Skip to main content

Unit 4 :- Obstacle in Chaining Numerical

1. P and Q are two points 367 m apart on the same bank of a river. The bearings of a tree on the other bank observed from P & Q are N36°25’E & N40°35’W, respectively. Find the width of the river if the bearings of PQ are S86°35’E.

2. P and Q are two points 367 m apart on the same bank of a river. The bearing of a tree on the opposite bank from P and Q are N30°E and N60°W respectively. Find the width of the river if the bearings of PQ are S86°35’E.

3.  A pond obstructs the chain line AB. Two lines AP and AQ are so chosen that APQ circumscribe the obstacles, and PBQ are on the same straight line. Find the distance AB when the measurements AP = 802 m, AQ = 100 m, PB = 451 m, and BQ = 501 m are given.

 4. A river is flowing from west to east. Two points, P and Q, are selected on the southern bank of the river such that the distance PQ is 100 meters. Point P is westwards. The bearings of a tree R on the northern bank are observed to be 38° and 338° respectively from points P and Q. Calculate the width of the river.

5. A survey line ABC crosses a river flowing from east to west. Points B and C represent the near and far bank of the river, respectively. A perpendicular BD, 50 meters long, is set out at B on the left side of line ABC. The bearings of DC and DA are given as 35°25’ and 125°25’ respectively. The chainages of points A and B are provided as 862 meters and 900 meters respectively. Find the chainage of point C.

6. Points A and B are two points 455 meters apart on the same banks of a river. The bearings of a pole on the other bank from A and B are N30°E and N60°W respectively. Find the width of the river.

7. A survey line ABC crosses a river, with points A and C being on the near and opposite banks respectively. A perpendicular AD, 40 meters long, is set out at A. The bearings of line AD and DC are 38°45’ and 278°45’ respectively. Find the width of the river.

8. A survey line ABC crosses a river, with points A and C being on the near and opposite banks respectively. A perpendicular BD, 50 meters long, is set out at the left side of the line. If the bearing of line DC is 35°24’ and the bearing of line DA is 125°25’, find the chainage of point C.

9.  A pond obstructs the chain line AB. Two lines AP and AQ are chosen such that APQ circumscribes the obstacles, and PBQ are on the same straight line. We are given the following measurements:

AP = 802 m
AQ = 100 m
PB = 451 m
BQ = 501 m
The task is to find the distance AB.

Labels:

Comments

Post a Comment

Popular posts from this blog

Unit 4 :- Chain Surveying (Obstacle in Chaining)

 Obstacle in Chaining Sometimes obstacle in chaining occurs while doing chain surveying. Obstacle in chaining are of three kinds: (a) When chaining is free but vision is obstructed. (b) When chaining is obstructed but vision is free. (c) When chaining an vision are both obstructed.   (a) When chaining is free but vision is obstructed :- Such a program arises when a rising ground or a jungle are present. Here the end station are not intervisible. There may be of two cases: i) 1st Case The end station may be visible from some intermediate point on the rising ground. In this case, reciprocal ranging is done and the chaining is done be the stepping method.   ii) 2nd Case (Random Ranging) The end station are not visible from intermediate point when a jungle are comes across chainline.   Let AB be the actual chain line which can’t be ranged and extended because of interruption by a jungle. Le t the chain line be extended upto R. A point ‘P’ is selected on the chain ...
1 comment
All Blogs

Unit-1 :- Trigonometrical Levelling (Surveying III)

Post a Comment
All Blogs

Unit 2 :- Accuracy and Errors

Accuracy Accuracy refers to the closeness of a measured value to a standard or known value. It depends on three factors: Precise instrument Precise method Good planning The degree of accuracy in chaining is expressed as a ratio called the chaining ratio. For example, if there is an error of 0.05 meters during the measurement of a total length of 500 meters, the chaining ratio is calculated as follows: \[ = \frac{{0.25 m}}{{500 m}} = \frac{{25 m}}{{50000 m}} = \frac{{1}}{{2000}} \] Some permissible limits of errors: For measurement with a steel band: ±8/1000 In normal conditions with a tested chain: ±8/1000 For rough work: ±8/50  Precision: Precision refers to the closeness of two or more measurements to each other. It is independent of accuracy and is often considered a degree of perfection. Discrepancy: This term represents the difference between two major measured values or quantities. Errors: Errors arise from the difference between a measured value and the actual value. These ...
Post a Comment
All Blogs