My Math Forum Deflection Angle of Closed Traverse: Traverse Theory

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October 31st, 2009, 07:56 AM   #1
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Deflection Angle of Closed Traverse: Traverse Theory

How do I calculate the deflection angle at B? The deflection at angle B is basically the line AB extended past B, and the angle created by B,B',C if that makes sense.

[attachment=1:2j71u0mv]Traverse Theory.JPG[/attachment:2j71u0mv]

[attachment=0:2j71u0mv]TT Cleared.JPG[/attachment:2j71u0mv]
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 Traverse Theory.JPG (32.8 KB, 2282 views) TT Cleared.JPG (15.6 KB, 2281 views)

 October 31st, 2009, 08:22 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,285 Thanks: 1681 2. Interior angle at B = 42°36' + 64°56' = 107°32'. Since it's an interior angle, not a deflection angle, the left/right question doesn't make sense. 1. Your definition of deflection angle builds in "clockwise" (i.e., "right"), rather than anti-clockwise, whenever the vertices are alphabetically ordered clockwise (at least when the polygon is convex). The deflection angle at B = 180° - 107°32' = 72°28'. 3. Bearing of CD is S64°56'E - 38°16' = S26°40'E. 4. Can you do this final part? (Check the meaning of "azimuth" if you don't already know it.)
 October 31st, 2009, 11:56 AM #3 Newbie   Joined: Oct 2009 Posts: 4 Thanks: 0 Re: Deflection Angle of Closed Traverse: Traverse Theory Ok, so in a clockwise approach: Az AB= 42°36' +180° Az BA= 222°36' -?B 107°32' Az BC= 115°04' +180° Az CB= 295°04' -?C 143°44' Az CD= 151°20' +180° Az DC= 331°20' -?D 38°16' Az DA= 293°04' -180° Az AD= 113°04' What am I doing wrong?
 October 31st, 2009, 01:19 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,285 Thanks: 1681 What the . . . ? The line is already annotated N70°42'W. Just subtract that angle from 360° to get the angle clockwise from due North, i.e., 289°18'.
 October 31st, 2009, 01:40 PM #5 Newbie   Joined: Oct 2009 Posts: 4 Thanks: 0 Re: Deflection Angle of Closed Traverse: Traverse Theory OH I think I see now, so would this be TRUE of azimuth and bearing relation: N.E. Quadrant: Bearing = Azimuth S.E. Quadrant: 180° - Azimuth = Bearing AND 180° - Bearing = Azimuth S.W. Quadrant: Azimuth - 180° = Bearing AND 180° + Bearing = Azimuth N.W. Quadrant: 360° - Azimuth = Bearing AND 360° - Bearing = Azimuth I tried to calculate it by going through the entire process, but should I not have produced the same answer anyway?
 October 31st, 2009, 02:42 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,285 Thanks: 1681 S.W. Quadrant: Azimuth - 180° = Bearing AND 180° + Bearing = Azimuth is what you probably intended. Now you should be able to spot the errors you made earlier.
 October 31st, 2009, 02:53 PM #7 Newbie   Joined: Oct 2009 Posts: 4 Thanks: 0 Re: Deflection Angle of Closed Traverse: Traverse Theory Will do. Thanks you've been an awesome amount of help.

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