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Angular Measurement . Requires three points From – Backsight reference At – Where the transit is located To – Foresight Established or Determined Units of Measure Degrees, Min, Sec Radians - rad = 180 degrees Gons – European, 400 Gon/circle. Direction of a Line. Same units as Angles
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Angular Measurement • Requires three points • From – Backsight reference • At – Where the transit is located • To – Foresight • Established or Determined • Units of Measure • Degrees, Min, Sec • Radians - rad = 180 degrees • Gons – European, 400 Gon/circle
Direction of a Line • Same units as Angles • Meridians – North/South Lines • True • Magnetic • Grid • Assumed • Azimuth • Bearing
Azimuth • Angles Measured Clockwise from North • Military and Astronomers use South • Angle between 0 and 360° • Back Az = Az ± 180° • AzBC = Back AzAB + <A
Bearings • Quadrants – NE, SE, SW, NW • Always < 90 • Written N23°15’W • Back Bearing – change directions • Bearing AB = N23°15’W • Back bearing AB = bearing BA=S23°15’E
Bearings to Azimuths • Each quadrant is different • NE: Bearing E from N Az = Bearing Angle (BA) NW NE • SE: Bearing E from S Az = 180 – BA • SW: Bearing W from SAz = 180 + BA • NW: Bearing E from NAz = 360 – BA SW SE
Problem 9-3, McCormac’s 4th Az OA = 141°16’ (SE Quad) B.A. = 180° - 141°16’= 38°84’ Bearing = S38°84’E Az OB = 217°23’ (SW Quad) B.A. = 217°23’ - 180° = 37°23’ Bearing = S37°23’W Az OC = 48°23’ (NE Quad) B.A. = Az; Bearing = N48°23’E
Problem 9-5, McCormac’s 4th Az OA = 17°22’16” Az OB = 180° - 70°18’46” = 109°41’14” Az OB = 180° + 11°8’52” = 191°8’52” Az OB = 360° - 76°30’52” = 283°29’8”
Traverse Directions • Given: • The direction of a side • The connecting angle • Find: • The direction of the adjacent side • Determine back azimuth • Add angle • Clockwise: (+) • Counterclockwise: (-)
C 98°27’ B 35°15’ 215°15’ A Traverse Azimuths Example • Az AB = 35°15’ • ABC = 98°27’ • Find AZ BC • Back Az = Az 180° • Az BA = 215°15’ • 215°15’ + 98°27’ = 313°42’
N N 250°42’ 70°42’ B 97°18’ A C Problem 9-7, McCormac’s 4th Az AB = 70°42’Az BA = 250°42’ ABC = 97°18’ Left (CCW) so Az BC = 250°42’ - 97°18’ = 153°24’ SE Quadrant:B.A. = 180° - 153°24’ = 26°36’ Bearing = S 26°36’E
Problem 9-7, McCormac’s 4th Az BC = 162°28’Az BA = 342°28’ BCD = 67°38’ Left (CCW) so Az CD = 342°28’ - 67°38’ = 274°50’ NW Quadrant:B.A. = 360° - 274°50’ = 85°10’ Bearing = N 85°10’W
Problem 9-17, McCormac’s 5th Az 34 = 351°50’00”Az 43 = 171°50’00”- 4 - 221°37’56” - 49°47’56” +360°00’00”Az 41 = 310°12’04” NW Quad, B.A. = 360° - Az Bearing 41 = N 49°47’56”W
Problem 9-17, McCormac’s 5th Az 41 = 310°12’04” Az 14 = 130°12’04”- 1 - 51°16’00”Az 12 = 78°56’04” NE Quad, B.A. = Az Bearing 12 = N 78°56’04”E Az 21 = 258°56’04”- 2 - 36°22’00”Az 23 = 222°34’04” SW Quad, B.A. = Az - 180° Bearing 23 = S42°34’04”E
Problem 9-17, McCormac’s 5th 3 = 360° - 221°37’56”- 51°16’00” - 36°22’00” = 50°44’04” Az 32 = 42°34’04”- 3 - 50°44’04” - 8°10’00” +360°00’00”Az 34 = 351°50’00” NW Quad, B.A. = 360° - Az Bearing 34 = N 8°10’00”W Closed traverse, Interior angles ’s = (n-2)180 = (4-2)180 = 360°