--- title: "Two peg, AB = 60 m, near A" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:two-peg-ab-60m-near-a-6m-inside" source: https://www.bctengineer.com/q/semester_3/surveying_i/two-peg-ab-60m-near-a-6m-inside site: BCT Engineering --- # Two peg, AB = 60 m, near A Surveying I · Chapter 5: Leveling - Asked in: 2075 Asoj, 4+2 marks - Video: [How to calculate the correct elevations of stations using Two Peg Test of Level|Two Peg Test Method](https://www.youtube.com/watch?v=J9P7l23-BD4) — Master Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/two-peg-ab-60m-near-a-6m-inside - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question The following staff readings were noted during a two peg test operation: | Instrument Station | Staff reading A | Staff reading B | Remarks | |---|---|---|---| | At mid point P | 1.585 | 1.287 | Distance between A and B = 60.00 m | | Near A, i.e. 6 m inside between A and B | 1.355 | 1.045 | | Compute the collimation precision. If error is there, compute the correct readings for A and B during II set up and describe the procedure for making the line of collimation horizontal. ## Short answer Collimation error +0.012 m over 48 m, precision 1:4000, line inclined downward. Correct readings: A = 1.3565 m, B = 1.0585 m. ## Answer **Data:** A and B are 60 m apart. Setup 1 at the mid-point P: A = 1.585 m, B = 1.287 m. Setup 2 near A, 6 m inside AB (so 54 m from B): A = 1.355 m, B = 1.045 m. #### Step 1: True difference of level (instrument at the mid-point) With equal sight lengths the collimation error cancels, so $$ \text{True difference (A} - \text{B)} = 1.585 - 1.287 = 0.298\ \text{m} $$ #### Step 2: Apparent difference from the second setup $$ \text{Apparent difference (A} - \text{B)} = 1.355 - 1.045 = 0.310\ \text{m} $$ #### Step 3: Error and collimation precision The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 6 m (near) and B = 54 m (far), so the difference of sight lengths is 48 m. $$ \text{apparent} - \text{true} = 0.310 - 0.298 = +0.012\ \text{m} $$ B is the far staff and is subtracted in (A $-$ B), so a larger apparent difference means the far reading on B is too small: the error in the far reading is $e = -0.012$ m. $$ \text{Error per metre of sight} = \frac{0.012}{48} = 0.00025\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{4000} $$ This is 25 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment. The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is **inclined downward**. #### Step 4: Correct readings for a horizontal line of sight (second setup) Error in the reading on A $= -0.00025 \times 6 = -0.0015$ m; error in the reading on B $= -0.00025 \times 54 = -0.0135$ m. $$ A = 1.355 - (-0.0015) = 1.3565\ \text{m}, \qquad B = 1.045 - (-0.0135) = 1.0585\ \text{m} $$ Check: A $-$ B $= 1.3565 - 1.0585 = 0.298$ m, equal to the true difference. #### Procedure to make the line of collimation horizontal 1. Keep the level at the second setup (near one peg) and centre the bubble with the foot screws. 2. Calculate the correct staff reading for a horizontal line of sight on the far staff (as found above). 3. Using the capstan-headed screws at the top and bottom of the diaphragm, move the horizontal cross-hair (loosen one screw and tighten the other) until it reads the correct value on the far staff. 4. Check that the bubble is still central. If it has moved, centre it with the foot screws and repeat the adjustment. 5. Repeat the two peg test; the error should now be within the permissible limit. **Answer:** collimation precision = 1 in 4000 (0.012 m in 48 m, i.e. 25 mm per 100 m), which is more than permissible; the line of collimation is inclined downward. Correct readings for the second setup: A = 1.3565 m, B = 1.0585 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the instrument near A, 6 m inside AB, the staff distances are: A. A = 6 m, B = 54 m B. A = 6 m, B = 60 m C. A = 30 m, B = 30 m D. A = 54 m, B = 6 m Answer: A. Sight difference is 48 m. (Key point: Distances) 2. e = 0.310 - 0.298 = 0.012 m over 48 m gives: A. 1 in 40000 B. 1 in 4000 C. 1 in 10000 D. 1 in 400 Answer: B. 0.012/48 = 0.00025 = 1/4000, worse than 1/10000. (Key point: Precision) 3. Instrument is near A and the apparent A - B (0.310) exceeds the true value (0.298). What does this show? A. Staff A reads too little B. There is no collimation error C. Far staff B reads too little, so the line of collimation is inclined downward D. Far staff B reads too much, so the line is inclined upward Answer: C. With the level near A, B is the far staff; A - B larger than true means B is read too small, so the sight is downward. (Key point: Direction) 4. Correct readings are: A. A = 1.3535, B = 1.0315 B. A = 1.355, B = 1.045 C. A = 1.3565, B = 1.0315 D. A = 1.3565, B = 1.0585 Answer: D. Errors -0.0015 and -0.0135 are removed by adding. 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