University Surveying Assignment 1: SVY2105, Semester 1, 2019

This document provides a comprehensive solution to SVY2105 Assignment 1, a civil engineering surveying assignment. The solution includes detailed calculations for the volume and standard deviation of a cylinder, considering measured height and diameter with associated errors. It also addre...Read More

Ans1.(a)
v+ ∆ v =π ( d+∆
2 )∗(h± ∆ h)
¿ π ¿
¿ π ¿
¿ π ¿
v+∆ v =3.99.012 ± 2.12%
∆ v=399.012∗2.12/100
∆ v=8.45 m3
Ans(c)
μ=1, σ =2
(1) P(1<x<6)
z-score = 1- 1−μ
σ = 1−1
2 =0
z-score2 = 1−μ
σ = 6−1
2 =2.5
0<z<2.5 = 0.99379-5.0000=0.49379

(2) P(x<-2.5)
z-score2 = 1−μ
σ = 1−(−2.5)
2 =1.75
=1- p(z>1.75)
=1 – 0.95994
= 0.04006
(3) P(-1.5<x<4.5)
z-score1=1.25
z-score2=1.75
=p(1.25<z<1.75)
=1- p(z>1.75) – p(z>1.25)
=1- 0.04006 -0.10565
=0.85429
(4) P(|x|>0.5)
=1-(-2<z<2)
=1-0.75
=0.25
(5) P(|X|<0.4)
=1-(-1.75<z<1.75)
=0.44066

Ans4
Angle 1 = 51°15´18" ± 8"
Angle 2 = 92°14´36" ± 7"
Angle 3 = 146°19´49" ± 6"
Angle 4 = 70°10´37" ± 8"
Angle 5 = 213°39´56" ± 8"
Angle 6 = 216°30´21" ± 6"
_______________________
788°127´271" ± 43"
--------------------------- = 131°41´46.167" ± 7.16"
6
The most probable values for the six angles=131°41´46.167" ± 7.16"
Error= 720°127´271" - 788°127´271"
= -68°127´271"
A:B:C:D:E:F =64:49:36:64:64:36 ------------------------(1)
A+B+C+D+E+F= -68°127´271" ----------------(2)
B=49/64 A ………………………………………(I)
C=36/64 A ………………………………………(II)
D=64/64 A ………………………………………(III)
E=64/64 A ………………………………………(IV)
F=49/64 A ………………………………………(V)
Put in equation (2)
A+ (49/64) A +(36/64) A +A+A +(36/64)A = -68°127´271"
A.488 = -68°127´271"
A.488 = -70.1769 °
A= -14°22´49.877"
A= -16.176944 degree
B = -12°22´ 40 .120"
D=--14°22´49.877"
C= --9°3´32.720"

E=-14°22´49.877"
F= -9°3´32.720"
Corrected angle
A =51.225° - 16.1769
A =35.0781
Corrected angle
F= 216.5058 – 9.0590 = 207.446°
B = 92.2433° - 12.2944° = 79.9489°
C = 146.33027° - 9.0590° = 137.27127°
D = 70.1769 ° -16.17694 =54.0009°
E= 213.6655° - 16.17694° = 197.4895°