BCS205S Engineering Surveying Assessment 2 - Solved Assignment

This document presents a comprehensive solution to an Engineering Surveying assignment, addressing multiple problems related to surveying principles. The solution includes detailed calculations and explanations for questions on accuracy, precision, probable error, temperature correction, slope correction, trigonometric applications, and chain surveying. It also covers staff reading, leveling, and the determination of probable elevation. Furthermore, the assignment incorporates solutions for deflection angles and area calculations. The document provides a step-by-step approach to each problem, ensuring clarity and aiding in the understanding of surveying concepts. The assignment covers all the important aspects of the Engineering Surveying course and the provided solutions are designed to help students understand the concepts and improve their grades.

ENGINEERING SURVEYING 0
Engineering Surveying

ENGINEERING SURVEYING 1
Question 1
First, we need to find accuracy for which we can take the average of each section which is
highlighted below:
For data set “A” average value = 299.275
Now the absolute value of length measurement can be obtained as:
11.975+ 11.225+ 6.775+ 7.525/4
Accuracy of measurement for data set A= 9.375
Prevision for data set A can be evaluated as:
The largest value in set A= 310.5
The lowest value in set A= 287.3
Range= 32.2
This range is for data set A and by repeating the same process we can find other values for
different data sets including B, C, and D.
Question 2
As per the given data, we need to find a probable error that can be obtained using the
below formula:
Probable error= [P(1-P)/n]1/2
Here, n refers to the sample size= 7 and P is probability= 50%
So, probable error= (0.5*0.5/7)1/2
Therefore, it is reported that the complete probable error= 0.488

ENGINEERING SURVEYING 2
Question 3
It is determined that the correction for temperature is (Ct) can be reported as:
Ct= L ( T - Ts) [1].
Here, L refers to the chosen length
Refers to the constant of thermal development
T refers to the temperature
Ts: refers to the temperature where the tape is standardized
Now, using the values provided in the question, correction for temperature can be obtained
which is highlighted below:
Ct = 30 x 0.000011 x (50 - 15) = 0.01155 m
It is determined that the real length of tape is given as:
30 + 0.01155 = 30.01155 meters
We now that modified part can be determined using below formula:
[2].
In which, Unhurried area = 455.4 x 295.7 = 134661.78 m2
Now, Corrected area can be obtained by using above value and highlighted below:
(134661.78) x (30.01155 / 30)2
Or, Corrected area= 134765.49 m2
Now, it is reported that precision in measurement is (0.01155 m per meter)2
= 0.00013 m2 per m2

ENGINEERING SURVEYING 3
= 1 m2 in 7496.1 m2
Question 4
It is determined that total slope correction can be represented as:
Cs = - h2 / 2L
Here, h is defined as slope height
L is Length measured
Now, using the provided values slope correction may be obtained which is highlighted
below:
Cs = - [(0.0625 x 280)2 / (2 x 280) + (0.085 x 390)2 / (2 x 390) + (0.06 x 410)2 / (2 x 410)]
= - [ 0.5468 + 1.40 + 0.738]
= - 2.685 m
So, it is determined that distance measured is given as:
280 + 390 + 410 = 1080 m
On the opposite side, corrected distance can be determined as:
1080 - 2.685 = 1077.315 m
Therefore, vertical height alteration from A to B is represented as:
(280 x 0.0625) + (0.085 x 390) + (0.06 x 410)
= 17.50 + 33.15 + 24.60
So, total vertical height is 75.25 m

ENGINEERING SURVEYING 4
Question 5
Using the concept of trigonometry this question can be solved easily and below figure
shows the functions and data provided in the question:
In angle CDE,
Tan@= DE/CE= 3.2/2.5……. (1)
Again, in angle ABC,
Tan@= AB/BC= AB/BE+ BC
tan@= h/2.5+ 6.5
Now, putting value of tan@ from above equation:
3.5/2.5= h/2.5+ 6.5
So, h= 12.6 meters
Question 6
It is found that designated length= 30 meters
Real length= 30.1 meters
Measured length= 480 meters

ENGINEERING SURVEYING 5
Let the correct length is L and error in one chain is I’-L
So, total error can be determined as:
(I’-L)*L’/I’
It is determined that correction= -error
So, correction= (30.1-30)*480/30= 1.6 meters
Moreover, real length of drain= 480+ 1.6= 481.6 meters
It is found that beak for 480 meters= $25300
Bill for 1 meters= $ 25300/480
Bill for 481.6 meters that means real distance is given as:
(25300/480)*481.6$
Question 7
As per the provided information, height= 15 meters
Angel x= 15.36 degree
Angle y= 54.42 degree
In triangle BAC,
Tany= AB/AC1
AC1= AB/tany= 15/tan(54.42)
AC1= 10.62 meters
Now, threesome BAC2

ENGINEERING SURVEYING 6
Tanx= AB/AC1+ C1C2
AC1+ C1C2= 15/tan15.36= 53.724 meters
Now, C1C2= 53.724- 10.62 meters
Therefore, C1C2= 43.10 meters
Question 8
Staff
reading
Height
of inst.
Evaluatio
n
Station BS IS FS HI RISE Fall RL Remarks
BM: A .75 - - 100.75 - - 100 Bench point
B 1.45 - 2.55 100.75 - 1.8 98.2 Changepoint
C - 1.65 - 99.65 - 0.2 98
D - 1.80 - 99.65 - 0.15 97.85
E 1.25 - 1.55 99.65 0.25 - 98.10 Changepoint
F - 1.50 - 99.35 - 0.25 97.85
G - 0.95 - 99.35 0.55 - 98.4
H - - 0.45 99.35 0.50 - 98.9
∑BS=
3.45
∑FS=
4.55
∑RISE=
1.3
∑Fall=
2.4
Now, ∑BS- ∑FS= last RL- First RL

ENGINEERING SURVEYING 7
3.45-4.55= 96.9-100
-1.10= -1.10 hence proved.
Moreover, ∑ Rise- ∑ Fall= last RL- First RL
1.3-2.4= 98.9-100
-1.10= -1.10 hence proved.
Question 9
After evaluating the provided information, it is found that most probable elevation of BM2
can be determined as:
Probable elevation of BM2= (566.3*3+ 569.2*4.5+ 570.5*5+ 570.9*6.5)/ 3+ 4.5+ 5+ 6.5
Probable elevation of BM2=
Question 10
Part A
Part B:
AB= 36.6 degree
BC= 180-84.3= 95.7 degree

ENGINEERING SURVEYING 8
CD= 360- 4.1= 355.9 degree
DE= 180+ 75.7= 255.7 degree
Part C: Deflection angles
It is found that there are major three Deflection angles including angle 1, angle 2 and angle
3.
So, angle 1= 180-84.3-36.6= 59.1 degree
Angle 2= 180-84.3= 95.7 degree
Angle 3= 180-75.7- 4.1= 100.2 degree

ENGINEERING SURVEYING 9
References
[1].J.L., Awange, A.H.M., Faisal Anwar, E., Forootan, H., Nikraz, K. Khandu and, J., Walker,
“Enhancing civil engineering surveying learning through workshops,” Journal of
Surveying Engineering, vol. 143, no. 3, p.05017001, 2017.
[2].L., Lin, R. Zhang and, Y., Shen, “Practical Research on Teaching Reform of
Engineering Surveying Course,” In 2018 3rd International Conference on Education,
E-learning and Management Technology (EEMT 2018), vol. 12, no. 6, pp. 12-18, 2018.