STEPS ON HOW TO CARRY-OUT A TWO PEG TEST AND CALCULATE IT, WITH A SOLVED CLEAR EXAMPLE

The only permanent adjustment check necessary for an automatic and digital level is to ensure that the compensator and diaphragm are set such that horizontal readings are taken when the circular bubble is centralized.
    If horizontal readings are not begin taken then a collimation error is present in the level.
The usual method of testing and adjusting a level is to carry out a two peg test which is carried out as follows:

1. On fairly level ground, hammer in two pegs, says A and B, at a maximum of 60m apart. let the distance be L meters.
2. Set-up the level exactly midway between the pegs(30m), say point C and level your instrument carefully.
3. Place a levelling staff at each peg (A and B) in turu obtain readings, says; S₁ and S₂.    

             Since AC=CB the error , X, in the reading S₁ and S₂ will be the same . this error is due to the collimation error, the effect of which is to incline the line of collimation by angle å, this gives:

            S₁ - S₂=(S'₁ + X)-(S'₂ + X)=S'₁ - S'₂

            = true difference in height between A and B

The assumption has been made that the line of collimation, as set by the compensator and diaphragm, lies above the true horizontal plane. even if this is not the case, it does not affect the calculation procedure since the sign of the collimation error is obtained in the calculation.

      4. Move the level closer to peg B, preferably 10m from the peg B to D(new instrument station) and take readings S₃ at B ( levelling staff at B) and S₄ at A ( levelling staff at A).

       Then you compute the apparent difference in height between A and B from ( S₃-S₄).

     Note: If the instrument is in adjustment you will get (S₁ - S₂)=(S₃ - S₄)

If there is any difference between the apparent and true values, then error has occured in a distance of L meters(60m) and hence :

       Collimation error (e) =(S₁ - S₂ )-(S₃ - S₄)m per L meters.

If the error is found to be less than + or - 3mm per 60m the level is not adjusted, instead any reading taken must be observed over equal or short lengths so that the collimation error cancels out or is negligible.

      5. To adjust the instrument at point D, the correct reading that should be obtained at A, S'₄ is computed from :

                          S'₄ = S₄-(collimation error ⅹ sighting distance)

A check on this reading is obtained by computing S'₃ and by comparing (S'₃ - S'₄) with the true difference in height.

     6. With the level still at D and having deduced the correct reading S'₄, the adjustment can be made by one of the two methods.

• For most instruments the cross hairs are moved using the diaghrapm adjusting screws until the reading S'₄ is obtained.
• In some levels , however , it is necessary that the compensator itself is adjusted. 

      7. The test should be repeated to ensure that the adjustment has been successful.

       WORKED EXAMPLE:

Question: The reading obtained from a two peg test carried out on an automatic level with a single level staff set-up alternately at two pegs A and B placed 50m apart were as follows:

1.  with the level instrument midway between A and B 

                                  staff reading at A = 1.283m

                                  staff reading at B = 0.860m

       

       2. With the level instrument positioned 5m from peg B on the same line AB: produced 

                                    staff reading at A = 1.612m

                                    staff reading at B =1.219m

calculate:

1. The collimation error of the level per 50m of sight.
2. The reading that should have been observed on the staff at point A from the level in position 5m from B 

SOLUTION:

     1. from the question we have: S₁ = 0.860m, S₂ = 1.283m, S₃ = 1.219m, S₄ = 1.612m

Collimation error (e) = (0.860 - 1.283) - (1.219 - 1.612) = -0.030m per 50m.

     2. For the instrument in position 5m from peg B. the reading that should have been obtained on the staff when held at point A is :

                    S'₄ = 1.612-[-0.030/50]55 = 1.645m

    This is checked by computing (S'₃ - S'₄) and by comparing with (S₁ - S₂) as follows:

                        S'₃= 1.219-[-0.030/50]5 = 1.222m

                     Hence 

                              (S'₃ - S'₄) = 1.222 - 1.645 = -0.423 = (S₁ -S₂).