"Given the following, calculate heading and groundspeed:
Track 270 Degree T; TAS 110KT; W/V 230/20; VAR 10 Degree E
Convert the track and wind to magnetic.
Complete the details you know:
TAS = 110
TR(M) = 260
W/V(M) = 310/20
HDG(M) = ________
GS = _______
1. Set TAS and mark W/V. Using an ASA E6-B flgiht computer, mark the wind on the wind face - set 310 against the "true index". Move the slide to put the TAS of 110 kt under the centre grommet then mark a small dot 20 kt below the grommet (at the 90 kt point) to represent the W/V.
2. Drift and HDG. Rotate the transparent face (inner direction ring) to set the TR of 260 against the true index. Read the drift; it is 9 degree left. We know the wind is from 310 so from out track of 260 turn the 9 degree into wind (towards 310). The heading looks like 269 but because we have turned the direction ring, the indicated drift is now 8 degree. So the refined HDG is 268; the drift is 8 degree left.
3. Track and groundspeed. The track is known as 260. To find the G/S, read it off under the wind mark; it is 96 kt."
Reference: Robson, David. "Chapter 12: Introduction to Navigation" Basic Aeronautical Knowledge (BAK). Darra: Aviation Theory Centre Pty, 2009. 360. Print.
I do not understand the sentence "The heading looks like 269 but because we have turned the direction ring, the indicated drift is now 8 degree. So the refined HDG is 268; the drift is 8 degree left."
and I have got no idea why the we need to refine the HDG to 268?
I hope someone can help me out. Thanks!
[Edited 2012-09-08 12:12:00]