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