A stadia measurement was undertaken with a theodolite having an internal focusing
telescope and stadia interval factor of 98.8. When a backsight was taken on a turning
point whose elevation is 205.62 m, the rod interval observed was 1.94 m; and with the horizontal hairs set at 1.50 m on the rod, the vertical angle reading was +13 45 . For a foresight to a control point, the rod intercept was 1.17 m; and with the horizontal hair reading of 1.60 m on the rod, the vertical angle observed was -7 18 . Determine the elevation of the line of sight at the instrument point and the elevation of the point on which the foresight was taken.
Answers
Step-by-step explanation:
stadia measurement was undertaken with a theodolite having an internal focusing
telescope and stadia interval factor of 98.8. When a backsight was taken on a turning
point whose elevation is 205.62 m, the rod interval observed was 1.94 m; and with the horizontal hairs set at 1.50 m on the rod, the vertical angle reading was +13 45 . For a foresight to a control point, the rod intercept was 1.17 m; and with the horizontal hair reading of 1.60 m on the rod, the vertical angle observed was -7 18 . Determine the elevation of the line of sight at the instrument point and the elevation of the point on which the foresight was taken.
The answer is 145.272m
Given:
stadia interval factor K = 98.8
For internal focussing telescope
addition constant. C = 0
a) back sight on focusing point
S = 1.94m
S₂= 1.50m
α= +13°45
L = Ks + C = 98.8 * 194 + 0
V = L(sin)α = 191.672x sin 13°45'
= 45.557m.
elevation of line of sight at Instrument point.
RL of TP + S₂-V
= 205-62 +1.50 - 45.557m
= 161.56m
b) ForeSight to Control point.
ForeSight to Control point.S = 1.17m
ForeSight to Control point.S = 1.17m
S₂ = 1.60m
ForeSight to Control point.S = 1.17m
S₂ = 1.60mβ = -7°18'
ForeSight to Control point.S = 1.17m
S₂ = 1.60mβ = -7°18'
L = Ks + C = 98 8 * 1.17 + 0
ForeSight to Control point.
S = 1.17m S₂ = 1.60m
β = -7°18'
L = Ks + C = 98 8 * 1.17 + 0= 115.596m
ForeSight to Control point.S = 1.17m
S₂ = 1.60m
β = -7°18'
L = Ks + C = 98 8 * 1.17 + 0 = 115.596m
V= L(sin β) = 115.596x sin(7°18)
= 115.596x sin(7°18)
= 14.688m
= 115.596x sin(7°18)
= 14.688m
elevation of the control point on which IS is taken
= 115.596x sin(7°18)
= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂) point
= 115.596x sin(7°18)
= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂) point
= 161.56-14-688-1.60
= 115.596x sin(7°18)= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂)
point= 161.56-14-688-1.60 = 145.272m
= 115.596x sin(7°18)= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂)
point= 161.56-14-688-1.60 = 145.272m
Hence,
= 115.596x sin(7°18)= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂)
point= 161.56-14-688-1.60 = 145.272m
Hence,
The answer is 145.272m
= 115.596x sin(7°18)= 14.688m
elevation of the control point on which IS is taken = (R.L. of the instrument -V-S₂)
point= 161.56-14-688-1.60 = 145.272m
Hence,
The correct answer is 145.272m
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