Question

Following observations were taken using a Tacheometer with staff held vertical. The additive and multiplying constants of the instrument are 0 and 100, respectively. The reduced level (RL) of the staff station R, in \(m\), is . (rounded off to two decimal places)

Instrument station	Staff station	Staff reading (m)	Vertical angle	Remarks
P	Q	1.4, 2.7, 3.9	7°	RL of Q = 102 m
P	R	2.1, 2.8, 3.6	5°

Hint:

Use the vertical intercept formula on the Q sighting to pin down the elevation of the instrument's line of sight, then apply that same elevation to the R sighting.

Answer 1

Correct Answer: 84.68

Step 1: Recall the tacheometric formulae for a vertical staff.
For an angle of elevation \(\theta\), with multiplying constant \(m\) and additive constant \(C\), the staff intercept is \(s=\) upper stadia reading minus lower stadia reading, and the vertical intercept above the line of collimation is \[ V=\frac{ms}{2}\sin(2\theta)+C\sin\theta \] Since the additive constant here is 0, this reduces to \(V=\frac{ms}{2}\sin(2\theta)\).

Step 2: Use station Q, whose RL is known, to fix the elevation of the line of sight.
For Q the readings are 1.4, 2.7, 3.9, so \(s_Q=3.9-1.4=2.5\ \text{m}\) and the middle hair reading is 2.7 m.
\(V_Q=\frac{100 \times 2.5}{2}\sin(14^\circ)=125 \times 0.2419=30.24\ \text{m}\).
Since the sight to Q is an elevation, RL of Q equals the instrument axis elevation \(H\) plus \(V_Q\) minus the middle hair reading: \(102=H+30.24-2.7\), so \(H=102-30.24+2.7=74.46\ \text{m}\).

Step 3: Apply the same instrument axis elevation to station R.
For R the readings are 2.1, 2.8, 3.6, so \(s_R=3.6-2.1=1.5\ \text{m}\) and the middle hair reading is 2.8 m.
\(V_R=\frac{100 \times 1.5}{2}\sin(10^\circ)=75 \times 0.17365=13.02\ \text{m}\).

Step 4: Find the RL of R.
RL of R \(=H+V_R-\text{middle reading}_R=74.46+13.02-2.8=84.68\ \text{m}\).

Final Answer:
The reduced level of station R is about \(84.68\ \text{m}\). \[ \boxed{84.68} \]