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

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

A theodolite