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Calculating Heading And Ground Speed  
User currently offlinemawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0
Posted (2 years 1 month 1 week 6 days 20 hours ago) and read 3993 times:

"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]

2 replies: All unread, jump to last
 
User currently offlinesaafnav From South Africa, joined Mar 2010, 281 posts, RR: 1
Reply 1, posted (2 years 1 month 1 week 6 days 18 hours ago) and read 3947 times:

When you turn heading, your drift will be different than the first heading. So if you need turn left 10° to maintain track, and you see you only have 8° drift after the compensation, you need to correct for that 2° difference.

When you turn the ring, the drift will differ to the new heading.

So, you turn the ring, assess the drift, adjust for it again, asses the drift, adjust again. Until the change you make is less than one degree.


Hope that helps.
Erich



On-board Direction Consultant
User currently offlineDavid L From United Kingdom, joined May 1999, 9524 posts, RR: 42
Reply 2, posted (2 years 1 month 1 week 5 days 20 hours ago) and read 3814 times:

Quoting mawingho (Thread starter):
"Given the following, calculate heading and groundspeed:
... W/V 230/20; VAR 10 Degree E
Quoting mawingho (Thread starter):
Convert the track and wind to magnetic.

Complete the details you know:
...
W/V(M) = 310/20

Should the unconverted wind data read "320/20"?


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