SURVEY: Methods Used to Set Out Right Angles and Perpendicular Lines in the Chain Surveying

METHODS USED TO SET OUT RIGHT ANGLES AND PERPENDICULAR LINES IN CHAIN SURVEYING

Chain surveying is the simplest method of surveying in which only linear measurements are made in the field. It is suitable for survey small areas in open grounds to obtain measurements for plotting exact description of boundaries of piece of land or for taking simple details. The principle of chain surveying therefore, provides a framework consists of a number of connected triangles. The following are the methods used to select right angles and perpendicular lines in chain surveying:

1. 3:4:5 METHOD

This method is used to set out a right angle from a certain point on the baseline. To set out right angles in the field the; measuring tape, two (2) ranging poles, pegs, and three persons (surveyors) are required.

The first person holds together between a thumb and fingers the zero mark and the twelve (12) metre mark of the tape. The second person holds the same way a three (3) metre mark of the tape and finally the third surveyor holds the eight (8) metre mark of the tape.

When all sides of the tape are stretched a triangle with a length of 3m, 4m, and 5m is formed and the angle near the third surveyor is a right angle. For instance;

Step 1: The baseline is defined by the poles A and B and a right angle has to be set out from peg C is on the baseline. Peg C is on the baseline.

Step 2: Three surveyors hold the tape the way it has been explained above; the first person holds the zero mark of the tape together with the 12m mark on the top of peg C. The second person holds the 3m mark in line with pole A and peg C on the baseline. The third surveyor holds the 8m mark and after stretching the tape, he/she places a peg at point D.

The angle between the line connecting peg C and peg D and the baseline is the right angle. Line CD can be extended by sighting ranging poles.

Another method is shown by the following figure:

O is the point on the chain line MN where a perpendicular line is to be erected. P and Q are two points at equal distances from O. the distance between OP and OQ can be conveniently chosen. Holding the ends of chain or tape at P and Q, take the middle point of the tape and pull it strait so that there are equal length of the tape on each side and the chain or tape is taut. Mark the point corresponding to R where the middle length of the tape rests. OR is perpendicular to the chain line at O, these can be proved easily as follows;

OP=OQ and RP=RQ by construction. In triangles ROP and ROQ, OP=OQ, RP=RQ while RO is a common side. The two triangles are congruent, therefore ROP = FCE = 90˚

Another method is shown in the following figure:

Select a point P at a convenient distance from O, the point on the chain line MN where the perpendicular is to be erected. Set the zero end of the tape at P and measure PO. Keeping the zero end of the tape at P and holding the tape at length PO, draw an arc with the tape to cut the chain line at Q. Mark the point Q with an arrow, and range the line QP and extend it to R such that QP=PR. Then RO is perpendicular to the chain line MN at O. This can be proved easily.

If the arc QO is extended, it will pass through R because PO=QO=PR. P is the centre of the arc. Arc QOR is the semicircle and hence        QOR is a right angle.

1. OPTIC SQUARE METHOD

The observer should stand on the chain line and approximately at the position where the perpendicular line is to be set. The optical square is held by the arm at the eye level. The ranging rod at the forward station B is observed through the unsilvered portion on the lower part of the horizon glass.

Then the observer looks through the upper silvered portion of the horizon glass to see the image of the object P.

Suppose the observer finds the ranging rod B and the image of the object P do not coincide. Then he/she should move forward or backward along the chain line until the ranging rod B and the image of P exactly coincide in figure below.

At this position the observer marks a point on the ground to locate the foot of the perpendicular.

The following methods are used when erecting a perpendicular line to a chain line from outside point

When the point is inaccessible, the following method can be adopted. The following figure can illustrate more with the support of words;

AB is the chain line and C is an inaccessible point from which a perpendicular line is to be drawn to AB. Select two convenient points on the chain line, points D and E; range the lines DC and EC. With the lines DC and EC marked on the ground, draw perpendiculars from D to EC and E to DC, DF and EG are the perpendiculars. Range the lines DF and EG to find their intersection; point H. If C is visible from H, range the line CH and continue the line to intersect AB at I. I is the foot of the perpendicular from C to AB. I is also the foot of the perpendicular line from C to AB.

Furthermore, the following illustration gives another method.

MN is the chain line, and O is the point outside the chain line from which a perpendicular line is to be drawn to it. Measure a line OP, P lying on the chain line and bisect OP at Q. Holding one end of the tape at Q and taking a length equal to QP or QO, swing the tape to cut the chain line at a second point R. R is the foot of the perpendicular line from O to MN. The proof is the same as given earlier. It is clear that the angle at R is an angle in a semicircle and hence is a right angle.

REFERENCES

Pritchard J. M. (1984), Practical Geography for African: Hong Kong.

Subramanian R. (2012), Surveying and Leveling: 2^nd Edition, Oxford University Press: New Delhi.