INTRODUCTION

Surveying is the art or science of making metric measurements of natural and man made
features in an effort to produce a map or plan. Mine surveyors plan the development and
future of mineral workings. The are two branches of surveying namely geodetic and
plane surveying. Plane surveying assume that the earths surface is a plane (flat), the
effects of the earths curvature is ignored as corrections are negligible due to the sense that
they are only concerned with small areas. Plane surveying calculation are done using
principles of plane geometry / trigonometry. In geodetic surveying the earths curvature is
taken into account as it covers large areas and is normally done to establish control for a
country and are very expensive to undertake. Surveying is divided into the following
groups---
Cadastral Survey -for the establishment of property boundaries.
Mining Survey - for the control of mining operations and volumetric
analysis (underground tunneling)
Hydrographic survey- these are those surveys done onshore or offshore with the
objective of fixing the position of the survey vessel to determine
depth of water and to map out sea surface topography.
Photogrammetric - done using aerial photographs taken with special cameras from
aircrafts.
Topographic Surveys –done to establish positions of natural and man made features over
a given area for the production of maps and plans. (Also show the
nature of the terrain)
Engineering Surveys -done with the objective of facilitating the implementation of
engineering projects i.e these are surveys done before, during and
after an engineering project.
SURVEY PRINCIPLE
Working from whole to part – „from whole”-means the laying of monuments to control
by very accurate means and “the part”- the survey of detail by less elaborate means but
without compromising accuracy. The area in question (country) is divided into
triangles/networks, which are further subdivided into smaller triangles, which can be
surveyed with lesser degree of accuracy.
ERRORS
In any measurement the 1ⁿis a deviation due to circumstances beyond the observers
ability- this is an error in the observation. In practice there is no exact measurement, all
measurements are prone to errors. In surveying errors are classified into three groups
namely gross errors- which are caused by the observer‟s inability, inexperience e.g
wrong bookings, systematic errors-these tend to follow a defined pattern such that if the
measuring process is repeated under the same conditions the same error will reoccur e.g
the effects of temperature change on the instrument and random errors-these are those
errors that remain after gross and systematic errors has been accounted for and they are
unavoidable and stochastic so probability models are used to correct them.
TERMS
Accuracy- is the closest approximation to the true value (which cannot be ascertained)
Precision- is the closeness of repeated observations to each other regardless of accuracy
i.e it is an index of accuracy.
Error- is the difference between the observed value and the true quantity of an
observation.
MPV-the most probable value ,it is the arithmetic mean .This is the closest
approximation to the true value that can be obtained from a set of data.

1
COORDINATES CALCULATIONS
Joins –the computation done to obtain the distance and direction between two points with
known coordinates.
Example
Point Coordinates
A +4730.521 +5960.610
D +5140.995 +6560.889
Calculate join AD?

Solution
Join AD D +5140.995 +6560.889
t = 34.21.52 A +4730.521 +5960.610
s = 727.203m Δy +410.474 Δx +600.279
where: t is the direction between the two points
s is the distance between the two points
t = arc tan( Δy/Δx )
s = √-( ( Δy ²+ (Δx)²)
Polar-the computation done to obtain the coordinates of the second point given its
distance and bearing from the first point of known coordinates.
Example
Point Coordinates
A +4730.521 +5960.61
Direction A-D=34.21.52
Distance A-D=727.203m
Find the coordinates of D given the above data?
Solution
Polar AD
t=34.21.52 A +4730.521 +5960.610
s= 727.203m Δy +410.474 Δx +600.279
D +5140.995 +6560.889
NB=To get Δy=s Sin t
Δx = s Cost
From the above computation it can be seen that one can check a join calculation by
calculating a polar and the reverse is true.

2
TACHEOMETRY
Example one
Zenith angles where considered from a station A to a staff held vertically at station C. Given the following
data, calculate the horizontal distance AC and the reduced level of C.
Instrument@ Staff@ Zenith angles Middle hair readings
A C 87.30.40 0.230m
B C 86.52.10 1.526m
Height of instrument @ = 1.650m
Reduced level of A = 1426.11m
Solution Reduction of vertical angles from the given zenith angles
A-C = 90.00.00-87.30.40 = 02.29.20
B-C = 90.00.00-86.52.310 = 03.07.50
S = Sb-Sa = 1.526-0.230 = 1.296
H = S / (tan2-tan1) = 1.296 / (tan03.07.50-tan02.29.20)
=115.443m
Check H = S / (Cot 1-Cot 2) = 1.296 / (Cot 87.30.40-Cot 86.52.10)
= 115.443m
The horizontal distance = 115.443m
V = vertical component = H tan1 = 115.443tan02.29.20 = 5.018m
Check V = H Cot 1 = 115.443Cot 87.30.40 = 5.018m
Height difference = Hi+V-t
= 1.650+5.018-0.230
= 6.438m
Reduced level = 1426.110+6.634
= 1432.548m
Example Two
The following tacheometric observations were made with a theodolite set-up @ A and the
staff held vertical. Foliage partially obstructed the view when the following readings
were taken.
Staff@ Horizontal circle readings Zenith angles Lower Middle Upper
B 30.45.00 87.20.00 2.377 2.565 2.753
C 115.13.00 92.45.00 1.533 1.956 Obstructed
Height of instrument = 1.450m
Reduced level of A = 100.000m
K = 100
C = 0
a) Determine distances AB, AC and BC?
b) Determine reduced level of C given the reduced level of A as 1431.820m?
Solution Vertical angles A – B = (90.00.00-87.20.00) = 02.40.00
Vertical angle A – C = (90.00.00-92.45.00) = -02.45.00
S = Upper hair reading minus lower hair reading
Distance AB = 100 S Cos² 1
= 100(0.376) Cos²02.40.00 = 37.519m
Distance AC = 100 S Cos²2
=100 (00.846) Cos² –02.45.00=84.405m
Calculating the obstructed value: middle hair reading = (upper –lower)/2
1.956 = (upper-1.533)/2

3
 Upper = 2.379

Diagram
C

115.13.00
R.d
84.28.00
A

30.45.00
B
a²=b²+c²-2bc Cos A
To find angle BAC Horizontal circle reading to the right minus the one to the left.
a = 89.001m
Reduced level of C = reduced level of A HI+V-L
V = H tan  /H Cot Z
RLc = 1431.82+1.450-4.054-1.956 = 1427.26m
Example Three
A tachometric exercise was done and the following data obtained.
Inst@ HI Staff Station Horizontal reading Vertical reading Lower Middle Upper
A 1.780 C 29.24.00 07.24.00 1.050 1.450 1.850
B 1.670 D 352.36.00 09.12.00 1.590 1.990 2.410
Coordinates
A +796.560 +850.620
B +296.120 +321.650
Calculate length and bearing CD?
Solution Distance AC = 100 S Cos² 1
= 100( 0.8)Cos² 07.24.00
` = 78.673m
Distance BD = 100 S Cos² 2
= 100(0.82)Cos²09.12.00
= 79.904m
POLAR AC
t = 29.24.00 A +796.560 +850.620
s = 78.673m Δy +38.621 Δx +68.541
C +835.181 +919.161
NB # t = direction and s = distance.

POLAR BD
t = 352.36.00 B +296.120 +321.650
s = 79.904 Δy -10.291 Δx +79.238
D +285.829 +400.888

Join CD
T = 226.40.03 D +285.829 +400.888
s=755.245m C +835.181 +919.161
Δy -549.352 Δx -518.273
Distance CD = 755.245m

4
 Direction CD = 226.40.03

Methods used in tacheometry

a) Substense system
-the parallactic angles are measured with greater accuracy using horizontal circle instead
of vertical circle, as the measured base is held horizontal instead or vertical. It provides
millimeter accuracy for tacheometric systems although distances tend to move in
proportion to the square of the distance measured.
b) Serial method (Subclined sections)
-When the distance is measured in a number of equal sections to reduce central errors of a
single observation, the resulting error can be found by dividing the error of a single
observation by square root of n cubed where n is the number of sections used.

Diagram

4

T2
T1 H1 H2 H3 H4 T3

1 2 4

H = T1-T3 = H1+H2+H3+H4
H = b/2 ( Cot1/2+ Cot2/2+ Cot3/2+ Cot4/2)
Where b = length of substence bar.
c) Auxiliary base method-does not divide the length to be measured and it saves time (
this compensate for lack of accuracy of this method).
Diagram
b

T3
P


T1 T2

H(T1-T3) = b/2 (Cot /2)

H (T1-T2) = H (T1-T3) Cotβ
= b/2 (cot/2)x Cotβ

5
Example four
The following data was obtained by stadia observations, vertical angle was +8.10.00,
staff intercept (S) was 2.5m, stadia interval factor was known to be 100 and the distance
from center of instrument to principal axis (C) was 0.75m Calculate the horizontal
distance (H) from the peg to staff and the vertical distance (V)?
Solution H = KS Cos²  +C Cos = 100x250 Cos²8.10.00+0.75Cos8.10.00
= 245.0+0.7 = 245.7m
V =1/2KS Sin² C Sin=1/2x100x2.50 Sin²8.10.00+0.75 Sin08.10.00
=35.3m
LEVELLING (Definations)
Datum-is a reference surface or point from which all points or measurements are taken
from or reference to.
Level surface- a surface of constant elevation that is perpendicular to a plumbline at
every point i.e having the same altitude above the MSL.
Backsight-is the reading observed after setting up the instrument, usually to a point of
know elevation.
Reduced level-is the elevation of a point above datum (MSL).
Bench mark- a permanent and monumented point with known elevation (above or below
datum).
Change point- these are points where both a backsight and a foresight reading are taken
to provide continuity in leveling.
Intermidiate sight-the readings taken between the backsight and the foresight before an
instrument is moved.
Example one
Given the following fieldbook calculate the reduced levels of all the pegs and adjusting
the error by the HPC method showing all checks?
BS IS FS Reduced level Remarks
0.731 96.667 BMA
1.642 4.381 Peg 1
3.080 Peg 2
2.430 Peg 3
0.943 3.112 Peg 4
4.070 Peg 5
3.610 Peg 6
3.420 Peg 7
4.130 Peg 8
3.481 89.015 BMB
Solution
BS IS FS HPC RL Corre Reduced level Remarks
0.731 97.398 96.667 +0.00 96.667 BMA
1.642 4.381 94.659 93.017 +0.002 93.019 Peg 1
3.080 91.579 +0.004 91.583 Peg 2
2.430 92.229 +0.004 92.233 Peg 3
0.943 3.112 92.490 91.547 +0.004 91.551 Peg 4
4.070 88.420 +0.006 88.426 Peg 5
3.610 88.880 +0.006 88.886 Peg 6
3.420 89.070 +0.006 89.076 Peg 7
4.130 88.360 +0.006 88.366 Peg 8
3.481 89.009 +0.006 89.015 BMB
∑=3.316 ∑=10.974

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 ∑BS-∑FS = Last RL-First RL
3.316-10.974 = 89.009-96.667 = -7,658
Misclosure = 89.009-89.015 = -0.006m
Adjustment = +0.006m
Explanation of the method
Getting HPC
Where there is a backsight you calculate the HPC i.e the first HPC =RL or BMA plus the
first backsight=96.667+0.731=97.398
The second HPC= RL of peg 1 plus second BS=93.017+1.642=94.659
The third HPC=RL of peg 4 plus the third BS=91.547+0.942=92.490
Getting reduced levels
First RL = HPC 1-FS
2rd RL = HPC2-IS
RD RL
3 = HPC2-IS
4TH RL = HPC2-FS
5th RL = HPC3-IS
th RL
6 = HPC3-IS
7th RL = HPC3-IS
th RL
8 = HPC3-IS
BMB RL = HPC3-FS
Corrections
Count the number of either BS or FS or Change points e.g in the example above its 3
Correction to first set up =1/3x0.006 = 0.002
Correction to second set up = 2/3x0.006 = 0.004
Correction to third set up =3/3x0.006 = 0.006
Then add the corrections to their corresponding initial RL to get the final RLs.
Example two
Reduce the following fieldbook using the rise and fall method showing all checks where
necessary.
BS IS FS FRL REMARKS
58.031 BMA
1.231 A
1.612 B
2.587 3.132 C
1.565 D
1.911 E
0.376 F
2.244 1.522 G
3.771 H
1.334 1.985 I
3.462 J
2.147 K
1.623 L
2.222 M
2.002 56.174 BMB

7
Solution
BS IS FS Rise Fall R Level Corr FRL Remarks
0.599 58.031 +0.00 58.031 REMARKS
1.231 0.632 57.399 0.005 57.404 BMA
1.612 0.381 57.018 0.005 57.023 A
2.587 3.132 1.520 55.498 0.005 55.503 B
1.565 1.022 56.520 0.010 56.530 C
1.911 0.346 56.174 0.010 56.184 D
0.376 1.535 57.709 0.010 57.719 E
2.244 1.522 1.146 56.563 0.010 56.573 F
3.771 1.527 55.036 0.015 55.051 G
1.334 1.985 1.786 56.822 0.015 56837 H
3.462 2.128 54.694 0.020 54.714 I
2.147 1.315 56.009 0.020 56.029 J
1.623 0.524 56.533 0.020 56.553 K
2.222 0.599 55.934 0.020 55.954 L
2.002 0.220 56.154 0.020 56.174 BMB
∑= ….. ∑=…… ∑=…… ∑=……

∑BS-∑FS = ∑R-∑F = Last RL–First RL

The two peg test
Two rigid points A and B are marked on the ground and the instrument is set up exactly
between them at point C. Readings are taken on the staff held at A and B and the
difference between them gives the difference in level of the two pegs. The equality in
length of the backsight and foresight ensures that any instrumental error is equal on both
readings .The instrument will then be moved so that it is outside the line of the pegs and
close to one peg-the instrument station being in line with pegs A and B. Readings are
again taken on the staff held at points A and B
As illustrated below .Any discrepancy between the level differences given by the first
readings compared to that from the second readings is due to instrument, being out of
adjustment.
Diagram
S1 S2

Δh AB = S2 – S1 (the height
differnce)

S4 – S3 = Δh AB
A B (apparent height difference)
S4 S3

A B

8
Example three
Readings obtained from a two peg test carried out with an automatic level with a single
level staff set up alternatively at 2 pegs A and B placed 50m apart were as follows –with
level midway btwn A and B staff readings at A was 1.283m at B was 0.860m With level
50m from B on line with AB staff readings at A was 1.612m and at B was 1.219m.
Calculate the collimation error per 50m of sight, the readings that should have been
observed on the staff at A from the level in position 5m from B.
Solution
True height difference = 1.283-0.860 = 0.423m
Apparent height difference = 1.612-1.219 = 0.393m
Collimation error (x) = apparent –true = 0.393-0.423 = -0.03m
Collimation error (e) per 50m = -0.03/50 = 0.0006rads
The difference in height btwn A and B from staff reading when level is 5m from B than
when midway btwn A and B. This implies that the staff reading 1.612m is too low and so
collimation error is downwards. With the instrument set 5m from B a horizontal line of
sight would bisect -staff at A at 1.612+0.0006x55 =1.645 = S‟3
-staff at B would be S‟4 =1.219+0.0006x55 =1.252
Check (S‟3-S‟4) = (S3-S4) = (1.645-1.252) = (1.612-1.219) 0.393
TRAVERSING
-Distance is measured using a tape or electromagnetic distance measurers
-Directions are observed using the theodolite
-Traversing is the coordinating of points i.e. the measuring of direction and distance from
a known point in any effect to coordinate the points.
TYPES OF TRAVERSES
1) Open traverse- start from appoint of known coordinates and ends at a point of
unknown coordinates, it is used to control underground tunneling.
2) Closed traverse- there are two types, they start and close at known points.
a)Loop traverse- starts and closes at the same known point.
b)Link traverse- starts at a point of known coordinates and ends at another point
of known coordinates.
Computational process
Method 1- Bowditch
Coordinate the following traverse by the Bowditch method
LEG DIRECTION LENGTH
AB 318.11.20 275.315
BC 59.42.30 386.475
CD 354.35.50 409.860
DE 15.37.35 315.045
EF 234.49.05 443.960

Coordinates of known points

A -7621.992 +2224.895
F -7788.496 +3080.624

9
Solution

LEG DIRECTION LENGTH Δy Δx STN Y COORDINATES X

AB 318.11.20 275.315 -183.546 +205.205 A -7621.992 +2224.895
BC 59.42.30 386.475 +333.709 +194.939 B ----------- -----------
CD 354.35.50 409.860 -38.591 408.039 C ------------ ------------
DE 15.37.35 315.045 +84.862 303.400 D ----------- ------------
EF 234.49.05 443.960 -362.860 -255.799 E ----------- ------------
F -7788.418 +3080.679
∑= ------- δy ----------- δx ----------

Directional misclosure = 54.48.40

Magnitude misclosure = 0.095m
Proportional misclosure = 1:19270/1:19000
Bowditch adjustment

LEG Corrections to Corrected STN Y Coordinates X

▲y ▲x ▲y ▲x A -7621.992 +2224.895
AB -0.012 -0.008 -183.558 +205.197 B -7805.550 +2430.092
BC -0.016 -0.012 -333.693 +194.927 C -7471.857 +2625.019
CD -0.017 -0.012 -38.608 +408.027 D -7510.465 +3033.046
DE -0.013 -0.009 +84.849 +303.391 E -7425.616 +3336.437
EF -0.019 -0.013 -362.880 -362.880 F -7788.496 +3080.624
------ ∑----- δy ---------- δx ----------

Example 2
Scale and swing method
Coordinate the following traverse by the scale and swing method.
LINE DIRECTION DISTANCE
PB 48.26.32 187.493
AB 63.37.20 204.436
BC 81.22.54 265.527
CD 75.18.38 141.397
DS 82.55.49 200.638
Coordinates of final points
P +1751.632 +2087.189
S 2673.426 +2402.825
Solution

LINE DIRECTION DISTANCE ▲y ▲x Stn Y Coodinates X

PB 48.26.32 187.493 140.299 124.378 P +1751.632 +2 087.189
AB 63.37.20 204.436 183.151 90.828 A ------------- -----------
BC 81.22.54 265.527 262.529 39.790 B ------------- ----------
CD 75.18.38 141.397 136.775 35.855 C ------------- ----------
DS 82.55.49 200.638 199.113 24.694 D ------------- ----------
∑=-------- S” 2673.499 +2402.734

10
Calculating joints
Join PS”
t = 71.06.16. S” +2673.499 +2402.734
s = 974.375396197m P +1751.632 +2087.189
Δy +921.867 Δx +315.545

Join PS
t = 71.05.53. S +2673.426 +2402.825
s = 974.33580604 P +1751.632 +2087.189
Δy +921.794 Δx +315.636

Scale factor = 974.33580604/974.375396197 = 0.999959368682

Swing = 71.05.53-71.06.16 = -00.00.23
NB To get new direction = old direction +Swing
To get new distance = old distance. scale factor

LINE N-Direction N-Distance N-Δy N-Δx STN Y Coordinates X

PA 48.26.09 187.485 140.279 124.388 P +1751.632 +2087.189
AB 63.36.57 204.428 183.134 90.845 A +1891.911 +2211.577
BC 81.22.31 265.516 262.513 39.817 B +2075.045 +2302.422
CD 75.18.15 141.391 136.766 35.869 C +2337.558 +2342.239
DS 82.55.26 200.630 199.102 24.715 D +2474.324 +2378.108
∑=-------- S +2673.426 2402.823

Example 3
Transit method
A link traverse was run between station A and X as shown in the diagram below. The
coordinates of the controlling station at the ends of the traverse are as follows
Point Y(m) X(m)
A +1769.15 +2094.72
B +1057.28 +2492.39
X +2334.71 +1747.32
Y +2995.85 +1616.18

242.53.40 Y
B 281.12.40 2 3

168.19.10 1 173.31.00
115.37.00 X
80.26.20
A 4
Note to scale
Distances
A – 1 = 208.26m
1 - 2 = 198.47m
2 – 3 = 326.71m
3 – 4 = 309.15m
4 - X = 224.79m
Calculate the coordinates of station 1,2,3 and 4 adjusting the misclosure by the transit
method?

11
Solution
Calculate Join A-B = Direction= 299.11.20 and distance = 815.414m
Calculate Join X-Y = Direction=101.13.10 and distance = 674.021m
Calculating angular misclosure
∑ of angles = (θ XY- θ AB ) + (n-1) 180
1061.59.50 # 1062.01.50
There is a misclosure of 00.02.00
Correction to each angle is 00.00.20
Apply the correction by the Bowditch method, which you are now familiar
with.

Line Corrections to Corrected Stn Y COORDINATES X

Δy Δx Δy Δx A +1769.15 +2094.72
A-1 0.02 0.01 ------------- 1 +1939.37 +2214.74
1-2 0.02 0.01 ------------- 2 +2071.67 +2355.93
2-3 0.02 0.02 ------------- 3 +2262.09 +2090.46
3-4 0.02 0.02 ------------- 4 +2120.54 +1815.65
4-X 0.02 0.00 ------------- X +2334.71 +1747.32
∑= -- ∑=--- δy ------- δx --------

Example 4
Coordinate the following traverse by any method
LEG QUADRANT LENGTH (M)
AB S 47.08.54 W 194.822
BC S 59.46.53 W 382.158
CD N 03.16.03 E 299.456
DA N 81.21.42 E 172.284

Coordinates of known point A –500.00 +2000.00

Solution
Hint ---Sketch the traverse
Calculate coordinates of B,C.D and adjust the misclosure ay method.

12
SHAFT PLUMBING
 The taking of survey control underground or down a slope.
Methods of transferring control
a) Wire surveys
- High tensile steel wires of optimum length of 300m.
- The lower end has a weigh of about 1kg.
- Leave the wire for 24hrs to allow oscillation to be negligible.
- The wire is immersed in a barrel of oil to dampen the oscillation.
Reasons for out of plumb of wire
i. Pendulum motion (oscillation).
ii. Magnetic infusion on the wire (magnetic influence).
iii. Displacement due to air currency.
b) Co-planning (alignment)

 red

Blue



Instrument station
Instrument should be in line with the wires red and blue and observations are taken.
c) Weisback triangle
W¹ a = b
r Sin q sin r

W²

q RO

I
Example
A double Weisback connects two coordinated points A and B
triangle to underground points U and V.
Using the data below;
i. Evaluate angles AQR and UPQ.
ii. Determine direction BA, AP, PQ, QU and UV
iii. Coordinate traverse APQUV
A +549,30 +1163,02
B +638,15 +1174,82

13
 V

A
B

Observed horizontal angles P

BAQ = 221.18.16
QAP = 29˝
QUP = 40˝
PUV = 218.49.10
Distances: - BP = 6,325: PQ = 8.00: QU = 6,952: UV = 89,632
Solution Join BA
262.26.06 A +549,30 +1163,02
89,630m B + 638,15 +1174,82
y = -88,85 x -11,80
AQˆP
6,325 = 8,000
Sin  Sin 29˝
 = 0.00.23 = AQP
UPQ 6,952 = 8,000
Sin  Sin 40˝
UPˆQ =  = 0.00.35
Direction AP
Angle BÂP = Angle BÂP – QÂP = 221.17.47
Direction AP = (82.26.06 – 221.17.47) + 360.00.00
= 221.08.19
Direction PQ
Direction PA = 41.08.19
APˆQ = 180 – (0.00.23 + 0.00.29”) = 179.59.08
Direction PQ = 179.59.08 + 41.08.19
= 221.07.27
Direction QU
Direction QP = 41.07.27
PQˆU = 180 – (0.00.35 + 0.00.40) = 179.58.45
Direction QU = 179.58.45 + 41.07.27
= 221.06.12
Direction UV
Direction UQ = 41.06.12
QÛV = 218.49.10 – 40” = 218.48.30
Direction UV = (41.06.12-218.48.30)+360.00.00 = 182.17.42

14
Open traverse computation
Leg Direction Distance Y X STN Coordinates
AP 221.08.19 6,325m A ------- ------
PQ 221.07.27 8,000m P ------- ------
QU 221.06.12 6,952m Q ------- ------
UV 182.17.42 89,632m U ------- ------
 = ----- V ------- ------
Weis Quadrilateral
TRIG

A
1 
B
TRIG 

6
2
3
C 5
4
D
1. Set two arbitrary points to make a quadrilateral with A and B.
2. Set Theo at points C and D, measure angles 2,3,4 and 5 and deduce 1 and 6.
3. From the theory of braced quadrilaterals let x =  +  = 3 + 4
 = Sin  = Sin 1 Sin 3 Sin 5
Sin  Sin 2 Sin 4 Sin 6
 + Cos x = Cot 
Sin x
Get  and hence 
The advantage of the quadrilateral is that it gives coordinates and bearings.
Gyro Theodolite
 Used to determine the direction of North.
 The modes of operation are (i) unclamped mode, (ii) clamped and non-spinning
mode.
 It is used to measure the reversal points (i.e. gyro scale).
 The type of gyro is defined by attachment e.g. tape suspended system where the
gyro is below the theodolite.
Questions
The following results were from an underground set up at C
Angles W¹ĈW² = 00.12.36
W²Ĉ b = 197.10.18
DCW ¹ = 162.37.06
Distance C- W¹ = 1,556m
C - W² = 2,617m
W¹ - W ² = 1,061m
Coordinates of W² = +6021,177 +1907,376
Bearing W² to W¹ = 45.27.23
Calculate the coordinates of C and bearing C to D

15
Solution
Diagram
W²

1,061 

W¹


1,556
2,617
0.02.36

162.37.06 C 197.10.18
D

Calculate coordinates of W¹ by taking polar W² –W¹

Polar W² – W¹
Dist = 1,061m W² +6021,177 +1907,376
Direction = 45.27.23 y 0,756 x 0,744
W¹ 6021,933 1908,120
Note y = Dist Sin Direction
x = Dist Cos Direction
Find angle  using sine rule
1,556 = 1,061
Sin  Sin 0.12.36
 = 0.18.29
ˆ = 180.00.00 – (0.18.29 + 0.12.36)
= 179.28.55
To check angle ^
2,617 = 1,061 = 0 .31.05 but W¹ C W² is
Sin  Sin 0.12.36
aΔ so and  cannot be 0.31.05 but 179.28.55 (180.00.00 – 0.31.05)
Direction W ²– C  Direction W² – W¹ - angle 
= 45.08.54
Polar W² –C
2,617 (Distance) W² +6021,177 +1907,376
45.08.54 (Direction) y 1,855 x 1,846
C 6023,032 1909,222
Check
Direction W¹ to C = ?
Direction W ¹ to W² = 180 + 45.27.23 = 225.27.23
Direction W¹ to C = 225.27.23 + 179.28.55
= 44.56.18
Polar W¹ to C¹
Distance = 1,556m W 6021,933 1908,120
Direction = 44.56.18 y 1,099 1,101
C 6023,032 1909,221

16
 Mean coordinates of YC (6023,032 + 6023,032)2
Yc = 6023,032
Xc = (1909,221 + 1909,222)/2
Xc = 1909,222
Coordinate of C +6023,032 1909,222
The diagram below shows an underground correction survey by the Weisbach method.

P  / 
Surface Q
Base line


B Ws

S  R A 

Underground  Wu 
Base line

Surface Observations
Angle PQWs = 112.29.36 Distance = 10,000m
Angle QWsB = 259.00.00 Distance = 5,000m
Angle AwsB = 0.01.20 Distance = 15,000m
Underground Observations
Angle BwuA = 0.01.50
Angle AwuR = 202.00.00 Distance Bwu = 4,000m
Angle WuRs = 239.20.21 Distance WuA = 14,00m
Calculate the bearing of the underground baseline RS if the coordinates of the polar of
the surface baseline are:
Y X
P +48964,38m +69886,75m
Q +48988,66 +62583,18m
Solution
Calculate join QP
And obtain direction QP and distance and also find Direction PQ.
Q

P

B

Ws
S  
R A

Wu
Underground
Base line

17
 Find direction Q - Ws
Calculate Polar Q - Ws
Find Direction Ws – B and take Polar Ws – B also find Direction Ws – Wu
and take Polar Ws – Wu
Find direction Wu to R and calculate Polar Wu – R. and find direction R – S
(Being the bearing direction of RS the underground baseline).
Curves
Question One


127.13.00 140.09.20

T²
T¹


R = 609,6m

Given bearing T¹B = 127.13.00 BEARING T²B = 140.09.20

Cal – Tangent length
Arc length
Chord length
Solution
 = 127.13.00 – 140.09.20
= 167.03.40
Tangent length = T¹ = T² = R Tan /2
= 5375,875m
Arc length = /360*2r
= 1777,451m
Chord length = 2R Sin /2
= 121,436m
Question Two
D

A
250
B
T
A surface track is to be changed in direction from TA to TB by means of a curve of
radius 250m, which is to be set out by 3,100m chords. The ends of the chords are to be
located by means of right-angled offsets from the line TA produced.
Calculate
 The distance from A to the offset points
 The length of the offsets to ends of the chords
 By what amount has the direction of the track changed.

18
Solution
B C
 

A 

C
b

9
 
A
 D

250

Question Three

of 110. The two straights are to be joined by a curve of radius of 100m. Using a
theodolite you are required to lay out point a, b, c and T2 on the curve starting at T1 and
finishing θT2 on the curve starting at T1 and finish at T2 so that chord T,a, AB, Bc, and
CT2 are equal. Calculate the layout data i.ie. length T1 A, T1B, T1C and T1T2 are the
respective angles.

Calculate   = 110/4 = 27.30.00

T1a = 2R Sin 1 = 200 Sin 13.45 = 97,537
T1B = 2R Sin 21 = 200 Sin 27.30.00 = 92,350
T1C = 2R Sin 31 = 200 Sin 41.15.00 = 13,869
T1T2 = 2R Sin 41 = 200 Sin 55.00.00 = 163.830
Sin /2 = s+b) (s-c)
Bc
S = 100 + 250 +250 = 300
2
Sin /2 = (300-250)(300-250)
250*250
Sin /2 = 50/250
/2 = 0.2
 = 11.32. 13
 = 23.04.26

19
Diagram

110

B
A  C
 

1 2 3 T2
4 
T1 

100m

110
Calculate Dist from A to offset point
A to 1 R Sin  = 250 Sin 23.04.26 = 97,979m
A to 2  R Sin 2 = 250 Sin 46.08.52 = 180,282m
A to 3  R Sin 3  = 250 Sin 69. 13.18 = 233,740m
Cal offset a,b, and C
a = R-R Cos  = 250 (1-cos 23.04.26) = 20,000m
b = R-R Cos 2  = 250 (1-cos 46.08.52) = 76,800m
c = R-R Cos 3  = 250 (1-cos 69.13.18) = 161,312m
Question 4
The direction of the track has been changed by 69.13.18
Two converging haulages are to be connected by a curve 80m in radius. Peg A is a
haulage which has a bearing of 240, B is on a haulage of bearing 210 given the
coordinates A +5132,680 –3274,420
B +4931,450 – 3026,850
Points M and N are tangent points
Calculate dist AM and BN

20
Solution
Calculate angle at tangent intersection point diff in bearings
-Cal length of tangent MP and NP = R tan /2
-Join AB (calculate HD and bearing AB)
-Calculate angles at A and B diff in bearing
-Calculate dist AP and BP by sine rule
-Calculate dist AM and BN
A -240
M

80m

210

Question 5

A C

Bearing =250º Bearing = 170º

O
A +3000,000 +7000,00
B +2100,000 +8000,00
A is a tangent point in east haulage whose coordinates are given above and direction of
east haulage 250
B is a peg in the North haulage C is a tangent point start of ----, Direct of North haulage
170 coordinate of B are given. Calculate
The coordinates of C
a) The radius of the curve
b) The length of the curve A
Solution
i. Calculate the angle of intersection (diff bearing)
ii. Join AB (horizontal distance and bearing)
iii. Cal angle CAB and ABC and BCA
iv. Cal dist AC and BC (sin rule), DC from A and B
v. Cal radius of curve
vi. Cal the length AC of the curve AC = /360

21
Question 6
In setting at the curve TCG we have to pass through C to the following information is
available
- The direction of tangent IG = 160.00.00
- Bearing tangent IT = 300.00.00
- Perpendicular dist from C to tangent IG i.e. CB = 32m
- Dist along IG = 80m I
- Cal the radius of curve

B
32m 80m


T G

Y/2

O
Solution
Cal angle Y
300 – 160 = 140
Angle Y = 180 – 140
= 40.00.00
Angle  = Tan  = 32/80 = 0.4 = 21.45.05
Cal Angle B = IOC
B = 90 - y/2 -  (right )
= 48.11.55
Cal  &  in  I C O
Sin  = IO/CO * Sin  but IO = R Sec y/2
Sin  = Sec y/2 sin 
Sin  = Sec 20 Sin 45.11.55
= 0.793301794
= 52.29.43
 = 180 – ( + )
= 79.18.22
Cal ---- IC IO & CO
IC = IB² + Bc²
= 80² + 32²
= 86,163m
IO = IC Sin  = 86, 163 Sin 48.11.8 Check radius = IO Sin 
Sin  Sin 79.18.22 Sin 
= 65,366m (R) 69,561 Sin 48.11.55 = 65,366m
Sin 52.29.43

22
Question 7
Diagram

280.12.00

O C

C is a point cross cut having a bearing of 280 12‟ 00” toward D, B is a point on the same
elevation as C and it has been decided to connect C to B by means of a curve CA of 40m
radius. Given coordinates
C +6539,580 +1476,520
B +6487,270 +1561,910
Cal the coordinate of A
Cal bearing CO
Bearing CD = 280 12.00 – 90.00.00
370.12.00
Coordinate of O from C
Bearing CO = 40,000m

23
Taping /Chain surveying

-The measurement of distances by the use of tapes or chains.

Standardization
-Comparison of tape length with its known length under different conditions.
Instruments used in taping
1. Clinometers- for determining vertical angle/ slope angles.
2. Thermometer- temperature measurement.
3. Ranging rods/poles for defining straight lines.
Methods used in taping
Surface taping
-Also known as ground taping
-Used when ground is fairly flat and the tape is allowed to rest on the ground and
therefore no sag correction is applied.
Diagram

A B

Step Taping
- The line AB is broken into different bays, which are then measured separately.
- It is an attempt to eliminate slope correction i.e. horizontal distance is measured
directly.
1 B

2
3

Catenary Taping
Tape

Ground surface

A B
-Tape is suspended between two points (AB) and tension is applied.
-For a number of bays it needs to be supported in between.
-It is better for rough ground and produce accurate result.
-But it is relatively slow and many people are required.
Example One
After completing a survey the 30m tape used was checked and found to be 30,023m long.
What is the length of line AB observed to be 125,510m long.?

Solution
T = M (t/n)
T = true length of line
M = measured length

24
 T = length of tape during measurement
N = nominal tape length
T = 125,510 (30,023/30)
= 124,606m
Example Two
A distance of 220,450m was measured with a steel tape of nominal length 30m. On
standardization the tape was found to be 30,003m Calculate the correct measured
distance.
Solution = 220,472m.
Corrections
There are five major corrections to be applied to measured distances namely,
temperature, slope, sag, tension and mean sea level correction.
Example One
A line was measured under the following conditions:
Line Distance Slope angle Field tension Field temperature No. of bays
PQ 30,003 00.21.40 150N 25 1

Tape Details
Standard tension = 120N
Standard temperature = 20C
Mass of tape/metre = 0.026kg/m
Cross-sectional area = 3,5mm²
Coefficient of expansion = 0.000011/C
Young‟s Modulus of Elasticity = 207 x 10 4 MN/m²
Radius of Earth = 6367km
Altitude of line PQ = 1950m
1Kg F = 9,81N
What is the corrected distance of PQ to four decimal places?
Solution
Temp C correction = + 0,0017m
Slope correction = -0,0006m
Tension correction = +0,0012m
Catenary correction = -0,0033m
Mean sea level correction = -0,0092m
 of correction = -0,0102
Mean sea level corrected dist of PQ = 30,003 – 0,0102
= 29,9928m
Example Two
Calculate the mean sea level distance of line AB from the following field data
Line Distance Slope angle Field temperature Field tension Catenary
AB 98,872m 3 40‟ 00 29C 120N 3Equal bays
Data
Std tempC = 26C
Std tension = 50N
Mass of tape = 0.17N/m
Young Modulus = 200 x 10²MNm²
Cross – sectional area = 2mm²

25
 Coefficient of expansion = 1,12,x 10ˉ5/C
Mean height of line = 1450
Radius = 6361km
Solution
Temp = +0,0034m
Slope = -0,2044m
Tension = +0,0175m
Sag = -0,00993m
MSL = -0,0228m
 = -0,2156m
MSL dist = 99,6564m
Example 3
During the measurement of a baseline in 4 bays the following information was obtained.
Bay Measured length Temp Diff in level (m) Tension
1 29,8986m 18,0 +0,064 178
2 29,9012m 18,0 +0,374 178
3 29,8824m 18,1 -0,232 178
4 29,9496m 17,9 +0,234 178

The tape has a mass of 0,026kg/m and a cross sectional area of 3,224mm². It was
standardized on the flat at 20C with a pull of 89N. Coefficient of expansion 0,000
0009/C. Young Modulus as 15,5 10 4 MN/m². Mean level of the base 26,89m MSl.
Determine absolute length of baseline reduced to MSL
Solution
Correction for temperature
Bay
1. = -0,00005
2.= -0,00005
3.= -0 Sin 52.29.43
4.= -0,00006
5.= -0,00002m
NB Do the same for the sag, slope, tension, and MSL corrections
Tension correction  = +0,0212m
 = -0,0092m
 = -0,0043m
 = -0,0005m
Total correction  = 0,0071m
Corrected length  = 119,6318 +0,0071
=119,6389m
Taping correction formulas
1.Temp C correction = L (TF–TS)
Where L = measured length
 = Coefficient of linear expansion
TF = measured field tempC
TS = Standard tension
2. Slope correction = L(Cos  - 1)
Where L = measured length
 = slope angle

26
NB The diameter/theodolite is used to measure the slope angle.

3. Sag/Catenary correction = - W² L³
24n2T²
Where W= weight per unit length of tape
L = measured length
n = number of bays
T = field tension
NB W and T must have same units, either kg or N(Newton).

4. Tension correction = L TF – TS
AE

Where L = measured length

TF = field tension
TS = standard tension
A = cross section area of tape
E = Young‟s Modulus of elasticity

5. M.S.L. correction = -L h
R+H

Where L = measured length

R = radius of earth
H = mean elevation of line

27
Area and Volume
Question One
Calculate the side widths and cross-sectional area of an embankment to a road with
formation width of 12,50m and side slopes 1 to 2 horizontal when the center height is
3,10m and the existing ground has a cross fall of 1 in 12.
Solution
W¹ = (b/2+mh) (k/k-m)
Where B = 12,50
K = 12
M =2
H = 3,10
W1 = 14,94m
W2 = (b/2 + mh)(k/k + m)
= 10,67m

Area = ½m 9b/2+mh) (w1 + w2) – b²/2

= 60,18m²
Diagram

w¹ w²

12,50m 1 in 2

1 in 2

centerline height = 3,10m

Question Two
A road has a formation width of 9,50m and side slopes of 1 in 1 in cut and 1 in 3 in fill.
The original ground had a cross fall of I in 5. If the depth of excavation at the centerline
is 0,5m, calculate the side width and the areas of cut and fill.
Solution
W1 = (k/k-n) (b/2 + nh)
b=9,50
k=5
n=1
h=0,5
m=3
=6,56m
W2= (k/k-m) (b/2 – mh)
=8,125m

Area of fill = ½ (b/2 – kh)2

(k – m)

= 1,27mm

28
 Area of cut = ½ (b/2 + kh)2
(k – n)

=6,57m²

Question Three
The fig below spot levels at 20m intervals over a site, which is to be excavated at 44,00m
to accommodate 3 tennis courts. Calculate the volume of materials to be removed
assuming that the excavation has vertical sides.
A46,1 B47,4 C48,4 D47,2

E47,8 F50,1 G47,7 H46,1

20m

J49,5 K50,1 L47,3 M44,8

20m

Solution

STATION DEPTH OF EXCAVATION NO OF SQUARES hx n

(n)
A 2,1 1 2,1
B 3,4 2 6,8
C 4,4 2 8,8
D 3,2 1 3,2
E 3,8 2 7,6
F 6,1 4 24,4
G 3,7 4 14,8
H 2,1 2 4,2
J 5,5 1 5,5
K 6,1 2 12,2
L 3,3 2 6,6
M 0,8 1 0,8
 97,0

NB At depth of excavation being 44,00m

Volume = 20² x  = 97,00m²
4

Diagram for Question 2 9,50 1 in 5

(link)

29
 Cut

1 in 1
Fill (1 in n)
0,5
1 in 3
(1 in m)

W2 W1

Question Four
Calculate the area of a plot, if the offsets, scaled from the plan at intervals of 10m are:

Offset O1 O2 O3 O4 O5 O6 O7 O8 O9 O10
Length 16,76 19,81 20,42 18,59 16,76 17,68 17,68 17,37 16,76 17,68

Solution
Using the trapezoidal rule
Area = W  ( O1 + O10)/2 +O2+O3+O4+O5+O6+O7+O8+O9
Area = 1622,9m²
= 0,1623hactares
Simpson‟s rule (taking only 7 offsets)
Area = x/3 (O1+O70+2(O3+O5)+4(O2+O6)
Trapezoidal rule
Assume that the area is a set of trapezoids
Area of the first trapezoid ABCD = h1+h2 x w
2
Area of second BEFC = h2+h3 x w² and so on.
2

The total area = w h1+h7 + h2 + h3 + h4 + h5 + h6

2

The area represents the area bounded by the broken line under the curving boundary.

Simpson rule
-It assumes a curved boundary.
-More accurate than the above.
-If the area were a parabola the formula would be exact.
-It requires an odd number of ordinates/offsets.
-Area = w/3 (h1 + h7) +4(h2 + h4 + h60 +2(h3 + h5)
-I.e. 1/3 the distance between ordinates multiplied by the sum of the first and last
ordinates plus twice the sum of the odd and 4 times the sum of the even.

Bonus
Calculate the area of the following figure given the coordinates of the corner, beacons as
follows: A 1000 1000
B 905 1135
C 1310 1540
D 1720 1400

30
 E 2000 1070
F 1630 795
G 1200 840
Using the following formula
n
2A=  –Yі+ı)
i =1
A/2 = 10009905-1200) +1135(1310-10000 +154091720-905) +1400
(2000-1310) +1070(1630-1720 +795(1200-2000) +840(1000-1630) = ---m²
Dips and Intersection of Planes
Terms
Handwall- roof of an excavation
Footwall- floor of an excavation
Development- this is the mining of an excavation necessary for the exploitation of the
ore body.
Drive- is a tunnel, which is developed on ore body and bradly flows the strike.
Haulage- a tunnel usually on waste which may run in any direction and is used
essentially for the movement of men, materials and rock by mechanical means
Raise- this is a tunnel with an inclination of +5 or over but not a shaft.
Strike- the direction of the line joining points of equal elevation on a hangwall or
footwall contact of an ore body
Dip- the true dip is the maximum deep of an ore body
Launder‟s Theorem
Tan
s 

s
 d B
Sin s
M T
w A


B
D
d
D
M w
T
Sin In ABT
AB = AT Sin D
In ABM
AB = AM Sin d
AM Sin d = AT Sin D =AB
Sin d = Sin D AT

31
 AM
AT = Cos  = Sin w

Sin d = Sin D Cos 

Sin d = Sin D Sin W
Tan In  ATB
AB/TB = Tan
AB = TB Tan D
ABM
AB/BM = Tan d
AB = BM Tan d
TB Tan D = BM Tan d = AB
Tan d = Tan D TB/BM
TB/BM = Cos C = Sin S
Tan d = Tan D Cos C
Tan d = Tan D Sin S
S = angle between minor dip and strike
Example
Calculate your major dip given minor dip 12 and the angle between the minor and major
dip as 65.30.00. Use the launder tan rule and check using the launder sin rule.
Solution Tan d = Tan D Cos C
Tan D = Tan d/Cos C
Sin d = Sin D Cos c
Tan D = Tan D Cos C
Tan 12 = Tan D Cos 65.30.00
Tan D = Tan 12/Cos 65.30.00 e.t.c.
Example (Typical)
A borehole was drilled from surface and intersected a reef body, which dips in a true
south direction at 45. The borehole was drilled in a south 30 west direction and the core
length was measured as 200cm long. Calculate the true width of the reef body. The
borehole dips at 70.
Solution W Strike E

30
True width = Core length x Cos (H  d) Cos D

D = 45º

Where
H = deflection of borehole from the vertical (hade) use (H  d) when
deflection of the borehole is is with the dip of the reef use (H - d)
when deflection is against dip of reef.

32
 Tan d = 45(Cos 30 (minor dip)
d = 40.53.36
TW = 200 Cos 40.53.36 + (90 - 70) Cos 45
Cos 40.53.36
TW = 91,004cm (TW =true width)

20

v – (width)
200cm

d  40.53.36

 = 90 -40 53.36 = 49.06.24

d = 180 – 49.06.24 = 130.53.36
 = 180 – 150.53.36 = 29.06.24

Vertical length = 200 Sin 29.06 24

Sin 130.53.36
= 128,699m
TW = vertical length x Cos 45
= 91,004m
Example A

51.02.00 C c²
c¹
80
d¹
B D 43.33.00
d²
E
F

Given minor dip AB is 52.02.00

AF IS 42.33.00
Horizontal angle BCF is 80.00.00
The direction of d² = 320.00.00 and direction of d¹ = 40.00.00
Calculate
i. The direction of strike

33
 ii. The direction of major dip
iii. Amount of major dip

Solution
C = d¹ – d²
= 40 – 320
= 80.00.00

Tan C¹ = Tan d² – Cot c

Tan d1 Sin c

= Tan 42.33.00 - Cot 80.00.00 = 0,577574087

Tan 51.02.00 Sin 80.00.00

Tan ‾¹ C¹ = 30.00.35

Tan C² = Tan d¹ - Cot C

Tan d² Sin c
= Tan 51.02.00 - Cot 80.00.00
Tan 42.33.00 Sin 80.00.00

Tan C² = 49.59.25

Check
C¹ + C² = 30.00.35 + 49.59.25
= 80.00.00 checked

Calculating major dip

Tan d¹ = Tan D Cos C¹ Tan d² = Tan D Cos C²
Tan D = Tan d1 Tan D = Tan d²
Tan C¹ Cos C²
Tan D = Tan 51.02.00 Tan D = Tan 42.33.00
Cos 30.00.35 Cos 49.59.28

D = 54.59.35 D = 54.59.34

D = 54.59.35

Calculating direction of true dip

Bearing CB = 40.00.00
-C = 30.00.35
Bearing true dip = 09.59.25
Bearing cf = 320.00.00

34
 +C² = 49.59.00
Bearing true dip = 09.59.25
Calculating direction of strike

Bearing true dip = 9.59.25 Bearing true dip = 09.59.25

- 90.00.00 + 90.00.00
Bearing EF = 279.59.25 Bearing EB = 99.59.25

Summary

Direction of strike = 279.59.25 and 99.59.25

Direction of true dip = 09.59.25
Amount of major dip = 54.59.35

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