UNIT- 1 TRIANGULATION
➢Difference between Geodetic and Plane Surveying
Triangulation
• Triangulation is the process of determining the location of a
point by measuring angles to it from known points at either end
of a fixed baseline, rather than measuring distances to the point
directly (Trilateration).
• Method of determining distance
geometry
based on the principles of
• A distant object is sighted from two well-separated locations.
• Utilizes geometric figures composed of triangles.
• Horizontal angles and a limited no. of sides (Base lines) are
measured.
• Base Line – The first line measured precisely.
• The length of the last triangle side is used to check with the
computed one.
• By using angles and base line length, triangles are solved
trigonometrically and the positions of stations are calculated.
Advantage
• Horizontal Control – Fixation of framework of triangles for a survey
• Suitable for impossible direct measurements
• Suitable for hilly areas and undulating area with building towers.
Disadvantage
Accumulation of errors in length and directions of lines
FIELD OPERATION
1. Reconnaissance
2. Station preparation
3. Base line measurement
4. Measurement of angles
5. Other specification that includes design of station and
signals, reduction and adjustment of observations, some
geodetic practices and office works.
• Important Formula in Trigonometry
• Sine law
• Cosine law
GEOMETRY OF FIGURES
1.Triangles
2. Quadrilateral
3. Polygons
1. Triangles: It is simple, rapid and economical
1. Triangles
➢It is the most economical and very rapid figure to be used when a narrow strip of
terrain is to be surveyed like river, valley, etc.
➢The accuracy of the survey is however compromised due to which frequent check
bases need to be provided.
2. Quadrilaterals
➢It is the most accurate and the strongest system. The various combinations of
sides and angles can be used to compute the lengths of the required sides and
checks can be made frequently.
3. Polygons
➢This system is used when a vast area in all directions is required to be covered.
The polygons may or may not have a central station.
➢Though this system provides checks on the accuracy of the work, it is not as
strong as the quadrilateral arrangement. The progress of the work is also quite
slow
2. Quadrilateral : These are best suited for hilly areas
Most accurate system as the number of
checks are more.
• Polygons:
• The progress of the work is slow because of more setting of the
instruments
• It provides better checks and better triangles can be selected than
quadrilateral
WELL CONDIT IONED TRIANGLE
➢ A well conditioned triangle is a triangle in which any error in angular measurement
has a minimum effect upon the computed lengths.
➢Mathematically, an isosceles triangle with base angle of 56° 14’ is preferred. From
practical considerations, an equilateral triangle may be treated as a well conditioned
triangle.
➢In actual practice, the triangles having an angle less than 30° or more than 120° is
not permitted.
CRITERIA FOR SELECTION OF TRIANGULAR
STATIONS
CRITERIA FOR SELECTION OF TRIANGULAR STATIONS
• Should be intervisible. For this purpose the station points should be
on the highest ground such as hill tops, house tops, etc.
• Easily accessible with instruments
• Form well-conditioned triangles.
• Located such that the length of the sights are neither too short nor
too long.
• In wooded country, the station should be selected such that the cost
of clearing and cutting, and building towers is minimum.
• No line of sight should pass over the industrial areas to avoid
irregular atmospheric refraction
Triangulation figures or system
• It may be defined as a system consisting of triangulation stations connected
by chains of triangles.
• The complete figure is called “ triangulation system or figure”
• The most common types of figure used in triangulation systems are
triangles, quadrilaterals and polygons
STRENGTH OF FIGURE
➢ It is a factor to be considered in establishing a triangulation system to
maintain the computations within a desired degree of precision.
➢Strength of figure, thus defined as a figure which gives the least errors
in the calculated length of last line in the system due to the shape of
triangle and the composition of figures.
➢It is observed that the sine of small angles change more rapidly than the
sine of large angles. Due to this reason, it must be ensured that the
smaller angles is not opposite to the side which needs to be computed.
➢The expression given by the U.S coast and Geodetic survey for
evaluation of the strength of figure.
•
Note: Smaller the value of R, greater the strength of Figure
( D = Δ A 2 + Δ AΔ B + Δ B 2 )
10 0
12 0
14 0
16 0
18 0
20 0
22 0
24 0
26 0
28 0
30 0
35 0
40 0
45 0
50 0
55 0
60 0
65 0
70 0
75 0
80 0
85 0
90 0
95 0
100 0
105 0
110 0
115 0
120 0
125 0
130 0
135 0
140 0
145 0
150 0
152 0
154 0
156 0
158 0
160 0
162 0
164 0
166 0
168 0
170 0
10 0
428
359
315
284
262
245
232
221
213
206
199
188
179
172
167
162
159
155
152
150
147
145
143
140
138
136
134
132
129
127
125
122
119
116
112
111
110
108
107
107
107
109
113
122
143
12 0
359
295
253
225
204
189
177
167
160
153
148
137
129
124
119
115
112
109
106
104
102
100
98
96
95
93
91
89
88
86
84
82
80
77
75
75
74
74
74
74
76
79
86
98
14 0
16 0
18 0
20 0
253
214
187
168
153
142
134
126
120
115
106
99
93
89
86
83
80
78
76
74
73
71
70
68
67
65
64
62
61
59
58
56
55
54
53
53
54
54
56
59
63
71
187
162
143
130
119
111
104
99
94
85
79
74
70
67
64
62
60
58
57
55
54
53
51
50
49
48
46
45
41
43
42
41
40
40
41
42
43
45
48
54
143
126
113
103
95
89
83
79
71
65
60
57
54
51
49
48
46
45
43
42
41
40
39
38
37
36
35
34
33
32
32
32
32
33
34
35
38
42
113
100
91
83
77
72
68
60
54
50
47
44
42
40
38
37
36
34
33
32
31
30
30
29
28
27
26
26
25
25
26
26
27
28
30
33
Table for determining relative strength of triangulations figures ( Values of d = δA 2 + δAδB + δB2)
22 0
24 0
26 0
28 0
30 0
35 0
40 0
45 0
50 0
91
81
74
68
63
59
52
47
43
39
37
35
33
32
30
29
28
27
26
25
25
24
23
22
22
21
21
20
21
21
22
23
25
27
74
67
61
57
53
46
41
37
34
32
30
28
27
25
24
23
22
22
21
20
19
19
18
18
17
17
17
17
18
19
21
22
61
56
51
48
41
36
32
29
27
25
24
23
21
20
20
19
19
18
17
17
16
15
15
14
14
14
14
16
16
19
51
47
43
37
32
28
26
24
22
21
19
18
17
16
16
15
14
14
13
13
12
12
12
12
12
13
15
16
43
40
33
29
25
23
21
19
18
17
16
15
14
13
13
12
12
11
11
10
10
10
10
10
11
13
33
27
23
20
18
16
14
13
12
11
10
10
9
9
8
8
7
7
7
7
7
7
8
9
23
19
16
14
12
11
10
9
8
7
7
6
6
6
5
5
5
5
5
5
5
6
16
13
11
10
9
7
7
6
5
5
4
4
4
4
3
3
3
4
4
4
11
9
8
7
6
5
4
4
3
3
3
3
2
2
2
2
3
3
55 0
60 0
65 0
70 0
75 0
80 0
85 0
90 0
8
7
5
5
4
3
3
2
2
2
2
2
2
2
2
2
5
4
4
3
2
2
2
1
1
1
1
1
1
1
4
3
2
2
1
1
1
1
1
1
1
1
2
2
1
1
1
1
0
0
0
1
1
1
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
RELATIVE STRENGTH OF TWO OR MORE POSSIBLE
SYSTEMS OF FIGURES IS
𝐷−𝐶
2
2 𝐷−𝐶
R=
∑δA + δAδB + δB )=
∑d
𝐷
𝐷
Distance angle A = angle opposite to known
side.
Distance angle B = angle opposite to the side
which is to be computed.
Hence, the difference courses of routes are
compared and the one which has the lowest
value of R, is chosen.
24
EXAMPLE-1
DONE
• In a triangle ABC, angles A,B,C were observed as 700,500 and 600.
The line AC was used as a base of known length. Calculate the
strength of figure.
EXAMPLE-1
• In a triangle ABC, angles A,B,C were observed as 700,500 and 600. The line AC
was used as a base of known length. Calculate the strength of figure.
Soln: C= ( n’ – s’ + 1) + (n - 2s + 3)
𝐷−𝐶
𝐷−𝐶
∑δA2 + δAδB + δB2 )=
∑d
𝐷
𝐷
n=number of lines=3
R=
s=total number of station=3
(d = δA2 + δAδB + δB2)
nʹ= number of lines observed in both directions(lines observed in one direction
only are usually shown by broken lines) =3
s’=number of stations occupied=3.
D = number of direction observed, excluding those for the known side of a given
figure= 6 - 2 = 4
C= ( 3 – 3 + 1) + (3 - 6 + 3)=1
For angle A = 500 and B = 700 , d= +5
Strength of figure, R =
𝐷−𝑐
𝐷
X ∑d =
4−1
4
X 5=3.75
26
EXAMPLE 2
DONE
• In the figure the side of AC was known and BD was computed; the
angles measured have been shown in the figure. Calculate the
strength.
B
430
A
470
240
660
690 C
560
340 210
D
EXAMPLE 2
• In the figure the side of AC was known and BD was computed; the angles
measured have been shown in the figure. Calculate the strength.
B
430
A
470
240
660
690 C
560
340 210
D
28
SOLUTION
• There can be several routes to compute the strength. The one which gives the
minimum strength will be opted.
B
n=6 n’=6 s=4 s’=4
C= ( n’ – s’ + 1) + (n - 2s + 3)
C= ( 6 – 4 + 1) + (6 – 2 x 4 + 3)=4
Strength of figure,
R=
𝐷−𝑐
𝐷
=
10 −4
10
A
470
560
240
660
690 C
BD computed
D= 6 x 2 – 2 =10
430
340 210
=0.6
D
B
A
ROUTE-1
A (non effective angle kept
very small)
B
B
C
a
Distance angle kept large
for
accurate computation of
sides
430
A
470
560
240
6600
69 C
C
340 210
D
• Using triangle ACD containing measured base AC and Triangle
ADB containing BD to be computed.
• For triangle ACD, the distance angle are 550, 690; and d=4
• For triangle ADB, the distance angle are 430, 1030; and d=4
•  ∑d = 4 + 4 = 8 and
• R1 = 0.6 x 8 = 4.8
30
B
430
ROUTE-2
A
470
240
660
690 C
560
• Using triangle ACD and DCB.
• For triangle ACD, the distance angle are 560, 550; and d=7
• For triangle DCB, the distance angle are 240, 1350; and d=17
340 210
D
•  ∑d = 7 + 17 = 24 and
• R2 = 0.6 x 24 = 14.4
31
B
430
ROUTE-3
A
470
240
660
690 C
560
340 210
• Using triangle ACB and ABD.
• For triangle ACB, the distance angle are 660, 670; and d=3
D
• For triangle ABD, the distance angle are 340, 1030; and
d=8
•  ∑d = 3+ 8 = 11 and
• R2 = 0.6 x 11 = 6.6
32
B
430
ROUTE-4
A
470
560
• Using triangle ACB and BCD.
• For triangle ACB, the distance angles are 670, 470; and d=7
240
6600
69 C
340 210
D
• For triangle BCD, the distance angle are 210, 1350; and d=26
•  ∑d = 7+ 26 = 33 and
• R2 = 0.6 x 33 = 19.8
• Hence the best route for calculating BD is the first one.
33
QUESTIONS
DONE
1. Compute the value of (D-C)/D for the triangulation figures below. All the
stations have been occupied and the directions observed are shown by the
arrows.
2. Compute the strength off the figure ABCD for all the routes by
which the length CD can be computed from the known side AB.
Assume that all the stations were occupied.
DO THIS LATER FOR PRACTICE
STATION MARKS
• The triangulation stations should be permanently marked on the ground so that
the theodolite and signal may be centered accurately over them.
• Stations are permanently fixed so that it can be identified for future reference.
• On rock, a hole of 10-15 cm depth is made in the rock and copper or iron bolt or
plate is fixed with cement. The plate should contain the information regarding the
station.
• On ground, two marks are used, one on the ground and the other below it. A large
stone fitted with a brass screw is buried in a hole of about 1m deep and embedded
in cement mortar. Another similar stone is placed on the ground. A pole is erected
at the station and a heap of conical stones placed around the pole (cairn).
SIGNALS AND TOWERS
➢A signal is a device erected over the station mark. It should be properly
centered over the station mark to avoid errors.
➢A good signal should fulfill the following criteria.
- It should carry flag on its top.
a. The signal should be clearly visible against any background.
b. It should be of sufficient height, vertical and accurately centered over the
station mark.
c. It should be of suitable size for accurate bisection.
d. It should be easy to erect in minimum time.
e. It should be free from phase error or should exhibit minimum phase error.
SIGNALS AND TOWERS
Classification of signals.
1. Opaque/daylight/non luminous signals. (for distances up to 30 km)
2. Luminous signals (for distances more than 30 km) H.W
1. Daylight signals.
➢These signals are of various types and following are some of the most
commonly used signals.
a. Pole signal- it consist of a round pole painted black and white in alternate
strips and supported vertically over the station mark on a tripod. They are
suitable for a distance of about 6 km.
SIGNALS AND TOWERS
b. Target signal- it consist of a pole carrying two squares or rectangular
targets placed at right angles to each other. The targets are generally made
of cloths stretched on wooden frames. They are suitable for a distance of
about 30 km.
c. Pole and brush signal- it consist of a straight pole about 2.5 m long with
a bunch of long grass tied symmetrically round the top making a cross.
The signal is erected vertically over the station mark by heaping a pile of
stones up to 1.7 m round the pole.
d. Stone cairn- it is a pile of heaped stone in a conical shape about 3 m
high. A cross shaped signal is erected over the stone heap.
e. Beacons- it consist of a red and white cloth tied round the three straight
poles.
SIGNALS AND TOWERS
➢A tower is a structure which is erected when the station or the signal or both
are to be elevated to make the observations possible from other stations in
case of problem of intervisibility.
➢Towers usually have two independent structures, the outer one and the inner
one.
➢The outer one is for supporting the observer or the signal and the inner one
is for only the instrument.
➢The inner and outer are made independent of each other so that the
movement of the observer does not disturb the setting of the instrument.
➢Towers are usually made of timber, steel or masonry depending on the
height.
INTERVISIBILITY OF TRIANGULAR STATIONS
• Case I – Intervisibility not obstructed by intervening ground
• Case II- Intervisibility obstructed by intervening ground
• Alternate method by captain G.T McCaw’s
INTERVISIBILITY OBSTRUCTED BY
INTERVENING GROUND
ALTERNATE METHOD BY C APTAIN G.T MCC AW’S
QUESTIONS
1. An observer standing on the deck of a ship just sees a lighthouse top
with his eyes at a height of 9 m. the top of the lighthouse is 64 m
above m.s.l. Find the distance of the observer from the lighthouse.
2. Two stations A and B , 80 km apart have elevations 15 m and 270 m
above m.s.l respectively. Ascertain if the stations are intervisible and
if not, calculate the minimum height of the signal at station B if the
line of sight is to be taken at least 3 m above the point of tangency.
3. There are two stations, P and Q at elevations of 200 m and 995 m
respectively. The distance of Q from P is 105 km. If the elevation of
a peak M at a distance of 38 km from P is 301 m, determine whether
Q is visible from P or not.
4. Solve question 3 using captain McCaw’s method.
SATELLITE STATION AND REDUCTION TO CENTER
➢Sometimes, to form well conditioned triangles and to obtain better visibility,
objects such as flag poles, towers, lighthouse, chimneys etc. are selected as
the triangulation station.
➢It is possible to sight these stations from other stations and make the
measurements but it is not possible to occupy these stations and make
measurements from there.
➢During this time, another station, very near to the inaccessible station is
chosen and measurements are made from there. Such a station is known as a
satellite station.
➢After carrying out the measurements from the satellite station, the true angle
which would have been observed from the inaccessible station is computed.
This is called reduction to center.
DIFFERENT POSITION OF SATELLITE STATION AND
REDUCTION TO CENTER
SATELLITE STATION –POSITION OF SATELLITE
STATION-WEST SIDE
TH E VALUE OF Ø IS TH E C ORREC TED VA L UE OF Ø WH EN
SA TELLITE STA TIONS “ S” IS TH E MA IN STA TION “C ”
In the general, the following four cases as shown
Case I: S towards the left of C
THE VALUE OF Ø IS TH E C ORREC TED VA L UE OF Ø WH EN
SA TELLITE STA TIONS “ S” IS TH E MA IN STA TION “C ”
In the general, the following four cases as shown
Case II: S towards the right of C
TH E VALUE OF Ø IS TH E C ORREC TED VA LUE OF Ø WH EN
SA TELLITE STA TIONS “ S” IS TH E MA IN STA TION “C ”
In the general, the following four cases as shown
Case III: S inside the triangle ABC
THE VALUE OF Ø IS TH E CORRECTED VALUE OF Ø WH EN
SA TEL L ITE STA TION S “ S” IS TH E MA IN STA TION “ C ”
In the general, the following four cases as shown
Case IV: S outside the triangle ABC
Questions
1. From a satellite station S, 15 m west of triangulation station A, the angles
measured to three stations are as below. The lengths of sides AC and AB
are 5806 m and 1633 m respectively. Calculate the angle BAC.
angle CSA= 35°12’55’’
angle BSC= 66°38’40’’
2. Directions were observed from a satellite station S, 80 m from C, with the
following results. The approximate lengths of AC and BC are 17 km and
24.15 km respectively. Compute the angle ACB.
A-00°00’00’’
B- 72°50’44’’
C- 299°22’00’’
Questions
3. From a satellite station S, 5.8 metres from the main triangulation station A, the
following directions were observed.
A – 0o 0’0’’
B - 132o 18’30’’
C - 232o 24’6’’
D - 296o 6’11’’
The lengths AB, AC and AD were computed to be 3265.5 m, 4022.2 m and 3086.4
m respectively. Determine the directions of AB, AC and AD.
4. A, B and C are the stations in a minor triangulation survey. A satellite station S is
set up near C such that AC and BC fall within triangle ASB. It is given that AC=7.5
km, BC=6.3 km, CS=40.5 m angle ASC= 62°10’40’’and angle ASB= 75°15’15’’.
Calculate angle ACB.
THE END