22/11/2019

hc&r = 0.0675K2
hr = (1/7)hc
K = distance in kilometers
hc&r = in meters
V
horizontal line
K of sight
hr
K
hc&r hc

actual line of
sight
R

How wide would a river be if a man 1.8 m tall h1 = 1.8 m

k1 k2
stands on the other bank of the river and still h2 = 30.5 m
could see a tower on the opposite bank of the
river which is 30.50 m high considering the
effect of curvature and refraction?
W
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D = k 1 + k2
h1 = 0.0675k12
1.5m = 0.0675k12 Three hills A, B and C are at elevations 135m, 146m and
k1 = 4.71 km 154m respectively above sea level. Distance AB is 3.9km
h1 = 1.5 m while distance BC is 3.1km.
h2 = 0.0675k22 k1 k2
18 m = 0.0675k22 h2 = 18 m a. Det. the effect of curvature and refraction between A and B.
k2 = 16.33 km b. Det. the effect of curvature and refraction between B and C.
c. What would be the height of tower to be constructed at hill
D = k1 + k2
D = 4.71 + 16.33 = 21.04 km
C so that the line of sight will clear hill B by 2.5m considering
the effect of the earth’s curvature and refraction?

a) h1 = 0.0675k12 b) h2 = 0.0675k22 STADIA FORMULA

HORIZONTAL SIGHT
c) = 0.0675(3.9)2 = 0.0675(3.1)2
C d= K s
G h1 = 1.03 m h2 = 0.65 m
G c f D=Ks+C
135 – h1 H y
i s
2.5 154 – h2
A H
C
2.5 m
146 153.35 B Elev = 154 m D
133.97 A
3.9 3.1 Elev = 146 m
Elev = 135 m INCLINED SIGHT
D E F H = K s cos2ϴ + C cos ϴ

D E F s

V = K s sin2ϴ + C sinϴ
V
2
c. What would be the height of tower to be constructed at hill C so ϴ V = Ks sinϴ cosϴ + C sinϴ
that the line of sight will clear hill B by 2.5m considering the effect
of the earth’s curvature and refraction? H
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ROD HAIR READINGS

Stadia Interval Factor (K) = 100
POSITION UPPER MIDDLE LOWER Stadia Constant (C) = 0.30
An engineer’s level with a stadia constant of 0.30 m A 1.330 1.175 1.020
was set up on the line between A and B. The following B 1.972 1.854 1.736
crosshair readings were observed:
ROD HAIR READINGS
POSITION UPPER MIDDLE LOWER
A 1.330 1.175 1.020
B 1.972 1.854 1.736

If the stadia interval factor of the level is 100,

determine the length of line AB. A C B

A transit with a stadia interval factor of 100.8 was set at C

on the line between A and B. The following data were
observed:
Vertical Hair Readings
A 15°25’ DCB
Angle Upper Middle Lower
DAC 9°07’
Rod A +150 25’ 1.972 1.854 1.732
Rod B -90 7’ 1.330 1.175 1.010

If the distance from the instrument to the focus is 0.38 m, C

determine the following: Vertical Hair Readings
a. Horizontal distance A and B Angle Upper Middle Lower B
Rod A +15025’ 1.972 1.854 1.732
b. Difference in elevation between A and B. Rod B -09007’ 1.330 1.175 1.010
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a) DAB = DAC + DCB

DAC = Ks cos2 θ + C cos θ
= 100.8 (1.972-1.732)cos2(15o25’) A transit with a stadia interval factor of 100.8 was set at C
+ 0.38 cos(15o25’) on the line between A and B. The following data were
DAC = 22.85 m observed:
DCB = Ks cos2 θ + C cos θ Vertical Hair Readings
= 100.8 (1.33-1.01) cos2(9o7’) + 0.38 cos(9o7’) Angle Upper Middle Lower
DCB = 31.82 m Rod A +150 25’ 1.972 1.854 1.732
A 15°25’
9°07’ Rod B -90 7’ 1.330 1.175 1.010
Vertical Hair Readings
Angle Upper Middle Lower If the distance from the instrument to the focus is 0.38 m,
Rod A +15025’ 1.972 1.854 1.732
Rod B -09007’ 1.330 1.175 1.010
determine the following:
C
a. Horizontal distance A and B
DAB = DAC + DCB = 22.85 + 31.82
DAB = 54.67m B
b. Difference in elevation between A and B.

a) DAB = DAC + DCB DAB = DAC + DCB = 22.85 + 31.82

DAC = Ks cos2 θ + C cos θ DAB = 54.67m
= 100.8 (1.972-1.732)cos2(15o25’) Elev.A + MHRA - VA = Elev.B + MHRB + VB
+ 0.38 cos(15 o 25’) Elev.DIFF = VA – MHRA + MHRB + VB
DAC = 22.85 m
Elev.DIFF = 6.30 – 1.854 + 1.175 + 5.11
MHRA DCB = Ks cos2 θ + C cos θ
VA Elev.DIFF = 10.731m
A 15°25’ = 100.8 (1.33-1.01) cos2(9o7’) + 0.38 cos(9o7’)
MHRA
9°07’ DCB = 31.82 m VA
A 15°25’
VB
b) VA = Ks sin θ cos θ + C sin θ 9°07’
Elev.Diff = 100.8(1.972-1.732)sin(15o25’)cos(15o25’) + 0.38 sin(15o25’) VB
C VA = 6.30m
MHRB
Vertical Hair Readings VB = Ks sin θ cos θ + C sin θ C
Angle Upper Middle Lower MHRB
Rod A +15025’ 1.972 1.854 1.732
= 100.8(1.33-1.01) sin(9o7’)cos(9o7’) + 0.38 sin(9o7’)
B
Rod B -09007’ 1.330 1.175 1.010 VB = 5.11m B
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Problem 5:
From the given data:
Station Occupied: A
Station Observed: B
Elevation of A = 93.13 m
0.86m

- End -
Height of telescope above station mark = 1.75
Rod reading at B = 1.64 m 1.64m
VB
Vertical angle = +5°24’ 5°24’ B
Stadia intercept = 0.86 m HAB
Stadia interval factor = 100 1.75
Stadia constant = 0.30
A El. 93.13
Compute the elevation of B.