When the instrument is stationed near point P, the staff readings at point P and Q are 1.800 m and 2.600 m, respectively. When the instrument is stationed near point Q, the staff readings at P and Q are 1.500 m and 2.400 m, respectively. R.L. of point P is 100 m. Find the R.L. of point
- A98.5 m
- B99.15 mCorrect
- C100.85 m
- D97.4 m
Solution
Concept
In reciprocal levelling, the true level difference is the average of the two apparent differences, which cancels collimation, curvature and refraction errors.
Formula
H = [(hq1 - hp1) + (hq2 - hp2)] / 2
Step-by-step solution
1. Instrument near P: difference = 2.600 - 1.800 = 0.800 m (Q is lower). 2. Instrument near Q: difference = 2.400 - 1.500 = 0.900 m (Q is lower). 3. True difference H = (0.800 + 0.900)/2 = 0.850 m. 4. RL of Q = 100 - 0.850 = 99.15 m.
Explanation
Q has larger staff readings, so Q is lower than P. Taking the average of the two differences removes the instrument and curvature errors. So the RL of Q is 100 - 0.85 = 99.15 m.
Why the other options are wrong
A: 98.5 m is wrong. It does not match the averaged difference of 0.85 m. C: 100.85 m adds the difference, but Q is lower than P. D: 97.4 m comes from adding the two differences instead of averaging them.
Shortcut
RLQ = RLP - average of the two differences = 100 - 0.85.
Common mistake
Adding the difference instead of subtracting it, or using only one set of readings.
Correct answer: B) 99.15 m