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Curs topografie generala - limba engleza - Ghe. Radulescu
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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