Survey Example Problem Interior Line Correction New Station Angle Length (L) Factor Interior Angle A 86° 25' 45" AB = 784.23 23" 86°26' 8" В 125° 59' 11" BC = 863.79 23" 125° 59' 34" 120° 60° 120° 45' 41" 60° 31' 55" C 45' 18" CD = 1107.53 23" 31' 32" DE = 1342.47 23" E 146° 16' 19" EA = 654.13 23" 146° 16' 42" ==========D Sum 539° 58' 05" Sum 4752.15' Sum 1' 15" Sum 540° Known Bearing: AB = N 38° 37' 22" W Sum of the IA = (n-2) * 180 = 540 New Decimal Latitude Departure = sin(B) *L Bearing Bearing(B) = cos(B) *L AB = N 38° 37' 22" W BC = S 87° 22' 12" W 38.6228° +612.6971 -489.5089 87.37° -39.6359 -862.8802 CD = S 28° 7' 53" W DE = N 88° 39' 48" E 28.1314° 88.6633° - 976.6959 -522.1951 +31.3167 +1342.1047 EA = N 54° 56' 30" E 54.9417° +375.7386 +535.4499 AB = N 38° 37' 22" W Sum of Lat = Sum of Dep = 3.4206' 2.9704' What is the Error of Closure? EC = SQRT (Sum of Lat.' + Sum of Dep.?) _4.5303' What is the Relative Accuracy? RA = Perimeter / Error of Closure _1048.9703 Would this be a good or bad survey job and why? Bad, Relative Accuracy needs to be 3000 or better
Survey Example Problem Interior Line Correction New Station Angle Length (L) Factor Interior Angle A 86° 25' 45" AB = 784.23 23" 86°26' 8" В 125° 59' 11" BC = 863.79 23" 125° 59' 34" 120° 60° 120° 45' 41" 60° 31' 55" C 45' 18" CD = 1107.53 23" 31' 32" DE = 1342.47 23" E 146° 16' 19" EA = 654.13 23" 146° 16' 42" ==========D Sum 539° 58' 05" Sum 4752.15' Sum 1' 15" Sum 540° Known Bearing: AB = N 38° 37' 22" W Sum of the IA = (n-2) * 180 = 540 New Decimal Latitude Departure = sin(B) *L Bearing Bearing(B) = cos(B) *L AB = N 38° 37' 22" W BC = S 87° 22' 12" W 38.6228° +612.6971 -489.5089 87.37° -39.6359 -862.8802 CD = S 28° 7' 53" W DE = N 88° 39' 48" E 28.1314° 88.6633° - 976.6959 -522.1951 +31.3167 +1342.1047 EA = N 54° 56' 30" E 54.9417° +375.7386 +535.4499 AB = N 38° 37' 22" W Sum of Lat = Sum of Dep = 3.4206' 2.9704' What is the Error of Closure? EC = SQRT (Sum of Lat.' + Sum of Dep.?) _4.5303' What is the Relative Accuracy? RA = Perimeter / Error of Closure _1048.9703 Would this be a good or bad survey job and why? Bad, Relative Accuracy needs to be 3000 or better
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Could I get assitance with the yellow highlighted part? I can not figure out how to determine Line BC? Thank you.
Transcribed Image Text:Survey Example Problem
Interior
Line
Correction
New
Station
Angle
Length (L)
Factor
Interior Angle
А
86°
25'
45"
AB = 784.23
23"
86°26' 8"
125° 59' 34"
120° 45' 41"
В
125°
59'
11"
BC = 863.79
23"
C
120°
45'
18"
CD = 1107.53
23"
D
60°
31'
32"
DE = 1342.47
23"
60° 31' 55"
E
146°
16'
19"
EA = 654.13
23"
146° 16' 42"
=========
===
Sum 539° 58' 05"
Sum 4752.15'
Sum 1' 15"
Sum 540°
Known Bearing: AB = N 38° 37' 22" W
Sum of the IA = (n-2) * 180 =
540
New
Decimal
Latitude
Departure
= sin(B) *L
Bearing
Bearing(B)
= cos(B) * L
AB = N 38° 37' 22" W
38.6228°
+612.6971
-489.5089
BC = S 87° 22' 12" W
CD = S 28° 7' 53" W
DE = N 88° 39' 48" E
87.37°
-39.6359
-862.8802
28.1314°
- 976.6959
-522.1951
88.6633°
+31.3167
+1342.1047
EA = N 54° 56' 30" E
54.9417°
+375.7386
+535.4499
AB = N 38° 37' 22" W
Sum of Lat =
Sum of Dep =
3.4206'
2.9704'
What is the Error of Closure?
EC = SQRT (Sum of Lat. + Sum of Dep.)
2
_4.5303'
What is the Relative Accuracy?
RA = Perimeter / Error of Closure
_1048.9703
Would this be a good or bad survey job and why? Bad, Relative Accuracy needs to be 3000 or better
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