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CR-3 ComputerWind Side
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5.FINDING WINDS IN FLIGHTGiven:
TAS 180 mph TC 175º TH 160º GS 144 mph
Find: Actual W/V
TC – TH = WCA
175º – 160º = 15º
WCA is 15º Left. Determine the “effective” TAS
ETAS = 174 mph
ETAS – GS = HW 174 – 144 = 30 mph
Head-wind component: 30 mph
Cross-wind component: 47 mph
Answer
Wind from 118º at 55 mph
6.TRUE COURSE (TRACK) AND GROUND SPEED
Given: TAS 156 mph MH 289º Var. 7ºW W/V 180º /40 mph
Find: TCGS
Left crosswind component: 39 mph
WCA: 14º
It now appears that the first WCA of 14º was 1º too much (the new one is 13º)
Back off 1º of the adjustment, making a true course reading of 295º
The crosswind component is still 36
The TC is 295º
The crab angle is greater than 10º (13º), it is necessary to use “effective TAS”
17 mph tailwind component
ETAS + TW = GS
152 +17 = 169 mph
True course = 295º
7.TRUE HEADING AND TRUE AIR SPEEDGiven:
TC 56º Desired GS 166 kts Wind 120º/45kts
Find: TAS TH
Desired GS: 166 kts
Headwind component: 20 kts
The TAS (or effective): 166 + 20 = 186 kts
The WCA is greater than 10º
Move the bottom disc till 186 is opposite 12º on the black scale (effective TAS)
Check to see that the 40-knot crosswind component on the outer scale is still close to 12º on the inner scale
Note that the TAS index points to 19
The wind is from the right TC + CA = TH or
…..
8.OFF-COURSE CORRECTION
Given: - Miles flown: 40 - Miles off course: 5 - Miles to destination: 160
Find: Degrees correction to heading to reach destination directly
7º is the number of degrees you must correct your heading in order to parallel your intended course
It is now necessary to find the number of degrees additional correction needed to reach your destination
You must decide whether the next correction should be 18º or 2º
Common sense will tell you that 2º
Remember the rule that 1º of drift will give approximately 1 mile off course in 60
Answer:
7º + 2º = 9º