Elevation and Depth of Coal Seams in Boreholes A, B, and C

This document provides information about the elevation and depth of coal seams in boreholes A, B, and C. It explains the vertical zonation of the coal seams and their locations relative to sea level.

Q1)
C2 = a2 + b2 – 2abcosC
C2 = 102.52 + 57.972 – 2*57.95*102.5cos175.5
C = 160.34
102.52 = 160.342 + 57.95*160.34*cosƟ
Ɵ = 2.820
The angle ACD = 100.180
C2 = 160.342 + 48.22- 2*48.2*160.34*cos100.18
C = 168.79 m
Slope = 90 – (4.5+5) = 80.50
Q2)
C = l∝¿m – ts)
= 154.2*11.2 * 10-6 *(23-17)
= +0.01036224 m
Q3)
C = L(Tn−Ts)
A∗E
= 30(78−100)
¿ ¿
= -0.00285
Q4)
Correction
+
-
Standardization
Tape is 29.997 at 75N tension
and 160
C = 0.997 m per 100 m
Temperature
C = 30*12*10-6*(14-16)
0.997

= -0.00072
Tension
C = 30(100−75)
¿ ¿
= +0.00083102
Slope
C = -d2/2l
= 0.462/2*30
= -0.00353
C =- W 2
24 [ 1
T m2 − 1
T s2 ]
W = wl = 0.0193*30*9.81
= 5.68 N
=- 5.682
24 [ 752−1002
752∗1002 ]
= +0.00313662
Height
C = hl/R
= 490∗30
6367.27∗103
= -0.00231
+0. 00083102
0. 00313662
0.00072
0.00353
1.00096764
-0.00656
0.00656
+0.9944
Measured length = +29.9824
Total correction = +0.9944
Corrected length = 30.9768
Q5)
Horizontal length = √(20.634 +1.78)2+ ( 18.612+0.54 )2
= 29.482 m

Q6)
Height = 10*sin700
= 9.4 m
Adjacent length = 10*cos70
= 3.42 m
Q7)
Let x be the distance near the foot of the tower
(x+42)tan45 = xtan60
X = 57.37
a) Height = (57.37+42)tan45
= 99.37 m
b) 57.37
Q8)
a
sinA = b
sinB= c
sinC
190.5
sin 22.25 = 466
sinC
sinC = 0.926248
c = 67.860
180 – (22.25 + 67.86) = 89.890
190.5
sin 22.25 = b
sin 89.890
b = 503.104
Q9)
∆ E=352.12

∆ N =551.06
Tan bearing AB = 032034’
corr Corrected angles
Summation of angles 183021’ -01’ 183020’
86045’ -01’ 86044’
329017’ -01’ 329016’
354036’ -01’ 354035’
306006’ -01’ 306005’
1260005’ 1260000’
Summation of angles should be equal to (2n+4)90
= (2*5 + 4)90 = 12600
The error is 05’ distributed as 01’
Bearing AB 032033’ …..N32034’E
Angle ABC = 183020’
…………
215054’
- 1800
………………
035054’ …………………N35054’E
Angle BCD = 86044’
…………..
122038’
+1800
…………….
302038’………N57022’W
Angle CDE = 329016’
………

631054’
- 5400
……………
Bearing DE 091054’………s88006’E
Angle DEF 354035’
………….
4460029’
- 1800
………………
Bearing EA 266029’ ………s86029’W
Angle EAB 306005’
………….
572034’
- 5400
………….
Bearing AB 032034’
Q10)
Distance AB = √ ( 1321.4+146.4 ) 2+ ( 8257.7−7542.7 ) 2
= 1632.69 m
Ɵ = tan-1(715/1467.8)
= 25.970
Hence bearing = 90 – 25.97 = 64.030
Q11)
Midpoint = [ 922.4 +58.6
2 , 1886.6 ± 495.4
2 ]

= 490.5E, -153.4N
Q12)
Gradient , G1 = 164−718
619−90 = 554/529
Gradient , G2 = 469−81
822−210 = 153/97
Equation 1
529y = 554x +329962
Equation 2
97y = 153x + 7977
At intersection y are equal
554x/529 +329962/529 = 153x/97 + 7977/97
-0.53006x = -541.509
X = 1021.6
Substituting in
97y = 153x + 7977
Y = 1693.63
1693.63N, 1021.6E
Q13)
BC = 167.7/cos21 = 179.63
CD = 222.5/cos18.5 = 234.62
Summation of xcosƟ = 0, Summation of xsinƟ = 0
38.7cos270 + 179.63cos184030’ + 23.62cos159015’ + 40.9cos40 + 25.3cos90 + xcosƟ= 0
xcosƟ = 398.477
38.7sin270 + 179.63sin184030’ + 23.62sin159015’ + 40.9sin40 + 25.3sin90 + xsinƟ= 0

xsinƟ= -173.93
tanƟ = sinƟ/cosƟ = -0.436
Ɵ = -23050’
= 358040’
Horizontal length = √398.47723 +173.932
= 353.53 m
B = 78.5 + 68.95 – 0.204
Total dip = 147.24 m
Q14)
B.S I.S F.S Rise Fall Height of
collimations
RL
0.738 113.047 112.309
1.094 0.356 111.953
1.713 0.619 111.334
2.265 0.552 110.782
0.942 2.685 0.42 111.304 110.362
1.100 0.158 110.204
1.533 0.433 109.771
-3.133 4.66 114.431
0.741 1.887 5.02 110.152 109.411
1.634 0.893 108.518
2.472 0.838 107.68
2.265 0.207 107.887
………….
2.421
6.837
……………..
6.837
…………
4.867
9.289
……………..
4.422
…………..
4.42
…………..
4.422
The final LEVEL = 107.887
FILLDEPTH = 112.309 – 107.887 = 4.22 m

Q15)

Q16)
B.S I.S F.S Rise Fall Height of
coll
RL
1.759 1217.798 1216.039
0.652 1.107 1217.146
1.689 0.091 0.56 1219.396 1217.707
0.430 1.259 1218.966
0.680 0.917 0.487 1219.159 1218.479
0.671 0.009 1218.488
2.243 0.494 0.177 1220.908 1218.665
0.682 0.561 1219.226
1.520 0.216 1.466 1222.212 1220.692
0.686 0.47 1220.222

Q17)
B.S I.S F.S Rise Fall Height of
coll
RL
1.194 48.395 47.201
2.322 1.128 46.073
0.707 1.615 47.688
2.167 2.177 3.37 2.663 47.192 45.025
2.856 1.692 0.485 48.356 45.500
2.246 2.246 0.610 48.356 46.110
2.657 0.411 45.699
3.148 2.166 0.491 46.19
1.292 1.856 47.064
Q18) close transverse, since we start and end with known points
Q19)
Angle Observed
value
corr Corrected
value
s d l ∂ d ∂ l
DAB 97041’ +5’ 97046’ AB 22.11 15.6314 15.6341 -0.098 +0.057
ABC 99055’ +5’ 99058’ BC 58.34 14.666 0.0 -0.065 +0.038
BCD 72022’ +5’ 72028 CD 39.97 0.0 -13.926 -0.062 +0.035
CDA 89059’ +5’ 89064’ DA 52.10 9.9323 -17.20 -0.088 +0.52
Total
=
172.52
∂ dl=−0.428
172.52 s=−2.484∗10−3 s
∂ l= 0.250
172.52 s=+1.449∗10−3 s
B = -30.3N, +219.76E
C = 523.9N, +397.5E
D = +522.6N, -1.2E
Q20)

1. The theodolite is set up with the surveyor and the point is nailed or stake, the point is
used in estimating the separation and edges.
2. The tripod is set and its stature checked to ensure that the theodolite is permitted to an
eye level, the mounting plate should be over the stake or nail.
3. The tripod legs will be drive to the ground to utilize all the sections around its legs
4. The theodolite is mounted by placing it on the tripod and sunk using mounting handle
5. The stature between the instrument and ground is measured to utilize different stations
references.
6. The theodolite is leveled by altering the tripod legs and utilization of the bull eye level,
the slight turning is made using leveling handles
7. Alteration of the vertical fall that is found at the theodolite base, the vertical plunge
ensures the instrument stays over stake or nail.
8. The line of sight is then for the point under estimation is pointed in the primary degree,
the handle is locked to ensure the theodolite keeps the point, record the vertical point and
the angle while utilizing the scope review that is found on the side of theodolite.
Q21)

Q22)
answer- Borehole A is located at = 1000m elevation from sea level
lower coal seam is occured at 450m depth
so the elevation of lower coal seam at borehole A from sea level is = 1000-450 = 550m
upper coal seam depth is = 50m
total vertical zonation of coal seam = lower coal seam depth - upper coal seam depth = 450-50 =
400m
Borehole B is located at = 700m elevation from sea level
lower coal seam is occured at 150 m depth
so the elevation of lower coal seam at borehole B from sea level is = 700-150 = 550m
Borehole C is located at = 700m
lower coal seam occurred at the depth of 250m
so the elevation of lower coal seam at borehole C from sea level is = 700-250 = 450m
it is given that vertical zonantion of coal seams is 400m
Q23)
a)
Tension
CT = L∆T/AE = 112.701∗(10∗9.80665)
3∗210∗103 = +0.0176
Temperature
Ct = LK∆t = 112.701 * 0.000011 * 10 = - 0.0124
Sag
Cs = LW 2
24 T 2 = 112.701∗12
24∗152 = -0.0210
Slope

Ch = h2/2L = 1
2∗30 ( 0.452 +0.302 ) + 0.452
2∗22.536 = -0.0154
+ -
+0.0176 -0.0124
-0.0210
-0.0154
+0.0176 -0.0488
Horizontal length of base (D) = measured length(M) + sum of corrections ©
= 112.701 + (-0.031)
= 112.670 m
Measured distance will be ;
M = D – C
= 112.670 – (-0.0311) = 112.701 m
b)
∆ T
AE = W 2
24 T 2
T 3A – TA2Ts - AE W 2
24 = 0
T 3A – 49TA2 -2524653 = 0
let
TA = (T + x)
(T + x)3 – 49T2(1 + 2x/T)2 – 2524653 = 0
Expanding
T3 + 3T2x – 49T2 – 98Tx- 2524653 = 0
X = 2524653−T 3+ 49 T 2
3 T@−98 T
Assuming T = 147 N
Then substituting
X = 75 N

Q24)
The sign of the corrections for measurements should be reversed when setting up EDM, long
tape will have negative corrections
Q25)
B.S I.S F.S Rise Fall Height of
collimations
RL
1.54 2.759 1.219
0.094 1.446 1.216
0.713 0.619 1.215
1.265 0.552 1.203
1.45 1.451 0.42 12.651 1.201
a) Horizontal distance = 12.651
b) RL = 1.201