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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Property of Lear Siegler

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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Property of Lear Seigler

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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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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Property of Lear Seigler

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

Property of Lear Siegler

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