property of lear siegler. 0° wind effect variation deviation property of lear sielger
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Property of Lear Siegler
0° WIND EFFECT
VARIATION
DEVIATIONProperty of Lear Sielger
Terminal Learning Objective
At the completion of this lesson the student will:
Action: Acting as a pilot, plan a flight
Condition: Given a VFR flight mission, required equipment, charts and publications
Standard: In Accordance With (IAW) Army Regulation (AR) 95-1, Federal Aviation Regulation (FAR) Part 91, and Field Manual (FM) 3-04.240
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Enabling Learning Objective (ELO) #1
Action: Solve for drift corrections necessary to maintain the true course
Condition: Given a True Course
Standard: IAW FM 1-240
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The proposed flight path measuredclockwise from true north.
TRUE COURSE
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True North(Meridian)
clockwise toflight path
The correction applied to prevent
drifting off course.
DRIFT CORRECTION
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wind 270°
DRIFT CORRECTIONNOT APPLIED
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wind 270°
Headwind = No Drift
Left Crosswind = Right Drift
DRIFT CORRECTIONNOT APPLIED
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wind 270°
True Course
Actual Flight Path
Right Crosswind = Left Drift
DRIFT CORRECTIONNOT APPLIED
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wind 270°
True Course
Actual Flight Path
DRIFT CORRECTIONNOT APPLIED
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wind 270°
Tailwind = No Drift
LEFT DRIFT CORRECTIONIS SUBTRACTED FROM TC
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TC = 360°
DC = 10°LEFT
TC = 360°DC = 10° L
360° - 10°= 350°
RIGHT DRIFT CORRECTIONIS ADDED TO TC
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TC = 360°
DC = 10°RIGHT
TC = 360°DC = 10°R
360° + 10° = 010°
TRUE HEADING
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The angle measured clockwise fromtrue north to the nose of the aircraft.
TC +/- DC = TH
TC
TH
DC
WIND 360
Questions?
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Enabling Learning Objective (ELO) #2
Action: Identify the effects of wind on an aircraft in flight
Condition: Given situational data
Standard: IAW FM 1-240
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TRACK
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THE ACTUAL FLIGHT PATH OVER THE GROUND,MEASURED CLOCKWISE FROM TRUE NORTH
TR
TN
DRIFT ANGLE
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THE ANGLE BETWEEN TRUEHEADING (TH) AND TRACK (TR)
TR
DATC
TH
DC
WIND 360
WIND IS MORE THAN PLANNING WINDS, DA > DC
DRIFT ANGLE
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THE ANGLE BETWEEN TRUEHEADING (TH) AND TRACK (TR)
TRDA
TC
TH
DC
WIND 360
WIND IS LESS THAN PLANNING WINDS, DA < DC
DRIFT ANGLE
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THE ANGLE BETWEEN TRUEHEADING (TH) AND TRACK (TR)
TR = TC
DA = DC
WIND EQUAL TO PLANNING WINDS, DA = DC
TH
DC
WIND 360
TC
TRACK EQUAL TRUE COURSE WHEN:
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1. NO WIND CONDITION
2. HEAD WIND CONDITION
3. TAIL WIND CONDITION
4. DRIFT CORRECTION IS CORRECT
Types of Airspeed
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Indicated Airspeed (IAS) - the airspeed readdirectly from the airspeed indicator.
IAS is not affected by wind.
Types of Airspeed
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Calibrated Airspeed (CAS) - indicated airspeedcorrected for instrument installation error.
CAS is not affected by the wind.
Types of Airspeed
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True Airspeed (TAS) - calibrated airspeed correctedfor error due to air density (altitude and temperature).
TAS is not affected by the wind.
1000’ PA, 13° C = 92 KTAS
4000’ PA, 7° C = 96 KTAS
90 KIAS
GROUND SPEED
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Ground Speed - speed of the aircraft over the ground.GS is TAS +/- wind speed. Used to calculate time
required to fly a certain distance.
TAS 92 KNOTS
AIRMASS MOVEMENT 10 KNOTS
TAS 92 KTS WIND SPEED - 10 KTS
GROUND SPEED = 82 KTS
GROUND SPEED
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Ground Speed - speed of the aircraft over the ground.GS is TAS +/- wind speed. Used to calculate time
required to fly a certain distance.
TAS 92 KNOTS
AIRMASS MOVEMENT 10 KNOTS
TAS 92 KTS WIND SPEED + 10 KTS
GROUND SPEED = 102 KTS
GROUND SPEED
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Headwinds - winds +/- 90° of the aircraft nose reduceground speed. TAS minus wind velocity.
GROUND SPEED
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Tailwinds - winds +/- 90° of the aircraft tail increaseground speed. TAS plus wind velocity.
Questions?
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Enabling Learning Objective (ELO) #3
Action: Complete the wind effect components of the VFR flight log
Condition: Given situational data
Standard: IAW the Student Handout
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Transport a passenger from Hanchey AHP31° 21’ N, 85° 40’ W to Sikes airport30° 47’ N, 86° 31’ W. The wind is
260° / 20 kts and will require a 9° DC.
1. Draw the true course.
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H
2. Determine wind direction in relation to the true course.
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H
WIND 260°
3. Complete the wind componentpart of the VFR flight log.
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H
WIND 260°
TC +/- DC = TH232° + 9°R = 241°
GS will be less thanTAS because of headwind.
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Fly from Skelly AHP, 31° 18’ N, 86° 07’ W to Crenshaw Memorial, 31° 51’ N, 86° 36’ W.The wind is 20015 and the DC is 12
Questions?
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Enabling Learning Objective (ELO) #4
Action: Select the definition of magnetic variation, agonic and isogonic lines and their application
Condition: Given situational data
Standard: IAW Fm 1-240 and the Student Handout
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VARIATIONThe angular difference between
true north (TN) and magnetic north (MN).
TN MN
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VARIATIONThe angular difference between
true north (TN) and magnetic north (MN).
TN MN
NOTE :Magnetic north is moving,
therefore magnetic variationchanges from one chart
edition to another.
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1° W
1° 30’ W
0° 30’ W
AGONIC AND ISOGONIC LINESRepresented on sectional charts as
dashed magenta lines, numbered and lettered, e.g. 1° W, in 1/2° (30’) increments.
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0°
AGONIC LINEA line connecting points of zero degree variation.There is only one agonic line in the United States.
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1° W
1° 30’ W
0° 30’ W
ISOGONIC LINESLines connecting points of equal magnetic variation.
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EASTERLY VARIATIONCompass needle points east of TN
and is subtracted. “EAST IS LEAST”
TN MN
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TH = 360°, compass is pointing 10° eastof TN. Must subtract 10° in order to fly to TN.
TC +/- DC = TH +/- VAR = MH360° + 0° = 360° - 10° = 350°
This example assumes a calm wind.TN MN
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MH = 350°, compass is pointing 10° eastof TN. Subtracted 10° in order to fly to TN.
TN MN
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WESTERLY VARIATIONCompass needle points west of TNand is added. “WEST IS BEST”
TNMN
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TH = 360°, compass is pointing 10° westof TN. Must add 10° in order to fly to TN.
TC +/- DC = TH +/- VAR = MH360° + 0° = 360° + 10° = 010°
This example assumes a calm wind.
TNMN
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MH = 010°, compass is pointing 10° westof TN. Added 10° in order to fly to TN.
TNMN
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1° W
1° 30’ W
0° 30’ W
APPLYING VARIATIONFind mid-point of TC, apply nearest
whole degree of variation.
A B
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1° E
0° 30’ E
1° 30’ E
1. Find mid-point of course leg.
A
B
2. Find nearest WHOLE degree of variation.
3. TC +/- DC = TH +/- VAR = MH 320° + 0° = 320° - 1°- 1° = 319°
EAST IS LEASTEAST IS LEAST
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1° W
1° 30’ W
0° 30’ W
1. Find mid-point of course leg.
A B
2. Find nearest WHOLE degree of variation.
3. TC +/- DC = TH +/- VAR = MH 090° + 0° = 090° +1°+1° = 091°
WEST IS BESTWEST IS BEST
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MAGNETIC HEADINGNow that you know how to figure out MH, the next
question is “What is it ?”
Magnetic Heading is measured from magneticnorth clockwise to the nose of the aircraft.
MN MHVAR
Questions?
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Enabling Learning Objective (ELO) #5
Action: Select the definition of deviation, use of the compass deviation card and completing VFR flight log
Condition: Given situational data
Standard: IAW Fm 1-240 and the Student Handout
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DEVIATIONCompass error caused by external magnetic
fields such as generators, radios, etc., causingthe compass needle to deviate from magnetic north.
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MN
N3 33
DEVIATIONCompass error caused by external magnetic
fields such as generators, radios, etc., causingthe compass needle to deviate from magnetic north.
No deviation,Compass Heading (CH) =Magnetic Heading (MH)
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MN N 33
DEVIATIONCompass error caused by external magnetic
fields such as generators, radios, etc., causingthe compass needle to deviate from magnetic north.
Deviation due to magnetic field caused by radio. Compass Heading (CH)
no longer equals Magnetic Heading (MH)
1. Determine MH, e.g. 010°.
2. Find number closest to MH in the “to fly” column.
3. Find corresponding CH in the “steer” column.
4. Apply difference (2°) to MH.
DETERMINE AND APPLY DEVIATION
TC +/- DC = TH +/- VAR = MH +/- DEV = CH 010° + 0° = 010° + 0° = 010° + 2° = 012°
? Is difference + or - ?
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Plan a flight from Lowe AHP 31° 21’ N, 85° 45’Wto Brewton airport
31° 03’N, 87° 04’W, winds 270°/15kts. DC is 9 degrees.
SOLUTION
TC256°
1. Draw course line and measure TC.
SOLUTION
TC +/- DC 256° + 9°
1. Draw course line and measure TC.
2. DC: 9°R or L.
SOLUTION
TC +/- DC = TH256° + 9° = 265°
1. Draw course line and measure TC.
2. DC: 9°R.
3. Compute TH.
SOLUTION
TC +/- DC = TH +/- VAR 256° + 9° = 265° + 2°
1. Draw course line and measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
SOLUTION
TC +/- DC = TH +/- VAR = MH 256° + 9° = 265° + 2° = 267°
1. Draw course line and measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
5. Compute MH.
SOLUTION
TC +/- DC = TH +/- VAR = MH +/- DEV 256° + 9° = 265° + 2° = 267° + 2°
1. Draw course line and measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
5. Compute MH.
6. Add/subtract DEV.
SOLUTION
TC +/- DC = TH +/- VAR = MH +/- DEV = CH 256° + 9° = 265° + 2° = 267° + 2° = 269°
1. Draw course line and measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
5. Compute MH.
6. Add/subtract DEV.
7. Compute CH.
For light background reading:
FM 3-04.240, Chapter 1, Paragraph 1-35 through 1-41 and Chapter 6, paragraph 6-1 through 6-7
Questions?
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End of Day 4
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