assignment of surveying technics on how to calculate latitude and departure even determination of the coordinate of second point

1. Nyarugenge campus
School of engineering
Department of civil, environmental and
geomatics engineering
Program: surveying and geomatics
engineering
Year of study: y2sge
Name: habumuremyi aime Emmanuel
Reg.number:219003056
Assignment number 2 of surveying technics
Submission date: 27 january 2021
Lecturer:mr mugabushaka adrien

2. Page 1
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
QUESTION 1
The table below illustrates the coordinates for two points A and
B measured on the field.
no Eastings(m) Northings(m)
1 EA=48964.38 NA=69866.75m
2 EB=48988.66m NB=62583.18m
We are required to determine a) the length and bearing of AB
b) the difference between azimuth(WCB) and quadrantal
bearing
ANSWER
a) I. DETERMINATION OF LENGTH (AB)
 Horizontal distance of line AB is given
by:ඥሺ𝑬𝑩 − 𝑬𝑨ሻ𝟐 + ሺ𝑵𝑩 − 𝑵𝑨ሻ𝟐
 Hor.distance 𝐻𝐴𝐵=ඥሺ𝐸𝐵 − 𝐸𝐴ሻ2 + ሺ𝑁𝐵 − 𝑁𝐴ሻ2
 𝐻𝐴𝐵=
ඥሺ48988.66 − 48964.38ሻ2 + ሺ62583.18 − 69866.75ሻ2
 𝐻𝐴𝐵=ඥሺ24.28ሻ2 + ሺ−7283.57ሻ2
 𝐻𝐴𝐵=√589.5184 + 530503991.94
 𝐻𝐴𝐵=√53050981.41
 𝐻𝐴𝐵=√53050981.41
 𝑯𝑨𝑩=7283.610469m
Therefore, length of line joining point A and B is
7283.610469m ≈7283.61m
ii)DETERMINATION OF BEARING OF AB
 normally bearing of AB is given by: 𝐭𝐚𝐧−𝟏
ቀ
𝒅𝒆𝒑𝒂𝒓𝒕𝒖𝒓𝒆𝒔
𝒍𝒂𝒕𝒊𝒕𝒖𝒅𝒆𝒔
ቁ

3. Page 2
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
where, departure(AB)=𝐸𝐵 − 𝐸𝐴 = ሺ48988.66 − 48964.38ሻm
=24.28m
Latitude(AB)=𝑁𝐵 − 𝑁𝐴 = ሺ62583.18 − 69866.75ሻ𝑚
=-7283.57m
 actually, since departure of AB is positive and latitude AB
is negative the line AB will be located in the 2𝑛𝑑
quadrant
means that line AB is directed from south to east.
 Therefore 𝐵𝐴𝐴𝐵=𝐭𝐚𝐧−𝟏
ቀ
𝟐𝟒.𝟐𝟖
−𝟕𝟐𝟖𝟑.𝟓𝟕
ቁ=0.190996499 as
decimals we have to compute into degrees =00
11′
28′′
 Hence the bearing of line AB =𝑺𝟎𝟎
𝟏𝟏′
𝟐𝟖′′
𝑬
 The whole circle bearing =1800
− 00
11′
28′′
=𝟏𝟕𝟗𝟎
𝟒𝟖′
𝟑𝟐′′

4. Page 3
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
ILLUSTRATION SHOWING ALL WORKINGS

5. Page 4
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
a) Difference between azimuth(WCB) and quadrantal
With respect to their definitions,
 Azimuth(WCB):is the horizontal angle made by a
line with the magnetic north in clockwise direction
whereas
 Qudrantal:is the horizontal angle made by a line with
magnetic north or south (whichever is closer from the
line) in the eastward or westward direction.
It is better to distinguish these types of bearing in tabular form
AZIMUTHAL(WHOLE CIRCLE
BEARING)
QUADRANTAL(REDUCED
BEARING)
 Clockwise angle from the
reference is only taken
 Both clockwise and anti-clockwise
angle from reference line are taken
 The value of the whole circle
bearing varies from 00
𝑡𝑜3600
 The value of the reduced bearing
varies from 00
to900
 That is horizontal angle made
by a line with true north or
magnetic north only
 That is horizontal angle by a line
with true or magnetic north or
south
 No direction is shown on the
value of angle
 Direction is shown is on the value
of angle(N…E)or (N…W)or
(S…E)or (S…W)
 Prismatic compass is
graduated on whole circle
bearing system
 Surveyors compass is graduated in
quadrantal system
 Examples:
260
, 1210
, 2450
, 3500
𝑒𝑡𝑐
 Examples:N260
E,S590
E,S650
W,
N100
W.
Q2.a surveyor took the following measurements in horizontal plane .the
coordinate of point Aሺ𝐸𝐴 = 48964.38𝑚, 𝑁𝐴 = 69866.75𝑚ሻbearing of
line ABሺ𝑊𝐶𝐵𝐴𝐵 = 2990
58′
46′′ሻand the horizontal distance of line AB
was 325.65m.calculate the coordinate of point B.

6. Page 5
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
ANSWER
Given :𝑁𝑐𝑜𝑜𝑟𝑑 1 = 69866.75𝑚
𝐸𝑐𝑜𝑜𝑟𝑑 1 = 48964.38𝑚
𝑊𝐶𝐵𝐴𝐵 = 2990
58′
46′′
= 𝑁600
01′
14′′
𝑊
𝐻𝑑𝐴𝐵=325.65m
We are asked to find the coordinate of point B.
 Practically, 𝑁𝑐𝑜𝑜𝑟𝑑 𝐵 = 𝑁𝑐𝑜𝑜𝑟𝑑 𝐴 + 𝐿𝑎𝑡𝐴𝐵
𝐸𝑐𝑜𝑜𝑟𝑑 𝐵 = 𝐸𝑐𝑜𝑜𝑟𝑑 𝐴 + 𝐷𝑒𝑝𝐴𝐵 and,
 𝐿𝑎𝑡𝐴𝐵 = 𝐻𝑑𝐴𝐵 ∗ 𝑐𝑜𝑠𝐵𝐴𝐴𝐵
=325.65m* 𝑐𝑜𝑠ሺ+600
01′
14′′ሻ=162.723811m
 𝐷𝑒𝑝𝐴𝐵 = 𝐻𝑑𝐴𝐵 ∗ 𝑠𝑖𝑛𝐵𝐴𝐴𝐵
=325.65m* 𝑠𝑖𝑛ሺ−600
01′
14′′ሻ=-282.07957m
 𝑁𝑐𝑜𝑜𝑟𝑑 𝐵 = 69866.75𝑚 + 162.723811𝑚 =
70029.47381𝑚 ≈ 𝟕𝟎𝟎𝟐𝟗. 𝟒𝟕𝒎
 𝐸𝑐𝑜𝑜𝑟𝑑 𝐵 = 48964.32𝑚 − 282.07957𝑚 = 48682.3004𝑚
≈ 𝟒𝟖𝟔𝟖𝟐. 𝟑𝟎𝒎
 Therefore the coordinate of point B is Bሺ𝐸𝐵, 𝑁𝐵ሻ
B(𝟒𝟖𝟔𝟖𝟐. 𝟑𝟎𝒎, 𝟕𝟎𝟎𝟐𝟗. 𝟒𝟕𝒎)

7. Page 6
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
ILLUSTRATION SHOWING ALL WORKINGS

8. Page 7
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
QUESTION 3
Compute the latitudes, departures and closing error. For the
following traverse conducted in Rwanda. Adjust also the
traverse using Bowditch’s rule.
line Length(m) W.C.B
AB 362.55 3260
50′
BC 218 390
08′
CD 163.22 1690
40′
DE 195.95 1130
10′
EA 278.53 2120
20′

9. Page 8
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
ANSWER
ILLUSTRATION COMES FIRST HERE

10. Page 9
Y2 SURVEYING AND GEOMATICS ENGINEERING SURVEYING TECHNICS ASSIGNMENT 2
I)finding latitudes and departures
 We know that latitude=length between two points*cos (ZA or BA)
 Departure=length between two points*sin (ZA or BA)
 From the above formulas the latitude and departure are computed
in the following table.
SIDE BEARING
Degree, minutes
LENGTH LATITUDE DEPARTURE
AB N300
10′
W 362.55 303.4843 -198.3425
BC N390
08′
E 218 169.0981 137.5857
CD S100
20′
E 163.22 -160.5727 29.2775
DE S660
50′
E 195.95 -77.0881 180.1494
EA S320
20′
W 278.53 -235.3441 -148.9701
∑sum ------------- PERIMETER=1218.25m -0.4225m -0.3m
Error in latitude represented by 𝑬𝑳 = −𝟎. 𝟒𝟐𝟐𝟓m
Error in departure represented by 𝑬𝑫𝒆𝒑 = −𝟎. 𝟑m
Closing error=√𝑬𝑳
𝟐
+ 𝑬𝑫𝒆𝒑
𝟐
= ඥሺ−𝟎. 𝟒𝟐𝟐𝟓ሻ𝟐 + ሺ−𝟎. 𝟑ሻ𝟐=√𝟎. 𝟐𝟔𝟖𝟓𝟎𝟔𝟐𝟓=0.5181m