Obstacle In Chain Survey And The Solution.

SLOPE SOLUTION i. STEPPING a) On ground which is of variable slope this is the best method and needs no calculation. b) The measurement is done in short lengths of 5-10m, the leader holding the length horizontal. c) The point on the ground below the free and of the band is the 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 the difficult job of holding the band taut, horizontal, and with the end vertically over the previous arrow. e) The leader has therefore to line him in. ii. MEASURING ALONG THE SLOPE a) Measurements of slope angle, Correct length = measured length × cos Where = angle of slope Thus the correction = -L (1-cos) b) AB represents one tape length, say 30m measured along the slope. c) Point C beyond B such that a plumb bobs at C will cut The horizontal through A at D, where AD is 30m on the horizontal. Now, AC = AD sec BC = AC – AD = AC – AD = AD sec - AD = AD - AD Where is a radian. Therefore, correction BC = AD say d) Slope can be expressed also as in 1 in n, which mean a rise 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

LAKES SOLUTION a) The network of lines is set out to surround the area in the manner shown. b) The base line AB is first scaled and plotted. c) From the shorter base AC, E is plotted and the line extended to F. d) From the new base EF, G is plotted and the line is extended to X. e) From the base DB, H is plotted and the line BH is also extended to X. f) The two separately plotted position of X should coincide to prove the accuracy of the plot.

RIVER SOLUTION i. WITOUT SETTING OUT A RIGHT-ANGLE a) Illustrates the solution where AB is the part of the ranged line which cannot be measured. b) A suitable point D is chosen and AD is measured and produced an equal distance to G. c) Another point C on the line is chosen; CD is measured and produced an equal distance to F. d) FG is now parallel to the ranged line AB and by producing FG to E, which also lines on BD produced, the triangle DGE is laid out equal to triangle ABD. e) The measure of GE produced the required measure of AB. ii. SETTING OUT A RIGHT-ANGLE a) Illustrate a similar solution using an optical square. b) At A a right-angle is set out and the line AC is measured with D its mid-point. c) At C another right-angle is set out towards E, which also lies on BD produced. d) The measure of CE produces the required measure of AB.

HILLS SOLUTION i. WHEN BOTH ENDS ARE VISIBLE FROM INTERMEDIATE POINTS ON A LINE a) A and B are terminal points of a survey line. b) These lines cannot be ranged directly because of the rising ground, but by taking up positions at C1, and D1, approximately on the line, both terminal stations can be seen from both points. c) From C1, D1 is ranged to D2 on line to B. d) D2 then ranges C1 to C2 on line to A e) C2 then ranges D2 again on line to B until the positions is reached when from C D can be seen to be on line to B, and from D C can be seen to be on line to A, when the whole line is properly ranged. ii. WHEN BOTH ENDS ARE NOT VISIBLE FROM ANY INTERMEDIATE POINT a) When it is impossible to adopt the method (1) the line may be range by means of the random line. b) Here a line AB’ is set out clear of the obstruction in such a way that a perpendicular from B may be dropped to the random line at B’ c) AB’ and B’B are measured and fro the similar triangles the perpendicular distance from C’ to C can be calculated if the distance AC’ is known d) Similarly when AD’ is known:

NARROW STRIP OF LAND SOLUTION a) Use the same principle with lakes. b) Although nowadays, this work would normally be undertaken by a theodolite transverse. The use of chained triangles is simply the linear method plotting the angles between the lines.

c) The triangles should be checked as usual.

BUILDING AND STRUCTURE SOLUTION i. WITHOUT SETTING OUT A RIGHT-ANGLE a) 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 angles ABC = 600 d) The lines BC is produced to D clear of the obstruction and another equilateral triangle is constructed as before. e) The line DF is then produced to G such that BD = DG, so that the triangle BDG is also equilateral. f) G now lies on the extension of AB, but the direction of the line cannot be establish until the third equilateral triangles GHK 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-ANGLE a) 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 and D is placed, such that AC = BD e) DC is now parallel to the line AB and can be extended past the obstruction to E and F. f) At both these points right-angles are set out to G and it H such that EG = FH =AC =BD. g) GH produced provides the extension of the line AB on the other side of the obstacle and the measured length of EC equals the obstructed length AG.

POND SOLUTION This can chain around i. ILLUSTRATE A SOLUTION WITOUT SETTING OUT A RIGHTANGLE a) The line AB is ranged, but the measurement of AB cannot be taken directly. b) A point C is set out clear of the obstruction and D and E are 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 1:2 may be used depending on surrounding obstructions. ii. ILLUTRATE THE SOLUTION OF THE SAME PROBLEM AN OPTICAL SQUARE a) Right-angle are set off the line at A and B, and C and D are marked such that AC = BD. b) CD is measured to give the length OF AB.