PROCEDURE:
Obstacles to Chaining:
During measurements, it is impossible to set out all the chain lines in a
straightforward method because of a variety of obstacles to chaining and ranging in
the field.
Obstacles to measurement:
The obstacles which do not obstruct the ranging (view) like ponds, rivers are known
as Obstacles to Measurement.
Obstacles to alignment:
The obstacles which we cannot see across, i.e. both the chaining and ranging are
obstructed, e.g. houses, stacks, etc. are known as Obstacles to Alignment.
Obstacles to measurement:
First Method:
Let ABCD be a chain line obstructed by a pond (Fig 1). Let BC be the obstructed
length. Two offsets BE and CF of equal lengths are made at B and C and chaining is
done along EF to measure the distance EF.
Now the required obstructed length BC is equal to the measured distance EF.
Therefore, BC = EF
Second Method:
Let AB be the obstructed length across the river (Fig 2). AC is laid off, of any
convenient length, perpendicular to the required distance AB.
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Now a perpendicular is laid off from C such that it meets the extended line of AB at
D.
Triangles ABC and ADC are similar triangles.
From the principle of similar triangles,
AB / AC = AC / AD
Therefore, obstructed length AB = AC2
/ AD
Third Method:
Let AB be a chain line obstructed by a river (Fig 3). A point I is assumed anywhere in
line with the required distance AB. A point H is taken in such a way that HJ = HI and
HK = HB.
Now a point L is established in line AH and at the same time in the line JK produced.
Triangles KHL and ABH are similar triangles and their corresponding sides are equal
to each other as the points K, B and I, J are equidistant either side from H.
Therefore, the obstructed length AB = KL
Obstacles to alignment:
First Method:
Let DE be the obstructed length across the building (Fig 4). A point C is assumed
arbitrarily. E and C are joined such that EC = CB. Now D and C are also joined such
that DC = CA.
Triangles CDE and CBA are similar triangles and their corresponding sides are equal
to each other as points BE and AD are equidistant either side from C.
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Therefore, obstructed length DE = BA
Second Method:
Let DE be the obstructed length across the building (Fig 5). A point F is established at
equal distances from D and E at any convenient distance. Points H and G are
established such that FH = FG.
Triangles FDE and FHG are similar triangles.
From the principle of similar triangles,
DE / DF = HG / HF
Therefore, obstructed length DE = (HG X DF) / HF
CALCULATIONS:
RESULT:
Obstacles to measurement:
Obstructed length from First Method = m
Obstructed length from Second Method = m
Obstructed length from Third Method = m
Obstacles to alignment:
Obstructed length from First Method = m
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Obstructed length from Second Method = m
PRELAB QUESTIONS:
1. What is obstacle to chaining?
2. What is obstacle to ranging?
3. What is the least count of metric chain?
4. What is indirect ranging and how it can be done?
LAB ASSIGNMNET:
You are required to conduct the experiment by adopting above procedure for a given
obstacle.
POSTLAB QUESTIONS:
1. How the offsets are taken in the field without cross staff?
2. What are the precautions are to be taken in indirect ranging?
3. What is obstacle to chaining and ranging both?
4. What is line ranger?