aviation mechanic general course-notes and faa figures

AVIATION MECHANIC GENERAL

COURSE NOTES AND FAA FIGURES

© 2013 King Schools, Inc.

All Rights Reserved KSI000A10

TABLE OF CONTENTS

COURSE NOTES

Mathematics .................................................................................................... 1

Aircraft Weight and Balance ............................................................................ 8

Reading Graphs and Charts ........................................................................... 13

Aircraft Drawings ............................................................................................ 15

Aircraft Materials and Processes .................................................................... 18

Aircraft Hardware ........................................................................................... 22

Physics ........................................................................................................... 24

Fluid Lines and Fittings .................................................................................. 26

Fuels and Fuel Systems ................................................................................. 28

Basic Electricity .............................................................................................. 29

Inspection Fundamentals ............................................................................... 42

Measuring Devices ......................................................................................... 50

Ground Handling, Safety, and Support Equipment ........................................ 52

LEARNING STATEMENT CODES

http://www.kingschools.com/AMELearningStatementCodes.asp#General

FAA FIGURES PAGES

FAA Figures .................................................................................... Appendix 1

FAA ADDENDUM A

FAA Figures .................................................................................. Addendum A

Appendix 1 FIGURES

FIGURE 1.—Equation....................................................................................1

FIGURE 2.—Equation....................................................................................2

FIGURE 3.—Equation....................................................................................3

FIGURE 4.—Circuit Diagram .........................................................................4

FIGURE 5.—Formula.....................................................................................5





FIGURE 10.—Battery Circuit .........................................................................10

FIGURE 11.—Circuit Diagram .......................................................................11




FIGURE 15.—Landing Gear Circuit ...............................................................15

FIGURE 16.—Fuel System Circuit .................................................................16

FIGURE 17.—Electrical Symbols ..................................................................17




FIGURE 21.—Electrical Symbols ..................................................................21

FIGURE 22.—Transistors ..............................................................................22

FIGURE 23.—Transistorized Circuit ..............................................................23

FIGURE 24.—Logic Gate ..............................................................................24



FIGURE 27.—Object Views ...........................................................................27

FIGURE 28.—Object Views ...........................................................................28

Return to Table of Contents

ssmith

Typewritten Text

FIGURES—Continued

FIGURE 29.—Object Views ........................................................................... 29

FIGURE 30.—Object Views ........................................................................... 30

FIGURE 31.—Sketches ................................................................................. 30

FIGURE 32.—Sketches ................................................................................. 32

FIGURE 33.—Material Symbols .................................................................... 33

FIGURE 34.—Aircraft Drawing ...................................................................... 34




FIGURE 38.—Performance Chart ................................................................. 38

FIGURE 39.—Electric Wire Chart ................................................................. 39

FIGURE 40.—Cable Tension Chart ............................................................... 40

FIGURE 41.—Performance Chart ................................................................. 41

FIGURE 42.—Aircraft Hardware .................................................................... 42

FIGURE 43.—Aircraft Hardware .................................................................... 43

FIGURE 44.—Welds .................................................................................... 44

FIGURE 45.—Welds .................................................................................... 45

FIGURE 46.—Precision Measurement ......................................................... 46




FIGURE 50.—Marshalling Signals ............................................................... 50

FIGURE 51.—Marshalling Signals ............................................................... 51

FIGURE 52.—Equation ................................................................................ 52

FIGURE 53.—Equation ................................................................................ 53

FIGURE 54.—Trapezoid Area ...................................................................... 54

FIGURE 55.—Triangle Area ......................................................................... 55

FIGURE 56.—Trapezoid Area ...................................................................... 56


FIGURES—Continued

FIGURE 57.—Triangle Area ........................................................................ 57

FIGURE 58.—Equation ................................................................................ 58

FIGURE 59.—Equation ................................................................................ 59

FIGURE 60.—Equation ................................................................................ 60

FIGURE 61.—Physics .................................................................................. 61

FIGURE 62.—Part 1 of 3 – Maintenance Data ............................................ 62

FIGURE 62A.—Part 2 of 3 – Maintenance Data .......................................... 63

FIGURE 62B.—Part 3 of 3 – Maintenance Data .......................................... 64

FIGURE 63.—Airworthiness Directive Excerpt ............................................ 65

FIGURE 64.—Resistance Total ................................................................... 66

FIGURE 65.—Scientific Notation ................................................................. 67

FIGURE 66.—Equation ................................................................................ 68

FIGURE 67.—Equation ................................................................................ 69

FIGURE 68.—Alternative Answer ................................................................ 70

FIGURE 69.—Equation ................................................................................ 71

FIGURE 70.—Alternative Answer ................................................................ 72

FIGURE 71.—Volume of a Sphere .............................................................. 73

Addendum A FIGURES

FIGURE 1.—Electric Wire Chart (Replaces Fig.39 from Appendix 1) ................1

FIGURE 2.—Precision Measurement (Replaces Fig.46 from Appendix 1) ........2

FIGURE 3.—Precision Measurement (Replaces Fig.49 from Appendix 1) ........3

FIGURE 4.—Triangle Area (Replaces Fig.55 from Appendix 1) ........................4

FIGURE 5.—Triangle Area (Replaces Fig.57 from Appendix 1) ........................5


COURSE NOTES


MATHEMATICS - NOTES 1

MATHEMATICS

POWERS

THE CUBE OF A NUMBER - is that number raised to the 3rd power. That means the number is used as a factor 3 times.

THE CUBE OF 64 - is written as 643.

64 x 64 x 64 = 262,144 (8384)

3-4 MEANS - 1 is to be divided by 3 a total of 4 times.

3-4 = 1/(3 x 3 x 3 x 3) = 1/81

SCIENTIFIC NOTATION - is a way to represent any decimal number with a number between 1 and 10 raised to specific power of ten.

EXAMPLE - Figure 65

3.47 x 104 = 34,700

2(410 ) = 2,097,152 (8540)

THE SQUARE ROOT OF A NUMBER - is a smaller number that, when multiplied by itself, will produce the larger number.

EXAMPLE - What is the square root of 1,746? (41.7852 - use electronic calculator.) (8380)

EXAMPLE - What is the square root of 3,722.1835? (61.0097 - use electronic calcula-tor.) (8382)

EXAMPLE - 8,019.0514 x 1/81 is equal to the square root of what number? (9801.) (8383)

8,019.0514 x 1/81 = 8,019.0514 ÷ 81 = 99 To find out what 99 is the square root of, you first need to square 99. 992 = 9,801

EXAMPLE – What is equal to the square root of 3844? (31(2) + 7 + (-3.5 x 2)) (8593)

31(2) + 7 + (-3.5 x 2) = 62 + 7 – 7 = 62

EXAMPLE - What is equal to the square root of (-1776) / (-2) – 632? (16) (8541)

]632)2(

)1776([ −−

− =

]632888[ − =

]256[ = 16

EXAMPLE – Figure 70. Which alternative answer is equal to 5.59? (1) (8594)

=÷+ 217)43()31(

=+217

)43()31(

=+289

56.657.5

59.502.057.5 =+

POWERS OF 10 For powers of 10, move the decimal to the right:

101 = 10 102 = 10x10 = 100

103 = 10x10x10 = 1,000

104 = 10x10x10x10 = 10,000

105 = 10x10x10x10x10 = 100,000

106 = 10x10x10x10x10x10 = 1,000,000

For negative powers of 10, move the decimal to the left:

101 = 10

100 = 1

10-1

= 1/10 = 0.110

-2 = 1/10x10 = 0.01

10-3

= 1/10x10x10 = 0.00110

-4 = 1/10x10x10x10 = 0.0001

10-5

= 1/10x10x10x10x10 = 0.0000110

-6 = 1/10x10x10x10x10x10 = 0.000001

etc.



EXAMPLE - What power of 10 is equal to 1,000,000? (10 to the sixth power - refer to table above.) (8379)

EXAMPLE - what power of 10 is equal to 1,000,000,000? (10 to the ninth power.) (8557)

EXAMPLE - What is the value of 10 raised to the negative sixth power? (0.000001 - refer to table above.) (8385)

EXAMPLE - What is an alternative answer that is equal to 16,300? (1.63 x 10 to the fourth power.) (8388)

1 6, 3 0 0.0 1 6, 3 0 0.0 x 100 1. 6 3 0 0 x 104 1. 6 3 x 104

EXAMPLE - What is the number 3.47 x 10 to the negative fourth power also equal to? (0.000347) (8387)

3. 4 7 x 10-4

0 0 0 3. 4 7 x 10

-4

0.0 0 0 3 4 7 x 100

0.0 0 0 3 4 7

EXAMPLE - What is the square root of 124.9924? (11.18 and also 1,118 x 10 to the negative second power.) (8389)

Square root of 124.9924 = 11.18 on the calculator. 11.18 x 10

0

1,118 x 10-2

EXAMPLE - What is the square root of 1,824? (42.708 and also 0.42708 x 10 to the second power.) (8393)

Square root of 1,824 = 42.708 on calculator. 42.708 x 10

0

0.42708 x 102

COMBINED ROOTS AND POWERS EXAMPLE - What is the square root of 16 raised to the fourth power? (256) (8390)

Square root of 16 = 4 on calculator. 4 x 4 x 4 x 4 = 256.

EXAMPLE - What is the square root of 4 raised to the fifth power? (32) (8386)

Square root of 4 = 2 on calculator. 2 x 2 x 2 x 2 x 2 = 32.

EXAMPLE - What is the result of 7 raised to the third power plus the square root of 39? (349.24) (8392) 7 x 7 x 7 = 343 on calculator. Square root of 39 = 6.24 on calculator. 343 + 6.24 = 349.24

CONVERTING FRACTIONS TO DECIMALS

Divide the numerator by the denominator.

EXAMPLE - What decimal is most nearly equal to a bend radius of 31/64? (0.4844) (8426) Divide 31 by 64 on calculator.

EXAMPLE - The radius of a piece of round stock is 7/32. What decimal is most nearly equal to the diameter? (0.4375) (8428) Two times the radius = diameter. 2 x 7/32 = 14/32 Divide 14 by 32 on calculator.

EXAMPLE - What decimal is equal to 39/32? (1.21875) (8410) Divide 39 by 32 on calculator.

EXAMPLE - What decimal is most nearly equal to 77/64? (1.2031) (8418) Divide 77 by 64 on calculator.

EXAMPLE - What decimal is equivalent of the fraction 43/32? (1.34375)

Divide 43 by 32 on calculator.

CONVERTING DECIMALS TO FRACTIONS

Work backwards from the answer choices converting the fractions to decimals.

EXAMPLE - Which fraction is equal to 0.025? A. 1/4. B. 1/400. C. 1/40. (8409) Answer choice "C", 1/40, is correct.

EXAMPLE - What is the fractional equivalent for a 0.0625-thick sheet of aluminum? (8416)

A. 1/32. B. 3/64. C. 1/16. For each answer choice convert the fraction to a decimal and match the results to the decimal in the question. Answer choice "C", 1/16, is correct.



EXAMPLE - A blueprint shows a hole of 0.17187 to be drilled. Which fraction size drill bit is most nearly equal? (8425)

A. 11/64. B. 9/32. C. 11/32.

For each answer choice convert the fraction to a decimal and match the results to the decimal in the question.

Answer choice "A", 11/64, is correct.

SOLVING EQUATIONS Use a calculator.

Account for negative signs and decimal points.

Do operations starting with innermost parentheses first.

EXAMPLE - Solve the equation:

( )[ ] =++−− 432634 (8439)

( )[ ][ ]

[ ]

4 3 6 5 44 3 30 4

4 3 264 78 82

− − + =− − + =

− − =+ =


−6[−9(−8 + 4)−2(7 + 3)] = (8440)

( ) ( )[ ]

[ ][ ]

− − − − =− − =

− = −

6 9 4 2 106 36 20

6 16 96

EXAMPLE - Solve the equation: ( ) ( )[ ]4 3 9 2 2× − + − × ÷ = (8432)

( ) ( )

( ) ( )

4 3 9 22

12 182

302

15

× − + − ×=

− + −=

− = −

EXAMPLE - Solve the equation: (64 × 3/8) ÷ (3/4) = (8433)

641

38

34

641

38

43

81

31

43

8 4 32

×⎛⎝⎜

⎞⎠⎟

÷ =

× × =

× × =

× =

EXAMPLE - Solve the equation: (32 × 3/8) ÷ (1/6) = (8434)

321

38

16

321

38

61

4 3 6 72

×⎛⎝⎜

⎞⎠⎟

÷ =

× × =

× × =

EXAMPLE - Solve the equation: 2/4(-30 + 34)5 = (8436)

10

15

12

11

15

14

21

=××

=××


−

− −=

4 125

6 36 (8438)

1254

366

1254

636

1254

16

12524

5 20

−÷ −

−=

−× −

−=

× −−

=

− = − .

EXAMPLE - Solve the equation: (-3 + 2)(-12 - 4) + (-4 + 6) x 3 = (8441)

(-1)(-16) + (2) x 3 =

16 + 6 = 22 EXAMPLE - Solve the equation. (8558) ( ) =−+ 1636100 ( ) 124610 =−+



EXAMPLE - Figure 60. Solve the equation:

( )( ) ( )( )− + − +

− ÷=

−5 23 2 3 6427 9

3

(8442)

( )( ) ( )18 2 13 3 3

8

3

36 827

336 0 296

335 704

3119

− +× ×

−=

− +

−=

− +−

=

−−

=

.

. .


( )

3 1 4 31 7 2

+ = (8391)

5 5678 6 5574289

121252289

0 04195

. .

. .

+ =

=


( )( ) ( )− + − +

=−35 25 7 16

25

2π (8437)

( )( )− − + ×× =

+ ×=

+ × =

+ =

10 7 3 1416 116 16

5

70 3 1416 1256

570 3 1416 0 0039

570 0 0123

514 00

.

( . )

( . . )

. .


( ) ( )( ) =++−240 3129664 (8381)

( )1 6 6 3

1 6 18

25 5

+ + × =

+ + =

=

AREA OF A TRIANGLE ( )Area base height= ×1

2

EXAMPLE - Figure 55. What is the area of the triangle shown? (8397)

( )Area in in

Area square inches

= ×

= =

12

4 3

122

6

. .

EXAMPLE - Figure 57. What is the area of the triangle formed by points A, B, and C?

A to B = 7.5 inches

A to D = 16.8 inches (8402)

( )Area in in

Area square inches

= ×

= =

12

16 8 7 5

1262

63

. . . .

AREA OF A TRAPEZOID

( )Area height sum of the bases= × 12

EXAMPLE - Figure 54. What is the area of the trapezoid? (8395)

( )

( )

Area ft ft ft

Area

Area square feet

= × +

= ×

= × =

5 12

12 9

5 12

21

5 10 5 52 5

. . .

. .

EXAMPLE - Figure 56. What is the area of the trapezoid? (8401)

( )Area ft ft ft

Area square feet

= × +

= × =

2 12

6 4

2 5 10

. . .

AREA OF A CUBE Use the formula for area to find the area of one side of the cube and then multiply by 6.

EXAMPLE - What is the surface area of a cube where a side (edge) measures 7.25 inches? (8592)

.in.sq375.3155625.526

)25.725.7(6area)wl(6

=×=×

=×



VOLUME OF A SPHERE EXAMPLE – Figure 71. What is the volume of a sphere with a radius of 4.5 inches? (8595)

VD6/1 3 =π

=×

6914159.3 3

=×

672914159.3

70.3816

22.2290 = cubic inches

VOLUME OF RECTANGULAR SHAPED FUEL TANKS

Volume length width depth= × × Convert the units as required.

EXAMPLE - A rectangular-shaped fuel tank measures 37-1/2 inches in length, 14 inches in width, and 8-1/4 inches in depth. How many cubic inches are within the tank? (8407)

Volume = 37.5 in. x 14 in. x 8.25 in. Volume = 4,331.25 cubic inches

EXAMPLE - A rectangular-shaped fuel tank measures 60 inches in length, 30 inches in width, and 12 inches in depth. How many cubic feet are within the tank? (8399)

Volume

Volume ft ft ftVolume cubic feet

= × ×

= × ×=

6012

3012

1212

5 2 5 112 5

. . . ..

EXAMPLE - A rectangular-shaped fuel tank measures 27-1/2 inches in length, 3/4 foot in width, and 8-1/4 inches in depth. How many gallons will the tank contain?

(231 cu. in. = 1 gal.) (8405)

Volume in in inVolume in in in

Volume cu in galcu in

Volume gallons

= × × ×= × ×

= ×

=

27 5 3 4 12 8 2527 5 9 8 25

2 042 1231

8 8

. . ( / ) . . .

. . . . .

, . . .. .

.

EXAMPLE - How many gallons of fuel will be contained in a rectangular-shaped tank which measures 2 feet in width, 3 feet in length, and 1 foot 8 inches in depth? (8404)

(7.5 gal. = 1 cu. ft.)

Volume ft ft ft

Volume ft

Volume cu ft

Volume cu ft galcu ft

Volume gallons

= × ×

= × ×

=

= ×

=

3 2 1 812

3 2 2012

10

10 7 51

75

. . .

.

. .

. . . .. .

EXAMPLE - What container size in cubic feet will be equal in volume to 60 gallons of fuel?

(7.5 gal. = 1 cu. ft.) (8400)

SizeSize cubic feet

= ÷=

60 7 58 0

..

CIRCUMFERENCE OF A CIRCLE EXAMPLE - What size sheet of metal is required to fabricate a cylinder 20 inches long and 8 inches in diameter? (8396)

Note: C = π D A. 20" x 25-5/32". B. 20" x 24-9/64". C. 20" x 25-9/64".

C = 3.1416 x 8 in. = 25.1328 in. Note: 9/64 = 0.140625 5/32 = 0.15625

Convert the fractions of the 25 inch lengths in the answer choices to decimals and choose the one just long enough to fabricate the piece.

Correct answer is "C", 20" x 25-9/64".

PISTON DISPLACEMENT Area of a circle = π × R2

R D=2

Engine Displacement =

π × ⎛⎝⎜

⎞⎠⎟

× ×bore Stroke2

2

No. of cylinders

THE TOTAL PISTON DISPLACEMENT OF A SPECIFIC ENGINE - is the volume displaced by all the pistons during one revolution of the crankshaft. (8394)



EXAMPLE - What is the piston displacement of a master cylinder with a 1.5-inch diameter bore and a piston stroke of 4 inches? (8403)

( )

Disp. Stroke No. of cylinders

Disp.

cubic inches

= × ⎛⎝⎜

⎞⎠⎟

× ×

= × ⎛⎝⎜

⎞⎠⎟

× ×

= × × ×= × × ×=

π D

DispDispDisp

2

31416 152

4 1

31416 0 75 4 131416 0 5625 4 17 0686

2

2

2

. .

. . .

. . .

. .

EXAMPLE - A four-cylinder aircraft engine has a cylinder bore of 3.78 inches and is 8.5 inches deep. With the piston on bottom center, the top of the piston measures 4.0 inches from the bottom of the cylinder. What is the approximate piston displacement of this engine? (8406)

( )

( )


Disp.

cubic inches

= × ⎛⎝⎜

⎞⎠⎟

× ×

= × ⎛⎝⎜

⎞⎠⎟

× − ×

= × × ×= × × ×=

π D

DispDispDisp

2

31416 3 782

8 5 4 0 4

31416 189 4 5 431416 3 5721 4 5 4202

2

2

2

. . . .

. . . .

. . . .

.

EXAMPLE - A six-cylinder engine has a bore of 3.5 inches, a cylinder height of 7 inches, and a stroke of 4.5 inches. What is the total piston displacement? (8408)

( )


Disp.

cubic inches

= × ⎛⎝⎜

⎞⎠⎟

× ×

= × ⎛⎝⎜

⎞⎠⎟

× ×

= × × ×= × × ×=

π D

DispDispDisp

2

31416 3 52

4 5 6

31416 175 4 5 631416 3 0625 4 5 6259 77

2

2

2

. . .

. . . .

. . . .

. .

RATIOS Divide the first term by the second term, and then reduce the resulting fraction to its lowest terms.

EXAMPLE - If the volume of a cylinder with the piston at bottom center is 84 cubic inches and the piston displacement is 70 cubic inches, what is the compression ratio? (8411)

Comp. Ratio = ( )( )Volume of piston at bottom

Volume of piston at top

( )Comp. Ratio = 8484 - 70

Comp. Ratio = 8414

=

Comp. Ratio = 6 to 1

61

EXAMPLE - What is the ratio of 10 feet to 30 inches? (8430)

10 1230

12030

41

41× = = = :

EXAMPLE - What is the ratio of a gasoline fuel load of 200 gallons to one of 1,680 pounds? (8435)

Note: Gasoline weighs 6.0 pounds per gallon.

200 61680

12001680

6084

57

× = = =

Ratio = 5:7

EXAMPLE - What is the speed ratio of a gear with 36 teeth meshed to a gear with 20 teeth? (8420)

5:91810

3620 ==

Note: The gear with the most teeth would be turning the slower of the two.

PROPORTION A proportion is a statement of equality between two or more ratios.

EXAMPLE - An airplane flying a distance of 750 miles used 60 gallons of gasoline. How many gallons will it need to travel 2,500 miles? (8419)

gallons 200750

002560=gal. ?

250060?750mi. 2500

gal. ?mi. 750

gal. 60

=××=×

=

GEAR SPEEDS Use the following proportion equation:

Teeth drivingTeeth driven

RPM drivenRPM driving

=



EXAMPLE - What is the speed of a spur gear with 42 teeth driven by a pinion gear with 14 teeth turning 420 RPM? (8421,8413)

14 teeth42 teeth RPM

RPM = 14 42042

RPM

=

× = =

?

?

RPM420

588042

140

FRACTIONS AS PERCENTAGE Convert the fraction to a decimal. Move the decimal two places to the right. Affix the percent symbol.

EXAMPLE - Express 5/8 as a percent. (8417)

5/8 = 0.625 = 62.5%

EXAMPLE - Express 7/8 as a percent. (8412)

7/8 = 0.875 = 87.5%

EXAMPLE - Sixty-five engines are what percent of 80 engines? (8427)

65/80 = 0.8125 = 81%

EXAMPLE - Maximum life for a certain part is 1100 hours. Recently, 15 of these parts were removed from different aircraft with an average life of 835.3 hours. What percent of the maximum engine life has been achieved? (8429)

%7676.01100

3.835 ==

OTHER PERCENTAGE PROBLEMS EXAMPLE - An engine of 98 horsepower maximum is running at 75 percent power. What is the horsepower being developed?(8424)

75% of 98 HP = 0.75 x 98 HP = 73.5 HP

EXAMPLE - If an engine is turning 1,965 RPM at 65 percent power, what is its maximum RPM? (8423)

To solve this problem you must presume that power is directly proportional to RPM.

Max RPM RPM100%

196565%

= ,

65 x Max RPM = 100 x 1,965

Max RPM = ×100 196565

,

Max RPM = 3,023

EXAMPLE - An engine develops 108 horsepower at 87 percent power. What horsepower would be developed at 65 percent power? (8414)

10865%

80 69

HP87% HP 87% = 108 HP 65%

? HP = 108 HP 65%87%

HP

=

× ×× =

?

?

.

HP

EXAMPLE - The parts department's profit is 12 percent on a new part. How much does the part cost if the selling price is $145.60? (8422)

Solution with profit based on cost: $145.60 = 100% cost + 12% cost $145.60 = 112% cost

Cost = $145.601.12

= $130.00


WEIGHT & BALANCE - NOTES 8

AIRCRAFT WEIGHT AND BALANCEPURPOSE

FAA REGULATIONS DO NOT REQUIRE - private aircraft to be weighed periodically or after any alteration. Their new weight and balance is normally calculated mathematically. (8158)

WEIGHT CHANGES OCCUR IN AGING AIRCRAFT – mainly because of repairs and alterations done over its lifetime. (8538)

EMPTY WEIGHT WHEN YOU’RE DOING AIRCRAFT LOADING COMPUTATIONS – you’ll need information from the weight and balance records of your aircraft to get the current empty weight and also the empty weight center of gravity (CG). (8597)

THE EMPTY WEIGHT OF AN AIRCRAFT INCLUDES - all operating equipment that has a fixed location and is actually installed.

THE AMOUNT OF FUEL - used for computing empty weight and corresponding CG is unusable fuel. (8176,8607)

FOR AIRCRAFT CERTIFICATED UNDER CURRENT AIRWORTHINESS STANDARDS (14 CFR PART 23) - all the oil contained in the supply tank is considered part of the empty weight. (This became effective March 1, 1978.) (8171)

USEFUL LOAD THE USEFUL LOAD OF AN AIRCRAFT - consists of the crew, usable fuel, passengers, and cargo. (8155)

THE USEFUL LOAD OF AN AIRCRAFT - is the difference between the maximum gross weight and the empty weight. (8170)

MAXIMUM WEIGHT THE MAXIMUM WEIGHT OF AN AIRCRAFT - is the maximum authorized weight of the aircraft and its contents. Or, in other words, empty weight plus useful load. (8163,8169)

THE MAXIMUM WEIGHT - as used in weight and balance control of a given aircraft, can normally be found in the Aircraft Specification or Type Certificate Data Sheet. (8173)

THE MAXIMUM TAKEOFF WEIGHT – will be exceeded in most modern aircraft if all the seats

are occupied, the full baggage weight is on board and all the fuel tanks are full. (8539)

ZERO FUEL WEIGHT ZERO FUEL WEIGHT - is the maximum permissible weight of a loaded aircraft (passengers, crew, and cargo) without fuel.(8167)

DATUM THE DATUM IS AN IMAGINARY VERTICAL PLANE - from which all horizontal measurements are taken for balance purposes.

IF THE REFERENCE DATUM LINE - is placed at the nose of an airplane rather than at the firewall or some other location aft of the nose, all measurement arms will be in positive numbers. (8166)

ARM THE ARM IS THE HORIZONTAL DISTANCE - that an object is located from the datum.

THE LENGTH OF THE ARM (DISTANCE) - is always given or measured in inches. (8161)

FUSELAGE STATION NUMBERS - are used to identify the arm distance in inches from the datum or some other point chosen by the manufacturer. (Station numbers are often used to identify the location of parts.) (8107)

MOMENT THE MOMENT IS THE PRODUCT - of a weight multiplied by its arm. Given any two items (weight, arm, or moment) the third number can be calculated.

EXAMPLE - If a 40-pound generator applies +1400-inch pounds to a reference axis, where is the generator located? (8182)

ARM = MOMENT inchesWEIGHT

= + = +140040

35

CENTER OF GRAVITY (CG) THE CG OF AN AIRCRAFT IS A POINT - about which the nose-heavy and tail-heavy moments are exactly equal.

IF AN AIRCRAFT WERE SUSPENDED FROM THIS POINT - it would remain level.



WHEN COMPUTING WEIGHT AND BALANCE - an airplane is considered to be in balance when the average moment arm of the loaded airplane falls within its CG range. (8153)

REMOVAL OF ANY ITEM OF USEFUL LOAD - will affect the center of gravity in proportion to its weight and its location. Since the item being removed is aft of the center of gravity, removing it will move the center of gravity forward. (8546)

MAC, LEMAC, AND TEMAC

THE CHORD OF A WING IS THE DISTANCE - from the leading edge to the trailing edge. The MAC or Mean Aerodynamic Chord occurs when the chord is through the center of the wing plan area of a sweptback wing. The CG range is a percentage of the MAC. This measurement is determined in inches aft of the datum or datum line. (8543)

THE FRONT OR LEADING EDGE - of MAC is LEMAC and it is the distance of the leading edge of the mean aerodynamic chord from the datum. The back or trailing edge of MAC is TEMAC and is the distance of the trailing edge of the mean aerodynamic chord from the datum. The center of gravity on this aircraft is 24% of MAC, which is the distance from LEMAC. (8544)

AIRCRAFT LEVELING TO OBTAIN USEFUL WEIGHT DATA FOR DETERMINING THE CG - it is necessary that an aircraft be weighed in a level flight attitude. (8160)

THERE ARE SEVERAL WAYS TO LEVEL AN AIRCRAFT. The method with the highest degree of accuracy is the spirit level. (8549)

AIRCRAFT WEIGHING PROCEDURE IF THE AIRCRAFT'S WEIGHT AND BALANCE RECORDS BECOME LOST - destroyed, or otherwise inaccurate, the only way the empty weight can be determined is by reweighing the aircraft. (8156)

WHEN WEIGHING AN AIRPLANE - for the purpose of finding the CG, the weighing points are projected on the floor with a chalk line. When measuring the distances or arms of these lines, the tape must be parallel to the centerline of the airplane. (8548)

PRIOR TO WEIGHING AN AIRCRAFT TO DETERMINE ITS EMPTY WEIGHT - remove all items except those on the aircraft equipment list, drain the fuel, and fill the hydraulic reservoir. (8154)

WEIGHING AN AIRCRAFT POSITIONED - on landing gear wheels requires the parking brake to be released to reduce the possibility of side loads that could give incorrect readings. (8596)

FINDING EMPTY WEIGHT THE WEIGHING POINTS USED - must be clearly indicated on the aircraft weighing form. (The arm values used in the computations are based on those locations.) (8165)

THE EMPTY WEIGHT OF AN AIRPLANE - is determined by subtracting the tare weight from the scale reading, and then adding the weight of each weighing point. (Tare weight consists of items like chocks, supports, etc.) (8168)

EMPTY WEIGHT CG EXAMPLE - What is the empty weight CG? (8186) GIVEN: Combined net wt. at main gears 3,540 lb. Arm of main gears 195.5 in. Net weight at nose gear 2,322 lb. Arm of nose gear 83.5 in. Datum line Forward of nose

SOLUTION: MOMENT = WEIGHT x ARM

CG TOTAL MOMENTTOTAL WEIGHT

=

WEIGH POINT WT. x ARM = MOMENT Main Gears 3,540 195.5 692,070 Nose Gear 2,322 83.5 193,887 TOTALS 5,862 885,957


CG

= = =

=

885 9575 862

1511

,,

. inches.

EMPTY WEIGHT CG WITH EXTRA ITEMS ON BOARD

EXAMPLE - What is the empty weight CG? (8184,8191) GIVEN: Datum forward of main gear 30.24 in. Tail gear to main gear distance 360.26 in. Net weight at right main gear 9,980 lb. Net weight at left main gear 9,770 lb. Net weight at tail gear 1,970 lb.

Items included in aircraft when weighed: Lavatory water tank 34 lb. at +352 Hydraulic fluid 22 lb. at - 8 Removable ballast 146 lb. at +380





=

THE RESULT OF A WEIGHT BEING ADDED OR REMOVED AND ITS LOCATION RELATIVE TO THE DATUM - determines whether the value of the moment is preceded by a plus (+) or a minus (-) sign. (8162)

IN A BALANCE COMPUTATION WHICH REQUIRES AN ITEM LOCATED AFT OF DATUM - to be removed, use (-)weight x (+)arm = (-) moment. (8183)

WEIGH POINT WT. x ARM = MOMENT Right Gear 9,980 30.24 301,795 Left Gear 9,770 30.24 295,445 Tail Gear 1,970 *390.50 769,285 Lavatory water** - 34 352.00 - 11,968 Remove ballast -146 380.00 - 55,480 TOTALS 21,540 1,299,077

* Note: Add distance between datum and main gear (30.24 in.) to distance between main gear and tail gear (360.26 in.) to find tail gear location of 390.50.

** Note: Hydraulic fluid is included in empty weight and should not be subtracted. Lavatory water and removable ballast are not part of empty weight and should be subtracted.

CG TOTAL MOMENT

TOTAL WEIGHTCG

= =

=

1299 07721540

60 31

, ,,

. inches.

EXAMPLE - What is the empty weight CG? (8178) GIVEN: Total empty weight as weighed 5,862 lb. Empty weight moment 885,957

On board the aircraft when weighed: Potable water 20 lb. at +84 Hydraulic fluid 23 lb. at +101



=

ITEM WT. x ARM = MOMENT Empty wt. 5,862 *151.14 885,957 Potable water** -20 84.00 -1,680 TOTALS 5,842 884,277

* Note: To calculate the original arm (CG), divide the empty weight moment by the empty weight.

** Note: Hydraulic fluid is included in empty weight and should not be subtracted. Potable water is not part of empty weight and should be subtracted.

CG TOTAL MOMENT

TOTAL WEIGHTCG

= =

=

884 2775 842

151365

,,

. inches.

EMPTY WEIGHT CG AFTER ENGINE CHANGE

EXAMPLE - What is the new empty weight CG? (8188)

GIVEN: Model B engine removed 175 lb. Model D engine installed 185 lb. Location of engine - 62.00-inch station Previous empty weight 998 lb. Previous empty weight CG 13.48 in.



=

ITEM WT. x ARM = MOMENT Empty Weight 998 13.48 13,453.04 B-engine rem. -175 - 62.00 +10,850.00 D-engine add +185 - 62.00 -11,470.00 TOTALS 1,008 12,833.04

NEW CG TOTAL MOMENTTOTAL WEIGHT

NEW CG

= =

=

12 833 041008

1273

, .,

. inches.

CG SHIFT AFTER ALTERATIONS EXAMPLE - How many inches has the new CG moved? (8174,8180,8181,8187)

GIVEN: Original empty weight 2,886 lb. Original empty weight moment 101,673.78 Remove 2 pass. seats 15 lb. each at +71 Install cabinet 97 lb. at +71 Install seat and safety belt 20 lb. at +71 Install radio equipment 30 lb. at +94



=



ITEM WT. x ARM = MOMENT Empty Weight 2,886 *35.23 101,673.78 2 Seats Rem. **-30 71.00 -2,130.00 Cabinet installed 97 71.00 +6,887.00 Seat installed 20 71.00 +1,420.00 Radio installed 30 94.00 +2,820.00 TOTALS 3,003 110,670.78

* Note: To calculate the original arm (CG), divide the empty weight moment by the empty weight.

** Note: Two seats at 15 lb. each is 30 lb.

NEW CG

TOTAL MOMENTTOTAL WEIGHT

NEW CG

= =

=

110 670 783 003

36 85

, .,

. inches.

DISTANCE CHANGE FROM OLD CG: New CG 36.85 Old CG -35.23 Change 1.62 inches aft

LEVER ARM THE THEORY OF WEIGHT AND BALANCE - is that of a lever that is in equilibrium or balance when it rests on a fulcrum in a level position.

THE DISTANCE OF ANY OBJECT FROM THE FULCRUM - is called the lever arm. (8157)

WEIGHT ADDITION THAT WILL NOT CHANGE CG

EXAMPLE - How far forward of the CG should a box weighing 20 pounds be placed so that the CG will not change, considering the two other boxes already added? (8179)

GIVEN: Box 1 10 lb. at 4 ft. aft of CG Box 2 5 lb. at 2 ft. aft of CG


ITEM WEIGHT x LEVER ARM = MOMENT Box 1 10 4 40 Box 2 5 2 10 Total Moments Added Aft 50

For the CG not to change, the moments forward and aft of the CG must be the same.

ITEM WEIGHT x LEVER ARM = MOMENT Boxes 1,2 +50 Box 3 20 -? -50

ARM = MOMENT

WEIGHT

feet forwardARM = − =5020

2 5.

BALLAST WEIGHT TO BRING CG WITHIN CG RANGE

EXAMPLE - What is the minimum weight of ballast needed to bring the CG within the CG range? (8177)

GIVEN: Loaded aircraft 4,954 lb. Loaded aircraft CG +30.5 in. CG range +32.0 in. to +42.1 in. Ballast arm +162 in.

SOLUTION: The weight (ballast) added causes a moment change in the entire aircraft equal to the moment change caused by adding the weight.

Step 1: Find distance CG is out of limits: Forward CG limit 32.0 Current CG - 30.5 1.5 inches

Step 2: Find the lever arm of the ballast: Ballast location 162.0 -Forward CG Limit - 32.0 130.0 inches

Step 3: Determine the required ballast weight:

ITEM WEIGHT x LEVER ARM = MOMENT CG shift req. 4,954 -1.5 -7,431 Ballast ? 130.0 7,431

WEIGHT MOMENTARM

= = 7 431130,

Ballast weight required is 57.16 pounds.



EMPTY WEIGHT CENTER OF GRAVITY RANGE

THE EMPTY WEIGHT CG RANGE - is the allowable travel within the CG limits.

IF THE EMPTY WEIGHT CG OF AN AIRPLANE - lies within the empty weight CG limits it is not necessary to calculate CG extremes. (8189)

WEIGHT AND BALANCE EXTREME CONDITIONS

WHEN IT IS NECESSARY TO COMPUTE THE MAXIMUM FORWARD LOADED CG - use the minimum weights, arms, and moments of the items of useful load that are located aft of the forward CG limit. (8190)

WHEN MAKING A REARWARD LOADED CG CHECK - to determine that the CG will not exceed the rearward limit during extreme conditions, use the minimum weights, arms and moments of the items of useful load that are located forward of the rearward CG limit. (8185)

ADVERSE LOADING CHECKS - are conducted using loads that are at or below the maximum gross weight of the aircraft.

HELICOPTER WEIGHT AND BALANCE WHEN COMPUTING WEIGHT AND BALANCE FOR A HELICOPTER - you must consider that it is computed generally the same as for airplanes. (8164)

THE CG RANGE IN SINGLE-ROTOR HELICOPTERS IS - more restricted than for airplanes. (8175)

IMPROPER LOADING OF A HELICOPTER - which results in exceeding either the fore or aft CG limits is hazardous due to the reduction or loss of effective cyclic pitch control. (The cyclic tilts the plane of the main rotor in the direction of desired horizontal movement.) (8172)


13

READING GRAPHS AND CHARTS

ENGINE PARAMETERS EXAMPLES - Figure 38. An aircraft reciprocating engine has a 1,830 cubic-inch displacement and develops 1,250 brake-horsepower at 2,500 RPM. What is the brake mean effective pressure? (8142)

Locate 1,250 BHP on the top of the chart. Drop vertically downward to the 1,830 Cubic Inch Displacement line. Now move horizontally to the right to the sloping line for 2,500 RPM, then vertically downward to read a value of 217 BMEP.

An aircraft reciprocating engine has a 2,800 cubic-inch displacement and develops 2,000 brake-horsepower at 2,200 RPM. What is the brake mean effective pressure? (8144)

Locate 2,000 BHP on the top of the chart. Drop vertically downward to the 2,800 Cubic Inch Displacement line. Now move horizontally to the right to the sloping line for 2,200 RPM, then vertically downward to read a value of 257.5 BMEP.

An aircraft engine has a 2,800 cubic-inch dis-placement, develops 2,000 brake-horsepower, and indicates 270 brake mean effective pres-sure. What is the engine speed (RPM)? (8143)

Locate 2,000 BHP on the top of the chart. Drop vertically downward to the 2,800 Cubic Inch Displacement line. Now move horizontally to the right to the vertical line for 270 BMEP. The intersection is at an engine speed of 2,100 RPM.

ELECTRIC WIRE CHART EXAMPLES - Figure 39.

TO SELECT THE PROPER SIZE OF CONDUCTOR -the following must be known:

1) The conductor length in feet.2) The number of amperes of current to be

carried.3) The amount of voltage drop permitted.4) Whether the current to be carried will be

intermittent or continuous, and, if continuous, whether it is a single conductor

in free air, or cables in a conduit or in a bundle.

What is the appropriate cable size of a 40-foot length of single cable in free air, with a continuous rating, running from a bus to the equipment in a 28-volt system with a 15-ampere load and a 1-volt drop? (8145)

First, determine the wire size based on voltage drop:

Start on the horizontal line in the 28V-1V drop column at the 40 feet length and move horizontally to the slanted 15-ampere line. Now move vertically down and read wire size between 10 and 12. The larger wire size No. 10 should be selected.

Second, determine the wire size based on overheating:

Move diagonally downward along the 15 ampere line until it intersects Curve 2 (Continuous Rating - Amperes Single Cable in Free-Air) and then vertically downward falling between No. 18 and No. 20. Selecting the larger size we would use No. 18 wire.

Comparing the wire sizes obtained under both conditions above, select No. 10 wire size as the smallest that will satisfy both conditions.

What is the minimum wire size of a single cable in a bundle carrying a continuous current of 20 amperes 10 feet from the bus to the equipment in a 28-volt system with an allowable 1-volt drop? (8147)

First, determine the wire size based on voltage drop:

Start on the horizontal line in the 28V-1V drop column at the 10 feet length and move horizontally to the slanted 20 ampere line. Now move vertically down and read just above No. 16 wire size, making No. 16 wire required.

Second, determine the wire size based on overheating:

Move diagonally downward along the 20 ampere line until it intersects Curve 1 and then vertically downward falling between wire sizes No. 14 and 12, making No. 12 size required.

GRAPHS & CHARTS - NOTES Return to Table of Contents

GRAPHS & CHARTS - NOTES 14

Comparing the wire sizes obtained under both conditions, select No. 12 wire as the smallest that will satisfy both requirements.

TO DETERMINE THE MAXIMUM LENGTH OF A CONDUCTOR - the following must be known:

1) The wire size of the cable.2) The number of amperes of current to be

carried.3) The amount of voltage drop permitted.4) Whether the current to be carried will be

intermittent or continuous, and, if continuous, whether it is a single conductor

in free air, or cables in a conduit, or in a bundle.

What is the maximum length of a No. 12 single cable that can be used between a 28-volt bus and a component utilizing 20 amperes continuous load in free air with a maximum acceptable 1-volt drop? (8148)

Locate No. 12 wire size at the bottom of the chart. Move vertically upward to the 20 ampere sloping line. From this point move horizontally to the left and read a length per 1-volt drop of 26.5 feet from the 28 Circuit Voltage column. Note that this length is verified by being above Curve 2.

What is the maximum length of a No. 16 cable to be installed from a bus to the equipment in a 28-volt system with a 25-ampere intermittent load and a 1-volt drop? (8146)

Locate No. 16 wire size at the bottom of the chart. Move vertically upward to an estimated 25 ampere sloping line (half way between 20 amps and 30 amps). At this point move horizontally to the left and read a length per 1-volt drop of 8 feet from the 28 Circuit Voltage column. Note that this length is verified by being on or above Curve 3, "Intermittent Rating-Amperes Maximum of 2 Minutes".

CABLE TENSION EXAMPLES - Figure 40.

What is the proper tension for a 3/16-inch cable (7 x 19 extra flex) if the temperature is 87 °F?

(8150)

At a temperature of 87 oF move vertically upward to the curve for a Cable Size of 3/16 7 x 19 and then to the right and read a Rigging Load value of 125 pounds.

What is the proper tension for a 1/8-inch cable (7 x 19) if the temperature is 80 °F? (8149)

At a temperature of 80 oF move vertically upward to the curve for a Cable Size of 1/8 7 x 19 and then to the right and read a Rigging Load value of 70 pounds.

FUEL CONSUMPTION EXAMPLES - Figure 41.

How much fuel would be required for a 30- minute reserve operating at 2,300 RPM? (8151)

At an engine speed of 2,300 RPM move vertically to the Propeller Load Specific Fuel Consumption curve then to the right and read a value of 0.46 lb/BHP/hr. Continue vertically upward at 2,300 RPM to the Propeller Load Horsepower curve and then to the left and read a value of 110 Brake Horsepower.

Fuel = 0.46 lb/BHP/hr. x 110 BHP x 1/2 hr. Fuel = 25.3 pounds

What is the fuel consumption with the engine operating at cruise, 2,350 RPM? (8152)

At an engine speed of 2,350 RPM move vertically to the Propeller Load Specific Fuel Consumption curve then to the right and read a value of 0.465 lb/BHP/hr. Continue vertically upward at 2,350 RPM to the Propeller Load Horsepower curve and then to the left and read a value of 119 Brake Horsepower.

Fuel = 0.465 lb/BHP/hr. x 119 BHP Fuel = 55.3 pounds per hour


DRAWINGS - NOTES 15

AIRCRAFT DRAWINGSWORKING DRAWINGS

WORKING DRAWINGS MAY BE DIVIDED INTO THREE CLASSES - they are: detail drawings, assembly drawings, and installation drawings. (8119)

DETAIL DRAWINGS A DETAIL DRAWING - is a description of a single part carefully and accurately drawn to scale and dimensioned. (8105,8136)

ASSEMBLY DRAWINGS AN ASSEMBLY DRAWING - is a description of an object made up of two or more parts.

INSTALLATION DRAWINGS AN INSTALLATION DRAWING - shows all the subassemblies or parts as brought together on the aircraft. (8138)

EXPLODED VIEW DRAWINGS - are often used in illustrated parts manuals. (8137)

A WIRING DIAGRAM - would show the wire size required for a particular installation.

PROJECTIONS AN ORTHOGRAPHIC PROJECTION - shows the exact size and shape of all the parts of complex objects. A number of views are necessary.

EXAMPLE - Figure 28. Which is the bottom view of the object shown? (The bottom view is "2".) (8106)

EXAMPLE - Figure 29. Which is the left side view of the object shown? (The left side view is "3".) (8109)

EXAMPLE - Figure 30. Which is the bottom view of the object shown? (The bottom view is "1".) (8111)

ONE-VIEW, TWO-VIEW, AND THREE-VIEW DRAWINGS - are the most common with orthographic projections. (For example, see Figure 32.) (8108)

WHEN THREE-VIEW PROJECTION IS USED - the top, front, and right side views are usually shown. (For example, see Figure 32.)

AN ISOMETRIC PROJECTION - is a type of drawing which shows the object inclined at an angle to the viewer.

EXAMPLE - Figure 27. In the isometric view of a typical aileron balance weight, which is the view indicated by the arrow? (The view indicated by the arrow is view "3".) (8104)

SKETCHES SKETCHES ARE SIMPLE, ROUGH DRAWINGS - that are made rapidly and without much detail.

ONE OR A COMBINATION OF SIX BASIC SHAPES - triangle, circle, cube, cylinder, cone, and sphere - will be used when sketching nearly any object. (8116)

SINCE SKETCHES ARE DRAWN WITHOUT THE USE OF DRAFTING INSTRUMENTS - using graph paper makes the layout process easier. (8120)

A SIMPLE WAY TO FIND THE CENTER OF A CIRCLE - is to draw two non-parallel chord lines across a circle. Then draw two bisecting lines perpendicular to the two chord lines. Extend those perpendicular lines until they intersect. That intersection is the center of the circle. (8551)

IF USED FOR MAKING A PART - a sketch must show all information to manufacture the part. (8114)

ON OCCASION, A MECHANIC - may need to make a simple sketch of a proposed repair to an aircraft, a new design, or a modification. (8118) (There is no requirement that repairs to an aircraft skin have a detailed dimensional sketch included in the permanent records.)

EXAMPLE - Figure 31. What are the proper procedural steps for sketching repairs and alterations? (8113,8117)

3 - Blocking in the views. 1 - Adding detail. 4 - Darkening the views. 2 - Adding the dimensions.

EXAMPLE - Figure 32. What is the next step required for a working sketch of the illustration? (Sketch extension and dimension lines.) (8115)


DRAWINGS - NOTES 16

SCHEMATICS ONE PURPOSE FOR SCHEMATIC DIAGRAMS - is to show the functional (not physical) location of components within a system. (8131)

SCHEMATIC DIAGRAMS INDICATE THE LOCATION - of components with respect to each other within the system, but do not indicate the location of individual components in the aircraft. (8112) EXAMPLE - Would a hydraulic system schematic drawing indicate the direction of fluid flow through the system? (Yes.) (8134)

A SCHEMATIC DIAGRAM - is best suited to troubleshoot a system malfunction because they show the location of the components in relation to each other. (8140)

PICTORIAL A PICTORIAL DIAGRAM – uses pictures to show electrical components instead of using the more familiar electrical symbols. (8139)

MEANING OF LINES A MEDIUM-WEIGHT DASHED LINE, ALSO KNOWN AS A HIDDEN LINE - is the type of line normally used in a mechanical drawing or blueprint to represent an edge or object not visible to the viewer. (A series of short dashes evenly spaced). (8103,8110)

AN EXTENSION LINE - is a light, solid line that extends from the point of the view to which a dimension refers.

EXAMPLE - Figure 35. Which is the extension line? (Line "3" is an extension line.) (8128)

OTHER DATA DIMENSIONS - are the means of conveying measurements through the medium of drawings. (For example, see Figure 34.) (8127)

ZONE NUMBERS ON AIRCRAFT BLUEPRINTS - are used to locate parts, sections, and views on large drawings. (Similar to the numbers and letters printed on the borders of a map). (8130)

ON WORKING DRAWINGS OF INDIVIDUAL PARTS - called detail drawings, the material specification is usually indicated in a note or in a parts list. It is not necessary, therefore, to repeat this information by use of a specific section-line symbol. In this case, the symbol for cast iron may be used for all materials. (8121)

EXAMPLE - Figure 33. Which material section-line symbol indicates cast iron? (Symbol 3.) (8122)

READING DIMENSIONS A MEASUREMENT SHOULD NOT BE SCALED FROM AN AIRCRAFT PRINT - because the paper shrinks or stretches when the print is made. (8136)

EXAMPLE - Figure 36. What is the diameter of the holes in the finished object? (1/2 inch.) (8129) Note 1. says: "Drill 31/64 inch, ream 1/2 inch."

EXAMPLE - Figure 34. Using the information, what size drill would be required to drill the clevis bolthole? (5/16 inch.) (8126)

Note: A clevis bolt is a special-purpose bolt used only where shear loads occur and never in tension. It is often inserted as a mechanical pin in a control system. 0.3125 in. hole = 5/16 in. drill

EXAMPLE - Figure 34. What would be the minimum diameter of 4130 round stock required for the construction of the clevis that would produce a machined surface? (1 inch.) A. 55/64 inch. B. 1 inch. C. 7/8 inch. (8125)

In order to machine the clevis pin, the starting dimension would have to be larger than 7/8 inch. Of the answer choices given, Choice "B", 1 inch, is the minimum diameter.

EXAMPLE - Figure 34. What is the dimension of the chamfer? (0.0625 x 45°.) (8123)

Note: A chamfer is the beveled corner or edge of an object. 1/16 x 45° = 0.0625 x 45°


DRAWINGS - NOTES 17

EXAMPLE - Figure 37. What is the vertical distance between the top of the plate and the bottom of the lowest 15/64" hole? (2.367) (8135)

Add together the following: From top of plate to center of first 15/64 hole 3/8 0.375 From center of first 15/64 hole to center of second 15/64 hole 7/8 0.875 From center of second 15/64 hole to center of third 15/64 hole 7/8 0.875 From center of third 15/64 hole to center of fourth 15/64 hole 1/8 0.125 From center of fourth 15/64 hole to bottom of that hole 15/128 0.117 2.367

TOLERANCES/ALLOWANCES TOLERANCE - is the difference between extreme permissible dimensions that a part may have and still be acceptable. (8141)

ALLOWANCE - is the difference between the nominal dimension of a part and its upper or lower limit.

EXAMPLE - What is the allowable manufacturing tolerance for a bushing where the outside dimensions shown on the blueprint are: 1.0625 + .0025 - .0003? (0.0028) (8133)

Tolerance = 0.0025 + 0.0003 = 0.0028

EXAMPLE - When reading a blueprint, a dimension is given as 4.387 inches + .005 - .002. What is the maximum acceptable size and what is the minimum acceptable size? (4.392, 4.385) (8132)

Maximum: 4.387 + 0.005 = 4.392 inchesMinimum: 4.387 + 0.002 = 4.385 inches

EXAMPLE - Figure 34. What is the maximum diameter of the hole for the clevis pin? (0.3175) (8124)

The hole in the clevis in which the pin fits has a diameter of: 0.3125 + 0.005 - 0.000 inch.

This tolerance makes the maximum diameter 0.3175 inch: 0.3125 + 0.005 = 0.3175


MATERIALS/PROCESSES - NOTES 18

AIRCRAFT MATERIALS AND PROCESSES

WELDING

A CHARACTERISTIC OF A GOOD GAS WELD - is that it tapers off smoothly into the base metal. (8283)

ONE CHARACTERISTIC OF A GOOD WELD - is that no oxide should be formed on the base metal at a distance from the weld of more than 1/2 inch. (8284)

CRACKING ADJACENT TO THE WELD - indicates a part has cooled too quickly after being welded. (8282)

WHEN EVALUATING A WELDED JOINT, - a mechanic should be familiar with the parts, proportions, and formation of the weld. (8553)

EXAMPLES - Figure 44.

Which illustration depicts a cold weld? (Illustration "2". A weld that appears rough and irregular and with its edges not feathered into the base metal is indicative of a cold weld.) (8279)

Which weld is caused by an excessive amount of acetylene? (Illustration "3". Slight bumps along the center and craters at the finish of the weld are caused by the boiling of the puddle due to excessive acetylene being used.) (8278)

IF HOLES AND A FEW PROJECTING GLOBULES ARE FOUND IN A WELD - remove all the old weld and reweld the joint. (8281)


What type of weld is shown at "A"? ("A" is a butt weld. It is made by placing two pieces of material edge to edge and welded without lapping.) (8285)

What type of weld is shown at "B"? ("B" is a double butt weld. A bead has been applied on both sides of the joint.) (8286)

What type of weld is shown at "G"? ("G" is a lap weld. Two pieces of metal are overlapped and welded at the joint.) (8287)

ON A FILLET WELD - the penetration requirement includes 25 to 50 percent of the base metal thickness. (8288)

CHEMICAL CORROSION THE RUST OR CORROSION - that occurs with most metals is the result of a tendency for them to return to their natural state. (8357)

THE TWO GENERAL CLASSIFICATIONS OF CHEMICAL CORROSION - are direct chemical attack and electrochemical attack.

DIRECT CHEMICAL ATTACK

DIRECT CHEMICAL ATTACK OR PURE CHEMICAL CORROSION - is an attack resulting from a direct exposure of a bare surface to caustic liquid or gaseous agents (battery acid, etc.).

CAUSTIC CLEANING PRODUCTS - used on aluminum structures have the effect of producing corrosion. (8355)

SPILLED MERCURY ON ALUMINUM - causes rapid and severe corrosion that is very difficult to control. (8378)

ELECTROCHEMICAL ATTACK

ELECTROCHEMICAL ATTACK (GALVANIC ACTION) IS RESPONSIBLE - for most forms of corrosion on aircraft structure and component parts.

CORROSION CAUSED BY GALVANIC ACTION IS THE RESULT OF - contact between two unlike metals. (8371,8542)

IN THE CORROSION PROCESS - it is the anodic area or dissimilar anodic material that corrodes. (The cathode does not corrode.) (8377)

IN THE GALVANIC OR ELECTRO-CHEMICAL SERIES FOR METALS - the most anodic metals are those that will give up electrons most easily. (8377)



A PARTIAL LIST - of the Electro-Chemical Series for metals arranged from most anodic to most cathodic follows: Magnesium Zinc Clad 2024 aluminum alloy Cadmium 7075-T6 aluminum alloy Stainless steel Gold

EXAMPLE - Of the following listed materials: Cadmium, 7075-T6 aluminum alloy, Magnesium, which is the most anodic? (Magnesium.) (8372)

EXAMPLE - Of the following listed materials: Zinc, 2024 aluminum alloy, Stainless steel, which is the most cathodic? (Stainless steel.) (8374) GALVANIC CORROSION IS LIKELY TO BE MOST RAPID AND SEVERE - when the surface area of the anodic metal is smaller than the surface area of the cathodic material. (8375)

GALVANIC ACTION CAUSED BY DISSIMILAR METAL CONTACT - may best be prevented by placing a nonporous dielectric between the surfaces. (8370)

A PASSIVE OXIDE FILM IS NOT AN ELECTROLYTE - and would tend to prevent electrical contact between anodic and cathodic areas. Therefore, it would not be a requirement for corrosion to occur. (8359)

INTERGRANULAR CORROSION INTERGRANULAR CORROSION (DELAMINATION OF THE GRAIN BOUNDARIES) - is a type of corrosion that attacks along the grain boundaries of an alloy, and commonly results from a lack of uniformity in the alloy structure.

A PRIMARY CAUSE OF INTERGRANULAR CORROSION - is improper or inadequate heat treatment. (8245,8365)

INTERGRANULAR CORROSION IN ALUMINUM ALLOY PARTS - cannot always be detected by surface indications. (8363)

EXFOLIATION IS THE LIFTING OR FLAKING - of the metal at the surface due to delamination of the grain boundaries caused by the pressure of corrosion residual product buildup. (8360)

FRETTING CORROSION FRETTING CORROSION IS MOST LIKELY TO OCCUR - when two surfaces fit tightly together but can move relative to one another. (8356)

CORROSION REMOVAL ALUMINUM WOOL - is used for general mechanical cleaning of aluminum surfaces.

FINE-GRIT ALUMINUM OXIDE - may be used to remove corrosion from highly stressed steel surfaces. (8364)

CORROSION SHOULD BE REMOVED FROM MAGNESIUM PARTS - with a stiff, nonmetallic brush. (8366)

CORROSION CONTROL ALUMINUM OXIDE - is a naturally protective film that forms on aluminum. It can be formed by electrolytic treatment at the factory (anodizing) or chemical treatment in the field (alodizing).

WHEN AN ANODIZED SURFACE COATING IS DAMAGED IN SERVICE - it can be partially restored by chemical surface treatment. (8349)

ALODINE - is a brand name of a chemical coating treatment applied for corrosion control. ALODIZING IS A NONELECTROLYTIC CHEMICAL TREATMENT FOR ALUMINUM ALLOYS - to improve paint-bonding qualities by leaving a slightly roughened finish, and increase corrosion resistance by leaving a hard, airtight oxide film. (8358,8361)

TO CLEAN AN ANODIZED SURFACE - you should use a mechanical cleaner that will not scratch and compromise the coating. Aluminum wool and fiber bristle brushes will be acceptable to use to clean an anodized surface. (8362)

THE INTERIOR SURFACE OF SEALED STRUCTURAL STEEL TUBING - is best protected against corrosion by a coating of hot linseed oil. (8373)

RESISTING STRESS CORROSION

ONE WAY OF OBTAINING INCREASED RESISTANCE TO STRESS CORROSION CRACKING - is by creating compressive stresses on the metal surface (by shot peening). (This must be overcome by tensile forces before the surface sees any tension stress.) (8376)



ENGINE STORAGE TO PREVENT CORROSION ON AN ENGINE BEING PREPARED FOR STORAGE - spray each cylinder interior with corrosion preventive mixture.

ON ENGINES PREPARED FOR STORAGE - it is important not to rotate the propeller shaft after the final spraying of corrosion-preventive mixture into the cylinders because the seal of corrosion preventive mixture will be broken. (8367)

SOAP AND WATER CLEANING BEFORE CLEANING PLASTIC SURFACES WITH SOAP AND WATER - (such as windows) flush the plastic surfaces with fresh water to prevent scratching. (8368) TO PREVENT RAPID DETERIORATION OF A TIRE WHEN IT COMES IN CONTACT WITH OIL OR GREASE - wipe the tire with a dry cloth followed by a washdown with soap and water. (8369)

CHEMICAL CLEANERS MAGNESIUM ENGINE PARTS MAY BE CLEANED - by washing with a commercial solvent, decarbonize (with a hot dichromate solution), and scrape or grit blast. (8348)

ALIPHATIC NAPHTHA IS THE SOLVENT RECOMMENDED - for wipedown of cleaned surfaces just before painting. (8350)

ALIPHATIC NAPHTHA CAN ALSO BE USED - to clean acrylics and rubber. (8353)

FAYED SURFACES CAUSE CONCERN IN CHEMICAL CLEANING - because of the danger of entrapping corrosive materials. (Fayed surfaces are ones that are tightly joined as in lap joints.) (8354) WHEN A FLAMMABLE AGENT - (such as a solvent), is used to chemically clean an aircraft, natural fibers like cotton are acceptable to be used for wiping cloths. (8352)

ORDINARY OR OTHERWISE NONAPPROVED CLEANING COMPOUNDS – should not be used for washing an aircraft because a condition called hydrogen embrittlement may result in the metal structures. (Hydrogen embrittlement happens when chemical cleaning compounds react with the metal and produce hydrogen gas that is then absorbed into the metal. This results in a loss of flexibility and cracks or stress corrosion can develop in the metal.) (8347)

AFTER CLEANING METAL OF ALL GREASE, PAINT, DIRT AND CORROSION, - the surface that remains should be cleaned with a metal cleaner until a film of water rests unbroken on the surface. This is a water break test. (8554)

STEEL

THE SOCIETY OF AUTOMOTIVE ENGINEERS (SAE) USES A NUMERICAL INDEX SYSTEM - to identify the composition of various steels.

THE FIRST DIGIT - indicates the basic alloying element.

THE SECOND DIGIT - indicates the percentage of the basic element in the alloy.

THE THIRD AND FOURTH DIGITS - indicate the percentage of carbon in the alloy in hundredths of a percent.

EXAMPLE - In the number "4130" designating chromium molybdenum steel, what does the first digit indicate? (The basic alloying element.) (8261)

STAINLESS STEEL IS GENERALLY USED - in the construction of aircraft engine firewalls. (8257)

HEAT TREATMENT CHANGING THE INTERNAL STRUCTURE OF A FERROUS METAL - is accomplished by heating the metal to a temperature above its upper critical point, holding it there, and then cooling it under controlled conditions.

THE REHEATING OF A HEAT TREATED METAL - such as with a welding torch, can significantly alter a metal's properties in the reheated area. (8251)

HARDNESS

HARDNESS REFERS TO THE ABILITY OF A METAL - to resist abrasion, penetration, cutting action, or permanent distortion.

HARDNESS MAY BE INCREASED BY - cold-working the metal and, with steel and certain aluminum alloys, by heat treatment.

REPEATEDLY APPLYING MECHANICAL FORCE - such as rolling, hammering, bending, or twisting to most metals at room temperature, causes a condition commonly known as cold working, or strain hardening. (8250)



TEMPERING TEMPERING IS A PROCESS - that relieves brittleness and/or internal strain. The steel is heated in a furnace to a specified temperature and then cooled in air, oil, water, or a special solution.

STEEL IS TEMPERED AFTER BEING HARDENED - to relieve its internal stresses and reduce its brittleness. (8252)

ANNEALING AND NORMALIZING ANNEALING AND NORMALIZING - remove the internal stresses in metal.

THE PRIMARY EFFECTS OF ANNEALING - steel and aluminum alloys are softening of the metal and a decrease in internal stress. (8246)

SLOW COOLING; LOW STRENGTH - is descriptive of the annealing process of steel during and after it has been annealed. (8255)

NORMALIZING IS A PROCESS - of heat treating iron-base metals only. (8249)

IT IS CONSIDERED GOOD PRACTICE TO NORMALIZE A PART AFTER WELDING - to relieve internal stresses developed within the base metal. (8280)

CASE HARDENING CASE HARDENING IS A HEAT-TREATING PROCESS OF METAL - which produces a hard, wear- resistant surface over a strong, tough core. (8247)

IN CASE HARDENING THE SURFACE OF THE METAL - is changed chemically by introducing a high carbide or nitride content. (8248)

ALUMINUM AND ALUMINUM ALLOYS ALUMINUM AND ALUMINUM ALLOYS ARE DESIGNATED BY - a four digit index system.

THE SYSTEM IS BROKEN INTO THREE DISTINCT GROUPS - 1xxx group, 2xxx through 8xxx group, and the 9xxx group.

THE FIRST DIGIT OF A DESIGNATION IDENTIFIES - the alloy type (but 1xxx = 99% pure aluminum).

EXAMPLE - The aluminum code number 1100 identifies what type of aluminum? (99 percent commercially pure aluminum.) (8274)

IN THE 2xxx THROUGH 8xxx GROUP, THE FIRST DIGIT INDICATES - the major alloying element used in the formation of the alloy.

2xxx = copper. 7xxx = zinc.

EXAMPLE - In the four-digit aluminum index system number 2024, what does the first digit indicate? (The first digit, 2, indicates copper is the major alloying element.) (8276)

"ALCLAD" DESIGNATES A METAL - consisting of pure aluminum surface layers on an aluminum alloy core. (8259)

EXAMPLE - What is the core material of Alclad 2024-T4? (The -T4 indicates the core material is heat treated aluminum alloy; the surface material is commercially pure aluminum.)(8273)

HARDNESS AND TEMPER DESIGNATIONS

A LETTER AND OFTEN A NUMBER COMBINATION - is placed after the alloy code to indicate the processes that have taken place and the degree of hardness. Basic designations are: F - As fabricated (no treatment) O - Annealed H - Cold worked or strain condition W - Unstable (temporary) condition T - Solution heat treated

EXAMPLE - Which of the following aluminum alloy designations: 3003-F 5052-H36 6061-O indicate that the metal has received no hardening or tempering treatment? (3003-F.) (8253)

QUENCHING QUENCHING IS THE PROCESS - of cooling heat-treated metal in a liquid bath. Rapid quenching is needed with aluminum alloys to keep the alloy grains very small and thus prevent intergranular corrosion.

PARTS HEAT TREATED IN A SODIUM AND POTASSIUM NITRATE BATH - are rinsed thoroughly in hot water to prevent corrosion.

RE-HEAT TREATMENT CLAD ALUMINUM ALLOYS CANNOT BE - heat treated repeatedly without harmful effects. (The pure aluminum and the aluminum alloy tend to intermix.) (8254)


HARDWARE - NOTES 22

AIRCRAFT HARDWARE

AIRCRAFT BOLTS MOST BOLTS USED IN AIRCRAFT STRUCTURES - are either general purpose AN (Air Force-Navy) or NAS (National Aircraft Standard) close tolerance bolts.

IDENTIFICATION AND CODING OF BOLTS

BOLTS ARE IDENTIFIED BY - the code markings on the bolt heads.

EXAMPLE - Aircraft bolts with a cross or asterisk marked on the bolthead identifies what kind of bolt? (AN standard steel bolts.) (8263)

EXAMPLE - A bolt with an X inside a triangle on the head identifies what kind of bolt? (NAS close tolerance bolt.) (8272)

EXAMPLE - Figure 42. Which of the bolthead code markings shown identifies a corrosion resistant AN standard steel bolt? (A single raised dash on the head, "3", identifies a corrosion resistant AN standard steel bolt.) (8262,8269)

CLEVIS BOLTS THE HEAD OF A CLEVIS BOLT IS ROUND - and is either slotted to receive a common screwdriver or recessed to receive a crosspoint screwdriver.

EXAMPLE - Figure 43. Which illustration is a clevis bolt? (Illustration "3".) (8267)

AN-TYPE CLEVIS BOLTS ARE USED IN AIRPLANES - only for shear load applications and never in tension. (8271)

A CLEVIS BOLT - used with a fork-end cable terminal is secured with a shear nut tightened to a snug fit, but with no strain imposed on the fork, and safetied with a cotter pin. (8270)

GRIP LENGTH AIRCRAFT BOLT GRIP LENGTHS - should be equal to the thickness of the material through which they extend. (The bolt grip length is the unthreaded portion.) (8264,8265)

EXAMPLE - A certain aircraft bolt has an overall length of 1-1/2 inches, with a shank length of 1-3/16 inches, and a threaded portion length of 5/8 inch. What is the grip length? (0.5625 inches.) (8415)

Grip Length = Shank Length - Thread Length Grip Length = 1-3/16 in. - 5/8 in. 1-3/16 = 19/16 Divide 19 by 16 on calculator Divide 5 by 8 on calculator Grip Length = 1.1875 in. – 0.625 in. Grip Length = 0.5625 inch

CLASSIFICATION OF THREADS THE CLASS OF A THREAD INDICATES - the tolerance allowed in manufacturing (whether the nut can be turned with the fingers or requires a wrench). CLASS 1 - is a loose fit. CLASS 2 - is a free fit. CLASS 3 - is a medium fit, and is what aircraft

bolts are usually manufactured with. CLASS 4 - is a close fit. (8275)

FIBER-TYPE LOCKNUT THE LOCKING FEATURE OF A FIBER-TYPE LOCKNUT - is obtained by the use of an unthreaded fiber locking insert. (8277)

A FIBER-TYPE, SELF-LOCKING NUT - must never be used on an aircraft if the bolt is subject to rotation. (8260)

INSTALLATION OF NUTS AND BOLTS ADVISORY CIRCULAR 43.13-1B, "AIRCRAFT INSPECTION AND REPAIR" - contains standards for protrusions of bolts, studs, and screws through self-locking nuts. (8534)

UNLESS OTHERWISE SPECIFIED OR REQUIRED - aircraft bolts should be installed so that the bolthead is upward, or in a forward direction. (8258)

UNLESS OTHERWISE SPECIFIED - torque values for tightening aircraft nuts and bolts relate to clean, dry, threads. (8256)


HARDWARE - NOTES 23

WHEN THE SPECIFIC TORQUE VALUE FOR NUTS IS NOT GIVEN - the recommended torque values can be found in Advisory Circular 43.13-1B. (8266)

A PARTICULAR COMPONENT IS ATTACHED TO THE AIRCRAFT STRUCTURE - by the use of an aircraft bolt and a castellated tension nut combination. If the cotter pin hole does not align within the recommended torque range, the acceptable practice is to change washers and try again. (8268)


PHYSICS - NOTES 24

PHYSICSPRESSURE AND TEMPERATURE

TEMPERATURE - is a measure of the kinetic energy of the molecules of any substance. (8482)

HEAT MAY BE TRANSFERRED - by convection, conduction, or radiation. Heat is not transferred by diffusion. (8467)

IF THE VOLUME OF A CONFINED GAS IS DOUBLED - (without the addition of more gas), the pressure will (assume the temperature remains constant) be reduced to one-half its original value. (Boyle's law.) (8475)

IF BOTH THE VOLUME AND THE ABSOLUTE TEMPERATURE - of a confined gas are doubled, the pressure will not change. (8485)

IF THE TEMPERATURE OF A CONFINED LIQUID IS HELD CONSTANT - and its pressure is tripled, the volume will remain the same. (8476)

THE BOILING POINT OF A GIVEN LIQUID - varies directly with pressure. (8466)

THE SPEED OF SOUND - in the atmosphere changes with a change in temperature. (8474)

HUMIDITY

ABSOLUTE HUMIDITY - is the actual amount of water vapor in a mixture of air and water. (8469,8483)

RELATIVE HUMIDITY - is the ratio of the water vapor actually present in the atmosphere to the amount that would be present if the air were saturated at the prevailing temperature and pressure. (8473)

DEWPOINT - is the temperature to which humid air must be cooled at constant pressure to become saturated. (8484)

HUMID AIR AT A GIVEN TEMPERATURE AND PRESSURE - is lighter than dry air at the same temperature and pressure.

EXAMPLE - Does 35 parts of dry air and 65 parts of water vapor weigh less than 50 parts of dry air and 50 parts of water vapor, or any other combination where the dry air content is more than 35 parts out of 100 parts? (Yes.) (8472)

THE TRUE LANDING SPEED OF AN AIRCRAFT IS GREATEST - under atmospheric conditions of high temperature with high humidity. (8478)

FLUID SYSTEMS BERNOULLI'S PRINCIPLE STATES - that the pressure of a fluid decreases at points where the velocity of the fluid increases. (8216)

THE GREATER THE PRESSURE DIFFERENTIAL BETWEEN UNMETERED PRESSURE AND METERED PRESSURE - the greater the rate of flow of liquid through the metering orifice.

EXAMPLE - Under which conditions will the rate of flow of liquid through a metering orifice (or jet) be the greatest (all other factors being equal)? (Orifice "C".)

A. Unmetered pressure: 18 PSI, metered pressure: 17.5 PSI, atmospheric pressure: 14.5 PSI. B. Unmetered pressure: 23 PSI, metered pressure: 12 PSI, atmospheric pressure: 14.3 PSI. C. Unmetered pressure: 17 PSI, metered pressure: 5 PSI, atmospheric pressure: 14.7 PSI. (8470)

The metering jet with the greatest pressure differential across it (Orifice C: 17 PSI - 5 PSI = 12 PSI) will have the greatest flow rate through it.

AERODYNAMICS

NEWTON’S FIRST LAW OF MOTION – also called the law of inertia states that a body (object) persists in its state of rest, or of motion in a straight line, unless acted upon by some outside force. (8606)

AN AIRPLANE WING - is designed to produce lift resulting from positive air pressure below the wing's surface and negative air pressure above the wing's surface (in addition to the downward deflection of air at the trailing edges of the wings). (8487)

ASPECT RATIO OF A WING - is defined as the ratio of the wingspan to the mean chord. (8489)

A WING WITH A VERY HIGH ASPECT RATIO - (in comparison with a low aspect ratio wing) will have a low stall speed. (8490)


PHYSICS - NOTES 25

IF ALL, OR A SIGNIFICANT PART OF A STALL STRIP IS MISSING ON AN AIRPLANE WING - a likely result will be asymmetrical lateral control at or near stall angles of attack. Stall strips ensure that the wing root areas stall first. (8486,8600)

WINGLETS ARE SMALL WING-SHAPED DEVICES – mounted at the wingtips and perpendicular to the wings.

WINGLETS ON AN AIRCRAFT'S WINGTIPS – increase the lift to drag ratio. Winglets reduce the strength of trailing vortices by lessening the amount of crossflow on the wings. The reduced vortices reduces drag and increases the wing's lift to drag ratio. (8491)

THE PURPOSE OF AIRCRAFT WING DIHEDRAL - is to increase lateral stability. (8488)

WORK AND POWER WORK IS DONE - when a resistance is overcome by a force acting through a measurable distance.

WORK = FORCE x DISTANCE W = F x D

EXAMPLE - How much work input is required to lower (not drop) a 120-pound weight from the top of a 3-foot table to the floor? (360 foot-pounds.) (8477) W = F x D = 120 lbs. x 3 ft. W = 360 foot-pounds

EXAMPLE - An engine that weighs 350 pounds is removed from an aircraft by means of a mobile hoist. The engine is raised 3 feet above its attachment mount, and the entire assembly is then moved forward 12 feet. A constant force of 70 pounds is required to move the loaded hoist. What is the total work input required to move the hoist? (840 foot- pounds.) (8468) W = F x D = 70 lbs. x 12 ft. W = 840 foot-pounds

POWER MEANS RATE OF DOING WORK - and is measured in terms of work accomplished per unit of time. (8480)

POWER

P F DT

= ×

= ×

FORCE DISTANCETIME

MACHINES A SHORTHAND METHOD OF FINDING THE MECHANICAL ADVANTAGE OF A PULLEY

SYSTEM - is to count the number of rope strands that move or support the movable block.

EXAMPLE - Figure 61. What is the amount of force applied to rope A to lift the weight? (15 pounds.) (8471) Four supporting ropes. Mechanical advantage = 4. 60 lbs./4 = 15 lbs. Each rope supports 15 lbs. Therefore, pull at rope A = 15 pounds THE INCLINED PLANE - is a simple machine that facilitates the raising or lowering of heavy objects by applying a small force over a long distance.

( )( )

( )( )

L R length of rampI height of ramp

weight of objectE force required

=

EXAMPLE - What force must be applied to roll a 120-pound barrel up an inclined plane 9 feet long to a height of 3 feet (disregard friction)? (40 pounds.) (8481)

LI

RE

E

=

=

= × =

9

120 40

ft.3 ft.

120 lbs.E

lbs. 3 ft.9 ft.

pounds

HYDRAULIC CYLINDERS PASCAL'S LAW STATES - that the force available in a hydraulic system is equal to the pressure in the cylinder multiplied by the cross sectional area of the piston. Force = Pressure x Area F = P x A EXAMPLE - What force is exerted on the piston in a hydraulic cylinder if the area of the piston is 1.2 square inches and the fluid pressure is 850 PSI? (1,020 pounds.) (8398,8479) F = P x A = 850 PSI x 1.2 sq. in. F = 1,020 pounds EXAMPLE - Approximately how much force will the actuator be able to produce when retracting, if a double-acting actuating cylinder in a 3,000 psi system has a piston with a surface area of three square inches on the extension side, and a rod with a cross-section area of one square inch attached to the piston on the other side? (8465) F = P x A F = 3,000 x (3-1) F = 6,000


FLUID LINES & FITTINGS - NOTES 26

FLUID LINES AND FITTINGSFLEXIBLE LINES

TEFLON OR BUTYL ARE HOSE MATERIALS - that can be used to carry a wide range of petroleum and synthetic fluids including phosphate-ester based hydraulic fluid (such as Skydrol). (8211)

FLEXIBLE HOSE USED IN AIRCRAFT PLUMBING - is classified in size according to the inside diameter. (8209)

EXAMPLE - A 3/8 inch aircraft high pressure flexible hose as compared to 3/8 inch metal tubing used in the same system will have what? (Equivalent flow characteristics.) (8218)

A GAS OR FLUID LINE MARKED WITH THE LETTERS "PHDAN" - is a line carrying a substance that may be PHysically DANgerous or hazardous to personnel. (8215)

GEOMETRIC SHAPES - that appear with color-coded bands are always black against a white background regardless of content. (They are used to identify line contents.) (8559)

A CERTAIN AMOUNT OF SLACK - must be left in a flexible hose during installation because, when under pressure, it contracts in length and expands in diameter. (8197)

FLEXIBLE LINES MUST BE INSTALLED - with a slack of 5 to 8 percent of the length. (8201)

EXAMPLE - The maximum distance between end fittings to which a straight hose assembly is to be connected is 50 inches. What is the minimum hose length to make such a connection? (52-1/2 inches.) (8202)

50 inches x 5% slack = 2-1/2 inches 50 inches + 2-1/2 inches = 52-1/2 inches

THE TERM "COLD FLOW" - is generally associated with rubber hose. It describes the deep, permanent impressions in the hose produced by the pressure of hose clamps or supports. (8198)

METAL TUBING IN A METAL TUBING INSTALLATION - tension is undesirable because pressurization will cause it to expand and shift. (8214)

EXCESSIVE STRESS ON FLUID OR PNEUMATIC METAL TUBING - caused by expansion and contraction due to temperature changes can best be avoided by providing suitable bends in the tubing. Preventing excessive stress on the tubing is the primary reason for adding bends. (8203,8605)

MINIMUM ALLOWABLE BEND RADII IS SOMETIMES DIFFERENT – for steel tubing than aluminum tubing. For tubing 1.5” OD or less, the minimum radius for steel is greater than for aluminum. (8599)

THE FLATTENING IN A BEND - must not be more than 25% of the original tube’s outside diameter and also be free of wrinkles and kinks.(8562)

ALUMINUM TUBING - or other soft metal tubing is cut best with a hand operated wheel-type tubing cutter. (8561)

SCRATCHES OR NICKS IN ALUMINUM ALLOY TUBING CAN BE REPAIRED BY BURNISHING - provided the scratch or nick does not appear in the heel of a bend, or if in a straight section is no deeper than 10% of the wall thickness. (8208,8210)

DEFECTS THAT ARE NOT ACCEPTABLE FOR METAL LINES ARE:

• Cracked flare, • Seams, • Dents in the heel of a bend less than 20%

of tube diameter, and • Dents in straight sections that are 20% of

wall thickness. (8598)

WHEN INSTALLING BONDED CLAMPS TO SUPPORT METAL TUBING - remove paint or anodizing from tube at clamp location. Note: Unbonded clamps are used for support when installing wiring. (8213,8217)

METAL TUBING FLUID LINES - are sized by wall thickness and outside diameter in 1/16 inch increments. (8193)

TO FIND THE INSIDE DIAMETER OF TUBING - subtract 2 wall thicknesses from the outside diameter.


FLUID LINES & FITTINGS - NOTES 27

EXAMPLE - A replacement oil line must be fabricated from 3/4-inch, 0.072 5052-0 aluminum alloy tubing. What is the inside dimension of this tubing? (0.606 inch.) (8204)

Outside diameter = 0.750 in. 2 x wall thickness = (2 x 0.072) = 0.144 in. Inside diameter = 0.750 - 0.144 = 0.606 in.

FLARES

AN (ARMY/NAVY) FLARE FITTINGS HAVE REPLACED - AC (Air Corp) fittings in newer aircraft. The fittings are shaped differently, but are sometimes (although not always) interchangeable.

AN (ARMY/NAVY) FLARE FITTINGS CAN EASILY BE IDENTIFIED - by the shoulder between the end of the threads and the flare cone. (8200)

THE COLOR - of an AN steel flared-tube fitting is black. (8199)

WHEN FLARING ALUMINUM TUBING - for use with AN fittings, the flare angle must be 37°. (8207)

IN MOST AIRCRAFT HYDRAULIC SYSTEMS - two-piece tube connectors consisting of a sleeve and nut are used when a tubing flare is required. The use of this type connector eliminates the possibility of reducing the flare thickness by wiping or ironing during the tightening process. (8205)

WHEN THE COUPLING NUT IS OVERTIGHTENED - on a flared tube, damage is most likely at the sleeve and flare junction. (8560)

THE FOLLOWING SEQUENCE OF STEPS - indicates the proper order you would use to make a single flare on a piece of tubing: 1. Slip the fitting nut and sleeve on the tube. 2. Place the tube in proper size hole in the

flaring block. 3. Center the plunger or flaring pin over the

tube. 4. Project the end of the tube slightly from the

top of the flaring tool, about the thickness of a dime.

5. Tighten the clamp bar securely to prevent slippage.

6. Strike the plunger several light blows with a light weight hammer or mallet and turn the plunger one-half turn after each blow.

(8194)

AN ADVANTAGE OF A DOUBLE FLARE ON ALUMINUM TUBING - is that it is more resistant to the shearing effect of torque. (8196)

COUPLING SIZES THE END DASH NUMBER OF AN-818 COUPLING NUTS INDICATES - in 1/16 inch increments the size of tubing on which the nut should be used.

EXAMPLE - What end dash number of an AN-818 coupling nut should be selected for use with 1/2-inch aluminum oil lines which are to be assembled using flared tube ends and standard AN nuts, sleeves, and fittings? (1/2 inch = 8/16, select an AN-818-8.) (8192)

MS FLARELESS FITTINGS DURING INSTALLATION, MS (MILITARY STANDARD) FLARELESS FITTINGS - are normally tightened by turning the nut a specified amount after the sleeve and fitting sealing surface have made contact, rather than being torqued. (8206)

HYDRAULIC LINES CORROSION-RESISTANT STEEL TUBING, ANNEALED OR 1/4H (1/4 HARD) - has the characteristics (high strength, abrasion resistance) necessary for use in a high-pressure (3,000 PSI) hydraulic system for operation of landing gear and flaps. (8212)

HYDRAULIC TUBING WHICH IS DAMAGED IN A LOCALIZED AREA - to such an extent that repair is necessary, may be repaired by cutting out the damaged area and utilizing a swaged tube fitting to join the tube ends. (8195)


FUELS/ FUEL SYSTEMS - NOTES 28

FUELS AND FUEL SYSTEMSCHARACTERISTICS OF AVIATION GASOLINE

CHARACTERISTICS OF AVIATION GASOLINE ARE - high heat value and high volatility. (8344)

VOLATILITY VOLATILITY IS A MEASURE OF THE TENDENCY OF A LIQUID SUBSTANCE - to vaporize under given conditions.

A FUEL THAT VAPORIZES TOO READILY - may cause vapor lock. (8341)

A FUEL THAT DOES NOT VAPORIZE READILY ENOUGH - can cause hard starting. (8346)

AN ABSORPTION OF HEAT - must accompany fuel vaporization (which can contribute to carburetor icing). (8339)

FUEL ADDITIVES TETRAETHYL LEAD IS ADDED TO AVIATION GASOLINE - to improve the gasoline's performance in the engine. (8345)

ETHYLENE DIBROMIDE IS ADDED TO AVIATION GASOLINE - to scavenge lead oxide from the cylinder combustion chambers. (8337)

DETONATION DETONATION IS AN ABNORMAL TYPE OF COMBUSTION - which occurs when the first portion of the charge burns in a normal manner but the last portion burns almost instantaneously.

CHARACTERISTICS OF DETONATION ARE: Rapid rise in cylinder pressure, Excessive cylinder head temperature, Decrease in engine power. (8340)

FUEL RATING ANTIKNOCK QUALITIES OF AVIATION FUEL ARE DESIGNATED BY GRADES - and the higher the grade, the more compression the fuel charge can stand without detonating.

TWO DIFFERENT SCALES - are used to designate fuel grades. Fuels below grade 100 are classified by octane numbers.

FUELS WHICH POSSESS GREATER ANTIKNOCK QUALITIES THAN 100 OCTANE - are classified by performance numbers. (8336)

THE MAIN DIFFERENCES - between grades 100 and 100LL fuel are lead content and color. (8343)

FUEL COLORS AVIATION GASOLINES ARE COLOR CODED - for purposes of identification. For example, 100LL fuel is blue. (8335)

TURBINE FUELS BOTH GASOLINE AND KEROSENE HAVE CERTAIN ADVANTAGES - for use as turbine fuel. Kerosene, however, has a higher heat energy per unit volume than gasoline (because kerosene weighs more per gallon). (8338)

JET FUEL NUMBER IDENTIFIERS - are type numbers and have no relation to the fuel's performance in the aircraft engine. (8342)

FUEL SYSTEM CONTAMINATION

JET FUEL IS MORE SUSCEPTIBLE TO CONTAMINATION THAN AVIATION GASOLINE - because jet fuel is of higher viscosity and therefore holds contaminants better. (8325)

AVIATION GASOLINE MIXED WITH JET FUEL CAN BE BURNED BY A JET ENGINE - but the tetraethyl lead in the gasoline forms deposits on the turbine blades reducing turbine engine efficiency. (8324)


ELECTRICITY - NOTES 29

BASIC ELECTRICITYCURRENT

ELECTRONS IN MOTION - make up an electric current. When the current flow is in one direction only, it is called direct current. Current that reverses itself periodically is called alternating current.

CURRENT FLOW - is measured in amperes.

MILLIAMPERE - is a measure of current equivalent to 0.001 ampere. (8030)

ONE WAY TO SHOW CURRENT IN AN ELECTRICAL CIRCUIT DIAGRAM – is the ground symbol. It shows that there is a return path for the current between the source of electrical energy and the load. (8604)

RESISTORS RESISTANCE is the property of a conductor that limits the flow of electric current.

THE RESISTANCE OF A CONDUCTOR - will decrease when its length is decreased or its cross-sectional area is increased. (8051)

A RESISTOR IS A CIRCUIT ELEMENT - designed to insert resistance in the circuit. It may be of low or of extremely high value, varying from only a few ohms up to several million ohms. They are generally classed as fixed, adjustable, or variable dependent on their construction and use. Typical fixed resistors are constructed of a small rod of a carbon compound.

AN ADJUSTABLE RESISTOR - is usually of the wire-wound type with a metal collar that can be moved along the resistance wire to vary the value of resistance placed in the circuit.

A VARIABLE RESISTOR - is arranged so that it can be changed in value at any time by the operator. Variable resistors are commonly known as rheostats or potentiometers.

EXAMPLE - Figure 17. Which of the components is a potentiometer? ("3" is a potentiometer.)(8065)

EXAMPLE - Figure 21. Which symbol represents a variable resistor? ("2 " is a variable resistor.) (8074)

VOLTAGE THE FORCE WHICH CAUSES ELECTRONS TO FLOW - through a conductor is electric pressure or electromotive force (emf). The practical unit for measurement of emf is the volt (V). The symbol for emf or electric pressure is the letter "E".

THE POTENTIAL DIFFERENCE BETWEEN TWO CONDUCTORS - which are insulated from each other is measured in volts. (8020)

ONE VOLT IS THE EMF - required to cause current to flow at the rate of 1 ampere through a resistance of 1 ohm.

1.0 KILOVOLT = 1,000 VOLTS.

EXAMPLE - 0.002 KV equals how many volts? (2.0 volts.) (8033)

Volts = Kilovolts x 1,000 = 0.002 x 1,000 Volts = 2.0

OHM'S LAW OHM'S LAW OUTLINES THE RELATIONSHIP - between voltage, current, and resistance in a direct current electrical circuit. Basically, Ohm's Law states that as voltage increases, current increases, and when the voltage decreases, the current decreases, if the resistance in the circuit remains constant.

THE OHM - is the unit used to measure resistance. In mathematical formulas, the capital letter "R", for "Resistance", is used.

AS AN EQUATION - Ohm's law is expressed as follows:

I = ER

where:

I = current in amperes, E = potential difference in volts, R = resistance in ohms.

If any two circuit quantities are known, the third may be found algebraically by:

I = ER

E = I R

R = EI

×



THE VOLTAGE DROP IN A CONDUCTOR - of known resistance is dependent on the amperage of the circuit. (8055)

USING THE OHMMETER EXAMPLE - Figure 7. If resistor R3 is disconnected at terminal D, what will the ohmmeter read? (The current path is broken and the ohmmeter would read infinite resistance.) (8026)

SERIES CIRCUITS THE SERIES CIRCUIT - is the most basic of electrical circuits. It is called a series circuit because the current can flow in only one possible path and must pass through all the circuit components, such as the battery and the resistor, one after the other, or "in series".

NO MATTER HOW MANY COMPONENTS ARE INCLUDED IN A SERIES CIRCUIT - the current is the same intensity throughout the circuit.

IN A SERIES CIRCUIT - the total resistance is equal to the sum of all the resistance’s in the circuit: R R R Rt = + + +1 2 3 etc.

EXAMPLE - A circuit has an applied voltage of 30 volts and a load consisting of a 10-ohm resistor in series with a 20-ohm resistor. What is the voltage drop across the 10-ohm resistor? (10 volts.) (8043)

I E

R= ampere

E = I R = 1 10 = 10 volts

=+

= =

× ×

3010 20

3030

1

EXAMPLE - If three resistors of 3 ohms, 5 ohms, and 22 ohms are connected in series in a 28-volt circuit, how much current will flow through the 3-ohm resistor? (0.93 ampere.) (8042) R R R Rt = + +1 2 3 = 3 + 5 + 22 = 30

I = ER

= 2830

= 0.93 ampere

EXAMPLE - A lead-acid battery with 12 cells connected in series (no-load voltage = 2.1 volts per cell) furnishes 10 amperes to a load of 2-ohm resistance. What is the internal resistance of the battery in this instance? (0.52 ohm.) (8085)

Total Voltage = 12 x 2.1 = 25.2 volts

R = E

I= 25.2

10= 2.52 ohms

R = 2.52 - 2.00 = 0.52 ohms

t

internal

PARALLEL CIRCUITS A PARALLEL CIRCUIT - is one in which two or more electrical resistance’s, or loads, are connected across the same voltage source.

THE PARALLEL CIRCUIT - differs from the series circuit in that more than one path is provided for current flow. The more paths added in parallel, the less opposition to flow of electrons from the source.

IN A PARALLEL CIRCUIT THE VOLTAGE ACROSS ANY RESISTANCE IN A GROUP - is equal to the voltage across any other resistance in the group. When the current flow and voltage are given for each resistor, the value of each resistor can be determined by Ohm's law.

EXAMPLE - Figure 11. What is the voltage across the 8-ohm resistor? (24 volts.) (8045)

In this parallel circuit, the applied voltage is the same across points C and F as it is across points D and G and across points E and H.

IN REFERENCE TO A PARALLEL CIRCUIT - the total current is equal to the sum of the currents through the individual branches of the circuit. (8039)

EXAMPLE - Figure 13. What is the total current flow in the circuit? (1.4 amperes.) (8049)

I

amperes

R112 0 4

12 0 2

12

14

= =

= =

= =

=

volts30 ohms

amperes

I volts60 ohms

amperes

I volts15 ohms

0.8 amperes

Total

R2

R3

.

.

.

THE FORMULA FOR THE TOTAL RESISTANCE IN A PARALLEL CIRCUIT - is:

R

R R R

t =+ +

11 1 11 2 3

IN A PARALLEL CIRCUIT - total resistance will be smaller than the smallest resistor. (8054)

IF ONE OF THREE BULBS IN A PARALLEL LIGHTING CIRCUIT - is removed, the total resistance of the circuit will become greater. (8047)



EXAMPLE - Figure 8. With an ohmmeter connected into the circuit as shown, what will the ohmmeter read? (The ohmmeter will read the resistance of R1 and R2 in parallel. The break in R3 gives it an infinite resistance and does not affect the ohmmeter reading. In this case, the ohmmeter will read 10 ohms.) (8027)

R

R R

R

R

R

R

R

R

t

t

t

t

t

t

t

=+

=+

=

=

= ÷

= ×

=

11 1

11

201

2012

2011

101 1

101 10

1

1 2

10 ohms

EXAMPLE - How many amperes will a 28-volt generator be required to supply to a circuit containing five lamps in parallel, three of which have a resistance of 6 ohms each and two of which have a resistance of 5 ohms each? (25.23 amperes.) (8018)

R

R

R

R

R

RER

I

t

t

t

t

t

t

=+ + + +

=+

=+

=+

=

=

=

=

=

116

16

16

15

15

136

25

112

251

0 5 0 41

0 9111

2811125 23

. .

..

..

ohms

I

I

amperes

EXAMPLE - Figure 6. If resistor R5 is disconnected at the junction of R3 and R4, what will the ohmmeter read? (3 ohms.) (8025)

R

R R R R

R

R

R

R

R

R

R

R

t

t

t

t

t

t

t

t

t

=+ +

+

=+ +

+

=+ +

=+ +

=

=

= ÷

= ×

=

11 1 1

11

1216

16 6

11

1216

112

11

122

121

1214

12113

3

1 2 3 4

1 13

1 31

ohms

Example - Figure 14. What is the total resistance of the circuit? (17 ohms.) (8050)

The total resistance can be found in two steps:

1. Total resistance in the parallel circuit is:

R

R

R

R

R

R

R

t

t

t

t

t

t

t

=+ +

=+ +

=

=

= ÷

= ×

=

114

16

112

13

122

121

1216

12112

2

1 12

1 21

ohms



2. Total resistance in the remaining series circuit is:

Rt = 2 + 5 + 10 = 17 ohms

EXAMPLE - Figure 12. What is the total resistance of the circuit? (21.2 ohms.) (8046)

The total resistance can be found in four steps:

1. R4 and R5 are in parallel:

ohms 4R141R

411R

411R

1231R

122

121

1R

61

121

1R

t

t

t

t

t

t

t

=

×=

÷=

=

=

+=

+=

2. The above 4 ohm combined resistance is in series with R2 for a total of 16 ohms:

Rt = R2 + 4 = 12 + 4 = 16 ohms

3. The above 16 ohms is in parallel with R3 giving a total of 3.2 ohms:

R

R

R

R

R

R

t

t

t

t

t

t

=+

=+

=

= ÷

= ×

=

11

1214

11

164

1615

161 5

161 16

53.2 ohms

4. This 3.2 ohms is in series with R1 for a total circuit resistance of 21.2 ohms:

Rt = 3.2 + 18 = 21.2 ohms

EXAMPLE - Figure 11. What is the total current flowing in the wire between points C and D? (3.0 amperes.) (8044) The total resistance between points C,D,G and C,D,E,H is:

amperes 3I824I

REI

ohms 8R5401R

4051R

4051R

401

404

1R

401

101

1R

R1

R1

1R

t

t

t

t

t

t

32

t

=

=

=

=

×=

÷=

=

+=

+=

+=

DIODES THE DIODE IS THE SIMPLEST - of electron tubes. It has two operating electrodes. One of these is the heated cathode, which emits the electrons, and the other is the plate or anode.

THE PRINCIPAL ADVANTAGE OF THE DIODE TUBE - is that it will permit the flow of current in one direction only, from the cathode to the anode. If an alternating current is applied to the cathode, the tube will conduct only during one-half of each cycle, that is, while the cathode is negative and the anode is positive. For this reason, diode tubes are often used as rectifiers.

A RECTIFIER IS A DEVICE - which allows current to flow in one direction but will stop, or oppose, current flow in the opposite direction. Rectifiers are, so to speak, one way gates for electrons.

DIODES ARE USED - in electrical power circuits primarily as rectifiers. (8040)



A ZENER DIODE - is a solid state device which will conduct electricity only under certain voltage conditions. Typical application for zener diodes, therefore, is as voltage regulators. (8077)

TRANSFORMERS A TRANSFORMER CHANGES ELECTRICAL ENERGY - of a given voltage into electrical energy at a different voltage level. It consists of two coils which are not electrically connected, but which are arranged so that the magnetic field surrounding one coil cuts across the other coil.

THE CURRENT IS STEPPED DOWN - by a 1 to 4 ratio in a voltage step-up transformer with a ratio of 1 to 4. (When voltage steps up, current steps down by the same ratio.) (8048)

POWER POWER IS DEFINED AS - the rate of doing work, and is equal to the product of the voltage and current in a dc circuit.

WHEN THE CURRENT - in amperes (I) is multiplied by the emf in volts (E), the result is power measured in watts (P): P = I x E

THE UNIT USED TO EXPRESS - electrical power is the watt. (8037)

A 24-VOLT SOURCE - is required to furnish 48 watts to a parallel circuit consisting of four resistors of equal value. Since the resistors are all in parallel across the 24-volt power source, each resistor will have a 24-volt drop across it. (8021)

A CABIN ENTRY LIGHT - of 10 watts and a dome light of 20 watts are connected in parallel to a 30-volt source. If the voltage across the 10-watt light is measured, it will be equal to the voltage across the 20-watt light. (8031)

EXAMPLE - A 48-volt source is required to furnish 192 watts to a parallel circuit consisting of three resistors of equal value. What is the value of each resistor? (36 ohms.) (8053)

I PE

EI

R

R R R

t

= = =

= = =

=+ +

19248

4

484

12

11 1 1

amperes

R ohmst

12 13

12 1 3

12 13

123

36

=

= ÷

= ×

=

=

R

RR

R

R ohms

EXAMPLE - What is the operating resistance of a 30-watt light bulb designed for a 28-volt system? (26 ohms.) (8038)

I P

E= = =

= = =

3028

107

28107

2617

.

..

amperes

R EI

ohms

EXAMPLE - A 24-volt source is required to furnish 48 watts to a parallel circuit consisting of two resistors of equal value. What is the value of each resistor? (24 ohms.)

GIVEN:

R EPt =

2 (8035)

SOLUTION:

( )

( )

R EP

R

R RR

RR

RR

RR R

R R

R

R

t

t

t

t

t

t

t

= = =

=+

=+

=

= ÷

= ×

=

=

=

2 22448

12

11 1

11 1

12

1 2

12

212

2

ohms

24 ohms

EXAMPLE - How much power must a 24-volt generator furnish to a system which contains the following loads? (8016)



UNIT RATING

One motor (75% efficient) 1/5 hp Three position lights 20 watts each One heating element 5 amp One anticollision light 3 amp (Note: 1 hp = 746 watts.)

Motor: P = ×15

74675%

= 199 watts

Position lights: P = 3 x 20 = 60 watts Heating element: P = IE = 5 x 24 = 120 watts Anticollision light: P = IE = 3 x 24 = 72 watts Total = 451 watts

EXAMPLE - Which of the following will require the most electrical power? (8015,8036)

(Note: 1 hp = 746 watts.)

1. Four 30-watt lamps arranged in a 12-volt parallel circuit. P = 4 x 30 = 120 watts

2. A 1/5-horsepower, 24-volt motor which is 75 percent efficient.

watts199 =%75

74651=P ×

3. A 24-volt anticollision light circuit consisting of two light assemblies which require 3 amps each during operation. P = IE = (2 x 3) x 24 = 144 watts

Circuit No. 2 at 199 watts requires the most power.

A 12-VOLT ELECTRIC MOTOR - has 1,000 watts input and 1 hp output. Maintaining the same efficiency, a 24-volt, 1-hp electric motor would also require 1,000 watts input power. (8017)

EXAMPLE - A 1-horsepower, 24-volt dc electric motor that is 80 percent efficient requires 932.5 watts. How much power will a 1-horsepower, 12-volt dc electric motor that is 75 percent efficient require? (994.7 watts.) (8019)

(Note: 1 horsepower = 746 watts

P = 1 HP 746 watts / HP

P = 994.7 watts

×75%

EXAMPLE - A 14-ohm resistor is to be installed in a series circuit carrying 0.05 ampere. How much power will the resistor be required to dissipate? (At least 35 milliwatts.) (8032)

E = IR = 0.05 x 14 = 0.7 volt P = IE = 0.05 x 0.7 = 0.035 watt 0.035 watt = 35 milliwatts

EXAMPLE - Figure 4. How much power is being furnished to the circuit? (2,645 watts.)(8023)

E = IR = 23 x 5 = 115 volts P = IE = 23 x 115 = 2,645 watts

EXAMPLE - How much current does a 30-volt motor, 1/2 horsepower, 85 percent efficient, draw from the bus? (14.6 amperes.) (8431)

(Note: 1 horsepower = 746 watts)

P

PE

= × =

= = =

12

74685%

438 8

438 830

14 6

.

. .

watts

I amperes

SUMMARY OF EQUATIONS OHM'S LAW

I ER

E I R

R EI

=

= ×

=

RESISTORS

( )( )parallel

R1

R1

R1

1R

series .ectRRRR

321

t

321t

++=

+++=

POWER

P I E

I PE

= ×

=



ALTERNATING CURRENT CIRCUITS ALTERNATING CURRENT - is defined as current which periodically changes direction and continuously changes in magnitude.

THERE ARE THREE VALUES - of alternating current which should be considered. They are:

Instantaneous, Maximum, and Effective.

THE INSTANTANEOUS VALUE - is the value of the induced voltage or current flowing at any instant.

THE MAXIMUM VALUE - is the largest instantaneous value.

THE EFFECTIVE VALUE - of alternating current is the same as the value of a direct current which can produce an equal heating effect.

IN AN AC CIRCUIT - the effective voltage is less than the maximum instantaneous voltage. (The effective is 0.707 times the maximum.) (8007)

UNLESS OTHERWISE SPECIFIED - any values given for current or voltage in an ac circuit are assumed to be effective values. (8010)

LINES OF FORCE MAGNETIC LINES OF FORCE - will pass most readily through iron (when compared with copper or aluminum). (8052)

INDUCTANCE WHEN AC CURRENT FLOWS IN A COIL OF WIRE - the rise and fall of the current flow sets up an expanding and collapsing magnetic field about the coil. A voltage is induced in the coil which is opposite in direction to the applied voltage. The property of a coil to oppose any change in the current flowing through it is called inductance.

TRANSFER OF ELECTRICAL ENERGY - from one conductor to another without the aid of electrical connections is called induction. (8041)

THE BASIS FOR TRANSFORMER OPERATION - in the use of alternating current is mutual inductance. (8003)

WHEN INDUCTORS ARE CONNECTED IN SERIES IN A CIRCUIT - the total inductance is (where the magnetic fields of each inductor do not affect the others) equal to the sum of the individual inductances. (8012) LT = L1 + L2 + L3 ...

WHEN MORE THAN TWO INDUCTORS - of different inductance’s are connected in parallel in a circuit, the total inductance is less than the inductance’s of the lowest rated inductor. (8013)

L

L L L

T =+ +

11 1 1

1 2 3...

INDUCTIVE REACTANCE INDUCTIVE REACTANCE - is the effect of inductance in an ac circuit and is measured in ohms, because, like resistance, it impedes the flow of current.

INDUCTIVE REACTANCE - is the opposition offered by a coil to the flow of alternating current (disregarding resistance). (8004)

AN INCREASE IN INDUCTANCE AND FREQUENCY - will cause an increase in the inductive reactance of a circuit. (8005)

CAPACITORS A CAPACITOR CONSISTS OF TWO CONDUCTORS - capable of holding an electric charge separated by an insulating medium.

THE AMOUNT OF ELECTRICITY A CAPACITOR CAN STORE - is directly proportional to the plate area and inversely proportional to the distance between the plates. (8008)

THE WORKING VOLTAGE OF A CAPACITOR - in an ac circuit should be at least 50 percent greater than the highest applied voltage. (8001)

WHILE CAPACITORS ARE NORMALLY FOUND IN AC CIRCUITS, - they can be used in dc circuits to smooth out any slight pulsations in current/voltage. (8547)

EXAMPLE - Figure 17. The electrical symbol at number 5 represents what kind of component? (A variable capacitor.) (8066)

WHEN DIFFERENT RATED CAPACITORS ARE CONNECTED IN SERIES IN A CIRCUIT - the total capacitance is less than the capacitance of the lowest rated capacitor. (8006)

C

C C C

T =+ +

11 1 1

1 2 3...



EXAMPLE - Figure 2. What is the total capacitance of a certain circuit containing three capacitors with capacitance’s of 0.02 microfarad, 0.05 microfarad, and 0.10 microfarad, respectively? (0.0125 µF.) (8009)

GIVEN:

321

T

C1

C1

C1

1C++

=

SOLUTION:

C

C

C

C F

T

T

T

T

=+ +

=+ +

=

=

11

0 021

0 051

0101

50 20 101

800 0125

. . .

. µ

WHEN DIFFERENT RATED CAPACITORS ARE CONNECTED IN PARALLEL IN A CIRCUIT - the total capacitance is equal to the sum of all the capacitance’s. (8011) CT = C1 + C2 + C3 ...

EXAMPLE - What is the total capacitance of a certain circuit containing three capacitors with capacitance’s of 0.25 microfarad, 0.03 microfarad, and 0.12 microfarad, respectively? (0.40 µF.) (8014)

GIVEN: CT = C1 + C2 + C3

SOLUTION: CT = 0.25 + 0.03 + 0.12 CT = 0.40 µF

CAPACITIVE REACTANCE CAPACITIVE REACTANCE - like inductive reactance, offers opposition to the flow of current and is also measured in ohms.

IMPEDANCE IMPEDANCE - which is the combination of resistance, inductive reactance, and capacitive reactance is the total opposition to current in an ac circuit.

THE UNIT FOR THE MEASURE OF IMPEDANCE - is the ohm.

THE TERM THAT DESCRIBES THE COMBINED RESISTIVE FORCES - in an ac circuit is impedance. (8002)

EXAMPLE - Figure 5. What is the impedance of an ac-series circuit consisting of an inductor with a reactance of 10 ohms, a capacitor with a reactance of 4 ohms, and a resistor with a resistance of 8 ohms? (Impedance is 10 ohms.) (8024)

GIVEN: ( )2CL

2 XXRZ −+= Z = Impedance R = Resistance XL = Inductive reactance XC = Capacitive reactance

SOLUTION:

( )Z

Z

Z

ZZ

= + −

= +

= +

==

8 10 4

64 6

64 36

10010

2 2

2

ohms

POWER

TRUE POWER - is the power consumed by the resistance of an ac circuit.

APPARENT POWER - is the power consumed by the entire ac circuit.

WHEN CALCULATING POWER - in a reactive or inductive ac circuit, the true power is less than the apparent power. (8022)



BATTERIES BATTERIES MAY BE CONNECTED - in parallel, in series, or in a combination of parallel and series.

FOR BATTERIES CONNECTED IN PARALLEL - (positive to positive to positive, etc. and negative to negative to negative, etc.) the total voltage is the same as any one battery.

FOR BATTERIES CONNECTED IN SERIES - (positive to negative to positive to negative, etc.) the voltages are added to obtain the total voltage.

EXAMPLE - Figure 10. What is the measured voltage of the series-parallel circuit between terminals A and B? (3.0 volts.) (8034)

There are two sets of batteries in series connected in parallel with each other. The voltage of batteries in series is added, giving a total of 3.0 volts for each pair of batteries. The voltage of batteries in parallel remains the same. The total voltage produced by the batteries, therefore, is 3.0 volts.

THE STATE-OF-CHARGE OF THE BATTERY - determines the amount of current which will flow through a battery while it is being charged by a constant voltage source. (8089)

LEAD-ACID BATTERIES A FULLY CHARGED LEAD-ACID BATTERY WILL NOT FREEZE - until extremely low temperatures are reached because most of the acid is in solution. (Sulfuric acid has a much lower freezing point than water.) (8088)

THE HYDROMETER READING - of a lead-acid storage battery electrolyte does not require a temperature correction if the electrolyte temperature is 80 oF. (A hydrometer correction for specific gravity is not required for electrolyte temperature between 70 oF and 90 oF.) (8087)

IF ELECTROLYTE FROM A LEAD-ACID BATTERY IS SPILLED IN THE BATTERY COMPARTMENT - neutralize the spilled battery acid by applying sodium bicarbonate solution to the affected area followed by a water rinse. (8086)

TO PREVENT SEDIMENT BUILDUP - a space is provided underneath the plates in a lead acid battery’s cell container so the sediment does not contact the plates and cause a short circuit. (8092)

NICKEL-CADMIUM BATTERIES THE ELECTROLYTE OF A NICKEL-CADMIUM BATTERY - is lowest when the battery is in a discharged condition.

THE ELECTROLYTE IS HIGHEST IN A NICAD BATTERY – when the battery is in a fully charged condition. (8096)

EXCESSIVE SPEWING IS LIKELY TO OCCUR DURING THE CHARGING CYCLE - if water is added to a nickel-cadmium battery when it is not fully charged. (8100)

NICKEL-CADMIUM BATTERIES WHICH ARE STORED FOR A LONG PERIOD OF TIME - will show a low liquid level because the electrolyte becomes absorbed in the plates. (8098)

THE STATE-OF-CHARGE - of a nickel-cadmium battery can be determined by a measured discharge. (That is, discharge the battery at a specified rate and measure its ampere-hour capacity, then recharge it.) (8099)

THE SERVICING AND CHARGING OF NICKEL-CADMIUM - and lead-acid batteries together in the same service area is likely to result in contamination of both types of batteries. (The electrolytes and their fumes are chemically opposite.) (8095)

THERE ARE TWO METHODS - of battery charging:

1. CONSTANT VOLTAGE - voltage constant, current varies.

USED IN THE AIR - generator puts out constant voltage.

ADVANTAGE - rapid charging rate.

DISADVANTAGE - high charging rate could cause over heating (NICAD).

2. CONSTANT CURRENT - current constant, voltage varies.

ADVANTAGE - controls charging rate.

DISADVANTAGE - lower rate of charge.

THE METHOD USED - to rapidly charge a nickel-cadmium battery utilizes constant voltage and varying current. (8091)



MULTIPLE BATTERIES - may be charged by either method:

1. CONSTANT VOLTAGE METHOD Batteries connected in parallel. May be different capacities. Must be same voltage.

2. CONSTANT CURRENT METHOD.

Batteries connected in series can be of different voltages.

Must be about the same capacity.

IF BATTERIES ARE SAME VOLTAGE AND CAPACITY- either method may be used.

EXAMPLE - Can several aircraft batteries with different voltages (but similar capacities) be connected in series with each other across the charger and charged using the constant current method? (Yes.) (8090)

EXAMPLE - Can several aircraft batteries of different ampere-hour capacity and same voltage be connected in parallel with each other across the charger and charged using the constant voltage method? (Yes.) (8090)

EXAMPLE - When charging several aircraft batteries of the same voltage and same ampere-hour capacity, must they be connected in series with each other across the charger and charged using the constant current method? (No.) (8090)

WHEN A CHARGING CURRENT IS APPLIED TO A NICKEL-CADMIUM BATTERY - the cells emit gas only toward the end of the charging cycle. (8102)

THE END-OF-CHARGE VOLTAGE OF A 19-CELL NICKEL-CADMIUM BATTERY - measured while still on charge, depends upon its temperature and the method used for charging. (About 1.45 volts to 1.58 volts per cell.) (8097)

IN NICKEL-CADMIUM BATTERIES - a rise in cell temperature causes a decrease in internal resistance. (8101)

THERMAL RUNAWAY - is the result of the high temperatures associated with nickel-cadmium batteries. (8550)

HEAT OR BURN MARKS ON THE HARDWARE - is an indication of improperly torqued cell link connections of a nickel-cadmium battery. (8093)

THE PRESENCE OF SMALL AMOUNTS OF POTASSIUM CARBONATE DEPOSITS - on the top of nickel-cadmium battery cells that have been in service for a time is an indication of normal operation. (8094)

NICKEL-CADMIUM BATTERY CASES AND DRAIN SURFACES - which have been affected by electrolyte should be neutralized with a solution of boric acid. (8351)

TROUBLESHOOTING AND DIAGRAMS ELECTRICAL CIRCUITS

THE CORRECT WAY TO USE AN OHMMETER - is to connect it in parallel with the unit to be evaluated with at least one end of that unit disconnected.

THE PROPER METHOD OF CONNECTING A TEST AMMETER - is to connect it in series with the load.

THE CORRECT WAY TO CONNECT A TEST VOLTMETER - in a circuit is in parallel with a unit. (8029)

EXAMPLE - Figure 9. How many instruments (voltmeters and ammeters) are installed correctly? (Two.) (8028)

The ammeter in series with the light and battery and the voltmeter in parallel with the light are properly installed. The ammeter in parallel with

the battery provides an unintended second path for the current. The other voltmeter is installed with reversed polarity.

EXAMPLE - Figure 20. Will troubleshooting an open circuit with a voltmeter as shown in this circuit permit the battery voltage to appear on the voltmeter? (Yes, since no current is flowing in the circuit, there is no voltage drop across the lamp). (8073)

IF AN AIRCRAFT AMMETER SHOWS A FULL CHARGING RATE - but the battery remains in a discharged state, the most likely cause is an internally shorted battery.

LANDING GEAR CIRCUITS IN AN ELECTRICAL CIRCUIT DIAGRAM - the ground reference point is considered to be at zero voltage. (8072)



A GROUND SYMBOL – in electrical circuit diagrams shows that there is a return path between the source of electrical energy and the load. (8604)


Must the control valve switch be placed in the neutral position when the landing gears are down to prevent the warning horn from sounding when the throttles are closed? (Yes.) (8068)

As shown, current for the warning horn through wires #14, #3, and #11 is interrupted at the control valve switch when the control valve switch is in neutral. If the control valve switch were not in the neutral position, ground for the horn would be supplied through wire #14, the left gear switch, wire #3, the right gear switch, wire #11, the control valve switch, wires #10 and #5, the closed throttle switch, and wire #6.

When the landing gears are up and the throttles are retarded, the warning horn will not sound if an open occurs in which wire: No. 2, No. 4, or No. 9? (Wire No. 4.) (8067)

The ground circuit for the warning horn is through wire #14, the left gear switch in the Up position, wire #4, the throttle switch in the Closed position, and wire #6.

Wire #2 is in the red warning light circuit.

Wire #9 is in the landing gear green light circuit.


Under which condition will a ground be provided for the warning horn through both gear switches when the throttles are closed? (Left gear up and right gear down.) (8069)

The ground circuit for the warning horn would then be through the right gear switch in the Down position, wire #5, the left gear switch in the Up position, wire #12, the closed throttle switches, and wire #11.

When the throttles are retarded with only the right gear down, the warning horn will not sound if an open occurs in which wire: No. 5, No. 6, or No. 13? (Wire No. 5.) (8070)

The ground circuit for the warning horn would be through the right gear switch in the Down position, wire #5, the left gear switch in the Up position, wire #12, the throttle switches in the Closed position, and wire #11.

Wire #6 is out of the warning horn circuit when the right gear switch is in the Down position.

Wire #13 is out of the warning horn circuit when the left gear switch is not in the Down position (and the control valve switch is in other than the Neutral position).

When the landing gears are up and the throttles are retarded, the warning horn will not sound if an open occurs in which wire: No. 5, No. 6, or No. 7? (Wire No. 6.) (8071)

The ground circuit for the landing gear warning horn is completed by the right gear switch in the Up position, wire #6, wire #12, the throttle switches in the Closed position, and wire #11.

Wire #5 is out of the warning horn circuit when the right gear switch is in the Up position.

Wire #7 is out of the warning horn circuit when the left gear switch is in the Up position.


The No. 7 wire is used to complete what circuit? (The PUSH-TO-TEST circuit.) (8058)

Wire No. 7 supplies power to both the red and green indicator lights from the bus through the 5-amp circuit breaker, and wires #18 and #17.

When either of the push-to-test lamps is pushed in, the circuit to ground is completed and the bulb will remain lighted as long as it is pressed in. With the landing gear retracted, the red indicator light will not come on if an open occurs in which wire: No. 7, No. 17, or No. 19? (Wire No. 19.) (8057) Power to the red indicator light is from the bus through the 5-amp circuit breaker, the up limit switch in the Closed position (shown Open in Figure 15), and wire #8. Wire #7 supplies power to the red and green indicator lights push-to-test circuit. Wire #17 is the power to the green indicator light push-to-test circuit. When the landing gear is down, the green light will not come on if an open occurs in which wire: No. 6, No. 7, or No. 17? (Wire No. 6.)(8059) Power for the green indicator light comes from the bus through the 5-amp circuit breaker, wire #6, the nose gear down switch, wire #5, the left gear down switch, wire #4, the right gear down switch, and wire #3. Wires No. 7 and No. 17 supply power to the green indicator light push-to-test function.



RELAY ACTIVATION TABLE

RELAYS POWER ON POWER ON POWER ON RIGHT TANK LEFT TANK NORMAL (8062) (8064) (8063) LTS NO YES NO RTS YES NO NO FCF NO NO NO PCO YES YES NO PCC NO NO YES TCO NO NO NO TCC YES YES YES FUEL PRESS SOLENOID YES YES NO FUEL TANK SOLENOID NO NO NO

FUEL SYSTEM CIRCUITS EXAMPLES - Figure 16. When electrical power is applied to the bus, which relays are energized? (Relays PCC and TCC are energized.) (8063) To energize relay PCC, current flows from the bus through the left-hand 5-amp circuit breaker, contacts 5,7,9, and 11, and then through the coil of relay PCC. At the same time to energize relay TCC, current flows from the bus through the right-hand 5-amp circuit breaker, contacts 18 and 20, and then through the coil of relay TCC. Since contact 12 will remain open, relay PCO will not be energized. Similarly, since contacts 19 will remain open, relay TCO will not be energized. EXAMPLE - With power to the bus and the fuel selector switched to the right-hand tank, how many relays in the system are operating? (Three relays, RTS, PCO, and TCC, will have operated.) (8062) With the fuel selector in the right-hand tank position, current will flow from the bus through the left-hand 5-amp circuit breaker, the fuel selector switch, and the coils of relay RTS. Actuation of relay RTS opened contacts 7 and closed contacts 8. Current will flow through contacts 5 and 8 and cause contacts 12 of the fuel pressure X-feed valve to close. Closed contacts 12 will allow current to actuate relay PCO.

Current from the bus will also flow through the right-hand 5-amp circuit breaker, contacts 18 and 20, and the coils of relay TCC to actuate that relay.

When the circuit is energized with the fuel tank selector switch selected to the left-hand position, what switches will change position? (Switches 5,6,11,12,13,15, and 16.) (8064) Current will flow from the bus through the left-hand 5-amp circuit breaker, the fuel selector switch in the left-hand tank position, and to the coils of relay LTS. Activation of relay LTS changes position of switches 5 and 6. With switch 5 now open, current is cut off through switches 7,9, and 11 resulting in de-energizing relay PCC. This changes the position of switch 15. (Switch 15 will snap open and then close again almost instantly). Current now flowing through closed switch 6 activates the fuel pressure X-feed valve and changes the position of switches 11 and 12. Closing of switch 12 energizes relay PCO which closes switch 13. At the same time, current is flowing from the bus through the right-hand 5-amp circuit breaker, switches 18 and 20, and to the coil of relay TCC. Activation of relay TCC changes the position of switch 16. What will be the effect if the PCO relay fails to operate when the left-hand tank is selected? (The fuel pressure crossfeed valve open light will not illuminate.) (8060) If relay PCO fails and does not operate, contacts 13 will not close and no power will get to and light the "Fuel Pressure Cross Feed Valve Open" warning light in the cockpit. Current is supplied to this warning light from the bus through the center 5-amp circuit breaker, and contacts 13 and 15.



Even if relay PCO fails to operate, the rest of the system will operate normally. The TCO relay will operate if 24-volts dc is applied to the bus and the fuel tank selector is in which position? (The crossfeed position.) (8061) With the fuel selector switch in the X-FEED position, current from the bus will flow through the left-hand 5-amp circuit breaker, the fuel selector switch, and to the coils of the FCF relay. When the FCF relay is energized, contacts 17 will close which will in turn drive contacts 19 closed. With contacts 19 closed, current can flow to relay TCO coils and operate it.

THERMAL SWITCH A THERMAL SWITCH - as used in an electric motor, is designed to open the circuit in order to allow cooling of the motor. (8056)

LOGIC GATES LOGIC SYSTEMS - involve the use of binary mathematics. The binary system of mathematics uses only two digits, 1 and 0. If a circuit is conducting, the signal is 1, if the circuit is not conducting, the signal is 0. A switch, transistor, or other unit can be used, therefore, as a "gate" to provide the desired signal for the required result. Examples of the symbolic depiction of gates are shown in Figures 24, 25, and 26.

EXAMPLE - Figure 25. The depicted logic gate output will be zero only with what inputs? (The symbol depicts an "AND" gate. All inputs must be a 1 to produce a 1 output. If one or more inputs are zero, the output will be zero.) (8083)

EXAMPLE - Figure 24. What is the relation between the inputs and output of the depicted logic gate? (The symbol depicts an "OR" gate. Any input being 1 will produce a 1 output.) (8082)

EXAMPLE - Figure 26. Which of the logic gate output conditions is correct with respect to the given inputs? (Gate 2 is correct. The symbol depicts an "EXCLUSIVE OR" gate which is designed to produce a 1 output whenever its input signals are dissimilar. Only gate 2 shows that arrangement.) (8084)

TRANSISTORS THE JUNCTION TRANSISTOR - is a three layered device in which the outer layers are one type of semiconductor, either P or N, and the center layer is the opposite type, N or P, respectively. Both junctions must be biased correctly to allow the transistor to conduct.

FORWARD BIASING - of a solid state device (such as a transistor) will cause the device to conduct. (8079)

IN AN N-P-N TRANSISTOR APPLICATION - the solid state device is turned on when the base is positive with respect to the emitter. (8076)

IN A P-N-P TRANSISTOR APPLICATION - the solid state device is turned on when the base is negative with respect to the emitter. (8075)

EXAMPLE - Figure 22. Which application is correct concerning bias application and current flow? (Illustration 1 is correct. The base of this N-P-N transistor is positive with respect to the emitter, and the base-emitter current adds to the collector-emitter current.) (8078) EXAMPLES - Figure 23. If an open occurs at R1, how will the light be affected? (The light cannot be turned off.) (8080)

Above a certain minimum base voltage the transistor will conduct, lighting the bulb. If R1 is open, then no matter where variable resistor R2 is set, it will always be at maximum voltage causing the light to stay on all the time.

If R2 sticks in the up position, how will the light be affected? (The light will be on full bright.) (8081)

If the variable resistor R2 is in the zero resistance setting (full up), then the base voltage will be maximum. The resulting maximum current flowing through the transistor will cause the light to be on full bright.


INSPECTION FUNDAMENTALS - NOTES 42

INSPECTION FUNDAMENTALSMECHANIC CERTIFICATION

FAR 65, Subpart D THE REQUIREMENTS FOR ISSUING MECHANIC CERTIFICATES - and associated ratings and the general operating rules for the holders of these certificates and ratings can be found in FAR Part 65, Subpart D. (8530)

65.18 ANY CHEATING - or other unauthorized conduct while taking any FAA test carries a maximum penalty of ineligibility for any certificate or rating for one year and a suspension or revocation of any certificate held. (8584)

65.71 A QUALIFIED MECHANIC - that does not read, write, speak, or understand English is eligible to apply for a mechanic certificate when a U.S. air carrier outside the United States employs him/her. (8590)

EXAMPLE – Why should an aircraft maintenance technician be familiar with weld nomenclature? (To compare welds with written (non-pictorial) description standards.) (8606)

65.13, 65.14 A TEMPORARY CERTIFICATE - is valid for 120 days. (8589)

AFTER SUCCESSFULLY COMPLETING - the required tests, a mechanic applicant is issued a temporary certificate. This period allows time for review of the application and supplementary documents. (8588)

65.15 ANY MECHANIC CERTIFICATE - (except Repairman) is valid until it is surrendered, suspended, or revoked. (8587)

43.12 THE MAXIMUM PENALTY - for falsification, alteration, or fraudulent reproduction of certificates, logbooks, reports, and records is suspension or revocation of any certificate held. (8585)

65.83 A CERTIFICATED MECHANIC SHALL NOT EXERCISE THE PRIVILEGES OF THE CERTIFICATE AND RATING - unless, within the preceding 24 months, the Administrator has found

that the certificate holder is able to do the work or the certificate holder has served as a mechanic under the certificate and rating for at least 6 months. (8531)

65.21 THE HOLDER OF A CERTIFICATE UNDER PART 65 - who has made a change in his permanent mailing address must notify the FAA Airmen Certification Branch in writing of the new address within 30 days. (8586)

1.1 THE DEFINITION OF MAINTENANCE - is overhaul, repair, parts replacement, inspection, and preservation. (8583)

PREVENTIVE MAINTENANCE – concerns simple preservation operations. It also means replacing small standard parts. It does not involve complex operations. (8602)

43.13 WHEN PERFORMING ANY KIND OF MAINTENANCE – on an aircraft, the work done should at least equal to its original or properly altered condition. (8603)

100-HOUR INSPECTIONS 65.87 A CERTIFICATED MECHANIC WITH A POWERPLANT RATING - may perform or supervise the 100-hour inspection required by the Federal Aviation Regulations on a powerplant or propeller or any component thereof, and may release the same to service. (8519,8529,8536)

65.85, 65.87 A CERTIFICATED AIRFRAME AND POWER-PLANT MECHANIC - is authorized to approve for return to service an aircraft, including certain aircraft being operated for hire (like pipeline patrol or banner towing), after a 100-hour inspection. (8464,8523)

43.3(d), 65.81 A PERSON WORKING UNDER THE SUPERVISION - of a certificated mechanic with an airframe and powerplant rating is not authorized to perform a 100-hour inspection. (The mechanic must actually perform the inspection.) (8524)



43.11 AFTER A MECHANIC HOLDING AN AIRFRAME AND POWERPLANT RATING COMPLETES A 100-HOUR INSPECTION - and before the aircraft is returned to service, the mechanic is required to make the proper entries in the aircraft's maintenance record. (8463) 43.11 THE PERSON APPROVING OR DISAPPROVING AN AIRCRAFT FOR RETURN TO SERVICE - is responsible for making the entry in the maintenance records after an annual, 100-hour, or progressive inspection. (8454) 65.81 CERTIFICATED MECHANICS - under their general certificate privileges, may perform 100-hour inspection of instruments (but not repair). (8525)

ANNUAL INSPECTIONS 43.15(c) EACH PERSON PERFORMING AN ANNUAL OR 100-HOUR INSPECTION - shall use a checklist that contains at least those items in FAR Part 43, Appendix D. (8461) 43.11(b) DURING AN ANNUAL INSPECTION - if a defect is found which makes the aircraft unairworthy, the person disapproving must provide a written notice of the defect to the owner. (8445) AC 43-9C, Part 43 DISCREPANCY LISTS - consist of items found during an inspection that could render the aircraft unairworthy. When the list is given to the aircraft owner/operator after an inspection it says in effect that except for the these discrepancies, the item inspected is airworthy. (8571) 43.3(b) IF AN AIRCRAFT OWNER WAS PROVIDED A LIST OF DISCREPANCIES - on an aircraft that was not approved for return to service after an annual inspection, any appropriately rated mechanic may correct the discrepancies. (8455) 43.11(a), 91.409(c) IF AN AIRCRAFT WAS NOT APPROVED FOR RETURN TO SERVICE - after an annual inspection and the owner wanted to fly the aircraft to another maintenance base, the owner must obtain a special flight permit. (8460)

MINOR REPAIRS 43.13 A REPAIR, AS PERFORMED ON AN AIRFRAME - means the restoration of the airframe to a

condition for safe operation after damage or deterioration. (8520) 43.13(b) THE INSTALLING PERSON OR AGENCY - is responsible for determining that materials used in aircraft maintenance and repair are of the proper type and conform to the appropriate standards. (8533) A MINOR REPAIR - generally is one that can be accomplished without any welding, riveting, or gluing.

FAR 43, Appendix A(b)(1) THE REPLACEMENT OF A DAMAGED ENGINE MOUNT - with a new identical engine mount purchased from the aircraft manufacturer is considered a minor repair. (8535)

THE REPLACEMENT OF A DAMAGED VERTICAL STABILIZER - with a new identical stabilizer purchased from the aircraft manufacturer is considered a minor repair. (8527) 65.81 FAA CERTIFICATED MECHANICS - may approve for return to service a minor alteration they have performed appropriate to the rating(s) they hold. (Major repairs and alterations require an IA's approval.) (8528) 43.9 WHEN A MINOR REPAIR IS PERFORMED - on a certificated aircraft, an entry in the aircraft's permanent records is required. (8457)

MAJOR REPAIRS A MAJOR REPAIR - is any repair that, if done improperly, might affect the structural strength or flight characteristics of the aircraft.

65.81 CERTIFICATED MECHANICS WITH A POWER-PLANT RATING - may not perform a major repair to a propeller even if they have the necessary equipment available. (8532,8591) FAR 43, Appendix A(b) THE REPLACEMENT OF FABRIC ON FABRIC-COVERED PARTS - such as wings, fuselages, stabilizers, or control surfaces is considered to be a major repair even though no other alteration or repair is performed. (8521) THE REPAIR OF PORTIONS OF SKIN SHEETS - by making additional seams (the splicing of skin sheets) is classified as a major repair. (8449,8522) OVERHAUL OF A HYDRAULIC PRESSURE PUMP - is considered an appliance major repair. (8447)



MAJOR ALTERATIONS A MAJOR ALTERATION - is a change in the basic design, not listed in the manufacturers' specifications, which could affect the structural strength or flight characteristics of the aircraft.

EXAMPLES - Figures 62, 62A, and 62B.

To which doubler part number(s) is the -100 in the title block (Area 1) applicable? (-101.) (8514)

Which doubler(s) require(s) heat treatment before installation? (-102.) (8512)

Using only the information given (when bend allowance, set back, etc. have been calculated) which doubler is it possible to construct and install? (The -101 doubler. The Process Specifications to construct the -102 doubler are not detailed in Area 3.) (8513)

How many parts will need to be fabricated by the mechanic in the construction and installation of one doubler? (For either a -100 or a -200 kit, you need to fabricate two clips and one doubler for a total of three parts.) (8576)

FAA FORM 337 FAR 43, Appendix B MAJOR REPAIRS AND MAJOR ALTERATIONS - are entered on an FAA Form 337 (as well as in the appropriate logbook). (8462)

AFTER MAKING A MAJOR REPAIR TO AN AIRCRAFT ENGINE - that is to be returned to service, two copies of FAA FORM 337 must be prepared: one copy for the aircraft owner and one copy for the FAA. (8458)

THE PERSON PERFORMING OR SUPERVISING - the work must prepare FORM 337. (8573)

WHEN A CERTIFICATED MECHANIC - (who may or may not be an IA) signs the form, he/she is certifying that the work was done under the requirements of 14 CFR Part 43. (8570)

WHEN FILLING OUT FORM 337 - you may need extra sheets. Extra sheets must show the aircraft nationality, registration mark and the date the work was completed. (8567)

FORM 337 FOR MAJOR REPAIRS AND ALTERATIONS - is only authorized for use on U.S. registered aircraft. (8575)

DATA THAT IS USED AS A BASIS FOR APPROVING MAJOR REPAIRS OR ALTERATIONS - for return to service must be FAA-approved prior to its use for that purpose. (8574)

DETERMINING INSPECTION INTERVALS

EXAMPLE - Given the statement below, at what intervals should the thrust bearing nut be checked for tightness? (Every 150 hours.) (8518)

"Check thrust bearing nuts for tightness on new or newly overhauled engines at the first 50-hour inspection following installation. Subsequent inspections on thrust bearing nuts will be made at each third 50-hour inspection."

From the above statement, thrust bearing nuts should be checked for tightness at 150 hour intervals.

EXAMPLE - Given the statement below, at what intervals will valve mechanism inspections be performed? (Every 100 hours.) (8517)

"A complete detailed inspection and adjustment of the valve mechanism will be made at the first 25 hours after the engine has been placed in service. Subsequent inspections of the valve mechanism will be made at each second 50-hour period."

From the above statement, valve mechanism inspections should be performed at 100 hour intervals.

MAINTENANCE ENTRIES ENTRIES IN AN AIRCRAFT’S MAINTENANCE RECORDS - must include certain required information. The regulations specify what information but do not specify any particular format whether from the manufacturer or the FAA. As long as there is continuity of the entries and required data, any format is acceptable. (8563)

91.417 MAINTENANCE, PREVENTIVE MAINTENANCE AND INSPECTIONS - must be recorded for each aircraft (which includes airframe) and for each engine, propeller, rotor and appliance of the aircraft. A 100-hour inspection must be recorded in each of the records. (8564)



43.11 AIRCRAFT OPERATING UNDER PART 91 - must have aircraft total time recorded in the maintenance record after completing any required inspection. (8565,8566)

THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES THE ACTION TAKEN - for a control cable showing approximately 20 percent wear on several of the individual outer wires at a fairlead is:

"Wear within acceptable limits, repair not necessary." (8451)

THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES THE ACTION TAKEN - for a 0.125-inch deep dent in a straight section of 1/2-inch aluminum alloy tubing is:

"Dented section removed and replaced with identical new tubing flared to 37%.” (8452)

THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES - a repair of a dent in a tubular steel structure dented at a cluster is:

"Welded a reinforcing plate over the dented area." (8453)

THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES THE REPLACEMENT - of several damaged heli-coils in a casting is:

"Eight 1/4 - 20 inch standard heli-coils were replaced. The damaged inserts were extracted, the tapped holes gauged, then new inserts installed, and tangs removed." (8450)

THE FOLLOWING TYPE ENTRY WOULD BE FOUND - on FAA Form 337, Major Repair and Alteration:

"Removed right wing from aircraft and removed skin from outer 6 feet. Repaired buckled spar 49 inches from tip in accordance with figure 8 in manufacturer's structural repair manual No. 28-1."

(This is a major repair.) (8448)

43.9(a) WHEN APPROVING FOR RETURN TO SERVICE - after maintenance or alteration, the approving person must enter in the maintenance record of the aircraft:

A description (or reference to acceptable data) of work performed, Date of completion, (not date begun), The name of the person performing the work (if someone else), Signature, Certificate number, and Kind of certificate held. (8456)

IF WORK PERFORMED ON AN AIRCRAFT - has been done satisfactorily, the signature of an authorized person on the maintenance records for maintenance or alterations performed constitutes approval for return to service only for the work performed (not for the entire aircraft). (8444)

91.417(a) THE AIRCRAFT OWNER IS RESPONSIBLE - for maintaining the required maintenance records for an airplane. (8459)

91.417, 91.419 WHEN OPERATING UNDER PART 91, - records of maintenance, preventive maintenance, alterations and 100-hour, annual and progressive inspections must be retained at least one year or until the work is repeated or superceded. These same records that must be retained and then transferred when the aircraft is sold. (8568,8569) IF AIRCRAFT MAINTENANCE RECORDS ARE LOST OR DESTROYED, - the records must be reconstructed. In order to do this total time-in-service of the airframe must be established. (8572)

INSTRUMENTS 23.1543 INFORMATION REGARDING INSTRUMENT RANGE MARKINGS - for an airplane certificated in the normal category would be provided in FAR Part 23. (8505)

23.1545 EXAMPLE - Given the following table of airspeed limits in an FAA-issued aircraft specification, what would be the high end of the white arc on the airspeed instrument? (139 knots.) (8516)

Normal operating speed 260 knots Never-exceed speed 293 knots Max. gear operation speed 174 knots Maximum flap extended speed 139 knots

The high end of the white arc is the maximum flaps extended speed: 139 knots.

65.81 A CERTIFICATED MECHANIC WITH AN AIRFRAME RATING - may not perform a minor repair to an airspeed indicator, even if they have the necessary equipment available. (They may not perform any repairs on any instruments.) (8532)

AC 43.13-1A INSTRUMENT REPAIRS MAY BE PERFORMED BY - an FAA-approved instrument repair station. (8537)



ATA MANUAL STANDARDS THE AIR TRANSPORT ASSOCIATION OF AMERICA (ATA) SPECIFICATION NO. 100 - establishes a standard for the presentation of technical data in maintenance manuals, and divides the aircraft into numbered systems and subsystems in order to simplify locating maintenance instructions. (8510)

AVIATION MAINTENANCE ALERTS AVIATION MAINTENANCE ALERTS - provide information about aircraft problems and suggested corrective actions. (8511)

AIRWORTHINESS DIRECTIVES FAR 39 FAA AIRWORTHINESS DIRECTIVES - are issued primarily to correct an unsafe condition. This is the means by which the FAA notifies aircraft owners and other interested persons of unsafe conditions and prescribes the condition under which the product may continue to be operated. (8446,8492,8494)

THE STATEMENT ON AN AD - that tells you how quickly the AD must be accomplished is the Compliance statement. (8577)

THE APPLICABILTY STATEMENT ON AN AD – tells you what the AD refers to – aircraft, aircraft engine, propeller or appliance. (8601)

An Airworthiness Directive, or AD, requires a specific action in order to comply with it. The action may take the form of an:

1. Inspection. 2. Part(s) replacement. 3. Design modification. 4. Change in operating procedure(s). (8578)

43.15 EACH PERSON WHO PERFORMS AN INSPECTION - required under Part 91, 123, 125, or 135 must determine whether the aircraft meets all applicable airworthiness requirements. This includes compliance with ADs. (8580)

91.417, 91.419 AD COMPLIANCE RECORDS - are required to be retained and, when the aircraft is sold, transferred with the aircraft. (8581)

EXAMPLE - Figure 63. An aircraft has a total time in service of 468 hours. The Airworthiness Directive given below was initially complied with at 454 hours in service. How many additional hours in service may be

accumulated before the AD must again be complied with? (186 hours.) (8515)

"Compliance required as indicated, unless already accomplished: I. Aircraft with less than 500-hours' total time in service: Inspect in accordance with instructions below at 500-hours' total time, or within the next 50-hours' time in service after the effective date of this AD, and repeat after each subsequent 200 hours in service. II. Aircraft with 500-hours' through 1,000- hours' total time in service: Inspect in accordance with instructions below within the next 50-hours' time in service after the effective date of this AD, and repeat after each subsequent 200 hours in service. III. Aircraft with more than 1,000-hours' time in service: Inspect in accordance with instructions below within the next 25-hours' time in service after the effective date of this AD, and repeat after each subsequent 200 hours in service."

The aircraft comes under Paragraph I. The AD was complied with at 454 hours. The next inspection is due after 200 hours time in service, or at 654 hours. The aircraft currently has 468 hours. 654 - 468 = 186 hours left before the next inspection is due. (The paragraphs have to do with when the first inspection is due. In all cases, subsequent inspections must be done at 200 hours after the previous inspection.)

PROPELLERS ARE INCLUDED - in the Airworthiness Directive system. (8506)

65.81, 65.87 IF AN AIRWORTHINESS DIRECTIVE REQUIRES - that a propeller be altered, a certificated powerplant mechanic could perform and approve the work for return to service if it is a minor alteration or a minor repair on an aluminum propeller. (8506,8526)

43.9 THE PERSON WHO COMPLIES WITH AN AIRWORTHINESS DIRECTIVE - or manufacturers' service bulletin must make an entry in the maintenance record of that equipment. (8443,8508)

AIRWORTHINESS CERTIFICATES FAR 21, Subpart H THE ISSUANCE OF AIRWORTHINESS CERTIFICATES - is governed by FAR Part 21, Subpart H. (Parts 23 and 25 cover Airworthiness Standards; Part 39 covers Airworthiness Directives.) (8498)



21.179 IF AN AIRWORTHY AIRCRAFT IS SOLD - the Airworthiness Certificate is transferred with the aircraft. (8497)

TYPE CERTIFICATE DATA SHEETS THE TYPE CERTIFICATE DATA SHEET - will reference the required equipment needed to maintain the validity of a standard Airworthiness Certificate. (8159)

THERE ARE MINIMUM STANDARDS SET BY THE FAA - regarding design, materials, workmanship, construction, and performance of aircraft, aircraft engines and propellers. When the standards are met or exceeded, a Type Certificate Data Sheet is issued. (8579)

THE SUITABILITY FOR USE OF A SPECIFIC PROPELLER - with a particular engine-airplane combination can be determined by reference to the Aircraft Specifications or Type Certificate Data Sheets. (8496)

PLACARDS REQUIRED ON AN AIRCRAFT - are specified in the Aircraft Specifications or Type Certificate Data Sheets. (8502)

THE LOCATION OF THE DATUM - is an item that would be contained in an aircraft Type Certificate Data Sheet. (8495)

CONTROL SURFACE MOVEMENTS - is information that is generally contained in Aircraft Specifications or Type Certificate Data Sheets. (8501)

MANY AIRCRAFT AND ENGINE SPECIFICATIONS - and some type certificate data sheets, carry coded information to describe the general characteristics of the item. 2 P C S M means a two place (number of seats), closed cockpit, sea, monoplane. (8582)

THESE ARE SOMETIMES USED AS AUTHORIZATION - to deviate from an aircraft's original type design:

1. FAA Form 337. 2. Supplemental Type Certificate. 3. Airworthiness Directive. (8555)

SUPPLEMENTAL TYPE CERTIFICATES FAR 21, Subpart E A SUPPLEMENTAL TYPE CERTIFICATE MAY BE ISSUED TO - more than one applicant for the same design change, providing each applicant shows compliance with the applicable airworthiness requirement. (8493)

THE FEDERAL AVIATION REGULATIONS REQUIRE APPROVAL - after compliance with the data of a Supplemental Type Certificate. (This is a major alteration.) (8504)

TECHNICAL STANDARD ORDERS (TSO's)

FAR 21, Subpart O ITEMS MANUFACTURED IN ACCORDANCE WITH A TECHNICAL STANDARD ORDER - still require approval for installation in a particular aircraft. (8493,8504)

FAA PUBLICATIONS SUCH AS - Technical Standard Orders, Airworthiness Certificates, Type Certificate Data Sheets, and Aircraft Specifications and Supplemental Type Certificates are all approved data. Not all manufacturer's data is approved. Advisory Circular 43.13-2A, for example, is not "approved" data either, because it is neither specific nor mandatory. (8507,8509)

PROPELLER TYPE CERTIFICATES TECHNICAL DESCRIPTIONS OF CERTIFICATED PROPELLERS - can be found in the Propeller Type Certificate Data Sheets. (8500)

AIRCRAFT LISTING TECHNICAL INFORMATION ABOUT OLDER AIRCRAFT MODELS - of which no more than 50 remain in service, or of which a limited number were manufactured under a type certificate and for which there is no current Aircraft Specification, can be found in the Aircraft Listing. (8499,8503)

MAGNETIC PARTICLE INSPECTION MAGNETIC PARTICLE INSPECTION - is a method of detecting invisible cracks and other defects in magnetic materials, such as iron and steel. It is not applicable to nonmagnetic materials.

IRON ALLOYS - can be inspected using the magnetic particle inspection procedure (but not aluminum, magnesium, copper or zinc alloys). (8229)

THE INSPECTION PROCESS - consists of magnetizing the part and then applying ferromagnetic particles, in either liquid or powder form, to the surface area to be inspected.

WET AND DRY PROCESS MATERIALS - are two types of indicating mediums available for magnetic particle inspection. (8228)



THE TESTING MEDIUM - that is generally used in a magnetic particle inspection utilizes a ferromagnetic material that has high permeability and low retentivity. (8225)

IN USING THIS METHOD OF INSPECTION - the location of the defect is indicated and the approximate size and shape are outlined.

MAGNETIC PARTICLE INSPECTION - is used primarily to detect flaws on or near the surface. (8219)

CONTINUOUS MAGNETIC PARTICLE INSPECTION IS USED MOST OFTEN - to inspect aircraft parts for invisible cracks and other defects. The continuous method is likely to reveal more than the residual method. (8223)

RESIDUAL MAGNETIZING INSPECTION - may be used with steels which have been heat treated for stressed applications. (It is less sensitive in detecting subsurface flaws.) (8226)

WHEN CHECKING AN ITEM WITH THE MAGNETIC PARTICLE INSPECTION METHOD - circular and longitudinal magnetization should be used to reveal all possible defects. (8234)

A FLAW THAT IS PERPENDICULAR - to the magnetic field flux lines generally causes a large disruption of the magnetic field. (8235)

CIRCULAR MAGNETIZATION - of a part can be used to detect defects parallel to the long axis of the part. (8243)

CONTINUOUS LONGITUDINAL MAGNETIZATION WITH A CABLE - will detect defects perpendicular to the long axis of the part. (8242)

A 45° CRACK CAN BE DETECTED - by magnetic particle inspection using either circular or longitudinal magnetization. (8231)

ONE WAY A PART MAY BE DEMAGNETIZED - after a magnetic particle inspection is by slowly moving the part out of an ac magnetic field of sufficient strength or by gradually reducing the strength of the field. (8230,8237) AN AIRCRAFT PART MAY BE DEMAGNETIZED - by subjecting it to a magnetizing force from direct current that is alternately reversed in direction and gradually reduced in strength. (8237)

DISCONTINUITIES IN NONDESTRUCTIVE TESTING A DISCONTINUITY - may be defined as an interruption in the normal physical structure or configuration of a part. (8244)

A DISCONTINUITY - may or may not affect the usefulness of a part. (8244)

FATIGUE CRACKS FATIGUE CRACKS - give sharp, clear patterns. They are found in highly stressed areas of parts that have been in service.

UNDER MAGNETIC PARTICLE INSPECTION - a part will be identified as having a fatigue crack when the discontinuity is found in a highly stressed area of the part. (8240)

INCLUSIONS INCLUSIONS ARE - nonmetallic materials, such as slag materials and chemical compounds, that have been trapped in the solidifying metal.

THE PATTERN FOR AN INCLUSION - is a magnetic particle buildup forming parallel lines. (8238)

DYE PENETRANT INSPECTION DYE PENETRANT INSPECTION IS A NONDESTRUCTIVE TEST - for defects open to the surface (cracks, etc.) in parts made of any nonporous material.

DYE PENETRANT INSPECTION - is a nondestructive test and will only detect defects open to the surface of any nonporous material. This requirement is the primary limitation of the dye penetrant method of inspection.

LIQUID PENETRANT INSPECTIONS - may be used on: 1. Ferrous metals. 2. Nonferrous metals. 3. Nonporous plastics. Liquid penetrant inspection is not suitable for use on porous plastics or smooth primer-sealed wood. (8220)

PENETRANT INSPECTION CALLS FOR VISUAL INSPECTION - of the part after it has been processed, but the visibility of the defect is increased by the addition of a dye. The dye may be either visible or fluorescent.

THE STEPS TO BE TAKEN WHEN PERFORMING A PENETRANT INSPECTION ARE:

1. Thorough cleaning of the metal surface. 2. Applying penetrant. 3. Removing penetrant with remover-emulsifier or

cleaner. 4. Drying the part. 5. Applying the developer. 6. Inspecting and interpreting results.



A PART WHICH IS BEING PREPARED FOR DYE PENETRANT INSPECTION - should be cleaned with a volatile petroleum-base solvent. (8239)

IN PERFORMING A DYE PENETRANT INSPECTION - the developer acts as a blotter to produce a visible indication. (8241)

IF DYE PENETRANT INSPECTION INDICATIONS ARE NOT SHARP AND CLEAR - the most probable cause is that the part was not thoroughly washed before developer was applied. (8236)

TO DETECT A MINUTE CRACK - using dye penetrant inspection requires a longer-than-normal penetrating time. (8233)

WHEN THE PENETRANT IS APPLIED - it must be ‘cured’ and this waiting period is called dwell time. If the size and shape of cracks are small, the dwell time will be longer because it takes more time to penetrate small discontinuities. (8552)

ULTRASONIC INSPECTION ULTRASONIC INSPECTION IS A NONDESTRUCTIVE TESTING METHOD - which employs electronically produced, high-frequency sound waves that will penetrate metals, liquids, and many other materials.

ULTRASONIC INSPECTION IS SUITABLE FOR - the inspection of most metals, plastics, and ceramics for surface and subsurface defects. (8221)

EDDY CURRENT INSPECTION THE PRINCIPLE OF EDDY CURRENT INSPECTION - is based on determining the ease with which a material will accept induced current.

EDDY CURRENT INSPECTION - is a nondestructive testing method which requires little or no part preparation, is used to detect surface or near-surface defects in most metals, and may also be used to separate metals or alloys and their heat-treat conditions. (8222)

RADIOGRAPHY X- AND GAMMA RADIATION’S - because of their unique ability to penetrate material and disclose discontinuities, have been applied to the radiographic (X-ray) inspection of metal fabrications and nonmetallic products.

THE PENETRATING RADIATION - is projected through the part to be inspected. The processed film shows a shadow picture of the object.

THE THREE MAJOR STEPS IN THE X-RAY PROCESS ARE:

Exposure to radiation. Processing of film. Interpretation of the radiograph.

THE ESSENTIAL FACTORS OF RADIOGRAPHIC EXPOSURE - are interdependent. These factors include, but are not limited to:

(a) Material thickness and density. (b) Exposure distance and angle. (c) Film characteristics.

(Processing of the film is not a factor for the exposure.) (8224)

METALLIC RING TEST AFTER A MECHANIC HAS COMPLETED - a bonded honeycomb repair using the potted compound repair technique, a metallic ring test is the nondestructive testing method used to determine the soundness of the repair after the repair has cured. (8227)

SURFACE CRACKS SURFACE CRACKS IN ALUMINUM CASTINGS AND FORGINGS - may usually be detected by: 1. Dye penetrant inspection. 2. Eddy current inspection. 3. Ultrasonic inspection. 4. Visual inspection. (8232)


MEASURING DEVICES- NOTES 50

MEASURING DEVICESCOMBINATION SET

A COMBINATION SET - is the tool used to find the center of a shaft or other cylindrical work. (Use the center head.) (8295)

DIVIDERS DIVIDERS DO NOT - provide a reading when used as a measuring device. (8291)

A MACHINIST SCALE - is the tool generally used to set a divider to an exact dimension. (8299)

DIAL INDICATOR A DIAL INDICATOR - is the tool that can be used to measure the alignment of a rotor shaft or the plane of rotation of a disk. (8289)

THICKNESS GAUGE A THICKNESS GAUGE - (feeler gauge) is used to measure the side clearances of piston rings. (8302)

THE CLEARANCE BETWEEN THE PISTON RINGS AND THE RING LANDS - is measured with a thickness gauge. (8305)

A THICKNESS GAUGE - is the tool used to measure the clearance between a surface plate and a relatively narrow surface being checked for flatness. (8293)

THE TWIST OF A CONNECTING ROD IS CHECKED - by installing push-fit arbors in both ends, supported by parallel steel bars on a surface plate. Measurements are taken between the arbor and the parallel bar with a thickness gauge. (8304)

SMALL-HOLE GAUGE TO ACCURATELY MEASURE THE DIAMETER OF A HOLE APPROXIMATELY 1/4 INCH IN DIAMETER - the mechanic should use a small-hole gauge and determine the size of the hole by taking a micrometer reading of the ball end of the gauge. (8297)

TELESCOPIC GAUGE A TELESCOPIC GAUGE AND MICROMETER - can be used for the dimensional inspection of a bearing in a rocker arm. (8303)

MICROMETER CALIPER A MICROMETER CALIPER - is the precision measuring tool used for measuring crankpin and main bearing journals for out-of-round wear. (8301)

TO DETERMINE PISTON PIN OUT-OF-ROUND WEAR - use a micrometer caliper to check different diameter positions on the piston pin. (8307)

A MICROMETER MAY BE USED - to check the stem on a poppet-type valve for stretch. (8306)

A GAUGE BLOCK - is the tool generally used to calibrate a micrometer or check its accuracy. (8300)

READING A MICROMETER THE LARGE MARKS ON THE LONGITUDINAL LINE - on the barrel of a micrometer represent tenths of inches (0.10, 0.20, etc.). Each small mark between the numbers represents 0.025 inches.

THE RELATIVELY NARROW MARKS ON THE THIMBLE - of a micrometer represent one-thousandths of an inch (0.001, 0.002, etc.).

THE VERNIER SCALE GRADUATIONS OF A MICROMETER - are each equal to 0.0001 inch (one ten-thousandths of an inch). (8294)

EXAMPLE - Figure 49. What is the measurement reading on the micrometer? (0.275 + 0.004 +0.0002 = 0.2792) (8298)

EXAMPLE - Figure 48. What is the reading on the micrometer? (0.300 + 0.0004 = 0.3004) (8296)

EXAMPLE - Figure 46. What is the measurement reading on the micrometer? (0.275 + 0.010 + 0.0001 = 0.2851) (8290)


MEASURING DEVICES - NOTES 51

VERNIER CALIPER A VERNIER CALIPER - is used for making measurements faster than with a micrometer caliper, and for bigger measurements than a micrometer can practically do.

ON A VERNIER CALIPER - each large number represents inches (not fractions of inches).

EXAMPLE - Figure 47. What is the measurement reading on the vernier caliper scale? (1.000 + 0.400 + 0.025 + 0.011 = 1.436 inches.) (8292)


GROUND EQUIPMENT - NOTES 52

GROUND HANDLING, SAFETY, AND SUPPORT EQUIPMENT

PISTON ENGINE STARTING IF A RADIAL ENGINE HAS BEEN SHUT DOWN - for more than 30 minutes, the propeller should be rotated through several revolutions to check for hydraulic lock. (8315)

WHEN STARTING AND OPERATING AN AIRCRAFT ENGINE ON THE GROUND – it should be heading into the wind primarily to help keep the engine cool. (In flight, ram air serves the same purpose.) (8321)

WHEN STARTING AN ENGINE EQUIPPED WITH A FLOAT-TYPE CARBURETOR - with an idle cutoff unit, place the mixture control in the FULL-RICH position.

TO CLEAR A FLOODED ENGINE EQUIPPED WITH A FLOAT-TYPE CARBURETOR OF EXCESSIVE FUEL - crank the engine with the starter or by hand, with the mixture control in cutoff, ignition switch off, and the throttle fully open, until the fuel charge has been cleared. (8318)

PRIMING TO PRIME A HORIZONTALLY OPPOSED FUEL INJECTED ENGINE - place the fuel control lever in the FULL-RICH position. (8316)

INDUCTION FIRES GENERALLY, WHEN AN INDUCTION FIRE OCCURS DURING STARTING - of a reciprocating engine, the first course of action should be to continue cranking and start the engine if possible. (8320)

THE MOST SATISFACTORY EXTINGUISHING AGENT - for use in case of carburetor or intake fire is carbon dioxide. (8313)

TURBINE ENGINE STARTING A HOT START OCCURS IN A TURBOJET ENGINE - when the fuel/air mixture is excessively rich. (8323,8556)

THE MOST IMPORTANT CONDITION - to be monitored during start after fuel flow begins in a turbine engine is the EGT, TIT, or ITT. (8317)

A HUNG START OCCURS IN A TURBOJET ENGINE - when the RPM remains below normal starting RPM. If a hung start occurs, shut the engine down. (8308)

A HUNG START IN A TURBOJET ENGINE - is often caused by the starter cutting off too soon. (8309)

APPROACHING AIRCRAFT WHEN APPROACHING AN IDLING TURBOJET ENGINE - the hazard area extends forward of the engine approximately 25 feet and aft of the engine approximately 100 feet. (8312,8322)

A PERSON SHOULD APPROACH OR LEAVE A HELICOPTER - in the pilot's field of vision whenever the engine is running in order to avoid the tail rotor. (8330)

TAXIING AIRCRAFT WHEN FIRST STARTING TO MOVE AN AIRCRAFT WHILE TAXIING - it is important to test the brakes. (8334)

WEATHERVANING IS THE TENDENCY DISPLAYED BY AN AIRPLANE – that wants to turn it into the wind. The greatest danger is when the wind is a direct crosswind for either nosewheel or tailwheel airplanes.

WEATHERVANING - is greatest with a direct crosswind because a tailwheel airplane has a bigger surface area behind the main gear (which acts as a pivot point) than the nosewheel airplane. (8327)

WITH A QUARTERING TAILWIND DURING TAXI – keeping the aileron and elevators down on the side from which the wind is blowing (upwind side) will help keep the wind from picking up that wing. (8328)

LIGHT GUN SIGNALS WHEN TAXIING (OR TOWING) AN AIRCRAFT - at an airport with an operating control tower and not in radio contact with the tower, the following light signals from the tower would apply:

Flashing Green - OK to move. Steady Red - Stop. Flashing Red - Move clear of runway or taxiway immediately. (8329)


GROUND EQUIPMENT - NOTES 53

Flashing White - Return to starting point. (8331)

Alternating Red and Green - OK to proceed but use extreme caution. (8332)

HAND SIGNALS EXAMPLE - Figure 50. Which is the signal to engage rotor on a rotorcraft? ("3" is ENGAGE ROTOR; "1" is START ENGINE; "2" is STOP ROTOR.) (8314)

EXAMPLE - Figure 51. Which marshalling signals should be given if a taxiing aircraft were in danger of striking some object? ("3" is EMERGENCY STOP; "1" is STOP; "2" is COME AHEAD.) (8319)

TOWING AIRCRAFT DURING TOWING OPERATIONS - a person should be in the cockpit to operate brakes. (8326)

WHEN TOWING AN AIRCRAFT WITH A STEERABLE NOSEWHEEL - the torque-link lock should be set to full swivel. (8310)

STOPPING AND TIEDOWN WHEN STOPPING A NOSEWHEEL-TYPE AIRPLANE - after taxiing, the nosewheel should be left pointed straight ahead. (8333)

NYLON OR DACRON ROPE - is preferred to manila rope because manila (hemp) rope has a tendency to shrink when it gets wet. (8311)


APPENDIX 1


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CT= 1/C1 + 1/C2 + 1/C3...1

Figure 1. Equation.

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1

CT= 1/C1 + 1/C2 + 1/C3

1

Figure 2. Equation.

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2

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LT= 1/L1 + 1/L2 + 1/L3...1

Figure 3. Equation.

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3

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23A

5Ω

Figure 4. Circuit diagram.

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4

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Z = R2+(XL−XC)2

Z = ImpedanceR = ResistanceXL = Inductive reactanceXC = Capacitive reactance

Figure 5. Formula.

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5

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Disconnected

Ω

R1

= 1

2 oh

ms

R2

= 6

ohm

s

R3

= 6

ohm

sR

4 =

6 o

hms

R5

= 6

ohm

s


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6

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BreakΩ40Ω

A B C D

R1

40Ω

R2

40Ω

R3


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7

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Break

Ω20Ω

R1

20Ω

R2

20Ω

R3


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8

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+

−

A

+

−

V

+

−

V

+ −

A


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9

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1.5V

1.5V

1.5V

1.5V

+−

A B

Figure 10. Battery circuit.

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10

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R1

= 8

ohm

s

R2

= 1

0 oh

ms

R3

= 4

0 oh

ms

24V

A

B

C

D

E

F G H


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11

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R1 =18 ohms R2 =12 ohms

R3

= 4

ohm

s

R4

= 1

2 oh

ms

R5

= 6

ohm

s

24V

Ra =1/R + 1/R 4 5

1

Rc =1/Rb + 1/R3

1

Rb =Ra + R2

Rt = Rc + R1


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12

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Et 30Ω 60Ω 15Ω

I1

It

I2 I3

R1 R2 R3

+

−

12V


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R1 =5Ω

R5 =10Ω

R2 =4Ω R3 =6Ω R4 =12ΩEt = 36V


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14

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Motor

Gear switchUp

Down

Gear safety switch

Up limitswitch

BU

S

Down limitswitch

Horn

Relay

GreenRed

NAV. switch bypass relay

Throttle switches (closed position)

Left geardown switch

NOTE: Switches shown gear down - on the ground

Nose geardown switch

Right geardown switch

#1

#3

#4#5#6

#7#8

#2

#13

#14

#16

#1920

5

#11

#17

#15

#12

#18

#10

Figure 15. Landing gear circuit.

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B A C D BACD

Fuel pressurecross feedvalve open

Fuel tankX-feed valve

Fuel pressX-feed valve

RTS relay

Fuel selector switchNorm

Close Open CloseOpen

LTSrelay

FCF relay

FCF

Caution warninglights in cockpit

Fuel tankcross feedvalve open

X-feed

RelayPCO

Drawing shown without electrical power to bus

All relays springloaded to positionshown

Relay PCC

BUS 24 VDC

Relay TCC

PumpX-feedopen

PumpX-feedclosed

TankX-feedopen

TankX-feedclosed

RelayTCORH tank

LH tank

1

3

2

5

5 6

7 8

9 10

11 12

13

17 18

19 20

14

15 16

5 5

4

Figure 16. Fuel system circuit.

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16

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G

+

−

1

3

26

7

8

9

10

11

5

4

Figure 17. Electrical symbols.

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17

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R G#1#13

#2

#3

#4

#5

#6

#7

#8

#9

#10

#11

#14

#12

Throttle switch (open)

Down

Warning horn

Left gear switch

Right gear switch

Control valve switch

28 V DC

Down

Neutral


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R G

Test

Warning horn

Throttle switchesLeft gear switchRight gear switch

Neutral

Down

Down

Controlvalveswitch

#1

#2

#3

#4

#5

#6

#7

#8

#9

#10

#11#13

#12

28 V DC


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VOpen


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20

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1

2

3

Figure 21. Electrical symbols.

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21

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Base

Current flowCollectorEmitter+

+

−

1 2

3

Base

Current flow

CollectorEmitter+

−

−

Base

Current flowCollectorEmitter+−

−

Figure 22. Transistors.

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Base

R2

R1

Collector

12VUpposition

Downposition

Emitter

Figure 23. Transistorized circuit.

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23

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Inputs Output

Figure 24. Logic gate.

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24

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Inputs Output


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25

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Inputs

1

11 1

0

0

1

1

0

Output Inputs Output

Inputs Output

1 2

3


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1 2 3

++

Figure 27. Object views.

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27

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Front view

1 2

3


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28

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Front view

1 2 3


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29

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Front view

1 2 3


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30

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3

1

2

4

1"

3"

1 1"2

1 1"2 1"

2

3"4

Figure 31. Sketches.

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Figure 32. Sketches.

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3

1 2

Figure 33. Material symbols.

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1/16x45°

1/2 DIA.

19/64

.3125 DIA.1/16 R.

7/8 DIA.

3/32

1 3/4

7/8

1/16 R.

15/32

.221±.003

19/32

.110±.001

12/32 R. Spherical

.665±.001 DIA.

+.005−.000

Figure 34. Aircraft drawing.

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Paint stripe

0.25

1

2

3

4


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Note 1.

¾"17"

¼"1"

¼"

¼"

¾"

¾"

1½"

Note 1.

+1/64"−3/64"

F

E

D G

J

C

B

AH

Notes:1. Drill 31/64 inch ream ½ inch.2. All tolerances ±1/32 unless otherwise specified.3. Finish all over 25


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36

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3

Drill

Drill4 Holes

116 11

16

1564

18

78

38

12

22132

78


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0 500 1000 1500 2000 2500 3000 3500 4000

0 25 50 75 100 125 150 175 200 225 250 275

Brake - Horsepower (BHP)

Brake Mean Effective Pressure (BMEP)CID = Cubic inch displacement

985

3000

2800

2600

2400

2200

2000

1800

1600

1400

1200

Engine speed 1000 RPM

1340

1830

2000

2800 CID

Figure 38. Performance chart.

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Electric Wire Chart

Wire Size

20 18 16 14 12 10 8 6 4 2 1 1/0 2/0 3/0 4/0

20 18 16 14 12 10 8 6 4 2 1 1/0 2/0 3/0 4/0

Curves:1. Continous rating-amperes cables

in conduit and bundles2. Continous rating-amperes single

cable in free-air3. Intermittent rating-amperes

maximum of 2 minutes.

Curve 1

Curve 2

Curve 3

1 1.5 2 3 4 5 6 7 8 9 10 15 20 30 40 50 60 70 80 90 100

125

150

175

200

300

400

Amperes

Circuit Voltage

Voltage Drop

115 200 14 28

800 100 200

600 75 150

400 700 50 100

360 630 45 90

320 560 40 80

280 490 35 70

240 420 30 60

200 350 25 50

160 280 20 40

120 210 15 30

100 175 12 25

80 140 10 20

72 120 9 18

64 112 8 16

56 98 7 14

48 84 6 12

40 70 5 10

36 63 4 9

32 56 8

28 49 7

24 42 3 6

20 35 2 5

4 7 .5 1

Wire

leng

th in

feet

for a

llow

able

vol

tage

dro

p

Figure 39. Electric wire chart.

See FAA Replacement Figure 1, Addendum A

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Temperature in degrees Fahrenheit

Values Include 10 Percent Structural Deflection

Design limit ri

g load

340

320

300

280

260

240

220

200

180

160

140

120

100

80

60

40

20

0

−65 −60 −50 −40 −30 −20 −10 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

1/4 7x193/16 7x195/32 7x19

1/8 7x193/32 7x71/16 7x7

Cable sizes

Rigging load in pounds

Figure 40. Cable tension chart.

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180

170

160

150

140

130

120

110

100

90

80

70

60

50

1800 2000 2200 2400 2600 2800

Full throttle spec. fuel cons

Propeller load spec. fuel cons.

Propeller load horsepower

Normal rated power

Full throttle power

Engine Speed - RPM

Bra

ke H

orse

pow

er

Spe

cific

fuel

con

s LB

/BH

P/H

R

.60

.55

.50

.45

Figure 41. Performance chart.

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1 2 3

Figure 42. Aircraft hardware.

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1 2 3

Figure 43. Aircraft hardware.

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43

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1 2

3 4

Figure 44. Welds.

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FE

D

G

C

BA

Figure 45. Welds.

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45

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0

01

234

567

10

15

20

1 2

Figure 46. Precision measurement.


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46

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4 5 6 7 8 9

2

1

0 5 10 15 20 25

Inch1

1000


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15

10

5

0

20

09876543210

0 1 2 3


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48

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0

01

234

567

5

10

15

1 2


See FAA Figure Replacement Figure 3, Addendum A

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49

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1 2 3

Figure 50. Marshalling signals.

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50

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1 2 3

Figure 51. Marshalling signals.

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51

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(√(−4)0+6+(√1296)(√3)2=4

Figure 52. Equation.

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52

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√31 + √43 =2 2

(17)2


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53

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9'

5'

12'

Figure 54. Trapezoid area.

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54

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7.5"3"

4"



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55

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4'

2'

6'


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56

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A

B

D

C



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57

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(-35 + 25) (-7) + (π) (16-2) =√25


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58

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-4 125 =-6 -36


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59

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(-5+23)(-2)+(3-3)(√64)=-27÷9


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60

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60 POUNDS

A

Figure 61. Physics.

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61

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REQ’d. PER ASSEM.

B ADD-200

A MAT’L THKNESS

LET. CHANGE BY Date Appr.

-200

-100

48453721

4845

372

1

DASH NUMBERS SHOWN

Unless otherwise noted

MS20470AD-4-4NAS1097-3-4NAS1473-3ANAS1097-4-5NAS1097-4-4 -103 -102 -101 Part number

Rivet RivetDomed Nutplate Rivet Rivet Clip Doubler Doubler

For continuation see zone

Break all sharp edges

All N/AUNIT WT.FIRST RELEASE

DWG. AREA

EFFTypeA/C

No. req.per

Airplane

.040sheet

2024-T3CLAD AL.

2024-T3CLAD AL.

PROJECT T. Smith

DESIGN R. Eamer

EngineerFAA D.E.R

DWG.Checker

DFTSMIN.

G. Winn

I. Wright

S. Linz

MAT’LDESCR

MAT’LSPEC. Zone

.040sheet 7075-0 AL.

.040sheetstocksize

Scale full992-148-XXX

Speedwind aircraftengineering section lastchance airport anytown OK 73125-1234

DASH NUMBERS OPPOSITENAME

111

-200-200-200

36TCP36P36P

001-All088-All001-087

1 The use of this document shall be restrictedto conveyance of information to customers ofvendors only. Neither classified nor unclassifieddocuments may be reproduced without the written consent of THE SPEEDWIND AIRCRAFT CORP.

Area 1

BREV.

Figure 62. Maintenance data - part 1 of 3.

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62

GENERAL NOTES - 100

1. ALL BENDS +/– .5 deg.2. All holes +/– .003.3. Apply Alodine 1,000.4. Prime with MIL-P-23377 or equivalent.5. Trim S-1 C just aft of the clip at STA. 355.750 and forward of the front face of the STA. 370.25 frame and remove from the airplane.6. Position the –101 doubler as shown. Install wet with NAS1097AD-4-4 and -4-5 rivets and a faying surface seal of PR 1,422. Pick up the rivet row that was in S-1 C and the aft rivets in sta. 370.25. Tie doubler into front frame with clips as shown using MS20470AD-4-4 rivets through the clips and the frame.7. Install 4 NAS1473-3A nutplates with NAS1097-3-4 rivets through the skin and doubler to retain the antenna.8. Strip paint and primer from under the antenna footprint.9. Treat skin with Alodine 1,000.10. Install antenna and apply weather seal fillet around antenna base.

AREA 2

GENERAL NOTES - 200

1. ALL BENDS IAW P.S. 1,000.2. All holes IAW P.S. 1,015.3. Heat treat –102 to –T6 IAW P.S. 5,602.4. Alodine IAW P.S. 10,000.5. Prime IAW P.S. 10,125.6. Trim S-1 C just aft of the clip at STA. 355.750 and forward of the front face of the STA. 370.25 frame and remove from airplane.7. Position the –102 doubler as shown. Install wet with NAS1097AD-4-4 and -4-5 rivets, and a faying surface seal IAW P.S. 41,255. Pick up the rivet row that was S-1 C and the aft rivets in STA. 370.25. Add two edge rows as shown. Tie doubler into front frame with clips as shown using MS20470AD-4-4 rivets through the clips and the frame. 8. Install 4 NAS1473-3A nutplates with NAS 1097-3-4 rivets through the skin and doubler to retain the antenna.9. Strip paint and primer from under the antenna footprint.10. Treat skin IAW P.S. 10,000.11. Install antenna and apply weather seal fillet around antenna base.

Note: P.S. = Process Specification IAW = in accordance with

AREA 3

Figure 62A. Maintenance data - part 2 of 3.

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63

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Station355.750

Station370.250

CenterlineStringer1 Center (S-1C)

View looking up

–101,–102

14.25

2.402.303.70

3.004.980

0.50

R-3T min.

Ø0.6250

Ø0.1875 4PL

Joggle as necessary

Trim as req’d for inst’l

R-3T min.

–1034.502.0 1.0

1.0

Size this area as required to clear S-1C.

Area 4

×

×

×

×

×××

×× ×

× ×××

××

× × × × × × ×

× × ×× × ×× × ×× × × × × ×

× × × × × × × ×× × × × × × × ×

×

+

×

×

×

×

×

×

×

+

Figure 62B. M

aintenance data - part 3 of 3.

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62

The following is the compliance portion of an Airworthiness Directive.

“Compliance required as indicated, unless already accomplished:

I. Aircraft with less than 500 hours total time in service: Inspect in accordance with instructions below at 500 hours total time, or within the next 50 hours time in service after the e�ective date of this AD, and repeat after each subsequent 200 hours in service.

II. Aircraft with 500 hours through 1,000 hours total time in service: Inspect in accordance with instructions below within the next 50 hours time in service after the e�ective date of this AD, and repeat after each subsequent 200 hours in service.

III. Aircraft with more than 1,000 hours time in service: Inspect in accordance with instructions below within the next 25 hours time in service after the e�ective date of this AD, and repeat after each subsequent 200 hours in service.”

Figure 63. Airworthiness directive excerpt.

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65

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Rt = E2/P

Figure 64. Resistance total.

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1. 3.47 × 104 = 34,700.2. 2(410) = 2,097,152.

Figure 65. Scientific notation.

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67

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–4 + 6 + 103 (√1296) =


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68

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(17)2 =√31 + √43


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69

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1. (4.631)5

3. 4.631 ×10-5

2. 4.631 × 105

Figure 68. Alternative answer.

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70

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(√100 + √36 – √16) =Figure 69. Equation.

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71

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1. (√31) + (√43) ÷ 172

3. (√31) + (√43) – 172

2. (√31) + √43) ÷ 172

Figure 70. Alternative answer.

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V= 1/6πD3

Figure 71. Volume of a sphere.

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73

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ADDENDUM A


Figure 1. Electric wire chart. (Replaces Figure 39 from Appendix 1.)

Electric Wire Chart

Wire Size

20 18 16 14 12 10 8 6 4 2 1 1/0 2/0 3/0 4/0

20 18 16 14 12 10 8 6 4 2 1 1/0 2/0 3/0 4/0

Curves:1. Continuous rating-amperes cables

in conduit and bundles2. Continuous rating-amperes single

cable in free-air3. Intermittent rating-amperes

maximum of 2 minutes

Curve 1

Curve 2

Curve 3

1 1.5 2 3 4 5 6 7 8 9 10 15 20 30 40 50 60 70 80 90 100125150

175

200

300

400

Amperes

Circuit Voltage

Voltage Drop

115 200 14 28

800 100 200

600 75 150

400 700 50 100 360 630 45 90 320 560 40 80 280 490 35 70 240 420 30 60 200 350 25 50

160 280 20 40

120 210 15 30

100 175 12 25

80 140 10 2072 120 9 1864 112 8 1656 98 7 1448 84 6 1240 70 5 1036 63 4 932 56 828 49 724 42 3 6

20 35 2 5

4 7 .5 1

Wire

leng

th in

feet

for a

llow

able

vol

tage

dro

p

Figure 39

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24

20

Figure 2. Percision measurement. (Replaces Figure 46 from Appendix 1.)

Figure 46

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Figure 3. Precision measurement. (Replaces Figure 49 from Appendix 1.)

24

010

5

Figure 49

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7.5"3"

4"

7.5"3"

4"

Figure 4. Triangle area. (Replaces Figure 55 from Appendix 1.)

Figure 55

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A

B

D

C

Figure 5. Triangle area. (Replaces Figure 57 from Appendix 1.)

Figure 57

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aviation mechanic general course-notes and faa figures

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