Transcript Var and Dev

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0°
WIND EFFECT
VARIATION
DEVIATION
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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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TRUE COURSE
The proposed flight path measured
clockwise from true north.
True North
(Meridian)
clockwise to
flight path
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DRIFT CORRECTION
wind 270°
The correction applied
to prevent
drifting off course.
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DRIFT CORRECTION
NOT APPLIED
wind 270°
Headwind = No Drift
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DRIFT CORRECTION
NOT APPLIED
Left Crosswind =
Right Drift
True Course
wind 270°
Actual Flight
Path
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DRIFT CORRECTION
NOT APPLIED
Right Crosswind =
Left Drift
Actual Flight
Path
wind 270°
True Course
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DRIFT CORRECTION
NOT APPLIED
wind 270°
Tailwind = No Drift
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LEFT DRIFT CORRECTION
IS SUBTRACTED FROM TC
TC = 360°
TC = 360°
DC = 10° L
DC = 10°
LEFT
360°
- 10°
= 350°
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RIGHT DRIFT CORRECTION
IS ADDED TO TC
TC = 360°
TC = 360°
DC = 10°R
DC = 10°
RIGHT
360°
+ 10°
= 010°
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TRUE HEADING
The angle measured clockwise from
true north to the nose of the aircraft.
TC +/- DC = TH
WIND 360
TH
DC
TC
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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
THE ACTUAL FLIGHT PATH OVER THE GROUND,
MEASURED CLOCKWISE FROM TRUE NORTH
TN
TR
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DRIFT ANGLE
THE ANGLE BETWEEN TRUE
HEADING (TH) AND TRACK (TR)
WIND 360
TH
DC DA
TC
TR
WIND IS MORE THAN PLANNING WINDS, DA > DC
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DRIFT ANGLE
THE ANGLE BETWEEN TRUE
HEADING (TH) AND TRACK (TR)
WIND 360
TH
DC
DA
TR
TC
WIND IS LESS THAN PLANNING WINDS, DA < DC
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DRIFT ANGLE
THE ANGLE BETWEEN TRUE
HEADING (TH) AND TRACK (TR)
WIND 360
TH
DC DA = DC
TR = TC
TC
WIND EQUAL TO PLANNING WINDS, DA = DC
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TRACK EQUAL TRUE COURSE WHEN:
1. NO WIND CONDITION
2. HEAD WIND CONDITION
3. TAIL WIND CONDITION
4. DRIFT CORRECTION IS CORRECT
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Types of Airspeed
Indicated Airspeed (IAS) - the airspeed read
directly from the airspeed indicator.
IAS is not affected by wind.
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Types of Airspeed
Calibrated Airspeed (CAS) - indicated airspeed
corrected for instrument installation error.
CAS is not affected by the wind.
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Types of Airspeed
True Airspeed (TAS) - calibrated airspeed corrected
for error due to air density (altitude and temperature).
TAS is not affected by the wind.
4000’ PA, 7° C = 96 KTAS
1000’ PA, 13° C = 92 KTAS
90 KIAS
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GROUND SPEED
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
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GROUND SPEED
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
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GROUND SPEED
Headwinds - winds +/- 90° of the aircraft nose reduce
ground speed. TAS minus wind velocity.
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GROUND SPEED
Tailwinds - winds +/- 90° of the aircraft tail increase
ground speed. TAS plus wind velocity.
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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 AHP
31° 21’ N, 85° 40’ W to Sikes airport
30° 47’ N, 86° 31’ W. The wind is
260° / 20 kts and will require a 9° DC.
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H
1. Draw the true course.
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H
2. Determine wind direction in relation to the true course.
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TC +/- DC = TH
232° + 9°R = 241°
H
GS will be less than
TAS because of headwind.
3. Complete the wind component
part of the VFR flight log.
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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
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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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VARIATION
The angular difference between
true north (TN) and magnetic north (MN).
TN MN
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VARIATION
The angular difference between
true north (TN) and magnetic north (MN).
TN MN
NOTE :
Magnetic north is moving,
therefore magnetic variation
changes from one chart
edition to another.
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AGONIC AND ISOGONIC LINES
Represented on sectional charts as
dashed magenta lines, numbered and
lettered, e.g. 1° W, in 1/2° (30’) increments.
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AGONIC LINE
A line connecting points of zero degree variation.
There is only one agonic line in the United States.
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ISOGONIC LINES
Lines connecting points of equal magnetic variation.
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EASTERLY VARIATION
Compass needle points east of TN
and is subtracted. “EAST IS LEAST”
TN
MN
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TH = 360°, compass is pointing 10° east
of TN. Must subtract 10° in order to fly to TN.
TC +/- DC = TH +/- VAR = MH
360° + 0° = 360° - 10° = 350°
This example assumes a calm wind.
TN MN
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MH = 350°, compass is pointing 10° east
of TN. Subtracted 10° in order to fly to TN.
TN
MN
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WESTERLY VARIATION
Compass needle points west of TN
and is added. “WEST IS BEST”
MN
TN
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TH = 360°, compass is pointing 10° west
of TN. Must add 10° in order to fly to TN.
TC +/- DC = TH +/- VAR = MH
360° + 0° = 360° + 10° = 010°
This example assumes a calm wind.
MN TN
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MH = 010°, compass is pointing 10° west
of TN. Added 10° in order to fly to TN.
MN
TN
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APPLYING VARIATION
Find mid-point of TC, apply nearest
whole degree of variation.
B
A
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B
1. Find mid-point of course leg.
2. Find nearest WHOLE
degree of variation.
A
3. TC +/- DC = TH +/- VAR = MH
320° + 0° = 320° - 1° = 319°
EAST IS LEAST
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1. Find mid-point of course leg.
2. Find nearest WHOLE degree of variation.
3. TC +/- DC = TH +/- VAR = MH
090° + 0° = 090° +1° = 091°
WEST IS BEST
B
A
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MAGNETIC HEADING
Now that you know how to figure out MH, the next
question is “What is it ?”
Magnetic Heading is measured from magnetic
north clockwise to the nose of the aircraft.
MN
VAR
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MH
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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DEVIATION
Compass error caused by external magnetic
fields such as generators, radios, etc., causing
the compass needle to deviate from magnetic north.
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DEVIATION
Compass error caused by external magnetic
fields such as generators, radios, etc., causing
the compass needle to deviate from magnetic north.
MN
3
N
33
No deviation,
Compass Heading (CH) =
Magnetic Heading (MH)
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DEVIATION
Compass error caused by external magnetic
fields such as generators, radios, etc., causing
the compass needle to deviate from magnetic north.
MN
N
33
Deviation due to magnetic field
caused by radio.
Compass Heading (CH)
no longer equals
Magnetic Heading (MH)
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DETERMINE AND APPLY DEVIATION
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.
? Is difference + or - ?
TC +/- DC = TH +/- VAR = MH +/- DEV = CH
010° + 0° = 010° + 0° = 010° + 2° = 012°
Plan a flight from Lowe AHP
31° 21’ N, 85° 45’W
to Brewton airport
31° 03’N, 87° 04’W,
winds 270°/15kts. DC is 9 degrees.
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SOLUTION
1. Draw course line and
measure TC.
TC
256°
SOLUTION
1. Draw course line and
measure TC.
2. DC: 9°R or L.
TC +/- DC
256° + 9°
SOLUTION
1. Draw course line and
measure TC.
2. DC: 9°R.
3. Compute TH.
TC +/- DC = TH
256° + 9° = 265°
SOLUTION
1. Draw course line and
measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
TC +/- DC = TH +/- VAR
256° + 9° = 265° + 2°
SOLUTION
1. Draw course line and
measure TC.
2. DC: 9°R.
3. Compute TH.
4. Add/subtract VAR.
5. Compute MH.
TC +/- DC = TH +/- VAR = MH
256° + 9° = 265° + 2° = 267°
SOLUTION
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.
TC +/- DC = TH +/- VAR = MH +/- DEV
256° + 9° = 265° + 2° = 267° + 2°
SOLUTION
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.
TC +/- DC = TH +/- VAR = MH +/- DEV = CH
256° + 9° = 265° + 2° = 267° + 2° = 269°
For light background reading:
FM 3-04.240, Chapter 1, Paragraph 1-35 through
and Chapter 6, paragraph 6-1 through 6-7
1-41
Questions?
End of Day 4
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