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°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 Adjusted Lengths and Directions 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°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°𝟏𝟗′ 𝐄 609.153 Bearing′bc = tan−1 587.673 = 𝐍 𝟒𝟔°𝟎𝟐′ 𝐄 788 Bearing′cd = tan−1 −328.403 = 𝐒 𝟔𝟕°𝟐𝟑′ 𝐄 215.817 Bearing′de = tan−𝟏 −1000.425 = 𝐒 𝟏𝟐°𝟏𝟎′ 𝐄 ′ Bearing′ef = tan−1 −1114.622 −125.713 = 𝐒 𝟖𝟑°𝟑𝟒 𝐖 ′𝐖 Bearing′fa = tan−1 −544.081 = 𝐍 𝟓𝟓°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°𝟑𝟎′ 𝐄 608.898 Bearing′bc = tan−1 587.117 = 𝐍 𝟒𝟔°𝟎𝟑′ 𝐄 786.789 Bearing′cd = tan−1 −328.403 = 𝐒 𝟔𝟕°𝟐𝟒′ 𝐄 218.313 Bearing′de = tan−𝟏 −1002.762 = 𝐒 𝟏𝟐°𝟏𝟕′ 𝐄 ′ Bearing′ef = tan−1 −1116.625 = 𝐒 𝟖𝟑°𝟒𝟒 𝐖 −122.662 ′𝐖 Bearing′fa = tan−1 −544.641 = 𝐍 𝟓𝟓°𝟐𝟕 375.006 THANK YOU