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Basic Problems in chaining - To erect a perpendicular to a chain line
To erect a perpendicular to a chain line from a point on it
I Method
AB is the chain line. It is required to erect a perpendicular to the chain line at point ‘C’ on it. Establish a point ‘E’ at a distance of 3m from ‘C’. Take 10m tape and put the zero (0) end of the tape at ‘E’ and the 10m end at ‘C’. The 5th and 6th meter marks of the tape are brought together to form a loop of 1m. The tape is now stretched tight by fastening the ends E and C. The point ‘D’ is thus established. Angle DCE will be 90º.
I Method
AB is the chain line. It is required to erect a perpendicular to the chain line at point ‘C’ on it. Establish a point ‘E’ at a distance of 3m from ‘C’. Take 10m tape and put the zero (0) end of the tape at ‘E’ and the 10m end at ‘C’. The 5th and 6th meter marks of the tape are brought together to form a loop of 1m. The tape is now stretched tight by fastening the ends E and C. The point ‘D’ is thus established. Angle DCE will be 90º.
II Method
Select E and F equidistant from C (Fig (b)). Hold the zero end of the tape at E, and 10m end at F. Pick up 5m mark, stretch the tape tight and establish D. Join DC.
Select E and F equidistant from C (Fig (b)). Hold the zero end of the tape at E, and 10m end at F. Pick up 5m mark, stretch the tape tight and establish D. Join DC.
Last modified: Friday, 29 April 2011, 8:57 AM