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
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Chapisha Maoni