How to Lay Out a Right Angle from a Point on a Chain Line.

To Lay Out a Right Angle from a Point on a Chain Line.

The operations to be considered in the next few pages are all based on simple geometrical propositions and are the equivalent on the ground of easy problems in geometrical drawing.

  Let AB, (Fig. 18), be a chain line, and let it be required to lay out a line from a point C on AB at right angles to AB.

  If an optical square is available, stand at C and, viewing a ranging pole at A or B directly through the square, signal to a man holding a ranging pole at D until the image of the latter in the instrument appears to coincide with the pole seen directly through the aperture in the square.

  If no optical square is available, the work can be done with the chain alone, although neither of the two methods to be described is as convenient as using an optical square. The principle of the first method depends on the well-known fact that, in a right-angled triangle, the sum of the squares on the two sides containing the right angle is equal to the square on the hypotenuse. Thus, with sides of 3, 4 and 5, we have 3² + 4² = 5², the sides of lengths 3 and 4 containing the right angle, and the side of length 5 being the hypotenuse.

Standing at C, lay out a point E in line with A so that the distance CE measures 4 units, and put arrows in the ground at E and C. Fastening one end of the chain at C, or getting an assistant to hold it there, get an assistant to hold the graduation 8 at E. Hold the chain at graduation 3 and move to the side of the line towards which the right angle is to be laid out until the two lengths of the chain are taut at some point D. Place an arrow at D. D is then a point on a line at right angles at C to the line AB, as the sides of the triangle ECD are equal to 5, 4 and 3, as shown in the diagram.

  Instead of taking short lengths of 5, 4 and 3, it is generally better

Another method is to lay out two equal distances CE and CF, to use a multiple of these figures. Thus, EC could be made 40 links, DC 30 links, and DE 50 links. each about 40 links long, on either side of C and both on the line AB (Fig. 19). Then, if the ends of the chain are held at E and F, an arrow is held at the centre of the chain, and the two lengths pulled equally taut, the arrow will be at a point G such that GC is at right angles

to AB.

  Any of the above described methods is good enough for ordinary chain survey work, but cases sometimes occur where the work will not be sufficiently accurate unless it is done with a theodolite. In this case, set up the instrument at C, sight on A or B, and lay off an angle of 90 on the horizontal circle. The line of collimation will then be perpendicular to AB.

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