Aviation Mechanic General Course-notes and Faa Figures

AVIATION MECHANIC GENERALCOURSE 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 6.—Circuit Diagram .........................................................................6 FIGURE 7.—Circuit Diagram .........................................................................7 FIGURE 8.—Circuit Diagram .........................................................................8 FIGURE 9.—Circuit Diagram .........................................................................9 FIGURE 10.—Battery Circuit .........................................................................10 FIGURE 11.—Circuit Diagram .......................................................................11 FIGURE 12.—Circuit Diagram .......................................................................12 FIGURE 13.—Circuit Diagram .......................................................................13 FIGURE 14.—Circuit Diagram .......................................................................14 FIGURE 15.—Landing Gear Circuit ...............................................................15 FIGURE 16.—Fuel System Circuit .................................................................16 FIGURE 17.—Electrical Symbols ..................................................................17 FIGURE 18.—Landing Gear Circuit ...............................................................18 FIGURE 19.—Landing Gear Circuit ...............................................................19 FIGURE 20.—Circuit Diagram .......................................................................20 FIGURE 21.—Electrical Symbols ..................................................................21 FIGURE 22.—Transistors ..............................................................................22 FIGURE 23.—Transistorized Circuit ..............................................................23 FIGURE 24.—Logic Gate ..............................................................................24 FIGURE 25.—Logic Gate ..............................................................................25 FIGURE 26.—Logic Gate ..............................................................................26 FIGURE 27.—Object Views ...........................................................................27 FIGURE 28.—Object Views ...........................................................................28 Return to Table of Contents 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 35.—Aircraft Drawing ...................................................................... 35 FIGURE 36.—Aircraft Drawing ...................................................................... 36 FIGURE 37.—Aircraft Drawing ...................................................................... 37 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 47.—Precision Measurement ......................................................... 47 FIGURE 48.—Precision Measurement ......................................................... 48 FIGURE 49.—Precision Measurement ......................................................... 49 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 Return to Table of Contents 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 Return to Table of Contents COURSE NOTES Return to Table of Contents MATHEMATICS EXAMPLE - What is equal to the square root of (-1776) / (-2) – 632? (16) (8541) 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. SCIENTIFIC NOTATION - is a way to represent any decimal number with a number between 1 and 10 raised to specific power of ten. [888 − 632 ] = [256] = 16 ( 31 ) + ( 43 ) ÷ 17 2 = ( 31 ) + ( 43 ) = 17 2 6.56 5.57 + = 289 5.57 + 0.02 = 5.59 EXAMPLE - Figure 65 3.47 x 104 = 34,700 (8540) THE SQUARE ROOT OF A NUMBER - is a smaller number that, when multiplied by itself, will produce the larger number. POWERS OF 10 For powers of 10, move the decimal to the right: EXAMPLE - What is the square root of 1,746? (41.7852 - use electronic calculator.) (8380) 101 = 10 102 = 10x10 = 100 3 10 = 10x10x10 = 1,000 4 10 = 10x10x10x10 = 10,000 5 10 = 10x10x10x10x10 = 100,000 6 10 = 10x10x10x10x10x10 = 1,000,000 EXAMPLE - What is the square root of 3,722.1835? (61.0097 - use electronic calculator.) (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 For negative powers of 10, move the decimal to the left: 1 10 = 10 0 10 = 1 -1 10 = 1/10 = 0.1 -2 10 = 1/10x10 = 0.01 -3 10 = 1/10x10x10 = 0.001 -4 10 = 1/10x10x10x10 = 0.0001 -5 10 = 1/10x10x10x10x10 = 0.00001 -6 10 = 1/10x10x10x10x10x10 = 0.000001 etc. 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 Return to Table of Contents = EXAMPLE – Figure 70. Which alternative (8594) answer is equal to 5.59? (1) 3-4 = 1/(3 x 3 x 3 x 3) = 1/81 2(410 ) = 2,097,152 ( −1776 ) − 632] ( −2) 1 MATHEMATICS - NOTES EXAMPLE - What power of 10 is equal to 1,000,000? (10 to the sixth power - refer to (8379) table above.) EXAMPLE - What is the result of 7 raised to the third power plus the square root of 39? (8392) (349.24) 7 x 7 x 7 = 343 on calculator. Square root of 39 = 6.24 on calculator. 343 + 6.24 = 349.24 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 (8385) table above.) CONVERTING FRACTIONS TO DECIMALS EXAMPLE - What is an alternative answer that is equal to 16,300? (1.63 x 10 to the fourth (8388) power.) Divide the numerator by the denominator. EXAMPLE - What decimal is most nearly equal (8426) to a bend radius of 31/64? (0.4844) 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 Divide 31 by 64 on calculator. EXAMPLE - The radius of a piece of round stock is 7/32. What decimal is most nearly (8428) equal to the diameter? (0.4375) EXAMPLE - What is the number 3.47 x 10 to the negative fourth power also equal to? (8387) (0.000347) Two times the radius = diameter. 2 x 7/32 = 14/32 Divide 14 by 32 on calculator. -4 3. 4 7 x 10 -4 0 0 0 3. 4 7 x 10 0 0.0 0 0 3 4 7 x 10 0.0 0 0 3 4 7 EXAMPLE - What decimal is equal to 39/32? (8410) (1.21875) Divide 39 by 32 on calculator. EXAMPLE - What decimal is most nearly equal (8418) to 77/64? (1.2031) EXAMPLE - What is the square root of 124.9924? (11.18 and also 1,118 x 10 to the (8389) negative second power.) Divide 77 by 64 on calculator. EXAMPLE - What decimal is equivalent of the fraction 43/32? (1.34375) Square root of 124.9924 = 11.18 on the calculator. 0 11.18 x 10 -2 1,118 x 10 Divide 43 by 32 on calculator. CONVERTING DECIMALS TO FRACTIONS EXAMPLE - What is the square root of 1,824? (42.708 and also 0.42708 x 10 to the second (8393) power.) Work backwards from the answer converting the fractions to decimals. Square root of 1,824 = 42.708 on calculator. 0 42.708 x 10 2 0.42708 x 10 choices EXAMPLE - Which fraction is equal to 0.025? A. 1/4. B. 1/400. C. 1/40. COMBINED ROOTS AND POWERS (8409) Answer choice "C", 1/40, is correct. EXAMPLE - What is the square root of 16 (8390) raised to the fourth power? (256) EXAMPLE - What is the fractional equivalent (8416) for a 0.0625-thick sheet of aluminum? Square root of 16 = 4 on calculator. 4 x 4 x 4 x 4 = 256. A. 1/32. B. 3/64. C. 1/16. EXAMPLE - What is the square root of 4 raised (8386) to the fifth power? (32) For each answer choice convert the fraction to a decimal and match the results to the decimal in the question. Square root of 4 = 2 on calculator. 2 x 2 x 2 x 2 x 2 = 32. Answer choice "C", 1/16, is correct. MATHEMATICS - NOTES 2 Return to Table of Contents EXAMPLE - A blueprint shows a hole of 0.17187 to be drilled. Which fraction size drill (8425) bit is most nearly equal? ⎛ 64 3 ⎞ 3 ⎜ × ⎟÷ = ⎝ 1 8⎠ 4 64 3 4 × × = 1 8 3 8 3 4 × × = 1 1 3 8 × 4 = 32 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. EXAMPLE - Solve the equation: (32 × 3/8) ÷ (1/6) = Answer choice "A", 11/64, is correct. ⎛ 32 × ⎜ ⎝ 1 3⎞ 1 ⎟÷ = 8⎠ 6 32 3 6 × × = 1 8 1 4 × 3 × 6 = 72 SOLVING EQUATIONS Use a calculator. Account for negative signs and decimal points. Do operations starting with innermost parentheses first. EXAMPLE - Solve the equation: 2/4(-30 + 34)5 = EXAMPLE - Solve the equation: 4 − 3[−6(2 + 3 ) + 4] = (8439) 4 − 3[ −30 + 4] = 4 − 3[ −26] = 4 + 78 = 82 −4 125 EXAMPLE - Solve the equation: −6 −36 (8440) 125 −36 ÷ −4 −6 125 −6 × −4 −36 125 −1 × 4 −6 −125 24 − 6[ 36 − 20 ] = − 6 [16 ] = − 96 EXAMPLE - Solve the equation: [ ( 4 × −3) + ( −9 × 2) ] ÷ 2 = ( 4 × −3 ) + ( −9 × 2) (8432) = 2 ( −12) + ( −18 ) = 2 − 30 = − 15 2 = (8438) = = = = −5.20 EXAMPLE - Solve the equation: (-3 + 2)(-12 - 4) + (-4 + 6) x 3 = 16 + 6 = 22 (8433) (8558) EXAMPLE - Solve the equation. ( 100 + (10 Return to Table of Contents (8441) (-1)(-16) + (2) x 3 = EXAMPLE - Solve the equation: (64 × 3/8) ÷ (3/4) = (8436) 1 4 5 × × = 2 1 1 1 2 5 × × = 10 1 1 1 EXAMPLE - Solve the equation: 4 − 3[ − 6( 5) + 4] = −6[−9(−8 + 4)−2(7 + 3)] = − 6[ − 9 ( − 4 ) − 2 (10 ) ] = (8434) 3 36 − 16 ) + 6 − 4 ) = 12 = MATHEMATICS - NOTES AREA OF A TRIANGLE EXAMPLE - Figure 60. Solve the equation: ( −5 + 23)( −2) + (3 −3 )( 64 ) −27 ÷ 9 = Area = (8442) EXAMPLE - Figure 55. What is the area of the (8397) triangle shown? 1 (18)( −2) + 3 × 3 × 3 ( 8) −3 = 1 ( 4 in.×3 in.) 2 12 Area = = 6 square inches 2 Area = 8 −36 + 27 = −3 −36 + 0.296 = −3 −35.704 = 119 . −3 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 EXAMPLE - Figure 53. Solve the equation: 31 + (1 7 43 ) 2 = ( −35 + 25)( −7) + π (16−2 ) 25 = Area = AREA OF A TRAPEZOID Area = height × 5 Area = 5 ft.× = 1 ( 21) 2 Area = 5 × 10.5 = 52.5 square feet EXAMPLE - Figure 56. What is the area of the (8401) trapezoid? 1 Area = 2 ft. × ( 6 ft. + 4 ft.) 2 Area = 2 × 5 = 10 square feet EXAMPLE - Figure 52. Solve the equation: (− 4 ) +6+ ( 1296 )( 3 ) 2 = 1 (12 ft. + 9 ft.) 2 Area = 5 × 1 ) 70 + ( 3.1416 × 256 = 5 70 + ( 3.1416 × 0.0039) = 5 70 + 0.0123 = 14.00 5 4 1 (sum of the bases) 2 EXAMPLE - Figure 54. What is the area of the (8395) trapezoid? (8437) 1 ( −10)( −7) + 3.1416 × 16 × 16 (8402) 1 (16.8 in. × 7.5 in.) 2 126 Area = = 63 square inches 2 (8391) 5.5678 + 6.5574 = 289 12.1252 = 0.04195 289 EXAMPLE - Figure 50. Solve the equation: 0 1 ( base × height) 2 AREA OF A CUBE (8381) Use the formula for area to find the area of one side of the cube and then multiply by 6. 1 + 6 + (6 × 3) = EXAMPLE - What is the surface area of a cube where a side (edge) measures 7.25 inches? (8592) 1 + 6 + 18 = 25 = 5 6 (l × w ) = area 6 (7.25 × 7.25) = 6 × 52.5625 = 315.375 sq. in. MATHEMATICS - NOTES 4 Return to Table of Contents 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 (8404) foot 8 inches in depth? VOLUME OF A SPHERE EXAMPLE – Figure 71. What is the volume of a (8595) sphere with a radius of 4.5 inches? 1/ 6 π D 3 = V (7.5 gal. = 1 cu. ft.) 3 3.14159 × 9 = 6 3.14159 × 729 = 6 2290.22 = 381.70 cubic inches 6 Volume = 3 ft. × 2 ft. × 1 20 ft. 12 Volume = 10 cu. ft. 7.5 gal. Volume = 10 cu. ft. × 1 cu. ft. Volume = 75 gallons Volume = 3 × 2 × VOLUME OF RECTANGULAR SHAPED FUEL TANKS EXAMPLE - What container size in cubic feet will be equal in volume to 60 gallons of fuel? Volume = length × width × depth Convert the units as required. (8400) (7.5 gal. = 1 cu. ft.) 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 (8407) cubic inches are within the tank? Size = 60 ÷ 7.5 Size = 8.0 cubic feet CIRCUMFERENCE OF A CIRCLE Volume = 37.5 in. x 14 in. x 8.25 in. Volume = 4,331.25 cubic inches EXAMPLE - What size sheet of metal is required to fabricate a cylinder 20 inches long (8396) and 8 inches in diameter? EXAMPLE - A rectangular-shaped fuel tank measures 60 inches in length, 30 inches in width, and 12 inches in depth. How many cubic (8399) feet are within the tank? Note: C = π D A. 20" x 25-5/32". B. 20" x 24-9/64". C. 20" x 25-9/64". 60 30 12 × × 12 12 12 Volume = 5 ft. × 2.5 ft. × 1ft. Volume = C = 3.1416 x 8 in. = 25.1328 in. Note: 9/64 = 0.140625 5/32 = 0.15625 Volume = 12.5 cubic feet 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. 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.) 8 ft. 12 Correct answer is "C", 20" x 25-9/64". PISTON DISPLACEMENT (8405) Area of a circle = π × R 2 D R= 2 Volume = 27.5 in. × (3 / 4 × 12) in. × 8.25 in. Volume = 27.5 in. × 9 in. × 8.25 in. 1 gal. Volume = 2,042 cu. in. × 231 cu. in. Volume = 8.8 gallons Engine Displacement = 2 ⎛ bore ⎞ ⎟ × Stroke × No. of cylinders ⎝ 2 ⎠ π ×⎜ THE TOTAL PISTON DISPLACEMENT OF A SPECIFIC ENGINE - is the volume displaced by all the pistons during one revolution of the crankshaft. (8394) Return to Table of Contents 5 MATHEMATICS - NOTES EXAMPLE - What is the piston displacement of a master cylinder with a 1.5-inch diameter bore (8403) and a piston stroke of 4 inches? Comp. Ratio = 84 6 = 14 1 Comp. Ratio = 6 to 1 Comp. Ratio = 2 ⎛ D⎞ Disp. = π × ⎜ ⎟ × Stroke × No. of cylinders ⎝ 2⎠ 2 . ⎞ ⎛ 15 Disp. = 3.1416 × ⎜ ⎟ × 4 × 1 ⎝ 2⎠ EXAMPLE - What is the ratio of 10 feet to 30 (8430) inches? Disp. = 3.1416 × (0.75) × 4 × 1 Disp. = 3.1416 × 0.5625 × 4 × 1 Disp. = 7.0686 cubic inches 2 10 × 12 120 4 = = = 4:1 30 30 1 EXAMPLE - What is the ratio of a gasoline fuel load of 200 gallons to one of 1,680 pounds? (8435) 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 (8406) engine? Note: Gasoline weighs 6.0 pounds per gallon. 200 × 6 1200 60 5 = = = 1680 1680 84 7 Ratio = 5:7 2 ⎛ D⎞ Disp. = π × ⎜ ⎟ × Stroke × No. of cylinders ⎝ 2⎠ EXAMPLE - What is the speed ratio of a gear with 36 teeth meshed to a gear with 20 teeth? (8420) 20 10 = =9:5 36 18 2 ⎛ 3.78 ⎞ Disp. = 3.1416 × ⎜ ⎟ × (8.5 − 4.0) × 4 ⎝ 2 ⎠ Disp. = 3.1416 × (189 . ) × 4.5 × 4 Disp. = 3.1416 × 3.5721 × 4.5 × 4 Disp. = 202 cubic inches 2 Note: The gear with the most teeth would be turning the slower of the two. 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 (8408) displacement? PROPORTION A proportion is a statement of equality between two or more ratios. 2 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) ⎛ D⎞ Disp. = π × ⎜ ⎟ × Stroke × No. of cylinders ⎝ 2⎠ 2 ⎛ 3.5 ⎞ Disp. = 3.1416 × ⎜ ⎟ × 4.5 × 6 ⎝ 2 ⎠ 60 gal. ? gal. = 750 mi. 2500 mi. 750 × ? = 60 × 2500 60 × 2500 ? gal. = = 200 gallons 750 Disp. = 3.1416 × (175 . ) × 4.5 × 6 Disp. = 3.1416 × 3.0625 × 4.5 × 6 Disp. = 259.77 cubic inches 2 RATIOS GEAR SPEEDS Divide the first term by the second term, and then reduce the resulting fraction to its lowest terms. Use the following proportion equation: Teeth driving RPM driven = Teeth driven RPM driving 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, (8411) what is the compression ratio? Comp. Ratio = 84 (84 - 70) ( Volume of piston at bottom) ( Volume of piston at top) MATHEMATICS - NOTES 6 Return to Table of Contents EXAMPLE - What is the speed of a spur gear with 42 teeth driven by a pinion gear with 14 (8421,8413) teeth turning 420 RPM? EXAMPLE - If an engine is turning 1,965 RPM at 65 percent power, what is its maximum (8423) RPM? To solve this problem you must presume that power is directly proportional to RPM. 14 teeth ? RPM = 42 teeth 420 RPM 14 × 420 5880 ? RPM = = = 140 RPM 42 42 Max RPM 1965 RPM , = 100% 65% 65 x Max RPM = 100 x 1,965 100 × 1965 , Max RPM = 65 Max RPM = 3,023 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. EXAMPLE - An engine develops 108 horsepower at 87 percent power. What horsepower would be developed at 65 percent (8414) power? (8417) 5/8 = 0.625 = 62.5% EXAMPLE - Express 7/8 as a percent. (8412) 108 HP ? HP = 87% 65% ? HP × 87% = 108 HP × 65% 7/8 = 0.875 = 87.5% EXAMPLE - Sixty-five percent of 80 engines? engines are what (8427) ? HP = 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) 108 HP × 65% = 80.69 HP 87% 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 $145.60 Cost = = $130.00 1.12 835.3 = 0.76 = 76% 1100 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 Return to Table of Contents 7 MATHEMATICS - NOTES AIRCRAFT WEIGHT AND BALANCE are occupied, the full baggage weight is on board and all the fuel tanks are full. (8539) PURPOSE 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) ZERO FUEL WEIGHT ZERO FUEL WEIGHT - is the maximum permissible weight of a loaded aircraft (passengers, crew, and cargo) without fuel.(8167) WEIGHT CHANGES OCCUR IN AGING AIRCRAFT – mainly because of repairs and alterations done over its lifetime. (8538) DATUM THE DATUM IS AN IMAGINARY VERTICAL PLANE - from which all horizontal measurements are taken for balance purposes. 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) 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) THE EMPTY WEIGHT OF AN AIRCRAFT INCLUDES - all operating equipment that has a fixed location and is actually installed. ARM THE ARM IS THE HORIZONTAL DISTANCE that an object is located from the datum. THE AMOUNT OF FUEL - used for computing empty weight and corresponding CG is unusable fuel. (8176,8607) THE LENGTH OF THE ARM (DISTANCE) - is always given or measured in inches. (8161) 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) 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) USEFUL LOAD MOMENT THE USEFUL LOAD OF AN AIRCRAFT - consists of the crew, usable fuel, passengers, and cargo. (8155) 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. THE USEFUL LOAD OF AN AIRCRAFT - is the difference between the maximum gross weight and the empty weight. (8170) EXAMPLE - If a 40-pound generator applies +1400-inch pounds to a reference axis, where is the generator located? (8182) 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) ARM = CENTER OF GRAVITY (CG) 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 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. THE MAXIMUM TAKEOFF WEIGHT – will be exceeded in most modern aircraft if all the seats WEIGHT & BALANCE - NOTES MOMENT +1400 = = +35 inches WEIGHT 40 8 Return to Table of Contents 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) 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) 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) 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) MAC, LEMAC, AND TEMAC 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) 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) 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 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) SOLUTION: MOMENT = WEIGHT x ARM TOTAL MOMENT CG = TOTAL WEIGHT 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) WEIGH POINT WT. x ARM Main Gears 3,540 195.5 83.5 Nose Gear 2,322 TOTALS 5,862 THERE ARE SEVERAL WAYS TO LEVEL AN AIRCRAFT. The method with the highest degree of accuracy is the spirit level. (8549) TOTAL MOMENT 885,957 = = TOTAL WEIGHT 5,862 CG = 151.1 inches. CG = 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) 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. 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) Return to Table of Contents = MOMENT 692,070 193,887 885,957 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 9 WEIGHT & BALANCE - NOTES * Note: To calculate the original arm (CG), divide the empty weight moment by the empty weight. SOLUTION: MOMENT = WEIGHT x ARM TOTAL MOMENT CG = TOTAL 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. 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) TOTAL MOMENT 884,277 = TOTAL WEIGHT 5,842 CG = 151365 . inches. CG = EMPTY WEIGHT CG AFTER ENGINE CHANGE IN A BALANCE COMPUTATION WHICH REQUIRES AN ITEM LOCATED AFT OF DATUM - to be removed, use (-)weight x (+)arm = (-) moment. (8183) WEIGH POINT Right Gear Left Gear Tail Gear Lavatory water** Remove ballast TOTALS EXAMPLE - What is the new empty weight CG? (8188) WT. x ARM = MOMENT 9,980 30.24 301,795 9,770 30.24 295,445 1,970 *390.50 769,285 - 34 352.00 - 11,968 - 55,480 -146 380.00 21,540 1,299,077 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. * 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. SOLUTION: MOMENT = WEIGHT x ARM TOTAL MOMENT CG = TOTAL WEIGHT ** 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. ITEM Empty Weight B-engine rem. D-engine add TOTALS CG = TOTAL MOMENT 1299 , ,077 = TOTAL WEIGHT 21540 , TOTAL MOMENT 12,833.04 = TOTAL WEIGHT 1008 , NEW CG = 12.73 inches. NEW CG = CG = 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 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 On board the aircraft when weighed: Potable water 20 lb. at +84 Hydraulic fluid 23 lb. at +101 SOLUTION: MOMENT = WEIGHT x ARM TOTAL MOMENT CG = TOTAL WEIGHT ITEM Empty wt. Potable water** TOTALS WT. 5,862 -20 5,842 x SOLUTION: MOMENT = WEIGHT x ARM TOTAL MOMENT CG = TOTAL WEIGHT ARM = MOMENT *151.14 885,957 84.00 WEIGHT & BALANCE - NOTES WT. x ARM = MOMENT 998 13.48 13,453.04 -175 - 62.00 +10,850.00 - 62.00 -11,470.00 +185 1,008 12,833.04 -1,680 884,277 10 Return to Table of Contents ITEM WEIGHT x Boxes 1,2 Box 3 20 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 94.00 +2,820.00 Radio installed 30 TOTALS 3,003 110,670.78 MOMENT WEIGHT − 50 ARM = = 2.5 feet forward 20 ARM = * Note: To calculate the original arm (CG), divide the empty weight moment by the empty weight. BALLAST WEIGHT TO BRING CG WITHIN CG RANGE ** Note: Two seats at 15 lb. each is 30 lb. NEW CG = TOTAL MOMENT 110,670.78 = TOTAL WEIGHT 3,003 EXAMPLE - What is the minimum weight of ballast needed to bring the CG within the CG range? (8177) NEW CG = 36.85 inches. GIVEN: Loaded aircraft Loaded aircraft CG CG range Ballast arm DISTANCE CHANGE FROM OLD CG: New CG 36.85 Old CG -35.23 Change 1.62 inches aft LEVER ARM 4,954 lb. +30.5 in. +32.0 in. to +42.1 in. +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. 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. Step 1: Find distance CG is out of limits: Forward CG limit 32.0 Current CG - 30.5 1.5 inches THE DISTANCE OF ANY OBJECT FROM THE FULCRUM - is called the lever arm. (8157) WEIGHT ADDITION THAT WILL NOT CHANGE CG Step 2: Find the lever arm of the ballast: Ballast location 162.0 -Forward CG Limit - 32.0 130.0 inches 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 Box 2 LEVER ARM = MOMENT +50 -? -50 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 10 lb. at 4 ft. aft of CG 5 lb. at 2 ft. aft of CG SOLUTION: MOMENT = WEIGHT x ARM WEIGHT = ITEM WEIGHT x LEVER ARM = MOMENT Box 1 10 4 40 Box 2 5 2 10 Total Moments Added Aft 50 MOMENT 7,431 = ARM 130 Ballast weight required is 57.16 pounds. For the CG not to change, the moments forward and aft of the CG must be the same. Return to Table of Contents 11 WEIGHT & BALANCE - NOTES EMPTY WEIGHT CENTER OF GRAVITY RANGE 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 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) THE CG RANGE IN SINGLE-ROTOR HELICOPTERS IS - more restricted than for airplanes. (8175) WEIGHT AND BALANCE EXTREME CONDITIONS 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) 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. WEIGHT & BALANCE - NOTES 12 Return to Table of Contents READING GRAPHS AND CHARTS 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 15ampere load and a 1-volt drop? (8145) 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) First, determine the wire size based on voltage drop: 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. 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. 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) 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. 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. Comparing the wire sizes obtained under both conditions above, select No. 10 wire size as the smallest that will satisfy both conditions. An aircraft engine has a 2,800 cubic-inch displacement, develops 2,000 brake-horsepower, and indicates 270 brake mean effective pressure. What is the engine speed (RPM)? (8143) 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 (8147) allowable 1-volt drop? 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. First, determine the wire size based on voltage drop: ELECTRIC WIRE CHART 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. 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. Return to Table of Contents 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. 13 GRAPHS & CHARTS - NOTES Comparing the wire sizes obtained under both conditions, select No. 12 wire as the smallest that will satisfy both requirements. 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) 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. 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. 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) FUEL CONSUMPTION 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. EXAMPLES - Figure 41. How much fuel would be required for a 30minute 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 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". GRAPHS & CHARTS - NOTES 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 14 Return to Table of Contents AIRCRAFT DRAWINGS AN ISOMETRIC PROJECTION - is a type of drawing which shows the object inclined at an angle to the viewer. WORKING DRAWINGS WORKING DRAWINGS MAY BE DIVIDED INTO THREE CLASSES - they are: detail drawings, assembly drawings, and installation drawings. (8119) 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) DETAIL DRAWINGS A DETAIL DRAWING - is a description of a single part carefully and accurately drawn to scale and dimensioned. (8105,8136) SKETCHES SKETCHES ARE SIMPLE, ROUGH DRAWINGS that are made rapidly and without much detail. ASSEMBLY DRAWINGS ONE OR A COMBINATION OF SIX BASIC SHAPES - triangle, circle, cube, cylinder, cone, and sphere - will be used when sketching nearly any object. (8116) AN ASSEMBLY DRAWING - is a description of an object made up of two or more parts. INSTALLATION DRAWINGS SINCE SKETCHES ARE DRAWN WITHOUT THE USE OF DRAFTING INSTRUMENTS - using graph paper makes the layout process easier. (8120) 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) PROJECTIONS 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) AN ORTHOGRAPHIC PROJECTION - shows the exact size and shape of all the parts of complex objects. A number of views are necessary. IF USED FOR MAKING A PART - a sketch must show all information to manufacture the part. (8114) EXAMPLE - Figure 28. Which is the bottom view of the object shown? (The bottom view is "2".) (8106) 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.) A WIRING DIAGRAM - would show the wire size required for a particular installation. EXAMPLE - Figure 29. Which is the left side view of the object shown? (The left side view is "3".) (8109) EXAMPLE - Figure 31. What are the proper procedural steps for sketching repairs and alterations? (8113,8117) EXAMPLE - Figure 30. Which is the bottom view of the object shown? (The bottom view is "1".) (8111) 3 - Blocking in the views. 1 - Adding detail. 4 - Darkening the views. 2 - Adding the dimensions. ONE-VIEW, TWO-VIEW, AND THREE-VIEW DRAWINGS - are the most common with orthographic projections. (For example, see Figure 32.) (8108) EXAMPLE - Figure 32. What is the next step required for a working sketch of the illustration? (Sketch extension and dimension lines.) (8115) WHEN THREE-VIEW PROJECTION IS USED the top, front, and right side views are usually shown. (For example, see Figure 32.) Return to Table of Contents 15 DRAWINGS - NOTES 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) 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) EXAMPLE - Figure 33. Which material sectionline 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) 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) 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." PICTORIAL A PICTORIAL DIAGRAM – uses pictures to show electrical components instead of using the more familiar electrical symbols. (8139) EXAMPLE - Figure 34. Using the information, what size drill would be required to drill the clevis bolthole? (5/16 inch.) (8126) MEANING OF LINES 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 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) 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) 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) 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. OTHER DATA DIMENSIONS - are the means of conveying measurements through the medium of drawings. (For example, see Figure 34.) (8127) EXAMPLE - Figure 34. What is the dimension of the chamfer? (0.0625 x 45°.) (8123) 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) DRAWINGS - NOTES Note: A chamfer is the beveled corner or edge of an object. 1/16 x 45° = 0.0625 x 45° 16 Return to Table of Contents 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 From center of first 15/64 hole to center of second 15/64 hole 7/8 From center of second 15/64 hole to center of third 15/64 hole 7/8 From center of third 15/64 hole to center of fourth 15/64 hole 1/8 From center of fourth 15/64 hole to bottom of that hole 15/128 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 0.375 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) 0.875 0.875 0.125 Maximum: 4.387 + 0.005 = 4.392 inches Minimum: 4.387 + 0.002 = 4.385 inches 0.117 2.367 TOLERANCES/ALLOWANCES EXAMPLE - Figure 34. What is the maximum diameter of the hole for the clevis pin? (0.3175) (8124) TOLERANCE - is the difference between extreme permissible dimensions that a part may have and still be acceptable. (8141) The hole in the clevis in which the pin fits has a diameter of: 0.3125 + 0.005 - 0.000 inch. ALLOWANCE - is the difference between the nominal dimension of a part and its upper or lower limit. This tolerance makes the maximum diameter 0.3175 inch: 0.3125 + 0.005 = 0.3175 Return to Table of Contents 17 DRAWINGS - NOTES AIRCRAFT MATERIALS AND PROCESSES ON A FILLET WELD - the penetration requirement includes 25 to 50 percent of the base metal thickness. (8288) WELDING A CHARACTERISTIC OF A GOOD GAS WELD is that it tapers off smoothly into the base metal. (8283) 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) 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) THE TWO GENERAL CLASSIFICATIONS OF CHEMICAL CORROSION - are direct chemical attack and electrochemical attack. CRACKING ADJACENT TO THE WELD indicates a part has cooled too quickly after being welded. (8282) DIRECT CHEMICAL ATTACK WHEN EVALUATING A WELDED JOINT, - a mechanic should be familiar with the parts, proportions, and formation of the weld. (8553) 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.). EXAMPLES - Figure 44. CAUSTIC CLEANING PRODUCTS - used on aluminum structures have the effect of producing corrosion. (8355) 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) SPILLED MERCURY ON ALUMINUM - causes rapid and severe corrosion that is very difficult to control. (8378) 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) ELECTROCHEMICAL ATTACK ELECTROCHEMICAL ATTACK (GALVANIC ACTION) IS RESPONSIBLE - for most forms of corrosion on aircraft structure and component parts. IF HOLES AND A FEW PROJECTING GLOBULES ARE FOUND IN A WELD - remove all the old weld and reweld the joint. (8281) CORROSION CAUSED BY GALVANIC ACTION IS THE RESULT OF - contact between two unlike metals. (8371,8542) EXAMPLES - Figure 45. IN THE CORROSION PROCESS - it is the anodic area or dissimilar anodic material that corrodes. (The cathode does not corrode.) (8377) 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) IN THE GALVANIC OR ELECTRO-CHEMICAL SERIES FOR METALS - the most anodic metals are those that will give up electrons most easily. (8377) 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) MATERIALS/PROCESSES - NOTES 18 Return to Table of Contents 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 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) EXAMPLE - Of the following listed materials: Cadmium, 7075-T6 aluminum alloy, Magnesium, which is the most anodic? (Magnesium.) (8372) CORROSION SHOULD BE REMOVED FROM MAGNESIUM PARTS - with a stiff, nonmetallic brush. (8366) EXAMPLE - Of the following listed materials: Zinc, 2024 aluminum alloy, Stainless steel, which is the most cathodic? (Stainless steel.) (8374) 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). 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) 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. GALVANIC ACTION CAUSED BY DISSIMILAR METAL CONTACT - may best be prevented by placing a nonporous dielectric between the surfaces. (8370) 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) 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) 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) 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. THE INTERIOR SURFACE OF SEALED STRUCTURAL STEEL TUBING - is best protected against corrosion by a coating of hot linseed oil. (8373) A PRIMARY CAUSE OF INTERGRANULAR CORROSION - is improper or inadequate heat treatment. (8245,8365) 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) 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) Return to Table of Contents 19 MATERIALS/PROCESSES - NOTES ENGINE STORAGE 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) 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) STEEL THE SOCIETY OF AUTOMOTIVE ENGINEERS (SAE) USES A NUMERICAL INDEX SYSTEM - to identify the composition of various steels. SOAP AND WATER CLEANING THE FIRST DIGIT - indicates the basic alloying element. 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) 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) 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) STAINLESS STEEL IS GENERALLY USED - in the construction of aircraft engine firewalls. (8257) HEAT TREATMENT ALIPHATIC NAPHTHA IS THE SOLVENT RECOMMENDED - for wipedown of cleaned surfaces just before painting. (8350) 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. 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) 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. 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) MATERIALS/PROCESSES - NOTES HARDNESS MAY BE INCREASED BY - coldworking 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) 20 Return to Table of Contents IN THE 2xxx THROUGH 8xxx GROUP, THE FIRST DIGIT INDICATES - the major alloying element used in the formation of the alloy. 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. 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) STEEL IS TEMPERED AFTER BEING HARDENED - to relieve its internal stresses and reduce its brittleness. (8252) ANNEALING AND NORMALIZING "ALCLAD" DESIGNATES A METAL - consisting of pure aluminum surface layers on an aluminum alloy core. (8259) 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) 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) SLOW COOLING; LOW STRENGTH - is descriptive of the annealing process of steel during and after it has been annealed. (8255) HARDNESS AND TEMPER DESIGNATIONS NORMALIZING IS A PROCESS - of heat treating iron-base metals only. (8249) 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 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) 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) IN CASE HARDENING THE SURFACE OF THE METAL - is changed chemically by introducing a high carbide or nitride content. (8248) ALUMINUM AND ALUMINUM ALLOYS QUENCHING ALUMINUM AND ALUMINUM ALLOYS ARE DESIGNATED BY - a four digit index system. 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. 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). PARTS HEAT TREATED IN A SODIUM AND POTASSIUM NITRATE BATH - are rinsed thoroughly in hot water to prevent corrosion. EXAMPLE - The aluminum code number 1100 identifies what type of aluminum? (99 percent commercially pure aluminum.) (8274) Return to Table of Contents 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) 21 MATERIALS/PROCESSES - NOTES AIRCRAFT HARDWARE 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) AIRCRAFT BOLTS MOST BOLTS USED IN AIRCRAFT STRUCTURES - are either general purpose AN (Air Force-Navy) or NAS (National Aircraft Standard) close tolerance bolts. 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 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) 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) 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) FIBER-TYPE LOCKNUT THE LOCKING FEATURE OF A FIBER-TYPE LOCKNUT - is obtained by the use of an unthreaded fiber locking insert. (8277) 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. A FIBER-TYPE, SELF-LOCKING NUT - must never be used on an aircraft if the bolt is subject to rotation. (8260) EXAMPLE - Figure 43. Which illustration is a clevis bolt? (Illustration "3".) (8267) INSTALLATION OF NUTS AND BOLTS AN-TYPE CLEVIS BOLTS ARE USED IN AIRPLANES - only for shear load applications and never in tension. (8271) ADVISORY CIRCULAR 43.13-1B, "AIRCRAFT INSPECTION AND REPAIR" - contains standards for protrusions of bolts, studs, and screws through self-locking nuts. (8534) 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) UNLESS OTHERWISE SPECIFIED OR REQUIRED - aircraft bolts should be installed so that the bolthead is upward, or in a forward direction. (8258) GRIP LENGTH UNLESS OTHERWISE SPECIFIED - torque values for tightening aircraft nuts and bolts relate to clean, dry, threads. (8256) 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) HARDWARE - NOTES 22 Return to Table of Contents WHEN THE SPECIFIC TORQUE VALUE FOR NUTS IS NOT GIVEN - the recommended torque values can be found in Advisory Circular 43.13-1B. (8266) Return to Table of Contents 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) 23 HARDWARE - NOTES PHYSICS THE TRUE LANDING SPEED OF AN AIRCRAFT IS GREATEST - under atmospheric conditions of high temperature with high humidity. (8478) PRESSURE AND TEMPERATURE TEMPERATURE - is a measure of the kinetic energy of the molecules of any substance. (8482) FLUID SYSTEMS HEAT MAY BE TRANSFERRED - by convection, conduction, or radiation. Heat is not transferred by diffusion. (8467) BERNOULLI'S PRINCIPLE STATES - that the pressure of a fluid decreases at points where the velocity of the fluid increases. (8216) 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) THE GREATER THE PRESSURE DIFFERENTIAL BETWEEN UNMETERED PRESSURE AND METERED PRESSURE - the greater the rate of flow of liquid through the metering orifice. IF BOTH THE VOLUME AND THE ABSOLUTE TEMPERATURE - of a confined gas are doubled, the pressure will not change. (8485) 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".) IF THE TEMPERATURE OF A CONFINED LIQUID IS HELD CONSTANT - and its pressure is tripled, the volume will remain the same. (8476) 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 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) 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. 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) 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) DEWPOINT - is the temperature to which humid air must be cooled at constant pressure to become saturated. (8484) 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) 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) PHYSICS - NOTES 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) 24 Return to Table of Contents 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) 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. 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) (L) length of ramp (R) weight of object = (I) height of ramp (E) force required THE PURPOSE OF AIRCRAFT WING DIHEDRAL - is to increase lateral stability. (8488) 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) WORK AND POWER WORK IS DONE - when a resistance is overcome by a force acting through a measurable distance. L R = I E 9 ft. 120 lbs. = 3 ft. E 120 lbs.×3 ft. E= = 40 pounds 9 ft. WORK = FORCE x DISTANCE W=FxD 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 footpounds.) (8477) W = F x D = 120 lbs. x 3 ft. W = 360 foot-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=PxA 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=PxA F = 3,000 x (3-1) F = 6,000 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 footpounds.) (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) FORCE × DISTANCE POWER = TIME F ×D P= T MACHINES A SHORTHAND METHOD OF FINDING THE MECHANICAL ADVANTAGE OF A PULLEY Return to Table of Contents 25 PHYSICS - NOTES FLUID LINES AND FITTINGS 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) FLEXIBLE 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) 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) 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) 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) 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) ALUMINUM TUBING - or other soft metal tubing is cut best with a hand operated wheel-type tubing cutter. (8561) GEOMETRIC SHAPES - that appear with colorcoded bands are always black against a white background regardless of content. (They are used to identify line contents.) (8559) 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) 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) DEFECTS THAT ARE NOT ACCEPTABLE FOR METAL LINES ARE: 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) • 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) 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 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. METAL TUBING IN A METAL TUBING INSTALLATION - tension is undesirable because pressurization will cause it to expand and shift. (8214) FLUID LINES & FITTINGS - NOTES 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) 26 Return to Table of Contents AN ADVANTAGE OF A DOUBLE FLARE ON ALUMINUM TUBING - is that it is more resistant to the shearing effect of torque. (8196) 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) COUPLING SIZES 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. 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. FLARES EXAMPLE - What end dash number of an AN818 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) 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) 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) 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) 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) WHEN THE COUPLING NUT IS OVERTIGHTENED - on a flared tube, damage is most likely at the sleeve and flare junction. (8560) 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) 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) Return to Table of Contents 27 FLUID LINES & FITTINGS - NOTES FUELS AND FUEL SYSTEMS TWO DIFFERENT SCALES - are used to designate fuel grades. Fuels below grade 100 are classified by octane numbers. CHARACTERISTICS OF AVIATION GASOLINE CHARACTERISTICS OF AVIATION GASOLINE ARE - high heat value and high volatility. (8344) FUELS WHICH POSSESS GREATER ANTIKNOCK QUALITIES THAN 100 OCTANE - are classified by performance numbers. (8336) VOLATILITY THE MAIN DIFFERENCES - between grades 100 and 100LL fuel are lead content and color. (8343) VOLATILITY IS A MEASURE OF THE TENDENCY OF A LIQUID SUBSTANCE - to vaporize under given conditions. FUEL COLORS A FUEL THAT VAPORIZES TOO READILY - may cause vapor lock. (8341) AVIATION GASOLINES ARE COLOR CODED for purposes of identification. For example, 100LL fuel is blue. (8335) A FUEL THAT DOES NOT VAPORIZE READILY ENOUGH - can cause hard starting. (8346) TURBINE FUELS AN ABSORPTION OF HEAT - must accompany fuel vaporization (which can contribute to carburetor icing). (8339) 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) FUEL ADDITIVES TETRAETHYL LEAD IS ADDED TO AVIATION GASOLINE - to improve the gasoline's performance in the engine. (8345) JET FUEL NUMBER IDENTIFIERS - are type numbers and have no relation to the fuel's performance in the aircraft engine. (8342) ETHYLENE DIBROMIDE IS ADDED TO AVIATION GASOLINE - to scavenge lead oxide from the cylinder combustion chambers. (8337) FUEL SYSTEM CONTAMINATION DETONATION JET FUEL IS MORE SUSCEPTIBLE TO CONTAMINATION THAN AVIATION GASOLINE because jet fuel is of higher viscosity and therefore holds contaminants better. (8325) 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. 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) 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. FUELS/ FUEL SYSTEMS - NOTES 28 Return to Table of Contents BASIC ELECTRICITY CURRENT VOLTAGE 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. 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". CURRENT FLOW - is measured in amperes. THE POTENTIAL DIFFERENCE BETWEEN TWO CONDUCTORS - which are insulated from each other is measured in volts. (8020) MILLIAMPERE - is a measure equivalent to 0.001 ampere. of current (8030) ONE VOLT IS THE EMF - required to cause current to flow at the rate of 1 ampere through a resistance of 1 ohm. 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) 1.0 KILOVOLT = 1,000 VOLTS. EXAMPLE - 0.002 KV equals how many volts? (2.0 volts.) (8033) RESISTORS Volts = Kilovolts x 1,000 = 0.002 x 1,000 Volts = 2.0 RESISTANCE is the property of a conductor that limits the flow of electric current. OHM'S LAW THE RESISTANCE OF A CONDUCTOR - will decrease when its length is decreased or its cross-sectional area is increased. (8051) 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. 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. THE OHM - is the unit used to measure resistance. In mathematical formulas, the capital letter "R", for "Resistance", is used. 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. AS AN EQUATION - Ohm's law is expressed as follows: E I= where: R 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. 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: EXAMPLE - Figure 17. Which of the components is a potentiometer? ("3" is a potentiometer.)(8065) E R E=I×R E R= I I= EXAMPLE - Figure 21. Which symbol represents a variable resistor? ("2 " is a variable resistor.) (8074) Return to Table of Contents 29 ELECTRICITY - NOTES THE VOLTAGE DROP IN A CONDUCTOR - of known resistance is dependent on the amperage of the circuit. (8055) E 25.2 = 2.52 ohms = 10 I Rinternal = 2.52 - 2.00 = 0.52 ohms Rt = USING THE OHMMETER PARALLEL CIRCUITS 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) 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. 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". 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. NO MATTER HOW MANY COMPONENTS ARE INCLUDED IN A SERIES CIRCUIT - the current is the same intensity throughout the circuit. EXAMPLE - Figure 11. What is the voltage across the 8-ohm resistor? (24 volts.) (8045) IN A SERIES CIRCUIT - the total resistance is equal to the sum of all the resistance’s in the circuit: R t = R1 + R 2 + R 3 + etc. 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 - 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) EXAMPLE - Figure 13. What is the total current flow in the circuit? (1.4 amperes.) (8049) E 30 30 I = = = = 1 ampere R 10 + 20 30 E = I × R = 1 × 10 = 10 volts 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 t = R1 + R 2 + R 3 = 3 + 5 + 22 = 30 IR2 IR3 THE FORMULA FOR THE TOTAL RESISTANCE IN A PARALLEL CIRCUIT - is: 1 Rt = 1 1 1 + + R1 R2 R3 E 28 I= = = 0.93 ampere R 30 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) 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) Total Voltage = 12 x 2.1 = 25.2 volts ELECTRICITY - NOTES 12 volts = 0.4 amperes 30 ohms 12 volts = = 0.2 amperes 60 ohms 12 volts = = 0.8 amperes 15 ohms Total = 14 . amperes IR1 = 30 Return to Table of Contents 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) 1 Rt = 1 1 + R1 R2 1 Rt = 1 1 + 20 20 1 Rt = 2 20 1 Rt = 1 10 1 Rt = 1 ÷ 10 10 Rt = 1× 1 Rt = 10 ohms EXAMPLE - Figure 6. If resistor R5 is disconnected at the junction of R3 and R4, what will the ohmmeter read? (3 ohms.) (8025) 1 Rt = 1 1 1 + + R1 R2 R3 + R4 1 Rt = 1 1 1 + + 12 6 6 + 6 1 Rt = 1 1 1 + + 12 6 12 1 Rt = 1 2 1 + + 12 12 12 1 Rt = 4 12 1 Rt = 1 3 1 Rt = 1 ÷ 3 3 Rt = 1 × 1 R t = 3 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) 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: Rt Rt Rt Rt Rt I I I Return to Table of Contents 1 1 1 1 + + 4 6 12 1 = 3 2 1 + + 12 12 12 1 = 6 12 1 = 1 2 1 = 1÷ 2 2 = 1× 1 = 2 ohms Rt = 1 1 1 1 1 1 + + + + 6 6 6 5 5 1 = 3 2 + 6 5 1 = 1 2 + 2 5 1 = 0.5 + 0.4 1 = 0.9 . ohms = 111 E = R 28 = 111 . = 25.23 amperes Rt = Rt Rt Rt Rt Rt Rt 31 ELECTRICITY - NOTES 2. Total resistance in the remaining series circuit is: 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) Rt = 2 + 5 + 10 = 17 ohms EXAMPLE - Figure 12. What is the total resistance of the circuit? (21.2 ohms.) (8046) The total resistance between points C,D,G and C,D,E,H is: The total resistance can be found in four steps: Rt = 1 1 + R 2 R3 1 Rt = 1 1 + 10 40 1 Rt = 4 1 + 40 40 1 Rt = 5 40 5 Rt = 1 ÷ 40 40 Rt = 1 × 5 R t = 8 ohms E I= R 24 I= 8 I = 3 amperes 1. R4 and R5 are in parallel: 1 1 1 + 12 6 1 = 1 2 + 12 12 1 = 3 12 1 = 1 4 1 = 1÷ 4 4 = 1× 1 = 4 ohms Rt = Rt Rt Rt Rt Rt Rt 2. The above 4 ohm combined resistance is in series with R2 for a total of 16 ohms: DIODES Rt = R2 + 4 = 12 + 4 = 16 ohms 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. 3. The above 16 ohms is in parallel with R3 giving a total of 3.2 ohms: 1 Rt = 1 1 + 12 4 1 Rt = 1 4 + 16 16 1 Rt = 5 16 5 Rt = 1 ÷ 16 16 Rt = 1× 5 Rt = 3.2 ohms 4. This 3.2 ohms is in series with R1 for a total circuit resistance of 21.2 ohms: ELECTRICITY - NOTES 1 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) 32 Return to Table of Contents 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) 12 = 1 3 R 3 R R 12 = 1 × 3 R 12 = 3 R = 36 ohms 12 = 1 ÷ 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. EXAMPLE - What is the operating resistance of a 30-watt light bulb designed for a 28-volt system? (26 ohms.) (8038) P 30 I= = = 107 . amperes E 28 E 28 = 26.17 ohms R= = I 107 . 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. 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.) WHEN THE CURRENT - in amperes (I) is multiplied by the emf in volts (E), the result is power measured in watts (P): P=IxE GIVEN: Rt = THE UNIT USED TO EXPRESS - electrical power is the watt. (8037) E2 P (8035) SOLUTION: E2 (24) = = 12 ohms P 48 1 Rt = 1 1 + R R 1 Rt = (1+ 1) 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) 2 Rt = A CABIN ENTRY LIGHT - of 10 watts and a dome light of 20 watts are connected in parallel to a 30volt source. If the voltage across the 10-watt light is measured, it will be equal to the voltage across the 20-watt light. (8031) 1 Rt = 2 R 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) R 2 R R Rt = 1× 2 R Rt = 2 R 12 = 2 R = 24 ohms Rt = 1÷ P 192 = = 4 amperes E 48 E 48 = 12 ohms Rt = = I 4 1 Rt = 1 1 1 + + R R R I= EXAMPLE - How much power must a 24-volt generator furnish to a system which contains the following loads? (8016) Return to Table of Contents 33 ELECTRICITY - NOTES UNIT One motor (75% efficient) Three position lights One heating element One anticollision light (Note: 1 hp = 746 watts.) RATING 1 HP × 746 watts / HP 75% P = 994.7 watts P= 1/5 hp 20 watts each 5 amp 3 amp 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 (8032) dissipate? (At least 35 milliwatts.) 1 746 × = 199 watts 5 75% 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 Motor: P = 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) EXAMPLE - Which of the following will require the most electrical power? (8015,8036) E = IR = 23 x 5 = 115 volts P = IE = 23 x 115 = 2,645 watts (Note: 1 hp = 746 watts.) EXAMPLE - How much current does a 30-volt motor, 1/2 horsepower, 85 percent efficient, (8431) draw from the bus? (14.6 amperes.) 1. Four 30-watt lamps arranged in a 12-volt parallel circuit. P = 4 x 30 = 120 watts (Note: 1 horsepower = 746 watts) 2. A 1/5-horsepower, 24-volt motor which is 75 percent efficient. 1 746 P= × = 199 watts 5 75% 1 746 × = 438.8 watts 2 85% P 438.8 I = = = 14.6 amperes E 30 P= 3. A 24-volt anticollision light circuit consisting of two light assemblies which require 3 amps each during operation. SUMMARY OF EQUATIONS OHM'S LAW E I= R E = I×R E R= I 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) RESISTORS R t = R1 + R2 + R3 + ect. (series ) 1 (parallel) Rt = 1 1 1 + + R1 R 2 R3 EXAMPLE - A 1-horsepower, 24-volt dc electric motor that is 80 percent efficient requires 932.5 watts. How much power will a 1horsepower, 12-volt dc electric motor that is 75 percent efficient require? (994.7 watts.) (8019) POWER P = I× E P I = E (Note: 1 horsepower = 746 watts ELECTRICITY - NOTES 34 Return to Table of Contents ALTERNATING CURRENT CIRCUITS ALTERNATING CURRENT - is defined as current which periodically changes direction and continuously changes in magnitude. 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) 1 LT = 1 1 1 + + ... L1 L 2 L 3 THERE ARE THREE VALUES - of alternating current which should be considered. They are: Instantaneous, Maximum, and Effective. INDUCTIVE REACTANCE THE INSTANTANEOUS VALUE - is the value of the induced voltage or current flowing at any instant. THE MAXIMUM VALUE instantaneous value. - is the 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. largest INDUCTIVE REACTANCE - is the opposition offered by a coil to the flow of alternating current (disregarding resistance). (8004) THE EFFECTIVE VALUE - of alternating current is the same as the value of a direct current which can produce an equal heating effect. AN INCREASE IN INDUCTANCE AND FREQUENCY - will cause an increase in the inductive reactance of a circuit. (8005) IN AN AC CIRCUIT - the effective voltage is less than the maximum instantaneous voltage. (The effective is 0.707 times the maximum.) (8007) CAPACITORS UNLESS OTHERWISE SPECIFIED - any values given for current or voltage in an ac circuit are assumed to be effective values. (8010) A CAPACITOR CONSISTS OF TWO CONDUCTORS - capable of holding an electric charge separated by an insulating medium. LINES OF FORCE 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) MAGNETIC LINES OF FORCE - will pass most readily through iron (when compared with copper or aluminum). (8052) INDUCTANCE THE WORKING VOLTAGE OF A CAPACITOR in an ac circuit should be at least 50 percent greater than the highest applied voltage. (8001) 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. 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? (8066) (A variable capacitor.) TRANSFER OF ELECTRICAL ENERGY - from one conductor to another without the aid of electrical connections is called induction. (8041) 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) 1 CT = 1 1 1 + + ... C1 C 2 C 3 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 ... Return to Table of Contents 35 ELECTRICITY - NOTES 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 (8009) microfarad, respectively? (0.0125 µF.) IMPEDANCE IMPEDANCE - which is the combination of resistance, inductive reactance, and capacitive reactance is the total opposition to current in an ac circuit. GIVEN: CT = THE UNIT FOR THE MEASURE OF IMPEDANCE is the ohm. 1 1 1 1 + + C1 C2 C3 THE TERM THAT DESCRIBES THE COMBINED RESISTIVE FORCES - in an ac circuit is impedance. (8002) SOLUTION: 1 1 1 1 + + 0.02 0.05 0.10 1 CT = 50 + 20 + 10 1 CT = 80 C T = 0.0125 µF 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 (8024) ohms.) CT = GIVEN: Z = R 2 + (XL − X C )2 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 ... Z = Impedance R = Resistance XL = Inductive reactance XC = Capacitive reactance SOLUTION: 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? (8014) (0.40 µF.) Z = 8 2 + (10 − 4) 2 Z = 64 + 6 2 Z = 64 + 36 GIVEN: CT = C1 + C2 + C3 Z = 100 Z = 10 ohms SOLUTION: CT = 0.25 + 0.03 + 0.12 CT = 0.40 µF POWER TRUE POWER - is the power consumed by the resistance of an ac circuit. CAPACITIVE REACTANCE APPARENT POWER - is the power consumed by the entire ac circuit. CAPACITIVE REACTANCE - like inductive reactance, offers opposition to the flow of current and is also measured in ohms. ELECTRICITY - NOTES WHEN CALCULATING POWER - in a reactive or inductive ac circuit, the true power is less than the apparent power. (8022) 36 Return to Table of Contents BATTERIES BATTERIES MAY BE CONNECTED - in parallel, in series, or in a combination of parallel and series. NICKEL-CADMIUM BATTERIES THE ELECTROLYTE OF A NICKEL-CADMIUM BATTERY - is lowest when the battery is in a discharged condition. 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. THE ELECTROLYTE IS HIGHEST IN A NICAD BATTERY – when the battery is in a fully charged condition. (8096) FOR BATTERIES CONNECTED IN SERIES (positive to negative to positive to negative, etc.) the voltages are added to obtain the total voltage. 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) EXAMPLE - Figure 10. What is the measured voltage of the series-parallel circuit between (8034) terminals A and B? (3.0 volts.) 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) 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 battery can be determined discharge. (That is, discharge specified rate and measure capacity, then recharge it.) 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) THE SERVICING AND CHARGING OF NICKELCADMIUM - 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) 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) THERE ARE TWO METHODS - of battery charging: 1. 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) CONSTANT VOLTAGE constant, current varies. - voltage USED IN THE AIR - generator puts out constant voltage. ADVANTAGE - rapid charging rate. DISADVANTAGE - high charging rate could cause over heating (NICAD). 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) 2. CONSTANT CURRENT constant, voltage varies. - current ADVANTAGE - controls charging rate. DISADVANTAGE - lower rate of charge. 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) Return to Table of Contents a nickel-cadmium by a measured the battery at a its ampere-hour (8099) THE METHOD USED - to rapidly charge a nickelcadmium battery utilizes constant voltage and varying current. (8091) 37 ELECTRICITY - NOTES 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. 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 19CELL 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) CONSTANT CURRENT METHOD. Batteries connected in series can be of different voltages. Must be about the same capacity. IF BATTERIES ARE SAME VOLTAGE CAPACITY- either method may be used. IN NICKEL-CADMIUM BATTERIES - a rise in cell temperature causes a decrease in internal resistance. (8101) AND THERMAL RUNAWAY - is the result of the high temperatures associated with nickel-cadmium batteries. (8550) 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 (8090) method? (Yes.) HEAT OR BURN MARKS ON THE HARDWARE is an indication of improperly torqued cell link connections of a nickel-cadmium battery. (8093) 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 (8090) method? (Yes.) 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) 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 (8090) method? (No.) 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 the battery provides an unintended second path for the current. The other voltmeter is installed with reversed polarity. 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. 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 (8073) across the lamp). 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) 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. EXAMPLE - Figure 9. How many instruments (voltmeters and ammeters) are installed correctly? (8028) (Two.) LANDING GEAR CIRCUITS 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 ELECTRICITY - NOTES IN AN ELECTRICAL CIRCUIT DIAGRAM - the ground reference point is considered to be at zero voltage. (8072) 38 Return to Table of Contents A GROUND SYMBOL – in electrical circuit diagrams shows that there is a return path between the source of electrical energy and the load. (8604) 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). EXAMPLES - Figure 18. 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. (8071) 5, No. 6, or No. 7? (Wire No. 6.) 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 (8068) the throttles are closed? (Yes.) 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. 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. 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. EXAMPLES - Figure 15. 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. (8067) 2, No. 4, or No. 9? (Wire No. 4.) 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 5amp circuit breaker, and wires #18 and #17. 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. 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. Wire #2 is in the red warning light circuit. 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? (8057) (Wire No. 19.) Wire #9 is in the landing gear green light circuit. EXAMPLES - Figure 19. Under which condition will a ground be provided for the warning horn through both gear switches when the throttles are closed? (8069) (Left gear up and right gear down.) 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. 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. 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 throttles are retarded with only the right gear down, the warning horn will not sound if an open occurs in which wire: No. 5, (8070) No. 6, or No. 13? (Wire No. 5.) 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. 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. Wires No. 7 and No. 17 supply power to the green indicator light push-to-test function. Wire #6 is out of the warning horn circuit when the right gear switch is in the Down position. Return to Table of Contents 39 ELECTRICITY - NOTES RELAY ACTIVATION TABLE RELAYS LTS RTS FCF PCO PCC TCO TCC FUEL PRESS SOLENOID FUEL TANK SOLENOID POWER ON RIGHT TANK (8062) POWER ON LEFT TANK (8064) NO YES NO YES NO NO YES YES NO YES NO NO YES NO NO YES YES NO NO NO NO NO YES NO YES NO NO When the circuit is energized with the fuel tank selector switch selected to the left-hand position, what switches will change position? (Switches (8064) 5,6,11,12,13,15, and 16.) FUEL SYSTEM CIRCUITS EXAMPLES - Figure 16. When electrical power is applied to the bus, which relays are energized? (Relays PCC and (8063) TCC are energized.) 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 (8060) will not illuminate.) 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. 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 (8062) operated.) 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. ELECTRICITY - NOTES POWER ON NORMAL (8063) 40 Return to Table of Contents Even if relay PCO fails to operate, the rest of the system will operate normally. 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. 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. 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) THERMAL SWITCH 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) A THERMAL SWITCH - as used in an electric motor, is designed to open the circuit in order to allow cooling of the motor. (8056) 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) 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. 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. 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) If R2 sticks in the up position, how will the light be affected? (The light will be on full bright.) (8081) 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) 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. 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 (8084) that arrangement.) Return to Table of Contents 41 ELECTRICITY - NOTES INSPECTION FUNDAMENTALS 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) MECHANIC 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) 65.21 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) 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) 1.1 THE DEFINITION OF MAINTENANCE - is overhaul, repair, parts replacement, inspection, and preservation. (8583) 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) 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) EXAMPLE – Why should an aircraft maintenance technician be familiar with weld nomenclature? (To compare welds with written (non-pictorial) description standards.) (8606) 100-HOUR INSPECTIONS 65.13, 65.14 A TEMPORARY CERTIFICATE - is valid for 120 days. (8589) 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) 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) 65.85, 65.87 A CERTIFICATED AIRFRAME AND POWERPLANT 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) 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) 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) 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 INSPECTION FUNDAMENTALS - NOTES 42 Return to Table of Contents 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) 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) 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) 65.81 CERTIFICATED MECHANICS - under their general certificate privileges, may perform 100hour inspection of instruments (but not repair). (8525) THE REPLACEMENT OF A DAMAGED VERTICAL STABILIZER - with a new identical stabilizer purchased from the aircraft manufacturer is considered a minor repair. (8527) 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) 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) 43.11(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) 43.9 WHEN A MINOR REPAIR IS PERFORMED - on a certificated aircraft, an entry in the aircraft's permanent records is required. (8457) 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) 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 POWERPLANT RATING - may not perform a major repair to a propeller even if they have the necessary equipment available. (8532,8591) 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) FAR 43, Appendix A(b) THE REPLACEMENT OF FABRIC ON FABRICCOVERED 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) 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 THE REPAIR OF PORTIONS OF SKIN SHEETS - by making additional seams (the splicing of skin sheets) is classified as a major repair. (8449,8522) 43.13 A REPAIR, AS PERFORMED ON AN AIRFRAME - means the restoration of the airframe to a OVERHAUL OF A HYDRAULIC PRESSURE PUMP - is considered an appliance major repair. (8447) Return to Table of Contents 43 INSPECTION FUNDAMENTALS - NOTES MAJOR ALTERATIONS 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) 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. DETERMINING INSPECTION INTERVALS EXAMPLES - Figures 62, 62A, and 62B. To which doubler part number(s) is the -100 in the title block (Area 1) applicable? (-101.) (8514) EXAMPLE - Given the statement below, at what intervals should the thrust bearing nut be checked for tightness? (Every 150 hours.) (8518) 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) "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. 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) 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." 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) From the above statement, valve mechanism inspections should be performed at 100 hour intervals. 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) 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) 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) 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) 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) INSPECTION FUNDAMENTALS - NOTES 44 Return to Table of Contents 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) 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: 91.417(a) THE AIRCRAFT OWNER IS RESPONSIBLE - for maintaining the required maintenance records for an airplane. (8459) "Wear within acceptable limits, repair not necessary." (8451) 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) THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES THE ACTION TAKEN - for a 0.125-inch deep dent in a straight section of 1/2inch 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: IF AIRCRAFT MAINTENANCE RECORDS ARE LOST OR DESTROYED, - the records must be reconstructed. In order to do this total time-inservice of the airframe must be established. (8572) "Welded a reinforcing plate over the dented area." (8453) INSTRUMENTS THE MAINTENANCE RECORD ENTRY THAT BEST DESCRIBES THE REPLACEMENT - of several damaged heli-coils in a casting is: 23.1543 INFORMATION REGARDING INSTRUMENT RANGE MARKINGS - for an airplane certificated in the normal category would be provided in FAR Part 23. (8505) "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) 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) 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.) Normal operating speed Never-exceed speed Max. gear operation speed Maximum flap extended speed (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) Return to Table of Contents 260 knots 293 knots 174 knots 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) 45 INSPECTION FUNDAMENTALS - NOTES accumulated before the AD must again be complied with? (186 hours.) (8515) 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) "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. AVIATION MAINTENANCE ALERTS II. Aircraft with 500-hours' through 1,000hours' 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. AVIATION MAINTENANCE ALERTS - provide information about aircraft problems and suggested corrective actions. (8511) AIRWORTHINESS DIRECTIVES 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." 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 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.) 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) PROPELLERS ARE INCLUDED Airworthiness Directive system. An Airworthiness Directive, or AD, requires a specific action in order to comply with it. The action may take the form of an: - in the (8506) (8578) 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) 43.15 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) 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) 91.417, 91.419 AD COMPLIANCE RECORDS - are required to be retained and, when the aircraft is sold, transferred with the aircraft. (8581) AIRWORTHINESS CERTIFICATES 1. 2. 3. 4. Inspection. Part(s) replacement. Design modification. Change in operating procedure(s). 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) 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 INSPECTION FUNDAMENTALS - NOTES 46 Return to Table of Contents 21.179 IF AN AIRWORTHY AIRCRAFT IS SOLD - the Airworthiness Certificate is transferred with the aircraft. (8497) THE FEDERAL AVIATION REGULATIONS REQUIRE APPROVAL - after compliance with the data of a Supplemental Type Certificate. (This is a major alteration.) (8504) TYPE CERTIFICATE DATA SHEETS TECHNICAL STANDARD ORDERS (TSO's) THE TYPE CERTIFICATE DATA SHEET - will reference the required equipment needed to maintain the validity of a standard Airworthiness Certificate. (8159) FAR 21, Subpart O ITEMS MANUFACTURED IN ACCORDANCE WITH A TECHNICAL STANDARD ORDER - still require approval for installation in a particular aircraft. (8493,8504) 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) 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) 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) PROPELLER TYPE CERTIFICATES PLACARDS REQUIRED ON AN AIRCRAFT - are specified in the Aircraft Specifications or Type Certificate Data Sheets. (8502) TECHNICAL DESCRIPTIONS OF CERTIFICATED PROPELLERS - can be found in the Propeller Type Certificate Data Sheets. (8500) THE LOCATION OF THE DATUM - is an item that would be contained in an aircraft Type Certificate Data Sheet. (8495) 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) 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) 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. 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. 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. (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) Return to Table of Contents WET AND DRY PROCESS MATERIALS - are two types of indicating mediums available for magnetic particle inspection. (8228) 47 INSPECTION FUNDAMENTALS - NOTES THE TESTING MEDIUM - that is generally used in a magnetic particle inspection utilizes a ferromagnetic material that has high permeability and low retentivity. (8225) A DISCONTINUITY - may or may not affect the usefulness of a part. (8244) IN USING THIS METHOD OF INSPECTION - the location of the defect is indicated and the approximate size and shape are outlined. FATIGUE CRACKS - give sharp, clear patterns. They are found in highly stressed areas of parts that have been in service. MAGNETIC PARTICLE INSPECTION - is used primarily to detect flaws on or near the surface. (8219) 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) FATIGUE CRACKS 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) INCLUSIONS INCLUSIONS ARE - nonmetallic materials, such as slag materials and chemical compounds, that have been trapped in the solidifying metal. 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) THE PATTERN FOR AN INCLUSION - is a magnetic particle buildup forming parallel lines. (8238) DYE PENETRANT INSPECTION WHEN CHECKING AN ITEM WITH THE MAGNETIC PARTICLE INSPECTION METHOD circular and longitudinal magnetization should be used to reveal all possible defects. (8234) DYE PENETRANT INSPECTION IS A NONDESTRUCTIVE TEST - for defects open to the surface (cracks, etc.) in parts made of any nonporous material. A FLAW THAT IS PERPENDICULAR - to the magnetic field flux lines generally causes a large disruption of the magnetic field. (8235) 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. CIRCULAR MAGNETIZATION - of a part can be used to detect defects parallel to the long axis of the part. (8243) 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) 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) 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. 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) 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. DISCONTINUITIES IN NONDESTRUCTIVE TESTING A DISCONTINUITY - may be defined as an interruption in the normal physical structure or configuration of a part. (8244) INSPECTION FUNDAMENTALS - NOTES 48 Return to Table of Contents A PART WHICH IS BEING PREPARED FOR DYE PENETRANT INSPECTION - should be cleaned with a volatile petroleum-base solvent. (8239) 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. 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) 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. TO DETECT A MINUTE CRACK - using dye penetrant inspection requires a longer-thannormal 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) 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. ULTRASONIC INSPECTION (Processing of the film is not a factor for the exposure.) (8224) ULTRASONIC INSPECTION IS A NONDESTRUCTIVE TESTING METHOD - which employs electronically produced, high-frequency sound waves that will penetrate metals, liquids, and many other materials. 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) ULTRASONIC INSPECTION IS SUITABLE FOR the inspection of most metals, plastics, and ceramics for surface and subsurface defects. (8221) EDDY CURRENT INSPECTION SURFACE CRACKS THE PRINCIPLE OF EDDY CURRENT INSPECTION - is based on determining the ease with which a material will accept induced current. 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) 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) Return to Table of Contents 49 INSPECTION FUNDAMENTALS - NOTES MEASURING DEVICES COMBINATION SET TELESCOPIC GAUGE A COMBINATION SET - is the tool used to find the center of a shaft or other cylindrical work. (Use the center head.) (8295) A TELESCOPIC GAUGE AND MICROMETER can be used for the dimensional inspection of a bearing in a rocker arm. (8303) DIVIDERS MICROMETER CALIPER DIVIDERS DO NOT - provide a reading when used as a measuring device. (8291) A MICROMETER CALIPER - is the precision measuring tool used for measuring crankpin and main bearing journals for out-of-round wear. (8301) A MACHINIST SCALE - is the tool generally used to set a divider to an exact dimension. (8299) TO DETERMINE PISTON PIN OUT-OF-ROUND WEAR - use a micrometer caliper to check different diameter positions on the piston pin. (8307) 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) 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) THICKNESS GAUGE A THICKNESS GAUGE - (feeler gauge) is used to measure the side clearances of piston rings. (8302) 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 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 RELATIVELY NARROW MARKS ON THE THIMBLE - of a micrometer represent onethousandths of an inch (0.001, 0.002, etc.). 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) 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) SMALL-HOLE GAUGE TO ACCURATELY MEASURE THE DIAMETER OF A HOLE APPROXIMATELY 1/4 INCH IN DIAMETER - the mechanic should use a smallhole gauge and determine the size of the hole by taking a micrometer reading of the ball end of the gauge. (8297) MEASURING DEVICES- NOTES 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) 50 Return to Table of Contents ON A VERNIER CALIPER - each large number represents inches (not fractions of inches). 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. Return to Table of Contents 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) 51 MEASURING DEVICES - NOTES GROUND HANDLING, SAFETY, AND SUPPORT EQUIPMENT 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) 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) A HUNG START IN A TURBOJET ENGINE - is often caused by the starter cutting off too soon. (8309) 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) 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) WHEN STARTING AN ENGINE EQUIPPED WITH A FLOAT-TYPE CARBURETOR - with an idle cutoff unit, place the mixture control in the FULL-RICH position. 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) 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) TAXIING AIRCRAFT WHEN FIRST STARTING TO MOVE AN AIRCRAFT WHILE TAXIING - it is important to test the brakes. (8334) PRIMING 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. TO PRIME A HORIZONTALLY OPPOSED FUEL INJECTED ENGINE - place the fuel control lever in the FULL-RICH position. (8316) 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) 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) 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) THE MOST SATISFACTORY EXTINGUISHING AGENT - for use in case of carburetor or intake fire is carbon dioxide. (8313) 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) 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) GROUND EQUIPMENT - NOTES 52 Return to Table of Contents Flashing White - Return to starting point. (8331) TOWING AIRCRAFT Alternating Red and Green - OK to proceed but use extreme caution. (8332) DURING TOWING OPERATIONS - a person should be in the cockpit to operate brakes. (8326) HAND SIGNALS WHEN TOWING AN AIRCRAFT WITH A STEERABLE NOSEWHEEL - the torque-link lock should be set to full swivel. (8310) 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) STOPPING AND TIEDOWN WHEN STOPPING A NOSEWHEEL-TYPE AIRPLANE - after taxiing, the nosewheel should be left pointed straight ahead. (8333) 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) Return to Table of Contents NYLON OR DACRON ROPE - is preferred to manila rope because manila (hemp) rope has a tendency to shrink when it gets wet. (8311) 53 GROUND EQUIPMENT - NOTES APPENDIX 1 Return to Table of Contents 1 CT= 1/C + 1/C + 1/C ... 1 2 3 Figure 1. Equation. Return to Appendix 1 1 CT= 1/C + 1/C + 1/C 1 2 3 Figure 2. Equation. 2 Return to Appendix 1 LT= 1/L + 1/L + 1/L ... 1 2 3 Figure 3. Equation. Return to Appendix 3 23A 5Ω Figure 4. Circuit diagram. 4 Return to Appendix Z = R2+(XL−XC)2 Z = R = XL = XC = Impedance Resistance Inductive reactance Capacitive reactance Figure 5. Formula. Return to Appendix 5 R5 = 6 ohms R4 = 6 ohms R3 = 6 ohms R2 = 6 ohms R1 = 12 ohms Disconnected Ω Figure 6. Circuit diagram. 6 Return to Appendix R1 R2 R3 40Ω 40Ω 40Ω Ω Break A B C Figure 7. Circuit diagram. Return to Appendix 7 D R1 R2 R3 20Ω 20Ω 20Ω Ω Break Figure 8. Circuit diagram. 8 Return to Appendix − A − + V V + + − + − A Figure 9. Circuit diagram. Return to Appendix 9 1.5V 1.5V 1.5V 1.5V − + A B Figure 10. Battery circuit. 10 Return to Appendix E D B A F G Figure 11. Circuit diagram. Return to Appendix R3 = 40 ohms R1 = 8 ohms 24V R2 = 10 ohms C 11 H R2 =12 ohms 1 1/Rb + 1/R3 Rc = R3 = 4 ohms Rb =Ra + R2 24V R5 = 6 ohms R1 =18 ohms R4 = 12 ohms 1 1/R + 1/R Ra = 4 5 Rt = Rc + R1 Figure 12. Circuit diagram. 12 Return to Appendix I1 − + Et 12V It I2 R2 Figure 13. Circuit diagram. Return to Appendix 60Ω 30Ω R1 13 I3 15Ω R3 R1 =5Ω Et = 36V R2 =4Ω R3 =6Ω R4 =12Ω R5 =10Ω Figure 14. Circuit diagram. 14 Return to Appendix Gear switch Up #13 #1 Relay Motor Down #14 #15 Gear safety switch Up limit switch #11 #2 Down limit switch #12 #19 5 #6 NAV. switch bypass relay #10 Throttle switches (closed position) #8 #7 #18 Red Nose gear down switch #5 #4 Left gear Right gear down switch down switch Figure 15. Landing gear circuit. Return to Appendix 15 #17 BUS 20 #3 #16 Horn Green NOTE: Switches shown gear down - on the ground Drawing shown without electrical power to bus BUS 24 VDC 5 All relays spring loaded to position shown 5 Fuel selector switch 4 Norm 1 X-feed RH tank 2 13 Pump X-feed closed 5 14 Relay TCO Relay TCC 15 16 Tank X-feed closed 6 Fuel pressure cross feed valve open LTS relay 7 Fuel tank cross feed valve open 8 Caution warning lights in cockpit RTS relay 9 B A 11 Close 18 17 10 FCF relay Fuel press X-feed valve Tank X-feed open Relay PCC LH tank 3 Pump X-feed open Relay PCO 5 C D D C 12 Fuel tank X-feed valve Open FCF A 19 Open B 20 Close Figure 16. Fuel system circuit. 16 Return to Appendix 5 6 2 7 4 3 8 9 10 + G 11 − Figure 17. Electrical symbols. Return to Appendix 17 1 28 V DC #13 #1 R #8 G #12 #7 Warning horn #9 Throttle switch (open) #2 Neutral #6 #5 #4 Down #3 #14 Down #10 #11 Right gear switch Control valve switch Left gear switch Figure 18. Landing gear circuit. 18 Return to Appendix 28 V DC #1 R G #2 #3 #9 Test #10 Warning horn #8 Neutral #4 Control valve switch #7 Down #12 Down Right gear switch #5 Left gear switch #6 Figure 19. Landing gear circuit. Return to Appendix #11 #13 19 Throttle switches Open V Figure 20. Circuit diagram. 20 Return to Appendix 1 2 3 Figure 21. Electrical symbols. Return to Appendix 21 Current flow Emitter Collector − + Current flow − Emitter + Collector + Base − Base 1 2 Current flow Emitter Collector − + − Base 3 Figure 22. Transistors. 22 Return to Appendix Emitter Base Collector Up position Down position Figure 23. Transistorized circuit. Return to Appendix 23 12V R2 R1 Inputs Output Figure 24. Logic gate. 24 Return to Appendix Inputs Output Figure 25. Logic gate. Return to Appendix 25 1 1 1 1 1 0 Inputs Output Inputs 1 Output 2 0 0 1 Inputs Output 3 Figure 26. Logic gate. 26 Return to Appendix + 1 2 3 Figure 27. Object views. Return to Appendix + 27 Front view 1 2 3 Figure 28. Object views. 28 Return to Appendix Front view 1 2 3 Figure 29. Object views. Return to Appendix 29 Front view 1 2 3 Figure 30. Object views. 30 Return to Appendix 1" 2 1 1 1" 2 1" 2 1" 12 3" 4 3" 3 4 Figure 31. Sketches. Return to Appendix 31 Figure 32. Sketches. 32 Return to Appendix 1 2 3 Figure 33. Material symbols. Return to Appendix 33 1/16 R. +.005 .3125−.000 DIA. 1/16x45° 7/8 DIA. 3/32 1 3/4 1/2 DIA. .221±.003 7/8 15/32 .665±.001 DIA. 19/32 19/64 .110±.001 12/32 R. Spherical 1/16 R. Figure 34. Aircraft drawing. 34 Return to Appendix 1 3 0.25 Paint stripe 2 4 Figure 35. Aircraft drawing. Return to Appendix 35 ¼" F Notes: 1. Drill 31/64 inch ream ½ inch. 2. All tolerances ±1/32 unless otherwise specified. 3. Finish all over 25 C ¼" Note 1. Note 1. E B D G ¾" 1½" J ¼" 1" ¾" ¾" +1/64" 17" −3/64" H A Figure 36. Aircraft drawing. 36 Return to Appendix 3 8 15 64 Drill 7 8 4 Holes 1 3 16 11 16 7 8 1 8 1 2 Drill Figure 37. Aircraft drawing. Return to Appendix 37 21 2 32 Brake - Horsepower (BHP) 0 500 1000 1500 2000 2500 3000 3500 4000 00 00 28 00 26 00 24 0 0 22 00 20 00 18 00 16 40 18 20 30 00 13 985 30 0 80 D CI 0 140 2 0 120 eed e sp 1000 RPM in Eng 0 25 50 CID = Cubic inch displacement 75 100 125 150 175 200 225 250 Brake Mean Effective Pressure (BMEP) Figure 38. Performance chart. 38 Return to Appendix 275 Circuit Voltage Wire length in feet for allowable voltage drop 115 14 28 800 200 100 200 600 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 28 49 24 42 20 4 Electric Wire Chart 20 18 16 14 12 10 8 6 5 1. 4 3 2 5 6 7 8 9 10 15 35 2 5 7 .5 1 30 0 5 0 40 50 60 70 80 90 10 12 155 170 20 ve 1 Cur ve 2 Cur e3 v Cur 20 18 16 14 12 Curves: 1. Continous rating-amperes cables in conduit and bundles 2. Continous rating-amperes single cable in free-air 3. Intermittent rating-amperes maximum of 2 minutes. 10 8 Wire Size Voltage Drop Figure 39. Electric wire chart. See FAA Replacement Figure 1, Addendum A Return to Appendix 20 0 30 0 40 7 6 2 1 1/0 2/0 3/0 4/0 Amperes 1 8 3 4 39 6 4 2 1 1/0 2/0 3/0 4/0 Values Include 10 Percent Structural Deflection 340 n sig De d oa gl t ri i lim 320 300 280 260 240 200 180 Cable sizes 1/4 7x19 3/16 7x19 5/32 7x19 1/8 7x19 3/32 7x7 1/16 7x7 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 Temperature in degrees Fahrenheit Figure 40. Cable tension chart. 40 Return to Appendix Rigging load in pounds 220 Normal rated power 180 170 Full throttle power 160 150 Brake Horsepower 140 Propeller load horsepower 130 120 110 100 90 Full throttle spec. fuel cons 70 .55 60 .50 50 .45 Propeller load spec. fuel cons. 1800 2000 2200 2400 Engine Speed - RPM Figure 41. Performance chart. Return to Appendix 41 2600 2800 Specific fuel cons LB/BHP/HR .60 80 1 2 3 Figure 42. Aircraft hardware. 42 Return to Appendix 1 2 3 Figure 43. Aircraft hardware. Return to Appendix 43 1 2 3 4 Figure 44. Welds. 44 Return to Appendix E F G A Figure 45. Welds. Return to Appendix C D 45 B 765 20 4 32 1 0 0 1 15 2 10 Figure 46. Precision measurement. See FAA Replacement Figure 2, Addendum A 46 Return to Appendix 2 4 5 0 6 5 7 8 10 15 1 Inch 1000 Figure 47. Precision measurement. Return to Appendix 47 9 20 1 25 15 0 9 8 7 6 5 4 3 2 1 0 10 5 0 1 2 3 0 20 Figure 48. Precision measurement. 48 Return to Appendix 15 765 4 32 1 0 0 1 10 2 5 Figure 49. Precision measurement. See FAA Figure Replacement Figure 3, Addendum A Return to Appendix 49 1 2 3 Figure 50. Marshalling signals. 50 Return to Appendix 1 2 3 Figure 51. Marshalling signals. Return to Appendix 51 (√(−4)0+6+(√4 1296)(√3)2= Figure 52. Equation. 52 Return to Appendix 2 2 √31 + √43 = 2 (17) Figure 53. Equation. Return to Appendix 53 9' 5' 12' Figure 54. Trapezoid area. 54 Return to Appendix 3" 7.5 " 4" Figure 55. Trapezoid area. See FAA Replacement Figure 4, Addendum A Return to Appendix 55 4' 2' 6' Figure 56. Trapezoid area. 56 Return to Appendix B C A D Figure 57. Trapezoid area. See FAA Replacement Figure 57, Addendum A Return to Appendix 57 -2 (-35 + 25) (-7) + (π) (16 ) = √25 Figure 58. Equation. 58 Return to Appendix -4 125 = -6 -36 Figure 59. Equation. Return to Appendix 59 -3 (-5+23)(-2)+(3 )(√64)= -27÷9 Figure 60. Equation. 60 Return to Appendix A 60 POUNDS Figure 61. Physics. Return to Appendix 61 4 8 4 5 37 2 1 Area 1 4 8 4 5 37 2 -100 -200 1 Rivet Rivet Domed Nutplate Rivet Rivet Clip Doubler Doubler MS20470AD-4-4 NAS1097-3-4 NAS1473-3A NAS1097-4-5 NAS1097-4-4 -103 -102 -101 Part number REV. .040 sheet .040 sheet .040 sheet stock size 2024-T3 CLAD AL. 7075-0 AL. 2024-T3 CLAD AL. MAT’L DESCR MAT’L Zone NAME SPEC. DASH NUMBERS SHOWN DASH NUMBERS OPPOSITE UNIT WT. DWG. AREA All N/A FIRST RELEASE For continuation see zone Unless otherwise noted REQ’d. PER ASSEM. B ADD-200 A MAT’L THKNESS LET. CHANGE 1 1 1 1 BY Date Appr. B Break all sharp edges No. req. per Scale full Airplane 992-148-XXX The use of this document shall be restricted to conveyance of information to customers of vendors only. Neither classified nor unclassified documents may be reproduced without the written consent of THE SPEEDWIND AIRCRAFT CORP. -200 36TCP 001-All -200 36P 088-All -200 36P 001-087 Type EFF A/C PROJECT T. Smith DESIGN R. Eamer Engineer FAA D.E.R G. Winn DWG. Checker I. Wright DFTSMIN. S. Linz Speedwind aircraft engineering section last chance airport anytown OK 73125-1234 Figure 62. Maintenance data - part 1 of 3. 62 Return to Appendix AREA 2 GENERAL NOTES - 100 1. 2. 3. 4. 5. ALL BENDS +/– .5 deg. All holes +/– .003. Apply Alodine 1,000. Prime with MIL-P-23377 or equivalent. 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 3 GENERAL NOTES - 200 Note: P.S. = Process Specification IAW = in accordance with 1. 2. 3. 4. 5. 6. 7. ALL BENDS IAW P.S. 1,000. All holes IAW P.S. 1,015. Heat treat –102 to –T6 IAW P.S. 5,602. Alodine IAW P.S. 10,000. Prime IAW P.S. 10,125. 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. 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. Figure 62A. Maintenance data - part 2 of 3. Return to Appendix 63 Figure 62B. Maintenance data - part 3 of 3. R-3T min. Joggle as necessary 0.50 4.980 3.00 3.70 2.30 0.1875 4PL 1.0 Ø 0.6250 Ø 4.50 2.40 2.0 1.0 –103 Trim as req’d for inst’l Size this area as required to clear S-1C. + 14.25 R-3T min. × Area 4 –101,–102 × × × × × × × × × × × × + × × × × × × × × × × × × × × × × × × × × × × × × × × × × × Station 355.750 × × × × × View looking up 62 × × × × × Centerline Stringer 1 Center (S-1C) × × Station 370.250 Return to Appendix 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 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.” Figure 63. Airworthiness directive excerpt. Return to Appendix 65 Rt = E2/P Figure 64. Resistance total. 66 Return to Appendix 1. 3.47 × 104 = 34,700. 2. 2(410) = 2,097,152. Figure 65. Scientific notation. Return to Appendix 67 –4 + 6 + 10 (√1296) = 3 Figure 66. Equation. 68 Return to Appendix √31 + √43 = 2 (17) Figure 67. Equation. Return to Appendix 69 1. (4.631)5 2. 4.631 × 105 3. 4.631 ×10-5 Figure 68. Alternative answer. 70 Return to Appendix (√100 + √36 – √16) = Figure 69. Equation. Return to Appendix 71 1. (√31) + (√43) ÷ 172 2. (√31) + √43) ÷ 172 3. (√31) + (√43) – 172 Figure 70. Alternative answer. 72 Return to Appendix V= 1/6πD3 Figure 71. Volume of a sphere. Return to Appendix 73 ADDENDUM A Return to Table of Contents Electric Wire Chart Circuit Voltage Wire length in feet for allowable voltage drop 115 200 14 28 800 100 200 600 75 150 400 360 320 280 700 630 560 490 50 45 40 35 100 90 80 70 240 420 30 60 200 350 25 50 160 280 20 40 120 210 15 30 100 175 12 25 80 72 64 56 140 120 112 98 10 9 8 7 20 18 16 14 48 84 6 12 40 36 32 28 24 70 63 56 49 42 5 4 3 10 9 8 7 6 20 35 2 5 4 7 .5 1 20 18 16 1 5 1. 14 12 8 6 4 2 1 1/0 2/0 3/0 4/0 Amperes 3 2 4 5 6 7 8 9 10 15 20 30 0 5 0 40 50 60 70 80 90 10 12 155 170 20 0 30 0 40 ve 1 Cur ve 2 Cur ve 3 Cur 20 18 16 14 12 Voltage Drop Curves: 1. Continuous rating-amperes cables in conduit and bundles 2. Continuous rating-amperes single cable in free-air 3. Intermittent rating-amperes maximum of 2 minutes 10 8 Wire Size Figure 1. Electric wire chart. (Replaces Figure 39 from Appendix 1.) Figure 39 Return to Addendum A 10 1 6 4 2 1 1/0 2/0 3/0 4/0 765 20 4 32 1 0 4 2 0 1 20 15 2 10 Figure 2. Percision measurement. (Replaces Figure 46 from Appendix 1.) Figure 46 2 Return to Addendum A 4 10 2 0 5 Figure 3. Precision measurement. (Replaces Figure 49 from Appendix 1.) Figure 49 Return to Addendum A 3 3"3" 7.5 7.5 " " 4"4" Figure 4. Triangle area. (Replaces Figure 55 from Appendix 1.) Figure 55 4 Return to Addendum A B C A D Figure 5. Triangle area. (Replaces Figure 57 from Appendix 1.) Figure 57 Return to Addendum A 5