Question solving, another technique 060 Navigation
Hi, I just wanted to show a technique i used which apparently is correct. It is for ATPL theory 060 navigation module.
The question states:
An aircraft T/O 2h before sunset, flying a track 090º(T), W/V 130º/20 kts, TAS 100kts. The correction he the admin used the computer, but i didn't. I found the GSo and GSh (Ground speed out and home), with cos(40º)=X/20, X the Head wind, X = 15kt. So Speed out is head wind so, GSo = 100-15 = 85 kts, and GSh back home is GSh = 100+15 = 115 kts. Now the only unvariable parameter is the time, we have 2 hours! And v=d/t (speed is distance over time), so t=d/v. In this case t = d/v1 +d/v2 indeed it is t= d/85+d/115, the time is both the same distance over the ground speed out PLUS + the same distance over the Ground speed home. then you can solve this equation:
t=2=(115d)/(9775)+(85d)/9775) finaly you get d = 97 NM!