6. CR-3 Wind Side 2/2
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6. CR-3 Wind Side 2/2

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Second part of the presentation shown in class abot the CR-3 Wind side

Second part of the presentation shown in class abot the CR-3 Wind side

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6. CR-3 Wind Side 2/2 6. CR-3 Wind Side 2/2 Presentation Transcript

  • CR-3 Computer Wind Side 2/2
  • 5.FINDING WINDS IN FLIGHT
    • Given:
      • 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:
      • TC
      • GS
  •  
  •  
  •  
    • Left crosswind component: 39 mph
    • WCA: 14º
  •  
  •  
  •  
    • It now appears that the first WCA of 14º was 1º too much (the new one is 13 º)
    • 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 SPEED
    • Given:
      • 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