A theodolite
Stations M, N, and P are for a right-angled triangle at station M as shown below. A theodolite whose constants are K= 100 and C=0 was used to determine the following chemometric data.
Instrument station M
Height of instrument I 1.410m
The reduced level of the station is 129.600m
| Target station | Vertical angle | Stadia hair readings | ||
| lower | middle | upper | ||
| N | -5040/ | 1.170 | 1.500 | 1.830 |
| P | 2030/ | 2.410 | 2.610 | 2.810 |
Calculate:
(a). Horizontal length MN and MP
(b). Horizontal length NP
(c). Reduced level of N and P
(d). Slope between p and N in form of 1 in n
Solution:
(a). Horizontal length MN (HMN)
HMN = CSCos2θ + KCosθ where C=100, S = 1.830-1.170 = 0.660m k = 0
HMN =100x 0.66Cos25040` = 65.36m
Horizontal length MP (HMp)
HMp = CSCos2θ + KCosθ
where C=100, S = 2.810-2.410 =0.400m
HMp =100x 0.40Cos22030` = 39.92m
(b). Horizontal length NP =√ ( MN2 + MP2)
=√ (65.362 + 39.922)
= 76.59m
( c ). Reduced level at N
Using RLs = RLI + i ±V-M
RLN = RLm + I ±V-m, Where RLm = 129.600m, I = 1.410m
M = 1.500m
V= ½(CSSin2θ) + ksinθ =1/2x 100×0.66Sin2(-5040`) = -6.49m
Therefore,
the Reduced level at N = 129.60 + 1.41 – 6.49- 1.5 = 123.020m
Reduced level at P =RLm + I ±V-m
Where RLm = 129.600m
I = 1.410m
M = 2.610m
V= ½(CSSin2θ) =1/2x 100×0.40Sin2(2030= 1.743m
Therefore, the Reduced level at P = 129.60 + 1.41 +1.743- 2.610 =130.143m
(d). Slope between p and N in form of 1 in n
Slope =∆y/∆x = (7.123)/(76.59) = 0.093 = 1in n = 1in10.75
