# Chapter 5 Traversing and Traverse Computations ```Chapter 5: Traversing and Traverse
Computations
Example: A cylinder weighing 400 lb is held against a smooth incline by means of a
weightless rod AB in Fig. P-309. Determine the forces P and N exerted on the
cylinder by the rod and the incline.
Solution by Force Polygon:
𝑳𝒂𝒕 = 𝒅 𝐜𝐨𝐬 𝜽
𝑫𝒆𝒑 = 𝒅 𝐬𝐢𝐧 𝜽
𝑳𝒂𝒕 = 𝒅 𝐜𝐨𝐬 𝜽
𝑫𝒆𝒑 = −𝒅 𝐬𝐢𝐧 𝜽
𝑳𝒂𝒕 = −𝒅 𝐜𝐨𝐬 𝜽
𝑫𝒆𝒑 = −𝒅 𝐬𝐢𝐧 𝜽
𝑳𝒂𝒕 = −𝒅 𝐜𝐨𝐬 𝜽
𝑫𝒆𝒑 = 𝒅 𝐬𝐢𝐧 𝜽
Error of Closure
The linear error of closure (LEC) is usually a short line of unknown length and
direction connecting the initial and final stations of the traverse. It is approximately
determined by plotting the traverse to scale, or more exactly by computing the
hypotenuse of a right triangle whose sides are the closure in latitudes and the closure
in departures, respectively. This quantity reflects the algebraic sum of all the
accumulated errors of measurement both in angles and distances when running the
traverse. The length of the linear error of closure and the angle that this line makes
with the meridian is determined by the following equations
𝑳𝑬𝑪 =
Where:
𝑪𝑳 𝟐 + 𝑪𝑫 𝟐 and
𝐭𝐚𝐧 𝜽 =
−𝑪𝑫
−𝑪𝑳
LEC = linear error of closure
CL = closure in latitude or the algebraic sum of north and south latitudes
CD = closure in departure or the algebraic sum of east and west departures
θ = bearing angle of the side of error
Sample Problem:
5
Sample Problem:
5
Sample Problem:
5
L. E. C. = −16.816
= 𝟐𝟒. 𝟔𝟖𝟔 𝐦
−1
Bearing = tan
24.686
1
𝟏
−3
R. P. =
= 4.937 𝑥 10 =
=
5000.13
202.549 𝟐𝟎𝟎
2
+ −18.073
2
−18.073
= 𝐒 𝟒𝟕&deg;𝟒′ 𝐖
−16.816
Compass Rule
The compass or Bowditch rule which was named after the distinguished
American navigator Nathaniel Bowditch (1773 - 1838), is a very popular rule for
adjusting a closed traverse. Of the several methods used for balancing latitudes and
departures, perhaps the most commonly used is this rule. It is simple to apply and at
the same time theoretically sound.
𝒄𝒍 = 𝑪𝑳
Where:
𝒅
𝑫
and
𝒄𝒅 = 𝑪𝑫
𝒅
𝑫
cl = correction to be applied to the latitude of any course
cd = correction to be applied to the departure of any course
CL = total closure in latitude or the algebraic sum of the north and south
latitudes (ƩNL + ƩSL)
CD = total closure in departure or the algebraic sum of the north and
south latitudes (ƩED + ƩWD)
d = length of any course
D = total length or perimeter of the traverse
After the latitudes and departures of the courses of a closed traverse have been
so adjusted, the bearings (or azimuths) of the courses and their lengths should also be
adjusted to correspond to the adjusted latitudes and departures. The following are the
equations used for this purpose.
𝑳′ =
Where:
𝑳𝒂𝒕′
𝟐
+ 𝑫𝒆𝒑′
𝟐
and
𝐭𝐚𝐧 𝜽 =
𝑫𝒆𝒑′
𝑳𝒂𝒕′
L’ = adjusted length of a course
Lat’ = adjusted latitude of a course
Dep’ = adjusted departure of a course
θ = adjusted horizontal angle between the reference meridian and a
course
𝐂𝐨𝐦𝐩𝐚𝐬𝐬 𝐑𝐮𝐥𝐞
CL
cl =
d
D
CL
16.816
=
= 3.363 𝑥 10−3 → 𝐀
D 5000.13
cab = 495.85 𝐀 = 1.668 m
cbc = 850.62 𝐀 = 2.861 m
ccd = 855.45 𝐀 = 2.877 m
cde = 1020.87 𝐀 = 3.433 m
cef = 1117.26 𝐀 = 3.757 m
cfa = 660.08 𝐀 = 2.220 m
Adj. Lat ab = 493.567 − 1.668 = 491.899 m
Adj. Lat bc = 590.534 − 2.861 = 587.673 m
Adj. Lat cd = −325.526 − 2.877 = −328.403 m
Adj. Lat de = −996.992 − 3.433 = −1000.425 m
Adj. Lat ef = −121.956 − 3.757 = −125.713 m
Adj. Lat fa = 377.189 − 2.220 = 374.969 m
𝐂𝐨𝐦𝐩𝐚𝐬𝐬 𝐑𝐮𝐥𝐞
CD
cd =
d
D
CD
18.073
=
= 3.615 𝑥 10−3 → 𝐁
D
5000.13
cab = 495.85 𝐁 = 1.792 m
cbc = 850.62 𝐁 = 3.075 m
ccd = 855.45 𝐁 = 3.092 m
cde = 1020.87 𝐁 = 3.690 m
cef = 1117.26 𝐁 = 4.038 m
cfa = 660.08 𝐁 = 2.386 m
Adj. Depab = 47.525 − 1.792 = 45.733 m
Adj. Depbc = 612.228 − 3.075 = 609.153 m
Adj. Depcd = 791.092 − 3.092 = 788.000 m
Adj. Depde = 219.507 − 3.690 = 215.817 m
Adj. Depef = −1110.584 − 4.038 = −1114.622 m
Adj. Depfa = −541.695 − 2.386 = −544.081 m
L. E. C. = −16.816
= 𝟐𝟒. 𝟔𝟖𝟔 𝐦
−1
Bearing = tan
24.686
1
𝟏
−3
R. P. =
= 4.937 𝑥 10 =
=
5000.13
202.549 𝟐𝟎𝟎
2
+ −18.073
2
−18.073
= 𝐒 𝟒𝟕&deg;𝟒′ 𝐖
−16.816
𝐂𝐨𝐦𝐩𝐚𝐬𝐬 𝐑𝐮𝐥𝐞
d′ab =
d′bc =
d′cd =
d′de =
d′ef =
d′fa =
491.899 2 + 45.733 2 = 𝟒𝟗𝟒. 𝟎𝟐 𝐦
587.673 2 + 609.153 2 = 𝟖𝟒𝟔. 𝟒𝟐 𝐦
−328.403 2 + 788 2 = 𝟖𝟓𝟑. 𝟔𝟗 𝐦
−1000.425 2 + 215.817 2 = 𝟏𝟎𝟐𝟑. 𝟒𝟒 𝐦
−125.713 2 + −1114.622 2 = 𝟏𝟏𝟐𝟏. 𝟔𝟗 𝐦
374.969 2 + −544.081 2 = 𝟔𝟔𝟎. 𝟕𝟖 𝐦
45.733
Bearing′ab = tan−1 491.899
= 𝐍 𝟎𝟓&deg;𝟏𝟗′ 𝐄
609.153
Bearing′bc = tan−1 587.673
= 𝐍 𝟒𝟔&deg;𝟎𝟐′ 𝐄
788
Bearing′cd = tan−1 −328.403
= 𝐒 𝟔𝟕&deg;𝟐𝟑′ 𝐄
215.817
Bearing′de = tan−𝟏 −1000.425
= 𝐒 𝟏𝟐&deg;𝟏𝟎′ 𝐄
′
Bearing′ef = tan−1 −1114.622
−125.713 = 𝐒 𝟖𝟑&deg;𝟑𝟒 𝐖
′𝐖
Bearing′fa = tan−1 −544.081
=
𝐍
𝟓𝟓&deg;𝟐𝟔
374.969
Transit Rule
The method of adjusting a traverse by the transit rule is similar to the method
using the compass rule. The main difference is that with the transit rule, the latitude
and departure corrections depend on the length of the latitude and departure of the
course respectively instead of both depending on the length of the course.
𝒄𝒍 =
Where:
𝑳𝒂𝒕
𝑪𝑳
Ʃ𝑵𝑳−Ʃ𝑺𝑳
and
𝒄𝒅 =
𝑫𝒆𝒑
𝑪𝑫
Ʃ𝑬𝑫−Ʃ𝑾𝑫
cl = correction to be applied to the latitude of any course
cd = correction to be applied to the departure of any course
CL = total closure in latitude or the algebraic sum of the north and south
latitudes (ƩNL + ƩSL)
CD = total closure in departure or the algebraic sum of the north and
south latitudes (ƩED + ƩWD)
𝐓𝐫𝐚𝐧𝐬𝐢𝐭 𝐑𝐮𝐥𝐞
cl =
CL
Lat
Lat
cab = 493.567 𝐀
cbc = 590.534 𝐀
ccd = 325.526 𝐀
cde = 996.992 𝐀
cef = 121.956 𝐀
cfa = 377.189 𝐀
CL
16.816
=
= 5.787 𝑥 10−3 → 𝐀
Lat
2905.764
= 2.856 m
= 3.417 m
= 1.884 m
= 5.770 m
= 0.706 m
= 2.183 m
Adj. Lat ab = 493.567 − 2.856 = 490.711 m
Adj. Lat bc = 590.534 − 3.417 = 587.117 m
Adj. Lat cd = −325.526 − 1.884 = −327.410 m
Adj. Lat de = −996.992 − 5.770 = −1002.762 m
Adj. Lat ef = −121.956 − 0.706 = −122.662 m
Adj. Lat fa = 377.189 − 2.183 = 375.006 m
𝐓𝐫𝐚𝐧𝐬𝐢𝐭 𝐑𝐮𝐥𝐞
cd =
CD
Dep
Dep
CD
18.073
=
= 5.439 𝑥 10−3 → 𝐁
Dep
3322.631
cab = 47.525 𝐁 = 0.259 m
cbc = 612.228 𝐁 = 3.330 m
ccd = 791.092 𝐁 = 4.303 m
cde = 219.507 𝐁 = 1.194 m
cef = 1110.584 𝐁 = 6.041 m
cfa = 541.695 𝐁 = 2.946 m
Adj. Depab = 47.525 − 0.259 = 47.266 m
Adj. Depbc = 612.228 − 3.330 = 608.898 m
Adj. Depcd = 791.092 − 4.303 = 786.789 m
Adj. Depde = 219.507 − 1.194 = 218.313 m
Adj. Depef = −1110.584 − 6.041 = −1116.625 m
Adj. Depfa = −541.695 − 2.946 = −544.641 m
𝐓𝐫𝐚𝐧𝐬𝐢𝐭 𝐑𝐮𝐥𝐞
d′ab =
d′bc =
d′cd =
d′de =
d′ef =
d′fa =
490.711 2 + 47.266 2 = 𝟒𝟗𝟐. 𝟗𝟖 𝐦
587.117 2 + 608.898 2 = 𝟖𝟒𝟓. 𝟖𝟓 𝐦
−327.41 2 + 788 2 = 𝟖𝟓𝟐. 𝟏𝟗 𝐦
−1002.762 2 + 218.313 2 = 𝟏𝟎𝟐𝟔. 𝟐𝟓 𝐦
−122.662 2 + −1116.625 2 = 𝟏𝟏𝟐𝟑. 𝟑𝟒 𝐦
375.006 2 + −544.641 2 = 𝟔𝟔𝟏. 𝟐𝟔 𝐦
47.266
Bearing′ab = tan−1 490.711
= 𝐍 𝟎𝟓&deg;𝟑𝟎′ 𝐄
608.898
Bearing′bc = tan−1 587.117
= 𝐍 𝟒𝟔&deg;𝟎𝟑′ 𝐄
786.789
Bearing′cd = tan−1 −328.403
= 𝐒 𝟔𝟕&deg;𝟐𝟒′ 𝐄
218.313
Bearing′de = tan−𝟏 −1002.762
= 𝐒 𝟏𝟐&deg;𝟏𝟕′ 𝐄
′
Bearing′ef = tan−1 −1116.625
=
𝐒
𝟖𝟑&deg;𝟒𝟒
𝐖
−122.662
′𝐖
Bearing′fa = tan−1 −544.641
=
𝐍
𝟓𝟓&deg;𝟐𝟕
375.006
THANK YOU
```