Skip to main content

Two peg test, AB 100 m, Q 10 m

Short answer

True diff: 1.587-1.289=0.298 m. Second setup apparent diff: 1.356-1.046=0.310 m. Error=0.012 m in 80 m (1/6667, 15 mm/100 m), exceeds permissible; line upward. Correct readings: A=1.3425 m, B=1.0445 m. Adjust diaphragm to read correct value on far staff.

The following staff readings were noted during a two peg test operation:

Instrument StationStaff reading A (m)Staff reading B (m)Remarks
At midpoint P1.5871.289Distance between A and B = 100.000 m
Near B, at Q point (10 m inside between A and B)1.3561.046

a) Check the collimation precision whether within permissible limit or not. b) If collimation error exists, compute the correct staff reading for A and B for the 2nd set up. c) Explain the procedures for making the line of collimation parallel to the axis of bubble if not.

a) Collimation precision

Data: A and B are 100 m apart. Setup 1 at the mid-point P: A = 1.587 m, B = 1.289 m. Setup 2 at Q, 10 m from B on the inner side (so Q is 90 m from A): A = 1.356 m, B = 1.046 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

True difference=1.587−1.289=0.298 m\text{True difference} = 1.587 - 1.289 = 0.298\ \text{m}

Step 2: Apparent difference from the second setup

Apparent difference=1.356−1.046=0.310 m\text{Apparent difference} = 1.356 - 1.046 = 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 = 90 m and B = 10 m, so the difference of sight lengths is 80 m.

e=apparent−true=0.310−0.298=+0.012 me = \text{apparent} - \text{true} = 0.310 - 0.298 = +0.012\ \text{m} Error per metre of sight=0.01280=0.00015 m/m,collimation precision=16667\text{Error per metre of sight} = \frac{0.012}{80} = 0.00015\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{6667}

This is 15.0 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 large and the error grows with distance (the far staff reads too much), so the line of collimation is inclined upward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A =0.00015×90=+0.0135= 0.00015 \times 90 = +0.0135 m; error in reading on B =0.00015×10=+0.0015= 0.00015 \times 10 = +0.0015 m.

A=1.356−(+0.0135)=1.3425 m,B=1.046−(+0.0015)=1.0445 mA = 1.356 - (+0.0135) = 1.3425\ \text{m}, \qquad B = 1.046 - (+0.0015) = 1.0445\ \text{m}

Check: A −- B =1.3425−1.0445=0.298= 1.3425 - 1.0445 = 0.298 m, equal to the true difference.

b) Correct readings

From Step 4: the correct staff readings for the second setup are A = 1.3425 m and B = 1.0445 m.

c) Making the line of collimation parallel to the axis of the bubble

  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: the error is 0.012 m in 80 m (1 in 6667, about 15 mm per 100 m), which is not within the permissible limit; the line of collimation is inclined upward. Correct readings: A = 1.3425 m, B = 1.0445 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Surveying I chapter 5 questions →