GHEORGHE M. T. RADULESCU

GENERAL TOPOGRAPHY

WORKBOOK OF PROBLEMS

FOREWORD

This workbook of problems is punctually presenting all the aspects that a user of

topographic methods and instruments could meet during his current practical

applications.

The first edition of the book, published in 1985, was considered extremely useful by

those who have studied it, the current edition being completed with other practical

applications, problems given to be solved.

I think that this workbook can represent the basis for testing the knowledge of those who

have studied this subject during some specialization programs (such as undergraduate

with attendance, without attendance or distance learning, post high school, or

postgraduate), this being, as far as I know, the first book that was published in this form

at national level.

The Author

II

TABLE OF CONTENTS

FOREWORD II

TABLE OF CONTENTS III

GENERAL TOPOGRAPHY 1

A. NOTIONS OF GENERAL TRIGONOMETRY 1B. THE TOPOGRAPHIC ELEMENTS OF THE TERRAIN 7

a. LINEAR ELEMENTS 7b. ANGULAR ELEMENTS 9

C. THE RELATION BETWEEN COORDINATES AND ORIENTATIONS 11a. COORDINATES → ORIENTATIONS 11b. ORIENTATIONS → COORDINATES 14

D. PROBLEMS SOLVED ON PLANS AND MAPS 15a. PLANIMETRY PROBLEMS 15

E. THE STUDY OF TOPOGRAPHIC INSTRUMENTS 31a. THE THEODOLITE 31b. THE TOPOGRAPHIC LEVEL 36

F. PLANIMETRY PROBLEMS 45a. THE DIRECT MEASURING OF DISTANCES 45b. THE INDIRECT MEASURING OF DISTANCES 50c. MEASURING ANGLES 58d. SURVEY OF THE DETAILS 65e. REPEATING DETAILS 77

G. LEVELING PROBLEMS 80a. GEOMETRIC LEVELING 80

III

GENERAL TOPOGRAPHY

A. NOTIONS OF GENERAL TRIGONOMETRY

1. Transform the following angular values into centesimal units:

a) 21º41’ 34”; b) 128º37′42″ + n″; c) 216º42′12” + nº; d) 294º56’43” – n’.

n represents the half-group order number

the solutions are presented for n=0.

Solution:

41 º 34 º21º41’34” = 21º + -------- + ---------- = 21º.692777 (1.1)

60 60 · 60

From the transformation rule it results that

10 g

α ‘g = ---------- αº. (1’.1) (2.1) 9º

10 gThus, α ‘g = ------- · 21º.692777 = 24g.103086 = 24 g · 10 c · 30 cc.86 9º

In what follows there is presented the solution of this exercise on a calculator (CASIO fx

– 120 type). There are indicated the keys ( ) of the calculator that intervene in the

solution.

Thus, the solution is 24 g10 c30 cc.86.

2. Transform the following angular values into sexagesimal units:

a) 16g 43c 66cc; b) 142g 52c 46cc + ncc; c) 221g 54c 68cc + ng; d) 316g 52c 16cc – nc

Solution:

16 g 43 c 66 cc = 16. g 4366.

According to the relation (A.1) αº = 0,9 α’ g

αº α g a(RAD)------- = ---------- = ----------------180º 200 g π

21°.692777 X 1 0 9 = 24

2 1 0,,, 4 1 0,,, 3 4 0,,, = display 21°.692777

display .103086g

1

We have αº = 0.9 g x 16.4366 = 14º.79294

Transform 0º.79294 in minutes: x’ = 60’ x 0.79294 = 47’.5764

Transform 0’.57640 in seconds: x”1 = 60” x 0.57640 = 34”.58

Thus, the solution is 14º.47’34”.58.

Solving the exercise on a calculator:

3. Find the trigonometric functions of the angles from the first quadrant corresponding to

the following angular values:

a) 94º16’21” + nº; b) 198º28’16” + nº; c) 298º18’43” + nº; d) 116g 62c 18cc + 2ng;

e) 222g 83c 24cc + ng; f) 384g 61c 22cc – ng.

Solution:

Table 1.3

a.

sin 94º16’21” =

+cos4º16’21”

cos = - sin

tg = - ctg

ctg = - tg

b.

sin 198º28’16” =

-sin 18º28’16”

cos = - cos

tg = + tg

ctg = + ctg

c.

sin 298º18’43” =

-cos 28º18’43”

cos = + sin

tg = - ctg

ctg = - tg

d.

sin 116g 62c18cc=

+cos 16g 62c18cc

cos = - sin

tg = - ctg

ctg = - tg

e.

sin 222g 83c24cc=

-sin 22g 83c24cc

cos = - cos

tg = + tg

ctg = + ctg

f.

sin 384g 61c228cc=

-cos 84g 61c22cc

cos = + sin

tg = - ctg

ctg = - tg

2

4. Compute the natural values corresponding to the trigonometric functions sin α, cos α,

tg α, ctg α for the following angular values:

a) 28º24’18” + nº; b) 96º16’26” + n’; c) 194º16’43” – n”; d) 284º51’18” –n’;

e) 46g51c83cc – n cc; f) 121g62c47cc + n g; g) 214g51c83 cc – ncc; h) 373g43c16cc – ng.

a) sin 28º24’18” = + 0.47570097

cos = + 0.87960706

tg = + 0.54081077

ctg = + 1.84907560

f) sin 121g62c47cc = + 0.94286134

cos = - 0.33318539

tg = - 2.82984000

ctg = - 0.35337687

The solution of the exercise on a calculator:

a)

In order to obtain the natural value for ctg, press the key after the value for tg

was displayed => ctg.2824’18” = display

f)

1/x1.84907560

3

=> ctg 121g 62c47cc = display

5. Which are the arguments iy of the specified trigonometric functions corresponding tp

the following natural values:

a. sin12 = 0.432116 + n (QUADRANT I) b. sin13 = 0.161722 – n (QUADRANT II)

c. sin14 = - 0.832217 + n (QUADRANT III) d. sin15 = - 0.732218 – n (QUADRANT IV)

e. cos22 = 0.221742 + n (QUADRANT I) f. cos23 = - 0.175263 + n (QUADRANT II)

g. cos24 = - 0.661722 – n (QUADRANT III)h. cos25 = 0.512215 + n (QUADRANT IV)

i. tg32 = 0.611542 + n (QUADRANT I) j. tg33 = - 0.935124 – n (QUADRANT II)

k. tg34 = 0.667315 – n (QUADRANT III) l. tg35 = - 0.721752 + n (QUADRANT IV)

m. ctg42 = 0.172243 + n (QUADRANT I) n. ctg43 = - 0.170450 – n (QUADRANT II)

o. ctg44 = 0.552117 – n (QUADRANT III) p. ctg45 = - 0.291060 + n (QUADRANT IV)

Solution:

Remark: iy will be expressed in centesimal units, and n will be applied to the last two

digits of the natural value.

a. arcsin 0.432116 = 28g 44c65cc.8 = 12

b. sin 13 = cos (13 -100g) = cos = 0.161722, = arcos 0.161722 =

89g65c90cc.4 => 13 = + 100g = 189g65c90cc.4

c. sin 14 = - sin (14 -200g) = - sin = - 0.832217, = arcsin 0.832217 =

62g58c57cc.3 => 14 = + 200g = 262g58c57cc.3.

d. sin 15 = - cos (15 - 300g) = - cos = -0.732218, = arccos 0.732218 =

47g69c70cc.4 => = + 300g = 347g69c70cc.4

The other exercises can be solved in a similar way.

The solution of the exercises on a calculator:

It is taken into consideration to compute the value corresponding to the first quadrant,

Ex. a. = arcos 0.161722

1/x -0.353337687

4

6. Represent the trigonometric circle, emphasizing the trigonometric lines in the four

quadrants. Specify the formulas for reducing to the first quadrant.

Figure 1.6. The trigonometric circle

Quadrant

Angle

Function

I

I

II

II

III

III

IV

IV

+ 100g + 200g + 300g

sini + sin + cos - sin - cos

cosi + cos - sin - cos + sin

tgi + tg - ctg + tg - ctg

ctgi + ctg - tg + ctg - tg

5

7. Present the variation graphs on the interval (0, 2) and the associated table

corresponding to the trigonometric functions sin, cos, tg and ctg.R

AD

6

4

3

2

2 3

3 4

5 6

7 6

5 4

4 3

3 2

5 3

7 4

11 3

2

Mon

oton

y (

inte

rval

)

Fun

ctio

n

0 30

45

60

90

120

135

150

180

210

225

240

270

300

315

330

360

sin 0 12

2 2

3 2

1 3 2

2 2

12

0 -12

- 2 2

- 3 2

-1 - 3 2

- 3 2

-1 2

0 2

cos 1 3 2

2 2

12

0 -12

- 2 2

- 3 2

-1 - 3 2

- 2 2

-12

0 12

2 2

3 2

1 2

tg 0 3 3

1 3 -3 -1 - 3 3

0 3 3

1 3 -3 -1 - 3 2

0

ctg + 3 1 3 3

0 - 3 3

-1 -3 + -3 1 - 3 3

0 - 3 3

-1 -3

6

8. Represent the topographic circle, emphasizing the trigonometric lines in the four

quadrants. Specify the formulas for reducing to the first quadrant.

B. THE TOPOGRAPHIC ELEMENTS OF THE TERRAIN

a. LINEAR ELEMENTS

9. Given LAB = 175.43 m+n (m), AB = 8g51c + nc, compute DAB.

DAB = LAB cos AB (1.9)

= 175.3 cos 8g51c

= 173.86 m.

According to figure 1.9 we shall determine:

7

Figure 1.19

It can be seen that in the triangle ABB’ the following relations can be stated:

ZAB DAB ZAB DAB

sin AB = ----------, cos AB = ------- , tgAB = ---------, ctgAB = -------- (2.9) LAB LAB DAB ZAB

LAB = D²AB + Z²AB and ZAB = ZB - ZA. The needed elements can be determined using

these relations, depending on the known (measured) elements.

10. Compute DAB ,ZAB, ZB, given the following:

LAB = 217,47 m + n (cm), AB = 12g17c + nc, ZA = 348.21 m.

Solution:

DAB = LAB cos AB = 217.47 m · cos 12g17c = 213.51 m; (1.10)

ZAB = LAB sin AB = 217.47 m · sin 12g17c = 41.32 m; (2.10)

ZB = ZA + ZAB = 348.21 m + 41.32 m = 389.53 m. (3.10)

11. Given: ZA = 361.14 m + n (cm), ZB = 363.22, AB = 5g42c + ng , compute: LAB, DAB.

Solution:

ZAB = ZB - ZA = 363.22 – 361.14 = 2.08 m;

DAB = ZABctgAB = 2.08 m · ctg · 5g42c = 24.37 m;

Vertical datumZA

DAB

LAB

ZAB

ZB

A

B

B’AB

8

LAB = DAB / cosAB = 24,37 / cos 5g42c = 24.46 m.

b. ANGULAR ELEMENTS

12. Which is the horizontal angle corresponding to the following gradations on the

horizontal circle of the theodolite:

CA = 117g51c + ng; CB = 247g58c.

= CB - CA = 247g58c - 117g51c = 130g07c (B.2)

13. Compute the value of the slope angle , when the values registered on the vertical

graduated circle on direction AB are:

a) VI = 83g51c + nc; b) VI = 112g63c

- nc; VII = 307g43c - nc; c) VII = 283g82c

+ nc;

d) VI = 88g62c + nc; VII = 311g39c; e) VI = 111g21c

- nc; VII = 288g79c.

9

14. LAB=184.52 m + n(m), I =1.47 m, s = 2.03, VI = 88g54c + nc; VII = 311g46c

are given.

Determine the vertical angle (’) corresponding to the aim B and the slope angle of the

terrain ().

h + i = S + ZAB (1.14)

ZAB

sin = ------- (2.14) LAB

hsin’=-------- (3.14) LAB

LAB sin’ + i - S

10

Thus: LAB sin + i = LAB sin + S => sin =------------------------- LAB

The angle ’ will be determined according to the principle that was used in the previous

problem:

100g00c - 88g54c + 311g46c - 300g00c

’ = -------------------------------------------------- = 11g46c

2

184.52m · sin11g46c + 1.46m – 2.03 m sin = -------------------------------------------------- = 0.17600772 184,52 m

= arcsin 0.17600772 = 11g26c36cc.7

C. THE RELATION BETWEEN COORDINATES AND

ORIENTATIONS

a. COORDINATES → ORIENTATIONS

15. Determine the values of the orientations ABI, AB

II, ABIII, AB

IV, corresponding to the

directions formed by the point A of known coordinates [XA = 116.43 m, YA = 124.55 m

+n(m)] with the points:

a. BI [XBI = 243.15 m + n(m), YB

I = 185.43 m];

b. BII [XBII = 91.17 m - n(m), YB

II = 175.43 m];

c. BIII [XBIII = 61.24 m , YB

III = 100.00 m – n(m)];

d. BIV [XBIV = 223.51 m , YB

IV = 85.22 m];

a. We start from the relation:

YAB

tgAB = --------- (1.15)

XAB

11

Depending on the sign of the components YAB, and XAB, respectively, we can

determine the quadrant in which the orientation AB is found.

Then, the angle corresponding to the first quadrant is determined.

Adding 100g, 200g or 300g depending on the quadrant, the value of the orientation AB is

determined:

a. YABI = YB

I -YA = 185.43 m – 124.55 m = 60.88 m;

XABI = XB

I -XA = 243.15 m – 116.43 m = 126.73 m;

YABI + 60.88

tgABI .= --------- = ----------- = + 0.4802929

XABI + 126.72

ABI .= arctg 0.48042929 = 28g51c22cc.1

b. YABII = YB

II -YA = 175.43 m – 124.55 m = 50.88 m;

XABII = XB

II -XA = 91.17 m – 116.43 m = - 25.26 m;

YABII +50.88

tgABII .= --------- = ----------- = - 2.01425178

XABII -25.26

tgABII .= - ctg (AB

II .- 100g ) = - ctg = - 2.01425178;

1tg = ---------------- = 0.49646226 => = arctg 0.49646226 2.01425178

Thus = 29g33c62cc.9 => ABII .= + 100g= 129g33c62cc.9

c. YABIII = YB

III -YA = 100.00 m – 124.55 m = - 24.55 m;

12

XABIII = XB

III -XA = 61.24 m – 116.43 m = - 55.19 m;

YABIII - 24.55 m

tgABIII .= --------- = ------------ = 0.44482696

XABIII -55.19 m

= arctg 0.44482696 = 26g64c53cc.2 => ABIII = + 200g = 226g64c53cc.2.

d. YABIV = YB

IV -YA = 85.22 m – 124.55 m = - 39.33 m;

XABIV = XB

IV -XA = 223.51m – 116.43 m = 107.08 m;

YABIV - 39.33 m

tgABIV .= --------- = ------------ = - 0.36729548

XABIV 107.08 m

tgABIV .= - ctg(AB

IV – 300g ) = - ctg = - 0.36739548

1tg = ----------------- = 2.72260361 => = arctg 2.72260361,

0,36729548

Thus = 77g59c10cc.5 => ABIV .= + 300g = 377g59c10cc.5.

Establishing the quadrant in which the orientation is found was performed based on the

data presented in table 1.15.

Table 1.15

The components of the natural value

The orientation quadrant iJ

I II III IV

YAB + + - -

XAB + - - +

The distances DABi are computed using the relation:

DABi = X²ABi + Y²ABi (2.15).

b. ORIENTATIONS → COORDINATES

16. The coordinates of the point A are [XA = 212.52 m – n(m), YA = 257.43 m], and the

distances between this point and the points CI, CII, CIII, and CIV are a. DACI = 112.51 m; b.

DACII = 81.32 m + n(m); c. DACIII = 125.45 m; and d. DACIV = 61.52 m – n(m);

respectively. The orientations are also known: a. ACI = 61g51c + ng; b. AC

II = 112g43c +

nc; c. ACIII = 217g51c; d. AC

IV = 343g61c - ng. Determine the coordinates of the points Ci.

13

a. In order to determine the coordinates (XCi, YCi), the following relations will be applied:

XACi = DACi · cosACi ; (1.16)

YACi = DACi · sinACi ;

XCi = XA + XACi ; (2.16)

YCi = YA + YACi ;

Thus XACI = 112.51 m · cos 61g51c = 63.95 m;

YACI = 112.51 m · sin 61g51c = 92.57 m;

XCI = 212.52 m + 63.95 m = 276.47 m;

YCI = 257.43 m + 92.57 m = 350.00 m;

b. XACII = 81.32 m · cos 112g43c = -15.78 m;

YACII = 81.32 m · sin 112g43c = 79.77 m;

XCII = 212.52 m - 15.78 m = 196.74 m;

YCII = 257.43 m + 79.77 m = 337.20 m;

14

c. XACIII = 125.45 m · cos 217g51c = -120.73 m;

YACIII = 125.45 m · sin 217g51c = - 34.07 m;

XCIII = 212.52 m – 120.73 m = 91.79 m;

YCIII = 257.43 m - 3407 m = 223.36 m;

d. XACIV = 61.52 m · cos 343g61c = 38.92 m;

YACIV = 61.52 m · sin 343g61c = - 34.07 m;

XCIV = 212.52 m + 38.92 m = 251.44 m;

YCIV = 257.43 m - 47.64 m = 209.79 m.

D. PROBLEMS SOLVED ON PLANS AND MAPS

a. PLANIMETRY PROBLEMS

Figure 1.17 represents a topographic plan, on the 1:1000 scale, on which, besides the

contours, there also appear the points A, B, C, and D, with respect to which numerous

problems with planimetric or altimetric (leveling) character will be solved.

17. The distance DAB will be determined using the graphical method.

Solution:

DAB = dAB · N (1.17) where:

dAB is the distance measured on the plan;

N: the scale denominator of the plan.

DAB = 97.4 mm x 1000 = 97400 = 97.40 m.

Remark: the precision of measuring a distance on the plan will be of 0.1 0.2mm.

X=200 m

X=150 m

X=100 m

Y=200 m

Y=25 0 m

Y=30 0 m

345 34

0

350

D

B

C

A

15

Figure1.17. Topographic plan

18. Determine the coordinates of the points A and B in the X0Y rectangular system.

From the point whose coordinates we want to determine, we draw the

perpendiculars towards the closest graticule corner (the point M in this case);

We measure the graphical values XMA , and YMA;

We compute the values corresponding to the situation in the terrain:

XMA = XMA · N (1.18)

YMA = YMA · N;

Determine the absolute coordinates of the point A, which the method of contours

was used for representing the relief.

XA = XM + XMA (2.18)

YA = YM + YMA

16

Thus: we measure XMA = 8.9 mm; YMA = 7.8 mm;

We compute XMA = 8.9 x 1000 = 8900 mm = 8.9 m;

YMA = 7.8 x 1000 = 7800 mm = 7.8 m;

The absolute coordinates of the point A will be:

XA = 100 m + 8.9 m = 108.9 m;

YA = 200 m + 7.8 m = 207.8 m.

19. On the part of a topographic map, which is represented in figure 1.19, determine the

geographic and rectangular coordinates of the point F.

a. Determining the geographic coordinates.

Latitude F = 4620’30” + ” = 4620’52”.

Longitude F = 2359’ + ” = 2359’44”.

Finding the values ”, ” – by means of linear interpolation, with respect to 30” (),

and 60” (), respectively, the linear correspondents of the arcs of 30” on the meridian and

of 60” on the parallel, respectively.

17

b. Determining the rectangular coordinates by repeating the point F to the nearest

graticule corner [N in this case (XN = 81,000 m; YN = 88,000m)].

In a similar manner as in the case presented in problem 18 we can obtain the values:

XF = XN + XNF = XN + XNF + N = 81,000 + 16.7 mm x 25,000 = 81,417.5 m;

YF = YN + YNF = YN + YNF + N = 88,000 + 12.8 mm x 25,000 = 88,320 m.

20. Compute the distance DAB using the analytical method.

Solution:

According to the relation (2.15): DAB = X²AB + Y²AB

18

XAB =XB - XA = 180.8 – 108.9 = 71.9 m;

YAB =YB - YA = 273.6 – 207.8 = 65.8 m;

DAB = 97.46 m.

It can be seen that the following condition is fulfilled DABGRAPHIC - DAB

ANALYTICAL ≤ T

(1.20), where, in this case, T = 0.2 mm x N = 0.2 m. (2.20)

21. Determine the orientation of the direction AB = AB using the graphical method.

Solution:

AB is measured using the centesimal protractor, obtaining:

AB = 47g20c.

22. Determine the value of the orientation AB using the analytical method.

Solution:

YAB 65.8tgAB = -------- = ---------- = 0.91515994 XAB 71.9

AB = arctg 0.91515994 = 47g18c17cc.

The solutions of the problems 21 and 22 fulfill the following condition:

AB GRAFIC - AB

ANALITIC = ≤ T (1.12), where T = 10c.

23. Determine the size of the surface ABCD using the analytical computation method.

Solution:

The coordinates of the points A, B, C, and D are known.

XA = 108.9 m

YA = 207.8 m

XB = 180.8 m

YB = 273.6 m

XC = 130.2 m

YC = 292.8 m

XD = 196.0 m

YD = 213.1 m

Apply the following relations:

D

2S = Xi (Yi +1 - Yi – 1) (1.23) i = A

D

2S = Yi (Xi -1 - Xi +1) (2.23)

19

i = A

Applying (D.6) we have to compute:

XA(YD – YC) + XD(YB – YA) + XB(YC – YD) + XC(YA - YB) S=------------------------------------------------------------------------ (1.23)’

2

We shall obtain S = 5030.035 m²;

The verification is performed applying (2.23) expanded:

YA(XC – XD) + YD(XA – XB) + YB(XD – XC) + YC(XB – XA)S=------------------------------------------------------------------------= 5030.035 m² 2

24. Determine SABCD using a trigonometric method.

Solution:

SABCD = SADB + SABC = SI + SII (1.24)

AD · AB sin DAB AB · AC sin BACSABCD = ------------------- + ----------------- (2.24) 2 2

The sides and angles involved in the relation (2.24) can be determined based on the

coordinates of the points A, B, C, and D.

AD = X²AD + Y²AD = 87.26 m;

AB = X²AB + Y²AB = 97.46 m;

AC = X²AC + Y²AC = 87.63 m;

DAB = AB - AD = 47g18c17cc - 3g86c90cc = 43g31c27cc;

BAC = AC - AB = 84g36c89cc - 47g18c17cc = 37g18c72cc;

20

87.26m · 97.46m · sin43g31c27cc 97.46m · 87.63m · sin37g18c72cc SABCD = ---------------------------------------- + ----------------------------------------

2 2

SABCD = 2674.91 m² + 2354.93 m² = 5029.84 m².

25. Determine the surface SABCD using the following geometric methods:

a. The numerical method;

b. The graphical method.

Solution:

a. Based on the coordinates, compute the sides of the two triangles that compose the

surface ABCD.

Thus:

DB = X²DB + Y²DB = 62.38 m;

BC = X²BC + Y²BC = 54.12 m;

Apply the relation S = p(p-a)(p-b)(p-c) (1.25)

a + b + cWhere p = --------------

2

Thus SABCD =123.55(123.55-87.26)(123.55-62.38)(123.55-97.46)

+119.61(119.61-97.46)(119.61-54.12)(119.61-87.63)

SABCD =2674.98 + 2354.99 = 5029.97 m²

b. Divide the polygon ABCD into two triangles: ADB and ABC, whose dimensions are

graphically determined:

AB · HADB AB · HABC

SABCD = SADB + SABC = ----------------- + ------------------ (2.25)2 2

97.40 · 54.90 97.40 · 48.40

21

SABCD = ------------------ + ------------------- = 2673.63 + 2357.08 =>2 2

SABCD = 5030.71 m.

26. Applying the graphical method of equidistant parallels, determine the surface ABCD.

Solution:

- On a transparent material (tracing paper) parallel and equidistant lines (a = 1cm)

were drawn;

- Overlap the tracing paper on figure ABCD, thus obtaining a series of geometrical

shapes (trapezes) whose are is determined using the well-known relations;

- In the end:

n

SABCD = A x Bi + Si (1.26) i = 1

Where: A = a · n (2.26)

n n

Bi = bi · N (3.26) i = 1 i = 1

22

The last relation (3.26) determines the areas from the ends, which will be added to the

obtained value.

For the case being presented:

SABCD = 1.0 cm x 1000 x [ (b1+b2+ ….bn )N] + 8 m² + 43 m² = 5030.43 m² a Bi S1 S2

27. Determine the area of the surface SABCD using the method of the network of equal

squares.

Solution:

SABCD = A²(n1 + n2) (1.27)

A = a · N = 1 cm · 1000 = 10 m;

n1 = 30 (the number of entire squares);

n2 = 20,3 (the number of approximate squares)

Thus, SABCD = 100 m² x 50.3 = 5030 m².

Remark: Problems 23-27 have the purpose to use practical example in order to concretize

some methods applied to determine the surfaces of plans and maps.

In practice, of course, the adequate method will be used for each case, depending on the

known elements, on the extent of the surface, on the plan scale, and on the surface

contour (sinuous, polygonal, etc.).

23

In the figures 1.28-1.33 the enumerated relief forms are geometrically presented. For

each case, trace the corresponding contours, at the specified interval (E).

28. The relief forms from figure 1.28, for E = 10m;

29. The relief forms from figure 1.29, for E = 10m;

30. The relief forms from figure 1.30, for E = 5m;

31. The relief forms from figure 1.31, for E = 5m;

32. The relief forms from figure 1.32, for E = 2m;

33. The relief forms from figure 1.33, for E = 10m;

The solutions are presented in the figures 2.28 – 2.33.

24

25

34. Determine the heights of the points A, B, C, and D on the topographic plan from

figure 1.17.

Solution:

Figure 1.34

The height of the point A is obviously equal to the height of the contour on which the

point is situated (ZA = 347 m).

The height of the point B can be obtained through linear interpolation

ZB = ZM + h (m) = 340 m + h (m); (1.34)

ZB = ZN – h1 (m) = 341 m – h1 (m);

d’h (m) = ----- (m) d

d – d’h1 (m) = ------- (m)

d

12We shall obtain: ZB = 340 + ---- (m) = 340.67 m. 18

26

35. Which is the value of the slope of the terrain between the points A and B?

Solution:

ZAB ZB – ZA

PAB = tg = ------- = ----------- (1.35) DAB DAB

340.67 – 347.00Thus PAB = -------------------- = - 0.0650

97.40

or PAB % = 100 PAB = -6.50% (2.35)

36. Which is the average of the slope in the area of the points A, B, C, and D?

Solution:

In the area of point D, we shall consider the contours situated on both sides of the point

(3-6 contours). EF represents the line from area D with the highest slope.

ZEF ZF – ZE

PD = ------- = ----------- (1.36) DEF dEF · N

350 – 345 5Thus PD = --------------- = -------- = 0.125 (or 12.5%) 0.04 x 1000 40 m

27

37. Determine the maximal and minimal slopes on the AB alignment.

Solution:

E EPmax = ---------- = ------------ (1.37)

Dmin dmin x N

E EPmin = ---------- = ------------ (2.37)

Dmax dmax x N

In the presented case:

1m 1mPmin = P56 = ---------------- = -------- = 0.0476 or Pmin % = 4.76%

0.021 x 1000 21 m

1m 1mPmax = P23 = ---------------- = -------- = 0.0909 or Pmax % = 9.09%

0.011 x 1000 11m

38. Trace a line of specified slope P0% = 5% + 0,n% between the points A and B.

28

Solution:

100 x 100 cmP0% = ----------------

d0 x 1000

10 cmP0% = --------

d0

10 cm d0 = -------- = 2 cm. 5%

With the computed value (d0) in the compass, starting from the point A, step by step,

trace one or more variants of the line P0%.

39. Trace the longitudinal profile of the AB alignment on the scale of distances 1:500 and

of heights 1:100.

Remark: according to the topographic plan from figure 1.17 and the notations from figure

1.37

29

40. Trace the topographic transversal profile, corresponding to the CD direction, on the

scale of distances and heights 1:500.

Note: the transversal profile was performed for the distance of 25m, towards left and

right respectively, from the AB alignment, on the CD direction.

30

E. THE STUDY OF TOPOGRAPHIC INSTRUMENTS

a. THE THEODOLITE

40. The design of a theodolite-tacheometer Theo 080 Carl-Zeiss Jena – ex. RDG is

presented in figure 1.41.

Indicate the name of each axis and main and secondary parts and specify the role of each

part.

Solution:

The constructive axes of the theodolite are:

VV: main rotation axis (vertical);

HH: secondary rotation axis (horizontal);

0: (reticule - lens) is the aiming axis of the telescope;

Cv: is the point of intersection of the three axes, named aiming center.

31

The main parts of the theodolite are:

1. The telescope;

2. The horizontal graduated circle (the bearing circle);

3. The vertical graduated circle (the clinometer);

4. The alidade;

5. The base.

The secondary parts (the accessories) are the following:

1’: device for approximate aiming;

1”: screw for clarifying (focusing) the aimed image;

1”’: screw for clarifying the reticule image;

5’: foot screws (three);

5”: screw for locking the device (the bearing circle) to the base;

6: level air bubble;

7: device (microscope) for centralized reading of the bearing circle and clinometer

gradations;

8: locking screw of the clinometer circle (and of the telescope);

8’: device for refined motion around the HH axis;

9: locking screw of the bearing circle;

9’: device for refined motion around the VV axis;

10: flap for locking the bearing circle on the alidade.

42. Draw the topographic telescope with internal focusing, specifying the name of the

composing parts.

32

Solution: (figure 1.42)

1: the lens tube;

2: the eyepiece tube;

3: the lens;

4: the eyepiece;

5: the reticule;

6: focusing lens;

7: focusing button;

8: the rack device;

9: reticule adjusting screws;

10: image forming when b is missing;

O1: optical center of the lens;

O2: optical center of the eyepiece;

: the center of the reticule;

XX: the geometric axis of the telescope;

0102: the optical axis of the telescope;

01: the aiming axis.

43. Determine the readings on the bearing circle and the clinometer, based on the image

from the field of the microscope with lines presented in figure 1.43.

33

Solution:

V (reading on the clinometer): 91g74c;

Hz (reading on the bearing circle): 114g94c

44. Present the schema of the microscope with lines (the image field) for the following

readings:

V: 394g28c - ngnc;

Hz: 217g51c + nc.

Solution:

The field of the microscope corresponding to the readings will be drawn similarly to the

image presented in figure 1.43.

Remark: the numbers written upside-down will be ignored.

45. Determine the readings on the bearing circle and the clinometer based on the image of

the scale microscope presented in figure 1.45.

34

Solution:

V: 84g86c90cc;

Hz: 218g13c70cc.

Remark: The field of the scale microscope allows estimating tens of seconds.

46. Present the schema of the image of the scale microscope for the following readings:

V: 372g51c20cc + ngnc;

Hz: 246g77c40cc + nc;

Solution:

Present the image corresponding to the specified readings similarly to the schema of the

scale microscope from figure 1.45.

47. Using schemas and explanations specify the steps for performing a measuring with

the theodolite. Emphasize the role and the importance of each step.

The solution of the problem will be drafted using the bibliography specified at the end of

the workbook.

48. Which are the verifications and adjustments of the theodolite, which are performed

before usage, and what do they consist of?

Remark: the same specification as in the previous problem.

b. THE TOPOGRAPHIC LEVEL

49. Specify the name, role and importance of each part that composes the rigid level NI

030 Carl Zeiss Jena (figure 1.49).

35

Solution:

1: the telescope of the level;

1’: the lens of the telescope;

1”: the eyepiece of the telescope;

1’”: capsulated reticule;

1iv: focusing screw;

2: air bubble level;

2’: refined foot screw;

2”: spherical level;

3: horizontal graduated circle (bearing circle);

3’: locking clamp for the motion around the vertical axis (VV);

3”: screw for refined motion around the vertical axis (VV);

3’”: the microscope for reading the angular values on the bearing circle;

4: the base of the level;

4’: three foot screws.

VV: the main rotation axis (vertical);

HH: the horizontal axis;

0: the aiming axis (with the condition 0 = HH);

NN: the directrix of the level air bubble.

36

50. The semiautomatic level NI 025 Carl Zeiss Jena is schematically presented in figure

1.50. Specify the name, role and importance of the parts enumerated in the schema.

Solution:

The name of the parts presented in figure 1.50 is similar to that from the previous case.

51. In figure 1.51 is presented the image obtained using a telescope-leveling device, a

centimetric measuring staff. Determine the readings corresponding to the three

stadimetric lines.

37

Solution:

The reading on the upper stadimetric line:

CS = 1879 (mm)

The reading on the level line:

CM = 1751 (mm);

The reading on the stadimetric line:

CJ = 1622 (mm)

Remark: the following verification must be performed:

CS + CJ

CM = --------- 2.

52. Sketch the image of the measuring staff corresponding to the following readings on

the measuring staff, based on the data presented in the previous problem:

CS = 2461 + n (mm);

CM = 2325 (mm);

CJ = 2189 – n (mm).

53. The measuring staff located on a leveling benchmark was aimed from a geometric

leveling station. The measurement was performed using a Ni 004 VEB Carl Zeiss Jena

device on a measuring staff with 3m invar (figure 1.53). Determine the corresponding

reading on the measuring staff and the micrometer.

Solution:

The reading consists in two parts:

a: reading on the measuring staff = 755;

b: reading on the micrometer = 56;

38

TOTAL: 75556

In order to determine the value expressed in meters:

- Subtract the constant K = 60650 from the total reading;

- Divide the previous value by 20.

Thus, there will be obtained a = 0.74530 m.

Remark: in the case when the reading is performed on the gradation on the left of the

measuring staff, the constant K is not subtracted.

54. Sketch the image of the invar measuring staff and of the micrometric drum

corresponding to the following reading: 69848 + n.

55. How are the verifications and adjustments of the rigid level performed? What about

the semiautomatic level?

The solution of this problem will be presented based on data obtained from the field

specialty literature.

56. Present the design of a self-reducing tacheometer DAHLTA 020 Carl Zeiss Jena,

specifying its components and main axis.

Solution:

39

The building axes of this kind of device are the same as in the case of the theodolite

(figure 1.41).

The characteristics of this device are:

- The image of the telescope is upright;

- The scale microscope (similar with the microscope of the Theo 020

theodolite) ensures a precision of 1C;

- A special board can be attached to the device, allowing tracing on scale the

planimetry and the contours, which the relief of the measured terrain is

represented by;

- The reticular plan is composed of a mobile part (which is used for measuring

the altitude differences) and a fixed part (needed to aim and to determine the

distances).

57. How are the readings on the Dahlta measuring staff recorded, in the case of

determining the altitude differences and the distances?

Solution: figure 1.57.

40

Readings:

- at the distance line Cd = 0.420;

- at the altitude line with constant K1 = + 10: CZ1 = 0.278;

- at the altitude line with constant K2 = + 20: CZ2 = 0.139.

58. If the measuring staff located in the point 48 was aimed from the station point 47 (Z 47

= 321.432 m) and there were recorded the values presented in figure 1.57, then compute

the horizontal distance between the two points and the absolute height of the benchmark

48.

Solution:

41

The horizontal distance: D47.48 = Cd · 100 (1.58)

Thus D47.48 = 40.20 m;

The altitude difference: Z47.48 = (i – v) + hm (2.58)

h1 = CZ1 x 10 = 2.78 m (3.58)

h2 = CZ2 x 20 = 2.78 m

h1 + h2 hm = ------------ = 2.78 m (4.58)

2

Thus, D47.48 = + 2.91 m;

The absolute height will be: Z48 = D47 + Z47.48; (5.58)

Z48 = 324.342 M.

59. Draft the design of the BRT 006 Carl Zeiss Jena tacheometer. Present the main and

secondary parts.

42

The device automatically reduces the distances at the horizon, allowing the direct

recording of horizontal distances.

60. Which are the procedures that a horizontal distance is recorded by, using the BRT

006 tacheometer-telemeter?

Solution:

1. Center and set horizontally the device in the station point;

2. In the aimed point fix a range pole or, depending on the case, a signal or an

aiming measuring staff;

43

3. (Figure 1.60 a): aim the signal;

4. On the distance scale record the value b (the variable basis);

5. Compute the horizontal distance.

61. The following values were recorded by aiming the range pole from the benchmark 61

in the station point 28 (Z28 = 328.561 m): L = 43.21 m, D = 1.24 m (the scaling

correction at the horizon). There was established the slope angle of the terrain = 15g

57c. Compute the horizontal distance and the height of the point 61.

Solution:

D = L - D (1.61)

Thus

D = 43.21 – 1.24 = 41.97 m.

Z’28.61 = L² - D² (2.61)

or

Z”28.61 = Dtg (3.61)

Will result:

Z’28.61 = 10.277 m;

Z”28.61 = 10.474 m

Z’ + Z”Z28.61 = -------------- (4.61)

2

thus, Z28.61 = 10.376 m.

44

The height of the point 61:

Z61 = Z28 + Z28.61 (5.61)

Z61 = 338.937 m.

F. PLANIMETRY PROBLEMS

a. THE DIRECT MEASURING OF DISTANCES

62. During a topographic work the horizontal distance between two points A and B was

measured, using the direct measuring method. Compute this distance, given the following

data of the measurement:

l0 = 50 m (the nominal length of the used tape);

l1 = 28.43 m (the distance recorded on the last tape);

n = 4 (the number of tapes applied);

The terrain is horizontal ( ≤ 5g).

Draft the schema corresponding to the measurement.

From the presented figure (figure 1,62) it results:

DAB = n · l0 + l1 (1.62)

In this case: DAB = 4 x 50 + 28.43 = 228.43 m.

45

63. The topographic points C and C are situated on a tilted terrain, with a known declivity

(). Knowing the measurement data, compute DCD and draft the corresponding schema.

Are given: l0 = 25 m;

l1 = 14.71 m;

n = 3;

= 9g21c.

Remark: the instruments used are the same as in the previous case.

From the schema: LAB = n · l0 + l1 (1.63)

DAB = LAB cos (2.63)

Replacing the problem data:

LAB = 3 x 25 + 14.71 = 89.71 m;

DAB = 89.71cos9g21c = 88.77 m.

46

64. Determine the horizontal distance between the points 21 and 22 situated on an

alignment, with the following successive declivities 1, 2 and 3, which are known. The

following are given:

l0 = 50 m;

n1 = 2; n2 = 1; n3 = 3;

l1 = 12.36 m; l2 = 16.52 m; l3 = 21.53 m;

1 = 16g31c. 2 = 12g52c. 3 = 7g67c.

Solution:

Compute the slanted distances:

L1 = n1 · l0 + l1 = 2 · 50 + 12.36 = 112.36 m;

L2 = n2 · l0 + l2 = 1 · 50 + 16.52 = 66.52 m;

L3 = n3 · l0 + l3 = 3 · 50 + 21.53 = 171.53 m.

The corresponding horizontal distances will be:

D1 = L1 · cos1 = 112.36 · cos 16g31c = 108.69 m;

D2 = L2 · cos2 = 66.52 · cos 12g52c = 65.24 m;

D3 = L3 · cos3 = 171.53 · cos 7g67c = 170.29 m.

The total distance D21.22 will be the sum of partial distances:

D21.22 = D1 + D2 + D3 = 108.69 = 65.24 + 170.29 = 344.22 m.

47

65. When measuring a distance using the correct method, the following values were

obtained:

l0 = 50 m;

l1 = 12.47 m;

n = 7;

3 = 12g51c;

lr = 50.007 m (the real length of the tape);

Fet = 3 daN/ mm² (the elongation force during calibration);

Fr = 3 daN/ mm² (the elongation force during measurements);

t0 = 20ºC (the temperature during tape calibration);

t1 = 28ºC (the temperature during measurements);

Asect = 10 mm² (the area of the cross section of the tape).

Compute the horizontal distance, applying the necessary corrections.

Solution:

The computation of the horizontal length consists in the following steps:

- Determine the slanted distance L:

L = n · l0 + l1 = 7 · 50 + 12.471 = 362.471 m (1.65);

- Compute the calibration correction, according to the following relation:

LCe = (lr - l0) ------ (2.65) l0

362.471Ce = (50.007 – 50) ----------- = 0.051 m 50

- Apply the calibration correction to the length L:

L’ = L + Ce (3.65);

L’ = 362.471 + 0.051 = 362.522 m;

- Determine the temperature correction:

L’Ct = lt ------ (4.65) l0

lt = l0 (t1 – t0) (5.65)

Thus: lt = 50 · 0.0115 (28 – 20) = 4.5 mm;

48

362.522 Ct = 4.6 ------------ = 33.4 mm = 0.033 m

50

- Then, correct the length L:

L” = L’ + Ct ; (6.65)

L” = 362.522 + 0.033 = 362.555 m;

- The computation of correction is performed using the relation:

L”(Fr - Fet )Cp = ----------------- (7.65)

E · Asect (cm²)

362.555 (4-3)In the case of our problem Cp = ----------------- = 0.002 m

2100000 · 0.1

- The correct slanted length will be:

L’’’ = L” + Cp (8.65)

L’’’ = 362.555 + 0.002 = 362.557 m.

The corresponding horizontal distance will be computed, as it is already known, from the

following relation:

D = L’’’ · cos (9.65)

Thus, in the end D = 362.557 cos 12g51c = 355.579 m.

b. THE INDIRECT MEASURING OF DISTANCES

66. Determine the horizontal distance between the points 43 and 44, given that the

following values were recorded using indirect tacheometric measuring:

CS = 1951 + n (m) i = 1,472 m (the height of the instrument);

CM= 1472 = 0g (the declivity angle);

CJ = 0993 – n (mm) K = 50 (the stadimetric constant).

49

- Verify the readings on the measuring staff:

CS + CJ

CM = ------------- (13) mm (1.66) 2

For this case:

1951 + 9931472 = --------------- = 1472 (mm) 2

- Note that the horizontal distance is directly recorded ( = 0g).

- Compute the horizontal distance: d = K · H = K (CS - CJ) (2.66);

- Thus, D = 50 (1.951 – 0.993) = 47.900 m.

67. The points 61 and 62 are situated on a tilted terrain. The following values were

obtained during the measuring of the distance between these points using the

tacheometric method:

CS = 2652 – n (mm) i = 1.537 m;

CM = 1537 (mm) = 9g61c + nc

CJ = 0422 (mm) K = 100

Compute the horizontal distance D61.62:

50

Verify the readings:

2652 + 0422--------------- = 1537 (mm) 2

Compute the horizontal distance:

D = KHcos² (1.67)

In this case:

D = 100 (2,652 – 0,422) cos29g61c = 217.957 m

68. Using the parallactic method – with basis at the end, a horizontal parallactic angle

was recorded having the value 7g31c + nc = (the difference between the horizontal

directions corresponding to the ends of the basis). If the aim on the basis was performed

at the height of the instrument and the measured declivity angle is zero – which is the

value of the horizontal distance between the device and the basis?

Solution:

51

From figure 1.68 it can be seen that:

Dctg----- = -------- and (b = 2m) 2 b ---- 2

From where D = ctg ----- (m) 2

7g31c Thus D = ctg ---------- = 17.399 m 2

69. Which is the horizontal distance between the points 76 and 77, if = 14g31c + nc and

the declivity angle of the aim is = 6g14c - nc?

52

In this case:

Lctg ----- = -------- and (1.69) 2 b ---- 2

Thus L = ctg ----- (m)

2

14g31c L = ctg ---------- = 8.86 m

2

And D = L cos (2.69)

From where D = 8.86 · cos 6g14c => D = 8.82 m.

70. In figure 1.70 there is presented the method of measuring the horizontal distance

between the points A and B, using the parallactic method with basis in the middle. Being

given the data of the measurement, determine DAB.

1 = 4g17c, 2 = 4g21c, 1 = 0g; 2 = 0g; b = 2m are given.

From the figure:

53

1 D1 = ctg----- (1.70)

2

2 D2 = ctg----- (2.70)

2

DAB = D1 + D2 (3.70)

Thus:

4g17c 4g21c

DAB = ctg -------- + ctg ---------2 2

DAB = 30.522 + 30.232 = 60.754 m

71. If the terrain is slanted, and the values recorded in the field are 1 = 2g17c, 2 = 2g22c,

1 = 12g43c +nc, 2 = 8g16c, b = 2 m, then determine the value of the horizontal distance

between the measured points 26 and 27.

As in the previous case, the horizontal distance consists of the two partial distances D1

and D2.

54

The aims towards the basis being slanted, D1 and D2 will be determined by the means of

the values L1 and L2 (the slanted distances).

1 2g17c

L1 = ctg ---- (1.71) L1 = ctg -------- = 58.670 m; 2 2

2 2g22c

L2 = ctg ---- (2.71) L2 = ctg ------- = 57.347 m; 2 2

D1 = L1cos1 (3.71) D1 = 58.670 · cos12g43c = 57.555 m;

D2 = L2cos2 (4.71) D2 = 57.347 · cos8g 16c = 56.877 m;

D26.27 = D1 + D2 (5.71) D26.27 = 57.555 + 56.877 = 114.432 m.

72. Using a helping basis CD, the elements needed for computing the horizontal distance

DAB were determined through the parallactic method. Based on the values of these

elements presented in what follows, compute the distance DAB.

1 = 2g62c , 2 = 7g16c + nc , b = 2 m, 1 = 0, 2 = 0.

Solution:

Compute DCB:

C

D

B

DC

B

DAB

Figure 1.72 The parallactic measuring of

distances with helping basis (horizontal

terrain). Plan schema

55

2 DCB = ctg ----- = 17.76 m (1.72)

2

1 DAB

ctg ----- = ---------- => 2 DCD

------- 2

DCD 1

DAB = ------- ctg ----- (2.72) 2 2

17.76 2g62c

DAB = --------- ctg --------- 2 2

DAB = 431.577 m.

73. The values needed to determine the distance D41.42 were taken using the method

presented before. They are:

1 = 2g84c + nc , 2 = 8g61c - nc , b = 2 m, 1 = 12g62c , 2 = 10g18c.

Solution: (see figure 1.73)

56

Remark: The plan image of the measurement appears in figure 1.72.

The computational steps are the following:

2 L2 = ctg ----- (1.73) L2 = 14.765 m; 2

DCD = L2 · cos2 (2.73) DCD = 14.765 · cos10g18c=14.577 m;

DCD 1 14.577 2g84c

L1 = -------- · ctg ----- = 17.76 m (3.73) L1 = --------.ctg -------- =326.707 m; 2 2 2 2

DAB = L1 · cos1 (4.73) DAB =326.707 · cos12g62c= 320.309m.

c. MEASURING ANGLES

74. Compute the horizontal angles specified in the schema of each problem. The

measuring method used in each case and the field operation method will be specified.

Station Aimed

point

Readings on the

bearing circle (c)

Horizontal angles

Remarks

Schemas

Code

1 2 3 4 5 6

S1

2(173 + n) · 41 ·

26

2

3

1 33 285 · 52 · 17

Solution:

Method: “reading differences”, one position of the telescope;

The number of angles measured from a station: one;

Computational method:

= C3 - C2 = 285g52c17cc - 173g41c26cc = 112g10c91cc (the value to be written in the

5th column).

75.

57

Table 1.75

Sta

tion

Aim

ed p

oin

t

Horizontal directions (readings

on the bearing circle) (c)

Horizontal

directions

Horizontal

angles

Remarks

Schemas

Position I (left) Position II (right) Means (M)

1 2 3 4 5 6 7 8

4

5 41 · 22 · 16 241 · 22 · 10+ncc

5

4 66

127 · 18 · 73-

ng327 · 18 · 75- ng

Solution:

Method: “reading differences”, two positions of the telescope;

The number of angles measured from a station: one;

Computational method:

22c16cc + 22c10cc

M5 = 41g (------------------------) = 41g22c13cc (column 5, the first line); 2

18c 73cc + 18c75cc

M6 = 127g (-----------------------) = 127g18c74cc (column 5, the second line); 2

= M6 – M5 = 127g18c74cc - 41g22c13cc = 85g96c61cc (column 7)

76.

Table 1.76

Sta

tion

Aimed point

Readings on the

bearing circle (c)

Horizontal

angles

Remarks

Schemas

Code

1 2 3 4 5 6

7

8 00 · 00 · 00

8

7 99 64 · 17 · 30 +nc

Solution:

Method: “zero in coincidence”, one position of the telescope;

The number of angles measured from a station: one;

58

Computational method:

= C9 – C8

= 64g17c30cc - 00g00c00cc = 64g17c30cc (column 5).

Therefore, the horizontal angle is directly measured.

77.

Solution:

Method: “zero in coincidence”, two positions of the telescope;

The number of angles measured from a station: one;

Computational method:

M11 = 00g00c00cc (column 5);

64c17cc + 64c23cc

M12 = 121g(-------------------------) =121g64c20cc (column 5); 2

= M12 - M11 = 121g64c20cc (column 7).

Table 1.77

Sta

tion

Aim

ed p

oin

t

Horizontal directions

(readings on the bearing

circle) (c)

Horizontal

directions

Horizontal

angles

Remarks

Schemas

Position I (left) Position II

(right)

Means (M)

1 2 3 4 5 6 7 8

10

11 00 · 00 · 00200 · 00 ·

00

11

10

12 12 121 · 64 · 17+ng

321 · 64 ·

23

78. Determine the vertical angles corresponding to the measured values presented in the

following tables. The method used, the characteristics of the measured angle and the field

operation method will be specified.

59

Table 1.78

Station Aimed

point

Readings on the

clinometer (zenithal

angle) (Z)

Vertical angle

(V or )

Remarks

Schemas

g | c | cc Code g | c | cc

1 2 3 4 5 6

21 i=1,32 22 i=1,32 98 · 17 · 00 +nc 21.22

Solution:

Method: determining a single vertical angle, from a station, using one position of the

telescope;

Measured angle: the declivity angle of the terrain (because i STATION = i AIM).

Computational method:

= 100g00c00cc – Z;

Thus, = 100g00c00cc - 98g17c00cc = 1g83c00cc (column 5).

79.

Table 1.79

Sta

tion

Aim

ed p

oin

t

Readings on the clinometer

(Z)

Vertical angle

(V or )

Remarks

Schemas

Position I

(ZI)

g c cc

Position II

(ZII)

g c cc

Cod

e

g c cc

1 2 3 4 6 7 8

23 i

= 1

,43

24 i

= 1

,43 86 · 28 · 50

+ng

313 · 72 · 00

- ng

23.24 The schema is exactly

the same as the one in

the previous case.

Solution:

Method: determining a vertical angle using two positions of the telescope;

Measured angle: the declivity of the terrain (because i STATION = i AIM).

Computational method:

60

I = 100g00c00cc - Z I = 13g71c50cc

II = Z II - 300g00c00cc = 13g72c00cc

I + II = ----------- = 13g71c75cc (column 6).

2

80.

Solution:

Method: determining a vertical angle using one position of the telescope;

Measured angle: vertical angle;

Computational method:

V = 100g – Z

Therefore V = 100g00c00cc – 64g12c00cc = 35g88c00cc (column 5).

Table 1.80

Station

I = (m)

Aimed

point

S = (m)

Readings on the

clinometer (zenithal

angle) (Z)

Vertical angle(V or )

Remarks

Schemas

g | c | cc Code g | c | cc1 2 3 4 5 6

25

i=1,

62

26

i=2,

02 64 · 12 · 00+ng

V25..26

81. The solution is the same as in the case of problem 79.

Table 1.81

Station Aimed

point

Readings on the

clinometer (Z)

Vertical angle

(V)

Remarks

Schemas

Position I

(ZI)

Position II

(ZII)

Code

1 2 3 4 6 7 8

61

8/ i

= 1

,46

9/ S

= 6

,21

43 · 21 · 16 356 · 78 · 90 V8.9

82. Data obtained in the field using the method of “the horizon tour” are presented in the

following table. Compute the horizontal angles , , and and the vertical angles

corresponding to each direction. Explain the method of operating in the field and the

computational steps.

62

Solution:

The working steps are the following:

Computing the values MI (1.82)

12c00cc +13c00cc

Mi1= 21g(---------------------)

2

= 21g12c50cc (column 5)

Computing the closing discrepancy error

e = Mf1 - Mi

1 (2.82)

Thus e = 21g14c00cc - 21g12c50cc

= 1c50cc

Computing the total correction C

C = - e = -1c50cc (3.82)

Determining the unitary correction Cu

C 1c50cc

Cu = ----- = --------- = - 30cc (4.82) n 5

n = the number of measured points

The corrections on directions will be:

63

C1 = 0 x Cu = 0c00cc;

C2 = 1 x Cu = - 0c30cc; (5.82)

C3 = 2 x Cu = - 0c60cc;

The directions Mi are determined as follows:

Mi = Mi + Ci (6.82)

For example M2 = 68g57c00cc + (- 0c30cc) = 68g56c70cc;

The directions reduced to zero: M0i = Mi – M1; (7.82)

For example:

M02 = M2 – M1;

The horizontal and vertical angles are computed as in the case of the previous problems

(74-81).

d. SURVEY OF THE DETAILS

83. Topographically describe the position of the topographic points represented in figure

1.83, using the graphical method.

Nr .11 Nr .13 Nr .15 Nr .17

Nr .8 Nr .10 Nr .12

Nr .2

Nr .14 Nr .16

Nr .1

Str

. Al. V

lah

uta

5

4 6

1 2 3 7 8

11 12 13 14 15

16 17 10 Str . Alba 13,10

Figure 1.83 The topographic description of benchmarks Scale 1:500

64

Solution:

The topographic description of the benchmark no.9 is presented in the next figure. The

distances between the topographic benchmark – and characteristic points (building

corners, different installations, etc.) are specified.

These distances are taken, depending on the case, from the field or from the available

documentation.

84. The method of direct intersection was used to determine the coordinates of the point

A with respect to the topographic benchmarks 1 and 2.

Given the following data:

The coordinates of the bearing points The elements measured in the field

X1 = 316.47 m + n(m);

Y1 = 125.48 m; P12 = = 24g17c53cc

X2 = 323.21 m + n(m); P21 = = 61g43c28cc + nc

Y2 = 392.54 m – n(m).

Compute (XA, YA)

65

Solution:

Basic orientation:

Y12 Y2 -Y1

tg12 = ------- = --------- X12 X2 -X1

tg12 = 39.623145

Therefore:

12 = 98g39c37cc

and 21 = 298g39c37cc = 12 + 200g

The orientations of the new sides:

1A = 12 - = 98g39c37cc - 24g17c53cc = 74g21c84cc;

2A = 21 - = 298g39c37cc + 61g43c28cc = 359g82c65cc;

Y1A YA -Y1

tg1A = ------- = --------- = (XA -X1) tg1A = YA -Y1; (2.84) X1A XA -X1

Y2A YA –Y2

tg2A = ------- = --------- = (XA –X2) tg2A = YA –Y2; (3.84) X2A XA –X2

Subtracting the second equation from the first one:

YA -Y1 -YA + Y2 = XA tg1A - X1 tg1A – XA tg2A + X2 tg2A

Y2 – Y1 + X1 tg1A - X2 tg2A

Thus: XA = -------------------------------------- (4.84) tg1A – tg2A

And: YA = Y1 + (XA – X1) tg1A; (5.84)

Or: YA = Y2 + (XA – X2) tg2A; (6.84)

Replacing the problem data the following values are obtained:

XA = 450.25 m; YA = 332.59 m.

85. The horizontal angles formed by the directions towards the points 3, 4, and 5 were

measured from the point B of unknown coordinates, using resection (the indirect method

or retrointersection, the map problem, Pothènot problem). The measurement data are

given:

The coordinates of the bearing points The elements measured in the field

66

X1 = 675.43 m + n(cm);

Y1 = 125.51 m;

X2 = 712.37 m – n (cm); 1B2 = = 53g13c21cc + ncc;

Y2 = 272.38 m + n(cm); 1B3 = = 123g61c87cc + ncc;

X3 = 525.82 m;

Y3 = 321.57 m – n(cm).

Compute the coordinates of the point B(XB,YB).

The specialty literature offers more than one solution to determine the coordinates of the

point P (Délambre, the Collins trigonometric method, etc.) We shall shortly present one

of these possibilities:

Compute 1: (1.85)

(Y2 – Y1)ctg + (Y1 – Y3)ctg + X3 – X2

1= --------------------------------------------------- (X2 – X1)ctg + (X1 – X3)ctg + Y2 – Y3

67

In what follows parse the following steps:

2= 1 + and tg2 = ………..

3= 1 + and tg3 = ………..

Y2 – Y1 + X1tg1 – X2tg2

X = ---------------------------------- (2.85) tg1 - tg2

Y = Y1 + (X – X1) tg1 or (3.85)

Y = Y2 + (X – X2) tg2 (4.85)

Y = Y3 + (X – X3) tg3 (5.85)

86. Compute the absolute coordinates of the points 21 and 22, using the compensation of

the planimetric traverse supported at the ends, presented in table 1.86.

68

Solution:

Solving the traverse is done in the following steps:

1. Determine cosiJ: cos12..21 = 0.9916958 are written

cos21..22 = 0.9870361 in column 6

cos22..14 = 0.9937838

2. Compute the horizontal distances using the relation: DiJ = LiJ · cosiJ; (1.86)

D12..21 = 54.20 x 0.9916958 = 53.750 m;

D21.22 = 52.10 x 0.9870361 = 51.425 m; column 10

D22.14 = 25.92 x 0.9937838 = 25.759 m.

3. The bearing orientations will be: (2.86)

Y12.13 Y13 –Y12 209.60-245.21 -35.61 tg12.13 = --------- = ----------- = tg12.13 = ------------------- = --------- X12.13 X13 –X12 677.90-620.73 + 57.17

tg12.13 = - 0.6228791, the angle 12.13 is in the IVth quadrant (-Y / +X);

Thus tg12.13 = - ctg = - 0.6228791 (where = tg12.13 – 300g);

1tg = ------ = 1.6054479 => arctg 1.6054479 = 64g53c56cc; ctg

From where 12.13 = + 300 g = 364 g 53 c 56 cc (The starting orientation)

Y14.15 Y15 –Y14 395.210 – 352.900 42.31tg14.15 = ---------- = ----------- => tg14.15 = ----------------------- = -------- X14.15 X15 –Y14 687.270 – 647.270 31.00

tg14.15 = 1.3648387, the angle 14.15 is in the Ist quadrant (+Y / +X);

69

Thus, 14.15 = arctg1.3648387 = 59 g 74 c 47 cc (The closing orientation).

4. Determine the orientations of the sides of the traverse: (column 8).

a. Temporary orientations:

12.21 = 12.13 + 1 - 400g = 364g53c56cc + 99g12c40cc - 400g = 63g65c96cc;

21.12 = 12.21 + 200g = 263g65c96cc;

21.22 = 21.12 + 2 - 400g = 263g65c96cc + 265g26c20cc - 400g = 128g92c16cc;

22.21 = 21.22 + 200g = 328g92c16cc;

22.14 = 22.21 + 3 - 400g = 328g92c16cc + 114g26c10cc - 400g = 43g18c26cc;

14.22 = 22.14 + 200g = 243g18c26cc;

14.15 = 14.22 + 4 - 400g = 243g18c26cc + 216g61c40cc - 400g = 59g79c66cc;

b. Computing the corrections:

The closing discrepancy error on the orientation e:

e = 14.15 COMPUTED - 14.15

GIVEN = 59g79c66cc - 59g74c47cc =5c19cc (3.86);

The total correction C:

C = - e = - 5c19cc (4.86);

The unitary correction Cu:

C - 5c19cc

Cu = ------ = ---------- = - 1c30cc (5.86); N 4

The corrections on the orientations:

C12.21 = 1 x Cu = - 1c30cc

C21.22 = 2 x Cu = - 2c60cc (6.86);

C22.14 = 3 x Cu = - 3c90cc

C14.15 = 4 x Cu = - 5c19cc

c. Correcting the orientations:

12.21 = 12.21 + C 12.21 = 63g65c96cc - 1c30cc = 63g64c66cc;

21.22 = 21.22 + C .21.22 = 128g92c16cc - 2c60cc = 128g89c56cc; (7.86)

22.14 = 22.14 + C 22.14 = 48g18c26cc - 3c90cc = 43g14c36cc;

Verification: 14.15 = 14.15 + C14.15 = 59g79c66cc - 5c19cc = 59g74c47cc = 14.15 GIVEN.

5. Determine the trigonometric function (the natural values) sin and cos for the

corrected orientations: (column 9)

70

sin12.21 = 0.8413404; cos12.21 = 0.5405055;

sin21.22 = 0.8987478; cos21.22 = - 0.4384658;

sin22.14 = 0.6270014; cos22.14 = 0.7790180;

sin14.15 = 0.8066533; cos14.15 = 0.5010249.

6. Computing the relative coordinates: (columns 11 and 12)

a. The relative raw coordinates XiJ ,YiJ

X12.21 = D12.21 · cos12.21 = 53.750 x 0.5405055 = 29.052 m; (8.86)

Y12.21 = D12.21 · sin12.21 = 53.750 x 0.8413404 = 45.222 m;

X21.22 = D21.22 · cos21.22 = 51.425 x (- 0.4384658) = - 22.548 m;

Y21.22 = D21.22 · sin21.22 = 51.425 x 0.8987478 = 46.218 m;

X22.14 = D22.14 · cos22.14 = 25.759 x 0.7790180 = 20.064 m;

Y22.14 = D22.14 · sin22.14 = 25.759 x 0.6270014 = 16.151 m;

b. Corrections of relative coordinates:

The closing discrepancy error on the coordinates eX, eY:

eX = XiJ - X12.14 = 26.568 m – 26.540 m = 28 mm; (9.86)

eY = YiJ - Y12.14 = 107.591 m – 107.690 m = - 99 mm; (10.86)

The total corrections CX, CY:

CX = - eX = - 28 mm; (11.86)

CY = - eY = 99 mm; (12.86)

The unitary corrections CuX, CuY:

CX - 28 mm - 28 mmCuX = ------- = ---------------------------- = ---------- = - 0.214 mm/1m; (13.86)

DiJ D12.21 + D21.22 + D22.24 130.934m

CY 99 mmCuY = ------- = ------------- = 0.756 mm / 1 m TRAVERSE (14.86). DiJ 130.934 m CORRECTION

The correction on relative coordinates:

CX 12.21 = CuX x D12.21 = - 0.214 mm/m x 53.75 m = - 12 mm; (15.86)

CY 12.21 = CuY x D12.21 = 0.756 mm/m x 53.75 m = 41 mm; (16.86)

CX 21.22 = CuX x D21.22 = - 0.214 mm/m x 51.425 m = - 11 mm;

CY 21.22 = CuY x D21.22 = 0.756 mm/m x 51.425 m = 39 mm;

71

CX 22.14 = CuX x D22.14 = - 0.214 mm/m x 25.759 m = - 5 mm;

CY 22.14 = CuY x D22.14 = 0.756 mm/m x 25.759 m = 19mm;

c. The correction of relative coordinates:

X12.21 = X12.21 + CX 12.21 = 29.052 – 0.012 = 29.040 m; (17.86)

Y12.21 = Y12.21 + CY 12.21 = 45.222 + 0.041 = 45.263 m; (18.86)

X21.22 = X21.22 + CX 21.22 = - 22.548 – 0.011 = - 22.559;

Y21.22 = Y21.22 + CY 21.22 = 46.218 + 0.039 = 46.257 m;

X22.14 = X22.14 + CX 22.14 = 20.064 – 0.005 = 20.059 m;

Y22.14 = Y22.14 + CY 22.14 = 16.151 + 0.019 = 16.170 m.

7. Determining the absolute coordinates: (columns 13 and 14)

X21 = X12 + X12.21 = 649.770 m; (18.86) Y21 = Y12 + Y12.21 = 290.473 m;

X22 = X21 + X21.22 = 627.211 m; (19.86) Y22 = Y21 + Y21.22 = 336.730 m;

Verification: Verification:

X14 = X22 + X22.14 = 647.27m. Y14 = Y22 + Y22.14 = 352.90m.

87. In order to measure a planimetric detail in the field through its characteristic points

(117 and 118), 21.22 was used as bearing side (traverse side). The survey was performed

using the method of polar coordinates.

If the coordinates of the bearing points 21 and 22 and the elements measured in the field

(angles and distances) are known, then determine the coordinates of the characteristic

points.

Te coordinates of the bearing points The elements measured in the field

X21 = 649.770 m + n (cm); 22.21.117 = 1= 128g51c + ng;

Y21 = 290.473 m; 22.21.118 = 1= 128g51c + ng;

X22 = 627.211 m; D22.117 = 46.52 m = D1;

Y22 = 336.730 m – n(cm). D22.118 = 61.27 m = D2.

72

Solution:

Compute the basic (bearing) orientation 22.21:

Y22.21 Y21 -Y22

tg22.21 = ---------- = ---------- X22.21 X21 -X22

290.473 – 336.730 - 46.257tg22.21 = ------------------------ = ------------= - 2.0504898

649.70 – 627.211 22.559

=>22.21 = 328g88c66cc.

Determine the orientations of the new sides (towards the radiated points):

22.117 = 22.21 + 1 - 400g = 41g 39c66cc; (2.87)

22.118 = 22.21 + 2 - 400g = 50g 03c66cc;

Compute the relative coordinates:

X22.117 = D1 · cos22.117 = 37.027 m; (3.87)

Y22.117 = D1 · sin22.117 = 28.163 m; (4.87)

X22.118 = D2 · cos22.118 = 43.300 m;

Y22.118 = D2 · sin22.118 = 43.349 m;

The absolute coordinates of the radiated points will be:

X117 = X22 + X22.117 = 627.221 + 37.027 = 664.238 m; (5.87)

Y117 = Y22 + Y22.117 = 336.730 + 28.163 = 364.893 m; (6.87)

X118 = X22 + X22.118 = 627.221 + 43.300 = 670.511 m;

Y118 = X22 + Y22.118 = 336.730 + 43.349 = 380.079 m.

Remark: Polar elements are enough in order to repeat the radiated points on topographic

plans (see the next problem).

73

88. The data obtained using the method of Cartesian-square coordinates for measuring

some planimetric details are presented in table 1.88. Compute the coordinates of the

measured points.

Solution:

In this case, the support side is parallel to the abscissa of the used local coordinate

system. The coordinates of the new points will be computed in the following way:

Table 1.88

Sid

e

Ori

gin

Mea

sure

d

Measured elements RemarksSchemas

Initial DataX

(m)

Y

(m)

28-2

9

28

1 6.27 10.52 +n(m) + X

+ Y

1 2

3 4 5 6

6 m

11 m

8 m

Y3 X328 29

The coordinates of the bearing points X(m) Y(m)

28 682.272 273.62229 682.272 343.657

2 6.27 18.64

3 12.17 21.73

4 12.17 37.84

5 21.58– n(m) 43.28

6 21.58– n(m) 61.74

Xi = X28 Xi (1.88)

Depending on the position of the detail with respect to the support side.

Yi = Y28 + Yi (2.88)

If the support side is not parallel to one of the coordinate axes – the square radiation is

computed in the same way as the polar radiation.

89. A series of details from the area were measured using the method of tacheometric

radiation from the station point 22 (figure 1.89). The traverse side 22.21 was used as

74

support basis. The elements measured in the field (table 1.89) will be used to compute the

absolute coordinates of the radiated points.

Solution:

In the case of tacheometric radiation, the distances are indirectly (tacheometrically)

obtained. The other elements, concerning the field and office operations, are similar to

those performed during planimetric radiation.

Table 1.89

Station Aim Bearing circle

Clinometer Measuring staff (mm)

CS

CM

CJ

Horizontal angle

Vertical angle

D(m)

Point

i= 1

.53

m

2

2

21 00 00 00 - - - - - -

12’ 71 27 00 100 00 00 1826 1520 1214 12

13’ 94 12 00 117 21 00 1641 1520 1400 13

14’ 101 16 00 81 16 00 1976 1530 1084 14

15’ 137 52 00 92 17 00 1715 1530 1312 15

21 00 01 00 - - - - - -

The working steps are the following:

Determine the horizontal angle i = Ci – C27 (1.89) where Ci = the reading on the

bearing circle towards some point i;

75

The vertical angle i: i = 100g - Vi (2.89)

Vi = the reading on the clinometer towards some point i;

The horizontal distance D26.i: D26.i = KHcos² .i (3.89)

K = 100 (the stadimetric constant)

Hi= (CS - CJ )i (4.89)

Knowing the horizontal angle and the horizontal distance, the absolute

coordinates of the radiated points will be obtained through the computations

described at planimetric radiation.

e. REPEATING DETAILS

90. Repeat the control network (planimetric traverse) of known coordinates (table 1.90)

on the 1:1000 scale, through Cartesian coordinates.

X (m) Y (m)

12 620.730 245.210

13 677.900 209.600

21 649.770 290.473

22 627.211 336.730 - n

14 647.270 + n 352.900

15 678.270 395.210 – n

Solution:

The repeating steps are the following:

- Trace the graticule of the plan on a sheet of paper (tracing paper, millimetric

paper);

- Trace, for each point, the corresponding axes (the abscissa, the ordinate)

(figure 1.90)

- Compute the coordinate differences, scaled down (x, y) with respect to the

coordinate axes that is the closest in value to the coordinates of the repeated

point;

- Mark by a chosen symbol (depending on the importance of the point) the

position on the plan, also writing down the number of the point (12, 13, etc.)

76

91. The points 12 and 13 were used as benchmarks for determining the coordinates of the

point 68 through direct intersection. Knowing the coordinates of the bearing points (table

1.90) and of the new point (X68 = 652.432 m + n(m), Y68 = 248.516 m), repeat this point

through Cartesian coordinates.

Solution:

The repeating is performed on the plan drafted in problem 90.

The repeating steps are the ones specified in the solution of that problem (no. 90).

After positioning the point on the plan, check with the protractor the angular elements

( and ) that were used for determining the absolute Cartesian coordinates of the

point obtained through direct intersection.

92. The coordinates of the point 72 were obtained using resection, having as bearing

points the points 12, 13 and 21. [X72 = 675.430 m, Y72 = 238.472 m + n (m)].

77

Repeat the point 72, using absolute Cartesian coordinates, on the topographic plan drafted

in problem 90.

Remark: the same specification as in the previous problem.

93. The coordinates of the point 117 (see problem 87) were determined through

planimetric radiation. Repeat this point through Cartesian coordinates on the topographic

plan drafted in the previous problems.

Remark: see problem no. 91.

94. Repeat the points 1-6 through Cartesian coordinates, computed in problem no. 88.

95. Repeat the points 12-15 through Cartesian coordinates, computed in problem no 89.

Remark: From the solution of problems 90-95 it can be seen that the repeating on the plan

is performed in a similar manner, regardless of the method used for computing the

absolute Cartesian coordinates of the points, regardless of the nature of the plan, and of

the scale of the plan. But there exists the possibility to verify the computation of the

coordinates and the repeating of the point on the plan, using initial data (angles,

distances).

96. The topographic point 96 was measured through angular intersection. Repeat it on the

topographic plan without computing its coordinates.

The coordinates of the bearing points 12 and 21 (see problem no. 86) and the angles

measured in the field are given:

96 = 21.12.96 = 31g46c + ng;

96 = 12.12.96 = 46g12c - ng;

Solution:

Figure 1.95 presents the centralized method for solving the problems no. 96-100. The

scale of the plan is 1:1000.

78

97. The point 97 was planimetrically measured through linear intersection. The

coordinates of the bearing points 21 and 22 were specified in the previous problems. The

distances D1 and D2 are given, being measured in the field:

D1 = D21.97 = 36.41 m;

D2 = D22.97 = 30.16 m + (n/4) m.

Repeat the points 97 on the topographic plan without computing its coordinates.

98. Repeat the points 117 and 118 on the 1:1000 plan using polar coordinates. The polar

elements needed for repeating were presented in the tutorial no. 87.

99. Problem 88 offers the data needed for repeating on the topographic plan some

objectives surveyed through square coordinates. Repeat these points on the 1:1000 plan

presented in figure 1.95 without computing their absolute Cartesian coordinates.

100. Repeat using the method of polar coordinates the points 12’, 13’, 14’, and 15’,

measured through the method of tacheometric radiation (problem no. 89). The

topographic plan presented in figure 1.95 will be used as repeating support.

G. LEVELING PROBLEMS

a. GEOMETRIC LEVELING

101. The absolute height of the point A is known and the elements needed for

determining the absolute height of the point B were measured through middle geometric

leveling.

Processing the data presented in table 1.101, determine the height ZB.

a. By the use of the altitude difference ∆ZAB.

b. By the use of the height of the instrument horizon Zi.

Table 2.101

Station Aimed point

Rod readingsab

(mm)

Altitude differences

∆ZAB

(m)

The height of the instrument

horizonZi

Absolute heights

Z(m)

Point

79

(m)1 2 3 4 5 6 7S1 A 1621 + n (mm) 350.375 A

B 0751 BFigure 1.101 Determining the relative heights (∆ZAB) or an absolute height (ZB)

through middle geometric leveling.

Solution, measuring staff, topographic level, measuring staff, vertical datum (N.M.N.)

a. ∆ZAB = a – b (1.101)

∆ZAB = ZB - ZA (2.101)

In this case:

∆ZAB = 1.621 – 0.751 = 0.870 m; (2.101)

ZB = ZA + ∆ZAB = 350.473 +

0.870

351.343 m

b. Zi = ZA + a (3.101)

Zi = ZB + b

Hence

Zi = 350.473 + 1.621 = 352.094 m

ZB = 352.094 – 0.751 = 351.343 m.

102. The method of end geometric leveling was used in order to determine the height of

the point C. Point A was taken into consideration as benchmark of know height. Based on

the data presented in table 1.102, determine the height of the point C:

a. By the use of the altitude difference ∆ZAC;

b. By the use of the height of the instrument horizon Zi.

Table 1.102

Stationi =

Aimed point

Rod readingsa = i

b(mm)

Altitude differences

∆ZAC

(m)

The height of the instrument

horizonZi

(m)

Absolute heights

Z(m)

Point

1 2 3 4 5 6 7A

i = 1.572m

C1572 350.473 A

0945 + n (m) C

80

Solution:

It can be seen that the previous relations do not change.

The height i of the instrument (level) in this station is considered instead of the rod

reading a in the point A.

Figure 1.102. End geometric leveling

a. ∆ZAc = i – b (1.102)

∆ZAc = ZA – ZC (2.102)

∆ZAc = 1.572 – 0.945 = 0.627 m (2.102)

ZC = ZA + ∆ZAc = 350.473 +

0.627

351.100 m

c. Zi = ZA + I (3.102)

d. Zi = ZC + b

Zi = 350.473 + 1.572 = 352.045 m;

ZC = 352.045 – 0.945 = 351.100 m.

103. The method of middle geometric leveling radiation – with one horizon of the

instrument, was used in order to determine the absolute heights of the points 11 and 12.

Processing the data from table 1.103, compute Z11 and Z12.

Table 1.103

Stationi =

Aimed point

Rod readingsa = ib1

b2 (mm)

Altitude differences

∆ZAC

(m)

The height of the instrument

horizonZi

(m)

Absolute heights Z

(m)

Point

1 2 3 4 5 6 7S2 A

1112

1547 350.473 A2063 110942 12

Figure 1.103. Middle geometric leveling radiation

Zi = ZA + a = 350.473 + 1.547 =>

Zi = 352.020 m;

Z11 = Zi – b11 = 352.020 – 2.063 = 349.957 m;

81

Z12 = Zi – b12 = 352.020 – 0.942 = 351.078 m.

104. The absolute heights of the points 14 and 15 were determined applying the method

of middle geometric leveling, with two horizons of the instrument (table 1.104). What

values do these heights have and how were they measured?

Table 1.104

Station Aimed point

Rod readings(mm)

The height of the

instrument horizon

Absolute heights

Z(m)

Point

Horizon I

Horizon II

ZIi

(m)ZII

i

(m)1 2 3 4 5 6 7 8S3

S’3

A 1624 1410 350.473 A14 1976 1762 1415 0604 0390 15

Figure 1.104 Middle geometric leveling radiation with two horizons of the

instrument.

Solution:

Z’i = ZA + a‘ (1.104)

Z”I = ZA + a“ (2.104)

Z’14 = Z’i – b’14 (3.104)

Z”14 = Z”i – b”14

Z’14 + Z”14

Z14 = -------------- (4.104) 2

The height of point 15 is determined similarly.

105. The data needed for computing the heights Z16 and Z17 were collected in the field

using the method of middle geometric leveling radiation with two horizons of the

instrument. Determine these heights by processing the data given in table 1.105.

Figure 1.105. Middle geometric leveling radiation with aims towards two bearing

points (of know heights)

Solution:

Z’i = ZE + a’

82

Z’i + Z”i

Zi = --------------- (average horizon) (1.105) 2

Z”i = ZF + a”

The absolute heights of the radiated points will result from the following relation:

ZK = Zi – bK (2.105)

Table 1.105

Station Aimed point

Rod readingsa’bi

a”(mm)

The height of the instrument

horizonZ’i

Zi

Z”i

(m)

Absolute heights

Z(m)

Point

1 2 3 4 5 6S4 E 1842 352.763 E

16 2076 1617 1243 17F 2092 352.510 F

106. In order to determine the absolute heights of some radiated points we could use the

methods of end geometric leveling, too. Of course, in this case the precision is smaller,

because the measurements are influenced by a series of errors, which are removed in the

case of middle leveling. There are situations when only the end leveling can be applied.

This is the reason why, in what fallows, we shall present some of the methods of this type

of geometric leveling.

Thus, stationing in the point 43 of known height, the values needed to compute the

radiated points 44 and 45 were determined (table 1.106). Compute these heights.

Table 1.106

Station

i = …

Aimed point

Rod readingsa = ib44

b45

(mm)

The height of the instrument

horizonZi

(m)

Absolute heights

Z(m)

Point

1 2 3 4 6 7

431.632

43 1632 361.273 4344 0751 4445 2072 45

83

Solution:

Zi = Z43 + i (1.106)

Z44 = Zi – b44 (2.106)

Z45 = Zi – b45

Therefore, there will be obtained:

Zi = 362.905 m;

Z44 = 362.154 m;

Z45 = 360.833 m.

Figure 1.106 End geometric leveling radiation

107. The method of end geometric leveling – with two horizons of the instrument, was

used in order to determine the heights of the points 74 and 75. Compute Z 74 and Z75

processing the data from table 1.107.

Table 1.107

Stationi‘ =i“ =(m)

Aimed point

Rod readings(mm)

The height of the

instrument horizon

Absolute heights

Z(m)

Point

Horizon I

Horizon II

ZIi

(m)ZII

i

(m)1 2 3 4 5 6 7 8

S’5

i = 1.264S”5

i = 1.373

M 1264 1373 362.172 M74 2656 2767 7475 0932 1040 75

Solution:

Z’i = ZM + I’; (1.107)

Z”i = ZM + I”;

Figure 1.107. End geometric leveling radiation with two horizons of the instrument

Hence Z’i = 362.172 +1.264

Z”i = 362.172+ 1.373

=>Z’i = 363.436 m;

Z’i = 363.545 m.

Z’74 = Z’i - b’74 = 360.780 m; (2.107)

84

Z”74 = Z’i - b”74 = 360.778 m.

Z’74 + Z”74

Z74 = ------------------ (3.107) 2

Z75 = 362.505 m is obtained analogously.

108. Compute the heights of the points 61 and 62, measured through middle geometric

leveling (table 1.108).

Table 1.108

Station Aimed point

ForwardBackward

Rod readings(mm)

Altitude differences

Hor

izon

tal

dis

tan

ces

DiJ (

m)

Abs

olut

e he

ight

s

Z (

m)

Backward Forward Coarse∆ZiJ

(m)

CorrectionsCDiJ

(mm)

Corrected∆ZiJ

(m)1 2 3 4 5 6 7 8 9S1 RN1

611751 1343 + n (mm) 54.43 A

S2 6162

2437 0975 121.72 14

S3 62RN2

1875 1947 76.43 15

Solution:

Figure 1.108. Middle geometric leveling traverse a. Section schema; b. Plan schema.

The computational steps are:

1. Determine the coarse altitude differences ∆ZiJ (column 5):

∆ZRN1.61 = a1 – b1 = 1.751 – 1.343 = + 0.408 m;

∆Z61.62 = a2 – b2 = 2.437 – 0.975 = + 1.462 m; (1.108)

∆Z62.RN2 = a3 – b3 = 1.875 – 1.947 = - 0.072 m

2. Compute the closing discrepancy error of the corrections:

RN2eAZ = ∑∆ZiJ - ∆ZRN1.RN2 = (∆ZRN1.61 + ∆Z61.62 + ∆Z62.RN2) – (ZRN2 – ZRN1) = RN1

= + 1.798 m – 1.804 m = - 0.006 m = - 6 mm. (2.108)

C∆Z = - e∆Z = 6 mm (total correction) (3.108)

The unitary correction will be:

85

C∆Z 6 mmCu∆Z = --------- = -------------- = 0.024 mm / 1 mm (4.108)

RN2 252.58 m∑∆ZiJRN1

The corrections on relative spatial coordinates (altitude differences) C∆ZiJ (column 6):

C∆ZRN1.61 = Cu∆Z x DRN1.61 = 0.024 mm / 1 m x 54.43 ≈ 1 mm;

C∆Z61.62 = Cu∆Z x D61.62 = 0.024 mm / 1 m x 121.72 m ≈ 3 mm; (5.108)

C∆Z62. RN2 = Cu∆Z x D62. RN2 = 0.024 mm / 1 m x 76.43 ≈ 2 mm.

Perform the verification:

RN2Cu∆Z = ∑∆ZiJ C∆ZiJ (6.108)

RN1

3. The corrected altitude differences ∆ZiJ (column 7):

∆ZRN1.61 = ∆ZRN1.61 + C∆ZRN1.61 = 0.408 + 0.001 = + 0.409 m;

∆Z61.62 = ∆Z61.62 + C∆Z61.62 = 1.462 +0.003 = + 1.465 m;

∆Z62.RN2 = ∆Z62.RN2 + C∆Z62.RN2 = - 0.072 + 0.002 = - 0.070 m.

Verifying this step is done in the following way:

RN2∑∆ZiJ = ∆ZRN1.RN2 (7.108)RN1

The absolute heights of the points 61 and 62 will be:

Z61 = ZRN1 + ∆ZRN1.61 = 354.231 + 0.409 = 354.640 m; (8.108)

Z62 = Z61 + ∆Z61.62 = 354.640 + 1.465 = 356.105 m.

Final verification: ZRN2 = Z62 + ∆Z62.RN2. (9.108)

109. Usually, the horizontal distances (DiJ) necessary to adjust the middle geometric

leveling traverse are determined indirectly (tacheometrically). In what follows, there will

be presented the operation procedure for this case. The unknowns of the problem are the

absolute heights of the points 86 and 87.

Table 1.109 offers the necessary data for determining these heights.

Solution:

Except the horizontal distances (columns), which are determined using the relation:

DiJ = K · H + 100 (CiS- Ci

J) (1.109)

86

the other elements – the plan section schema, the computational steps, etc. – are similar to

those from the previous case.

It should be specified that, for a given leveling, the distance that represents the weight in

adjustment results as sum of the distances D’iJ (device – backward point) and D”iJ (device

– forward point).

Thus: DiJ = D’iJ + D”iJ. (2.109)

Table 1.109

Sta

tion

Aimed point

Rod readings(mm)

Altitude differences

Hor

izon

tal d

ista

nce

sD

iJ (

m)

Ab

solu

te h

eigh

ts

Z(m

m)

Poi

nt

Backward Forward Coarse Corrections CorrectedCS CS ∆ZiJ

(m)C∆ZiJ (mm) ∆ZiJ (m)

CM = a CM = bCJ CJ

1 2 3 4 5 6 7 8 9 10

S1

RN7

1942RN71582

1220

862651

8623452040

S2

861652

8615001348

872062 + n

871902 + n1742 + n

S3

870970

8706940420

RN8

1872RN81646

1420

110. A possibility to increase the precision of middle geometric leveling traverse is to use

the double horizon of the instrument. This method was used to collect the necessary data

87

for determining the absolute heights of the points 112 and 113 (table 1.110). Process the

data from the mentioned table and determine Z112 and Z113.

Table 1.110

Sta

tion

Aimed point

ForwardBackward

Rod readings Altitude differences

Hor

izon

tal

dis

tan

ces

DiJ (

m)

Ab

solu

te

hei

ghts

Z(m

m)

Poi

nt

Horizon I

Horizon II

Coarse Corrections Corrected

a‘b‘

(mm)

a“b“

(mm)

∆Z’ + ∆Z”∆ZiJ = -------------

2(m)

C∆ZiJ

(m)

∆ZiJ

(m)

1 2 3 4 5 6 7 8 9 10S1 RN9 1651 1843 112.43 354.752 RN9

112 0972 1166 112S2 112 2647 2839 78.17 112

113 0960 + n

1162 + n

113

S3 113 1751 1941 72.63 113RN10 3043 3236 355.830 RN10

Solution:

Figure 1.110. Middle geometric leveling traverse with two horizons of the

instrument. a. Section schema; b. Plan view.

The working steps are the same as in the case of using only one horizon (problem 108).

The particularity of the method consists in the computation of coarse altitude differences.

For example:

∆Z’RN9.112 = a’1 – b’1 = 1.651 – 0.972 = + 0.679 m (1.110)

∆Z”RN9.112 = a”1 – b”1 = 1.843 – 1.166 = + 0.677 m;

∆Z’ + ∆Z” 0.679 + 0.677∆ZRN9.112 = --------------- = ------------------- = 0.678 m (2.110)

2 2

111. In the case of middle geometric leveling traverse with two horizons, too, it is

efficient to determine tacheometrically the distances necessary for the adjustment (see

problem 109). The data needed to compute the horizontal distance D iJ can be collected for

88

one horizon or for both. It is taken into account that the precision required for the values

DiJ is small (± 1 m). Table 1.111 contains the data required to compute the absolute

heights of the points 127 and 128.

Hint: As it can be seen, there are no fundamental differences between the values of the

horizontal distances for the two horizons. Thus, we can work with the stadimetric data

(CS, CJ) given by the first horizon. The horizontal distances will be computed similarly to

those determined in problem 109, and the absolute heights will be deduced by the

indications from problems 110 and 108.

Table 1.11

Sta

tion

Aimed point

Rod readings(mm)

Altitude differences

Hor

izon

tal d

ista

nce

sD

iJ (

m)

Ab

solu

te h

eigh

ts Z

(m)

Poi

nt

Backward Forward Coarse Corrections CorrectedCS CS ∆ZiJ

(m)C∆ZiJ (mm) ∆ZiJ (m)

CM = a’ b‘

CM = a“ b”

CJ CJ

1 2 3 4 5 6 7 8 9 10

S1

RN22

1621 1616351.637 – n (mm)

1470 14651319 1314

127

1872 18651700 1695

1530 1525

S2

1272076 2066

1754 + n 1744 + n1432 1422

128

1940 19291620 1609

1300 1290

S3

1281857 1900

1580 16241302 1345

RN23

0976 1019352.4300701 0744

0424 0467

89

112. For determining the heights of the points 261, 262 and 263, there was only one

benchmark of known height in the area. The necessary data for determining the absolute

heights Z261, Z262 and Z263 (table 1.112) were collected through a geometric leveling

traverse in closed circuit, using the benchmark RN27 as start and end point. Compute

these heights.

Remark: The computational steps are the same as in the case of middle geometric

leveling, supported at the ends (problem 108). The only difference consists in the

computation of the error:

RN27

e∆Z = ∑∆ZiJ (1.112)

RN27

Table 1.112

Sta

tion

Aimed point

ForwardBackward

Rod readings Altitude differences

Hor

izon

tal

dis

tan

ces

DiJ (

m)

Ab

solu

te

hei

ghts

Z

(m)Backward Forward Coarse Corrections Corrected

ai = …(mm)

bi = …

(mm)∆ZiJ (m)

C∆ZiJ

(m)

∆ZiJ

(m)1 2 3 4 5 6 7 8 9

S1 RN27 1961 – n (mm) 2874

121.43 367.122261

S2 261 07511242

71.15262

S3 262 16811047

83.43263

S4 263 20521277 143.17

367.122RN27

Figure 1.112. Middle geometric leveling traverse in closed circuit. a. Section schema;

b. Plan schema

113. In order to place an arena in a given area, there were performed a series of

measurements needed for location studies. The leveling survey of the aimed area was

performed through the method of small squares, with corners radiated through middle

geometric leveling (figure 1.113).

One leveling station (with two horizons) was enough for determining the required

heights. Only one leveling benchmark RN43 was in the area. The terrain not being rough,

90

there were used squares with 25 m sides. The readings performed in horizon I station can

be found at the numerator of the ratio from the corner of each square, and the horizon II

readings are at the denominator.

Figure 1.113. Surface leveling through small squares. The method of middle

geometric leveling radiation. Z43 = 360.270 m.

Solution:

The solving steps for this case are the following:

1. Compute the average of the readings for each point:

For example:

2042point RN43 -------- = 1831.5 ≈1832 mm;

1621

1871point 1 --------- = 1660.5 ≈1661 mm.

1450

etc.

2. Determine the average height of the aiming plan Zi:

Zi = Z43 + a43 (1.113)

Where Z43 = 360.270 m;

a43 = the average reading (1832mm);

Hence Zi = 362.102 m.

3. Compute the heights of the radiated points:

Z1 = Zi – b1 (2.113)

b1= the average reading (1661 mm) => Z1 = 360.441 m

The other heights are determined similarly.

4. Draft a new schema (figure 2.113) in which the heights of the measured points are

specified.

Figure 2.113. The heights of the corners of the small squares.

The schema transformed in this way has mainly two purposes:

- Tracing the contours (which are dealt with in problem 125);

- Determining the embankment volume (problem 161).

91

114. The method of large squares – middle geometric leveling – is used for the leveling

survey of large surfaces. Therefore, measure a surface of around 8 ha, for placing some

industrial objectives.

Figure 1.114 presents the elements measured in the field, which should be processed in

order to obtain the embankment volume (digging – filling up) in order to bring the terrain

to the specified foundation height.

Figure 1.114 Surface leveling through large squares

The following are given:

Benchmark of known height: ZA1 = 350.000;

Foundation height: ZF = - 2.50 m + n (cm);

± 0.00 height: Z + 0.00 = 350.500 m.

Solution:

Consider the main traverse in closed circuit:

A1.A2.A3.A4.B4.C4.D4.D3.D2.D1.C1.B1.A1, which is computed and adjusted through

the known method (problem 112).

After determining the corrected absolute heights of the mentioned points, compute the

heights of the intermediary points (B2, B3; C2, C3). For that, consider the secondary

routes B1 ÷ B4 and C1 ÷ C4 or A2 ÷ D2 and A3 ÷ D3, which, as it results from the

schema, are traverses supported at the ends. The computation of these traverses is

performed by the model presented in problem 108.

Taking into account the purpose of such works, we consider that the step of heights

adjusting is not indispensable.

In the next step, draft a schema that contains the heights of the corners of the squares

(similar to the one presented in figure 2.113). Based on this schema, compute the

embankment volume.

115. Based on the data presented in table 1.115 draw the longitudinal profile of the

measured area, between the points 46 and 47, on the distance scale 1:1000 and height

scale 1:100.

92

The heights of the points from the profile were obtained through middle geometric

leveling traverse (problem 108), and the distances between the points were measured

directly.

Hint: the longitudinal profile will be drawn by the model presented in problem 39.

Table 1.115

Point Absolute height Z(m)

Horizontal distancesDiJ (m)

Section

1 2 3 4

46 351.47238.76 46.121

121 352.16324.73 + n 121.122

122 350.07531.64 122.123

123 349.117 + n (m)

37.15 123.124124 352.106

32.43 124.4747 353.272

Figure 1.115. The leveling survey of a surface – through the longitudinal profile –

combined with transversal profiles. a. Section schema; b. Plan schema.

116. From the stations S1 ÷ S5 there were performed the necessary measurements for

drafting the transversal profiles. Based on the data presented in table 1.115, draft the

transversal profile 121 (PT.121), by the model presented in problem 40. The distance and

height scale is equal to 1:250.

Solution:

For each station there are two aims towards points of known heights (computed by

traversing). In this case:

Z’i = ZBENCHMARK46 + a’ (1.116)

1 (m)n * = ---------- · n

5

93

Z”i = ZBENCHMARK121 + a”

Z’i + Z”i Zi = ------------- (2.116)

2

The heights of the radiated points, from the transversal profile will be:

Z1= Zi – b1 (3.116)

Point Absolute height Z(m)

Horizontal distancesDiJ (m)

Section

1 2 3 4

1 351.7638.63 1.2

2 350.87510.72 2.121

121 352.1637.57 121.3

3 350.425 + n*

4.22 + n* 3.44 351.621

6.17 4.55 352.017

Hint: the transversal profile will be drafted by the model from figure 1.40.

117. The absolute plan coordinates of the points A and B are:

XA = 785.21 m + n (m); XB = 851.36 m;

YA = 572.43 m – n (m); YB = 675.26 m.

Determine the height of point B, through trigonometric leveling, knowing that:

i = 1.63 m + n (cm); S = 3.05 m – n (cm);

ZA = 352.47 m; φ’ = 12g51c12cc + ncc

Solution:

DAB = √∆X2AB + ∆Y2

AB

DAB = √(XB - XA)2 + (YB -YA)2

DAB = 122.27 m.

Figure 1.117. Trigonometric leveling on large distances

Theodolite, butterfly signal beacon

From figure 1.117 there results the equality:

94

h + i = s + ∆ZAB => ∆ZAB = h + i – s; (2.117)

hBut tgφ’ = -------- (3.117) => h = DAB tgφ’ => ∆ZAB = h + i – s (2’.117)

DAB

Will result: h = 24.343 m and ∆ZAB = 22.923 m.

In the end: ZB = ZA + ∆ZAB = 375.393 m. (3.117)

118. Determine the height of point B, given that:

XA = 621.58 m; XB = 521.26 m; φ = - 8g43c27cc – nc;

i = 1.27 m + n (cm); S = 2.72 m; ZA = 357.21 m.

Solution:

In this case:

h + s = i + ∆ZAB (1.118)

∆ZAB = h + s – i (1’.118).

The other values are determined similarly to the previous case.

Figure 1.118. Trigonometric leveling on large distances with descending aim.

119. Which is the value of the absolute height of the point B, given that the absolute

height of the point A is ZA = 347.21 m and the following values were measured in the

field, through trigonometric leveling:

i = 1.46 m, s = 4.52 m – n (cm), φ = 8g61c + nc, DAB = 161.23 m.

Solution:

Proceed as in the case of problem 117, this time the horizontal distance DAB being known

(through direct or indirect measurement).

120. Aiming, in point B, the sign that marks on the range pole the height of the

instrument from station A, the declivity angle of the terrain between this two points φ =

10g58c19cc – ncc was measured. There were also measured i = 1.63 m, DAB = 143.15 m + n

(m). Compute ∆ZAB, given ∆ZA = 364.172 m.

Solution:

In this case s = i (the aiming height is equal to the height of the instrument in the station.

Replacing that, the relation from problem 1.117 can be used).

95

121. If φ’ = - 6g61c23cc + ncc and the other data are presented in problem 119, which is the

absolute height of point B?

Solution:

Use the schema and relations presented in problem 118.

122. If φ = - 8g12c61cc – ng and the other data are presented in problem 120, which is the

absolute height of point B?

Solution:

In this case s = i; thus, with this change, the relations presented in problem 118 can be

used.

123. Compute the height of point B through the method of tacheometric leveling,

knowing the following elements: ZA = 343.262 m + n (m); φ’ = 8g61c27cc, the rod readings

CS =1971 mm, CM = 1752 mm, CJ = 1533 mm, K = 100, i = 1.752m.

From the figure it results:

∆ZAB

tgφ = ---------- => ∆ZAB = DAB tgφ (1.123)DAB

L = KH’ (2.123)

cosφ = H’/H (3.123)

=> H’ = H cosφ (3’.123)

Thus L = KH cosφ (2’.123)

Since DAB = L cosφ (4.123)

=> DAB = KH cos2φ and replacing it in the relation

=> ΔZAB = KH sinφ cosφ (1’.123)

Figure 1.123. Tacheometric leveling

For the problem data: ΔZAB = 100 (1.971 – 1.5330 · sin8g61c27cc · cos8g61c27cc) = 5.854

m.

In the end, it will result ZB = ZA + ΔZAB = 349.116m (5.123)

124. Fill in the following table of tacheometric measurements:

96

Solution:

ωi = αi – α23 = αi;

φi = 100g – Vi; (1.124)

DiJ = KHcos2φ; (2.124)

H = CS – CJ; (3.124)

K = 100;

Zi = Z22 + ΔZi22.

It can be seen that, having the side 22.23 as basis, the radiated points are determined both

planimetrically and by leveling.

Table 1.124.

Sta

tion

Aim

ing

poi

nt

Readings on the

bearing circle

αi

Position I

Readings on the

clinometer Vi

Position I

Rod readings

Hor

izon

tal

angl

e ω

i

Ver

tica

l an

gle

φi cos2φ

Hor

izon

tal

dis

tan

ces

DiJ (

m)

Alt

itu

de

dif

fere

nce

s ∆

ZiJ

Ab

solu

te h

eigh

ts

Z

Poi

ntCS

CM sinφ·cosφCJ

g c g c (mm) - (m) (m) (m)1 2 3 4 5 6 7 8 9 10 11 12

I =

1.4

32 m

, k =

100

22

23 00.00 - - - - - 47.61 - 350.437 22

27 37g12c 92g76c1751

2715901439

28 42g63c 112g31c1648

2814321216

29 61g17c 124g71c1682

2914321181

23 00.02 - - - - - - - 350.617 23

125. Interpolate the contours on the height plane presented in figure 1.125.

Solution:

The simplest interpolation method uses the isograph as working instrument (figure

2.125).

97

The isograph consists of a piece of tracing paper (30 x 10 cm), on which parallel lines are

drawn. These are numbered increasingly starting from the base line.

Figure 1.125. Height plan (without parametric details)

Scale 1:1000

Using the isograph, the interpolation is done in the following way (figure 3.125):

- Put the isograph with the corresponding gradation on the first point (1);

- Rotate the isograph around this point (fixed with a pin), until it reaches in

front of the next point (2) with the corresponding gradation;

- Intersect the ruling of the isograph with the alignment 12, obtaining the

intermediary height points.

Proceed similarly for the other pairs of points.

Attention: perform interpolation between the pairs 12, 23, 34, 56, … but not between 15,

26, 16, etc.!

Then, unite the points with equal heights through contours.

Figure 3.125. Contour plan derived from the height plan.

98