HOW TO AVOID OBSTACLES IN
CHAIN SURVEY
1…..SLOPE
SOLUTION
.STEPPINGa)
On
ground which is of variable slope this is the bestmethod and needs no calculation.
b)The measurement is done in short
lengths of 5-10m,
theleader holding the length
horizontal
.c)The point on the ground below the free and of the band isthe best
located by plumb bob.
d)It
will be seen that it is easier
to work downhill when‘stepping’ than to work uphill, the follower then
haring thedifficult job of holding the
band taut, horizontal, and withthe end vertically over the previous arrow
.e)The leader has therefore to line him in.ii.MEASURING ALONG THE SLOPEa)Measurements of slope angle,Correct
length = measured length × cosWhere = angle of slope Thus the
correction = -L (1-cos)b)AB represents
one tape length, say 30m measured alongthe slope.
F)
Point
C beyond B such that a plumb bobs at C will
cut Thehorizontal through A at D, where
AD is 30m
on thehorizontal.Now,AC = AD secBC = AC – AD = AC – AD = AD sec - AD= AD - ADWhere is a radian. Therefore, correction
BC = AD
sayd)Slope can be
expressed also as in 1 in n, which mean arise of 1 unit for n units horizontally, for small angles =radians.e)Slope can also be expressed in terms of difference in level,h,
between two points.f)Finally, Pythagoras’ Theorem may be used
2…LAKES
SOLUTION
a)The
network of lines is set out to surround
the area in themanner shown.
b)The
base line AB is first scaled and plotted
.c)From the shorter base AC, E is plotted and the lineextended to F.
d)From the new base EF, G is plotted and the line isextended
to X.e)From the base DB, H is plotted
and the line BH is
alsoextended to X.f)The two separately plotted position
of X should coincide toprove the
accuracy of the plot.
3….RIVER
SOLUTION
i.WITOUT SETTING OUT A RIGHT-ANGLE
a)Illustrates the solution where AB is the part of theranged
line which cannot be measured
.b)A suitable point D is chosen and AD is measured andproduced an equal distance to G.
c)Another point C on the line
is chosen; CD is measuredand
produced an equal distance to F
.d)FG is now parallel to the ranged line AB and byproducing FG to E, which also lines on BD produced,
thetriangle DGE is laid out equal to
triangle ABD.e)The measure of GE produced the required measure of AB.ii.SETTING
OUT A RIGHT-ANGLEa)Illustrate a similar solution using an optical square.
b)
At
A a right-angle is set out and the line AC is measuredwith D its mid-point.c)At C another right-angle is set out towards E,
which alsolies on BD produced.d)The measure of CE produces the required measure of AB
4..HILLS
SOLUTION
i.WHEN BOTH ENDS ARE VISIBLE FROM INTERMEDIATEPOINTS ON A LINEa)A and B are terminal points of a survey line.
b)
These
lines cannot be ranged directly because of therising
ground, but by taking up positions at C
1
,
and D
1
,approximately on the line, both terminal
stations can beseen from both points.
c)
From C
1
, D
1
is
ranged to D
2
on
line to B.
d)
D
2
then ranges C
1
to C
2
on
line to A
e)
C
2
then ranges D
2
again
on line to B until the positions isreached
when from C D can be seen to be on line
to B, andfrom D C can be seen to be on
line to
A, when the wholeline is properly ranged.ii.WHEN BOTH ENDS ARE NOT VISIBLE FROM ANYINTERMEDIATE POINTa)When it is impossible to adopt the method (1) the linemay be range by means of the random line.b)Here a line AB’ is set out clear of the obstruction in sucha way that a perpendicular from B may be dropped tothe random line at B’c)AB’ and B’B are measured and fro the similar trianglesthe perpendicular distance from C’ to C can becalculated if the distance AC’ is knownd)Similarly when AD’ is known:
5….NARROW STRIP OF LAND
SOLUTION
a)Use the same principle with lakes
.b)Although nowadays, this
work would normally beundertaken
by a theodolite transverse. The use of chainedtriangles is simply the linear method plotting the angles
c)The triangles should
be checked as usual.
6…BUILDING AND STRUCTURE
SOLUTIONi
.WITHOUT SETTING OUT A RIGHT-ANGLEa)Proceeds as far
as A and can go
no farther.b)From the base AB a point C is set out where AB = AC
=BC.
c)
This
results in the equilateral triangle ABC where anglesABC = 60
0
d)The
lines BC is produced to D clear of the obstruction andanother
equilateral triangle is constructed as before.e)The line DF is then produced to G such that BD = DG, sothat the triangle BDG is also equilateral.f)G now lies on the extension of AB, but the direction of theline
cannot be establish until the third equilateral trianglesGHK is set out.g)Once
this is done HG produced provides the extension of the
line AB on the other side of the building.h)The obstructed length AH = BD – (AB + GH) because BD =DG = GB by construction.i)ii.SETTING OUT RIGHT-ANGLEa)Again
the ranged line ends at A and can go no farther.
b)
At
A a right –angle is set out.c)C is placed
clear of the obstruction.d)Going
back to a point B another right-angle is set out andD is placed, such that AC = BDe)DC is now parallel to the line AB and can be extended pastthe obstruction to E and F.f)At both these points right-angles
are set out to G and it Hsuch
that EG = FH =AC =BD.g)GH produced provides
the extension of the
line AB on theother side of the obstacle and the measured
length of ECequals the obstructed length
AG.
7..POND
SOLUTION This can chain aroundi.ILLUSTRATE A SOLUTION WITOUT SETTING OUT A RIGHT-ANGLEa)The line AB is
ranged, but the measurement of AB cannotbe taken directly.b)A point C is set out clear of the obstruction and D and Eare placed midway along
the lines AC and CB respectively.c)ED is measured
and twice this distance
gives the length of AB.d)Other ratios for similar
triangles such as 1:3
instead of 1:2may be used depending on
surrounding obstructions.ii.ILLUTRATE
THE SOLUTION OF THE
SAME PROBLEM ANOPTICAL SQUAREa)Right-angle are set off the line at A and B, and C and D aremarked such that AC = BD.b)CD is measured to give the length
OF AB
SORRY….THE DIAGRAM UNABLE TO
SEEN
WRITTEN BY SHAMILY
2013
