What is Local Attraction in Surveying? - Civil Engineering

Surveying involves the measurement of lot of readings. Local attraction is one of the important concept of surveying. Bearings of lines are measured and they depends on magnetic meridian which gets affected by magnetic forces at site. So, calculations need to be done separately to rectify the readings of bearings.

Without a further delay, let's how all these calculations can be done properly so as obtained correct readings of bearings of line.

What is Local Attraction?

• A magnetic meridian at a place is established by a magnetic needle which is uninfluenced by other attracting forces.
• Some of the sources of this effect are: magnetite in the ground, wire carrying electric current, steel structures, railroad rails, underground iron pipes, keys, steel-bowed spectacles, metal buttons, axes, chains, steel tapes etc.​
• The local attraction at a particular place can be detected by observing the fore and back bearings of each line and finding its difference.​
• If Difference between Fore and Back bearing = 180 degrees for a station then, that station is free from Local attraction.

Elimination of Local Attraction:

• First check whether any one of the station is free from local attraction or not. This can be checked from difference of fore bearing and back bearing.
• In case there are such stations, use their bearing values and find others’ values.
• If there are no such stations free from local attraction, then, included angles will be taken into calculation to rectify all bearings values.

There are two types of questions in this topic and we have discussed here one-one from both types. Make sure to watch it carefully for better understanding.

Type(1)- There are stations free from Local Attraction

(Q) The following bearings were observed while traversing with a compass. Determine the corrected Bearings of lines.

Solution:

• Observed back bearing of CD (i e 209°10’ is correct as station D is free from local attraction). 
• Therefore, the correct fore bearing of CD will be 209°10’ - 180° = 29°10’, while the observed bearing is 29°45’. 
• The correction required at C equals to -35’.
• As in case of line BC, back bearing at C = 277°5’ - 35’ = 276°30’ and fore bearing = 276°30’ - 180° = 96°30’. Therefore, correction required for station B = -15’.
• Similarly, do it for station A.

Type(2) - No such stations are free from Local Attraction

(Q) The following bearings were observed in running a closed traverse. Determine the correct magnetic bearings of the lines.

This can be solved using include angles. Try this on your own now, I will upload the solution and will mention it here.