A flight benefits from a strong tail wind which was not forecast On arrival at destination a straight in approach and immediate landing clearance is ?
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The centre of gravity of an aeroplane is at 25% of the Mean Aerodynamic Chord This means that the centre of gravity of the aeroplane is situated at 25% of the length of ?
To measure the mass and CG position of an aircraft it should be weighed with a minimum of ?
An aeroplane has a mean aerodynamic chord mac of 134 5 inches the leading edge of this chord at a distance of 625 6 inches aft of datum define location of centre of gravity of aeroplane in terms of percentage mac if mass of aeroplane acting vertically through a balance arm located 650 inches aft of datum The mean aerodynamic chord in relation to leading edge. Leading edge mac at 625 6 inches aft of datum the centre of gravity at 650 inches aft of datum 650 625 6 = 24 4 inches from leading edge mac (24 4 / 134 5) x 100 = 18 10 %. 
The empty mass of an aircraft recorded in The weighing schedule is amended to take account of changes due to modifications of aircraft. The empty mass of an aircraft recorded in weighing schedule is amended to take account of changes due to modifications of aircraft normally aircraft weighed before entering in service every 4 years individual aircraft every 9 years fleet masses every time a modification takes place (equipment installed or uninstalled) as weighing procedure long complex operator allowed to adjust empty mass of an aircraft on weighing schedule between to weighing procedure. 
The maximum quantity of fuel that can be loaded into an aircraft's tanks given as 400 us gallons if fuel density specific gravity given as 0 79 mass of fuel which may be loaded The weighing schedule is amended to take account of changes due to modifications of aircraft. 1 ug gallons = 3 7854 litres 400 usg x 3 7854 = 1514 litres 1514 x 0 79 = 1196 kg. 
The maximum quantity of fuel that can be loaded into an aircraft's tanks given as 2200 l if fuel density specific gravity given as 0 79 mass of fuel which may be loaded The weighing schedule is amended to take account of changes due to modifications of aircraft. 2200 x 0 79 = 1738 kg. Question 69-8
The basic empty mass 4960 kg dry operating mass 5220 kg and zero fuel mass 6040 kg if take off mass 7630 kg useful load The weighing schedule is amended to take account of changes due to modifications of aircraft. Useful load = maximum take off mass dry operating mass useful load = 7630 kg 5220 kg = 2410 kg. Question 69-9
The basic empty mass 4800 kg dry operating mass 5050 kg and zero fuel mass 6210 kg if take off mass 8010 kg useful load The weighing schedule is amended to take account of changes due to modifications of aircraft. Useful load = the mass of aircraft including everything everyone contained within it at commencement of take off the total mass of an aircraft ready a specific type of operation excluding all usable fuel traffic load useful load = take off mass dry operating mass useful load = 8010 kg 5050 kg = 2960 kg. Question 69-10
Longitudinal cg location normally expressed As a percentage of mac from its leading edge. Useful load = the mass of aircraft including everything everyone contained within it at commencement of take off the total mass of an aircraft ready a specific type of operation excluding all usable fuel traffic load useful load = take off mass dry operating mass useful load = 8010 kg 5050 kg = 2960 kg. Question 69-11
Given basic empty mass bem 1 200 kgbalance arm of bem 3 0 mcentre of gravity at bem 25% of mac length of mac 2 0 m in mass and balance section of flight manual following information given position arm front seats 2 5 mrear seats 3 5 mrear hold 4 5 mfuel tanks 3 0 m the pilot and one passenger embark each weighs 80 kg fuel tanks contain 140 litres of petrol with a density of 0 714 kg/l the rear seats are not occupied taxi fuel negligible the position of centre of gravity at take off As a percentage of mac from its leading edge. Com_en/annexe php?annexe=com031 226 jpg basic mass cg was at 25% of mean aerodynamic chord (mac) with an arm of 3 m the length of mac 2m so cg at 0 5 m from leading edge mac take off mass cg = moment/mass = 4300 / 1460 = 2 945 m with a cg located at 2 945 m take off it at 0 445 m from leading edge mac 0 445/2 x 100 = 22 25% mac hwoarangitf please how do you found '0 445'm from leading edge mac? thank you at basic mass cg at 25% of mac a chord length of 2m 3m aft of datum 25% of 2m 50 cm thus leading edge of mac at 2 5m aft of datum cg location at take off will be at 2 945 2 5m = 0 445m aft of leading edge of mac com_en/annexe php?annexe=com031 226b jpg. Question 69-12
Given following data distance from datum to centre of gravity 12 53 mdistance from datum to leading edge 9 63 mlength of mac 8 00 mcalculate centre of gravity in % mac mean aerodynamic chord As a percentage of mac from its leading edge. Position of cg on mac = distance from datum to centre of gravity distance from datum to leading edge position of cg on mac = 12 53 9 63 = 2 9 m 8 m the lenght of mac center of gravity located on mac at 2 9 m 8 m 100% of length 2 9m at ? 2 9 x 100 / 8 = 36 25 %. Question 69-13
At reference to determine dry operating index doi a dom of 35000kg and a %mac of 14% 229 As a percentage of mac from its leading edge. Position of cg on mac = distance from datum to centre of gravity distance from datum to leading edge position of cg on mac = 12 53 9 63 = 2 9 m 8 m the lenght of mac center of gravity located on mac at 2 9 m 8 m 100% of length 2 9m at ? 2 9 x 100 / 8 = 36 25 %. Question 69-14
Which of following combinations of compartment centroid and maximum load correct 230 As a percentage of mac from its leading edge. take care we are talking in inches in kilograms. Question 69-15
Given following data calculate cg as a %mac mean aerodynamic cord when 12000 n of last minute cargo added to a hold 10 m from datum aum all up mass 460000 nlemac leading edge mac 14 m from datummac 4 6 mcurrent cg 15 4 m from datum As a percentage of mac from its leading edge. mass x arm = moment 460000 n x 15 4 m = 7084000 nm 12000 n of last minute cargo added to a hold 10 m from datum 12000 n x 10 m = 120000 nm new mass is 472000 n new moment is 7204000 nm cg position now 7204000 / 472000 = 15 26 m in percentage of mean aerodynamic chord it lemac at 14 m therefore cg 1 26m from lemac 1 26 / 4 6 x 100 = 27 5% mac. Question 69-16
What are standard masses used crew 85 kg flight crew 75 kg cabin crew including hand baggage. Eu ops 1 615 mass values crew (a) an operator shall use following mass values to determine dry operating mass 1 actual masses including any crew baggage or 2 standard masses including hand baggage of 85 kg flight crew members 75 kg cabin crew members or 3 other standard masses acceptable to authority b) an operator must correct dry operating mass to account any additional baggage the position of this additional baggage must be accounted when establishing centre of gravity of aeroplane. Question 69-17
For medium range twin jet datum point located 235 54 inches forward of front spar. In bottom of appendix you can read '2 1 datum point 540inches forward of front spar(fs) '. Question 69-18
The chemical fluids used to charge aircraft toilets are counted as Part of variable load. In bottom of appendix you can read '2 1 datum point 540inches forward of front spar(fs) '. Question 69-19
for this question use annex ecqb 031 054 v2015 01 calculate basic empty mass and centre of gravity 240 Part of variable load. Ecqb03 july 2016 sorry we haven't yet recovered annex we get only question the correct answer (this nice isn't it?) !!!. Question 69-20
for this question use annex ecqb 031 057 v2015 05 using loading manifest in annex determine basic empty mass 240 Part of variable load. Ecqb03 july 2016 3735 lb 15 lb 390 120 = 3210 lbs. Question 69-21
The bem of an aircraft the basic empty weight of an aircraft without crew and items of removable equipment it found in Weighing schedule at last date of weighing. Ecqb03 july 2016 3735 lb 15 lb 390 120 = 3210 lbs. Question 69-22
The mass of an aircraft 2000 kg and 400 kg of freight added to a hold 2 m aft of present cg position movement of cg Weighing schedule at last date of weighing. Change of cg = (mass added x distance from hold to cg) / new total mass change of cg = (400 x 2) / (2000 + 400) = 0 33 m we add cargo aft thus cg will move aft. Question 69-23
Once mass and balance documentation has been signed prior to flight Acceptable last minute change to load must be documented. See easa air ops regulation no 965/2012 cat pol mab 105 mass balance data documentation. Question 69-24
Using loadsheet on annex determine underload and difference between maximum allowed tom and actual tom the underload 248 377 kg the difference 7 7 kg. Underload = allowed tl total tl underload = 15000 kg 11230 kg = 3770 kg difference mtom actual tom actual tom = om + tl actual tom = 44400 kg + 11230 kg = 55630 kg difference = mtom actual tom difference = 62800 kg 55630 kg = 7170 kg. Question 69-25
A location in aircraft which identified a number designating its distance from datum known as 377 kg the difference 7 7 kg. Station a location in aircraft which identified a number designating its distance from datum. Question 69-26
The centre of gravity of a body that point Through which sum of forces of all masses of body considered to act. Station a location in aircraft which identified a number designating its distance from datum. Question 69-27
Assuming gross mass altitude and airspeed remain unchanged movement of centre of gravity from forward to aft limit will cause Increased cruise range. With a forward cg aircraft 'nose heavy' he has a nose down moment thus downforce on tail on a steady flight must increase the total aircraft weight increases therefore more weight = more drag = more power = reduced maximum cruise range with an aft cg downforce on tail on a steady flight decreased less weight = less drag = less power = increased cruise range. Question 69-28
Moment balance arms are measured from a specific point to body station at which mass located that point known as Increased cruise range. The datum an imaginary vertical line chosen on longitudinal axis of aircraft specified the designer from which all horizontal cg measurements are made in most cases datum located in vicinity of aircraft nose usually firewall. Question 69-29
The centre of gravity of an aircraft that point through which total mass of aircraft said to act the weight acts in a direction Parallel to gravity vector. the sum of aircraft weight acts through center of gravity towards earth's center. Question 69-30
When an aircraft stationary on ground its total weight will act vertically Through its centre of gravity. Centre of gravity (cg) that point through which force of gravity said to act on a mass always acts parallel to gravity vector. Question 69-31
The weight of an aircraft which in level non accelerated flight said to act Vertically through centre of gravity. Centre of gravity (cg) that point through which force of gravity said to act on a mass always acts parallel to gravity vector. Question 69-32
The distance from datum to centre of gravity of a mass known as The moment arm or balance arm. Balance arm the distance from datum to centre of gravity of a mass. Question 69-33
During take off you notice that a given elevator input aeroplane rotates much more rapidly than expected this an indication that The centre of gravity may be towards aft limit. If a very small up deflection of horizontal stabilizer generates a downward force on tail sufficiently strong to raise airplane nose this an indication that centre of gravity may be towards aft limit. Question 69-34
The mass displacement caused landing gear extension Creates a longitudinal moment in direction (pitch up or pitch down) determined the type of landing gear. The mass displacement caused landing gear extension creates a longitudinal moment in direction (pitch up or pitch down) determined the type of landing gear. Question 69-35
The datum used balance calculations chosen on longitudinal axis of aircraft But not necessarily between nose the tail of aircraft. The datum an imaginary vertical line specified the designer from which all horizontal c of g measurements are made img /com_en/com031 158 jpg in most cases datum located in vicinity of aircraft nose usually firewall. Question 69-36
The centre of gravity the Point where all aircraft mass considered to be concentrated. Centre of gravity (cg) that point through which force of gravity said to act on a mass always acts parallel to gravity vector. Question 69-37
For purpose of aeroplane mass and balance calculations datum point defined as A fixed point from which all balance arms are measured it may be located anywhere on aeroplane's longitudinal axis or on extensions to that axis. The datum point the point on aircraft designated the manufacturer from which all centre of gravity measurements calculations are made in most cases datum located in vicinity of aircraft nose usually firewall it not necessarily a point 'on' aircraft. Question 69-38
In calculations with respect to position of centre of gravity a reference made to a datum the datum A reference plane which chosen the aircraft manufacturer its position given in aircraft flight or loading manual. The datum that point on longitudinal axis (or extension thereof) from which centre of gravity of all masses are referenced its position determined the manufacturer found in loading manual operating data manual. Question 69-39
The mass of an item multiplied it's distance from datum it's A reference plane which chosen the aircraft manufacturer its position given in aircraft flight or loading manual. Arm (moment arm) the horizontal distance in inches from reference datum line to center of gravity of an item moment the product of weight of an item multiplied its arm moments are expressed in pound inches (lb in) total moment the weight of airplane multiplied the distance between datum the cg. Question 69-40
The moment an item The mass of item multiplied it's distance from datum. The moment the product of mass the balance arm.
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