δZ AB tgα = or δZ AB = d ⋅ tgα (9.11) d In the case when l is known: 126
  • 132. δZ AB sin α = or δZ AB = l ⋅ sin α (9.12) l Usually, topographic devices record the value ω (the complement of the verticalangle α), called zenithal angle. As result, some formulas can be applied, that would notrequire transforming ω in α by the relation: α = 100 g − ω (9.13)and these are: δhAB = l ⋅ cos ω = d ⋅ ctgω (9.14) b) Far trigonometric leveling On large distances, the aim is performed to a signal of known height (S), and thedistance d is determined from the known coordinates of the two points (station point andaimed point) (figure 9.5). Figure 9.5. (vertical datum) It can be seen that: h + i = S + δZ AB (9.15) Hence: δZ AB = h + i − S But: h = d ⋅ tgα = d ⋅ ctgω (9.15’) And as result: δZ AB = d ⋅ tgα + (i − S ) (9.15”) δZ AB = d ⋅ ctgω + (i − S ) (9.16) 127
  • 133. the height of point B being computed still with the formula (9.8).9.1.2.2. Tacheometric leveling Having a much lower precision in comparison with the previous methods,precision that decreases with the increase of the distance (d) and of the slope (ϕ), thetacheometric leveling is used especially in embankment works. In tutorial 5 there was presented the method for computing distances using thetacheometric procedure, and we shall see in the sequel how are computed altitudedifferences. The computational principle is the same as in the case of trigonometric leveling –with the difference that in this case the distances are determined indirectly(tacheometrically) and therefore determining the three digits that represent the readings atthe stadia hairs (upper and lower) and at the level hair (middle) on the measuring staffbecomes necessary (figure 9.6). Figure 9.6. (vertical datum) Reviewing the relations of tachymetry: L = K ⋅ H (9.17) Where L’ – the slanted distance from the device to the measuring staff, K = the stadimetric constant (50, 100 or 200), 128
  • 134. H’ = an imaginary value, representing a generator number that would have been determined on a fictitious measuring staff, that we assume is normal on the aim in the point M. Since we know only H, determined from the relation: H = CS − CJ (9.18) CS and CJ being the rod readings on the measuring staff situated in the point B, arelation between H’ and H can be determined, noticing that the angle (with the apex inM) that would have been formed between the two measuring staffs (the fictitious and thereal one) is still the angle α – determined using the tacheometer. Therefore: H cos α = 2 H from where: H = H ⋅ cos α (9.19) 2 And then, from the relation (9.17): L = K ⋅ H ⋅ cos α (9.20) But it can be seen that: cos α = d / L (9.21) Hence d = L⋅ cos α = KH ⋅ cos 2 α (9.22) By analogy, using the formula (9.15): h + i = δZ AB + S where h can be deduced with the formula (9.15’): h = d ⋅ tgα And reviewing: δZ AB = KH ⋅ sin α ⋅ cos α + (i − S ) (9.23)and since usually i = S (hence α = ϕ ), we have: δZ AB = KH ⋅ sin α ⋅ cos α = KH ⋅ sin ϕ ⋅ cos ϕ (9.24) It is assumed that a precision of de ± 3 – 16 cm at one hundred meters of levelingcan be obtained for slopes of 3g – 25g.9.2. APPLICATIONS 129
  • 135. a) Determining altitude differences and heights by means of middle geometric leveling We shall consider known the field data, so we have to compute the heights ofsome points, using the studied formulas. Exercise #1 In table 1 is given the data measured in the field. Fill in the empty spaces, basedon the formulas of middle geometric leveling.Table 1 1 2 3 4 5 6 7 Studied Aimed Rod readings Heights δZAB Point point point Backward Forward Z (m) 1432 - A 1261 337.210 A 1091 S1 - 1243 B 1073 B 0902 Having the three rod readings (upper, middle, lower), it is necessary to verify thecorrectness of measurements. As we know: CS + CJ = CM with a maximal permissible error of 1-2 mm. Therefore, 2for the measurement from point A, we have: 1432 + 1091 = 1261.5 ≅ 1261mm 2 And for the point B: 1243 + 0902 = 1072.5 ≅ 1073mm 2 In order to compute the altitude difference, only the middle readings (at the levelhair) are used. Hence, δZ AB = a − b = 1.261 − 1.073 = 0.188m (will be written in column 5).Knowing the height of the point A: ZA = 337.210, the height of B is determined: 130
  • 136. Z B = Z A + δZ AB = 337.210 + 0.188 = 337.398m (will be written in column6, on the row of point B). From station S2 there were measured more points, from among which point A ofknown height (backward readings) and the points B, C, D, and E of unknown heights(forward readings) (figure 9.7). Determine the heights ZB, ZC, ZD, and ZE, using the method of the horizon of thedevice (Zi) and write them in table 2. Only the corrected readings at the level hair are given. First, compute Zi, using the formula Z i = Z A + a , where a is the rod reading in thepoint A. Therefore: Z i = 350.432 +1.436 = 351.868m . Then, applying the formula: Z B = Z i − b we shall have: Z B = 351.868 − 1.021 = 350.847 mand then, with the same formula we shall obtain Z C = 350.336 m, ZD = 350.432 m, ZE =350.011 m.Table 2 1 2 3 4 5 6 7 Studied Aimed Rod reading (mm) Zi (m) Heights Z Point point point Backward Forward (m) A 1436 - 350.432 B - 1021 S1 C - 1532 D - 1436 E - 1857 b) Determine the heights by means of end geometric leveling 131
  • 137. Exercise #3 In table 3 is given the data measured in the field for two points A and B (A havingknown height). Fill in the empty cells of the table.Table 3 1 2 3 4 5 6 7 Station Aimed Rod readings (mm) δZ Heights Point point point Backward Forward (m) Z (m) A 354.420 A A - - i=1.52 m - 1732 B B Therefore: δZ AB = i − b = 1.520 − 1.732 = −0.212m And then: Z B = Z B + δZ AB = 354.420 − 0.212 = 354.208m Stationing in the point A, the backward rod readings are missing, the height of thedevice (i) being assimilated with a mean reading in the point A. Exercise #4 The point A of given height (ZA) and the data measured in the field for the pointsB, C, D, and E are given (figure 9.8 and 9.9).Figure 9.8. Plan schema Figure 9.9. Section schema for the point B(station) (vertical datum) 132
  • 138. Determine the heights of these points, using the method of the horizon of thedevice (Zi), and fill in table 4. From figure 9.9 and using formula (9.9), we deduce: Z i = Z A + i = 353.472 + 1.473 = 354.945mTable 4 1 2 3 4 5 6 Station Aimed Forward rod Heights Zi (m) Point point point reading (mm) Z - - 353.472 A B 1021 B A C 1473 C i = 1.473 D 1957 D E 2021 E For point B, from the same figure and using formula (9.10): Z B = Z i − b = 354.945 −1.021 = 353.824m The heights of the points C, D, and E are determined similarly. c) Determining the heights by means of near trigonometric leveling Exercise #5 (figure 9.10) The data needed to compute the height of the point B was measured stationing instation A, using the method of trigonometric leveling, and was written in table 5. Figure 9.10. Section schema (vertical datum) 133
  • 139. Applying the corresponding formulas, compute the height ZB and fill in table 5. Table 5 1 2 3 4 5 6 7 8 9Station Aimed Readings on the Vertical Horizontal δZAB Heights Point point point clinometer angle α distance (m) Z (m) Position Position d (m) I ωI II ωII A 351.430 Ai = 1.54 B 92g37c 307g61c 102.54 m B In tutorial 4 (measuring angles with the theodolite), we saw that: 100 g − I + II − 300 g α= 2 So that for the presented case we have: 100 g 00 c − 92 g 37 c + 307 g 61c − 300 g 00 c α= = 7 g 62 c 2 Applying formula (9.11): δZ AB = d ⋅ tgα = 102.54m ⋅ tg (7 g 62c ) = 12.33m And then, applying formula (9.12): Z B = Z A + δZ AB = 351.430 + 12.33 = 363.760m d) Determining the heights by means of far trigonometric leveling Exercise #6 The data needed to compute the height of the point B was measured from station A (figure 9.11). 134
  • 140. Figure 9.11. Section schema It should be specified that in table 6 is given directly the computed vertical angleα. Compute the height of the point B and fill in table 6. Based on formula (9.15”) and analyzing figure 9.11, we deduce: δZ AB = d ⋅ tgα + (i − S ) = 272.54 ⋅ tg (8 g 31c ) + (1.47 − 2.12) = 35.129m And from formula (9.23) we’ll have: Z B = Z A + δZ AB = 353.382 + 35.129 = 388.511mTable 6 1 2 3 4 5 6 7 Station Aimed Vertical Horizontal δZAB Heights Point point point angle α distance d (m) Z (m) (m) A B 8g31c 272.54 353.382 A B i = 1.47m S = 2.12m e) Determining heights by means of tacheometric leveling Exercise #7 The data measured in the field from station A towards the points B and C aregiven in table 7 (figure 9.12 and 9.13). Figure 9.12. The section schema for point B 135
  • 141. Figure 9.13. The section schema for point C Fill in table 7, computing the heights ZB and ZC. It can be seen that in the case of point B, the aim at the level hair is made at thesame height (i) as the height of the device in station A. Therefore, S B = i and the formula(9.23) becomes: δZ AB = KH B ⋅ sin α B ⋅ cos α B (9.23’) Knowing that K=100 and determining H B = CS B − CJ B = 2140 − 1120 = 1.02m ,we shall have: δZ AB = 100 ⋅ 1.02 ⋅ sin 3g 21c ⋅ cos 3g 21c = 5.134m , resulting: Z B = Z A + δZ AB = 357.486 + 5.134 = 362.62mTable 7 1 2 3 4 5 6 7 Station Aimed Rod readings Vertical δZ (m) Heights Z Point point point CS angle α (m) CM CJ 2140 B 1630 3g21c 357.486 A A 1120 i = 1.63m 1830 B C 1425 6g54c 1020 C 136
  • 142. In the case of point C, the aim has height different from (i). Thus, applying theformula (9.23) we obtain: δZ AC = KH C ⋅ sin α C ⋅ cos α C + i − S C (9.23”) Where: H C = 1830 −1020 = 0.81m Hence: δZ AC = 100 ⋅ 0.81 ⋅ sin 6 g 54c ⋅ cos 6 g 54c + 1.630 − 1.425 = 8.468m , Obtaining in the end: Z C = Z A + δZ AC = 357.486 + 8.468 = 365.954m 137
  • 143. 10. GEOMETRIC LEVELING TRAVERSE WITH RADIATIONS The content of the tutorial: Complex method of geometric leveling, the traversewith radiation has two purposes, namely: thickening the altimetric control network andsurveying the characteristic points needed to draft a height plan of a given territory. In the first case, we are dealing with proper traverse, and in the second case, withleveling radiations. In what follows, we present the theoretical problems and the practical solution ofa case of traverse with radiations, by means of middle geometric leveling.10.1 THE GENERAL CONDITIONS OF A GEOMETRIC LEVELING TRAVERSE In order to obtain an adequate precision, the following is recommended: - The maximal length of the traverse should not exceed 2 km; - The length of a span – at least 10 m, at most 150 m; - The slope of the chosen route should be as small as possible; - The aiming radius should not pass at a distance less than 0.3 m from the ground; - Reading the value at the level hair, on the measuring staff, will be performed with at least two horizons of the device; - The measuring staffs will be placed perfectly vertical, with the use of the plumb-bob wire or of a spherical level. There are many types of geometric leveling traverses, distinguished by their shapein the plan, by the way how they are supported on points of known heights, or by themeasurement control method. 138
  • 144. We shall discuss in the sequel the most known and used case: the simplegeometric leveling traverse, supported at both ends on points of known height.10.2. RECOGNIZING AND PREPARING THE ROUTE OF THE TRAVERSE Before beginning the measurements, it is necessary to parse the route of thetraverse, observing the following: - Establishing and recognizing the bearing points; - Establishing the intermediary points and the way to designate them; - Verifying the visibility between the measured points; - Performing a schema of the route. In the same time, or after the terrain recognition operations mentioned above arefinished, the terrain is prepared in the following way: - Designate the points whose height will be determined, which will remain in the field as height marks; - Temporarily designating (perhaps using leveling broaste) the intermediary points; - Measuring the distances between the points using direct methods (or indirect methods – optical, with the telescope of the instrument); - Designating the station point situated as close as possible to the middle of the level (maximal deviation of 2 m); - Fixing the points that are to be radiated (situated at most at 150 ± 200 m from the station).10.3. FIELD WORKS – MEASUREMENTS THAT ARE PERFORMED IN THE CASE OF SIMPLE MIDDLE GEOMETRIC LEVELING TRAVERSE, SUPPORTED AT THE ENDS 139
  • 145. We shall present in the sequel an example from this category, mentioning thefollowing: - The field measurements are performed in only one direction; - For measurement control and for computing the distances that will intervene for compensating the errors produced by measurements, there are performed readings at the three horizontal cross hairs (two stadia hairs, upper and lower, and the level hair); - In each station point, measurement control is performed, verifying the data written in the measurement book; Therefore, the points A and B of known height are given (figure 10.1) and theheights of the points of the traverse 1, 2, 3, 4 and of the radiated points 1001 – 1009 arerequired. The measured data will be written into the LEVELING BOOK (table 1). In each station, the following field operations are performed: - Placing the level in a station situated at equal distances (aims) to the measured points (e.g. S1); - Performing the measurements concerning the points that are part of the traverse, and writing them into the book. In the case of station S1, proceed in the following way: - Aim the measuring staff – backward (from the point A) and perform the readings at the three cross hairs: CS = 1532 CM = 1021 CJ = 0510 CS + CJ - Verify using the known formula: = CM ; 2 - Write it in the leveling book, in column 3; - Aim the measuring staff – forward (from point 1), reading the values: CS = 1449 CM = 0937 CJ = 0427 140
  • 146. - Verify these values and write them into the book (in column 7). Next, perform measurements that aim the radiated points each at a time, aiming the measuring staffs from the points 1001, 1002, and 1003, and proceeding in the same way as in the case of the points A and 1, the data being written in column 5. Figure 10.1 (elevation, vertical datum) 141
  • 147. Table 10.1Work ……………… LEVELING BOOKComputed by ……… Date ……………Operator …………… Verified by ……………… Weather ………… Rod readings Temporary Permanent Aiming plan height Zi altitude altitude Schema, remarks Backward Intermediary Forward differences differences Corrections Ch Point number Distances Di Aimed point Heights Z Station Recorded Recorded Recorded Mean Mean Mean + - + -1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1532 A 1021 1021 352.436 A 0510 1449 1 0937 0937 352.522 1 0427 1474 353.457 204.40S1 1001 1072 1072 0.084 - +2 0.086 - 352.385 1001 0670 1291 1002 0943 0942 352.515 1002 0592 1864 1003 1532 1533 351.924 1003 1203 1393 1 0873 0873 352.522 1 0351 1648 2 1126 1126 352.271 2 353.395 208.80 0602S2 - 0.253 +2 - 0.251 1418 1004 0922 0923 352.472 1004 0430 1682 1005 1261 1261 352.134 1005 0840S3 2 1862 0.277 - +2 0.275 - 352.271 2 1320 1320 142
  • 148. Work ……………… LEVELING BOOKComputed by ……… Date ……………Operator …………… Verified by ……………… Weather ………… Rod readings Temporary Permanent Aiming plan height Zi altitude altitude Schema, remarks Backward Intermediary Forward differences differences Corrections Ch Point number Distances Di Aimed point Heights Z Station Recorded Recorded Recorded Mean Mean Mean + - + -1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 0778 1587 3 1043 1043 352.550 3 0498 1332 353.591 217.30 1006 0843 0843 352.752 1006 0353 1852 1007 1420 1420 352.171 1007 0986 1732 3 1231 1230 352.550 3 0728 1932 353.780 199.70S4 4 1436 1436 - 0.206 +2 - 0.204 352.436 4 0939 1512 1008 1200 1200 352.580 1008 0887 1658 4 1208 1208 352.346 4 0758 1327 353.554 179.90S5 B 0878 0878 0.330 - +2 0.332 - 352.678 B 0428 1964 1009 1643 1643 351.911 1009 1322 0.691 0.459 0.697 0.455 ∑Di = 1010.1 m ∑Ch = +10 mm +0.232 +0.232 After finishing these measurements, the level is disassembled and transported in S2 – where the same rules are applied. 143
  • 149. The measuring staff from A is taken into the point 3, and the measuring staff from2 stays in place. ∑Z δ = 0.691 − 0.459 = +0.232m ∆Z AB = Z B − Z A = 352.678 − 352.436 = +0.242m eh = ∑δZ − ∆Z AB = +0.232 − 0.242 = −10m T = ±20 LKM = ±20 1 = ±20mm eh < T Ch = −eh = +10mm mm Ch + 10mm ChP = = ≅ 1mm / 100m of traverse. Lmm 1010m10.4. OFFICE OPERATIONS The computation begins with the traverse, still in table 10.1, solving successivelythe following problems: a) Computing the mean readings – in general the mean reading recorded in column 3 (and 7 respectively) is kept – under the circumstances that the CS + CJ formula − CM ≤ 1mm is satisfied; if the difference is of 2-3 mm, 2 then the mean of the terms on the left hand side of the inequation is computed, this value being recorded in column 4, and 8, respectively. b) Computing the temporary altitude differences (δz’), using the relation: δZ A1 = mA − m1 δZ12 = m1 − m2 δZ 23 = m2 − m3 (10.1) δZ 34 = m3 − m4 δZ 4 B = m4 − mB 144
  • 150. Where mA, m1’, …, m4’ are situated in column 4, on the row corresponding to thepoint A, 1, …, 4, and m1, …, mB are situated in column 8, on the row corresponding tothe points 1, …, B. c) Each page of leveling is controlled using the relation: ∑col.4 − ∑col.8 = ∑ Z δ Or: the sum of mean readings backward minus the sum of mean readingsforwards must be equal to the sum of temporary altitude differences, which are written onthan page in column 10 (the positive ones) and in column 11 (the negative ones). d) Compute the length of the spans, using the relation: d1 = K ( CS A − CJ A ) = 100(1.532 − 0.510 ) = 102.20m d1 = K ( CS1 − CJ1 ) = 100(1.449 − 0.427 ) = 102.20m and so on Computing the length of the levels: D1 = d1 + d1 = 102.2 + 102.2 = 204.4m (will be written in column 12) Then compute the total length of the traverse: Lkm = ∑Di = D1 + D2 + ... + D5 = 1.010km e) Compensate the altitude differences based on the condition: B ∑δZ A i = ∆Z AB (10.2) Where: B ∑δZ = ∑col10 + ∑col11 − 0.691 − 0.459 = +0.232m A i And ∆Z AB = Z B − Z A = 352.678 − 352.436 = +0.242m Compute the total error (et) from the relation: B et = ∑δZ i − ∆Z AB = +0.232 − 0.242 = −10mm (10.3) Awhich should be less than the tolerance: T = ± mm Lkm 20 (10.4) T = ±20mm 1 = ± mm 20and therefore et < T. 145
  • 151. Determine the total correction Ct from the relation: Ct = −et (10.5) Ct = −(−10mm) =10mm Compute the partial corrections: Ct Ci = Di (10.6) Lkm km 10 Hence C A1 = 0.204 ≅ 2mm (which will be written in column 13). 1.010 Control using the relation: ∑C i = Ct (10.7) The correctness of the computation: ∑C i =10mm = Ct Compute the corrected altitude differences: δZ A1 = δZ A1 + C1 (10.8) δZ A1 = +0.084 + 0.002 = +0.086 (will be written in column 14 or 15, based on the sign). Verify if: B ∑δZ A i = ∆Z AB (10.9)and compute step by step the coordinates of the points: Z1 = Z A + δZ A1 (10.10) Z1 = 352.436 + 0.086 = 352.522m (will be written in column 16) f) Compute the altitudes of the radiated points with the use of the height of theaim plan (Zi) computed for each station with the relation: ZiS = Z A + mA (10.11) 1 Hence Z i S1 = 352.436 + 1.021 = 353.457 m (column 9), by analogy, we shallhave: Z i S 2 = Z1 + m1 and so on. Knowing the height Z i compute the heights of the radiated points S2 Z1001 = Z i S − m1001 1 146
  • 152. Thus: Z1001 = 353.457 −1.072 = 352.385m (10.12) Where m1001 is the mean reading (from column 6). Compute for each station the relation: Z R S i = m ⋅ Z i S i − ∑ mi Where Z R is the sum of the heights of the radiated points from that station, Si n – the number of radiated points from that station, Z i S i - the height of the aiming plan of the station, mi – the sum of the mean readings (column 6) from that station, for the radiated points. In general, because determining the heights of the radiated points has no control, aspecial care is required for this operation. 147
  • 153. 11. LEVELING OF PROFILES AND SURFACES The content of the tutorial: In order to draft topographic height plans, it isnecessary to determine the characteristic points of the terrain. In the case of works needed for traffic routes, the method of profiles is applied,and in the care of surfaces on which buildings, hydrotechnical works, etc. are designed,that is, surfaces having two sides comparable in value, one of the methods of surfaceleveling is used. In this tutorial we present, through practical examples, the method for performingsuch works.11.1 LONGITUDINAL AND TRANSVERSAL LEVELING THROUGH PROFILES It has the purpose to perform the longitudinal and transversal profile of a band ofterrain on which there will be designed and performed a traffic route. The surveillances that are performed, should be adequate from the point of viewof precision, because they will be used for the project and for its field application.11.1.1 THE CONDITIONS OF THE ROUTE a) The conditions of the longitudinal leveling route The longitudinal leveling is performed through a geometric leveling traverse,which follows the axis of the future traffic route, taking into account all the characteristicpoints if the terrain, in the same time respecting the conditions of a leveling traverse. 148
  • 154. b) The conditions of the transversal profile The necessary measurements for drafting transversal profiles are performedtogether with the longitudinal leveling, and in the case when this is not possible, thenthey are executed separately. The transversal profiles are normal to the longitudinal axis, placed at each 50 m or100 m, in general, depending on the configuration of the terrain (slope changes) and onthe nature of the designed work.11.1.2. RECOGNIZING THE TERRAIN AND PREPARING THE ROUTE Recognizing the terrain and preparing the route is performed, in general, as in thecase of geometric leveling traverses, to which the following issues are added: - The connection points on the axis of the route are placed at a distance of 100 m from each other and are numbered with the number of the hectometer that they represent (e.g. no.21, no.22, etc.). - Additionally, some intermediary points are chosen at the slope changes of the terrain, which are denoted with the number of the hectometer, to which the number of meters from the hectometer to the intermediary point is added (e.g. 21+2).11.1.3. PERFORMING FIELD MEASUREMENTS A longitudinal profile with four levels and three transversal profiles were chosenas example (figure 11.1). The field operations are carried on as in the case of traversing with radiations,considering the intermediary points and those from the transversal profile, as radiatedpoints. In table 11.1 is given the data from the terrain – the computed mean rod readings(the way to obtain them was studied in tutorial 10). 149
  • 155. The distances from the longitudinal profile are measured directly with the tape,and those from the transversal profiles are measured with the tape or the measuring reel.11.1.4. OFFICE OPERATIONS11.1.4.1. Computing heights It begins with computing the traverse, after which, depending on the permanentheights of the points of the traverse, the height of the aiming plan of the instrument iscomputed, for each station, depending on which the heights of the intermediary pointsand of those from the transversal profile are computed. In table 11.1 we have written the permanent values of the heights, thecomputational method not being important (similar to tutorial 10) here, but the way howthe data is used being of interest.11.1.4.2. Drafting profiles The profile is an imaginary vertical section through the topographic surface of theearth, performed with the use of the data obtained from the measurements carried on inthat area. a) Drafting the longitudinal profile It is carried on throughout the following steps: - Choose the horizontal distance scale (1:500, 1:1000, 1:2000); - The height scale is usually chosen 10-100 times larger than the distance scale, in order to better emphasize the relief; - On a sheet of paper of convenient size (it is recommended to use ruled paper), trace the axes of the profile: the 0X axis – horizontal, on which the lengths will be designated, and the 0Z axis – of heights; 150
  • 156. - Choose a convenient horizon, which would coincide with the 0X axis and would be situated with 1-5 m below the smallest height of the longitudinal profile; - Draw the ruling of the profile, with the header presented in figure 11.2 (depending on the destination of the profile, it can be modified); - Transform on the scale the distances between the pegs and between the pegs and the intermediary points; - Designate on the 0X axis their position, at the corresponding abscissa; - Write in row 1 the name of the pegs; - In the second row, write the height of each point that appears in the profile; - The partial distances are the distances computed between two neighboring points on the profile (row 3); - The cumulated distances are: equal to the total distance from the first point of the profile (no.21 in this case) to some point; - The slope of the terrain is computed between the points where the declivity changes visibly, being equal to the ratio between the altitude difference between these points and the distance between these points; - The heights of all points that appear in the profile are surveyed with respect to the standard horizon, on the chosen scale; - Uniting the points obtained previously, the longitudinal profile will result (exaggerated 20 times, in this case – by the ratio between the two chosen scales – of heights and of distances). b) Drafting the transversal profile It is performed in the same way as the longitudinal profile, keeping only the firstthree rows (figure 11.3). In this figure the first transversal profile was exemplified – the profile from thepeg no.21+83. The scale was chosen larger in the case of the longitudinal profile (it is within thevalues 1:10 – 1:500, usually 1:100, 1:200) and it is unique for distances and heights. 151
  • 157. TRANSVERSAL PROFILE NO.21-83 Scale 1:500 PEGS FIELD HEIGHTS DISTANCES Figure 11.3.11.2. SURFACE LEVELING We shall discuss in the sequel two similar methods of geometric leveling foraltimetric surveillance of surfaces, namely: - The method of small squares, and - The method of large squares.11.2.1. SURFACE LEVELING THROUGH THE METHOD OF SMALL SQUARES Applicability: for terrains with surfaces smaller than 4 ha, non-rough, with slopesless than 5%, having a good visibility. 152
  • 158. a) Field works Choose a base AB on the surface that is to be surveyed, and trace a network ofsquares with the side of 10-30 m, with respect to that base (figure 11.4). Figure 11.4. The size of the square side is chosen depending on the roughness degree of theterrain and the necessary precision. In the points A and B draw normals with the topographic square, on whichmeasure sides (15.59 or 48.12, etc.), obtaining in the end the points C and D. The side AB is divided into sections of equal length. In the end there are obtained squares that cover the entire surface of interest. The points A, B, C, and D are set out, and the points 1, 2, … are pegged out withstakes. Place the device in the middle of the surface, in station S1, from where areperformed all the rod readings in the corners of the squares (the aim should not exceed150 m), considering them as radiated points. The measured data is written in a leveling book, as the one from table 11.1. b) Computing the heights is performed using the method of the horizon of the device, taking as starting point either a benchmark of known height, or, if 153
  • 159. there is no such benchmark, one of the corners of the graticule (e.g. A), to whom an arbitrary height is assigned (e.g. ± 100,00)11.2.2. SURFACE LEVELING THROUGH THE METHOD OD LARGE SQUARES It is applied in the case of plain type relief – less rough, with middle and largesurface. a) Field works As in the case of the previous procedure, on the surface there is traced a networkof squares with sides of 50-200 m. In this case, the normals are traced with topographic devices. Stationing in S 1 aimeach corner of the square, repeating the operation in the stations S2, …, S10. Theprocessed data will be written in the leveling book. It should be specified that it is notnecessary to station in the squares from the center of the surface, their corners beingmeasured from the other points. In order to verify the correctness of the measurements, it is recommended that, asthe work is carried on, the schema of the squares to be drafted (figure 11.5), based on theread values (at the level hair) for each point, on the measuring staff. The control is performed confronting the values read from two distinct stations,for two points: 1731 + 2047 = 1917 + 1862 within the margin of 304 mm. 154
  • 160. Figure 11.5. The relation is obtained as result of the fact that the altitude difference betweentwo points is the same, regardless of the station from which we measure it. b) Computing the heights is performed in the following way: - Assign to the point 1a an arbitrary height (for example, Z = 100.00m ); - Consider the polygon of the vertices (1a-1d, 1d-5d, 5d-5a, 5a-1a) as a fundamental polygon, computing it and compensating it as such; - Consider the secondary traverses of geometric leveling supported at the ends: 1b-5b, 1c-5c. It can be seen that as result of the computations above, we have obtained theheight of each point. It should be mentioned that these heights are standard heights, and not absoluteheights with respect to the fundamental benchmark.11.3. USING THE DATA OBTAINED THROUGH SURFACE LEVELING The operations mentioned above can have to final purposes: a) Determining the relief on topographic plans, and b) Determining the embankment volume that appears when carrying out an investment. a) Representing the relief on topographic plans We obtain a height plan, if we draw a series of points whose height is determinedthrough one of the methods presented above, or using other methods (tutorial 10), on atopographic plan, and writing next to each point the value of its height. 155
  • 161. Though it is useful in various circumstances, the height plan presents theadvantage that it does not expressively represent the relief. But processing it, the contour plan is obtained. The operation of obtaining thecontours is called interpolation and can be performed in several ways. We have chosen asexample the interpolation of contours using the isograph. The isograph is simple to be built, made of a piece of tracing paper (of about 30 x10 cm), on which some equidistant lines are traced (at 3-5 mm), numbered increasingly,starting from the smallest height on the plan, up to the largest height (figure 11.6). Figure 11.6. The isograph The interpolation principle between two points of known heights (ZA = 356.00 andZB = 351.00) is the following (figure 11.7): Figure 11.7. Using the isograph - Place the isograph with the line that has the value 356 over the point A; - Rotate the isograph around the point A, adjusting the line 351 over B; - At the intersection of the alignment AB with the other lines of the isograph there will be obtained the points of heights 352, …, 355, which will be designated on the plan. This operation is repeated between all the pairs of neighboring points, obtainingpoints of fix heights. Uniting the points of same height will generate the contour that hasthe value of those points (figure 11.8). Figure 11.8. Interpolating contours Then, these points will be erased, and the value of the height will be written onthe contour, and in the end, the value of the height will be written on the main contour350, 355, 340, etc. 156
  • 162. b) Determining the embankment volume that appears when carrying out an investment (the embankment cartogram) In order to execute a construction, to get from the height of the natural terrain tothe height of the arranged platform or to the height of the bottom of the general digging,it is necessary to perform a volume of digging and filling up, synthetically calledembankment (or ground movement). The embankment volume must be known exactly, one method to obtain this databeing the cartogram. The way the cartogram is performed and processed will bepresented practically in the example solved within the homework of the tutorial.11.4. THE HOMEWORK OF THE TUTORIAL Problem #1: The data from table 11.1, obtained by performing the levelingtraverse with longitudinal and transversal profiles between the points 21 and 25 will bemodified as follows: Z 21 = 350.217 = n / 2 (mm) (rounding the height at mm). The rod readings +23, l+12, r+27, l+15, +80, r+14 will be modified adding n(mm) to the values presented in column 6. From the rod readings +83, l+22, r+10, r+25, +15 from column 6 the value n(mm) will be subtracted. The other data remains unchanged. Table 11.1 will be processed, the elevated and the plan schemas of the traverse(figure 11.1) and the longitudinal (figure 11.2) and transversal (figure 11.3) profiles willbe drafted on the specified scales. Problem #2: In table 11.2 is given the data collected in the field from the levelingof a surface through small squares (with sides of 20 m), in order to draw the cartogram ofthe embankments and the point 101 of known height (Z101 = 350.25 m – n(cm)). Figure 11.9. The schema of the measurement (problem #2) Requirements: 157
  • 163. a) Process the data from table 11.2, computing the heights of the square corners (see table 11.3). b) Draw the cartogram of the embankments on the scale 1:500, computing their volume, in the case when the terrain is to be arranged at the height of 349.00 + n(cm) (see cartogram 1). Problem #3: In table 11.4 is given the data collected from the field from theleveling of a surface through large squares (columns 1-6). a) Process the table, computing the heights of the points that represent the corners of the graticule (see table 11.4, columns 7 and 8). b) Interpolate the contours on the height plan previously computed (see figure 11.11) Figure 11.10. The schema of the measurement (problem #3) Figure 11.11. Interpolating the contours Explanations concerning solving the problems: Problem #1: Will be solved according to the explanations presented atlongitudinal and transversal leveling through profiles. Problem #2: - Zi (column 7) – the height of the aiming plan is obtained adding the rod reading CM101=1621 mm from the known benchmark (101) to its height Z 101 = 350.25 m. - Z1 (column 8) – the height of the natural terrain in point 1 is obtained subtracting the rod reading CM1 = 1617 mm from this point from Z i. The heights of the other points will be obtained similarly. - The altitude differences (∆ZAm-1) between the height of the natural terrain and the height of the arranged platform (ZAm=349000 m) are obtained subtracting ZAm from the height of each point (columns 10 and 11). - On cartogram 1, write the computed height of the natural terrain in each graticule corner, and above it, the altitude differences from columns 10 and 11: 158
  • 164. Figure (altitude difference, point number, height of natural terrain) - The mean altitude difference (which is written in the upper cell from the center of each square) is computed as the arithmetic mean of the altitude differences of the four corners of the squares: −1.227 −1.140 −1.139 −1.188 = −1.174m 4 We mention that on the cartogram, the altitude differences are written withopposite sign than in the columns 10 and 11, because in this case they refer themovement of the ground. Figure The other cell (-11,740 m3) represents the digging volume (or filling up volume,in other cases) needed in order to get to the height ZAm = 349.000 m on the surface of thatsquare. Thus, it will be obtained: -1.174m · 10,000m2 = -11,740m3.where 10,000m2 represents the surface of the square (having sides of 100 m). These values are summed up on columns, and are written in the row that ends thecartogram. By summing up the values on this row, the total ground movement of thestudied area is obtained (in this case –264,620 m3, therefore, this quantity will be dug). Problem #3: In this case, the height of the point at (Z 1a = 348,210 m) is given by the absolutesystem. Since the computation of the leveling traverses was discussed in tutorial 10, inthis case we have computed each square separately (similar to the example presented incolumn 9). In figure 11.11 there were written the heights of each point, and then thecontours were interpolated (on the vertical: 1a with 2a, 2a with 3a, and so on). 159
  • 165. Figure (altitude difference, point number, height of natural terrain) - The mean altitude difference (which is written in the upper cell from the center of each square) is computed as the arithmetic mean of the altitude differences of the four corners of the squares: −1.227 −1.140 −1.139 −1.188 = −1.174m 4 We mention that on the cartogram, the altitude differences are written withopposite sign than in the columns 10 and 11, because in this case they refer themovement of the ground. Figure The other cell (-11,740 m3) represents the digging volume (or filling up volume,in other cases) needed in order to get to the height ZAm = 349.000 m on the surface of thatsquare. Thus, it will be obtained: -1.174m · 10,000m2 = -11,740m3.where 10,000m2 represents the surface of the square (having sides of 100 m). These values are summed up on columns, and are written in the row that ends thecartogram. By summing up the values on this row, the total ground movement of thestudied area is obtained (in this case –264,620 m3, therefore, this quantity will be dug). Problem #3: In this case, the height of the point at (Z 1a = 348,210 m) is given by the absolutesystem. Since the computation of the leveling traverses was discussed in tutorial 10, inthis case we have computed each square separately (similar to the example presented incolumn 9). In figure 11.11 there were written the heights of each point, and then thecontours were interpolated (on the vertical: 1a with 2a, 2a with 3a, and so on). 159
  • 166. Figure (altitude difference, point number, height of natural terrain) - The mean altitude difference (which is written in the upper cell from the center of each square) is computed as the arithmetic mean of the altitude differences of the four corners of the squares: −1.227 −1.140 −1.139 −1.188 = −1.174m 4 We mention that on the cartogram, the altitude differences are written withopposite sign than in the columns 10 and 11, because in this case they refer themovement of the ground. Figure The other cell (-11,740 m3) represents the digging volume (or filling up volume,in other cases) needed in order to get to the height ZAm = 349.000 m on the surface of thatsquare. Thus, it will be obtained: -1.174m · 10,000m2 = -11,740m3.where 10,000m2 represents the surface of the square (having sides of 100 m). These values are summed up on columns, and are written in the row that ends thecartogram. By summing up the values on this row, the total ground movement of thestudied area is obtained (in this case –264,620 m3, therefore, this quantity will be dug). Problem #3: In this case, the height of the point at (Z 1a = 348,210 m) is given by the absolutesystem. Since the computation of the leveling traverses was discussed in tutorial 10, inthis case we have computed each square separately (similar to the example presented incolumn 9). In figure 11.11 there were written the heights of each point, and then thecontours were interpolated (on the vertical: 1a with 2a, 2a with 3a, and so on). 159