Question

Following is the page of a field book used for levelling. Few readings marked with ‘-’ are illegible. The Reduced Level (RL) of the Temporary Bench Mark (TBM) is ................. m (Rounded off to 2 decimal places). All the readings are in m. 

Hint:

When entries are missing, use turning-point equality: the RL obtained from the previous setup’s foresight must equal the RL obtained from the next setup’s backsight. Then continue with \( \text{HI}=\text{RL}+\text{BS} \) and \( \text{RL}=\text{HI}-\text{FS} \).

Answer 1

Solution (Height of Instrument method, reconstructing the illegible entries): Setup 1: At BM, RL(BM)$=100.00$ m and HI$_1=101.50$ m. Back sight on BM is \( \text{BS}_1=\text{HI}_1-\text{RL(BM)}=101.50-100.00=1.50 \) m (illegible in the book). Foresight to Turn Point 1 (TP1) is shown in the next row as \(\text{FS}_1=2.00\) m, so \[ \text{RL(TP1)}=\text{HI}_1-\text{FS}_1=101.50-2.00=99.50\ \text{m}. \] Setup 2: With BS$_2=3.50$ m on TP1 and HI$_2=103.00$ m (given), the same RL(TP1) checks: \[ \text{RL(TP1)}=\text{HI}_2-\text{BS}_2=103.00-3.50=99.50\ \text{m}. \] Foresight to Turn Point 2 (TP2) is \(\text{FS}_2=2.50\) m (row 3), hence \[ \text{RL(TP2)}=\text{HI}_2-\text{FS}_2=103.00-2.50=100.50\ \text{m}(\text{matches table}). \] Setup 3: BS$_3=1.50$ m on TP2, so \[ \text{HI}_3=\text{RL(TP2)}+\text{BS}_3=100.50+1.50=102.00\ \text{m}. \] Final foresight to TBM is \(\text{FS}_3=0.50\) m, therefore \[ \boxed{\text{RL(TBM)}=\text{HI}_3-\text{FS}_3=102.00-0.50=101.50\ \text{m}.} \]