1 / 14

Angular Measurement

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

yuri
Download Presentation

Angular Measurement

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. Direction of a Line • Same units as Angles • Meridians – North/South Lines • True • Magnetic • Grid • Assumed • Azimuth • Bearing

  3. Azimuth • Angles Measured Clockwise from North • Military and Astronomers use South • Angle between 0 and 360° • Back Az = Az ± 180° • AzBC = Back AzAB + <A

  4. 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

  5. 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

  6. 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

  7. 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”

  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: (-)

  9. 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’

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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°

More Related